BOUGHT WITH THE INCOME FROM THE SAGE ENDOWMENT FUND THE GIFT OF Henrg W, Sage 1891 l^.m%.n /f4^P'^ Cornell University Library arV19267 Higher algebra / 3 1924 031 243 987 olin.anx The original of tliis book is in tlie Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031243987 HIGHER ALGEBRA BT GEORGE EGBEET FISHEE, M.A., Ph.D. AND ISAAC J. SCHWATT, Ph.D. ASSISTANT PROFESSORS OF MATHEMATICS IN THE UNIVERSITY OF PENNSYLVANIA o^^o PHILADELPHIA FISHEE AND SCHWATT 1901 K COPTEI&HT, 1901, bt pishee and schwatt. NottoooB IPreas J. S. Cushins a Co. — Berwick & Smith E. FlemiDg & Co- Norwood Mass. U.S.A. PREFACE. The present book is an enlargement of the authors' Ele- ments of Algebra. To the end of Chapter XXVIII. it is identical with the latter and the School Algebra. Some revision of the later chapters in the Elements has been incor- porated in this book, and a number of new chapters have been added. The scope of the books, amply justified by their successful use in high and normal schools and colleges, is stated in the preface to the Elements : "The aim has been to make the transition from ordinary Arithmetic to Algebra natural and easy. Ko efforts. have been spared to present the subject in a simple and clear man- ner. Yet nothing has been slighted or evaded, and all diffi- culties have been honestly faced and explained. New terms and ideas have been introduced only when the development of the subject made them necessary. Special attention has been paid to making clear the reason for every step taken. Each principle is first illustrated by particular examples, thus preparing the mind of the student to grasp the meaning of a formal statement of the principle and its proof. Directions for performing the different operations are, as a rule, given after these operations have been illustrated by particular examples. "The importance of mental discipline to every student of mathematics has also been fully recognized. On this account great care has been taken to develop the subject in a logical iv PREFACE. manner. Eigorous, but, as a rule, simple, proofs of all princi- ples have been given. "If mathematics is to develop the reasoning power of the student and to teach him to think logically, it is better to omit a proof altogether than to give as a proof logically incorrect statements, thus training the mind of the student in illogical thinking. "Concrete illustrations, such as receiving and paying out money, going north and going south, have their proper places, but cannot be said to constitute proofs. If Algebra, like Arithmetic, treats of number, then the laws governing the operations with numbers should be derived from the properties of and the relations between them. "The subject-matter in the book has been printed in two sizes of type. The matter in smaller type consists of the formal proofs of principles and of the more difficult portions of each topic treated. The matter given in the larger type is logically complete (except for the proofs of principles), and can be taken up as a first course in the subject. " To economize space the exercises have been put in smaller type, and not the explanations and solutions of illustrative examples in the text. It is regarded as more important that the student should have these, which he is to study, most clearly represented rather than the examples which he is to copy and then work from his paper. "The attention of teachers is especially invited to the fol- lowing features of the book : " The introductory chapter and the development in Chapter II. of the fundamental operations with algebraic numbers. "The use of type-forms in multiplication and division (Chapter VI.) and in factoring (Chapter VIII.). PREFACE. V "The application of factoring to the solution of equations (Chapters VIII. and XXI.). By the early introduction of this method it has been possible to give problems which lead to quadratic equations before the formal treatment of that topic. " The solutions of equations based upon equivalent equations and equivalent systems of equations (Chapter IV., etc.). This method is of extreme importance, even to the beginner. The ordinary way of treating equations is illogical, leads to serious errors, and is therefore also pedagogically wrong. " Thus, no one will dispute that if both sides of an equation be multiplied by the same algebraical number an equation is obtained ; but whether it is legitimate to assume that the solu- tions of this equation are the solutions of the given equation is quite another matter. " The treatment of irrational equations (Chapter XXIII.). " The special suggestions given in the first chapter on prob- lems (Chapter V.), and applied subsequently to assist the student in acquiring facility in translating the verbal language of the problem into the symbolic language of the equation. " The discussion of general problems (Chapter XI.) and the interpretation of positive, negative, zero, indeterminate, and infinite solutions of problems (Chapter XII.). "The outline of irrational numbers (Chapter XVIII.). " The brief introduction to imaginary and complex numbers (Chapter XX.). " The exercises are voluminous. The aim has been not only to give examples for sufficient drill in the applications of the principles, but to include also many which tend to develop the thinking power of the student, rather than to develop him in a treadmill way. vi PREFACE. "The authors take pleasure in acknowledging their 'indebted- ness to their colleagues in secondary schools and colleges for many helpful suggestions and criticisms which have been of much assistance to them in preparing this book.'' A similar plan has been followed in preparing the later chapters. It is hoped that they will be found to be not only teachable, but also thoroughly sound. This book not only gives a thorough course in the more advanced subjects required for admission to universities and scientific schools, but treats each subject with sufficient ful- ness for the course in the first year in college. G. E. F. I. J. S. University of Pennsylvania, Philadelphia, Pa. TABLE OF CONTENTS. CHAPTER I. Introduction. PAGE § 1. General NnMEER . ■ 1 Axioms 7 Fundamental Principles 7 Problems solved by Equations 10 § 2. Algebraic (Positive and Negative) Numbers ... 12 Relations between Positive and Negative Numbers and Zero 15 Positive and Negative Numbers are Opposite Numbers . 16 CHAPTER II. The Four Fundamental Operations with Algbbkaio Numbers. § 1. Addition op Algebraic Numbers 19 Addition of Numbers with Like and Unlike Signs . . 19 The Associative and Commutative Laws for Addition . 21 Property of Zero in Addition 23 § 2. Subtraction op Algebraic Numbers 24 Successive Additions and Subtractions .... 25 The Associative and Commutative Laws for Subtraction . 27 Removal of Parentheses 28 Insertion of Parentheses 29 Property of Zero in Subtraction 30 § 3. Multiplication op Algebraic Numbers . . . 30 The Commutative Law for Multiplication .... 34 The Associative Law for Multiplication . . . 36 vii VUl CONTENTS. § 4. Division op Algebkaic Numbers . The Commutative Law for Division The Associative Law for Division § 5. One Set oe Signs for Quality and Operation § 6. Positive Integral Powers Properties of Positive Integral Powers PAGE 38 41 42 43 45 47 CHAPTER III. The Fundamental Operations with Integral Algebraic Expressions. i 1. Definitions . . ( 2. Addition and Subtraction . Addition and Subtraction of Multinomials . i 3. Multiplication ... Principles of Powers Degree. Homogeneous Expressions . The Distributive Law for Multiplication Multiplication of a Multinomial by a Monomial Multiplication of Multinomials by Multinomials Zero in Multiplication . Division Division of Monomials by Monomials . Distributive Law for Division Division of a Multinomial by a Monomial Division of a Multinomial by a Multinomial Infinites , . . . 49 51 54 56 56 59 61 62 64 67 68 69 69 70 71 77 CHAPTER IV. Integral Algebraic Equations. § 1. Identical Equations § 2. Conditional Equations . Equivalent Equations . ... Eundamental Principles for Solving Integral Equations §3. § 4. Linear Equations in One Unknown Number 80 80 82 83 86 Problems CONTENTS ix CHAPTER V. PAGE 89 CHAPTER VI. Typb-Eokms. § 1-. Ttpe-Eoems in Multiplication 100 The Square of an Algebraic Expression . . . 100 Product of the Sum and Difference of Two Numbers . .102 The Product (» + «)(» + &) 103 The Product (ax + 6) (ca; + d) 104 The Cube of an Algebraic Expression 105 Higher Powers of a Binomial . .... 106 § 2. Ttpe-Forms in Division .... 107 Quotient of the Sum or the Diflerence of Like Powers of Two Numbers by the Sum or the Difference of the Numbers . 107 The Remainder Theorem 110 CHAPTER VII. Parentheses. Removal of Parentheses 113 Insertion of Parentheses . 115 CHAPTER VIII. Factors and Multiples op Integral Algebraic Expressions. § 1. Integral Algebraic Factors 116 The Fundamental Formula for Factoring . . . .117 Trinomial Type-Forms 119 Binomial Type-Forms 125 Special Devices for Factoring 180 Symmetry 131 Method of Substitution 133 § 2. Highest Common Factors 136 H. C. F., by Factoring 136 H, C. F., by Division . 138 § 3. Lowest Common Multiples 147 L. C. M., by Factoring . . 147 L. C. M., by Means of H. C. F 148 Relation between H. C. F. and L. C. M 150 § 4. Solution of Equations by Factoring 150 X CONTENTS. CHAPTER IX. Fractions. PAOB Reduction of Fractions to Lowest Terms .... 156 Reduction of Fractions to a Lowest Common Denominator . 158 Addition and Subtraction of Fractions 161 Multiplication of Fractions 166 Powers of Fractions 168 Reciprocal Fractions 169 Division of Fractions 169 Complex Fractions 172 Continued Fractions 173 Factors of Fractional Expressions 174 Indeterminate Fractions 175 Some Principles of Fractions 175 CHAPTER X. Fractional Equations in One Unknown Number. Problems 185 CHAPTER XI. Literal Equations in One Unknown Number. General Problems 191 CHAPTER XIL Interpretation of the Solutions or Problems. Positive Solutions 196 Negative Solutions 197 Zero Solutions 198 Indeterminate Solutions 199 Infinite Solutions 199 The Problem of the Couriers 200 CHAPTER XIII. Simultaneous Linear Equations. § 1. Systems of Equations . 203 § 2. Equivalent Systems 205 CONTENTS. XI § 3. Systems op Linear Equations Elimination by Addition and Subtraction Elimination by Substitution Elimination by Comparison The General Solution of a System of Two Linear Equations in Two Unknown Numbers .... Linear Equations in Tliree or More Unkiiown Numbers Number of Solutions of a System of Linear Equations § 4. Systems of Fractional Equations .... FAGE 207 207 210 211 '213 214 216 218 CHAPTER XIV. Problems which lead to Simultaneous Linear Equations. Discussion of Solutions 233 CHAPTER XV. Indeterminate Linear Equations 235 CHAPTER XVI. Evolution. Iv^ Definitions and Principles Number of Roots Principal Roots . Evolution Principles of Roots Roots op Monomials Square Roots op Multinomials Cube Roots op Multinomials Higher Roots Roots op Arithmetical Numbers Square Roots Cube Roots .... 239 240 241 242 242 244 246 249 251 252 252 255 xii CONTENTS. CHAPTEE XVII. Inequalities. Principles of Inequalities Absolute Inequalities Conditional Inequalities . Problems A Property of Fractions . A Property of Powers PA61! 258 261 262 264 265 265 CHAPTER XVIII. Irrational Numbers. The Fundamental Operations with Irrational Numbers . . 271 CHAPTEE XIX. Surds. Classification of Surds . . ... 275 Reduction of Surds . . . . . 276 Addition and Subtraction of Surds 278 Eeduction of Surds of Different Orders to Equivalent Surds of the Same Order . . . . . 279 Multiplication of Surda , . 280 Division of Surds . . . . 283 SurdPaotors .... 284 Eationalization .... . 286 Properties of Quadratic Surds . . . 288 Evolution of Surd Expressions ... . 290 Square Eoots of Simple Binomial Surds . . . 290 Approximate Values of Surd Numbers 291 CHAPTEE XX. Imaginary and Complex Numbehs. Imaginary Numbers . ... 294 Complex Numbers . . . 298 Complex Factors ... . . . 301 Square Eoot of a Complex Number 301 CONTENTS. xiii CHAPTER XXI. Quadratic Equations. PAQB Pure Quadratic Equations 305 Solution by Factoring . . 305 Solution by Completing the Square . . . 307 General Solution 309 Fractional Equations that lead to Quadratic Equations . . 310 Theory of Quadratic Equations 313 Nature of the Roots of a Quadratic Equation . . 316 Maxima and Minima . . . ... 318 Problems ... 322 CHAPTER XXII. Equations of Highek Degree than the Second . . . 327 CHAPTER XXIII. Irrational Equations. Special Devices ... 333 CHAPTER XXIV. Simultaneous Quadratic and Higher Equations. § 1. Simultaneous Quadratic Equations . . . . 386 § 2. Simultaneous Higher Equations 348 § 3. Problems 351 CHAPTER XXV. Ratio, Proportion, and Variation. § 1. Ratio 354 § 2. Proportion 356 S 3. Variation .... 362 XIV CONTENTS. CHAPTER XXVI. Doctrine of Exponents. PAGE Zeroth Powers 367 Negative Integral Powers 368 Fractional (Positive or Negative) Powers 370 CHAPTER XXVII. Progressions. §1- Series ... ...... 379 sa- Arithmetical Progression 879 384 Problems . 386 gs. Geometrical Progression ... 389 Sum of an Infinite Geometrical Progression 393 Geometrical Means . 394 Problems 895 §4. Harmonical Progression 398 Problems 400 CHAPTER XXVIII. The Binomial Theorem for Positive Integral Exponents 402 CHAPTER XXIX. Logarithms. Irrational Powers 408 Solution of the Equation 6"^ = a 411 Logarithms 418 Principles of Logarithms ... .... 4] 5 Systems of Logarithms . . ... 418 Properties of Common Logarithms . . . 419 To find the Logarithm of a Given Number 422 To find a Number from its Logarithm . . . 426 Cologarithms ...... . . 429 Applications . ... .... 430 Exponential Equations . . 434 Logarithmic Equations ... ... 436 Compound Interest and Annuities . . . 437 CONTENTS. XV CHAPTER XXX. Permutations and Combinations. i 1. Definitions i 2. Permutations . i 3. Combinations . Greatest Values of nOr i 4. Two Important Principles \ 5. Problems CHAPTER XXXI. Probability. Mortality Table PAGK 440 441 444 446 448 449 453 CHAPTER XXXII. Variables and Limits. § 1. Variables , 45g Functions ... 456 § 2. Limits 457 Infinites and Infinitesimals 459 Fundamental Principles of Limits 460 Indeterminate Fractions 462 CHAPTER XXXIIL Infinite Series. Methods of Comparison 468 The General Term of a Series 472 Series having Negative Terms .... . . 475 The Ratio of Convergenoy 476 CHAPTER XXXIV. The Binomial Theorem fob Any Rational Exponent. § 1. The Binomial Theorem for Positive Integral Exponents 481 Properties of Binomial Coefficients ...'.. 482 § 2. The Binomial Theorem for Any Rational Exponent . 483 Roots of Arithmetical Numbers 487 XVI CONTENTS. CHAPTER XXXV. Undetekmined Coefficients. § 1. Method of Undetekmined Coefficients § 2. Expansion op Certain Functions into Infinite Sekibs Rational Fractions Surds § 3. Reversion of Series § 4. Partial Fractions Tlie General Term PAGE 488 489 490 492 492 493 498 CHAPTER XXXVI. Continued Fractions. To convert a Common Fraction into a Terminating Continued Fraction . . 503 To reduce a Terminating Continued Fraction to a Commuu Fraction 504 Properties of Convergents . . . 507 Limit to Error of Any Convergent . . 509 To reduce a Quadratic Surd to a Continued Fraction 510 Application of Convergents .... . . 512 To reduce a Periodic Continued Fraction to an Irrational Numter 513 CHAPTER XXXVII. Summation op Series. By Undetermined Coefficients . By separating Terms into Partial Fractions Recurring Series . ... Convergency of a Recurring Series . Method of Finite Differences . Piles of Cannon Balls Interpolation ... CHAPTER XXXVIII. The Exponential and Logarithmic Series. The Exponential Series . . . The Logarithmic Series . .... Computation of Logarithms 515 516 518 523 524 528 531 534 536 537 CONTENTS. xvii CHAPTER XXXIX. Determinants. PAGE Minors . . . .... 543 Principles of Determinants . 545 Solution of Linear Simultaneous Equations . . . .551 Determinants of Higher Order 553 CHAPTER XL. Theory op Equations. General Form . 556 Synthetic Division . . 557 Number of Roots 559 Depression of Equations . 560 Relation between the Roots and Coefficients . . . 561 Symmetrical Eunotions 563 Formation of an Equation from its Roots . . 563 Real and Rational Roots . 565 Surd Roots ... . . . 566 Imaginary Roots . . 56() To transform an Equation into Another whose Roots are Any Multiples of the Roots of the Given Equation . . . 567 To transform an Equation into Another whose Roots are Those of the Given Equation with Signs Changed . . 569 Descartes' Rule 569 Limits to the Roots ... 572 Newton's Method 575 Derived Functions . ... ... 577 Multiple Roots 578 Graphic Representation 580 Greatest and Least Terms in/ (x) 587 Principle of Continuity . . ..... 588 To find an Equation whose Roots are Less than the Roots of a Given Equation by a Definite Number .... 591 Horner's Method of Approximation .... 593 Roots of JsTumbers 599 Sturm's Theorem 600 Roots nearly Equal 604 Reciprocal Equations . .... 606 Binomial Equations 609 xviii CONTENTS. PAGE General Solutions 610 To transform an Equation into Another in which a Particular Term is Wanting 611 Cube Roots of Unity . . 612 Cardan's Solution of the Cubic 612 General Solution of the Biquadratic 614 Table op Logarithms i-xviii CHAPTER I. INTRODUCTION. Algebra, like Arithmetic, treats of number. But the mean- ing of number, and the mode of representing it, are extended in passing from ordinary Arithmetic to Algebra. §1. GENERAL NUMBER. 1. In ordinary Arithmetic all numbers have particular values and are represented by definite symbols, the Arabic numerals, 1, 2, 3, etc. The symbol 7, for instance, stands for a group of seven units. In Algebra, however, such symbols as a, h, x, y, are used to represent numbers which may have any values whatever, or numbers whose values are, as yet, unknown. Just as we speak of 10 miles, of 95 dollars, etc., in Arith- metic ; so in Algebra we speak of a miles, meaning any num- ber of miles or an unknoivn number of miles; of x dollars, meaning any number or an unknown number of dollars, etc. For the sake of brevity, we shall say the number a, or simply a, meaning thereby the number denoted by the symbol a. 2. The symbols of Arithmetic, 1, 2, 3, etc., are retained in Algebra with their exact arithmetical meanings. The numbers represented by letters are, for the sake of distinction, called Literal or General Numbers. Other symbols than letters might be used to represent general numbers, but letters are more con- venient to write and to pronounce. 3. The operations of Addition, Subtraction, Multiplication, and Division are denoted by the same symbols in Algebra as in Arithmetic. 1 2 ALGEBRA. [Ch. 1 4. The Symbol of Addition, +, read plus, is placed between two numbers to indicate that the number on its right is to be added to the number on its left. E.g., just as 5 + 3, read five plus three, means that 3 is to be added to 5; so a + h, read a plus b, means that b is to be added to a. 5. The Symbol of Subtraction, — , read minus, is placed be- tween two numbers to indicate that the number on its right is to be subtracted from the number on its left. E.g., just as 5 — 3, read five minus three, means that 3 is to be subtracted from 5 ; so a — 6, read a minus b, means that b is to be subtracted from a. In a chain of additions and subtractions the operations are to be performed successively from left to right. E.g., 7 + 4-3 + 2 = 11 -3 + 2 = 8+2 = 10. 6. The Symbol of Multiplication, x, read multiplied by, or times, means that the number on its left is to be multiplied by the number on its right. ^.g'., just as 5x3, read five multiplied by three, or three times five, means that 5 is to be multiplied by 3 ; so a x 6, read a multiplied by b, or 6 times a, means that a is to be multi- plied by 6. A dot (•) is frequently used, instead of the symbol x, to denote multiplication ; as a • 6 f or a x &. The symbol of multiplication between two literal numbers, or one literal number and an Arabic numeral, is frequently omitted. E.g., the product x x y X «, or x ■ y • «, is usually written, xyz. The product a x 6, or a • 6, is written, a 6. It will be proved later that a x b = b x a, or ab = ba. On this account the product a 6 is usually written 6 a, the Arabic numeral being placed first. But the symbol of multiplication between two numerals cannot be omitted without changing the meaning. E.g., if in the indicated multiplication, 3 x 6, or 3 • 6, the symbol, x , or •, were omitted, we should have 36, not 18. § 1] GENERAL NUMBER. 3 7. The Symbol of Division, -=-, read divided by, is placed between two numbers to indicate that the number on its left is to be divided by the number on its right. E.g., just as 10 -^ 5, read ten divided by five, means that 10 is to be divided by 5 ; so a -;- 6, read a divided by b, means that a is to be divided by b. Id. a chain of multiplications and divisions the operations are to be performed successively from left to right. E.g., 12x2-=-3x4 = 24-f-3x4 = 8x4 = 32. 8. The use of letters to represent general numbers may be illustrated by a few simple examples. Ex. 1. If a boy has 3 books and is given 2 more, he has 3 + 2 books. If he has a books and is given 5 more, he has a + 5 books. If he has m books and is given n more, he has m + n books. Ex. 2. If a boy has 5 oranges and gives away 2, he has left 5 — 2 oranges. If he has p oranges and gives away 7, he has left p — 7 oranges. If he has u oranges and gives away v, he has left u — v oranges. Ex. 3. If a man buys 5 city lots at 120 dollars each, he pays 120 X 5 dollars for the lots. If he buys a lots at 150 dollars each, he pays 150 a dollars for the lots. If he buys u lots at V dollars each, he pays vu dollars for the lots. Ex. 4. If a train runs 60 miles in 2 hours, it runs 60-7-2 miles in 1 hour. If it runs a miles in 5 hours, it runs a-i-5 miles in 1 hour. If it runs p miles in q hours, it runs p-7- q miles in 1 hour. Ex. 5. If a pupil buys 2 note books at 10 cents each and 3 note books at 12 cents each, he pays 10 x 2 + 12 x 3 cents for all. If he buys a note books at m cents each and b note books at n cents each, he pays ma + nb cents for all. Ex. 6. If, in a number of two digits, the digit in the units' place is 3 and the digit in the tens' place is 5, the number is 10 X 5 + 3. If the digit in the units' place is a and the digit in the tens' place is b, the number is 10 & + a. 4 ALGEBRA. [Ch. I Ex. 7. Just as 2 = 1 + 1, and 3 = 1 + 1+1, so 2 a = a + a, and Sa = a + a + a. Therefore, just as 3 + 2 = 5, so 3a + 2a = 5a. In like manner, ^x + ^x = ^x. 9. Observe that in the preceding examples the reasoning is the same whether the numbers are represented by letters or by Arabic numerals. The results of these operations are numbers in all cases, whether letters or numerals, or both, are involved. Thus, the result of adding 6 to a, a + &, is a number, just as 5 + 3, or 8, is a number. Likewise, a + b — c, ab — c, read is greater than, is used to indicate that the number or expression on its left is greater than that on its right ; as 7 > 5. § 1] GENERAL NUMBER. 7 The Symbol of Inequality, <, read is less than, is used to indicate that the number or expression on its left is less than that on its right ; as 3 < 4 + 2. Axioms. 14. An Axiom is a truth so simple that it cannot be made to depend upon a truth still simpler. Algebra makes frequent use of the following mathematical axioms : (i.) Every number is equal to itsdf. E.g., 7 = 7, a = a. (ii.) The whole is equal to the sum of all its parts. E.g., 7 = 3 + 4, 5 = 1 + 1 + 1 + 1 + 1.' (iii.) If two numbers be equal, either can replace the other in any algebraic expression in which it occurs. E.g., If a + 6 = c, and b = d, then a + d = c, replacing 6 by d. (iv.) Two numbers which are each equal to a third number are equal to each other. E.g., If a = 6, and c = b, then a = c. (v.) The whole is greater than any of its parts ; and, con- versely, any part is less than the whole. E.g., 3 + 2 > 2 and 2 < 3 + 2. Fundamental Principles. 15. The following principles can be inferred directly from the axioms : (i.) If the same number, or equal numbers, be added to equal numbers, the sums will be equM. (ii.) If the same number, or equal numbers, be subtracted from equal numbers, the remainders will be equal. (iii.) If equal numbers be multiplied by the same number, or by equal numbers, the products will be equal. (iv.) If equal numbers be divided by the same number {except 0), or by equal numbers, the quotients will be equal. 8 ALGEBRA. [Ch. I 16. Literal numbers, as has been stated, are used to repre- sent numbers which may have any values whatever, or numbers whose values are, as yet, unknown. But it is frequently neces- sary to assign particular values to such numbers. Substitution is the process of replacing a literal number in an algebraic expression by a particular value. See axiom (iii.). Ex. 1. If in a + &, a = 3 (read a has the value 3) and 6 = 5, then a-|-& = 3-|-5 = 8, or a-F& = 8. Notice that the last step involved an application of axiom (iv.). For we have a + b = 3 + 5, and 3 -|- 6 = 8 ; therefore, by axiom (iv.), a-\- & = 8. Ex. 2. If in a — (& -I- c), a = 11, 6 = 2, and c = 3, we have a - (6 -I- c) = 11 - (2 -1- 3) = 11 - 5 = 6. Ex. 3. If, in a -I- 6 - 2 a + 3 6 - c, we let a = 6, b = 11|, c = -I, we have a-|-6-2a-|-36-c = 6-hlli-2x6 + 3x 11^-1 = 6+-2/-12 + ^-| = 38i Observe that in the work of the last example, the expres- sion a + 6 — 2a-|-36 — cistobe understood on the left of the symbol, =, in the second line. Ex. 4. If, in the last example, a = 3, 6=1, and c = 1, we have a-f6-2a-|-36-c = 3 + l-6 + 3-l = 4-6-|-3-l. We cannot further reduce 4— 6 -|- 3—1, since we are unable, as yet, to subtract 6 from 4. BXEECISES III. What are the values of the following expressions when a = 6, 6 = 4, c = 2: I- a + b. 2. a-b. 3. ab. 4. a -^ b. 5. a - 6 + c. 6. a-b-c. 1. abc. 8. a -^ b x c. 9. a + ib-c). 10. a-(6 + c). 11. (a - &)c. 12. c-=-(a-6). 13. [o+(6_c)]o. 14. [a-(6-c)]-=-6. 15. (o-6)(c-l). 16. (12 -a) -^(7 -ft). 17. [15-(7-a)]x[(25-6)-(16-c)]. § 1] GENERAL NUMBER. 9 17. Some of the advantages of using literal numbers are shown by the following example : Ex. The two equations 7^7 7 7 11^.11 11 11 are particular examples of the following arithmetical principle : The sum of two fractions which have a common denominator is a fraction whose denominator is that common denominator, and whose numerator is the sum of the two given numerators ; or, Isf num. 2d num. _ Isi num. + 2d num. com. den. com. den. com. den. This principle can be stated still more concisely if the terms of the fractions, which may be any numbers whatever, are rep- resented by three symbols, say a, b, c. We then have a b _ a + b c c c This equation states by means of signs and symbols all that is contained in the verbal statement of the principle. It is thus a symbolic statement of a general principle, and includes all particular cases that result from assigning particular values to a, b, c. 18. Notice the following advantages thus secured by intro- ducing general numbers : (i.) General laws and relations can be expressed with great brevity, and yet include all that the most general verbal state- ments can express. (ii.) Such symbolic statements mass under the eye the various operations involved, and thus enable the eye to assist the under- standing and memory. BXBECISBS IV. Express ui algebraic language (i.e., by means of the signs and symbols of Algebra) the following principles of Arithmetic : 1. If a, b, and c are any three numbers, their sum diminished by any one of them is equal to the sum of the other two. 10 ALGEBRA. [Ch. I If z is the result of subtracting y from x, express in algebraic language the following principles of subtraction : 2. The minuend is equal to the subtrahend plus the remainder. 3. The subtrahend is equal to the minuend diminished by the remainder. If z is the result of multiplying x by y, express in algebraic language the following principles of multiplication : 4. The multiplicand is equal to the product divided by the multiplier. 5. The multiplier is equal to the product divided by the multiplicand. If a is exactly divisible by 6, and q is the quotient, express in algebraic language the following principles of division : 6. The dividend is equal to the divisor multiplied by the quotient. 7. The divisor is equal to the dividend divided by the quotient. If — is any fraction, and m is any integer, express in algebraic language 6 the following principles of fractions : 8. If the numerator of a fraction is multiplied by any integer, the value of the fraction is multiplied by that integer. 9. If the denominator of a fraction is multiplied by any integer, the value of the fraction is divided by that integer. Problems solved by Equations. 19. Another advantage of using literal numbers is shown by the following problem : Pr. The older of two brothers has twice as many marbles as the younger, and together they have 33 marbles. How many has the younger ? The number of marbles the younger brother has is, as yet, an unknown number. Let us represent this unknown number by some letter, say x. Then, since the older brother has twice as many, he has a; x 2, or 2 a;, marbles. The problem states, in verbal language : tJie number of marbles the younger has plus the number the older has is equal to S3; in algebraic language : a; + 2 a; = 33, or 3 a; = 33. Dividing by 3 [Art. 16 (iv.)], x = 11, the number of marbles the younger has. The older has 2 a;, = 22. Check : a; + 2 a; = 11 + 22 = 33. § 1] GENERAL NUMBER. H EXBBCISBS V. 1. What number added to three times itself gives 28 ? 2. Divide 69 into two parts so that the greater shall be twice the less. 3. If twice a number be added to three times the number, the sum will be 63. What is the number ? 4. If three times a number be subtracted from five times the number, the remainder will be 6. What is the number 1 5. Divide 150 into two parts so that the less is one-fifth of the greater. 6. A and B together have f 180, and A has five times as much as B. How many dollars has each ? 7. In a school are 120 pupils ; m the second grade are twice as many as in the first, and in the third three times as many as in the first. How many pupils are in each grade ? 8. A, B, and C together invest $8000 ; A invests twice as much as B, and B five times as much as C. How many dollars does each invest ? 9. Divide 30 into three parts, so that the second shall be one-half of the first, and the third one-third of the second. 10. Divide 52 into three parts, so that the second shall be one-half of the first, and the third one-fourth of the second. 11. Divide 85 into three parts, so that the first shall be four times the second, and one-third of the third. 12. If one-fourth of a number be subtracted from one-third of the number, the remainder will be 6. What is the number ? 13. Three boys. A, B, and C, have together 27 pencils. B has twice as many as A, and C twice as many as A and B together. How many pencils has each ? 14. Three boys. A, B, and C, have together 32 pens ; B has one-third as many as A, and C three times as many as A and B together. How many has each ? 15. Three boys. A, B, and C, have together 16 note books ; A has five times as many as B, and the number that A has more than B is twice the number that C has. How many note books has each ? 20. General numbers are most frequently represented by the italicized letters of the English alphabet. But letters of other alphabets are sometimes employed, and there is often an advantage in using the same letter with some distinguishing marks to represent different numbers in the same discussion. 12 ALGEBRA. [Ch. I We add a list of the more common symbols. Greek letters: a, /3, y, 8, etc., read alpha, beta, gamma, delta, etc. ; ■with prime marks : a', a", a"', a'"', read a prime, a two prime, a three prime, a n prime; with subscripts : a^, a^, a^, etc., read a sub-one, a sub-two, a sub- three, etc., or simply a one, a two, a three, etc. §2. POSITIVE AND NEGATIVE NUMBERS, OR ALGEBRAIC NUMBERS. 1. A still greater extension of the idea of number in passing from Arithmetic to Algebra is arrived at by the following con- siderations : In ordinary Arithmetic we subtract a number from a greater or an equal number. We are familiar with such operations as 7-5 = 2, 6-5 = 1, 6-5 = 0. (i.) But such operations as 4 — 5, 3 — 5, etc., (ii.) have not occurred in ordinary Arithmetic and cannot be car- ried out in terms of arithmetical numbers. Por, from an arithmetical point of view, we cannot subtract from a number more units than are contained in that number. In general, the indicated operation a — b can, as yet, be performed only when a is greater than b. But if a and b are to have any values whatever, the case in which a is less than 6, that is, in which the minuend is less than the subtrahend, must be included in the operation of subtraction. 2. Now observe that, as the minuend in equations (i.) de- creases by 1, 2, or more units (the subtrahend remaining the same) the remainder decreases by an equal number of units. When the minuend is equal to the subtrahend, the remainder is 0. If then, as in the indicated operations (ii.), the minuend becomes less than the subtrahend by 1, 2, or more units, the remainder must decrease by an equal number of units, and therefore become less than by 1, 2, or more units. §2] ALGEBRAIC NUMBERS. 13 The operation of subtracting a greater number from a less is therefore possible only when numbers less than zero are introduced. We then have from (i.) and (ii.) : Min. - Subt. = Eem. 7-5 = 2 6-5 = 1 5-5 = (iii.) 4 — 5 = a number one unit less than 3 — 5 = a number two units less than , 3. Numbers less than zero are called Negative Numbers. Numbers greater than zero are, for the sake of distinction, called Positive Numbers. Positive and negative numbers are called Algebraic or Rela- tive Numbers. 4. The Absolute Value of a number is the number of units contained in it without regard to their quality {i.e. whether positive or negative). A positive number may be indicated by placing a small sign, +, to the left and a little above its absolute value ; as, +5, +10, +16 ; read positive 5, positive 10, positive 16. A negative number may be indicated by placing a small sign, -, to the left and a little above its absolute value ; as, ~6, ~10, ~16 ; read negative 5, negative 10, negative 16. We must, as yet, carefully distinguish these symbols of quality, + and ~,-from the (larger) symbols of operation, + andi — . 5. Equations (iii.) can now be written as follows : Min. — Subt. = Eem. pos. 7 —pos. 5 =pos. 2 pos. 6 —pos. 5 =pos. 1 pos. 5 —pos. 5 = o\ or ] +5 — +5 = }- (iv.) pos. 4 — pos. 5 = neg. 1 pos. 3 — pos. 5 = neg. 2 , A negative remainder does not mean that more units have been taken from the minuend than were contained in it ; such Vlin. +7 - Subt. = Eem - +5 = +2 +6 - +5 = +1 +5 -+5= +4 - +5 = -1 +3- - +5 = -2 14 ALGEBRA. [Ch. I a remainder indicates that the subtrahend is greater than the minuend by as many units as are contained in the remainder. Thus, in +10 - +15 = "5 and +87 - +92 =-5, the remainder, "6, indicates that the subtrahend is, in each case, 5 units greater than the minuend. 6. The results of the preceding articles, restated briefly, are : A positive number is a number greater than zero, by as many units as are contained in its absolute value. E.g., +2 is two units greater than 0. A negative number is a number less than zero by as many units as are contained in its absolute value. E.g., ~3 is three units less than 0. Zero is the result of subtracting a number from an equal number. E.g., = +7 - +7 = -5 - -5 = +n - +fl = -n - 'n, ■wherein n denotes any absolute number. Since zero can be neither greater nor less than itself, it is neither a positive nor a negative number. It stands by itself, as the number from which positive and negative numbers are counted. 7. The Sign of Continuation, •■•, read and so on, or and so on to, is used to indicate that a succession of numbers continues vsrithout end, as 1, 2, 3, •••, read, one, two, three, and so on; or that the succession continues as far as a certain number which is written after the sign •■•, as 1, 2, 3, •••, 10, read one, two, three, as far as, or to, 10. We may now write the series of algebraic numbers : ... -4, -3,-2,-1,0,+!, +2, +3, +4,... In this series the numbers increase from left to right, and decrease from right to left ; or a number is greater than any number on its left and less than any number on its right. The numbers of ordinary Arithmetic are the absolute values of the positive and negative numbers of Algebra. §2] ALGEBRAIC NUMBERS. 15 Relations between Positive and Negative Numbers and Zero. 8. From the results of the preceding article, we obtain the following general relations : (i.,) Of two positive numbers, that number is the greater which has the greater absolute value; and that number is the less which has the less absolute value. (ii.) Of two negative numbers, that number is the greater which has the less absolute value; and that number is the less which has the greater absolute value. For example, ~3 >~5, or ~5 < ~3, since "5 is five units less than 0, and ~3 is only three units less than 0. 9. Although negative numbers arise through the extension of the operation of subtraction, it is- necessary to treat them as numbers apart from this particular operation. As in Arithmetic, so in Algebra, any integer is an aggregate of like units. Just as 4 = 1 + 1 + 1 + 1, so +4 =+1 ++1 ++1 ++1, and "4 =-1 +-1 +-1 +-1. Just as I = I + 1, so +(f)=+(i)++a), and -(i)=-a)+-a). Since letters are to represent numbers which may have any values whatever, they can represent either positive or negative numbers. Thus, in one case a may have the value +2, in •another case the value ~7 ; in the first case the absolute value of a is 2, in the second case the absolute value of a is 7. EXBRCISBS VI. 1. What is the absolute value of +8 ? 01 -11 ? Of +(2 + y) ? For what values of x do the following expressions reduce to : 2. CB - +3 ? 3. a; - -18 ? i. x-+a? 5. a; - -(o + 6) ? What values of a make the first members of the following equations identical with the second members : 6. o - +7 = +2 ? 7. a - +7 = -2 ? 8. a - +7 = -5 ? What are the results of the following indicated operations : 9. +17- +2? 10. -1-2 -+30? 11. +19 -+25? 12. +(i)-+(|)? 13. +Ca + 2)-+2? 14. +n-+(n + 3)? 15. +9 - +(re + 9) ? 16 ALGEBRA. [Ch. 1 How many units is each of the following numbers greater or less than : -3? 18. -(i)? 19. +x? 20. -y^ 16. +10? 17. Which number is greater, 21. +3 or -5 ? 22. -12 or -5 ? 23. or -3 ? 24. -5 or +4 ? 25. +(5+a)or+a? 26. -(5 + a) or -a? 27. -(6 + a) or +(2 + a) ? 28. -(a + 1) or -{a - 1) ? 29. -(a + 1) or +(a - 1) ? Positive and Negative Numbers are Opposite Numbers. 10. The student is familiar with the principle of subtraction in Arithmetic that the remainder added to the subtrahend is equal to the minuend. This principle, like all principles of arith- metical operations, is retained in Algebra. Consequently, continuing equations (iv.). Art. 5, we have : Subt. + Eem. = Min. +5 +-2 =+3] (-)' ^^^ +5+-4=n +5 +-5= 0. Min. — Subt, = Kem. +3 -+6=-2- +2 -+6=-3 +1 -+5 =-4 -+5=-6. (^i-) 11. The last of equations (vi.), +5 + "5 = 0, furnishes an important relation between positive and negative numbers : The sum of a positive and a negative number having the same absolute value is equal to zero; i.e., two such numbers cancel each other when united by addition. E.g., +1 + -1 = 0, +3 + -3 = 0, -17^ + +17^ = 0. In general, +n +-n = 0. For this reason, positive and negative numbers in their rela- tion to each other are called opposite numbers. When their absolute values are equal, they are called equal and opposite numbers. 12. Any quantities which in their relation to each other are opposite, may be represented in Algebra by positive and negative numbers ; as credits and debits, gain and loss. Ex. 1. 100 dollars credit and 100 dollars debit cancel each other. That is, 100 dollars credit united with 100 dollars debit is equal to neither credit nor debit ; or, §2] ALGEBRAIC NUMBERS. 17 100 dollars credit + 100 dollars debit = neither credit nor debit. If credits be taken positively and debits negatively, then 100 dollars credit may be represented by +100, and ioO dollars debit by -100. Their united effect, as stated above, may then be represented algebraically thus : +100 +-100 = 0. The result, 0, means neither credit nor debit. Similarly for opposite temperatures. Ex. 2. If a body is first heated 10° and then cooled down 8°, its final temperature is 2° above its original temperature ; or, stated algebraically, +10 +-8 =+2. The result, +2, means a rise of 2° in temperature. Similar reasoning applies to opposite directions. Ex. 3. If a man walks 10 miles due north, and turning, walks 14 miles due south, he is then 4 miles south of his starting point; or, +10 +-14 =-4. The result, -4, means that he is now 4 miles south of his starting point. 13. It is evidently immaterial which of two opposite quanti- ties is taken positively and which negatively in any particular problem. Thus, ,we might call distances south positive and dis- tances north negative. We have only to interpret results differently. BXBECISES VII. State algebraically in two ways each of the following relations (by Art. 13): 1. 100 dollars gain and 20 dollars loss is equivalent to 80 dollars net gain. 3. 250 dollars gain and 250 dollars loss is equivalent to neither gain nor loss. 3. A rise of 16° in temperature followed by a fall of 22° is equivalent to a fall of 7°. 18 ALGEBRA. [Ch. 1 4. If a man ascends from the foot of a ladder 20 steps, and then descends 7 steps, he is 13 steps up. 5. If a man ascends from the foot of a ladder 10 steps, and then descends 10 steps, he is at the foot of the ladder. 6. If a man walks 150 feet to the right, then 50 feet to the left, and then 75 feet to the right, he is finally 175 feet to the right of his original position. 7. Two men, A and B, run a race. The first minute A runs 5 feet more than B, the second minute A runs 8 feet less than B ; in the two minutes A runs 3 feet less than B. 14. We have deiined negative numbers as numhers less than zero ; that is, as the result of enlarging our conception of the operation of sub- traction. We afterward find, as we have seen, that they often have a meaning when applied to practical problems. Yet even if they had not, we should be justified in introducing them in order to make our principles general. Sometimes, indeed, negative results indicate an impossibility. E.g., a men are at work on a building, and 6 men quit work. How many are still working ? Evidently a — b. If 6 >a, the negative result has no meaning, and indicates that we have stated an impossibility. A man has a dollars and pays out 6 dollars. How many dollars has he left ? Evidently a-h. When 6 > a, the negative result has no meaning, if the money be regarded as actually handled. But in dealing with book accounts, it is quite possible that the debits shall exceed the credits ; a state which would, as we have seen, be indicated by a negative result (if credits be taken positively and debits negatively) . So, too, when applied to opposi- tion in direction, etc., negative results are as intelligible as positive results. In fact, there is no more objection to the use of negative numbers than to the use of fractions, for each kind of number may indicate an impos- sible state. For instance, there are a men in a company, which is divided into 6 equal groups. How many men are there in each group? Evi- dently a -=-6. If a be not exactly divisible by 6, the result is as impossible as taking 6 men from a men, when 6 > a. In Arithmetic we proceed to prove all the laws of fractions, without inquiring whether they can be applied in all cases. So, in Algebra, we shall proceed to operate with and upon negative numbers without inquir- ing whether or not they will always have a meaning in particular problems. CHAPTER II. THE POUR FUNDAMENTAL OPERATIONS WITH ALGEBRAIC NUMBER. § 1. ADDITION OF ALGEBEAIC NUMBERS. 1. Addition of one algebraic number to another is the process of uniting it with the other into one aggregate. As in Aritlimetic, the one number is said to be added to the other, and the result of the addition is called the Sum. Addition of Numbers T;7ith Like Signs. 2. Ex. 1. Add +3 to +4. The three positive units, +3, when united by addition with the four positive units, +4, give an aggregate of four plus three, or seven, positive units. That is, +4 + +3 = +(4 + 3) = +7. In like manner Ex.2. -4 + -3 = -(4 + 3) = -7. Ex. 3. +2 + +(f) = +(2 + 1) = +2|. These examples illustrate the following principle : To add one algebraic number to another, with like sign (i.e., both numbers positive or both negative), add arithmetically the absolute value of the one number to the absolute value of the other, and prefix to the sum the common sign of quality. Or, stated symbolically, +o++6=+(a + 6) (i.). -a+-6=-(a + 6) (ii.). Addition of Numbers with Unlike Signs. 3. Ex. 1. Add -2 to +5. The two negative units, "2, when united by addition with the five positive units, """5, cancel two of the Jive positive units 19 20 ALGEBRA. [Ch. II (Chap. I., § 2, Art. 11). There remain then five minus two, or three, positive units. That is, +5 +-2 =+(6 -2) =+3. Ex. 2. Add +2 to -5. The two positive units, +2, when united by addition with the five negative units, ~6, cancel two of the five negative units. There remain then five minus two, or three, negative units. That is, -5 ++2 =-(5 - 2) =-3. Observe that in both examples the sum is of the same quality as the number which has the greater absolute value, and that the absolute value of the sum is obtained by subtracting the less abso- lute value, 2, from the greater, 5. Ex. 3. +2 +-(f) =+(2 - f) =+1^. These examples illustrate the following principle : To add one algebraic number to another, with unlike sign, sub- tract arithmetically the less absolute value from the greater, and prefix to the remainder the sign of quality of the number which has the greater absolute value. Or, stated symbolically, +a +~6 =+(a — 6), when a > 6 (iii.), +0 +~6 ="(6 — o), when a< 6 (iv.). 4. The proofs of principles (i.)-(iv.) are as follows : In (i.), the positive units and parts of positive units represented by +6,- when united by addition with the positive units and parts of positive units represented by +a, give an aggregate of positive units and parts of posi- tive units represented by +(a + 6). In lilce manner (ii.) can be proved. In (iii.), the negative units and parts of negative units represented by —6, when united by addition with the positive units and parts of positive units represented by +a, cancel an equal number of positive units and parts of positive units. There remain positive units and parts of positive units represented by + (a — 6). In like manner (iv.) can be proved. Addition of Three or More Numbers. 5. To unite three or more algebraic numbers by addition, add the second to the first, to that sum add the third, again to that sum the fourth, and so on. Ex.1. +2 ++3 ++7 =+5 ++7 =+12. Ex.2. +11 +-8 ++2 =+3 ++2 =+5. § 1] ADDITION OF ALGEBRAIC NUMBERS. 21 EXERCISES I. 5"ind the results of the following indicated additions : 1. +11 ++5. 2. -9 + -5. 3. -2|+-3J. 4. -16 + +7. 5. +16 + -7. 6. -3f + +l^. - 7. +5J + -2f. 8. + -5. 9. +7 + -5 + +8. 10. -8 + +11 + -3. 11. -81 + +70 + -ISO + +12. Add 12. +17 to +5. 13. -6 to -27. 14. +13 to +a. 15. -18 to -&. 16. +10 to -5. 17. -20 to +6. 18. +20 to -6. 19. -(|) to +(f). 20. +11 to -a, when a > 11. 21. +11 to -a, when o < 11. 22. -17 to +x, when x > 17. 23. -17 to +x, when x < 17. 24. -2 to +8 + -4. 25. +18 to -2 + +35. 26. "5 to -11 + +15. Find the results of the following indicated additions, first uniting the numbers within the parentheses : 27. +7 +(+8 + -3). 28. +11 +(-12 + +2). 29. (+2 + -3) + (-11 + +12). ' 30. (+5 + -8) + (-12 + +3). What is the value of a + &, 31. When a = +5, 6 = +3? 32. When a = -7|, 6 = -3|? 33. When a = +71, 6 = -53? 34. When a = +25, 6 = -34? 35. When a = +2 + -3, & = -8 + +7 ? 36. When a = -5 + +3, 6 = +11 + -4 ? What is the value of a + 6, wherein a = m + n and b =p + q, 37. When m = -1, n = +2, p = -3, g = +4 ? 38. When m = +8, n = -3, p = -5, q = -9? The Associative and Commutative Laws for Addition. 6. In the preceding articles the process of addition has been carried out from left to right from number to number. E.g., +7 +-3 ++5 +-8 =+4 +-+5 +"8 =+9 +-8 = +1. But the result is the same if two or more successive numbers be associated in performing the additions. Kg., +7 +-3 + (+6 +-8) =+7 +'3 +'3 =+4 +-3 =+1. +7 + (-3 ++5) +-8 =+7 ++2 +-8 =+9 +-8 =+1. 22 ALGEBRA. [Ch- II This example illustrates the following principle : The Associative Law. — The mm of three or more numbers is the same in whatever way successive numbers are grouped- or associated in the process of adding. Or, stated symbolically, a + b +c = a + (b + c); a + b + c+d=a+b + ic+d)=a + (b+c+(/)=a + (b+c)+d. 7. In an indicated addition, the number on- the right of the sign + is to be added to the number on its left. .E.^., In +5 +-3, =+2, ~3 is added to +5 ; while in -3 ++5, =+2, +6 is added to -3. But the result is the same, whichever of the two numbers, +5 and ~3, be added to the other. That is, +5 +-3 =-3 ++5. This example illustrates the following principle : The Commutative Law. — The sum of two or more numbers is the same in whatever order they may be added. Or, stated symbolically, a+b=b+a a+b+c+d=b+a+d+c = d + c + b + a, etc. 8. The proof of the principles enunciated in Arts. 6 and 7 is as follows : The total number of units and parts of units, positive and negative, in the given numbers is the same in whatever way they may be grouped or arranged ; a given number of positive units will cancel an equal number of negative units, and vice versa ; and a given number of parts of positive units will cancel an equal number of like parts of negative units, and vice versa. Therefore the final result will be the same, whatever order or way of associating the units and parts of units may be used. 9. Since a + b = b + a, the two numbers a and b, when united by addition, are given the common name Summand. 10. The Associative and Commutative Laws may be applied simultaneously. E.g., -2 ++4 +-3 ++1 = (-2 +"3) + (+4 ++1) ="5 ++5 = 0. In general, a + b + c = a + (c + b) = c + (a + b), etc. §1] ADDITION OF ALGEBRAIC NUMBERS. 23 11. In adding three or more numbers, some of wMch are positive and some negative, the Commutative and Associative Laws enable us to employ the following method : Add all the numbers of one sign, then all the numbers of the opposite sign, and add the two resulting sums. E.g., -8 ++3 +-5 ++7 ++3 =-8 +-5 ++3 ++7 ++3 =-13 ++13 = 0. EXERCISES II. Find, In three different ways, by applying the Commutative Law, the values of : 1. +18 + -4 + +2. 2. +12 + -13 + +1. 3. -20 + -3 + +17. Find, in the most convenient way, the values of : 4. -998 + +500 + -2. 5. +333| + -125 + +e6f . Find, in the most convenient way, the value of a-\-b + c+ d, 6. When a = -5, 6 = +100, c = -95, d = +4. 7. When o = -763, 6 = +1000, c = -237, d = ~Z. Find the values of the following expressions by the method of Art. 11 : 8. +3 +-4 +-6 ++9 ++2. 9. -5 ++7 ++19 +-15 +-22. 10. -13 ++5 +-15 ++8 +-4. n. -(|)++(|)+-(J)+-(|)++(i). 12. In ordinary Arithmetic to add a number to any number . increases the latter. E.g., 7 + 4 = 11, and 11 > 7. But such is not always the case in adding one algebraic number to another. E.g., +7 ++4 =+11, and +11 > +7 ; but +7 +-4= +3, and +3<+7. Property of Zero in Addition. 13. We have +3 ++2 +-2 =+3. But +3 ++2 +-2 =+3 + (+2 +-2), by Assoc. Law, =+3 + 0, since +2 +-2 = 0. Therefore, by Axiom (iv.), +3 + =+3.- In general, JV++a +"a = JV. But N++a+-a = N+{+a+-a)=N+0. Therefore, by Axiom (iv.), N + Q = N. (i.) 24 ALGEBRA. [Ch. 11 §2. SUBTRACTION OF ALGEBRAIC NUMBERS. 1. Subtraction is the inverse of addition. In addition two numbers are given, and it is required to find their sum. In subtraction the sum and one of the numbers are given, and it is required to find the other number. As in ordinary Arithmetic, the given sum is called the Minuend, the given number the Subtrahend, and the required number the Remainder. Ex. 1. Subtract "2 from +9 +"2. We have (+9 +-2) — ~2 =+9, by definition of subtraction. That is, if from the sum of two numbers either of the numbers be subtracted, the remainder is the other number. In general, if the given sum be a + b, we have, by the defi- nition of subtraction, (a + 6) - 6 = a (i.), and (a + b) - a = b (ii.). 2. The miquend is, as a rule, a single number, and does not appear as a sum of two numbers, one of which is the given subtrahend. We must, therefore, derive from the definition of subtraction a principle which will enable us to subtract any one number from any other. Ex. 1. Subtract +6 from +7. In +7— ■'"5, the minuend, +7, is to be expressed as the sum of two numbers, one of which is +5. But +7 =+7 + (-5 ++6), by § 1, Art. 13, = (+7 +""6) ++5, by Assoc. Law. Therefore, by definition of subtraction, +7 -+6 = [(+7 +-5) ++6]-+5 = +7 +-5 =+2. That is, to subtract +5 from +7 is equivalent to adding ~5 to +7. Ex. 2. Subtract "5 from +7. We have +7 - -6 = [(+7 + +5) + -6] - -5 = +7 ++6 =+12. §2] SUBTRACTION OF ALGEBRAIC NUMBERS. 25 That is, to subtract ~5from +7 is equivalent to adding +5 to +7. These examples illustrate the following principle : To subtract one number from another number, reverse the sign of quality of the subtrahend, and add. Or, stated symbolically, N-+b = N+-b (i.), N--b = N++b (ii.). E.g., +2 - +3 = +2 + -3, = -1. "2 - +3 = "2 + "3, = "5. +2 --3 =+2 ++3, =+5. -2 --3 =-2 ++3, =+1. 3. The proof of the principle enunciated in Art. 2, is as follows : Let N denote any number, positive or negative. Then in N-+b, -ZVis to be expressed as the sum of two numbers, one of which is +6. We have N= N + {-h ++6), by § 1, Art. 13, = (iV+-6) + +6, by Assoc. Law. Therefore, by the definition of subtraction, iV-+6=[(iV" + -&) + +6]-+6 = iV+-6. In like manner iV — -6 = i\r++6. Successive Additions and Subtractions. 4. Successive subtractions are carried out by applying the principle of subtraction at each step of the process. E.g., +7 --2 -+4 --8 =+9 -+4 --8 =+5 --8 =+13. In like manner successive additions and subtractions are performed. E.g., +9 +-5 -+2 ++4 =+4 -+2 ++4 =+2 ++4 =+6. 5. The following definition is based upon the principle that every operation of subtraction is equivalent to an operation of addition. An Algebraic Sum is an expression which consists of a chain of indicated additions and subtractions. E.g., a — b, x + y ^z, etc., are algebraic sums. EXERCISES III. Find the results of the following indicated subtractions : 1. (+2 ++9)- +2. 2. (+4 +-5) --5. 3. (-7 +-11) --11. 4. (-a +-7)— -7. 8. (-»»++»)— -m. 6. {-m++n) — +n. 26 ALGEBRA. [Ch. II 7. +18 -+5. 8. -28 -+17. 9. +7 -+8. 10. +11 -+25. 11. +41 --50. 12. -30 --35. 13. -66 --45. 14. +44 --52, 15. -2++3-+4. 16. +11 -+13 +-12. 17. -7 --11 +-2. 18. (-2--6) + (-6++3). 19. (+ll--8)-(-2-+3). Subtract : 30. +3 from +a, when a > 3. 21. +3 from +a, when a < 3. 22. -17 from -«, when x > 17. 23. -17 from -x, when x < 17. 24. -11 from +«. 25. +14 from -y. What is the value of a — &, 26. When a = +4, 6 = +3 ? 27. When a = +5, & = +6 ? 28. When a = -7, & = +8 ? 29. When o = -4, & = -9 ? 30. When a = "2 + -3, 6 = +11 - +4 ? 31. When a = +5 - -4, 6 = -18 -+11 What is the value of a — 6, wherein a = m + n and h =p — q, 32. When m = -1, n = +2, p = -3, q = +i1 S3. When m = +5, n = -6, p = -11, g = -12 ? 6. The following examples illustrate the meaning of results in the subtraction of algebraic numbers. Ex. 1. Two men, A and B, starting from the same point, P, walk at different rates in the same direction, A 8 miles to the point Q, B 11 miles to the point M. How far is B then from A ? +11 P oUZ ' o o -B(-B) Fig. 1. As we have seen in Ch. I., distances in one and the same direction may be represented by numbers of the same sign. Let distances toward the right be taken positively, as in Fig. I, and consequently distances toward the left negatively. The distance of B from A is then represented by QB, and QIi= PR-PQ = +11 - +8 = +3. The positive result shows that B is 3 miles to the right of A. In general, however far either may walk, the distance of B from A will always be obtained by subtracting A's distance from the .starting point from B's distance from the same point. §2] SUBTRACTION OF ALGEBRAIC NUMBERS. 27 Ex. 2. If A walks 8 miles to the left and B 11 miles to the left, theii distances from P are both negative, as in Fig. 2. -11 (£) U S' — o " oP Fig. 2. We then have qB = PB-Pq = -11 - -8 = -3. The negative result shows that B is 3 miles to the left of A. In a similar way other variations of the problem may be interpreted. EXERCISES IV. 1. A's assets are a dollars and B's are 6 dollars. What number expresses the excess of A's assets over B's, if assets be taken positively ? What number, if assets be taken negatively ? What are the meanings of the results of Ex. 1, 2. When a = 3500, & = 2750 ? 3. When a = 2000, 6 = 2000 ? 4. When a = 2600, 6 = 3000 ? 5. The temperature in Chicago on a certain day was a° and in Phila- delphia 6°. What number expresses the excess of temperature in Chicago over that in Philadelphia 1 What is the meaning of the result of Ex. 5, taking temperature above zero positively, 6. When a = +90, & = +68 ? 7. When a = +65, 6 = +98 ? 8. When a = -12, 6 = -4 ? 9. "When a = "5, 6 = -8 ? 10. When a = +5, & = -2 ? 11. When a = -6, 6 = +3 ? The Associative and Commutative Laws for Subtraction. 7. If a number, preceded by the sign +, or the sign — , stand first in a chain of additions and subtractions, or first within parentheses, it may be regarded as added to, or subtracted from, 0. Thus, ++8 -+3 = + +8 -+3, -+3++8 = 0-+3++8. Siace every operation of subtraction is equivalent to an operation of addition, it follows that the Associative and Com- mutative Laws which were proved for addition hold also for subtraction, and for successive additions and subtractions. 28 ALGEBRA. [Ch. li Ex. ++8 -+3 = ++8 +-3, since -+3 = +"3 = +~3 +''"8, by Comm. Law = -+3 ++8, since +-3 = -+3. Observe that in changing the order of the operations the sign of operation, + or —, must be transferred with each number. The method of applying the Associative Law depends upon a proper use of parentheses, which will be taken up in the next article. EXERCISES V. Find in three different ways, by applying the Commutative Law, the values of : 1. +8 + -3 - +4. 2. -17 - +12 + -5. 3. +28 - -14 + -2. 4. -31 - -17 + -36 + +46 - -11 - +19 + -49 + +11. 5. -45 + +81 - -15 - +12 + -5 - -9 + -8 + +4. Find, in the most convenient way, the values of : 6. +103 - -12 - +3. 7. -799 - -11 + -1. Removal of Parentheses. 8. We have +9 + (+5 ++6) =+9 ++5 ++6, since to add the sum +6 ++6 is equivalent to adding succes- sively the single numbers of that sum. Again, +9 + (+5 -+6) =+9 + (+5 +-6), since -+6 = +-6, =+9 -(-+6 +"6, removing parentheses, =+9 ++5 -+6, since +"6 =-+6. This example illustrates the following principle : (i.) When the sign of addition, +, precedes parentheses, they may be removed, and the signs of operation, + and — , within them be left unchanged; that is, /t+(+a + b) = W + a + b, a + (+ a - b) = N + a ~ b, etc. We have +9-(+5 + '-6)=+9-+5-+6, since to subtract the sum +6 ++6 is equivalent to subtracting successively the single numbers of that sum. Again, +9 - (+5 -+6) =+9 - (+.5 +"6), since -+6 = +-6, = +9 — +5 — ~6, removing parentheses, =+9 -+6 ++6, since --6 =++6. §2] SUBTRACTION OF ALGEBRAIC NUMBERS. 29 This example illustrates the folio-wing principle : (ii.) When the sign of subtraction, —, precedes parentheses, they may he removed, if the signs of operation within them be reversed from + to —, and from — to + ; that is, f/~(+a + b)=H-a~b, N~{+a — b) = N-a + b, etc. For, -ZV" + (+a+ +6)=iV++a + +6, since to add the sum +a + +6 is equivalent to adding successively the single numbers of that sum. iV + (+a-+6)=iV+(+a + -6), since -+6= +-6, =: N++a + -b, removing parentheses, = N'+ +a — +&, since + -6 = — +6. Evidently the preceding proof does not depend upon the signs of quality of the numbers within the parentheses, nor upon how many numbers are Inclosed. In a similar manner (ii.) is proved. Insertion of Parentheses. 9. The insertion of parentheses is the converse of the process of removing them. (i.) An expression may be inclosed within parentheses pre- ceded by the sign of addition, if the signs of operation, + and —, preceding the numbers inclosed within the parentheses remain unchanged. E.g., ++7 -+5 +-3 --4 = ++7 +(-+5 +"3 --4) = ++7-+6+(+-3--4) = ++7-+5+-3+(--4). (ii.) An expression may be inclosed within parentheses pre- ceded by the sign of subtraction, if the signs of operation preceding the numbers inclosed within the parentheses be reversed, from + to — and from — to -\-. Kg., ++7 -+5 +-3 --4 = ++7 -(++5 -"3 +"4) = ++7-+5-(--3+-4) = ++7-+5+-3-(+-4). The insertion of parentheses is a direct application of the Associative Law. 30 ALGEBRA. [Ch. II BXEBCISES VI. Find the values of the following expressions, first removing parentheses ■. 1. +12 + (+4 + -6). 2. -15 +(-6- +2). 3. +28 +(-5- +6). 4. +18 +(-2 + -3 --5). 5. +11 - (+12 + -5). 6. +15 - (-6 - +2). 7. -17 -(-3 --5). 8. -21 -(-4 + -5 --6). 9. m + (m —p), when m = -4, n = -6, p = +5. 10. x — {y — z), when x = +3, !/ = -4, 2 = +5. Insert parentheses in the expression +8 —-5 +-3 — +7, 11. To inclose the last three numbers preceded by the sign +; pre- ceded by the sign — . 12. To inclose the last two numbers preceded by the sign +; pre- ceded by the sign — . 13. To inclose the first and third numbers preceded by the sign + ; preceded by the sign — . 10. In ordinary Arithmetic, to subtract a number from any number decreases the latter. E.g., 7-4 = 3, and 3 < 7. But such is not always the case in subtracting one algebraic number from another. Thus, +7 -+4= +3, and +3<+7; but +7 --4 =+11, and +11 > +7. Property of Zero in Subtraction. 11. From § 1, Art. 13, we have N+(i = N. If, therefore, from N, which is the sum of N and 0, be sub- tracted either N or 0, the remainder is or N, respectively, by the definition of subtraction. Thatis, N-N = 0, SLTid. N-0 = N. (i.) §3. MULTIPLICATION OF ALGEBRAIC NUMBERS. 1. As in Arithmetic, the number multiplied is called the Multiplicand, the number that multiplies the Multiplier, and the result the Product. In ordinary Arithmetic, multiplication by an integer is defined as an abbreviated addition. Thus, to §3] MULTIPLICATION OF ALGEBRAIC NUMBERS. 31 multiply 4 by 3, tlie number 4 is used three times as a sum- mand; or 4x3 = 4 + 4 + 4. Now the number 3 stands for an aggregate of three units ; or 3 = 1 + 1 + 1. We thus see that, just as 3 is obtained by taking the unit, 1, three times as a summand, so the product 4 x 3 is obtained by taMng 4 three times as a summand. 2. We are thus naturally led to the following definition of multiplication : Tlie product is obtained from the multiplicand just as the mul- tiplier is obtained from the positive unit. The above definition is an extension of the meaning of arith- metical multiplication when the multiplier is an integer, and gives an intelligible meaning to arithmetical multiplication when the multiplier is a fraction. Thus, f is obtained from the unit, 1, by taking one-third of the latter twice as a summand ; or 2 — 1 -4- i. In like manner, to multiply 6 by f, we take one-third of 6 twice as a summand ; or 5x-| = | + | = J^. 3. There are two cases to be considered in the multiplica- tion of algebraic numbers. (i.) The Multiplier Positive, — Ex. 1. Multiply +4 by +3. By the definition of multiplication, the product, +4 X +3, is obtained from +4 just as +3 is obtained from the positive unit. But +3 is obtained from the positive unit by taking the latter three times as a summand; or +3=+l++l++l. Consequently the required product is obtained by taking +4 three times as a summand; or +4 X +3 =+4 ++4 ++4 =+(4 + 4 + 4)=+(4 x 3)=+12. 32 ALGEBRA. [Ch. U Ex. 2. Multiply -4 by +3. By the definition of multiplication, we have -4 x+3 =-4 +-4 +-4 =-(4 + 4 + 4) = -(4 x 3) = -12. (ii.) The Multiplier Negative. — Ex. 3. Multiply +4 by "3. By the definition of multiplication, the product, +4 X -3, is obtained from +4 just as ~3 is obtained from the positive unit. But -3 =-1 +-!+-!= -+1-+1-+1; that is, "3 is obtained by subtracting the positive unit, +1, three times in succession from 0. Consequently, the required product is obtained by subtracting the multiplicand, +4, three times in succession from ; or +4 x-3 = -+4 -+4 -+4 = +-4 +-4 +"4 =-(4 x 3). Ex. 4. Multiply -4 by S. By the definition of multiplication, we have -4 X -3 = --4 --4 --4 = ++4 ++4 ++4 =+(4 x 3). 4. In Art. 3 the examples were limited to the multiplication of integers having the same or opposite signs. But the essential part of the results therein obtained is the sign of the product. Since this sign depends only upon the signs of the multi- plicand and multiplier, and not upon their absolute values, the sign of the product in each example would have been the same as above, if the multiplicand and multiplier, either or both, had been fractions. These examples illustrate the following Rule of Signs for Multiplication : The product of two numbers having like signs is positive; and the product of two numbers having unlike signs is negative. Or, stated symbolically, +a X +b =+(ab), -a x +b =-(ab), -a x~b =+(a6). +a x~b =-(ab). §3] MULTIPLICATION OF ALGEBRAIC NUMBERS. 33 5. The proof of the principle enunciated in Art. 4 is as follows : The product +a x +6, wherein a and 6 are, as yet, limited to integral values, is obtained from +a just as +6 is obtained from the positive unit. Bat +6 is obtained from +1 by taking the latter 6 times as a summand; or, +6 =+1 ++1 ++1 + ...6 summands. Therefore the required product is obtained by taking +a as a summand b times; or +^ x+b =+a++a-\-+a + -b summands =+(a + a + a + ••• 6 summands) =+(a6). Consequently, _ +a x+6 =+(ab). In like manner, the other products are proved. Since the essential part of the above proof is the sign of the product, the results hold when a and 6 have fractional values. Continued Products. 6. The results of the preceding articles may be applied to determine the value of a chain of indicated multiplications, i.e., of a continued product. E.g., +ax+b X +c =+(a6) x +c =+(abo), +a X +6 X ~c =+(a6) x ~c =-(abc), +a x~b X ~c =~(ab) x ~c =+(abc), ~a x~b X ~c ='*'(ab) x ~c =-{abc). These equations illustrate a more general rule of signs : A continued product which contains no negative number, or an even number of negative numbers, is positive; one that contains an odd number of negative number's is negative. In practice the sign of a required product may first be deter- mined by inspection, and that sign prefixed to the product of the absolute values of the numbers in the continued product. E.g., the sign of the product (+2), X (-3) x (-7) X (+4) X (-5) is negative, since it contains three negative numbers ; the product of the absolute values is 840. Consequently, (+2) X (-3) X (-7) X (+4) X (-5) =-840. 34 ALGEBKA. [Ch. II EXERCISES VII. Find the values of the following indicated multiplications : 1. +2 X +3. 2. +5 X +7. 3. -3x+7. 4. -11 X +9. 8. +5 X -6. 6. +7 X -4. 7. -7 X-9. 8. -22 x-6. Find the values of : 9. (+15x-4) + (-12x+l). 10. (+16x-3)-(+5x+7). 11. (-lx+ll)-(-22x+2). 12. (-52x-2)-(+12x-3). Find the values of the following continued products : 13. -2 X +4 X -3. 14. -5 X -6 X +7. IS. +12 x -2 x -5 x -4. What is the value of (a — 6) (c + d) , 16. When a =-2, 6 =+4, c=-5, d = +Q1 17. When a =+5, 6 =-8, c=-9, d=+10? What is the value of abed, 18. When a=-2, &=+3, c=-4, d=+5? 19. When a ="7, 6 =+2, c=-3, d=-5? The Commutative La^v for Multiplication. 7. In an indicated multiplication, the number -which follows the symbol of multiplication is the multiplier. Thus, in -4 x+3 =-4 +-4 +-4, =-12, the multiplier is +3 ; while in +3 x-4 = -+3-+3-+3-+3, =-12, the multiplier is ~4. But the result is the same, whichever of the two numbers, +3 or -4, is used as the multiplier.. This example illustrates the following principle : The Commutative Law. — Tlie product of two numbers is the same, if either be taken as the multiplier and the other as the multiplicand; or, stated symbolically, a X b = b X a. 8. It follows from the rule of signs. Art. 4, that the signs of the multiplier and multiplicand may be interchanged with- out affecting the sign of the product. E.g., +3 X -4 =-(3 x 4) =-12, and -3 x +4 =-(3 x 4) =-12. This principle is called the Commutative Law of Signs. § 3] MULTIPLICATION OF ALGEBRAIC NUMBERS. 35 "We have, therefore, to prove the Comnmtative Law only for the multiplication of absolute numbers. Take first a particular example : 4x3 = 3x4. Consider the following arrangement of units in rows and columns : 1111 1111 1111 The total number of units in this arrangement is obtained either by multiplying the number in each row, 4, by the num- ber of rows, 3, giving 4x3; or by multiplying the number of units in each column, 3, by the number of columns, 4, giving 3x4. Consequently, 4x3 = 3x4. In general, consider the following arrangement of units in rows and columns : a units in each row b rows The total number of units in this arrangement is obtained either by multiplying the number in each row, a, by the number of rows, 6, giving a X 6 ; or by multiplying the number in each column, &, by the number of columns, a, giving b x a. Consequently, a x b = b x a, wherein a and 6 are absolute integers. Consider next the case in which a and 6 denote absolute fractions. 4 2^ 4 5 3 5x3 + 4 5x3 4+4_4x2 by the definition of multiplication, 5x3 2x4 5x3 , since 4x2 = 2x4 and 5 x 3 r: 3 x 5, 3x5 _ 2+2+2+2 _ 2,2,2,2 3x5 : + + ; 3x5 3x5 3x5 : + = 2xi 3x5 3 5 36 ALGEBRA. [Ch. II Similar reasoning can be applied to the product of any two absolute fractions. Consequently, the Commutative Law for multiplication, a X b = b X a, holds for all values of a and b, positive or negative, integral or fractional. The Associative Law for Multiplication. 9. In finding the value of a contimied product in Art. 6, the indicated operations were performed successively from left to right. E.g., (+4 X +3) X -2 = +12 x -2 = -24. But the same result is obtained if +3 be first multiplied by "2 and then +4 be multiplied by the product. E.g., +4 x (+3 X -2) = +4 x "6 = -24. This example illustrates the following principle : The Associative Law. — The product of three numbers is the same in whichever way two successive numbers are grouped or associated in the process of multiplying; or, stated symbolically, (ab)c = a{bc). 10. For the reason stated in Art. 8, it is sufficient to prove this law for absolute numbers. Consider tlie following arrangement of 6's in rows and columns : c columns of 6's a rows of 6's b b b b b b b b b Each row contains b x c units; and, since there are a rows, the total number of units is (6 X c) X a, or a X (6 X c), by Art. 7. But each column contains 6 x a, or a x 6 units ; and, since there are c columns, the total number of units is (a X 6) X c. Therefore (a x 6) x c = a x (6 x c). §3] MULTIPLICATION OF ALGEBRAIC NUMBERS. 37 In the above representation, the values of a, 6, and c are limited to absolute integers. It can be shown, however, as in Art. 8, that the law holds also for absolute fractional values of a, 6, c. M.q., g v> 4 ^ 6 _ 2 X 4 6 _ 2 X 4 X 6 _ 2 X (4 X 6) _ 2 /4 6\ "'357 3x5^7~3x5x7~3x(5x7)~3^U 7J' We conclude, therefore, that the Associative Law for multiplication holds for all values of a, 6, c, positive or negative, integral or fractional. 11. The Associative and Commutative Laws may be extended as follows : The value of a product of three or more numbers remains the same if, in performing the indicated multiplications, the order of the numbers be changed, or if two or more numbers be associated in any way. E.g., +2 X -3 X -4=+2 x "4 x -3=-4 x +2 x -3=etc. +2 X -3 X -4 X +5=+2 x (-3 x -4) x +5=+2 x "3 x ("4 x +5) = etc. In general, abc = acb = bca = bac = cab = cba. abed = a(bcd) = a(bc)d = ab{cd). The two laws may be applied simultaneously. E.g., +2 x -3 X -4 X +5=+2 X (+5 X -3) x -4=-3 x ("4 x +2 x +5)=etc. In general, abed = a(cb)d = e(adb) = etc. 12. Since the multiplier and the multiplicand can be inter- changed without affecting the value of the product, they are both given the common name Factor. Thus, a and 6 are the factors of the product ab. For a similar reason, each number in a continued product is called a factor of the product. Thus a, b, c, and d are factors of abed. In subsequent work we shall, for convenience in writing, frequently place the multiplier on the left. EXERCISES VIII. 1. Express the sum, +4 + +4 + +4, as a sum of summands each equal to +3. 2. Express the sum, -5 + -5 -1- -5 + -5, as a sum of summands each equal to -4. 38 ALGEBRA. [Ch. II 3. Express the sum, -(f) + -(|) + -(f) + -(|) + -(f) + -(i). as a sum of summands each equal to -(f). 4. Express the sum, +9 + +9 + +9 + ■■■ a summands, as a sum of summands each equal to +a. 5. Express the sum, -5 + -5 + ■■■x summands, as a sum of sum- mands each equal to -x. Find, in the most convenient way, the values of : 6. -17 X +5 X -2. 7. +38 x -2| x -4. 8. -139 x -3 x +33J. 9. +228 X +250 x -4. 10. -139 x -8 x +12^. 11. -17 x -16f x -3. Find, In the most convenient way, the value of abc : 12. When a=-4, 6=+33J, c=-9. 13. When a=+19, 6 = -66|, c=-S. Find, in the most convenient way, the value of abed, 14. When a = -37|, 6 = +5, c = -3, d = +8. 15. When a = -12|, 6 = -16, c = +33^, d = -6. §4. DIVISION OF ALGEBRAIC NUMBERS. 1. Division is the inverse of multiplication. In multiplication two factors are given, and it is required to find their product. In division the product of two factors and one of them are given, and it is required to find the other factor. As in ordi- nary Arithmetic, the given product is called the Dividend, the given factor the Divisor, and the required factor the Quotient. ^.gr., Since -28=-4x+7, therefore, -28--+7=-4, and -28 -=--4 =+7. 2. Prom the definition of division we infer the following principle : If the product of two factors be divided by either of the factors, the quotient is the other factor. In general, if the given product be a x 6, we have, by the definition of division, (a X 6) H- 6 = a, and (a x 6) -=- a = 6. (i.) 3. The dividend is, as a rule, a single number and does not appear as the product of two factors, one of which is the divisor. §4] DIVISION OF ALGEBRAIC NUMBERS. 39 Since the absolute value of the product of two factors is equal to the product of their absolute values, it follows, from the definition of division, that the absolute value of the quotient is equal to the quotient of the absolute values of the dividend and the divisor. By the definition of division, the equations of § 3, Art . 4, may be written +(ab)-i-+a=+b; -(a6)-f--a =+6; +(ab)-h-a =-b; -(a6)-=-+a =-«. Prom these equations, we derive the following Rule of Signs for Division : Like signs of dividend and divisor give a positive quotient; unlike signs of dividend and divisor give a negative quotient. E.g., +8-f-+2=+4; -8 -^-2 =+4; -8 -=-+2 =-4; +8h--2=-4. 4. In a chain of indicated divisions, the operations are to be performed successively from left to right. E.g., -16 -=-+4 --2 =-4 --2 =+2 ; +210 -r--3 -^-2 -f-+6 =-70 -=--2 -=-+5 =+36 -f-+5 =+7. Likewise, in a chain of indicated multiplications and divi- sions, the operations are to be performed successively from left to right. E.g., -375 X +3 -=--6 x +2 ^-9 =-1125 ^-5 x +2 -=--9 = +225 x+2 -^-9 =+450 ^-"9 ="50. 5. In a succession of additions, subtractions, multiplica- tions, and divisions, the multiplications and divisions are first to be performed, and then the additions and subtractions. E.g., +2 X -3 -F -4 x +5 = -6 -F -20 = -26. When a different order of performing the operations is pro- posed, the required order must be indicated by the insertion of parentheses. E.g., +2 X (-3 + -4) X +5 = +2 X -7 X +5 = "70, not -26, as before. 40 ALGEBRA. [Ch. II 6. From the definition of division, we have Quotient x Divisor = Dividend. Since the quotient of a divided by 6 is a h- 6, we have (a -7- 6) X 6 = a, or a H- 6 X 6 = a. (1) From (1) and Art. 2 (i.), we derive the following principle : If a given number be first divided and then multiplied by one and the same number, or be first multiplied and then divided by one and the same number, the result is the given number ; or, stated symbolically, /Vx6-^6 = /V-4-6x6 = /Vx+1=/Vh-+1=/V. That is, x 6 ^ 6 = -T- 6 X 6 = x +1 = -H +1, (2) whatever number be placed on the left of the two indicated operations. E.g., -11 X +3 -^ +3 = -11, +11 ^ "5 x -5 = +11. EXERCISES IX. Find the values of the following indicated divisions ; 1. (-27 X +3)-^+3. 2. (-27 X +3)- -27. 3. (+2| x +5f ) -=- +2f . 4. +27 H- +9. 5. +81 H- -9. 6. -33 -=- +11. 7. "105 -=- -7. Find the values of : 8. (-16-=-+2) + (+18--3). 9. (-24 H- -8) -(-36 -=-+6). 10. (-15 -. -5) - (+100 ^ -25) + (-200 ~ +8) . Find the values of : 11. +210 ^ -5 ^ -7 -^ +2. 12. +375 h- -5 x +2 -=- -3. 13. -280 ^ -4 X +2 X -3 -- -42. 14. +15 x -6 -f- -2 h- +5 x +4. Find the values of the following expressions : 15. +180 -=- -36 + -4 X -2. 16. -25 x +4 - +36 -=- -12. 17. +48 X -2 - -96 -- -24. 18. -7 + +15 ^- +3 x -2 - -28 -i- +4. What is the value of (a — 6) -f- (c + d), 19. When a = +18, 6 = -2, c = +3, d = +2? 20. When a = -23, 6 = +5, c = -4, d=-S? What is the value of o -=- 6 x c h- cZ, 21. When a = +125, 6 = -5, c = +i, d = "10 ? 22. When a = -49, 6 = -7, c = +18, rf = -2 ? §4] DIVISION OF ALGEBRAIC NUMBERS. 41 The Commutative IiavT- for Division. 7. In a chain of divisions, or of multiplications and divisions, the successive operations are to be performed, as has been stated, in order from left to right. E.g., -14 --+2 X -7 =-7 X -7 = +49. +8 -=--4 -^+2 =-2 -^+2 =-1. But, if the operations in the above examples be performed in a different order, the symbol of operation, x or -=-, being carried with its proper constituent, we have -14 X -7 -=-+2 =+98 H-+2 =+49, as above. +8 -=-+2 -=--4 = +4^-4= -1, as above. This example illustrates the Commutative Law : (i.) To multiply any number by a second number and then to divide the product by a third number, gives the same result as first to divide the given number by the third number and then to multiply the res^dting quotient by the second number; and vice versa ; or, stated symbolically, IVxb-r-c = IV^cxb, or x6-^c=-=-cx6. (ii.) If a given number be divided successively by two numbers, the result is the same whichever of the tioo divisions is first per- formed ; or, stated symbolically, N-irb^c = N^c^b, or -^-6-^c = -=-c^6. These principles are proved as follows : (1.) If in iV X 6 -=- c, N be replaced by iV-=- c x c, = N, by (2), Art. 6, we have iVx6-f-c = iV-i-cxcx6-f-c = iV-f-cx6xc^c, since xcx6 = x6xc, = JV-T- c X 6, since x c ^ c = x +1, by (2), Art. 6. In like manner, it can be shown that the Commutative Law holds for any number of successive multiplications and divisions. 42 ALGEBRA. [Ch. II The Associative Law for Division. 8. By Art. 4, "32 x +4 -=--2 ="128 --2 =+64, while, -32 X (+4 -- -2) = "32 x "2 = +64. Likewise, +32 -=- -4 x +2 = "8 x +2 = "16, while, +32--(-4-=-+2) = +32-=--2=-16. And, -32 -=- -4 -=- -2 = +8 -f- -2 = "4, while, -32 -=- (-4 x "2) = "32 -- +8 = -4. These examples illustrate the Associative Law : (i.) A chain of multiplications and divisions may be inclosed within parentheses preceded by the symbol of multiplication, if the symbols of operation, x and -^, preceding the numbers in- closed within the parentheses be left unchanged ; or, stated sym- bolically, /Vxa^6 = /irx(a^6). (ii.) A chain of multiplications and divisions may be inclosed within parentheses preceded by the symbol of division, if the symbols of operation, x and -=-, preceding the numbers inclosed within the parentheses be reversed from x to -=- and from -5- to X ; or, stated symbolically, yV^aH-6=;V^-(a x6), and/l^-f-a X 6 = yl^-^(a-=-6). The proof is as follows : K in JV-H a -=- 6, N be replaced by iV-=-(a x 6) x (a x 6), = JV, by (2), Art. 6, we have N'^a^b = N'-h(axb)x(axb)^a^b = N^(axb)xbxa-^a^b, since xax6=x6xa, = iV -H (a X 6) X 6 -7- 6, since x o h- a = x +1, = iVT-(a X 6), since x & -h 6 = x +1. In like manner the other principles are proved, and all can be extended to include any number of successive multiplications and divisions. 9. An even number is one whose absolute value is exactly divisible by 2 ; as 4, 6, etc. § 5] THE SIGNS OF QUALITY AND OPERATION. 43 Since, by the Commutative Law, 2 n is always an even number when n is an integer. EXERCISES X. Find, in the most convenient way, the values of : 1. -25x-12h-+5. 3. -100 X -7 H- -25. 3. -1000 x-11 -h+125, 4. +33J---20 x+3. 5. -30-r--9 x-12. 6. -10 -=-+17 x -34. Find, in the most convenient way, the value of a -^ 6 -^ c x d, 7. When a = +170, 6 = -3, c = +17, d = -6. 8. When a = -125, 6 = -7, c = +25, d=-li. Find the values of the following expressions, first removing the paren- 9. +25 x(+12h--4). 10. -20h-(-5--+2). 11. + 100-^ (+25 x -2). 12. -600-4-(-200-=--25x+3^-4). 13. +300-;- (-150-=-+6x +8^-4). § 5. ONE SET OF SIGNS FOB QUALITY AND OPERATION. 1. In conformity with the usage of most text-books of Algebra we shall in subsequent work use the one set of signs, + and — , to denote both quality and operation. For the sake of brevity the sign + is usually omitted when it denotes quality ; the sign — is never omitted. Thus, instead of +2, we shall write -f 2, or 2 ; instead of ~2, we shall write — 2. 2. We have used the double set of signs hitherto in order to emphasize the difference between quality and operation. It should be kept clearly in mind that the same distinction still exists. We now have N++2 = N+{-\-2) = N+2, omitting the sign of quality, +; iV+-2 = iV+(— 2), wherein + denotes operation, and — denotes quality. JV— +2 = iV— (+2) = -?7'— 2, omitting the sign of quality, +; M —~2 = N — (— 2), wherein the first sign, — , denotes opera- tion, the second sign, — , denotes quality. 44 ALGEBRA. [Ch. II 3. In the chain of operations (+2) + (-5) -(+2) -(-11) the signs within the parentheses denote quality, those without denote operation. That expression reduces to (+2) -(+6) -(+2) + (+11), or 2 — 5 — 2 + 11, dropping the sign of quality, +. In the latter expression all the signs denote operation, and the numbers are all positive. 4. The following examples illustrate the double use of the signs + and — . Ex. 1. +4 ++3 = + 4 + (+3) = 4 + 3 = 7. Ex.2. -5++2 = -6 + (+2)=-6 + 2 = -3. Ex. 3. +7--6 = + 7 - (- 5) =+ 7 + (+ 6) = 7 + 5 = 12. Ex. 4. -4 X +3 = - 4 X (+ 3) = -4 X 3 = - 12. Ex. 5. -4 X -3 = - 4 X (- 3) = 12. BXEBCISBS XI. Mnd the values of the expressions in Bxx. 1-8, first changing them into equivalent expressions in which there is only the one set of signs, + and — : 1. +2 + +3. 2. +14 + -9. 8. -8 + +5. 4. -4 + +5. 8. +8 - +4. 6. +5 - -2. 7. -8 - +3. 8. -2 - "8. Find the values of the expressions in Exx. 9-14, first changing them into equivalent expressions in v?hich there is only one set of signs, + and — , and then removing the parentheses : 9. +5 +(+4 + +3). 10. +7 +(-5 + +3). 11. -8 -(-12 + +4). 12. +17 -(+5 + -8). 13. +5 - (-3 + -4) + (-2 - +8) . 14. -7 - (-5 - -8) + (+6 - -7) . Find the values of the expressions in Exx. 15-22, first changing them into equivalent expressions in which there is only one set of signs, + and — : 15. +3 X +4. 16. +18 X -4. 17. "9 x +11. 18. -4 x -7. 19. +18 H- +2. 20. +35 H- - 5. 21. -18 - +3. 22. -96 -=- -6. §6] POSITIVE INTEGRAL POWERS. 45 What are the values of a + 6 - c + <^ - e and a - (6 - c + d - e), 23. When a = 3, 6=-5, c=-8, d = -9, e = 7? 24. When a = - 9, 6 = 6, c = - 9, d = - 11, e = - 12 ? What is the value of a (6 — c + d), 25. When a = 5, &=-9, c = -ll, fl! = 8? 26. When a = -2, & = 11, c = -12, d = -9? 27. When a = -13, & = -2, c = 3, d = -3? What is the value of a h- (6 + c — d + e), 28. When a = -5, 6=-3, c = 4, d = -5, e=-7? 29. When a = 12, & = - 2, c = - 9, d! = 13, e = 28 ? Find the results of the following indicated operations : 30. 4-7. 31. - 3 + 11. 32. 18 - 22. 33. 1-55. 34. 2 - 4 + 6. 35. 5 _ 8 - 2. 36. -2-3 + 17-25-18 + 1. 37. 1-4 + 5-6 + 8-11. 38. -7x11. 39. 15x(-8). 40. (-9)x(-17). 41. (-9)x(-8). 42. 18 ^(-2). 43. (-15)-=- 5. 44. (-72)-h(-12). 45. ( - 96) x 12 -=- 4. 46. 45-(-5)x3. 47. (-2)x(-3) + (-4)x5-7 x(-3). 48. 5 x(-8)-16 -=-(-4) + 2 x(-5). §6. POSITIVE INTEGRAL POWERS. 1. A continued product of equal factors is called a Power of that factor. Thus, 2 X 2 is called the second power of 2, or 2 raised to the second power; oMi is called the third power of a, or a raised to the third power. In general, aaa ■■• to n factors is called the nth power of a, or a raised to the nth power. The second power of a is often called the square of a, or a squared; and the third power of a the cube of a, or a cubed. 2. The notation for powers is abbreviated as follows : a^ is written instead of aa; a" instead of aaa; a" instead of aaa ••• to w factors. 3. The Base of a power is the number which is repeated as a factor. I!.g., a is the base of a^, a% •■■, a". 46 ALGEBRA. [Ch. II The Exponent of a power is the number which indicates how many times the base is used as a factor, and is written to the right and a little above the base. E.g., the exponent of a' is 2, of a^ is 3, of a" is n. The exponent 1 is usually omitted. Thus, a^ = a. An exponent must not be confused with a subscript. Thus, a^ stands for the product aaa; while a^ is a notation for a single number. 4. The definition of a power given above requires the esj- ponent to be a positive integer. In a subsequent chapter this definition will be extended to include powers with nega- tive and fractional exponents. ISTotice that the words positive integral refer to the exponent and not to the value of the power, which may be negative or fractional. E.g., (-2)^ = (-2)(-2)(-2) = -8; (|)^ = |x| = f 5. The base of a power must be inclosed within parentheses to prevent ambiguity : (i.) When the base is a negative number. Thus, (-5)2=(-5)(-6)=25; while -52 = -(5 x 5) = -26. (ii.) WJien the base is a product or a quotient. Thus, (2 X 5)'== (2 X 5) (2 X 5)(2 x 5) = 1000; while 2 X 5' = 2(5 X 5 X 6) = 250. Likewise f|Y=| x | = ^, while ?!=2 x^ 4 vSy 3 3 9 3 3 3 (iii.) When the base is a sum. Thus, (2 + 3)^ = (2 + 3)(2 + 3) = 5x5=25; while 2 + 32=2 + 3x3 = 2+9 = 11. (iv.) When the base is itself a power. Thus, (28)2= 23 X 2« = (2 X 2 X 2) (2 X 2 X 2) = 64, while 23* = 22x8= 29 = 2x3 v ■? -2 x2 x2x2 x2x2 = 512. §6] POSITIVE INTEGRAL POWERS. 47 EXERCISES XII. Express the following powers in the abbreviated notation : 1. 2x2x2. 2. (-3)(-3)(-3). 3. aaaaaa. 4. (_xy)(xy)(xy). 6. (- 3a)(- 3a)(- 3a)(- 3ffi). 6. 6 • 6 • 6 ••■ to 8 factors. 7. 5 x 5 x 5 ••• to m factors. 8. (— a6)(— a&)(— a&) ••• to n factors. 9. 32 X 32 X 32 X 32 X 32. 10. X" ■ X" ■ x" — to m factors. Express the following powers as continued products : 11. 25. 12. (-3)7. 13. {5xy. 14. (-62)6. 15. 4"'. 16. xf. 17. (aby. 18. (-7a;)». 19. (22)8. 20. (-3*)2. 81. (a*)2. 32. [(,-xyyY- Express the following powers in the abbreviated notation : 23. {a + x){a+x){a + x)(a + x). 24. (aaa — b)(_aaa — l>){aaa — 6) ••. to 17 factors. Express in algebraic notation : 25. The sum of the squares of a and 6. 26. The square of the sum of a and b. 27. The sum of the cubes of x, y, and z. 28. The cube of the sum of x, y, and z. Properties of Positive Integral Po-wers. 6. (i.) All {even and odd) powers of positive bases are positive. E.g., 2^ = 2x2x2 = 8. 3*= 3 x 3 x 3 x 3 = 81. In general, (+ a)" = (+ a) (+ a) (+ a) •■• n factors. = + a" w even or odd. (ii.) Even powers of negative bases are positive; odd powers of negative bases are negative. Notice that the words even and odd refer to the exponents. Also, that, for all integral values of m, 2 w is even (§ 4, Art. 9), and hence that 2 ii + 1 or 2 w — 1 is odd. E.g., (_2y=(-2)(-2)(-2)(-2) = 16; (-6)^ =(-5) (-5) (-5) = -125. 48 ALGEBRA. [Ch. 11 In general, (— a)^ = (— a)(— «)(— a) ••• 2 « factors, = + (aaa ■■■ 2n factors), by § 3, Art. 6, = + a"". And, (- a)^+^ = (- a) (- a) (- a) - (2n + 1) factors, = —lcMa ••• (2w + l) factors], by § 3, A.rt. 6, = _ a^+\ 7. If a = b, then a" = b". This principle follows directly from the axioms. BXBBCISES XIII. Find the values of the following powers : 1. 23. 2. 32. 3. (-3)5. 4. -36. 5. (-2)«. 6. -26. 7. (-5)*. 8. -5*. 9. (-a;)*. 10. (-x)'. Express as powers of 2 : 11. 8. 12. 64. 13. 512. 14. 4096. Express as powers of — 3 : 15. 9. 16. 81. 17. -243. 18. 729. Express as products of powers of 2 and 3 : 19. 24. 20. 144. 21. 1536. 22. 2916. Determine, by inspection, the signs of the following powers • 23. (-5)". 24. (-7)™. 25. (-3)2*. 26. [-iyp+K 27. (-7)"-!, when wis even. 28. (- 3)», when re is even. Find the values of the following expressions : 29. 22 + 32. 30. (2 + 3)2. 31. (33_53). 33. (3-5)'. 33. 5x43. 34. (5x4)3. 35. 2(- 7)3. 36. [2(- 7)]3. 37. 2 X 3* - (3 X 2)4. 38. (5 x 6)2 - 6 x 52. 39. 2' - (- 2)'. 40. 2924 -(-29)24. 41. [9(-4)]ii + 36". 42. (-37)13 + 3713. When a = 1, & = — 3, c = 2, find the values of : 43. a". 44. 6». 45. ah'. 46. (aby. 47. (6«)=. 48. b"'. 49. a2 + 62 + c2. 80. (a6)s +(6c)3 +(ac)3. Find the values of the following sums when re = 5, a = 2, 6=— 3: 51. 12 + 22 + $2 + ... + ra2. 52. (1 + 2 + 3+ .■. + }i)3, 63. a + a2 + a3+ ■•• + a». 54. b + b^ + b^ + — + b''. CHAPTER III. THE FUNDAMENTAL OPERATIONS 'WITH INTEGRAL ALGEBRAIC EXPRESSIONS. §1. DEFINITIONS. 1. An Integral Algebraic Expression is an expression which involves only additions, subtractions, multiplications, and posi- tive integral powers of literal numbers ; that is, in which the literal numbers are connected only by one or more of the sym- bols of operation, +, — , x, but not by the symbol -;-. E.g., 1+x+a?, 6 a% -I- f cd?, etc. The word integral refers only to the literal parts of the expres- sion. At the same time, the letters are not limited to integral numerical values, but, as always, may have any values whatever. E.g., a + b is algebraically integral ; but when a = ^, 6 = f , wehave «-F6=i+|=li. 2. Coefficients. — In a product, any factor, or product of fac- tors, is called the Coefficient of the product of the remaining factors. E.g., in 3 abc, 3 is the coefficient of a&c,, 3 6 of ac, etc. A Numerical Coefficient is a coefficient expressed in figures. E.g., in —Sab, — 3 is the numerical coefficient of ab. A Literal Coefficient is a coefficient expressed in letters, or in letters and figures. E.g., in 3 ab, a is the literal coefficient of 3 b, and 3 a of 6. The coefficients + 1 and — 1 are usually omitted. 3. The sign + or the sign — , preceding a product, is to be regarded as the sign of its numerical coefficient. 49 50 ALGEBRA. [Ch. Ill Thus, +3 a means the product of positive 3 by a ; —5x means the product of negative 5 by x. In particular, + a means the product of positive 1 by a, and — a means the product of negative 1 by a, unless the contrary is stated. BXEBCISBS I. What is the coefficient of a; in 1. 2 a;? 2. -3a;? 3. 5 ax? 4. -76a;? 5. (a + 6) a;? 6. If the sum, a+ a + a + a,be represented as a product, what is the coefficient of a ? 7. If the algebraic sum, —6 — 6 — 6 — 6 — 6, be represented as a product, what is the coefficient of — 6 ? Of 6 ? 8. If the sum, 2 ax + 2 ax + 2 ax + ■■■to y summands, be represented as a product, what is the coefficient of 2 aa; ? Of 2 a ? Of ay ? 4. Terms. — In the expression 4 a — 3 &, the sign — means operation, i.e., subtraction. But, since 4a — 36 = 4a + (— 3&), the given expression is the result of adding 4 a and — 3 &. Upon these considerations are based the following definitions : The Terms of an algebraic expression are the additive and subtractive parts of that expression. £l.g^, the terms of 4 a — 3 & are 4 a and — 3 6. The Sign of a Term is the sign of quality, + or — , of its numerical coeflB.cient. E.g., the sign of the term 4 a is +, of — 3 6 is — . A Positive Term is one whose sign is + ; as 4 a. A Negative Term is one whose sign is — ; as — 3 &. 5. Like or Similar Terms are terms which do not differ, or which differ only in their numerical coefficients. E.g., in the expression +3a + 6ab — 5a + 7ab, +3a and — 5a are like terms ; so are + 6 ab and + 7 ab. Unlike or Dissimilar Terms are terms which are not like. E.g., + 3 a and +7 ab in the above expression. §2] ADDITION AND SUBTRACTION. 51 6. A Monomial is an expression of one term ; as a, — 7 be. A Binomial is an expression of two terms ; as —2 a' + 3 be. A Trinomial is an expression of three terms. E.g., a + b-c, -da" +,7 b^ -5 c\ A Multinomial * is an expression of two or more terms, in- cluding, therefore, binomials and trinomials as particular cases. E.g., a + b\ a^ + b — (?, ab + bc — cd^ ef. §2. ADDITION AND SUBTRACTION. 1. Like Terms can be united by addition and subtraction into a single like term. Just as 2 = 1 + 1, so 2 xy = xy + xy; just as — 3 = — 1 — 1 — 1, so — S xy = — xy — oay — xy. Therefore, just as 2 - (- 3) = 2 + 3 = 6, so 2xy—(—3xy)=l2 — {—3)']xy = 5xy. That is, to add or subtract like terms, add or subtract their numerical coefficients and annex to that result their common literal part. Ex. 1. Add — 7 a6 to 4 ab. We have 4 a& + (- 7 ab) = [4 + (- 7)] ab=-3 ab. Ex. 2. Eind the sum of 3 a, — 5 a, 8 a, — 4 a. By the Commutative Law for addition 3a + 8a + (-5a) + (-4a)=[3 + 8 + (-5) + (-4)]a = 2a. Ex. 3. Subtract —Sx'y from — 7 aPy. We have -7x'y-{-5afy)=-7x^y + 5x'y = (-7 + 5)!ic!'y=-2x'y. BXEBCISBS II. Add 1. 2 a to 3 a. 2. -TytoSy. 3. 4 6 to -9 6. 4. _6c to -5c. 5. -4xno^a;3. 6. Jj^aUoJa^. * The word Polynomial is frequently used instead of Multinomial. 52 ALGEBRA. [Ch. Ill Subtract 7. 2 a from 5 a. 8. 3 6 from — 5 &. 9. 10 m from — m. 10. -9ylTom2y. 11. -16x2 from -5 a;^. 12. -3o6from7a6. 13. - 7 ^"-'(j/ - z) from 2 jc-^d/ - «). Find the sum of 14. a, 2 a, -3 a. 15. - ab, - 3 ab, - 7 a6. 16. Sx", - 4 K", - 9 x». 17. 2 a262, - a%% - 6 a^ft^. 18. — 5 ax"*, 7 ax", - 9 ax", 3 ax". 19. (x» + 2/»), - 3 (x" + ir), 4 («" + r)> - 7 (x" + !/"), 15 (x» + y"). Simplify the following expressions : 20. 5x-2x + 4x. 21. -9&-26-36. 23. -7TO+4m-5m. 23. - x2" + 7 x2» + 2 x2» - 5 x2». 24. a% -2ba^-3 a^b + 4 ba^. 25. - (a + 6 - c) + 2 (a + 6 - c) + 11 (a + 6 - c) - 7 (a + 6 - c). 26. 3x + (-5x). 27. -10a + (-12a). 28. 12x2 -(-7x2). 29. 2a -[-4a -(-6a)]. 30. m + [2m - (3m - 4m)]. 31. 6!/ -[5 2/ -4 2/ -(-3 2/ + 2 2/)] -2/. 32. X- [x-2x- (x-3x) - (x-4x)]. 2. Unlike Terms are added and subtracted by writing them in succession, each preceded by the sign + if it is to be added, by the sign — if it is to be subtracted. Ex. 1. Add 3 & to 2 a. We have 2a + 3b. Ex. 2. Add -3xHo2 f. We have 2f ■\-{-~3v?) = 2f -3x\ Ex. 3. Subtract — 11 m, from 2 n. We have 2 n — (— 11 m) = 2 w + 11 m. Such steps as changing +(—3 9;^) into — 3a^, and —(—11m) into + 11 m should be performed mentally. 3. A multinomial consisting of two or more sets of like terms can be simplified by uniting like terms. Ex. 1. 2a-36-5a + 46 = 2a— 5a — 3& + 4:&, by the Commutative Law, = — 3 a + 6, by the Associative Law. i,2] ADDITION AND SUBTRACTION. 53 EXERCISES III. Add 1. a to 1. 2. - 3 to 2 K. 3. - a; to — y. 4. x^ to - 2/2. 5. a^ to a. 6. - 7 s to - 3 x. 7. — 2 a6 to a^. 8. — xy^ to — x'^y. 9. — 3 m"? to 2 ra". Subtract 10. X from 1. 11. — a from 2. 12. -2m from -3b 13. — 3? from 3 k. 14. — ix from — cy. 15. 7 x™ from — 2 y". Add 16. 1, - X, xK 17. - 3, 2 X, -Zy. 18. - a6, - ac, - ad. 19. Subtract — 3 a;^ from the sum of 2 and — 4 x. 20. Add — 5 x^ to the result of subtracting — 2 x from 0. Simplify the following expressions by uniting like terms : 21. a + l + a-1. 22. 2x + 5 + 3x-7. 23. -hmn + Zn-i nm -Qn. 24. 3 x^ - 4 y^ + 2 x^ - 6 y'^. 25. a + 6-3a + c-46 + 6a-5c-8a-3c + 116. 26. 6 TO* - cm* + 3 + 9 cm* - 8 m^ + 5 m* - m'^c + 11. 27. 9 a6* - 6x - 13 a6* - a*6 + 3 6x - 2 a6* + 10 a*6 - 2 6x - a6's 28. 3 (a + m) - 4 (a + m) - 2 (o + m') + 8 (a + m). 29. (a + a)8 - 2 (a + 3)3 + 2 (a + a') + 7 (a + a)s - 5 (a + «»). Simplify the following expressions, first removing parentheses : 30. a+l-(2-3a). 31. 5x - (- 2?/ + 3x). 32. 2m + 3ra - (5m - 4»i) — (— 3 m + 7 ra). 33. l-[a3-2-(-2a3-3)]. 34. x2 - 2/2 + [- 3 x2 - 2 2/2 - (2 x2 - 3 2/^)]. 35. 2x2/ + 52/s - (2x2/ - Zyz)-\_ixy - (3»y -2yz) + 5yz']. Find the values of the expressions in Exx. 30-35, 36. When a = 1, x = 3, ^ = — 5, s = 10, m = 4, n = — 7. 37. When a = -3, x = 6, 2/ = -7, z = 8, m=-l, ?i=-2. Simplify 38. a+(a + l)+(a + 2) + (a + 3). 39. x + (x + 2) + (x + 4)+ - +(x + 10). 40. 2TO+(2m-l) + (2m-2)+ -+(2m-9). 41. Find the sum of 7 terms, the first term being x^, and each succeea ing term being 1 less than the preceding term. 42. Find the sum of 6 terms, the first term being m + n, and each succeeding term being p less than the preceding term. 54 ALGEBRA. [Ch. Ill Addition and Subtraction of Multinomials. 4. Ex. 1. Add -2a + 3&to3a-5&. We have {3a - 5b) + (- 2 a + 3b) = 3a - 5b - 2a + 3b, = a-2b. Ex. 2. Subtract — 2 a + 3 6 from 3 a — 5 b. We have (3a- 5b) - (-2a + 3b) =3 a- 5b +2 a -3b, = 5a-8b. In adding multinomials, it is often convenient to write one underneath the other, placing like terms in the same column. Ex. 3. Find the sum of -ix^ + 3y^-8z% 2a^-3»^ and 2y^ + 5z\ We have - 4.x' + 3y''-8z^ 2v? -3z^ 2y^ + 5z^ -2x^ + 5y^-6z^ It is evidently immaterial whether the addition is performed from left to right, or from right to left, since there is no carry- ing as in arithmetical addition. Ex. 4. Subtract —2a + 3b from 3a — 5b. Changing mentally the signs of the terms of the subtrahend, and adding, we have 3a-5b -2a+3b 5a-8b Ex. 5. Subtract 2a^-3z= from -4:X^ + 3y% and from the result subtract 2y^ + 5 »l When several multinomials are to be subtracted in succession, the work is simplified by writing them with the signs of the terms already changed. We then have -4:x' + 3y' -2^ ^-Sg" -2f-5z^ -6x2+ yi_2z' §5§2] ADDITION AND SUBTRACTION. 55 BXBBCISBS IV. Add 1. a + 4 to a — 4. 2. —x + ytox—y. 3. 7 a - 4 6 to - 3 a + 2 6. 4. 2x^ -xyto - x'^ + y^. 6. a;2 + K + 1 to a;2 _ a; + 1. 6. 2 a2 - 3 a6 - 62 to - a2 + 5 a6 + 2 b^. Subtract 7. a - 1 from o + 1. 8. a — 2 6 from 0. 9. 8 a — 3 & from 7 a — 2 6. 10. a; — j^ from x + y. 11. -xhi- xy^ from a;' + j^s. 12. a^fi + aft' from o* + 6*. 13. a8 - 2 a^ - a - 2 from a - 5. 14. - a^ + 7 a;s + 3 a;2 + 3 a; - 9 from 0. 15. a;8 - 3 a;2j^ + 3 xy^ - y^ from x^ + ^x?y + Z xy'^ + yK 16. 2x* - 3a;S - 7 a;2 + 3 a; + 1 from 2 + 4a;-6a;2-2a;3 + 3x*. 17. a;* — a;3 + a;2 — a; + 1 from x^ + x^ + I. 18. - 4a;6 - |a;3 + |a;2 - 1 from x^ -^\x^ -x^-\. 19. -3(a; + 2/)2 + 4(x + 2/)-7from -(x + j/)^ - 7(a; + ?/)+ 3. Find the sum of 20. 7a-9&-c, 5a-36-2c, 2a + 3&-5c. 21. 3a;2-5a; + l, 7a;2 + 2x-3, -x2-2a;-3. 22. x^-ax-'ra'^, 2x^ + 3ax-4a.^, x^ + ax + 2 a^. 23. 3o2-4a6 + 62, a'' - 2ab -2b^, 2a^ - Sab + ib^. 24. a2-2a6+2&2, 2a^-3ab + b\ a^ + 5ab-b^. 25. 2 xV + 4 a;32,2, _ 5 xV + 2 a;V - 3 xV, 4 xV _ 5 xV - 6 x^jf'. 26. 3a-26+5c, a + 6-c, -2a + 56-3c, -2a + 6-c. 27. xs + 2 x2 - 3 X + 1, 2 x' - 3 x2 + 4 X - 2, 5 x' + 4 x^ + 5, 6x8-5x2-4x-3. 28. 5a'-3a2 + 2a, a^ _ ^2^ a2 _ „ + 1, a3-2a2-a-2. 29. 2a + b-(_c+d), a+(b-c)-d, a+6-(c-(J). 30. a + 26+c, 2a-(6-c)-d, 3a + 6-(2c + d). 31. 3(a + &)-4(a+6)2 + 5(a + 6)^ (a + 6)2 - 2(a + 6)^ -(a + 6)8 + 2(a + by - (a + 6). 32. 7(x2+J/2)_3(x2-y2)+2xj/, 2(x2-^2)_4a;j/, 3(x2 + j,2)_(a;2_2,2), 33. Subtract the sum of a^ + ab + 6^ and ab from 2 a^ + 3 a6 + 2 6'^ 34. Subtract the sum of o" and 6" + c^ from the sum of 6^ and a^ — c^ IQ ALGEBRA. [Ch. Ill 86. How much does m^ + m^ exceed ni^ — n^? 36. How much does 1 — x" exceed 2 — Sx^? 37. What expression must be added to 2a-36 + 4c to give 4 a + 26-2c? 38. What expression must be added to xy+xz + yz to give x^+y^+z^ ? 39. What expression must be subtracted from a^ + ab + 6^ to give a2_2ff& + 62? 40. What expression must be subtracted from x^ — 2xy + y^ to give x^ + 2xy + y^? 41. What expression must be added to x^ + a: + 1 to give ? ■ Ifx = 2a-36 + 4p, y = -Sa + 2b-7c, z = 9a-76 + 6c, find the values of i2. x + y + z. iZ. x-y + z. H. x + y- z. ib. x-y-z. Given the four expressions : X = 5 a2 _ 3 a6 + b^-3ac + 2bc+ c% 2/ = 2a2 + 5a6-362 + 2ac-46c + 3c2, z = 4a2-7a6 + 562-4ac-56c+ c% M = 2 a^ + 9 a6 - 8 &2 + .3 ac + 3 6c + 2 c2, find the values of i6. x+y- z + u. VI. x + y — z — u. iS. x — y — z-u. § 3. MULTIPLICATION. Principles of Powers. 1. Products of Powers. Ex. xxV = (x) (xx) (xxx) = xxxxxx = o;^ = a!^+^+^ This example illustrates the following principle : The product of two or more powers of one and the same basi is equal to a power of that base whose exponent is the sum of the exponents of the given powers; or, stated symbolically, a'^a" = 0""+" ; a^d^aP — «'"+»+»' ; etc. For, a^a" = {aaa ••• to m factors) (aaa ••■ to « factors) = aaa ■•■to (jn + n) factors = «?»+". ^mgngp _ (ara")ap = a"'+''ai' = a'"+"+''. § 3] MULTIPLICATION. 67 EXERCISES V. Express each of the following products as a single power : 1. 2 X 22. 2. (- 7)S(- 7)6. 3. (- 2)32*. 4. 5*(-5)6. 6. a^aK 6. {- xyxK 7. (a6)2(a6)s. 8. a'^aH^. 9. (- k)(- a;)2(- k)3. 10. a^aPo'. 11. a'»-'a'»+2. 12. a;"'+"a;'»-3". 13. (Cfi + 62)8(a2 + 62)5. 14. (,. ^ j,)™^^; + y)i. 2. Powers of Powers. Ex. (ay = a*a*a*a'a' = a*+^+^+*+* = a*x= = a". This example illustrates the following principle : A power of a power of a given base is equal to a power of that base whose exponent is the product of the given exponents; or, stated symbolically, (a*")" = a""" ; [(o")"]" = a""" ; etc. For, (a™)" = a'^a'^a'^ ■ • . to ra factors —- Qm+m+m-\- — to n aummands ^ Q,^^"^, Likewise, [(a")"]!" =(a"'")'' = ''°"'''; and so on. EXERCISES VI. Find the values of the following powers : 1. (32)3. 3. 328. 3. (48)2. 4. [(-2)3]*. 5. (-28)5. Simplify the following powers : 6. (11*)5. 7. [(-18)5]6. 8. [(28)2]*. 9. ((j8)4 10. (-a^)3. 11. [(-a;)*]3. 12. [(a6)2]6. 13. [(a;2)6]7. 14. [(-?i2)6]3. 15. (a;>.)2n. 16. (a;8)2». 17. [(K + y)']^ Express the following powers as powers of 2 : 18. [(23)2]*. 19. (23^)*. 20. 85. 21. 3225. Express the following powers as powers of 32 : 22. 38. 23. 32». 24. 9*. 25. 27«. Express the following powers as powers of 53 : 26. 515. 27. 581. 28. 1256. 29. 25^. Simplify the following powers : 30. (a2a3)*. 31. (xH^y. 32. [(- a;)2a;']8. 33. [(-c)8c*]5». 58 ALGEBRA. [Ch. Ill Write the squares and the cubes of : 34. a^. 35. -aK 36. (x^x^)'. 87. [(-3/)=2/*]^ 38. x + y. 89. (a - 6)2. 40. - (a + 6 - c)K Write 41. The fourth power of a. 42. The ath power of 4. Write the sum of ten terras, the first term being x, 43. When each term is the square of the preoec/ing term. 44. When each term is the mth power of the preceding term. Simplify 45. 3(a3)*+2(a*)s-4(a2)6. 46. 3(a^)i-2(a*)^-5{a^oy+7[(a'^y]^ 3. Like and Unlike Powers. — Two powers are said to be like or unlike according as their exponents are equal or unequal, whether or not their bases are equal. Thus, a^, V are like powers ; a', a?, a* are unlike powers. 4. Products of Like Powers. Ex. a*6V = {(mad) (bbbb) (cccc) = (a6c) (a6c) (a&c) (abc), by the Commutative Law, = (abcy, by the definition of a power. This example illustrates the following principle : (i.) ITie product of like powers of two or more given bases is the like power of the product of the bases; or, stated symbolically, a™*" = (a*)" ; a'^b'^c'^ = {abcf, etc. For, a^ft" = (,aaa ■■• to m factors) (666 ••• to m factors) = (a6)(a6) (ah) ■•• to m factors, by the Commutative Law, = (afi)", by the definition of a power. In like manner the principle can be extended to the product of any number of like powers. (ii.) The converse of the principle is evidently true : (aby = a"*"* ; {abcY = a'^b^'c^, etc. E.g., (xyf^x'f; (2a'by=2\ayb'' = 8aV. § 3] MULTIPLICATION. 69 EXERCISES VII. Express the following products of powers as powers of products : 1- 78x53. 2. (8)ix(-3)<. 3. aW. 4. (-x)y. 5. (- ayb^{- ey. 6. 0^(6 + c)^. 7. aSfiSc". 8. a;i2j/%i8. Express the following powers of products as products of powers, re- ducing powers of any numerical factors : 9. (xyy. 10. (-2m)5. 11. {-2xyy. 12. (_abcy. 13. (a268c)4. 14. (^-Sx^y. 15. (^-x^y^zy. 16. (mW)P. "Write 17. The square of twice a. 18. Twice the square of a. 19. Four times the square of the difference between x and y. 20. The square of four times the difference between x and y. Given two numbers, a and 6, write in algebraic language : 21. The square of the first number, plus twice the product of the two numbers, plus the square of the second number. 22. The cube of the first number, plus three times the product of the square of the first by the second, plus three times the product of the first by the square of the second, plus the cube of the second. 23. Write in algebraic language the verbal statements in Exx. 21 and 22, when the given numbers are 2 a and — 3 6. Degree. Homogeneous Expressions. 5. An integral term which is the product of n letters is said to be of the nth degree, or of n dimensions. Thus, the Degree of an Integral Term is indicated by the sum of the exponents of its literal factors. E.g., Sab is of the second degree; 2se'y, =2xxy, is of the third degree. The Degree of a Multinomial is the degree of that term which is of highest degree. E.g., the degree of a?y + xf — 3?fz is the degree of 3?fz ; i.e., the sixth. 6. It is often desirable to speak of the degree of a term, or of an expression, in regard to one or more of its literal factors. 60 ALGEBRA. [Ch. Ill E.g., the term avfy^ is of the fifth degree in x and y, of the first degree in a, of the second degree in x, of the third degree in y, etc. The expression ax' + 2 hxy + cy'^ is of the second degree in X, in y, and in x and y. 7. A Homogeneous Expression in one or more letters is an expression all of whose terms are of the same degree in these letters. E.g., a^ + 2ah + V is homogeneous in a and 6. 8. If the terms of a multinomial be arranged so that the exponents of some one letter increase, or decrease, from term to term, the multinomial is said to be arranged to ascending, or descending, powers of that letter. The letter is called the letter of arrangement. E.g., a' + 3 a^6 + 3 aV + W is arranged to descending powers of a, which is then the letter of arrangement ; or to ascending powers of h, which is then the letter of arrangement. EXERCISES vni. What Is the degree of 2 a^ft^a;*?/* 1. In a ? 2. In X ? 3. In 6 and ?/ ? 4. In a, 6, x, and y 1 What is the degree of the expression aW — 6 a%Vy + 5 abx''^ 5. In a; ? 6. In ?/ ? 7. In a ? 8. In 6 ? 9. Arrange 2 a; — 3 a;^ + 7 — 2 a;* + 3 a;^ to ascending powers of a: ; to descending powers of x. 10. Arrange 3 y — 7 xj/' + 5 x'y^ + 4 x^y^ to ascending powers of x ; to ascending powers of y. 11. Arrange 29 a^ft* + 4 je _ 30 ajs^s + 25 a*62 _ 12 ab^ to descending powers of a ; to descending powers of 6. Multiplication of Monomials by Monomials. 9. Ex. 1. 3 a X 6 & = 3 X 5 X a X & = 16 a6. Ex.2. 2a;x(-42/2) = 2(-4)a;/ = -8a!2/l Ex. 3. f a^ X 6 a62 X 11 6° = I X 6 X 11 X a^aWh^ = 44 a«6l Ex. 4. 3 a^+ift^ X 5 a56»-i = 3x5 a'»+'a%%»-' = 15 «'»+*&>+'• § 3] MULTIPLICATION. 61 These examples illustrate the following method : The product of two or more monomials is obtained by multiply- ing the product of their numerical coefficients by the product of their literal factors. EXERCISES IX. , Multiply 1. 3 a by 4. 3. -5 by -2 a. 8. 7by-5K2. 4. 2 a by 3 a. 5. 2 J a; by - 5 xK 6. - 3 a" by - 4 a. 7. - 2 a6 by 5 ab. 8. 3 a'^b by - 7 aS^. 9. 4 ft^c by - 3 &W. 10. 3 abc by ab^A 11. - 3 x^z by xy^. 12. a%xy by aft^V- 13. 2(a + 6)8 by - 3(a + 6)2. 14. 7J a'Ca; - y)^ by 6 a8(a; - y)^. 15. 12 0*6"* by - f ab". 16. 3 a^-^e"*' by - 12 a^b^. Simplify the following continued products : 17. Sabx5bcx6ao. 18. - 1 x^'y x (- 2 y^z) x 3 xz". 19. — axy X 1 abx^z X 2 bx^yz. 20. x^y"+^x6x'^y'''x(—x^^~^). 21. (2 aa;S)^ x (5 a62a;^)2 x (- 2 a^ajSj/S^s. 22. (1 -xyxZabx 4(1 - xy x (- 2 a^c). The Distributive Law for Multiplication. 10. If the indicated operation within the parentheses in the product, 4(2 + 3), be first performed, we have 4(2 + 3) = 4x5 = 20. But if each term within the parentheses be multiplied by 4 and the resulting products be then added, we have 4 X 2 + 4 X 3 = 8 + 12 = 20, as above. Therefore 4(2 + 3) = 4 x 2 + 4 x 3. This example illustrates the following principle : The Distributive Law. — The product of a multinomial by a monomial is obtained by multiplying each term of the multinomial by the monomial and adding algebraically the resulting products. That is, a (6 + c — -* - 2 oJ--! - 4 «!» by 2 a»i>-s 4- 3 «»-*. S3. 3 o»-%i!« - a«-^ + a' by a^ix"-! - 3 xn+i - 2 ax". SS. 2 x(a2 + fta)3 _ 3 9.2(^2 + 62)2 + 4 ajS^aS ^ js) by X (o9 + 6")'' - 2 (a2 + 68) . 34. (X + 2/)»+2 + 3 (X + y)''+^ - 5 (x + j;)" by 6 (x + 2/)»+i + 4 (X + !/)"- 2 (x + j/)—*. Perform the following indicated operations : 35. (x-2)(x + 3)(x-4). 36. (x - 3)(x - 5)(x -7). 37. (2x-3j/)(4x + i/)(x + 5j/). 38. (xy — 2z)(3s — 4:xy){z + 5xy). 39. (2OT2 + 3m-2)(m-l)(2m + 3). 40. (x2 + 4x-l)(x2-2x+ l)(x + 2). 41. (a2-a + l)(a2 + o + l)(a«-a2 + l). 42. (a™ + 6"') (a" + &») (a? - 62"). 43. (1 + x" + x") (3 - 2 x» + X™) (5 x"-^ — 3 x"-i). 14. T%e converse of the Distributive Law evidently holds; that is, ab + ac — ad = a(b + — d), etc. E.g., ax + bx= (a+b)x, 2 ay — 3 by = (2 a — 3 b)y. 15. If the coefficients of the multiplicand and multiplier, arranged to a common letter of arrangement, be literal, it is frequently desirable to unite the terms of the product which are like in this letter of arrangement. 3] MULTIPLICATION. 67 Ex. Multiply x+ ahj x+ b. We have x +a X +b a^ + ax bx + ab a^ + ax + bx + ab = ai' + (a + b)x + ab,hj Art. 14. BXEBCISES XII. Arrange the values of the following products to descending powers of X, uniting like terms in x : 1. (3? + ax + b)(x + a'). 2. (x^ - am? + bx~ c)(x-b). 3. [a;2 + (a- 6)x — a6](a;-c). 4. lx^-(_a + b')x + abJ{x^+(o-d)x-cd]. 5. (_px^ + qx^ + rx + s) {ax'^ + 6a; + c). Zero in Mnltlpllcatlon. 16. Since N-Q = N{b - 6), by definition of 0, = Nb-Nb = Q, we have /l^ • = and • /IT = 0. That is, a product is if one of its factors be zero. 17. The words is not equal to, does not have the same value as, etc., are frequently denoted by the symbol Tt. E.g., 7 =#: 2, read seven is not equal to 2. 18. It follows, conversely, from Art. 16 : If a product be 0, one or more of its factors is 0. That is, if Px Q = 0, then either P = and Q ^ ; or Q = OandP:?!=0; orP=Oand Q = 0. EXEBCISBS XIII. 1. What is the value of 2 (a — 6), when b = a? a. What is the value of (a + 6) (c - (?), when c = (J ? 3. What is the value of (6 + c) (a + 6 - c), when c = a + b? 4. What is the value of (a;^ - 9) (x* - 7 a;S + 2 a; - 9) , when x = 3? When a; = - 8 ? 68 ALGEBRA. [Ch. Ill If P X § X -B = 0, what can we infer, 5. "WhenPgtO? 6. WhenQi=0? 7. When P ^fc and JJ zjt ? For what values of x does each of the following expressions reduce to : 8. z(x-2)? 9. (a;-4)(x + 7)? 10. (x-l)(x-a)? 11. (x - 6) (x + 8)(x2 - 25)? 12. X (x - a) (x - 6) (x - c) ? §4. DIVISION. 1. One power is said to be higher or lower than another according as its exponent is greater oi less than the exponent of the other. E.g., a* is a higher power than a^ or b^, but is a lower power than a^ or 6'. 2. Quotient of Powers of One and the Same Base. Ex. a' -7- a^= (aaaaaaa) -=- (aaa) = (aaaa) x (aaa) -^ (aaa), by Assoc. Law, = aaaa = a* = a'~^- This example illustrates the following principle : (i.) The quotient of a higher power of a given base by a lower power of the same base, is equal to a power of that base whose exponent is the exponent of the dividend minus the exponent of the divisor; or, stated symbolically, a'" -i-a" = a"'-", when m> n. For a" -j- a'^ = {aaa ••• to m factors) -=- (aaa ••■ to n factors) = \_aaa ■•■ to (m — Ji) factors] x {aaa ■■■ to n factors) -i-{aaa •■• to n factors) = aaa ■■. to (m — n) factors, = a"*-". (ii.) a™ -f - a" = 1, when m = n. E.g., a' H i-a==l. EXERCISES XIV. Express each of the following quotients as a single power : 1. 23^2. 2. x5^x2. 3. (-5)'h-(-5)*. 4. (-6)5-63. 5. (-a)9^a*. 6. (a6)5 -=-(a6)2. 7. (-x)'h-(-x)*. 8. (-»;!/)ii-i-(-X2/)'. 9. a-'-i-a^. 10. 3"-=-3» 11. a"+i -=- a. 12. x»+'' -H x». 13. 6=^+3 -=- 6«+i. 14. a" -=- a"-'. 16. a^" h- a"-'. 16. (a + 6)5 -(a + 6)' 2. 17. (xj/ - 1)2"-^ -i- (xj/ - l)"-3. §4] DIVISION. 69 ■6^ Division of Monomials by Monomials. 3. Ex.1. 12a-^4 = 12-^-4 Xffl, =3a. Ex.2. - 27 x' ^ 3 ocF = (-27-^ 3) X (a? -=-x'), =-9afi. Ex. 3. 15 a?b^ -h(-5 ab^ = [15 -=- (- 5)] x (a' -f- a) x (6' h = -3aK These examples illustrate the f ollowing method : The quotient of one monomial divided by another is the quotient of their numerical coefficients multiplied by the quotient of their literal factors. Divide 1. 6 a by 3. 4. 5 a^ by 2 ic. 7. 4 a6 by - 2 a. 10. 6 x^y by 5 jc*. 12. 7 o'l^oc^^ by - 5 a'^b^(fi. 14. 15(a + 6)by3(a + 5). 16. 10 a2»65 by - 5 a»6'. 18. x^''-^y^+^ by x'+h/^-K Simplify 20. aH^ -^ ( — ax") x 2 oa;y. EXERCISES XV 2. 12 a; by - x. 6. 9a;3by -Sx^. 8. 6 a6c by — 3 ac. 3. — 15 TO by 3 TO. 6. - 11 a' by - 5 a". 9. JaS6by3a26. 11. - 15 aSft' by - 3 aft^. 13. I m^n^ps by — I TO^n'^^. 15. 25 a;2(a; + l)s by - 5 x{x + 1)K 17. - 27 a^'+ij/S" by - 9 xj/^™. 19. o"-i6"-2 by a"-36n-*. 21. 35 a;V« x 2 ai^s -=- (7 a;V«^)- 22. (l^n-lJ,r»+Ic">+» -f- ffln&wc™ -^ a"-^6c"-3. 23. 6 si?»+i2/»-i ^ ( — x'^-hf-") X (3 aiV^^) . The Distributive Law for Division. 4. If the indicated operation within the parentheses in the quotient (8 + 6) h- 2 be first performed, we have (8 + 6) ^ 2 = 14 -T- 2 = 7. But if each term within the parentheses be first divided by 2 and the resulting quotients be then added, we have 8-^2 + 6-7-2 = 4 + 3 = 7, as above. Therefore (8 + 6)-5-2 = 8-5-2 + 6-r-2. This example illustrates the following principle : 70 ALGEBRA. [Ch. Ill Distributive Law. — The quotient of a multinomial by a mo- nomial is obtained by dividing each term of the multinomial by the monomial and adding algebraically the resulting quotients; that is, (a + b — o)^d = a-^d-\-b^d — c^d. For, since -=-(Jxd=-^l, we have (a + 6 — c)->d = (aH-dxd + 6-j-(Jx(i — c-^dx(J)-f-(? = (a -=-(J+6-^d-c-f-d)xt?H-c;, by §3, Art. 14, =^a-i-d-^h-i-d — c-i-d, since x (J -^ (J = x 1. 5. It follows, conversely, from the Distributive Law that a^d + b-7-d — c-i-d = (a-\-b — c)^d. Zero in Division. 6. Since ^ iV= (a — a)-^ iV, by definition of 0, = a-i-iV-a-hiV"=0. We have 0^N = 0, when N^O. Observe that this relation is proved only when N^lO. 7. It follows, conversely, from Art. 6 : If a quotient be 0, the dividend is 0. That is, if M-i- JV= 0, then M=0. Division of a Multinomial by a Monomial. 8. The division of a multinomial by a monomial is a direct application of the Distributive Law. Ex. 1. Divide 6 a^ - 12 a; by 3 a;. Wehave {6x'-12x) ■i-3x = 6a^-i-3x-12x-h3x = 2a;-4. Ex. 2. Divide - 105 a'&= - 75 a^&= + 27 a'b* by - 15 a%. We have (- 105 aV - 75 aV + 27 a^b*) -?- (- 15 a'b) = (- 105 a'&=) -- (- 15 a%) - 75 aV -=- (- 15 a'b) + 27a%^-=-(-15o»6) = 7ab + 5b^-^¥. § 4] DIVISION. 71 EXERCISES XVI. Divide 1. 5 + 10aby5. 2. 4a + 86by-4. 3. ax + bx hyx. 4. 3o«-6o6by -3a. 6. 21 a»6 - 14 aft" by - 7 06. 6. 8 am» - 2 a»tj» + 4 oSro" by 2 am. 7. 26(a + 6)»-2a(a + 6) by 6(0+6). Simplify 9. 2a»-(o«-8a)^-o. 10. (Bx-ix^)-i-^x-(-2x^y + nxy)^xy. 11. (ab - a% + 3a86) H-a6 - (4aS _ ^a^) -^ 2a. Divide 9aV - 6 a'x* + 12 a^x^ by 18. 3a2. 13. -3*8. 14. , and the divisor, d, are algebraic expressions. Ex. Divide ay' + 3x + 2 by .r + 1. We have {3^+3x+2)-i-(x+l)=x+[(x'+3x+2)-x{x+r)^^(x+l) (1) =x+(a:^+3x+2-x^-x)-r-(x+l) (2) =a;+(2a;+2)--(a!+l) (3) = a;+2 + [(2a;+2)-2(a;+l)]-(a;+l) (4) =a;+2-|-0-=-(a;+l) =x+2, since 0-=-(a;-fl)=0. § 4] DIVISION. 73 We take the quotient of the term containing the highest po-wer of X in the dividend by the term containing the highest power of a; in the divisor as the partial quotient at each step. The work may be arranged more conveniently thus : a,-2 + 3 a; + 2 as + l x + 2, quotient. a? + X ■■• x{x+ 1) to be subtracted from x^ + Sx + 2; see (1) and (2) above. 2 a; + 2 • ■ • Remainder to be divided by a; + 1 ; see (3) above. 2 a; + 2 • • ■ 2(a; + 1) to be subtracted from 2 a; + 2 ; see (4). 11. The method of applying the principle of Art. 9 to the division of multinomials, as illustrated by this example, may be stated as follows : Arrange the dividend and divisor to ascending or descending powers of some common letter, the letter of arrangement. Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the divisor by this first term of the quotient, and sub- tract the resulting product from the dividend. Divide the first term of the remainder by the first term of the divisor, and write the result as the second term of the quotient. Multiply the divisor by this second term of the quotient, and subtract the product from the remainder previously obtained. Proceed with the second remainder and all subsequent remainders, in like manner, until a remainder zero is obtained, or until the highest power of the letter of arrangement in the remainder is less than the highest power of that letter in the divisor. In the first case the division is exact ; in the second case the quotient at this stage of the work is called the quotient of the division, and the reminder the remainder of the division. Ex. 1. Divide a'6 -15b* + 19 aW + a* - 8 a^b^ hj a^-BW + S ab. Arranging dividend and divisor to descending powers of a, we have 74 ALGEBRA. [Ch. Ill a*+ a^b - S a'b' + 19 ab^- 15 b*\ a' + Sab-Bb^ a^ + Sa^b-ba^b" -2a^b-6d'b' + 10a^ aS_2ad + 36'' 3aV+ 9ab^-15b* Ex. 2. Divide 8oi? — f by 2xy + ia^ + f. Arranging the divisor to descending powers of x, we have 8 a;' — 2/^ 4:3^ + 2xy + y^ 8a? + 4:a?y + 2xy' 2x — y — 4 a^ — 2 952/' — 2/^ — 4 a^y — 2 asy" — t^ Observe that the remainder after the first partial, division is arranged to descending powers of x. Ex. 3. Divide 12 a"+i + 8 a" - 45 a"-! + 25 a"-^ by 6 a - 5. We have 12 a"+» + 8 a" - 45 a»-i + 25 a""' | 6ffl-5 12a"+i-10a'' 2a" + 3a''-^-6a"-'' 18 a" — 45 a"-^ 18a"-15a"-» -30a"-' + 25a»-^ -30a"-i + 25a"-2 Ex. 4. Divide sc' + (a + 6 + c)a!^ + (a6 + ac + 6c)a; + a6c hy a^ + (a + b)x + ab. We have x' + (a + 6 + c)«^ + (a6 + ac + &c)a; + a6c a^+(a + b )a^+ ab x cx^ + (ao +bc)x + abc cv? + {aa -\-b(i)x-\- dbc ar' + (a + 6)a! + a6 X + c § 4] DIVISION. 75 EXERCISES XVII. Find the values of the following indicated divisions : 1. (a;2 + 2a; + l)-7-(a; + l). 2. («»+ 11a; + 30)-=-(a + 5). 3. (a;2_x-90)-=-(a; + 9). 4. (cc" - 5 a; + 6)-f-(a! - S). 5. (4 a;2- 12 a; + 9) -4- (2 a; -3). 6. (2ot2 - 3m + 1)-H(m - 1). 7. (2a2 + o_6)-f-(2o-3). 8. (3 a;* - 13 k - 10) -;- (8 a; + 2). 9. (6a;*-10-lla!a)-!-(2a!!>-5). 10. (2a;2+6a» + 7aa;)-f-(2a;+3a). 11. (a2- 2 06 + 6")-=- (0-6). 12. (35a;« + a;j/-88j/«)-h(7a:-llj/). IS. (a;a + 5| a;;/ + 8J ^) -^(x + hy). 14. aaa + |a6 + Tfe6'') + ao + i5). 15. (a2 -li,ax,y~ 248 xY) -4- (o + 9a;j^). 16. (8 xV - 65 x.ys'^ - 63 «*) -:- {xy - 9 »a). 17. (6wS-7m2a; + 2na;2)-^(-a; + 2n). 18. (a;*y + 6a;5-2!«;W.^(3j.a4.2a;j,). 19. (-19o2x2 + 3a;* + 4oa;8)H-(ix-o). 20. (4x3_3a;2-24a;-9)-s-(a;-3). 21. (3x5-13a:2 + 23a;-21)-i-(3a!-7). 22. (3a;*-3a;3-2a;2-a!-l)H-(8a;2+l). 23. (o8 - 3 a26 + 3 a&2 - &») ^ (o - 6). 24. (o6-6a*+9a2-4)H-(a2-l). 25. (21 o86 + 20 6* - 22 a%'^ - 29 0*62) -=- (3 0^6 - 5 6^). 26. (4 a;«2/5 _ j a;iy -i- 12 a;^ _ 11 ifiyV) ^ (4 a;^?/^ - zy^) . 27. (a;S + 8a;2 + 9a;-18)-4-(x2 + 5a;-6). 28. (a;* + a;3-4a;2 + 5a;-3)^(a;2 + 2a;-3). 39. (6 a;* - a;8 - 11 x2 - 10 a; - 2) -=- (2 05= - 3 a; - 1). 30. (a;s-l)-i-(a;2 + a; + l). 31. (o^ + 8)-;-(o2 - 2a + 4). 32. (125a;P-642/8)-=-(5a;2-42^). 33. (aSa;^ + 2/5)-(aa; + «/). 34. (a;* + a;2 + l)^(a;2-x + l). 35. (o*a;5+64a;)-=-(4oa;+a2x2+8). 36. (24a;8 + 25a;-a;6)-f-(5 + a; + 9!8 + 5a;2). 37. (4o* - 25c* - 30 62c2 - 9 6*) -;- (2 a^ + 50= + 362). 38. (27 a;* - 6c2a;2 + \<^)^{c^ -6cx + 9a;2). 39. (8 aV + 32 08 + Jw") -4- (4 a« + w^ + 4 a^). 76 ALGEBRA. [Ch. Ill 40. (16 a<62 + 9 a2j4 _ 12 a«b« -8a^b + 3a')-^ (a* + 3 a^ft^ - 2 a^b) . 41. (28 a^c - 26 aV - 13 a^c^ + 15 a^c*) h- (2 a^c'^ + 7 a^c - 5 ac') . 42. (81 z« - 90 6*2* + 81 ¥z^ - 20 68) ^ (9 gi + 9 fi^gS - 5 6«)- 43. (x3 + 2/8 + 3x2/- 1) -=-(» + 2/ -1). 44. (a3 + &3 + c3 - 3 a6c) h- (a + 6 + c). 45. (a2 + 2a6 + &2-a;2 + 4sc2/-4^2)^(a,4 6 _j; + 2 2/). 46. (a2 + 2 ac - &2 _ 2 6(i + c2 - (P) ^ (a + c - & - d). 47. (32 aS + b^) h- (16 a* - 8 o^ft + 4 a^fiz -2ab^ + b*) . 48. (81 x8 - 16 2/8) H- (27 afl + 18 x42/^ + 12 xY + 8 «/«) . 49. (^a2 - ijia6 + 9ac + 262 - 6c)H-(|a - 3& + |c). 50. (28x2_43|2/2+140 2/3-112s2)-H(7x + 842/-14a). 51. (I a%+6 acd-i 6c2+16 c^d-^ a6d+ J bcd-S cd:^) h- (| a + 4 c-2 d). 52. (f aH -1^ aH^ + 1| aV + ^ ax* - x^) ^ (| aS - | «% + j x^) . Find the values of the following indicated divisions : 53. [(6 + c)x2 _ box + x3 - 6c(6 + c)] h- (x^ - 6c). 54. [xs + (a + 6 + c)x2 + (a6 + ac + 6c)x + abc] -=- (x + 6). 55. [x' + (a + 6 — c)x2 + (06 — ac — 6c)x — abc\ -=- (x — c). 56. [a6c - 62(a + c)+ a2(6 + c)+o\a + 6)]-t- (a5 + oc + 6c). 57. \_a{a - l)x8 + (a^ + 2 a - 2)x2 + (3 o^ - a8)x - a*] ^ (ax^ +-2 x - a^). 68. [x5 —(1 +m)x* +(1 + m + re)*' — (m + B + p)x2+(j) + re)x— ^j] -[x2-(x-l)]. 59. [(10a2 + 29 - 34a)x +(5 - 2ffl2 + Za? - 8a)x= + Sa^ + 21 - 26a + (17 - 22 a + 4 a5)x2] -=- [(a - 2)x+ (a^ - 1 + a)x2 + 2 a - 3], Find the values of the following indicated divisions : 60. (6x3»-25x2» + 27x»-5)h-(2x''-5). 61. (6xS»- 11x<" + 23xS"+ 13x2''-3x'' + 2)^(3x" + 2). 62. (6 x2"+i - 29 x2» + 43 x^-'- - 20 x^"-^) -=- (2 x» - 5 x"-i) . 63. (1 + a6« - 2 a^^) -f- (3 a^' + 2 o^^ + 2 a»^ + a*« + 1). 64. [2 a? (6 + c)2» - J] - [a(6 + c)» + i]. 65. [8 (x - 2/)s» - x8] ^ [4 (x - 2/)2» + 2 x(x - 2/)" + x^]. 12. In the equation D -r- d = q + {D — qd) -i- d, D — qd is the remainder at any stage of the work, and q is the corre- sponding partial quotient. If, for brevity, we let B stand for the re- mainder at any stage, we have, D-i-d = q+H^d. (1) § 4] DIVISION. 77 That is, the result of dividing one number by another is equal to the partial quotient at any stage, plus the remainder at this stage divided by the given divisor. E.g.^ 29--6 = 4 + 5-=-6=4 + |; (a;2 - a; + 2) -H (a; + 1) = (K _ 2) + 4 -H (a; + 1). 13. If both members of the equation D^d = q + I{^d be multiplied by d, we have, by Ch. I., § 1, Art. 15 (iii.), D^dxd=iq+R~-d)d = qd + B~dxd = qd + S, since -^d x d=~ 1. Therefore, D = qd + ft. That is, the dividend is equal to the product of the quotient at any stage by the divisor, plus the remainder at this stage. E.g., 29 = 4 X 6 + 5, and aj2 - a; + 2 = (k - 2)(a; + 1) + 4. EXBECISBS XVIII. Find the remainder of each of the following indicated divisions, and verify the work by applying the principle of Art. 13 : 1. (a;2-7a;+ll)-f-(a;-3). 3. (Sx^ + 5 a; - 9) ■- (a; - 4). 3. (aS - 17 a;2 + 15 a; - 13) -=- (2 a; - 5). 4. (5a;5-7a;2 + 2a;-l)-f-(a;2-7a: + 3). 5. (6 n5a;8 + 12 nH^ - 14 n*3fi + Ji^ - 1) -=- (2 a;' - n). 6. (12 65 + 8 &2c8 -2b^c-i be* - 38 fe^c^) -=- - (2 c^ + 6 6c - 4 62). 7. (4 c2"a;3» — 13 (fi«x^ + 14 c*^» - 2 c^") -=- (c"a;2» - 2 c^'x" + c'"). Infinites. 14. The following considerations lead to an important mathematical concept. Observe that the quotients 1-^(1 -.9) =l-=-.l =10, 1 -^(l-.99) =1h-.01 =100, 1 -f- (1 - .999) = 1 -V- .001 = 1000, etc., increase as the divisors decrease, the dividend remaining the same. If the divisor be still further decreased, the dividend remaining the same, the quotient will be still further increased. Thus, 1 -r-(l — .999999999)= 1 -^ .000000001 = 1000000000, etc. It is evident that, by taking the divisor sufficiently small (and positive), we can make the quotient as great as we please. If the divisor become 78 ALGEBRA. [Ch. Ill less than any assigned number, however small, the quotient will become greater than any assigned number, however great. That is, If the dividend be positive, and remain the same, as the divisor decreases below any assigned positive number, however small, i.e. , becomes more and more nearly equal to 0, the quotient increases beyond any assigned positive number, however great. The symbol, +00, read a Positive Infinite Number, or a Positive Infinite, is used as an abbreviation for the words, a number greater than any assigned positive number, however great. The principle enunciated above can be expressed symbolically thus : ^-/V■^0 = + oo. (1) 15. It is important to observe that the symbol, + 00 , does not stand for one definite number. It stands for any number which is greater than any assigned positive number, however great, but which can be still further increased. Therefore one infinite number can be greater or less than another infinite number. Likewise, equation (1), Art. 14, is to be understood only as, expressing the fact that, as the divisor becomes more and more nearly equal to 0, the quotient increases beyond any assigned positive number, however great. 16. We can also arrive at the conception of a positive infinite number by taking the dividend and the divisor both negative. Thus, -l-(l-l.l) = -lH-(-.l)=10, _ 1 ^(1 _ 1.0001) = - 1 -h(- .0001)= 10000, etc. 17. In a similar manner, we gain the conception of a Negative Infinite Number, or a Negative Infinite. Thus, -l-=-(l-.9) = -l-=-.l=-10, - 1 -V- (1 - .9999) = - 1 ^ .0001 = - 10000, etc. We therefore have — /V -^ = — oo. 18. The numbers which we have hitherto used in this book are, for the sake of distinction, called finite numbers. In subsequent work we shall assume that the numbers involved are finite, unless the contrary is expressly stated. 19. Observe that the quotients 1 -- 10 = .1, 1 -H 100 = .01, 1 H- 1000000 = .000001, etc. ; decrease as the divisors increase. It is evident that if the divisor become greater than any assigned number, however great, the quotient will become less than any assigned number, however small. That is. If the dividend remain the same, as the absolute value of the divisor increases beyond any assigned number, however great, the quotient de- § 4] DIVISION. 79 creases in absolute value below any assigned number, however small, i.e., becomes more and more nearly equal to 0. Or, stated symbolically, /V-=-(±oo) = 0. 20. If ilf -=- iV = 0, and iV":^ oo, then by Art. 7, Jf = 0. 21. The consideration of other relations which involve and oo is deferred. It is important to notice that the relation X a = of § 3, Art. 16, was proved only toe the case in which a is finite. EXBBCISBS XIX. MISCELLANEOUS EXAMPLES. Given a = 2x^-3x^ + 5x^-1, b = x^ + nx^ -3x + &, e = T x'^ + 5x» - 9x^ - Gx, d= 5x* -7 x^ + Sx- 1, find the values of 1. a + b + c — d. 2. a — b + c — d. 3. a — b — c + d. 4. a + b — 1(c-d). 5. b -e + Sx(a + d). 6. d - c - 59;2(ffl- 6). 7. (o + 6)(c-d). 8. (a-d)(b-c). 9. {a -b + c)d. Multiply : 10. 4 a*x^+« - J aV"*^ + 10 a;"-i by ^ a^^"-^ + 7^ a;"-'. 11. 4 o"»+i62 + a^-^b — 2 a'^+^b' by 8 a^B^ — a™+36' — 5 a^+^ft*. 12. 5 an+S'S'-i - 2 a»-'6'+i + 3 a»+5'6' -^ + on+'ft' by a"+'6' + 4 fls"-*6'^-' — 2 a^-^b'+^. 13. a;*(a;2 + 2)''-3 + 2 a;2(iBa + 2)a.-i + ^x^ + 2)'«+i by a;'(a;2 + 2)"-^ - 4 9;8(a;''' + 2)3»-i + 8 a;(a;2 + 2)*"+^ Divide : 14. 2a;i'' - ,075a;S + g.eSaj' - 1.05 a;6 - 19.25a;8 + 8.6x^ by 2.5a^-3a;5 + .5x'-.15a;*. 15. 6 a4»+i6"'+« - I a<''B™+6 + | a<»-i6'»+* + f ate-sftm+s _ ^ a*"-s6"'+'' by 3 a^^+ift^+i - J a2»6'». 16. 15 a^^^-*b^+'' + 14 a6''-"'-*6^+* — ^ a9»-4&6p+i by 5 a^i-^BS-? + 6 a^+^-^fe?. What is the value of (jc + l)(a; + 2)(a; + 3)--(a; + n), when IT. x = l, n = S? 18. x = -2, n = 5? 19. a! = 8, ra = 4? 20. What is the value of (TO _ 5)»-i(m - 4)"-2(n - 5)»^3 - (» - 2)''-*(m - 1)"-", when n = 6 ? 21. What is the value of s(s - a)(s - &)(s - c), when s = i(a + 6 + c), a = 5, & = 6, c = 9 ? CHAPTER IV. INTEGRAL AIiGEBRAIC EQUATIONS. An equation has been defined (Ch. I., § 1, Art. 12) as a state- ment that two numbers or expressions are equal. We must now distinguish between two kinds of equations. §1. IDENTICAL EQUATIONS. 1. Examples of the one kind are : (a + b){a-b)=d'-b\ (1) (a2_ 62) -=.(«- 6) = (a + 6)2 -=-(« + 6). (2) The first member of (1) is reduced to the second member by performing the indicated multiplication. Both members of (2) are reduced to the common form, a + b,hj performing the indicated divisions. 2. An Identical Equation, or simply an Identity, is an equar tion one of whose members can be reduced to the other, or both of whose members can be reduced to a common form, by performing the indicated operations. 3. Notice that identical equations are true for all values that may be substituted for the literal numbers involved. E.g., if a = 6 and 6 = 3, equation (1) becomes 8 X 2 = 25 - 9, or 16 = 16 ; and equation (2) becomes 16 -- 2 = 64 -- 8, or 8 = 8. We need not further discuss identical equations, since we have constantly dealt with them in the preceding chapters. §2. CONDITIONAL EQUATIONS. 1. Examples of the second kind are : a! + l = 3. (1) 9^-1 = 8. (2) x + y = 5. (3) 80 §2] CONDITIONAL EQUATIONS. 81 The first member of (1) reduces to the second member, when x = 2. It seems evident, and we shall later prove, that a; + 1 reduces to 3 ordy when a; = 2. The first member of (2) reduces to the second member, when a; = + 3 and when a; = — 3. We shall later prove that ^ — \ reduces to 8 only when a; = + 3 or — 3. The first member of (3) reduces to the second member, when a; = 1 and 2/ = 4, when a; = — 3 and 2/ = 8 ; but not when a; = 6 and 2/ = 6, when a; = — 4 and y = 8. Therefore, equation (3) is true for many pairs of values of x and y, but not for all pairs of values chosen at random. 2. Such equations impose cx>nditions upon the values of the literal numbers involved. Thus, equation (1) imposes the con- dition that if 1 be added to the value of x, the sum will be 3. A Conditional Equation is an equation one of whose members can be reduced to the other only for certain definite values of one or more letters contained in it. Whenever the word equation is used in subsequent work, we shall understand by it a conditional equation, unless the con- trary is expressly stated. 3. The Unknown Numbers of an equation are the numbers whose values are fixed or determined by the equation. The Known Numbers of an equation are the numbers whose values are given or known. In the equation a^ — 1 = 8 the unknown number is x, and the known numbers are 1 and 8. In the equation x + y = 3 the unknown numbers are x and y ; the known number is 3. The unknown numbers are usually represented by the final letters of the alphabet, x, y, %, etc., as in the above examples. 4. An Integral Algebraic Equation is an equation whose mem- bers are integral algebraic expressions in the unknown number or numbers. The known numbers may enter in any way whatever. E.g., 3sje^ — 4: = 2x, and | a: + 5 2/ = f , are integral equations. 82 ALGEBRA. [Ch. IV 5. The Degree of an integral equation is the degree of its term of highest degree in the unknown number or numbers. 6. A Linear or Simple Equation is an equation of the first degree. E.g., a; + 1 = 6 is a linear equation in one unknown number ; 2a; + 3y = 5isa linear equation in two unknown numbers. • 7. A Solution of an equation is a value of the unknown num- ber, or a set of values of the unknown numbers, which if substituted in the equation, converts it into an identity. E.g., 2 is a solution of the equation a; + 1 = 3, since, when substituted for x in the equation, it converts the equation into the identity 2 + 1 = 3. Solutions of the equation aj^ — 1 = 8 are + 3 and — 3. The set of values 1 'and 2, of x and y, respectively, is a solu- tion of the equation a; + j/ = 3. 8. To Solve an equation is to find its solution. The process of solving an equation is also frequently called the solution of the equation. An equation is said to he satisfied hy its solution, or the solu- tion is said to satisfy the equation, since it converts the equation into an identity. 9. When the equation contains only one unknown number, a solution is frequently called a Root of the equation. E.g., 3 and — 3 are roots of the equation a;^— 1 = 8. § 3. EQUIVALENT EQUATIONS. 1. We shall now give some principles upon which the solu- tion of integral equations depends. But it is to be kept in mind that the final test of the correctness of a solution, no matter how obtained, is that it shall satisfy the given equation. Consider the equation f x — 5 = 1. (1) Adding 5 to both members, by Ch. I., § 1, Art. 16 (i.), we have fa; — 5-t-5 = l-|-5, or |a; = 6. (2) §3] EQUIVALENT EQUATIONS. 83 Multiplying both members of (2) by 3, by Ch. I., § 1, Art. 15 (iii.), we have 2 a; = 18. (3) Dividing both members of (3) by 2, by Ch. I., § 1, Art. 15 (iv.), we have a: = 9. (4) It is evident that equations (1), (2), (3), and (4) are satisfied by the same root 9. 2. Two equations are equivalent when every solution of the first is a solution of the second, and every solution of the second is a -solution of the first. E.g., equations (1), (2), (3), and (4) of Art. 1. 3. The methods of solving integral equations depend upon principles which enable us to change a given equation into an equivalent equation whose solution is more easily obtained than that of the given one. This process is called transforming the equation, or the transformation of the equation. Fundamental Principles for solving Integral Equations. 4. In the principles of equivalent equations which we shall now prove, the solutions are limited to finite values. 5. The transformations made in the example of Art. 1 illus- trate the following principles : (i.) Addition and Subtraction. — If the same number or eaypres- sion be added to, or subtracted from, both members of an equation, the derived equation will be equivalent to the given one. (ii.) Multiplication. — If both members of an equation be mul- tiplied by one and the same number, not 0, or by an expression which does not contain the unknown number or numbers, the derived equation will be equivalent to the given one. (iii.) Division. — If both members of an equation be divided by one and the same number, not 0, or by an eoiypression which does not contain the unknown number or numbers, the derived equa- tion will be equivalent to the given one. 84 ALGEBRA. [Ch. IV (i.) Let P= Q be the given equation, and iV be any number or expression. Then the equation wherein the upper signs, +, go together and the lower signs, — , go together, is equivalent to the given one. For any solution of the given equation makes P equal to Q. There- fore, by Ch. I., § 1, Art. 15 (i.) and (ii.), that solution makes P ± N equal to Q ± N, and hence is a solution of the derived equation. Conse- quently, no solution is lost by the transformation. But the given equation is obtained from the derived equation by sub- tracting the number or expression which was added, or by adding the number or expression which was subtracted, in forming the derived equa- tion. Therefore any solution of the derived equation is a solution of the given equation, and no solution is gained by the transformation. Conse- quently, the two equations are equivalent. (ii.) It is more convenient to prove this principle when all the terms of the equation are in the same member, say the first. The latter equa- tion is, as we have seen, equivalent to the given one. Th'en any solution must reduce the first member to 0. Let P = be the given equation, and JV be any number, not 0, or any expression which does not contain the unknown number or numbers. Then the equation i\r.p=JV. 0=0 is equivalent to the given one. For any solution of the given equation must reduce P to 0, and, there- fore, by Ch. III., § 3, Art. 16, must also reduce N- F to 0. Hence it is also a solution of the derived equation. That is, no solution is lost by the transformation. Any solution of the derived equation must reduce JV • P to 0. But N is not 0, and, since it does not contain the unknown number or numbers, it cannot reduce to for any value of the unknown number or numbers. Consequently, by Ch. III., § 3, Art. 18, any solution of the derived equa- tion must reduce P to 0, and hence is a solution of the given equation. That is, no solution is gained by the transformation. Consequently, the two equations are equivalent. In a similar way (iii.) is proved. 6. It is important to notice that the solution of an equation does not rest simply on the principles used in Art. 1. For, by these principles we should be permitted to multiply both members of the equation by 0, or to multiply or divide both §3] EQUIVALENT EQUATIONS. 85 members by an expression which contains the unknown numbers. If the multiplier were 0, any value of the unknown number would be a solution of the derived equation, but not of the given equation. E.g., 2 a; - 6 = has the root 3, while (2 a; - 6) x = is evidently satisfied by 1, 2, 3, 4, etc., without end. If the multiplier contain the unknown number or numbers, values of the unknown number or numbers will reduce the multiplier to 0, and therefore the first member of the derived equation to 0, without reducing the first member of the given equation to 0. E.g., 2 a; - 6 = has the root 3, while (2 a; - 6)(a;-2)=0 is satisfied not only by 3, since (6 — 6) x 1 = x 1 = 0, but also by 2, since (4 - 6) (2 - 2) = (- 2) x = 0. But 2 is not a solution of the given equation. That is, in multiplying both members of the given equation by a; — 2, we have gained a root 2. The derived equation is, therefore, not equivalent to the given one. If the divisor be an expression which contains the unknown number or numbers, one or more solutions are lost. E.g., the equation a;^ — 1 = 2 (a; + 1) is satisfied by the two roots — 1 and 3. Dividing members by as + 1, we obtain a; — 1 = 2. This equation is satisfied by 3, but not by — 1. The derived equation is, therefore, not equivalent to the given one. Applications. 7. The following applications of the preceding principles will simplify the work of solving an equation. (i.) Any term may be transferred from one member of an equation to the other, if its sign be reversed from + to —, or from — to +. E.g., 2x — 4: = x + l and 2a; — a!=l + 4 are equivalent equations. This step is equivalent to adding 4 to, and subtracting x from, both members of the given equation. 86 ALGEBRA. [Ch. IV (ii.) Hie same term, or equal terms, may be dropped from both members of an equation. E.g., 2a;-3 + 8 = a; — 3 and 2a! + 8 = £(; are equivalent equations. This step is called cancellation of equal terms. (iii.) The signs of all the terms of an equation may be reversed. E.g., 5a! — 3 = 9 — a; becomes — 6a; + 3 = — 9 + a!, when both members are multiplied by — 1. 8. The preceding principles apply to integral equations of any degree. In this chapter we shall confine our attention to linear equations in one unknown number. § 4. LINEAR EQUATIONS, IN ONE UNKNOWN NUMBER. 1. Ex. 1. Solve the equation 17 a; + 6 = 10 a; + 27. Transferring 10 x to the first member and 6 to the second member, we have 17a;-10a; = 27-6. Uniting like terms, 7 a; = 21. Dividing by 7, x = 3. Chech. — Substituting 3 for x in the given equation, we obtain the identity 51 + 6 = 30 + 27. This check is to test the accuracy of the work, and not the equivalence of the equations. Ex. 2. Solve the equation 14 — 8 a; = 19 — 3 a;. Transferring terms, — 8 a; + 3 a; = 19 — 14. Uniting like terms, — 5 a; = 5. Dividing by — 5, a; = — 1. Ex. 3. Solve the equation |^ (a; + 6) — -^ a; = ^(3 a; — 1) + 1. Multiplying both members by 12, the lowest common multi- ple of the fractional coefficients, we obtain 6(a; + 5)-4a; = 3(3a;-l) + 12. Removing parentheses, 6a; + 30 — 4a; = 9a! — 3+12. §4] LINEAR EQUATIONS. 87 Transferring and uniting terms, — 7 a; = — 21. Dividing by — 7, x = 3. Ex. 4. Solve the equation 3|(a; + 1) + 4|- (aj + 1) = 16. Uniting terms in the first member, without clearing of frac- tions or removing parentheses, we have 8(a; + l) = 16. Dividing by 8, a; + 1 = 2 ; whence a; = 1. 2. The following general directions will be found useful in preparing an equation for solution : (i.) Remove any fractional coefficients by multiplying both sides of the equation by the L.C.M. of their denominators. (ii.) Remove any parentheses. (iii.) Transfer ail terms containing unknown numbers to one member of the equation, usually to the first member, and all the terms containing Icnovm numbers to the other member. (iv.) Unite like terms. An equation thus prepared for solution is called the Normal Form of that equation. The preceding suggestions apply also to an integral equation of any degree. If the equation be linear in one unknown number, the solution is completed by dividing both members by the coef&cient of the unknown number. EXERCISES. Solve each of the following equations : 1. a; + 2 = 3. 2. 15 - a; 4. Ja; = 6. 5.-2 = 7. 5a; = 15. 8. 11 = 10. fa; -5 = -8. 11. i(a; + 5) 13. 5 a: + 7 = 11 + 4 X. 15. ix + %=-\x-\. 17. -3a;-7=-4a;-7. 19. 36 -9z = 116 + lis. s=-27. 3. 17 = 9 - a;. -*«. 6. ^a; = 0. -22 a;. 9. 4 a; = - 16. 5) =4. 12. i(a;-6) = 7. 14. 5a;- -7 = : 4 a; -t- 3. 16. 7a; -1-8 = 4 a; -)- 15 -1- 2 a;. 18. 15 a; -8: = 20 a; - 8 - 4 a;. 20. 61- 5s,: = 7 2/ 4- 85. 88 ALGEBRA. [Ch. IV 21. 8 a; -18 = a; + 12 -3 a;. 22. 5x + 11 = 16 - 3 k - 4 k. 23. 5a; + 7 -3a; = 8a;- 5a; + 9. 24. -79;-2+33;=-x-4x+3. 25. 15a;+4+7x=14x+5 + 7x. 26. 3x - 5 - 9x = 2x - 7 - 9x. 27. x-7 = ^x + |x. 28. ^x+Jx = ix-7. 29. fx-fx + ^=-Jx. 30. -x + Jx + |x = ll. 31. 2x-(5x+5) = 7. 32. 7x- (3x-ll) = 4. 33. 3 X - 7 - (5 x + 17) = 0. 34. 3 (x + 1) = - 5 (x - 1). 35. i(x + 3) = Tij(3x+ 16). 36. |(5x - 2) - 6 = K^a; - 3). 37. 4 X - 2 (2 - x) = 6. 38. 6 x - [7 x - (8 x - 18)] = 16. 39. i(x-2) + i-[x-K2*-l)]=0. 40. 3J[28-Qx + 24)] = 3i(2i + ix). 41. 2(x + l)-3(x+l) +9(x+l)+18 = 7(x + l). 42. (2 X + 7) (x - 3) = (x - 3) (2 X + 8). 43. (x + l)(x + 2)=(x-3)(x-4). 44. x2 - X [1 - X - 2 (3 - x)] = X + 1. 45. (s + 1) (x + 1) = [111 - (1 - x)] X - 80. 46. 3-x = 2(x-l)(x + 2) + (x-3)(5-2x). 47. 5 (3 X - 5) - 17 - 8 (3 X - 5) - 2 (3 X - 5) = 3. 48. - 17 (7 X - 83) + 28 (7 X - 83) - 34 = 12 (7 x - 83). 49. (16x + 5)(9x + 31) = (4x+ 14)(36x+ 10). 60. (5x-2)(3x-4) = (3x+5)(5x-6). 61. x(x + 2)+x(x + l) = (2x-l)(x + 3). Find tlie remainder of each of the following divisions, and hence the value of m which will make the dividend exactly divisible by the divisor : 52. (9x2-3x + m)--(x-l). 63. [4 x8 - 2 x2 + X - |(m + 1)] H- (2 X + 3). 64. [7x2-(»i-l)x + 3] -r-(x+2). 65. [4 x3 - 24 x2 + (36 - m) x - 15] -f- (2 x - 5). 66. [21 x3 - 23 x2 + (15 - m) x - 8] -=- (3 x - 2). 57. [x8-5x2 + 3x-(m-4)]^(x-5). 68. [2x8-5(m-l)x2 + 4x-2OT]H-(x-2). CHAPTER V. PROBLEMS. 1. A Problem is a question proposed for solution. Pr. 1. The greater of two numbers is three times the and their sum is 84. What are the numbers ? This problem involves the given number 84 and two required numbers. The statements of the problem impose two condi- tions upon the values of the required numbers : ( i. ) The greater number is three times the less. (ii.) TJie sum of the two numbers is 84. To solve the problem, it is necessary first to translate these relations or conditions from the verbal language of the prob- lem into the symbolic language of Algebra, i.e., to express them by means of algebraic signs and symbols. Let X stand for the less required number. Then, by the first condition, the greater number is, in verbal language : three times the less ; in algebraic language : 3 x. Consequently, the required numbers are represented by x (the less) and 3 x (the greater). The second condition is, in verbal language : the less number plus the greater is equal to 84; in algebraic language : a; -f- 3 a; = 84. This equation is called the equation of the problem. From this equation we obtain x = 21, the less number. Therefore 3x= 63, the greater number. Notice that this problem could have been solved by letting X stand for the greater number, and consequently ^x for the less. The resulting equation would then have been ^X + X=84:. Whence x = 63, the greater number ; and ^ a; = 21, the less. 90 ALGEBRA. [Ch. V This metliod leads to an equation in wliich the unknown number is one of the required numbers of the problem. Pr. 2. rind two consecutive integers whose sum is 163. In this problem the conditions are not both explicitly stated. The first condition is contained in the words, two consecutive integers. Let x stand for the less number. Then, by the first condition, the greater number is, in verbal language : the less number plus 1 ; in algebraic language : x + 1. The required numbers are thus represented by x (the less) and a; + 1 (the greater) . The second condition is, in verbal language : the less number plus the greater is 163 ; in algebraic language : x + (x + l) = 163, the equation of the problem. From this equation we obtain x = 81, the less number ; and therefore a; + 1 = 82, the greater. Pr. 3. A is 40 years old and B is 10 years old. After how many years will A be three times as old as B ? Let X stan.d for the required number of years, after which A will be three times as old as B. The condition of the problem involves other unknown num- bers than the required number. These we first express in terms of the required and given numbers. In X years the number of years in A's age will be 40 + a; ; the number of years in B's age will be 10 + x. The condition of the problem is, in verbal language : the number of years in A's age x years hence is equal to three times the number of years in B's age X years hence ; in algebraic language : 40 + a; = 3 (10 + a;). From this equation we obtain x = 5, the required number. In 5 years A will be 45 years old, and B will be 15 years old. Notice that the numbers used in the solution are abstract numbers ; 40 is the number of years in A's age, not A's age. PROBLEMS. 91 Pr. 4. At an election at which 943 votes were cast, A and B were candidates. A received a majority of 65 votes. How many votes were east for each candidate ? Let X stand for number of votes cast for A. Then, by the first condition, the number of votes cast for B is, in verbal language : 943 minus the number cast for A ; in algebraic language : 943 — x. The second condition is, in verbal language : the number of votes cast for A exceeds the number cast for B by 65; in algebraic language : x — (943 — x) = 65. From this equation we obtain x = 504, whence 943 — x = 439. Pr. 5. Fifteen coins, dollars and quarter-dollars, amount to f 7.50. How many coins of each kind are there ? We take one dollar as the unit, and express parts of dollars as fractional parts of this unit. Let X stand for the number of dollars. Then, by the first condition, the number of quarter-dollars is, in verbal language : 15 minus the number of dollars ; in algebraic language : 15 — «. The second condition is, in verbal language : the number of dollars plus one-fourth of the number of quarter-dollars is 7^ ; in algebraic language : x + ^(15 — x) = 7^. From this equation we obtain x — 5; whence 15 — a; = 10. Evidently the total value of the coins is 5 + J^ dollars, or f 7|. Both conditions refer to abstract numbers ; the first condition to the number of coins, the second to the number of dollai;s. Pr. 6. A drove of sheep and goats, 200 animals in all, is to be sold. A offers to pay $ 1.25 for each sheep and f 1.60 for each goat ; B offers to pay $ 1.50 for each animal. The owner of the drove accepts B's offer because he finds that it will net him $ 22 more than A's offer. Find the number of sheep and goats in the drove. Let X stand for the number of sheep. 92 ALGEBRA. [Ch. V Then, by the first condition, the number of goats is, in verbal language : 200 minus the number of sheep ; in algebraic language : 200 — x. The second condition involves other unknown numbers than the required numbers. We must express the number of dollars in A's offer and the number of dollars in B's offer in terms of the required and given numbers. The number of dollars in A's offer is, in verbal language : the number of sheep multiplied by the num- ber of dollars offered for each sheep, plus the number of goats multiplied by the number of dollars offered for each goat ; in algebraic language : 1.26 x + 1.6 (200 — a;). The number of dollars in B's offer is, in verbal language: the number of animals multiplied by the mimber of dollars offered for each animal; in algebraic language : 200 x 1.5. The second condition is, in verbal language : the number of dollars in B's offer minus the nimiber of dollars in A's offer is 22 ; in algebraic language: 200xl.5-[1.25a;+1.6(200-a!)]=22. Prom this equation we obtain x = 120, the number of sheep ; whence 200 — a; = 80, the number of goats. Notice again that both conditions refer to abstract numbers. Pr. 7. A box contains a certain number of pencils, of which one-third are red, one-sixth are blue, and 15 are black. How many of the pencils are red, and how many are blue ? This problem can be solved more readily by assuming for the unknown number of the equation another number than one of the required numbers. Let X stand for the total number of pencils. Then, by the first condition, the number of red pencils is \ x, and by the second condition, the number of blue pencils is \ x. Finally, by the third condition, \x + \x-\-15 = x; whence x = 30. Therefore, ^ a; = 10, the number of red pencils, and ^x = 5, the number of blue pencils. PROBLEMS. 93 Pr. 8. A number is composed of two digits whose sum is 8. If the digits be interchanged, the resulting number will exceed the original number by 18. What is the number ? In accordance with the suggestion in Pr. 7, we assume one of the digits of the required number, not the required number, as the unknown number. Let X stand for the digit in the units' place. Then, by the first condition, the digit in the tens' place is, in verbal language : 8 minus the digit in the units' place ; in algebraic language : 8 — aj. Therefore the original number is 10 (8 — a;) + a; ; the second number (when the digits are interchanged) is 10 a; + (8 — x). The second condition of the problem, then is, in verbal language: the second number is equal to the original number plus 18 ; in algebraic language : 10 a; + (8 — a;) = 10 (8 — a;) + a; + 18. Whence, x = 5, the digit in the units' place ; and 8 — a; = 3, the digit in the tens' place. The original number is 10(8 — a;) + a;, =35; the second number is 10 a; + 8 — a;, = 53, and 53 — 35 = 18. Pr. 9. A carriage, starting from a point A, travels 35 miles daily; a second carriage, starting from a point B, 84 miles behind A, travels in the same direction 49 miles daily. After how many days will the second carriage overtake the first? At what distance from B will the meeting take place ? Let X stand for the number of days after which they meet. Then the number of miles traveled by the first carriage is 35 X, and the number of miles traveled by the second carriage is 49 X. The condition of the problem is, in verbal language : the number of miles traveled by the first carriage is equal to the number of miles traveled by the second carriage minus 84; in algebraic language : 35 a; = 49 a; — 84. Prom this equation, we obtain a; = 6. 94 ALGEBRA. [Ch. V The distance traveled by the first carriage is 210 miles, and the distance traveled by the second carriage is 294 miles. They therefore meet 294 miles from B. Pr. 10. A man asked another what time it was, and received the answer : " It is between 5 and 6 o'clock, and the minute- hand is directly over the hour-hand." What time was it ? At 6 o'clock, the minute-hand points to 12 and the hour- hand to 5. The hour-hand is therefore 25 minute-divisions in advance of the minute-hand. Let X stand for the number of minute-divisions passed over by the minute-hand from 5 o'clock until it is directly over the hour-hand between 6 and 6 o'clock. By the first condition, which is implied in the problem, the number of minute-divisions passed over by the hour-hand is, in verbal language : tJie number of minute-divisions passed over by the minute-hand minus 25 ; in algebraic language : x — 25. The second condition, which is also implied in the problem, is, in verbal language : the number of minute-divisions passed over by the minute-hand is 12 times the number of minute-divi- sions passed over by the hour-hand; in algebraic language : a; = 12 (a; — 25). From this equation we obtain x.= 27^y. Consequently, the two hands coincide at 27^ minutes past 5 o'clock. 2. The beginner will find some suggestions for translating the conditions of a problem into algebraic language helpful. (i.) Observe what are the numbers whose values are required. It will, in general, be possible to continue the solution of the prob- lem by representing one of these numbers by a letter, and operat- ing upon or by that letter as if it were a known number. See Prs. 1, 2, 3, 4, 5, 6, 9, and 10. (ii.) Every problem which can be solved must state, implicitly or explicitly, as many conditions as there are required numbers in the problem. PROBLEMS. 95 (iii.) Tlie numbers involved in the statements will, in general, be not only the required numbers, but also other unknown num- bers which must be expressed in terms of the required numbers. See, in particular, Prs. 3, 6, 8. ' (iv.) Eatress concisely in verbal language each given coiidition. It is frequently necessary to modify the statement in order to adapt it to translation into algebraic language. See Prs. 2 and 10. (v.) Translate each verbal statement of a condition into alge- braic language. All but one of the conditions will give expressions for required numbers. The last condition will give the equation of the problem. (vi.) There are problems in which other numbers than the re- quired numbers can be used to better advantage in applying the conditions of the problems, and from which the required numbers can be readily found. See Prs. 7, 8. 3. In applying the suggestions of Art. 2, it is important to remember that the letter x always represents an abstract num- ber. The beginner must nerer put x for distance, time, weight, etc., but for the number of miles, of hours, of pounds, etc. Keep in mind also that in any one equation the magnitudes of all concrete quantities of the same kind must be referred to the same unit ; if x refer to a certain number of yards, then all other distances must likewise represent numbers of yards, not of miles or of feet. EXERCISES. 1. If twice a number be added to 18, the sum will be 82. What Is the number ? 2. If 4 be subtracted from five times a number, the remainder will be 11. What is the number ? 3. If one-fourth of a number be diminished by 5, the remainder will be 2. What is the number f 4. The sum of two consecutive even numbers is 34. What are the munbers ? 5. Find the number whose double exceeds its half by 6. 6. Find three consecutive odd numbers whose sum is 57. 7. If 48 be added to a number, the sum will be equal to nine times the number. What is the number ? 96 ALGEBRA. [Ch. V 8. If $120 be divided between A and B so that A shall receive 1 20 more than B, how many dollars will each receive ? 9. A sum of $ 2500 is divided between A and B. B receives f 4 as often as A receives $ 1. How much does each receive ? 10. Divide 75 into two parts, such that three times the first part shall be 15 greater than seven times the second. 11. Divide 190 into three parts so that the second shall be three times the first, and the third five times the second. 13. A father's age exceeds his son's by 18 years, and the sum of their ages is four times the son's age. What are their ages ? 13. A man bought a horse, a carriage, and harness for $320. The horse cost five times as much as the harness, and the carriage cost twice as much as the horse. How much did each cost ? 14. The deposits in a bank during three days amounted to $ 16,900. If the deposits each day after the first were one-third of the deposits of the preceding day, how many dollars were deposited each day ? 15. A merchant, after selling one-third, one-fourth, and one-sixth of a piece of silk, has 15 yards left. How many yards were there in the piece ? 16. If two trains start together and run in the same direction, one at the rate of 20 miles an hour, and the other at the rate of 30 miles an hour, after how many hours will they be 250 miles apart ? 17. A teacher proposes 16 problems to a pupil. The latter is to receive 5 marks in his favor for each problem solved, and 3 marks against him for each problem not solved. If the number of marks in his favor exceed those against him by 32, how many problems will he have solved ? 18. A merchant paid 30 cents a yard for a piece of cloth. He sold one-half for 35 cents a yard, one-third for 29 cents a yard, and the remainder for 32 cents a yard, gaining $ 18.15 by the transaction. How many yards did he buy ? 19. Two men start from points 100 miles apart and travel toward each other, one at the rate of 15 miles an hour, and the other at the rate of 10 miles an hour. After how many hours will they meet, and how far will their point of meeting be from the starting point of the first ? 20. A father is 32 years old, and his son is 8 years old. After how many years will the father's age be twice the son's ? 21. Divide 130 into five parts so that each part shall be 12 greater than the next less part. 23. A, traveling at the rate of 20 miles a day, has four days' start of B, who travels at the rate of 25 miles a day in the same direction. After how many days will B overtake A ? PROBLEMS. 97 23. A sum of money is equally divided among four persons. If $ 60 more be divided equally among six persons, the shares will be the same as before. How many dollars are divided ? 34. Atmospheric air is a mixture of four parts of nitrogen with one of oxygen. How many cubic feet of oxygen are there in a room 10 yards long, 5 yards wide, and 12 feet high ? 25. A merchant paid $ 6.15 in an equal number of dimes and five-cent pieces. How many coins of each kind did he pay ? 26. A man has $4.75 in dimes and quarters, and he has 5 more quarters than dimes. How many coins of each kind has he ? 27. A leaves a certain town P, traveling at the rate of 21 miles in 5 hours ; B leaves the same town 3 hours later and travels in the same direction at the rate of 21 miles in 4 hours. After how many hours will B overtake A, and at what distance from P ? 28. The circumference of the front and hind wheels of a wagon are 2 and 3 yards, respectively. What distance has the wagon moved when the front wheel has made 10 revolutions more than the hind wheel ? 29. The sum of two numbers is 47, and their difference increased by 7 is equal to the less. What are the numbers ? 30. The sum of three consecutive even numbers exceeds the least by 42. What are the numbers ? 31. The sum of the two digits of a number is 4. If the digits be inter- changed, the resulting number will be equal to the original one. What is the number ? 32. A father is three times as old as his son, and 10 years ago he was five times as old as his son. What is the present age of each ? 33. One barrel contained 48 gallons, and another 88 quarts of wine. From the first twice as much wine was drawn as from the second ; the first then contained three times as much wine as the second. How much wine was drawn from each ? 34. A child was born in November. On the 10th of December the number of days in its age was equal to the number of days from the 1st of November to the day of its birth, inclusive. What was the date of its birth? 35. A regiment moves from A to B, marching 20 miles a day. Two days later a second regiment leaves B for A, and marches 30 miles a day. At what distance from A do the regiments meet, A being 350 miles from B ? 36. The sum of two digits of a number is 12. If the digits be inter- changed, the resulting number exceeds the original one by three-fourths of the original number. What is the number ? 98 ALGEBRA. [Ch. V 37. Three boys, A, B, and C, have a number of marbles. A and B have 44, B and C have 43, and A and C have 39. How many marbles have they all, and how many marbles has each ? 38. The tail of a fish is 4 inches long. Its head is as long as its tail and one-seventh of its body, and its body is as long as its head and one- half of its tail. How long is the fish, and how long are its head and its body? 39. A father divided his property equally among his sons. To the oldest son he gave $ 1000 and one-seventh of what remained ; to the second son he gave $2000 and one-seventh of what was then left; to the third son he gave $ 3000 and one-seventh of the remainder ; and so on. What was the amount of his property, and how many sons had he ? 40. A man, wishing to give alms to several beggars, finds that in order to give 15 cents to each one, he must have 10 cents more than he has ; but that if he were to give 12 cents to each one, he would have 14 cents left. How many beggars are there ? 41. A train runs from A to B at the rate of 30 miles an hour; and returning runs from B to A at the rate of 28 miles an hour. The time required to go from A to B and return is 15 hours, including 30 minutes' stop at B. How far is A from B ? 42. A cistern has 3 taps. By the first it can be emptied in 80 minutes, by the second in 200 minutes, and by the third in 5 hours. After how many hours will the cistern be emptied, if all the taps be opened ? 43. A cistern has 3 taps. By the first it can be filled in 6 hours, by the second in 8 hours, and by the third it can be emptied in 12 hours. In what time will it be filled if all the taps be opened ? 44. An inlet pipe can fill a cistern in 3 hours, and an outlet pipe can empty it in 9 hours. After how many hours will the cistern be filled if both pipes be open half the time, and the outlet pipe be closed during the second half of the time ? 45. In my right pocket I have as many dollars as I have cents in my left pocket. If I transfer $6.93 from my right pocket to my left, I shall have as many dollars in my left pocket as I shall have cents in my right. How much money have I in my left pocket ? 46. A servant is to receive $ 170 and a dress for one year's services. At the end of 7 months she leaves her place and receives f 95 and the dress. What is the value of the dress ? 47. A farmer found that his supply of feed for hie cows would la«t only 14 weeks. He therefore sold 60 cows, and his supply then lasted 20 weeks. How many cows had he ? 48. At 6 o'clock the hands of a clock are in a straight line. At wbat time between 7 and 8 o'clock will they be again ia a straight line ? At what time between 9 and 10 o'clock ? PROBLEMS. 99 49. A cistern has 3 pipes which can empty it in 6, 8, and 10 hours respectively. After all three pipes have been open for 2 hours they have discharged 94 gallons. What is the capacity of the cistern ? 50. At what time between 10 and 11 o'clock are the minute-hand and the hour-hand of a clock at right angles to each other ? Eind two solu- tions. At what time between 12 and 1 o'clock ? 51. At what time between 3 and 4 o'clock will the minute-hand of a clock be 5 minute-divisions in advance of the hour-hand ? At what time 17 minute-divisions ? A watch has the second-hand attached at the same point as the hour- and the minute-hand: 52. At what time between 1 and 2 o'clock is the second-hand over the minute-hand ? At what time between 8 and 9 o'clock ? 53. At what time between 11 and 12 o'clock does the second-hand of a watch bisect the angle between the hour- and the minute-hand ? At what time between 4 and 5 o'clock ? 54. A woman sells J an apple more than one-half of her apples. She next sells J an apple more than one-half of the apples not yet sold, and then has 6 apples left. How many apples had she at first ? 55. A steamer and a sailing vessel are both to sail from M to N. The steamer sails 40 miles every 3 hours, and the sailing vessel 24 miles in the same time. The sailing vessel has traveled 13J miles when the steamer sails, and arrives at N 6 hours later than the steamer. How long is the steamer in sailing from M to N, and how far is M from N ? 56. A wall can be built by 20 workmen in 11 days, or by 30 other workmen in 7 days. If 22 of the first class work together with 21 of the second class, after how many days will the work be completed ? 57. In a certain family each son has as many brothers as sisters, but each daughter has twice as many brothers as sisters. How many children are in the family ? 58. A merchant's investment yields him yearly 33|% profit. At the end of each year, after deducting 1 1000 for personal expenses, he adds the balance of his profits to his invested capital. At the end of three years his capital is twice his original investment. How much did he invest ? 59. I have in mind a number of six digits, the last one on the left being 1. If I bring this digit to the first place on the right, I shall obtain a number which is three times the number I have in mind. What is the number ? 60. A dog caught sight of a hare at a distance of 50 dog's leaps. The dog makes 3 leaps while the hare makes 4 leaps, but the length of two dog's leaps is equal to the length of 3 hare's leaps. How many leaps will the hare make before the dog overtakes him ? CHAPTER VI. TYPE-FORMS. We shall in this chapter consider a number of products and quotients which are of frequent occurrence. They are called Type-Forms. § 1. TYPB-PORMS IN MULTIPLICATION. The Square of an Algebraic Expression. 1. By actual multiplication, we have (a + bf = (a + b)(a + b) = a' + 2ab + b\ That is, tlie square of the sum of two numbers is equal to the square of the first number, plus tioice the product of the two num- bers, plus the square of the second number. E.g., {2x+5yy= (2 xy + 2 (2 a;) (6 y) + (5 yf = i ocF + 20 xy + 25 y\ 2. By actual multiplication, we have (a - by ={a-b){a-b)=a'-2ab + b\ That is, the square of the difference of two numbers is equal to the square of the first number, minus twice the product of the two numbers, plus the square of the second number. E.g., (3 X - 1 yy= (3 xy -2{Sx) (7 y) + (7 yy = 9x^-4:2xy + 4:9y\ Observe that this type-form is equivalent to that of Art. 1, since a — b = a+(—b). E.g., (3x-7 yy= (3xy + 2{3x)(- 7y) + i-7 yy = 9 a^ — 4:2 xy + 4:9 y^, as above. The signs of all the terms of an expression which is to be squared may be changed without changing the result. For, {a-by = [-(b-a)f^{b-ay. 100 § 1] TYPE-FORMS IN MULTIPLICATION. 101 3. We have, by Art. 1, (a + b + cy=l{a + b)+cy = (a + by + 2(a + b)c + d' = a' + 2ab+b^ + 2ac + 2bc + (?. Therefore (o + 6 + c)^ = a^ + 6^ + c^ + 2 a6 + 2 ac + 2 6c. In like manner {a + b - of = a" + b'' + c^ + % ab -2 ac -2 be. (a-b- cy = a^+b^ + c''-2ab-2ac + 2 be. By repeated application of this principle we can obtain the square of a multinomial of any number of terms. We have (a + b + c + ay ^[(a + b + cf + df = a" + b^ + c" + 2 ab + 2 ac + 2 be + 2{a + b + c)d + d' = a'' + b^ + - X + l) = i{x' + 1)+ x]l(a? + 1)- x-\ ^Ix' + Vf-x' = d' + 2a? + l-x^ = x^ + a? + l. Ex. 3. {x-y + %){x + y-z) = lx-{y- «)] [a; + (2/ - z)] — ^—{y — z)2 = 0^ — (2/^ — 2 2/2 + 3^) = a? — y^ — z^ + 2yz. EXERCISES II. Find, without performing the actual multiplications, the values of 1. (a;-|-2)(x-2). 2. (2 a - 3) (2 a + 3). 3. (a-|)(o + f). 4. {bx + iy)(l>x-iy). 8. (3a2 + J(jj6)(3a52_ia6). g. (-^ 3*2+ 7) (3^2 + 7). 7. (2aa;2^3„25;-)(-2ax2-i-3a2x). 8. (3 a" + 7 6"')(3 n" - 7 6«). 9. (5 mn2 + 2 m'^n) ( - 8 mn^ + 2 m^n) . 10. ( - 5 x»+i + 9 x«-i) (5 x''+i + 9 »»-!) . 11. i:a2 + 6(a + 6)][a2-6(o + 6)]. § 1] TYPE-FORMS IN MULTIPLICATION. 103 12. (a; + 2/ + 6)(cc + 2/-5). 13. (4a-3 6 -7)(4a - 3 & + 7). 14. (K2 + ^2 + ;52)(_a;2 + j,2 4.22). 15. (a^ - a6 + 62) (a" + 06 + 62). 16. (a;2 + 2a;-l)(a;2_2a;-l). 17. {x* -x^ + l)(x* +31? -1). 18. (-«2_62+3)(o2-&2 + 3). 19. (o2-62-c')(a2 + 62 + c2). 20. (l+2a + 36 + 4c)(l + 2a-36-4c). 21. (a+6 + c-d)(a + 6-C+dr). 22. («» - 3 o^a; + S ax^ - mfi) (_a? + 3 a^a; + 3 a»2 + »»). 28. (3? - ■f - ^ - vfi^infl ■{■ ifi + z-^ - 'Ufl). 24. (-s3 + a!i2-2a;-l)(a;3-f-a;2 + 2;e-l). Simplify the following expressions : . 25. (l + a;)2-(l-K)(l + a;). 26. (2x -{-Zyfiix-Zyy. 27. (1 - a6)2 (1 + a6)2. 28. (a;-3)(a;- l)(x + l)(x + 3). 29. (a,-x){a + x) (^cfl + «2) (o* _|_ -54) . 30. (S2 _ 1) (a;9 + 1) (a;l 4. 1) (a;2 + 1). 31. (a2 + 2 a6) (0.2 - 2 ah) (a^ + 16 ai6*) (a« + 4 a262). 32. (x3 -a; + l)(a;2 + a; + 1) (a* - a" + 1). 33. (a + 6 - c)(a + c - 6)(6 + c - «)(« + 6 + c). 34. (a - 6) (a + 6 - c) + (6 - c)(6 + c - a) + (c - a)(c + a - 6). The Product (* + a) (x + 6). 5. By actual multiplication, we have (x + a){x + b) = x^ + {a + b)x + ab; {x + a)ix-b)=x^+{a-b)x- ab; (x - a) (x- b)=x^-ia + b)x + ab. That is, the product of two binomials, having the same first term, is equal to the square of this term, plus the product of the algebraic sum of the second terms by the common first term, plus the product of the second terms. Ex. 1. (a; + 3) (a; + 5) = as^ + 8 a; + 15. Ex. 2. {ax — 6) (ax — c) = aV — (b + c)ax + be. Ex. 3. (a + & + 5) (a + & - 3) = [(a + 6) + 6] [(« + &) - 3] = {a + by + 2(a + b)-15. 104 ALGEBRA. [Ch. VI BXBBCISES III. Find, without performing the actual multiplications, the values of the following indicated products : 1. (x + 7)(x + 4). 2. (x.-5)(a;-4). 3. (a-6)(a + 8). 4. (x-3)(x + 2). 5. (5 + 6)(7 + 6). 6. (3-x){i-x). 7- (6 + 2/)(2/-3). 8. (2x+l)(2x + 3). 9. (aJ/ + l)(aj/ + 5). 10. {3xy -8)(3xy -7). 11. (5a -36)(5a-76). 12. (2x!/ + 3z)(2x2/ - 4»). 13. (x2-7)(x2-5). 14. (2x3 -11) (2x8 + 4). 15. (x"'+i-4)(x"'+i-7). 16. (x + y- 3)(x + !/-5). 17. (ax + 6j/-c)(ax+ 62/-2c). 18. (x^ - 2x + 7)(x2 - 2x - 3). The Product (a;ir + 6)(c;r + rf). 6. By actual multiplication, we obtain (ax + b) (ex + d) = aex^ + (ad +bc)x + bd. In this type-form that part of the multiplication which gives the middle term of the type-form may be represented concisely by the following arrangement : cx + d ax + b (ad + be) X The products of the terms connected by the cross lines are called cross-products, and their sum is the middle term of the given trinomial. That is, the product of two binomials, arranged to powers of a common letter, is equal to the product of the first terms, plus the sum of the cross-products, plus the product of the last terms. Ex. 1. (7x--5y)(2x-\-3tj)=7x-2x+(7-3-5-2)xy-5y-3y = 14 x^ + 11 xy — 15 y\ EXERCISES IV. Pind, without performing the actual multiplications, the values of the following indicated products : 1. (3o-H)(5a-)-2). 2. (7x - 3)(3x - 1). 3. (5x -I- 7)(.3x-2). 4. (2x-9)(5x-M). 5. (2x-l-15)(4x-5). 6. (U u-3)(9a-h7). §1] TYPE-FORMS IN MULTIPLICATION. 105 7. (2a+&)(3a-&). 8. (2a - 6)(3a + &). 9. {Sx- y)(2x-y). 10. (7a + 36)(5a-26). 11. (Gx ~1 y-)(3x + 2y). 12. (5a;-3z)(2a; + 53). 13. (ly + 2u)(8y -7 u). 14. (2ab-x){3ab + x). 15. (5mji - 3j3)(6m« + 7p). 16. (9m2-3)(8m2 + n). 17. (3x^ + 5y^)(2x^ - 3y^). 18. [3(a + 6)+5][5(a + 6)-2]. 19. [2(a; - 2/)- 7][3(a; - «/)+ 2]. The Cube of an Algebraic EzpresMon. 7. By actual multiplication, we have (a + 6)« = o5 + 3a^6 + 3a62 + 6^ (1) and (a-by = a'--3a^b + Zab^-b^ (2) That is, the cube of the sum of two numbers is equal to the cube of the first number, plus three times the product of the square of the first number by the second, plus three times the product of the first number by the square of the second, plus the cube of the second. A similar statement can be made for (2). Ex. 1. {Bx + Syy=(5xf + S(5x)\3y) + 3(5a;) {Zyf + {Zyf = V2,Ba? + 225x'y + imxy^ + 27f. Ex.2. (x + 2y-Zzf=\_(x + 2y)-(Zz)y = (x + 2yf-^(x+2y)\Bz) + Z{x + 2y){3zy-{Zzf = 3?+&x'^y+12xy^+%f-^x'z-B&xyz - 36 yh + 27 xz^ + 54 yz^ - 27 ^. Observe that (2) is similar to (1), since (a-by=\:a+(-b)f. Also that (a -by = -(b- ay. BXBECISES V. Write, without performing the actual multiplications, the values of 1. (a + iy. 2. (a -2)3. 3. (2 a; + 3)3. 4. (s-iyy. 5. (a + 2 by. 6. (3 a -6)3. 7. iax + byy. 8. (2x-3yzy. 9. (x^ + Sxy. 106 ALGEBRA. [Ch. VI 10. (a%» - 6 any. 11. (2 xm^ + 5 x^i^y. la. (lo^jc- jaa!»)s. 13. (7 x" + 2 a;"-!)'. 14. (a-ft - 5 a6"+i)». 15. (a'^ + a + 1)'. 16. (3-a;-«3)s. 17. (^id' - x + iy. 18. What is the vatae ol k^ - 3 k^ + 3 x - 1, when x = m + 1 ? 19. What is the value of a' - 2 x^ + 3 x - 1, when x = ?/ - 2 ? Verify the following identities : 20. (a + 6 + c)S - 3 (a + &)(& + c)(c + a) - «' + 6' + c^. 21. (a - 6)8 + 3 (a - b)\a + 6) + (a -f 6)= + 3 (a - 6)(o + 6)^ = 8 n^. 22. (a + 6 + c)s - (6 + c - a)8 - (c + a - 6)3 - (a + 6 - c)3 = 24 a6c. Higher Powers of a Binomial. 8. By actual multiplication, we have (a + by = a' + ia'b+ 6a'b'+ iab' +b*; {a-by = a'-'ta'b+ Ba^^- ^ab^+h*; (a + by = a' + 5 a*b + \OaW + 10a^6' + bab' + b'; (a -by = a'-5 a*b + lOa'b' - lOaW + 5ab'-¥; (a + by = a^ + 6a'b + l5a'b^ + 20a^b^ + 15a^b* + 6ab' + b'; (a -by = a^-6a'b + 15 a^b" - 20 aW + 15 a'b* -6ab' + b\ The result of performing the indicated operation in a power of a binomial is called the Expansion of that power of the binomial. In the preceding expansions the following laws are evident : (i.) TJie number of terms exceeds the hinomial exponent by 1. (ii.) The exponent of a in thefii-st term is equal to the binomial exponent, and decreases by 1 from term to term. (iii.) The exponent of b in the second term is 1 and increases by 1 from term to term, and in the last term is equal to the binomial exponent. (iv.) The -coefficient of the first term is 1, and that of the second term, except for sign, is equal to the binomial exponent. (v.) 27*6 coefficient of any term after the second is obtained, except for sign, by multiplying the coefficient of the preceding term by the exponent of a in that term, and dividing the product by a member greater by 1 than the exponent of b in that term. §2] TYPE-FORMS IN DIVISION. 107 E.g., the coefficient of th.e_^fth term in the expansion of (a + 6)Ms 10 X 2 ^ 4 :^ 5. (vi.) TJie signs of the terms are all positive when the terms of the binomial are both positive; the signs of the terms alternate, + and — , whe^i one of the terms of the binomial is negative. Observe, as a check ; (vii.) The sum of the exponents of a and b in any term is equal to the binomial eoeponent. (viii.) The coefficients of two terms equally distant from the beginning and the end of the expansion are equal. In a subsequent chapter the above laws will be proved to hold for any positive integral power of the binomial. Ex. 1. (2 as - ^yy = (2a;)* - l{2x)\By) + Q(2x)\5yy -4(2a;)(32/)^ + (32/y = 16 a;* ^ 96 a'y + 216 a;^2/2 - 216 a;j/3 + 81 2/*. EXBHOISBS VI. Find the values of the following powers : 1. (1 + a;)*. 2. (3-4a)«. 8. (2x + ZyY. 4. (3m-2n)5. 6. (2a +1/. 6. (3a!-22/)«. 7. (jcfl + aby. 8. (7fi-x)K 9- (infin-mrfif. 10. (a»+i — &)*. 11. (a"-i6 + db'^y. 13. (aw+ia;"-! - a^-lx^+J)'. 13. Verify the identity [(a - 6)2 + (6 - c)2 + (c - a)2]2 = 2[(a - 6)* + (6 - c)* + (c - a)*]. §2. TTPB-rORMS IN DIVISION. Quotient of the Sum or the Difference of Like Powers of Two Numbers by the Sum or the Difference of the Numbers. X. By actual division, we have (a^-60-H(a + A)==a-6 and (a^-62)^(a =- A) = a + 6. That is, the difference of the squares of two numbers is exactly dimisible by the sum of the numbers, and also by the difference of the numbers, taken in the same order; the quotient in the first case is the difference of the two numbers, tahen in the same order, and in the second case is the sum of the two numbers. 108 ALGEBRA. [Ch. VI Ex.1. (9 - 25 a^)- (3 + 5 a;) =.3 -5 a;. Ex.2. (16x*-81y^-^(Ax'-9f) = ia-' + 9f. Ex. 3. (a;^ ^y^ — z^ — 2xy)-i-(x — y — z) = (ocF — 2xy + y^ — z^)-^(x — y — z) = [(« - yf - 2']h- [(« -y)-z2 — x-y + z. (i-) (ii.) (iii.) 2. By actual division, we have r (a= + 6=) ^{a + b) = a^-ab + 6^; (a= + 6«) - (a + 6) = a^ - a'b + a^A^ - 06^ + b\ • (a^ _ b^) ^{a-b) = a^ + ab + b'; '.{a'- 6=) - (a - 6) = a^ + i^b + a^A^ + 06== + b\ r (a* - b') - (a + 6) = a^ - a^6 + a6^ - b'; (a*- b') -=- (a - 6) = a^ + a^b + ab^ + b\ The above identities, and the identities in Art. 1, illustrate the following principles, which will be proved in Art. 6 : (i.) The sum of the like odd powers of two numbers is exactly divisible by the sum of the numbers. (ii.) The difference of the like odd powers of two numbers is exactly divisible by the difference of the numbers, taken in the same order. (iii.) The difference of the like even powers of two numbers is exactly divisible by the sum, and also by the difference of the numbers, taken in the same order. (iv.) When the divisor is a sum, the signs of the terms of the quotient alternate, + and — . (v.) When the divisor is a difference, the signs of the terms of the quotient are all +. (vi.) In the first term of the quotient the exponent of a is less by 1 tha7i its exponent in the dividend, and decreases by 1 from term to term. (vii.) The exponent of b is 1 in the second term of the quotient, and increases by 1 from term to term. Observe that the quotient is homogeneous in a and b, of degree less by 1 than the degree of the dividend. § 2] TYPE-FORMS IN DIVISION. 109 Ex. 1. (8a? + j^)^(2x + i) = (2o:y-(2x)(^) + ay Ex. 2. (32 y'o - «'») -f- (2 ?/^ - o^) = (2 fy + (2 2/2)»(a;3) + (2 2^2)2(a^)a + (2 y2)(^3 + (^4 = 16 2/« + 8 2/W + 4 2/ V + 2 2/ V + x^l Notice that in the type-forms each term, beginning with the second, is equal to the preceding term xh -i- a when the divisor is a — 6, and x {—h)-r-a when the divisor is a + &. Thus, in Ex.2: 16 ^ = 32 2/1° -=- 2 2/= ; 8 2/W = 16 2/* X a;' -f- 2 2/^ 4 2/*a!6 ^ 8 2/ V X »^ -^ 2 y2 . 2 y^^? = 4. y^x^ x a? ^ 2 y" ; !»" = 2 2/ V X «= -^ 2 y\ It is important to notice that, as we shall prove in Art. 6, the sum of the like even powers of two numbers is not exactly divisible by either the sum or the difference of the numbers. E.g., a* + 6* is not divisible either hy a + 6 or by a — 6. BXBECISBS VII. Find the values of the following quotients, without performing the actual divisions: 1. (x^-l)-^ix-l). 2. (25-a;2)--(5 + a;). 3. (4ffi2-9)^(2a-3). 4. (i- x''y^)^(i + xy). 6. (16x^-9y^)^(4x-Sy). 6. (64 a^fi^ - 121 c^) -=- (8 a6 + 11 c). 7. (x4-1)h-((k2 + 1). 8. (4a*-62)-=-(2a2-6). 9. (a;2«-l)-^(a:»-l). 10. (a*» - 16 6i<')H-(a2" + 4 68). 11. (a;2"+2-4)-=-(a;"+i + 2). 12. (a^" - 6*»+'') ^ («*" - 62»+2). 13. [(a + 6)2-l]-( + y^]. 19. (s* + 2 kV + 2/* - 2^ - 2 sm - m^) -h (a;2 + !/2 + m + 2). 20. (a2 _ 62 + 2 6« - 2 ax + x2 - a2) H- (a - a; - 6 + s). 21. (a;82/S-|-l)-f-(a;i/ + l). 22. (1 - aS)-^(l - a). 33. (a;» + 125)^(a; + 5). 24. (8 a' - 27)h-(3 - 2 a). 25. (a6-l)-5-(o2-l). 26. (ai^ + 27)H-(a« + 3). 27. (8mi5m8-i)i2)-H(2m«M-p*). 28. (««"- 1) -=-(«»- 1). 29. (343a;3"-3_2/8») -- (7a;"'-i-2/2»). 30. [(a; + yy -efi]-r-(x + y- z^). 110 ALGEBRA. [Ch. VI 31. (x' + Sy' + z' + 6 xhj + 12 xy^) -i- (x + 2 ?/ + z). 33. (a;<-l)^(a;-l). 33. (1 - 16 a*) -=- (1 + 2 a). 34. (^K<- 16!/*)--(^a;-2j/). 35. {?,\ x'^ ~ \Q y^) -^ (fi x'^ + 2 y'^). 36. (x8yi2_256zi6)^(a;y-4zi)- 37. (»*« - 2/*'')^(x" + r)- 38. (a5 + !)--(« + 1). 39. {Z2x^ + y^)^(2x + y). 40. (1 - x5) - (x - 1). 41. (243 a666 _ (.6) ^ (3 a6 - c). 42. (xiV + 32zi6)^(a;2j, + 2g8). 43. (ai»6i5c2» - d^s) h- (a^feSc* - d^). 44. (xi5» - 1) ^ (K" - !)• 45. (1 + a^™) h- (1 + a™). 46. (a6 _ 68) -H (a - 6). 47. {a' + 6') -^ (a + 6). 48. (a" + l)H-(a+l). 49. (ai2 _ l)-f-(a _ 1). 50. (ai»-^65)-=-(a2-i6). 51. (a' - x")^(a - x^). 52. (x3)_ i)^(x5 + l). 63. (a'W» +&"")-;- (a^K" + 62«). Of what divisions are the following expressions the quotients : 54. x2 + X + 1. 55. a^-ab + b^. 56. x' - x^ + X - 1. 51. a* + a^+ a^ + a+ 1. 58. x^y^ + mx^y^ + m^xy + m'. 69. xf"-^ + x'"-' + x'^ ^ -\ l-x+1. The Remainder Theorem. 3. If 3 x' — 4 x^ — 6 X + 7 be divided by x — 2, we obtain a partial quotient 3 x^ + 2 x — 2 and a remainder 3. If now 2 be substituted for x in the given expression, we obtain 3 X 28 - 4 X 22 - 6 X 2 + 7 = 3, the above remainder. This example illustrates the following principle : If an expression, arranged to powers of a letter of arrangement, say x, be not exactly divisible by x — a, the remainder of the division is equal to the result of substituting a for x in the given expression. Let the given expression be of the form Ax" + .Bx»-i 4 i-Ux + V, in which « is a positive Integer. Let Q stand for the partial quotient of the division by x — a, and S for the remainder. Then, by Ch. III. , § 4, Art. 13, we have Ax» + .Bx»-i + ••• + Ux + V=Qix-a)+ B. If now a be substituted for x in the last equation, we obtain Aa" + -Ba»-i + ••■ + Ua + V=Q(a- a) + M = Q-0 + li = M. That is, the remainder, B, of dividing the given expression by x — a, is equal to the result of substituting a for x in the expression. §2] TYPE-FORMS IN DIVISION. HI 4. From the principle of the preceding article we derive the following : If an expression, arranged to powers of a letter of arrangement, say x, be exactly divisible by x — a, the result of substituting a for x in the given expression is ; and conversely. For if the division be exact, the remainder Sn 0, and therefore the result of the substitution is 0. E.g., 3 a;^ — 4 a;2 _ 6 j; ^. 4 is exactly divisible by a; — 2. Substituting 2 for x, we obtain 3x23-4x22 -6x2 + 4 = 0. 5. If an expression be arranged to descending powers of a letter of arrangement, the following is a convenient method of substituting a par- ticular value for the letter of arrangement. Ex. 1. Substitute 2 for a; in 3 sc^ - 4 a;^ _ 6 k + 7. We have 3 a:' = 3 a; • as- = 6 a;^, when a; = 2 j therefore 3 a;^ - 4 x^ = 6 x^ - 4 a;^ = 2 x^, when x = 2 ; then 2 a;2 = 2 X ■ X = 4 X, when x = 2 ; and 4x — 6x = — 2x = — 4, when x = 2 ; finally —4 + 7=3, the result of the substitution. Ex. 2. Is x8 + x^ — X + 7 exactly divisible by x + 2 ? Since x + 2 = x— (— 2), we substitute — 2 for x. We then have x3=-2x2; -2x2+ x2 = -x2 = 2x; 2x-x = a; = -2; -2 + 7 = 5. Therefore a,^ + x^ — x + 7 is not exactly divisible by x + 2, and the remainder of the division is 5. EXERCISES VIII. Prove that the following dividends are exactly divisible by the corre- sponding divisors, without performing the divisions : 1. (x2-3x + 2)--(x-2). 2. (x2-3a; + 2)-^(x-l). 3. (x3-18x-35)h-(x-5). 4. (x^ + 2x2 -x - 2)-4-(x + 1). 5. (x8 + 21x + 342)-h(x + 6). 6. (2x3 + Sx^ + 3x + l)-i-(x+ J). 7. (x6-6a;*-19x2 + 84)H-(x2-3)(x2 + 4). 8. (x« + 4x4 + x2-6) + (x2-l)(x2 + 2). 9. (2 X* + 4 ax3 - 5 aV - 3 a'x + 2 a*) -=- (x - a). 10. (x^ + 3aV + 5a'x + ai)^ix + a). Find the remainders of the following indicated divisions, vrithout per- forming the divisions : 11. (2x3-7x2 + 6x- 15)h-(x + 4). 12. (5x«-llx2 + 2x-7)^(x-2). 13. (17x8-2x2 + 4x-3)-^(x-|). 112 ALGEBRA. [Ch. VI 14. Prove that (x + 1)"" +(x — 1)"* is exactly divisible by x when m is odd. 15. Prove that (a + 6 + c)^ — (a^ + 6^ + c^) is exactly divisible by (a + 6)(a + c)(6 + c). 16. Prove that xiff + yiz'' + ««a?' — x'j/' — yzi — s''x« is exactly divisi- ble by (x - !/)(x - z){y - z). 6. We are now prepared to prove the following principles enunciated in Art. 2 : (i.) a"+6" is exactly divisible by a + b, but not by a — b, when n is odd. The quotient is (jn-l _ an-2J, ^ a»-362 _ ... ^ aj2Jn-3 _ (jftn-2 ^ J,i-1_ (ii.) a" — 6" is exactly divisible by a—b, but not by a + b, when n is odd. The quotient is a'>-i + a»-26 + a»-862 ^ [. a^fen-s + ab»-^ + ft"-!. (iii.) a" — 6" is exactly divisible by a + b, and by a — b, when n is even. The quotient is, when a + 6 is the divisor, a»-i _ a"-^b + a^-W a^ft-'-s + ab"-^ - 6"-' ; and when a — 6 is the divisor, an-i _|_ (j»-26 + a»-362 -\ [- a^ftn-s + a6''-2 + 6"-i. (iv.) a" + &" is not exactly divisible by either a + b or a — b, when n is even. For if — 6 be substituted for a in a" + 6", we obtain (— &)" + 6" = 0, only when n is odd. Therefore, o» + 6" is exactly divisible by a + 6, only when n is odd. If 6 be substituted for a in o" + 6", we obtain 6" + 6", ^t 0. Therefore, a" + 6" is not divisible by a — 6. In like manner the other principles can be proved. It is evident that the forms of the quotients considered in this article, obtained by actual division, conform to principles (iv.)-(vii.). Art. 2. CHAPTER VII. PARENTHESES. 1. The use of parentheses has been briefly discussed in Ch. II., § 2, Arts. 8-9. It is frequently necessary to employ more than two sets of parentheses in the same chain of opera- tions, and to distinguish them the following forms are used : Parentheses, ( ) ; Brackets, [ ] ; Braces, | | . A Vinculum is a line drawn over an expression, and is equiv- alent to parentheses inclosing it. E.g., (a + b)(c — d)=a + b-c — d. If more forms of parentheses than the above are needed, one or more of them is made larger and heavier. Removal of Parentheses. 2. The principles given in Ch. II., § 2, Art. 8, are to be applied successively when several sets of parentheses are to be removed from a given expression. In removing parentheses we may begin either with the inmost or with the outmost. 3. The following examples will illustrate the method of removing parentheses, beginning with the inmost: Ex.1. 4a-S3a+[2a-(a-l)]S = 4a-{3a+[2a-a-f-l]l = ia — l3a + a + l\ = 4a-4a-l = -l. Ex. 2. [b" -Ka" + b)a-(a'-b)b- a\a - b)\f = [62 _ ja' + a& - a?b -f 6^ _ «» + a^&|]» J[W-\ab + b''\f = [62 _ a& - WJ = (- abf = - a,%\ 113 114 ALGEBRA. [Ce. Vll 4. The method of removing parentheses, beginning with the outmost, may be illustrated by the following example. Notice that the part which is free from parentheses is sim- plified at the same time that the next parentheses are removed. a? - [2 a? -[3 a^ +(4: x"- 5 0^-1)^1 = a^ — 2a!'+ [3 a;^+ (4 a;^ — Sar' — 1)], removing braces, = — a)^ + 3a!^+(4a^— 5a^— 1), removing brackets, = 2 a;^ + 4 a;^ — 5 9!^ — 1, removing parentheses, = 6a!^ — 5a^+l, removing vinculum, = a^ + l. EXERCISES I. Simplify the following expressions by removing parentheses : 1. a + 26-[6a-{3 6-(6a-6 6)}]. 2. 2x-{32/-[4x-(52/-6a;)]}. 3. {50 - [35 - (10 - X) k] x}x. 4. 4a;-{[a;-3(2-a;)]sc-4}2. 5. 6 a - [7 ffi - (8 a - (9 a - 10 a - 6)}]. 6. a - {5 6 - [a - (3 c - 3 6) + 2 c - (a - 2 6 - c)]}. 7. 12 a -13(10 [7 (4 a -3) -6] -9 a}. 8. x-{x + y -[x + y + z-(x + y + z + v)}}. 9. 10 - 2 (a: - 5 [3 - 2 a; - 6 (4 a; - 7)] - 3 (5 - 2 a;)}. 10. 7 a" - {2 a" - [a" - 3 a" + (5 a™ - 2 a") - 4 a"] - 2 a"). 11. {[(x + y'')x - (2y - l)-\x - («fl - 2y)x - xY?- 12. [(x-yy + 6 xy] - [(a;^ + 2 xy) - {a;^ - [2 x^ - (4 xy -y^):\} -(-x2-2a;!/)]. 13. I a^x + f ax2 -{^ax^-[-^ aH - {\ ax^ - a)] }. Find the values of the expressions in Exx. 1-13, 14. When a=: — 3, 6=4, c = — 5, m = 2, »i = l,x = 8, 2/ = — 9, « = 7, « = -4. Solve the following equations : 16. (X + 3)2 = x2 + 15. 16. (2x + l)2-8 = (2x-l)a. 17. (5 + x)2 + 9 = 3x(9 + Jx). 18. (x + 1)2 = [6 - (1 -x)]a;-2, 19. (3-ix)2+(fx-5)2=(|x)=. 20. 2a;2 + 17x=(8 + 2x)2-67 -x(3 + 2x). 21. (2x3 + 1 _ j;)(2x2 - 1 + x)= 1 + x2(2x + l)(2x - 1). PARENTHESES. II5 22. (X - l)(a;2 + x + 1) - 6(x'- - 1) = - (2 - xy. 23. 56 a;3 - 4 a;(4 a - 7)2 = 164 a;2 - (2 x - 5)s. 24. 6-{5-(4-{3-[2-(l-x)]})j = 4. 25. 2[8-2{6-2(5-2x -1)}] = 8. 26- |[f{|(6-x)-x}-x]-x = 4. 27. 4{4[4(4 X - 3) - 3] - 3} - 3 = 1. 28. -4-4{4-4[4-4(4-x)]} = 44. 29. -4x-(5x-[6x-{7x-(8x-9)}]) = -10. Insertion of Parentheses. 5. The principles for inserting parentheses in a given ex- pression were proved in Ch. II., § 2, Art. 9. Ex. 1. Express 4(a; — y)+y — x as a product, of which one factor is a; — y. We have 4,(x — y)+ y — x=: 4:(x — y) — (x — y)= 3{x — y). The sign + or — before a pair of parentheses can evidently be reversed from + to — , or from — to +, if the signs of the terms within the parentheses be reversed. Ex.2. 7(a;^l)-3(l-a;) = 7(a!-l)+3(a!-l) = 10(a;-l). EXERCISES II. Write each of the following expressions as a product, of which the ex- pression within the parentheses is one of the factors : 1. 3(a-&)-a + 6. 2. b(x^ - y) - x^ + y. S. 3m,-5n-i(5n-3m). 4. 1 - a" + 3(0" - 1). 5. 5(x2 - X + 1) - x2 4- a; - 1. 6. x - y - z - 6(y + z - x). Write each of the following expressions as a single product, of which the expression within the first parentheses is a factor : 7. (2x-l)-3(l -2x). 8. 2(2m-3«) + (8ra-2m). 9. 6(x2_ 2,2)+ 2(2/2 -x2). 10. 7(x2/-s)-(«-x!/). Simplify the following expressions without removing the parentheses : 11. (a - 6)c + (6 - a)c. 12. 5(x - y)z + 5(2/ - x)«. 13. (l-x)(l+x2) + (x- l)(l + x2). 14. 9(X2/ + 3) (a - 5) + IQcy + 3) (5 - «). CHAPTER VIII. FACTORS AND MULTIPLES OF INTEGRAL ALGEBRAIC EXPRESSIONS. § 1. INTEGRAL ALGEBRAIC FACTORS. 1. Factors have already been defined in multiplication (Ch. II., § 3, Art. 12). The factors were there given, and their product was required. The converse process, given a product to find its factors, is equally important. 2. A product of two or more factors is, by the definition of division, exactly divisible by any one of the factors. An Integral Algebraic Factor of an expression is an integral expression by which the given one is exactly divisible. E.g., integral factors of 6 c^x are 6, a^x, 3 a;, 2 a?, etc. ; integral factors of a? — &^ are a + h and a — h. The word integral, here as in Ch. III., § 1, Art. 1, refers only to the literal parts of the expression. E.g., \ a and | x are integral algebraic factors of 6 d?x. 3. A Prime Factor is one which is exactly divisible only by itself and unity. E.g., the prime factors of 6 a^x are 2, 3, a, a, x. A Composite Factor is one which is not prime, i.e., which is itself the product of two or more prime factors. E.g., composite factors of 6 a\ are 6, ax, 2 a, 3 ax, etc. 4. Any monomial can be resolved into its prime factors by inspection. E.g., the prime factors of 4 a%^ are 2, 2, o, a, a, b, b. 116 §1] INTEGRAL ALGEBRAIC FACTORS. 117 The Fundamental Formula for Factoring. 5. A multinomial whose terms contain a common factor can be factored by applying the converse of the Distributive Law for Multiplication. From Ch. III., § 3, Art. 14, we have ab + ac -ad = a{b-\-c — d). (1) That is, if the terms of a multinomial contain a common factor, -the multinomial can be written as the product of the common factor and the algebraic sum of the remaining factors of the terms. The relation (1) niay be called the Fundamental Formula for Factoring. Ex. 1. Factor 2se'y-2 xy^ The factor 2 xy is common to both terms ; the remaining factor of the first term is x, that of the second term is — y, and their algebraic sum is x — y. Consequently 2 ae'y — 2 ooif = 2 xy (x — y). Ex. 2. ab^ + abc + b^c = b (ab + ac + be). 6. In the fundamental formula the letters a, b, c, d may stand for binomial or multinomial expressions. Ex. 1. Factor a{x — 2y)+b{x — 2y). The factor x — 2y is common to both terms ; the remaining factor of the first term is a, that of the second term is b, and their algebraic sum is a + 6. Consequently a(x — 2y) + b(x — 2y) = (x — 2y)(a + b). Ex. 2. Factor 1 — a + x(l — a). We have 1 — a + «(! — a) = (l — a)(l + »)■ Ex. 3. Factor (x - y)(a' + b") -{x + y){d' + If). We have {x - y){a' + b'^-{x + y)(a' + b^ = (a' + b%(x -y)-{x + y)] It frequently happens that the parts of a given expression have a common factor except for sign. 118 ALGEBRA. [Ch. VIII Ex. 4. Factor x'(l — m)— y'(m — l). Since 1 — m and m — 1 differ only in sign, i.e., m — 1 = —(1 — in), we may take either as the common factor. Taking 1 — m as the common factor, we have x' (1 — m) — y^ (m — 1) = (1 — m)(a^ + y^. The fundamental formula must often be applied more than once. Ex. 5. Factor by(x — a) — bx(y — a). Taking out b, b[y (x — a) — x(y — a)^ = b (ax — ay). Taking out a, we have ab (x — y). EXERCISES I. Factor the following expressions : 1. 5 a; + 5. 2. ax - a. 3. 4a8-6. 4. -x.«-x\ 5. a^b-ab^. 6. 2 an -in''. 7. 3 K* - 2 x8. 8. 12 a8&3 _ 3 ^22,2. 9. jq a*a;2 _ 15 aV. 10. 3 a6 + 6 ac - 12 ac?. 11. 70 xy -98y^- 140 ^/a. 12. ^ ax + il 6x2 + J x. 13. 6 ^x* - 15 a^jaiS + 18 a%^'x^ 14. 8 a2?i5x5 _ 10 are^x' + 4 a%W. 15. 45 m%=p + 90 mVp - 75 m^np". 16. 28 0^630-84 aS6«c2 + 98 o*6«c3. 17. 21 xh/*z■'+136xY^*-^'>'^Y^*■ 18. 7(a + 6)-14. 19. a-^(a + x)+x2(o + x). 20. 3a(a-l)-3(a-l). 21. 2(ji + 1)2 _ 4(n + 1). 22. a(x-l)-x + l. 23. mCg -p)-(p - 2). 24. - 2 a» + 4 a^^+ 6 a*«. 25. a»+' - a + 0"-^ 26. 6 m"+i - 3 m"+2 + 9 njo+s. 27. 5»+s - 125 x + 625 x^. 28. aa;" - 6x"+i + cx»+2. 29. 2''+i - 8 x 2"-! + 16. Grouping Terms. 7. When all the terms of a given expression do not contain a common factor, it is sometimes possible to group the terms so that all the groups shall contain a common factor. § 1] INTEGRAL ALGEBRAIC FACTORS. 119 Ex. 1. Factor 2a + 2b + ax + bx. Factoring the first two terms by themselves, and the last two terms by themselves, we obtain 2{a + b) + x(a + b) = (a + b) (2 + x). Ex. 2. 3? — xy — xz + yz = x(x — y) — z(x — y) = (x — y){x — z). Ex. 3. {2a-bf + ^ax -2bx = {2 a -bf + 2 x(2 a -b) = (2a-6)[2a-6 + 2'a;]. EXERCISES II. Factor the following expressions : 1. ac + ad + be + hd. 2. 2 aa; - 3 6?/ - 2 at; + 3 hx. 3. 20 ad - 35 &d - 8 ax + 14 hx. 4. hax-ox-bay + cy. 5. a^ - a^e + ac^ - c». 6. cfi-x^+ x-1. 7. 18ra2a;-12x-9m2 + 6. 8. 3k* - x' + 6« - 2. 9. 3 c* - 3 c% + crfi - n'. 10. 3(t? + mfi -6nH-2 x'. XI. 6a^-6a.^y + 2ay^-2y^ 12. ^ty + ^xy - ^tz - ixz. 13. 12a86*-4a26''-4a268 + 12aW 14. a* - a'n'' + a% - an^ + m^ _ a^s. 15. x« - ax3 + 3 aV _ 2 a'^ftx^ + 2 a'ftx - 6 a*&. 16. ax + by + cz + bx + cy + az + cx + ay + 62. 17. ax - 62/ + cz - 6x - c^ - az - ex + 02/ + 62. 18. ax + 62/ + C2 - 6x - cy + 02 + ex - at/ - bz. 19. ax + 62/ + C2 - 6x + cj; - a2 - ex - at/ + &2. 20. 6 X" + 8 x»-i - 9 x''-2 - 12 x»-8 + 3 x"-< + 4 x"-*. Use of Type-Forms in Factoring. 8. If an expression be in the form of one of the type-forms considered in Ch. VI., or if it can be reduced to such a form, its factors can be written by inspection. Trinomial Type-Forms. 9. From Ch. VI., § 1, Arts. 1 and 2, we have a' + 2ab + b' = (a + by, a'-2ab + b' = (a-by. 120 ALGEBRA. [Ch. Vm From these identities we see that a trinomial which is the square of a binomial must satisfy the following conditions : (i.) One term of the trinomial is the square of the first term of the binomial. (ii.) A second term of the trinomial is the square of the second term of the binomial. (iii.) Tlie remaining term of the trinomial is twice the product of the two terms of the binomial. Ex. 1. Factor a^ + 6 a; + 9. a? is the square of x, 9 is the square of 3, and 6 a; = 2 • a; • 3. Therefore a;^ + 6 a; + 9 = (a; + 3)^. Ex. 2. Factor — 4 a.^ + 4 a^ + 2/1 4 a^ is the square of 2x, or of —2a;; 'if is the square of y, or of — y. Since the middle term in the given expression is negative, one term of the binomial is negative, the other positive. Therefore — 4 a;?/ + 4 a;^ + 2/^ = (2 a; — j/)^ = (— 2a! + yy. Ex. 3. 60a;2/-36a;='-252/2 = -(36a;=-60a;2/ + 25y') = — (6 a; — 5 j/)^ Ex. 4. {n" - 2 nxf + 2 («V -2na?) + a^ = {n' - 2 nxf + 2 x'{n^ - 2 nx) + od^ = {n'-2nx + af)'^= [(w -xfy = (n - a;)*. EXERCISES III. Factor the following expressions : 1. a;2 - 2 a; + 1. 2. a^ + 6 a + 9. 3. 2/2 + 12?/ + 36. 4. a^- 10a + 26. 5. 4iK2-12a;+9. 6. 9a2+ 30a + 25. 7. 20a;-4a;2-25. 8. B6x-ix^-8l. 9. 16 a" + 40 a6 + 25 62. 10. i9x^ -28xy + iy^. 11. 9x^^-30xyg^ + 25zK 12. 2ixy -9x^ -16y^. 13. a* -2a^ + x^. 14. x*-2 xV + V*- 15. aH^-iacH + icf: 16. 2 a^a^ - a* - x*. 17. (o + a;)2 + 2(a + a;) + l. 18. (k - 4)^ - 4(a - 4) +4. § 1] INTEGRAL ALGEBRAIC FACTORS. 121 19. (2a;-9)2-6(9-2a;)+9. 20. 4 a:^" - 12 »» + 9. 21. 36 a»+2 - 48 a» + 16 a—a. 22. iax + 2a'' + 2x^. 23. 6 a^x^ - 3 aV - 3 a!h:. 24. 16 a^b^ + 9 c^ + 24 ab'c*. 25. a2"-a-2a— V+i + Ka"*^. 26. (o^ + 2a6 + 62)c+ (a+ 6)(P. 27. a;!/-a;3- (y2-2j/a + a2). 28. a'^ + 2an + rfi - ap -pn. 39. 2 a + ad - d2 - 4 d - 4. 30. a^ + 2 a& - 4 ac - 4 6c + 4 c^. 31. n!'x-xy-n^y + 2n^y''-yK 32. (a - c)3 + 2a2c -4ac2 + 2cS. 10. From Ch. VI., § 1, Art. 6, we have jr^ + (a + 6) Jr + a6 = (;r + a) (jr + 6). When a trinomial, arranged to descending powers of some letter, say x, can be factored into two binomials, in both of which the first term is the letter of arrangement, it must satisfy the following conditions : (i.) One term of the trinomial is the square of the letter of arrangement, i.e., of the common first termofthe binomial factors. (ii.) The coefficient of the first power of the letter of arrange- ment in the trinomial is the algebraic sum of two numbers whose product is the remaining term of the trinomial. (iii.) These two numbers are the second terms of the binomial fouytors. Ex.1. Factor £B= + 8a; + 15. The common first term of the binomial factors is evidently x. The second terms are two numbers whose product is 15, and whose sum is 8. By inspection we see that 3 + 5 = 8 and 3x5 = 15; that is, the second terms of the binomial factors are 3 and 6. Consequently, ar" + 8 a; + 15 = (a; + 3) (a; + 6). Ex. 2. Factor a^ — 7 a; + 12. The common first term of the binomial factors is x. The second terms are two numbers whose product is 12, and whose sum is — 7. Since their product is positive, they must be both positive or both negative ; and since their sum is negative, they must be both negative. 122 ALGEBRA. [Ch. VIL The possible pairs of negative factors of 12 are — 1 and - 12, - 2 and - 6, - 3 and - 4. But since _ 3 + (- 4) = - 7, the second terms of the binomial factors are — 3 and — 4. Consequently a^ - 7 a; + 12 = (« - 3) (a; - 4). Ex. 3, Factor aV + 5 ax — 24. The common first term of the binomial factors is ax. The second terms are two numbers whose product is — 24, and whose sum is 5. Since their product is negative, one must be positive and the other negative; and since their sum is posi- tive, the positive number must have the greater absolute value. The possible pairs of factors of — 24 are — 1 and 24, — 2 and 12, - 3 and 8, - 4 and 6. But since —3 + 8 = 5, the second terms of the binomial factors are — 3 and 8. Consequently aV + 5 aa; — 24 = {ax — 3) (ax + 8). Ex. 4. Factor x^-3xy-28 y\ The common first term of the binomial factors is x. The second terms are two numbers whose product is — 28 if, and whose sum is — 3 y. It is evident that both of these terms contain y as, a. factor. Therefore we have only to find their numerical coefficients. Since their product is negative, one must be positive and the other negative ;. and since their sum is negative, the negative number must have the greater absolute value. The possible pairs of factors of — 28 are 1 and — 28, 2 and — 14, 4 and — 7. But since 4 + (-7) = — 3, the second terms of the binomial factors are 4 y and — 7 y. Consequently a;^ — 3 xy — 28 2/^ = (a; + 4 j/) (a; — 7 2/). EXERCISES IV. Factor the following expressions : 1. a2-3a; + 2. 2. x^ - a; - 2. Z. x^ + x-&. 4. a;2 + 3a; + 2. h. x'^ + x-'i. 6. x'^-x-Q. 7. a;2-5a; + 6. 8. x^-&x^-b. 9. a;2-4a;-60. § 1] INTEGRAL ALGEBRAIC FACTORS. 123 10. x^ + 1x-30. 11. x2 + 12a; + 32. 12. a;2-3a;-40. 13. 3:2 -12 a; + 35. ^4 x«-nx^ + 72x. 15. a;2 + 13a;-30. 16. 6 a; - a;2 _ xK 17. 35 + 2 a; - x\ 18. ce* - 5 a;^ - 24. 19. a;« + 8a;2+15. 20. a;* - 24 a;^ + 63. 21. 3x6+39a;3+66. 22. a;6-x3-50. 23. a;^" + 6 x« - 112. 24. a;2»-16a;»+55. 25. x^ -i- (a + b)x + ab. 26. a;^ -(m + n)a; + ran. 27. a;2 + (p - g)a; - pg. 28. a;^ + (3 r - 2 s)a; - 6 rs. 29. ax2 + 7 a^a; + 6 a'. 30. x^ + 2xy-16yK 31. x2_4ax_i2a2. 32. a;^ - 7 aa; + 12 a^. 33. 2 aa;V - 26 oa;V + 84 aaii/*. 34. a;^ - llaim + 30m2. 35. x%2 + 12xz- 13. 36. a^ft^ - 7 a6 + 10. 37. mW-20mre + 99. 38. (a + 6)2 + 7(a + 6) + 6. 39. (a -6)2 +7(0! -6) +12. 40. (to + m)2 + 2(nt + m)- 15. 11. From Ch. VI., § 1, Art. 6, we have (ax + b) (ex + (/) = aox^ + {ad +bc)x + bd. A trinomial which can be factored by this type-form must satisfy the following conditions : (i.) One term of the trinomial is the product of the first terms of its binomial factors. (ii.) A second term of the trinomial is the product of the second terms of its binomial factors. (iii.) The remaining term of the trinomial is the sum of the products of the first term of each binomial factor by the second term of the other. Ex. 1. Factor 6 a^ + 19 a; + 10. The first terms of the required binomial factors are factors of 6 a^, the second terms are factors of 10, and the sum of the cross-products is 19 x. The factors of 6 a^ are x and &x,2x and 3 x ; and the factors of 10 are 1 and 10, 2 and 5. The following arrangements represent possible pairs of fac- tors: a!+l a; + 10 x + 2 x + B X X X X 6 a; -+10 6 a; -1-1 6 a; +-5 6a!-|-2 16 a! 61a; 17 a; 32 a; 124 ALGEBRA. [Ch. VIII 2a! + l 2a; + 10 2a! + 2 2a; + 5 X X X X 3 a; + 10 3a; + l 3a; + 5 3a; + 2 23 a; 32 a; 16 a; 19 a; Since the sum of the cross-products in the last arrangement is equal to the middle term of the given trinomial, we have 6 ar" + 19 a; + 10 = (2 a; + 5) (3 a; + 2). Ex. 2. Factor 5x^-6xy-8y\ The factors of 5 a;^ are x and 5 x, and the factors of — 8 ^^ are y and —8y, —y and Sy,2y and —42/, —2y and 4 y. x — 2y Since the sum of the cross-products in the arrange- Xment on the left is equal to the middle term of the _ , . given trinomial, we have Sx + iy " ' -6xy 5 x^ — 6 xy - 8 y' = (x — 2 y)(5 X + iy). Observe that the reason given in Art. 10, Ex. 4, for rejecting at sight some factors of the last term of the trinomial does not hold in the above example. For, although the middle term, — 6xy,is negative, the negative factor of —8y^ is less in ab- solute value than the positive factor. Ex. 3. Factor 10 a^ + a^b - 21 &l The factors of 10 a* are a^ and 10 a", 2 a? and 6 a" ; and the factors of - 21 W are 6 and - 21 6, - & and 21 6, 3 & and - 7 6, -3 6 and 7 &. 2 a^ + 3 & Since the sum of the cross-products in the arrange- Xment on the left is equal to the middle term of the _ z _ 7 J, given trinomial, we have ~~^ 10 a* + a?h - 21 6^ = (2 a^ + 3 6)(5 a?-l h). Ex. 4. Factor - 15 a;^ ^ 22 a; - 8. We have - 15 a;^ + 22 a; - 8 = - (15 a.-^ - 22 a: + 8) = -(3a;-2)(5a;-4) = (2 - 3 a;) (5 a; - 4). §1] INTEGRAL ALGEBRAIC FACTORS. 1'25 12. The following directions may be observed in factoring trinomials which come under this type-form : (i.) Wlien all the terms of the trinomial are positive, only posi- tive factors of the last term are to be tried. (ii.) When the middle term is negative and the last term is positive, the factors of the last term must be both negative. (iii.) When the middle term and the last term are both negative, one factor of the last term must be positive, the other negative. (iv.) Select that pair of factors of the last term which, by cross- multiplication, gives the middle term of the trinomial. EXERCISES V. Factor the following expressions : 1. 6x2 + a -12. 2. 6a;2-aj-12. 3. 36 a;^ + 32 cc - 12. 4. 36a;2 + a;_12. 5. 35 x^ + 16 a; - 12. 6. 35x2-13x-12. 7. 2x2 + 6x + 2. 8. 10 + 16x + 6x2. 9. 6 + 13x-63x2 10. 3x2 + 13x + 12. 11. 40 + 2x-2x2. 12. 25x3 + 25xS - 6x. 13. 36x1-18x2-10. 14. 12x-6x2-90x8. 15. 10x2 + 7x-33. 16. 8 X* - 19 a;2 - 15. 17. 40 + 6 x - 27 x^. 18. 49 x^ - 35 x + 6. 19. 64x2-92x + 30. 20. 6-19x + 15x2. 21. 6x2-41x-56. 22. 30 x* - 89 X + 35. 23. 18x^-3xy- 45 j/^. 24. 3 a" - 5 a6 - 2 b\ 25. 18 x« + 3 x^y - 10 2/2. 26. a&x2 -(a2 - 62)x - a6. 27. 5 a*x2 - 4 a^xz - 96 22. 28. - 10 a* + 7 0252 + 12 6*. 29. 4 x2 - X2/ - 3 2/2. 30. 10 0= + 11 a6 - 6 62. 31. 9x2»-4x»-5. 32. 2 a;2'+2 - 3 x'+i - 2. 33. 6x2™+x'»r-152/*'. 34. 10(a + &)2 + 7 c(a + &)- 6 c2. 35. 7(x - 2/)= - 37«(x - 2/)+ 10 32. 36. 6(x2 + 2/2)2 _ 9(a;2 + 2/2)22 _ 15 ^4. 37. 2(a2 - c2)2 _ 4 6(a2 - c2) - 6 &2. Binomial Type-Porms. 13. From Ch. VI., § 1, Art. 4, we have That is, the difference of the squares of two numbers can be written as the product of the sum and the difference of the numbers. Ex.1. aV-\b'=(axy-(^by = (ax-{-^b){ax-^b). 126 ALGEBRA. [Ch. VUl Ex.2. 32 m*n - 2 m= = 2 }i(16m*-m=) = 2 n [(4 my - n''\ = 2 w (4 m^ + n)(4 m^ — n). Ex. 3. 3?-A.xy + 'iy^-'dz'' = {x-2yf-(3zf = (x-2y + dz){x-2y-Zz). In factoring a given expression the type-form must fre- quently be applied more than once. Ex. 4. 4 aV - {a" -W + c^y = (2 ac + a^ - 62 + c2) (2 ac-a' + b''- c") = [(a + c)2 - 6^] [6^ - (a - c)^] = (a -1- c + 6) (a + c - &)(6 + a - c)(6 - a + c). 14. The difference of the like, or unlike, even powers of two numbers can always be written as the difference of the squares of two numbers, and should therefore first be factored by applying this type-form. Ex. a*-6«=(ay-(6y = (a' + b^(a'-b^ • =la' + b^la + b)(a-b). BXEECISES VI. Factor the following ' expressions : 1. x^-1. 2. 4-a2. 3. a2 - x22/2. 4. 25x2-9. 5. 36 a2 _ 49 62. 6. 4 x2 - yi. 7. 862 _ 142. 8. 572 - 432. 9. 372 - 272. 10. 81 a« - 16. 11. fa2&2-|fc2d2. 12. 16 08 - 25 ftV. 13. a^b*c^ - i. 14. iaH'^-ji^ifi. 15. a2»-l. 16. (fin _ yim_ 17. a;2»'+2 _ 4. 18. 9 a2»62 - 4 c^. 19. (m - to)2 - 1. 20. C2- (0-6)2. 21. 9-(3-x)2. 22. (4x-3)2- -16x2. 23. (a- 6)2 -( C - d)2. 24. (5a; -2)2- -(4x- 3)2. 25. (3x^ -5)2. -(2xy-6)2 26. (x2 + X + 1)2 -(x2-x+ 1)2. 27. a* - aS + a _ 1. 28. 7 - 112 X*. 29. 16 x* - yK 30. a^ ~ 6^. 31. 1 - 256 x8j/8. 38. x" - ?/". 33. ^is _ 1. 34. 5 a2 - 180 64. 35. 75 a%* - 108 d^dK 36. 243 65^6 _ 75 57. §1] INTEGRAL ALGEBRAIC FACTORS. 127 37. f ab2 _ I ac*. 40. a^'' — b*". 43. a2»+36^ - a562»+2. 46. x^+Sx^-a^-Zx. 49. a^-a^n + an^—n^. 52. a^ — n^ + 2 np — pK 54. a* - 2 a63 - 6* + 2 a^b. 56. x't/ — a;^^ + x^y + xy^. 58. 2 (a6 + C(?) - (a^ + &'^ - c^ - d^; 60. 4a262-(a2 + 62-c2)2. 62. a« + 4 0^0-4 6H4 6(i + 4c2- 38. J a;?/* - 3^, xa«. 41. 144 a;" — x"+2. 44. m*'-%6™+2 - 1. 47. a^~x^ + a-x. 50. x^-2xy + y^-z^ 53. i32_j,! 55. a2 + 6 39. a^.. _ 62m. 42. 4 a3''+3 — ffo+i. 45. a2 - 62 + (a + &)c. 48. x^ — xz — yz — y^. 61. a2 ^. 2 6c - 62 _ c2 -4z-4. - c2 - d2 + 2 (a6 + c(i). 57. a2 + 62 - c2 - d2 _ 2(a6 - C(2). 59. a^-b^ + 2bz~2ax + x^~ z\ 61. a^^ - a^"- - 2 a^-- - ai*. 63. 4(a(^+6c)2-(a2-62-o2+d2)2 64. (a + n) (a2 - a;2) - (« - x) (a^ - »i2). 65. (n - a;)(5 ra2 - 4 a;2) - (3 a;2 - 4 n2) (j; _ „). 66. (a+ 6)2-1 -2(a + 6 + 1). 67. (a -26)2- 9 -3(a- 26 + 3). 15. From Ch. VI., § 2, Art. 2 (i.) and (ii.), we derive a' - 6^ =(a - 6) (a^ + a6 + 6^). Ex.1. »' + 8 2/= = k3+(2 2/)= = («; + 2 2/)[x2-a;(2t/)+(2 2/)T =(a; + 2 2/)(a!2-2a?i/ + 4 2/2). Ex. 2. (1 - a;)' - 8 a^ =(1 - a;)^ -(2 a;)' = (1 - a; - 2 a;) [(1 - a;)^ +(1 - a!)(2 a;)+(2a!7] = (l-3a!)(l + 3a;0. Ex. 3. = (8 af* + 2/^) [(8 a^)2 - (8 ^ {f) + {ff-\ = [(2 a!)3 + /] (64 a;" - 8 a;^2/^ + j/^) = (2a! + 2/) (4«2 - 2 a;?/ + 2/')(64 a!« - 8 a^/ + /). Ex. 4. o«-729fi« = (a')'-(27 6y = (a' + 27 6')(a3-27&') = (a+3 6)(a2-3a&+9&2)(a_3 6)(a2+3a&+9 62) 128 ALGEBRA. [Ch. VIII 16. Trom Ch. VI., § 2, Art. 2 (i.) and (ii.), we infer : (i.) The sum of the like odd powers of two numbers contains the sum of the numbers as a factor. (ii.) TJie difference of the like odd powers of two numbers con- tains the difference of the numbers as a factor. Ex.1. a? + f = (x + y){oi^-a?y + a?y^-xf + f). Ex. 2. x' -y' = {x-y){a^ + 3?y + xY + 3?f + 3?y' -\-oc^ + f). 17. The sum of the like even powers of two aumbers, whose exponents are divisible by an odd number, except 1, can be factored by applying the type-forms of Arts. 16 and 16. Ex. a;i2 + 2/12 =(a;4)3 +(2^)3 ={:>^ + y')[_{o^r-{^)(f)+{ff'] EXERCISES VII. Factor the following expressions : 1. x» + 1. 2. x^ - 8. 3. a^ + 27. 4. 64a;8-l. 5. Sx^-^. 6. 8x^-27. 7. 125 ay + 8. 8. 3 a2- 24 0(5. 9, 27a-a!*6«. 10. 2a;'!/5 + 432!,2 n. aS + 243. 12. ofi-\-t- 13. vfi - 64. 14. x^ + 2/9. 15. x^ - 1. 16. 27a;8-j/9. 17. a^b^ + &i^(fi. 18. n^x^-y^z^. 19. aP - ftw. 30. xio + 2/10. 21. x^^ _ 1. 22. a;"j/i* - 1. 23. x" + 2/". 24. a}^ + fti^. 25. 1 - x^\ 26. 1 - o". 27. x^^ + 2/". 28. a^ - 63". 29. 8 x^^ - 729 2/»+32«. 30. 27 - (3 + 2 xy. 31. (2 a + a;)' + (a - 2 xy. 32. 4 - a;^ + 4 a;' - a;^. 33. x^-y0-2x^y + 2 xy". 34. (a + 6)' - (c + d)8. 35. x5-x'-a;2 + l. 36. x^ _ g _ 6a;2 + 12x. 37. a'-ia^c-i ac^ + c". 88. «« + 5 » n, can now be extended to the case in which the dividend is a lower power than the divisor. E.g., £f = ^^ = J_ = i. fl™ 1 In general, — = , when m + -°-^} Powers of Fractions. 18. From the principle for multiplying fractions we have : A power of a fraction is a fraction whose numerator is the like power of the numerator of the given fraction, and whose denominator is the like power of the denominator; or, stated symbolically, ray a" bj b" wherein n is, as yet, a positive integer. ^' ' \ c* J {cy c" ■ The converse of the principle evidently holds ; that is, b" \b^ Ex.2. (<^-5x + 6y ^f^-5x + 6y,_^y FRACTIONS. 169 BXBROISBS VIII. Simplify the following expressions : ■ V 3a; y ' 3x ■ ' ^ Tfi^cV ' ' V3a:V/ " 5 / 2af2f2^y_ g /'a!\"- 7 / 2a% y" „ /2aV^\«+i V hV^&yj' '\bj' '\ 56V/ ' ' \ 36V / g (a' - 1)° 10 (a:°-l)° ^ Ca:«-jf^)» jg (x'-5a:+6)3 (a; -1)6 ■ (x2+a;+l)8 " (k^+j,!)^' ' (a;-2)8 13. f«+IVx*i^. u. (^ + 7)* ■ 15 {^'-^y'y\ \h + \j a^ + \ (a;2 + 6a;-7)4 ' (ifi - y^y^ Ig / g" + db + }fi y ^ I v? ~xy + y^ y_ ' \ X3 + 2/5 ) { as_63 j \x^-2xy + y^) \2x^-bx + 3) ' 18 (^^-z% ( d + z ^J d^-dz + g^ y^ ■ d-" + z3 d - U^ + roduct of the dividend and the reciprocal of the divisor; or, stated symbolically, ?-=-?- = ? X - bd'h c B, a — x b + x a — x^,a + x_ a' — a^ ^■9-> 1 Z "^ ZTTZ ~ T. — Z. ^ a + X b — X b + x b'^ — x^ 170 ALGEBRA. [Ch. IX We have - -^ - = (a -h 6) -i- (c -4- d), by definition of a fraction, b d = a-i-b-i-cxd, since -r- (c ^ d) = -^ c x d, = a^bxd-i-c, since -hcx d-—xd-i-c, = (a -^ 6) X (d -7- c), since x d-f-c = x(d-^c), = - X -, by definition of a fraction. 6 c •n, 4(a''-a&) . 6 a ^ 4a(a-6) (a-6)(a + 6) ^' ■ (a + by ' a?-^ (a + hy 60 ^ 2(a-6y 3(a + 6)' Ex.2, (a' - 6' - c^ + 2 &c) -r- °^ + ^ ~ '^ ^ ^ a+6+c = (a+6-c)(a-& + c) x 5^+|+« a+6— c = (a-6+c)(a + 6 + c). 22. If the numerator and denominator of the dividend be multiples of the numerator and denominator of the divisor, respectively, the following principle should invariably be used : The quotient of one fraction divided by another is a fraction whose numerator is the quotient of the numerator of the first fraction divided by the numerator of the second, and whose de- nomirwLtor is the quotient of the denominator of the first fraction divided by the denominator of the second; or, stated symboli- cally, a . c a-\-c E.g., b d b-ird ■X _{a^ — a?)-i-{a — x) _a + x b-x (b" - x") ^ (b - x) b + x - = (a -^ 6) -^ (c T- d), by definition of a fraction, We have b ■ d' = a^b-i-cxd, since -;- (c -=- d) = -=- c x d, = a-!-c-^6xd, since -^6-=-c=-HC-r-6, = (a -7- c) -^ (6 -=- d), since -f- 6 x d = -f- (6 -4- d), _ a — c ^ ^y definition of a fraction. 6 -H d FRACTIONS. 171 23. Observe that a fraction is divided by an integral ex- pression, which is a factor of its numerator, by dividing its numerator by the expression. Ex.l. glz:J!^(a-6)^ («'-^^)^(«-S) ^i±j. xy ^ ' xy xy Also that a fraction is divided by an integral expression, which is not a factor of its numerator, by multiplying its denominator by the expression. Ex.2. g^+J!^(« + 6)= «^ + 6- . xy xy{a + b) EXEBCISBS IX. Simplify the following expressions : J 27 a°6* . 9 gSft^ 2 gSfts . gUb* 5. g2 _[, 7 a + 12 . a: + 4 eCgiJ - 6^)' . 3(a + 6) a;2 + 2a;-15 ■ a; + 5' ' 7(a;8 - 1) ' (1 - a;) " ,^ 2 gs - 2 aft" . g" - 62 ^ a'-6a: + 8 . a;-4 g + 26 ■2g + 4 6' ' z^ + 2x + 1 ' x+ l' g a;2+j/2_2a;y_z2 a-y+g jjj 4c5dx-9dy . 30x-6y g2-9+4 62+4 g6 ■ g+2 6-3' ' 20a6a;-10 62a; ' 20a2a;-5 62x' jj g2 - (5 - c)2 . g - 5 + c j2 a:° - 1 . a" + a; + 1 (g2 - 62)2 ■ (i4 _ 64 ■ ■ x2 - g2 ■ a; - g ' J3 1 + w-ro8- re* . re2-l ^^ l-2a: . l-2a: + a:2-2g8 1 - g2 ■ g2 - l' ■ 1 - a;8 ' 1 + 2 a; + 2 x2 + a;^' ^j g2 + g5 , g86 + g6° + 2 g262 jg 1-a: . l-x^ «2^ft2 ■ a* — 6* ' x^+x'^—x^'x^—x?—2x^—x (g + 2 6)g8 -(2 g + 6)6^ , (g + 5)' g*6* ■ g'62 + a^b* jg a;2 + 2a;-3 . a:2 + 4a: + 3 ,^ a;'' + l 'a;2-2a;-3a;2-4a; + 3 a;8-l' a;4 + a;2y2 4, yi n/st .[■ y(2 X + y) g° + y» x2 + 2;2 ^ z8-y8 ■ a;2--2/(2x-2/)' a + x a — x 1 + i X a + y 1-i 172 ALGEBRA. [Cii. JX Complex Fractions. 24. A Complex Fraction is a fraction whose numerator and denominator, either or both, are fractional expressions. 2 ■ 5 a — y Observe that the line which separates the terms of the com- plex fraction is drawn heavier than the lines which separate the terms of the fractions in its numerator and denominator. If no distinction be made between the lines of division, the indicated divisions are to be performed successively from above downward. E.g., |=2--3-=-4-=-6=2--(3x4x6), by Ch. II., § 4, Art. 8, ^=2-60 = ^; 2 while I = |x|=f 25. Complex fractions are simplified by applying succes- sively the principles already established for simple fractions. Ex. 1. -^-^— = i — - H- (1 - a;) = ti:^ by Art. 23. 1— a; X ^ ' X ■' m^ + ii' m? + n^ — mn Ex. 2. m^ + M^ n m — w 1 n _ 1 m 7n? -n^' m{'m? + n'- ■mn) mr + w ^^ (m + n)(m - n) , ^ ^_ m — n (»* + 7i)(m/' — mn + n^) We might have simplified the complex fraction in Ex. 2 by first multiplying both its terms by mn, the L. C. D. of the fractions in them, instead of uniting these fractions before FRACTIONS. 173 multiplying by mn. This fraction would then first have reduced to m? + mn^ — m^n _ m (m^ + m^ — mn) , ■ - ^ -^ • £IS oiDOVG. m—n m—n Continued Fractions. 26. A Continued Fraction is a fraction whose numerator is an integer, and whose denominator is an integer plus (or minus) another fraction whose numerator is an integer, and whose denominator is an integer plus (or minus) a third fraction, etc. 1 1 1 1 22 Ex. 1. 2+^_ 2 + A 2 + U M 65 * T -1- • Observe that in this reduction the work proceeds from below upward. Ex 2 3 3 3 1 ~ , 1 ~ ,3-x 2 , «_+l 4 4 • 3-x 3-x 3 4 3a; + 3 x + 1 4 EXERCISES X. Simplify the following expressions : a + ^ a ^- -^ ^+^ 1. _^. 2. _±±J. 3. ^^ll , J, , bo „ , ax X X — 1 OH a + — — -- a a — X x+1 " .+ 1 a — a* a + X a + ax a~x X 1- 63 1 l + a; 62 'a' 4. iLl^i 5. — ^^. 6. a+^,- 1 «_ 1-— i- a + - 1- a l + a: a x—L- ^•rr^^F «• i_6 «•=» 1 + , + !^ « + 1 a I -X 174 ALGEBRA. [Cii. IX iO. , 11. x^ + l 1 ^ a + x 2x 2a; — 12 ._ x x — a x + 2 ■ d' + x^ i-x a + x X — y 13. " ^ I X 1— 2x x — a 14. 15. x^ + ay 1+ ± 1 ^ l + x + -2^ 1 L_ \ ~ x 1 — X I n a+x a—x a+1 , a— 1 ,„ n + X n — x ,_ a — x a + x a — 1 a+1 16. 17. z • 18. Ti -. n . n 4 ax a + I a — 1 n — X n + X a^ — x^ a — la+1 19. — ^^^— 20. sc + 1 ^— rr— a + — ^— a; + 2 - ^ ^ + ^ , x:' , 1 a-i — x + 21. x-2 x+2 2x 23. ix^-x? + 4x-8 a n — X . ax n a n^ — nx a ^n-x^^ a x+2 x^ I x + 1 x* + xP 24. x^ + x + l + \ X X' a + 2b 2a + b a¥ a^b /I 1\ /I l\ a' + o' ftV [l)2 (;2J \(fi ci) a2g2 a^ + ab a*- a-3a^ + Sa^ „~ a2 + 62 ^_ a»b-b* a^b + ab^ + 2 a^fi^ a* + a^ - 2 aa a« - 6* a^tfi + ab^ + b* Factors of Fractional lixpressions. 27. Fractional Expressions can be factored by the same methods as were employed in factoring integral expressions, Ex.1. }c^ + Jc + n+- = lc'fl + -^+n[l + -\ Ex.2. FRACTIONS. 175 96^ \3b^ Asb J Ex.3. a^ + ±- + l = a^ + 2 + ±-l= x + - =(:+l+')(:^l-'-} EXERCISES XI. Factor the following expressions ; 1. 2. --i- 3. 64a;i' ^ 125 !/8 4. x-^ + J-. 5. x^+l+-^- 6. '^^+1 + 4/ 7. x' + \ + 3x X 8. lfi + X + - + -- X *3 9. 10. x^+x+2+^ X -i- 11. K4+4a;2+6+4 x^ + 1 a;*" 12. xa-l + .-l. Indeterminate Fractions. 28. By Art. 5, a fraction is a number which, multiplied by the de- nominator, gives the numerator. Therefore the fraction J is a number which, multiplied by 0, gives 0. But by Ch. III., § 3, Art. 16, any num- ber, multiplied by 0, gives 0. Therefore the fraction g may denote any number whatever. For this reason, it is called an Indetermmate Fraction. Some Principles of Fractions. 29. The following principles will be of use in subsequent work : then each of these fractions is equal to the fraction a»i -I- 6»{! + ens + ••• E.g., adi + bdi + cds + ••• 2_4_5 X 2-f6 X 4 3 6 5x3-1-6x6 Let the common value of the given fractions be v. Then from ^ = v, ^= = ^, ^=«,etc., di cfa «3 we have rai = div, n% = d^o, hs = d^v, etc. 176 ALGEBRA. ' [Ch. IX Multiplying these equations by a, b, c, etc., respectively, and adding corresponding members of the resulting equations, we have an\ + hrii + ens + ••• = o,^iV + S'^s'' + cdau + ••• = {adi + bdi + c^s + •••)"■ Therefore ad\ + 6^2 + cda + ••• di 6,2 a ni + Sma + crasH _ ^ _ Ml _ »? _ gtg_ (ii.) In particular, Ml _ «2_re3 _ __ _ ni + W2 + TOs + ••• ^ di ^2 da di + di + d-i + ■■• E.g., 2 = 4 = 2 + 4. "' 3 6 3 + 6 (iii. ) If the fraction - he in its lowest terms, then —, wherein p is a positive integer, is in its lowest terms. For by Ch. VIII., § 2, Art. 13 (vii.), m'' and d? are prime to each other when n and d are prime to each other. n N (iv.) If two fractions, — and — , whose terms are positive integers, be n d D equal, and if - be in its lowest terms, then iV"= kn, D = kd, wherein k is d a positive integer. E.g., ^^ = f, and 10 = 5 X 2, 15=5 X 3. From :^=^, wehaveiV = ^. (1) D d d Since N is an integer, this equation shows that nD is exactly divisible by d ; that is, that it contains d as a factor. Also, since - is in its lowest d terms, n and d are prime to each other. Therefore, by Ch. VIII., § 2, Art. 13 (iv.), cZ is a factor of D ; that is, D = kd, wherein & is a positive integer. Substituting kd for B in (1), we have N=^ = kn. d (v.) If two fractions - and — , whose terms are positive integers, be d D equal, and each he in its lowest terms, then N = n and D = d. This principle follows directly from (iv.). FRACTIONS. 177 BXEBCISES XII. MISCELLANEOUS EXAMPLES. Simplify tlie following expressions : 1 Vl+«/ 2 \a + b ) \y + al a + h a + b ax + x,^\ a-xj \ a-l) 5. a + 6 /1 1\ 6 + c /l 1\ g /a + b ^ a-b\ . /a+b a-b\ ah \a bj be \c b) ' [c+d c-d) ' \o-d c+d)' a+l b+l ■ 1+f 1+5 1+1 b a b a a b 9. m 5-. 10. f5±^ + ?Vf^±^ ^^ l_TO + m'+ "^ V x+2/ yl\y x+yj 1+m 11. (^+«)(^__^)_(^+^)(^__„y Va + a; J \a — x J \a + x /\a — x j 12. -^^ L_. 13. °'-"\x ^ ^ ^ _i._A + J. <» + * a; — 1 a; + 1 a'' aa a;^ 14. *= ^. (-It 15. /' 2a:+y I 2y-a; a;^ \ . g' + y' \ a; + j^ x — y x^ — y^Jx'^ — y^ 16. M+laxf— ^— +-^— V ixy + 6ay \ax + ay 2x + 2yJ 17 / »-! >t + l \ /I n 1 \. Vk+1 ?i-l/ \2 4 4»y a6+ 1 Ig a + n a^ + rfi + ian ,„ & 1 a a" ■ • a I 1 &(a6c + a + c)' a+n o2 - n" 6c + 1 178 ALGEBRA. [Ch. IX a + 20. c + ab c + d a — 1} + d ^ a+ b ed 6 + c + d c + d a + b a+\ 6+1 c+i 21. ^x jx J. &+i c+i a+i b c 1 a; + a a; x'^+ a^ X x^ + a^ a + X 1 g — a: 23. a a^ + ax 2x cfi — x^ \ iax J a a^ + x'^ a a^ + xi^ Lj)2 q^ P + q\p qJ-i 25. rf^+iuf^-i+i^ix-m-. L\y^ xj \y^ y x/J x + y 26 rf 2a; '^^ \ ■ ( ^ a: Vf g,Q gSy-j/* . / x^ + a:8y + a:2yii . (^ yy^ \ (x2 - ^2)8 M^ '^xj i , 1 + n — n^ — n* 30. - x?/'2 + x^ 1 , 1 — x-1 1 - 3 X + x2 31. 32. 33. Sx + Cx-l)-' x^-l x-1 1 - 2 X + x^ - 2 x3 l + 2x + 2x2 + x' 1 +x + x2+ hx»-i + 1 -X 1 + 2 X + x2 _ ,£±2 . ^8 \ 3 + x»y x"+2 + 3 x2 6 x2" - 24 2x x2b+3 + 6x"+3 + 9x« 3x» + ( FRACTIONS. 179 In each of the following expressions make the indicated substitution, and simplify the result : 34. Inf^^l^y, letm = «+^. \m-bj 2 1-.2 - + — , let X en n — ex e e^ — n 36. In 1 + ^^ + "'^ ~ °' . let o + 6 + c = 2 s. 2 6c 37. In^fl-^Uj^fl-!?), leta = m + n. n\ al mV a/ 38. Ing^+i-ra(ic + l)- "(^'-l)-^ ], leta; = ^^ X \_ X A a Verify each of the following identities : 39. «(a - ») _SI6+£l^ ^^ ^hen x = a-h. a 40 a^^.:^a)+6(£:^&)+£(^i^^^^^henx = a + 6 + c. 6 + c a + e a + b 41. £±l« + «iz2« = _ia&_ .whence =-«^ 26-a; 26 + a; 4 62-a;2 a + 6 42. (1+ a;) (1 + 1/) (1 + 2) = (!-») (1-2/) (1-3), when a + 6 6 + c c + a 43. 62-a;2 = i-s(s-o)(s-6)(s-c), when c^ ^ _ 5^ + c^ - o^ and a + 6 + c = 2 s. 2c CHAPTER X. FRACTIONAL EQUATIONS IN ONE UNKNOWN NUMBER. 1. A Fractional Equation is an equation whose members, either or both, are fractional expressions in the unknown number or numbers. TP 3 2 „ , 4-2a; „ ^' x + 2 x + 1' x + 1 Observe that we cannot speak of the degree of a fractional equation. The term degree, as used in Ch. IV., § 2, Art. 6, applies only to integral equations. 2. The principles of equivalent equations established in Ch. IV., § 3, hold also for fractional equations. Ex. 1. If both members of the equation _!- = —?— (1) x+2 x+1 ^^ be multiplied by (x + 2)(x + 1), the L. C. D. of the denomi- nators of its fractional terms, we obtain the integral equation 3(x + l) = 2(a; + 2); (2) whence x = l. The root 1 of the derived equation (2) is found, by substitu- tion, to be a root of the given equation. Ex. 2. If both members of the equation ^+r^=— xT-3 (1) ar — 1 1 — X x + 1 be multiplied by x' — 1, we obtain the integral equation -2x'-x{x + l)=-x(x-r)-3(3^- 1), or (x + 1) (a; - 3) = 0. (2) 180 FRACTIONAL EQUATIONS. 181 Now observe that it was not necessary to multiply by sis' — 1, = (a; + 1) (a; — 1), to clear the given equation of fractions. For, if the terms in the second member be transferred to the first member, we have a;^ — 1 1 — as 1+x or, uniting terms, ar' — 2a; — 3 ^ ^ a? — l or, canceling a; + 1, = 0. a; — 1 Clearing the last equation of fractions, we have a;-3 = 0; (3) whence a; = 3. The root 3 of the derived equation (3) is found, by substi- tution, to be a root of the given equation. Had we solved equation (2), we should have obtained the additional root — 1, which, as will be proved later, is not a root of the given equation. This root, which does not satisfy the given equation, and which was introduced by multiplying both members of the given equation by the unnecessary factor a; + 1, is a root of the equar tion obtained by equating this factor to 0. 3. The roots of a fractional equation are found by solving the integral equation derived from it by clearing of fractions. The equivalence of the given equation and the derived integral equation is determined by the following principle : If both members of a fractional equation, in one unknown number, be multiplied by an integral expression which is neces- sary to clear the equation of fractions, the integral equation thus derived will be equivalent to the given fractional eqvution. Thus, in Art. 2, Ex. 1, equation (2) is equivalent to (1); and in Ex. 2, equation (3) is equivalent to (1), while (2) is not equivalent to (1). 182 ALGEBRA. [Ch. X Let :^=0 (1) be the given fractional equation when all its terms are transferred to the first member, added algebraically, and the resulting fraction reduced to its lowest terms. In deriving equation (1) from the given fractional equation, terms were transferred from one member to the other, by Ch. IV., § 3, Art. 7 (i.) ; and then only indicated operations were per- formed. Therefore equation (1) is equivalent to the given fractional equation. Clearing (1) of fractions, we have the integral equation iV=0. (2) Any root of (1) reduces — to 0. But any value of x which reduces "^ to must reduce JV to (Ch. III., § 4, Art. 7), and hence is a root of the derived equation. That is, no solution is lost by the transformation. Any root of the derived equation reduces N to 0. But, since — is a fraction in its lowest terms, JV and D have no common factor, and there- fore cannot both reduce to for the same value of x (Ch. VIII., § 4, Art. 2). Consequently, any value of x which reduces iV to must JV reduce — to (Ch. III., § 4, Art. 6). That is, no root is gained by the transformation. Therefore the derived integral equation is equivalent to equation (1), and hence to the given fractional equation. 4. If all the terms of a fractional equation be transferred to its first member, be united into a single fraction, and this frac- tion be reduced to its lowest terms, the integral equation ob- tained by then clearing of fractions is equivalent to the given fractional equation. But it is not necessary, nor advisable, to make this transformation before clearing of fractions. If any new root be introduced, it will, as we have seen, be a root of one of the factors of the L. C. D., equated to 0, and can there- fore be rejected at sight. Ex. Solve the equation I^JLlO = 5^ + 35. ^ a; - 2 12 6 Multiplying by 12 (a; - 2), 84 a; -I- 120 = 5 a^ - 10 a; -I- 70 a; - 140. FRACTIONAL EQUATIONS. * 183 Transferring, uniting terms, and factoring, (a;-10)(5a!+26) = 0. Whence a; = 10 and x = — s^. Since neither 10 nor — -2^- is a root of the L. C. D. equated to 0, that is, of 12 (a; — 2) = 0, both 10 and — ^ are roots of the given equation. 5. The following suggestions will simplify the work of solving many fractional equations: (i.) If any fraction be not in its lowest terms it should be reduced. (ii.) The form of an equation will frequently suggest grouping and uniting some fractional terms. This reduction should always be made if two or more fractions have a common denominator. 1111 Ex. 1. Solve the equation ■2 X — i aj — 6 Uniting the fractional terms in each member separately, and dividing by — 2, 1 ^ 1 (x -2)(x- 4) (a; - 6) (a; - 8)" Clearing of fractions, (so -6)(x- 8) = {x - 2) (a; - 4). Therefore a^-14a; + 48 = w^-Ba; +8. Whence — 8 a; = — 40, or a; = 5. Since 5 is not a root of any factor of the L. C. D. equated to 0, it is a root of the given equation. (1) Ex 2 Solve the equation a^ 1 + 10, a;-l a!-l Transferring and uniting terms. x'-l — 10 x-1 Eeducing to lowest terms. x + 1 = 10, Whence X. = 9. (2) 184 ALGEBRA. [Ch. X The integral equation (2) is equivalent to the given equation, and therefore 9 is the required root. Had the given equation been cleared of fractions by multi- plying by 35 — 1, the root 1 of a; — 1 = would have been introduced. (iii.) An improper fraction should in some cases first be reduced to a mixed expression. Ex. 3. Solve the equation ^^ + ^^ = 2. x—i x—o Reducing the improper fractions, we obtain l+-^ + H-^ = 2, x—i x—5 x—A x—o Clearing of fractions, a; — 5 + 05 — 4 = 0. Whence a; = f . BXBECISES I. Solve the following equations : 1. 1? = 4. X 2. 5-§ = X = 2. 3. 5^-^ = 3. x + 1 4. x-1 2 a;+l 3 5. x-1 x-3 X 4 ^2 <-t- '■ 5-1-^- 7. a; — 2 x — 5 2a;-5 2a;-2 8. 25 x-i 10 -0. 3 X 3x „ 3x-4 ■ x+ 1 x + 2 10. ^-1 + 1-1. a;+ 1 X jj 2x + l 2x + l_o ■ (x + 2)2 x + 2 12. 5x 1 ,7 6 13 1 , 2 13 6x 3x + l 9a; + 3 "'2'x + 2 8 4x + 8 14. 3 |x + J f _3 ^4* 15 4 4 _ 5 4 J + a; i + a; "■ (x+l)2 x(x+l)2 2x(x+l) 16. ^ + ^ =- l-3a; l-5a; 4 2a;- 1' 1^ x-7 2X-15 1 ■ x+ 7 2x-6 2(x+7) 18. 2a;+l 8 2x-l 2x+l 19 ^ 1 ^ ^^ 2a; -1 4a;»-l x + 2'x + 3 x2 + 5x + 6 20. x-3 12 -2x 3a;- -27 „j 2X + 19 17 3 _Q a;2 _ 9 a;2 - 36 x^ - 81 5 x^ - 5 x^ - 1 FRACTIONAL EQUATIONS. 185 7 _|_ 8 _ 37 -9a; a;2-l a;2_2x + l x' -x^-x + 1 7 _| 3 15 6X + 30 4 a; -20 2a;2-50 24. 4^ + 4 1_^ a;» + 8 5a;2 - 10 a; + 20 a; + 2 28. 1 + 4 _ 5 x2 + 2x+l a; + 2a;2 + a;3 2a; + 2fl;3 2g _1 1 x + 3 _ 6 ■ a;-l 2(a;+l) 2(a;2 + 1) ~a;* - l' 27.^^ ^=^ L-. 28. 6 9 _ 1 a; — 2 x-i X — 6 a;-8 a; — 5 a;- 3 a;-7 a; — 1 29. _l_ + ^_ = _7_+_J_. 30. _1 2_+_ 2 - 1 x-9 x-i x-1 a;-ll a;-13 a;-15 a;-18 a;-19 31 a: + 8 a; + 6a: + 4 _ a:+5 x + 2 x + H a; — 5 a; — 6 a; — 7 a; — 5 x — 6 x — 7 ,,, x + 5_x + 3 x + 1 _ x + 2 _ X — 1 , X x-7 X— 8 X— 9 x-7 x-8 x-9 Problems. 6. Pr. 1. A niimber of men received $120, to be divided equally. If their number had been 4 less, each one would have received three times as much. How many men were there ? Let X stand for the number of men. Then each man received 120 dollars. If their number had been 4 less, each one would ^ 120 have received dollars. « — 4 Therefore, by the condition of the problem, we have 120 „ 120 , „ ■ = 3 X ; whence a; = 6. 35 — 4 X Pr. 2. The value of a fraction when reduced to its lowest terms is ^. If its numerator and denominator be each dimin- ished by 1, the resulting fraction will be equal to ^. What is the fraction ? The numerator of the required fraction must be a multiple of 1, and the denominator the same multiple of 5. Let X stand for this multiple. The required fraction is 5x 186 ALGEBRA. [Ch. X By the condition of the problem, we have 5a!-l 6' Whence x = 5. The required fraction is t^, = \. EXERCISES II. 1. What number added to the numerator and denominator of f will give a fraction equal to | V 2. The sum of two numbers is 18, and the quotient of the less divided by the greater is equal to |. What are the numbers ? 3. The denominator of a fraction exceeds its numerator by 2, and if 1 be added to both numerator and denominator, the resulting fraction will be equal to |. What is the fraction ? 4. The sum of a number and seven times its reciprocal is 8. What is the number ? 5. The value of a fraction, when reduced to its lowest terms, is ^. If its numerator be increased by 7 and its denominator be decreased by 7, the resulting fraction will be equal to |. What is the number ? 6. What number must be added to the numerator and subtracted from the denominator of the fraction -j^, to give its reciprocal ? 7. If J be divided by a certain number increased by J, and J be sub- tracted from the quotient, the remainder will be J. What is the number ? 8. A train runs 200 miles in a certain time. If it were to run 5 miles an hour faster, it would run 40 miles further in the same time. What is the rate of the train ? 9. A number has three digits, which increase by 1 from left to right. The quotient of the number divided by the sum of the digits is 26. What is the number ? 10. A number of men have |72 to divide. If $144 were divided among 3 more men, each one would receive 1 4 more. How many men are there ? 11. It was intended to divide -J by a certain number, but by mistake ^ was added to the number. The result was, nevertheless, the same. What is the number? 12. A steamer can run 20 miles an hour in still water. If it can run 72 miles with the current in the same time that it can run 48 miles against the current, what is the speed of the current ? 13. A man buys two kinds of wine, 14 bottles in aU, paying $9 for one kind and $ 12 for the other. If the price of each kind is the same, how many bottles of each does he buy ? FRACTIONAL EQUATIONS. 187 14. A can do a piece of -work in 10 days, B in 6 days ; and A, B, and C together in 3 days. In how many days can C do the work ? 15. A and B together can do a piece of work in 2 days, B and C together in 3 days, and A and C together in 2^ days. In how many days can A, B, and C together do the work ? 16. The circumference of the hind wheel of a carriage exceeds the cir- cumference of the front wheel by 4 feet, and the front wheel makes the same number of revolutions in running 400 yards that the hind wheel makes in running 500 yards. What is the circumference of each wheel ? 17. In a number of two digits, the digit in the tens' place exceeds the digit in the units' place bji2. If the digits be interchanged and the re- sulting number be divided by the original number, the quotient will be equal to f|. What is the number ? 18. In a number of three digits, the digit in the hundreds' place is 2 ; if this digit be transferred to the units' place, and the resulting number be divided by the original number, the quotient will be equal to J{. What is the number ? 19. In one hour a train runs 10 miles further than a man rides on a bicycle in the same time. If it takes the train 6 hours longer to run 255 miles than it takes the man to ride 63 miles, what is the rate of the train ? 30. Two engines are used in different places in a mine to pump out water. The one pumps 11 gallons every 5 minutes from a depth of 155 yards ; the other pumps 31 gallons every 10 minutes from a depth of 88 yards. The engines together represent the power of 54 horses. Each engine represents the power of how many horses ? 21. A cistern has three pipes. To fill it, the iirst pipe takes one-half of the time required by the second, and the second takes two-thirds of the time required by the third. If the three pipes be open together, the cis- tern will be filled in 6 hours. In what time will each pipe fill the cistern ? 33. A and B ride 100 miles from P to Q. They ride together at a uniform rate until they are within 30 miles of Q, when A increases his rate by ^ of his previous rate. When B is within 20 miles of Q, he in- creases his rate by J of his previous rate, and arrives at Q 10 minutes earlier than A. At what rate did A and B first ride ? 33. A circular road has three stations. A, B, and C, so placed that A is 15 miles from B, B is 13 miles from C in the same direction, and C is 14 miles from ^ in. the same direction. Two messengers leaving A at the same time, and traveling in opposite directions, meet at B. The faster messenger then reaches A, 7 hours before the slower one. What is the rate of each messenger ? CHAPTER XI. LITERAL EQUATIONS IN ONE UNKNOWN NUMBER. 1. The unknown numbers of an equation are frequently to be determined in terms of general numbers, i.e., in terms of numbers represented by letters. The latter are commonly represented by the leading letters of the alphabet, a, b, c, etc. Such numbers as a, b, c, etc., are to be regarded as known. E.g., in the equation x + a^=b, a and h are the known num- bers, and X is the unknown number. From this equation we obtain x = b — a. 2. It is important to notice that tlie assumption that x, y, z, etc., are the unknown numbers of an equation, and that a, 6, c, etc., are the known numbers, is arbitrary. In the equation x + a = b, either « or 6 could be taken as the unknown number. If a be taken as the unknown number, we have a = b — x; if 6 be taken as the unknown number, we have b —x + a. 3. A Ifumerical Equation is one in which all the known num- bers are numerals ; as2x + 3 = 7; 4a; — 32/ = 7. A Literal Equation is one in which some or all of the known numbers are literal ; as 2 aw + 3 6 = 5 ; ax + by = c. 4. Ex. 1. Solve the equation ^^1^ + ^^ = - ^"' ~ ^^' - b a 2ab Clearing of fractions, 2ax-2a^ + 2bx-2b' = -a'' + 2ab-b^ Transferring and uniting terms, 2(a + b)x = a' + 2 ab + b". Dividing hj2(a + b), x = ^^^^. Notice that the above equation, although algebraically frac- tional, is integral in the unknown number x. The equations which follow are fractional in the unknown number. 188 LITERAL EQUATIONS. 189 Ex.2. Solve the equatioa Liz^ ^ ^a' - 1 ^_ 2. 6a! ax Clearing of fractions, a — a'x=b^x — b +2 abx. Transferring and uniting terms, (a' + 2 ab + b^x = a + b. Dividing by a^ + 2 a& + b', x = — — a + b Ex. 3. Solve the equation 1 C X a — b {x — b){x — c) {x — b){x — c) Uniting fractions with common denominator, 1 , G —X a — b (x — b){x—c) = 0. Eeducing to lowest terms, = 0. a — b X — b Clearing of fractions, a; — 6 — a + 6 = 0. Whence a; = a. Had we cleared of fractions at once, we should have intro- duced the root c of the factor x— c equated to 0. 5. A linear equation in one unknown number has one, and only one, distinct root. Any linear equation in one unknown number can be reduced to the form ax = b. If both members of this equation be divided by a, when o # 0, we obtain x = -- a Since this value of x satisfies the equation, we conclude that every linear equation has at least one root. Let us assume that the equation ax = b has two distinct roots, and let us denote them by ri and r^. Then, since they must both satisfy the given equation, we have ari = & (1), and ara = b (2). Subtracting (2) from (1), we obtain a(n -ri) = 0. 190 ALGEBRA. [Ch. XI Since a i^fc 0, we have, by Ch. III., § 3, Art. 18, ri — J'2 = ; whence ri = ra. Therefore the assumption that the equation has two distinct roots is untenable. Hence the truth of the principle enunciated. 6. Observe that, in Art. 5, we have no authority for dividing both members of the equation ax = b by a, when a = 0. But if we assume that - still gives the solution of a the equation when a = 0, the value of x will be indeterminate (f ) or infinite (oo), according as b = oi b ^0. Evidently, when a = and 6=0, any finite value of x will satisfy the equation ; while, when a = and 6 ^fc 0, no finite value of x will satisfy the equation. EXEECISBS I. Solve the following equations : 1. a — X = c. 2. mx + a = b. 3. a — bx = c. 4. mx = 7ix + 2. 5. X — ax + 1 = bx. 6. ax+ bx — x = 0. 7. S ax — 5 ab + 6 ax — 1 ac = 2 ax + 2 ab. 8. 4 o2 - 2 afta; + &2 + 3 aH = 5a^-b^x + 2 aH. 9. (2 a - 6> = 4 a2 - 3 a(6 + x). 10. a(a; + a)- 6(x- 6) = 3aa;+(a - &)2. 11. x{x + ffl) + a;(a; + 6) - 2 (x + a) (a; + 6) = 0. 12. (0 + !c) (6 + x)-(c -»;)((? -x) = 0. 13. (3a-x)(a- 6)+2aa; = 4 6(a + a;). 14. a + _ = c. 15. - + 6 = - + a. 16. — - — = -,- X ex X — a i 17 ^4-k — ^ — ^. 18 5_+_^_E_+J;. 19 x+a__2__x^-a ax a b X b+x b + 1 2 x+a 2 gQ 6x + a 3x— & _n 21 a + x _ g — X „„ x — a _ x + 6 4x+6 2x— a 6 + a 6— a x — 2 x — 3 23 t" — ^ ~ ft'^ 04 x+a6 _ o^+a6 + 6'' „- a; + a _ (2 x + a)' ' 6 x-a^' ' x-06 a'^-ab + b"' ' x-b~{2x-by 26 ai^ + g^ a: _ 1 37 a(x + 1)- 6(g - 1) _ g^ ■ 4x''-a2 2x + a 4' ' 6(x + 1)- a(x - 1) fts' 88. a^-b° ^ aCx-6'') + 5(a^-x) , gg 1 , 1 ^ ^Q. a'+6' a(x— 6^)— 6(a^— x) ab — ax bc—bx ac—ax 30. ^=^+^lI^+£z:C^l_^l^l. 3j_ g-ft^ft^-^^c-^^Q 2 6c 2oc 2ab a b x — I « — 2 y. — 3 LITERAL EQUATIONS. 191 32 ^~ " 4. ^ — i I * — c _ Sx b+c a+c a+b a+b+c 33 a + X a — x _1 3 a 34. 35. a^ + ax + a;2 a^ — ax + ofl x(a* + a'x^ + x^) l-2ax^ l + 2ax^ _ 4 abx^ 1 + 2 6*2 1 - 2 6x2 4 62a;4 _ i a^ + ia a _ 1 a;2 + a; — a'^+a x + a x — a + 1 36 ^~ ^ + 2 a2(l - a:) _ 2 a:- 1 _ l-x ' a-1 a*-l 1-a* l + a' a^ + a; a^ - k _ 4 a6x + 2 a^ - 2 6^ 37. 38. 39. 40. b^ + x &* - a:^ ax + x'^ a" — (fix + w)? _ 1 a* + oH + aa;2 + a;^ a* + 2 oM + x* a + x a^ — X 2a + X a* a; - 2 a a'^ -x~ aH + iax -2a^ -x^' 2 (a: — a) c — x ■ 2 ax + x^ a^ — ac + ex — 2ax + x^ a; — (a + c) 1-T^ , a + x-- ^l I g' ^ 1 ^3 «L_+*_i 2ax a/j_l\ a^ ■ a + x (a + x)2 x\ a/ General Problems. 7. A problem in which the given numbers have particular values can be generalized by assuming literal numbers for the given numbers. A problem so stated is a general problem, and its solution is called a general solution. A particular solution can be obtained from the general solu- tion by assigning to the literal numbers in the latter, particular numerical values. We will now obtain general solutions of some of the prob- lems already solved in" Oh. V. Ch. v., Pr. 1. The greater of two numbers is m times the less, and their sum is s. What are the numbers ? Let X stand for the less required number. Then mx stands for the greater. By the condition of the problem, we have X + mx = s ; whence, x= , the less number, and mx=— — , the greater. 1+m l+m 192 ALGEBRA. [Ch. XI If m = 3 and s = 84 (as in the particular problem), we have ^* : 21, and mx = 3x21 = 63. 1+3 When the numbers are equal, m = 1, and we obtain s s a; = - and mx = -> 2 2 for all values of s ; that is, either of the two numbers is half their sum. Ch. v., Pr. 9. A carriage, starting from a point A, travels m miles daily ; a second carriage, starting from a point B, p miles behind A, travels in the same direction n miles daily. After how many days will the second carriage overtake the first, and at what distance from B will the meeting take place ? Let X stand for the number of days after which the carriages meet. Then the number of miles traveled by the first carriage will be mx; the number traveled by the second will be nx. Therefore, by the condition of the problem, mx =nx — p\ whence x = — — — > n — m the number of days after which the carriages meet. The distance traveled by the first carriage is — ^— miles, •' n — m and the distance traveled by the second carriage is — ^— miles., ■' n—m They, therefore, meet — ^-— miles from B. ■' ' n — m Ch. v., Pr. 10. One man asked another what time it was, and received the answer : " It is between n and m + 1 o'clock, and the minute-hand is directly over the hour-hand." What time was it? At n o'clock the minute-hand points to 12 and the hour-hand to n. The hour-hand is therefore 5 n minute-divisions in ad- vance of the minute-hand. Let X stand for the number of minute-divisions passed over by the minute-hand from n o'clock until it is directly over the hour-hand between n and n + l o'clock. LITERAL EQUATIONS. 193 Then thfe number of minute-divisions passed over by the hour-hand is equal to the number of minute-divisions passed over by the minute-hand, minus 5 n ; that is, to a; — 5 n. But since the minute-hand moves 12 times as fast as the hour-hand, we have x = 12(x — 5n); whence x = — -^. Consequently, the time was -rrr- minutes past n. If w = 1, it was ff , or 5 j^ minutes past 1. If n = 2, it was ^i, or lOi^ minutes past 2. Etc. If w = 11, it was -^j or 60 minutes past 11 ; i.e., 12 o'clock. If w = 12, it was ^^, or 65-j^ minutes past 12 ; i.e., 5^ minutes past 1. Notice that the two hands coincide at 12 o'clock, but not between 12 and 1. SXBBCISES II. Find the general solution of each of the following problems, and from this solution obtain the particular solution for the numerical values assigned to the literal numbers in the problem. 1. Find a number, such that the result of adding it to n shall be equal to n times the number. Let n =2 ; 5. 2. Divide a into two parts, such that — of the first, plus _ of the TO n second, shall be equal to 6. Let a = 100, 6 = 30, m = 3, ra = 5. 3. Find a number, such that the sum of the results of subtracting it from a and from b shall be equal to c. Let a = 3, 6 = 6, c = 5. 4. One boy said to another: "Think of some number, add 3 to it, multiply the sum by 2, add 4 to the product, divide the result by 2, sub- tract 1 from the quotient, multiply the difference by 4, add 4 to the product, divide the result by 4, tell me the result, and I will tell you the number you have in mind." If the result be d, what number does the boy think of ? 5. A sum of d dollars is divided between A and B. B receives 6 dol- lars as often as A receives a dollars. How much does each receive ? Let d = 7000, a = 8, 6 = 2, 194 ALGEBRA. [Ch. XI 6. A father's age exceeds his son's age by m years, and the sum of their ages is n times the son's age. What are their ages ? Let m = 20, M = 4 ; m = 25, n = 7. 7. If two trains start together and run in the same direction, one at the rate of mi miles an hour, and the other at tlie rate of mi miles an hour, after how many hours will they be d, miles apart ? Let d, = 200, mi = 35, m2 = 30. 8. A farmer can plow a field in a days, and his son in 6 days ; in how many days can they plow the field, working together ? Let a = 10, 6 = 15. 9. A pupil was told to add m to a certain number, and to divide the sum by re. But he misunderstood the problem, and subtracted n from the number and multiplied the remainder by m. Nevertheless he obtained the correct result. What was the number ? Let m — 12, n = 13. 10. What time is it, if the number of hours which have elapsed since noon is m times the number of hours to midnight ? Let m = J. 11. A starts from P and walks to Q, a distance of d miles. At the same time B starts from Q and walks to P. If A walk at the rate of m miles a day and B at the rate of n miles a day, at what distance from F do they meet, and how many days after they start ? Let m = 20, re = 30, d = 600. 12. Two friends, A and B, each intending to visit the other, start from their houses at the same time. A could reach B's house in m minutes, and B could reach A's house in re minutes. After how many minutes do they meet ? Let m = 12|, n = lOJ. 13. Two couriers start at the same time and move in the same direc- tion, the first from a place d miles ahead of the second. The first courier travels at the rate of m\ miles an hour, and the second at the rate of im miles an hour. After how many hours will the second courier overtake the first ? Let d = 15, mi = 17, nii = 20. From the result of the preceding example find the results of Exx. 14-16. 14. At what rate must the second courier travel in order to overtake the first after h hours ? Let d = 18, mi = 15, ft = 3. 15. At what rate must the first courier travel in order that the second may overtake him after ft hours ? Let d = 12, mj = 22, -ft = 3. 16. How many miles behind the first courier must the second start in order to overtake the first after ft hours ? Let mi = 18, ma = 21, ft = 4. 17. In a company are a men and 6 women ; and to every m unmarried men there are re unmarried women. How many married couples are in the company ? Let a = 13, 6 = 17, m = 3, re = 5. 18. The annual dues of a certain club are at first a dollars. Subse- quently the yearly expenses increased by d doUairs, while the number of LITERAL EQUATIONS. 195 members decreased by n. In consequence the annual dues were increased by 6 dollars. How many members were originally in tlie club ? Let a = 25, d = 315, m = 7, and 6 = 2. 19. A father divided his property equally among his sons. To the oldest he gave d dollars and - of what remained ; to the second son he 1 ** gave 2 d dollars and - of what was then left ; to the third son he gave 3 d 1 " dollars and - of the remainder ; and so on. What was the amount of his n property ? Let d = 1500, n = 11 ; d = 2000, w = 6. 20. Two couriers start from the same place and move in the same direction, one h, hours after the other. The first one travels at the rate of mi miles an hour, and the second at the rate of m2 miles an hour. After how many hours will the second courier overtake the first ? Let A = 2, mi = 15, ma = 20. From the result of the preceding example, find the results of Exx. 21-23. 21. At what rate must the second courier travel in order to overtake the first after H hours ? Let IT = 6, A = 2, mi = 12. 22. At what rate must the first courier travel in order that the second may overtake him after H hours ? Let S=4, h = 1, m2 = 20. 23. How many hours after the first courier starts must the second start in order to overtake the first after H hours ? Let H—Q, mi = 14, ma = 22. 24. Two boys run a race from Ato B,ii, distance of d yards. The first runs a yards a second ; after reaching B, he turns and runs back at the same rate to meet the other boy, who runs 6 yards a second. How many seconds after they start does the faster runner meet the other ? Let d = 253, a = 2.5, b = 2.1. 25. An accommodation train leaves A every h hours, and runs to B at the rate of m miles an hour. At the same time an express train leaves B and runs to A at the rate of n miles an hour. What time elapses after an express train meets an accommodation train until it meets the next accommodation ^rain ? Let h =3, m = 20, n = 40. 26. At what time between n and n + 1 o'clock will the hands of a clock be in a straight line ? Let n = l ; 2 ; 3 ; ... to 12. 27. At what time between n and re + 1 o'clock are the minute-hand and the hour-hand of a clock at right angles to each other ? Let re = 1 ; 2 ; 3 ; ... to 12. 28. At what time between re and n -I- 1 o'clock does the second-hand bisect the angle between the hour-hand and the minute-hand ? Let re = 1 ; 2; 3; .,• tQ 12. CHAPTER XII. INTSBPRETATION OF THE SOLUTIONS OF PROBLEMS. 1. In solving equations we do not concern ourselves with the meaning of the results. When, however, an equation has arisen in connection with a problem, the interpretation of the result becomes important. In this chapter we shall interpret the solutions of some linear equations in connection with the problems from which they arise. Positive Solutions. 2. Pr. A company of 20 people, men and women, proposed to arrange a fair for the benefit of a poor family. Each man contributed $ 3, ajid each woman $ 1. If f 55 were contributed, how many men and how many women were in the company ? Let X stand for the number of men ; then the number of women was 20 — X. The amount contributed by the men was 3 x dollars, that by the women 20 — a; dollars. By the condition of the problem, we have 3 a; + (20 - a;) = 55 ; whence x = 17 J. The result, 17J, satisfies the equation, but not the problem. For the number of men must be an integer. This implied condition could not be introduced into the equation. The conditions stated in the problem are impossible, since they are inconsistent with the implied condition. If the problem be generalized, its solution will show how the given data can be modified so that all the conditions, expressed and implied, shall be consistent. The generalized problem may be stated thus : A company of m people, men and women, proposed to arrange a fair for the benefit of a poor family. Each man contributed a dollars, and each woman 6 dollars. If n dollars were contributed, how many men and how many women were in the company ? The solution of the equation of this problem is n — bm a ~ b 196 INTERPRETATION OF SOLUTIONS. 197 In order that x may be an integer, n — bin must be exactly divisible by a — 6. Thus, if, in the given problem, the number of people were 21 instead of 20, the other data being the same, we should have ^ ^ 65 - 1 X 21 ^ 34 ^ ^^ 3-1 2 If all the conditions of a problem, expressed and implied, be consistent, a positive solution will satisfy these conditions and therefore give the solution of the problem. Negative Solutions. 3. Fr. A father is 40 years old, and his son 10 years old. After how many years will the father be seven times as old as his son ? Let X stand for the required number of years. Then after x years the father will be 40 + s years old, and the son 10 + a; years old. By the condition of the problem, we have 40 + a; = 7(10 + x), whence x=-5. (1) This result satisfies the equation, but not the condition of the problem. Tor since the question of the problem is " a/{e»' how many years ? " the result, if added to the number of years in the ages of father and son, should increase them, and therefore be positive. Consequently, at no time in the future will the father be seven times as old as his son. But since to add — 5 is equivalent to subtracting 5, we conclude that the question of the problem should have been, " How many years ago ? " The equation of the problem, with this modified question, is : 40 - a; = 7(10 - a) ; whence x = 5. (2) Notice that equation (2) could have been obtained from equation (1) by changing x into — x. 4. The interpretation of a negative result in a given problem is often facilitated by the following principle : If — xhe substituted for x in an equation which has a negative root, the resulting equation will have a positive root of the same absolute value ; and vice versa. E.g. , the equation x+ l= — x — S has the root — 2 ; while the equation — a;4-l=a; — 3 has the root 2. In general, the equation ax = b has the root — (1) And the equation — ax=zb (2) -• If the root - be negative, then the root — - a a a and vice versa. 198 ALGEBRA. [Ch. XII 5. Pr. Two'pooket-books contain together $ 100. If one-half of the contents of one pocket-book, and one-third of the contents of the other be removed, the amount of money left in both will be $70. How many dollars does each pocket-book contain ? Let X stand for the number of dollars contained in the first pocket- book ; then the number of dollars contained in the second is 100 — x. When one-half of the contents of the iirst, and one-third of the contents of the second are removed, the number of dollars remaining in the first is -, and in the second f (100 — x) . By the conditions of the problem, we have ix + 1(100 - x) = 70, whence x = - 20. Substituting — x f or x in the given equation, we obtain -^x-|-|(100-|-x)=70, or |(100-|-x)-ix = 70. This equation corresponds to the following conditions : If X stand for the number of dollars in one pocket-book, then 100 -f- x stands for the number of dollars in the other ; that is, one pocket-book contains $ 100 more than the other. The second condition of the prob- lem, obtained from the equation, is : two-thirds of the contents of one pocket-book exceeds one-half of the contents of the other by $70. Therefore the modified problem reads as follows : Two pocket-books contain a certain amount of money, and one contains $ 100 more than the other. If one-third of the contents be removed from the first pocket-book, and one-half of the contents from the second, the first will then contain $ 70 more than the second. How much money is contained in each pocket-book ? 6. These problems show that the required modification of an assump- tion, question, or condition of a problem which has led to a negative result, consists in making the assumption, question, or condition the opposite of what it originally was. Thus, if a positive result signify a distance toward the right from a certain point, a negative result will signify a distance toward the left from the same point ; and vice versa ; etc. Zero Solutions. 7. A zero result gives in some cases the answer to the question ; in other cases it proves its impossibility. Pr. A merchant has two kinds of wine, one worth | 7.25 a gallon, and the other f 5.50 a gallon. How many gallons of each kind must be taken to make a mixture of 16 gallons worth $ 88 ? Let X stand for the number of gallons of the first kind ; then 16 — cc will stand for the number of gallons of the second kind. INTERPRETATION OF SOLUTIONS. 199 Therefore, by the condition of the problem, we have 7.25 X + 5.5 (16 - k) = 88 ; whence a; = 0. That is, no mixture which contains the first kind of wine can be made to satisfy the condition. In fact, 16 gallons of the second kind are worth $88. Indeterminate Solutions. 8. Pr. A merchant buys 4 pieces of goods. In the second piece there are 3 yards less than in the first, in tlie third 7 yards less than in the first, and in the fourth 10 yards less than in the first. The number of yards in the first and fourth is equal to the number of yards in the second and third. How many yards are there in the first piece ? Let % stand for the number of yards in the first piece ; then the num- ber of yards in the second piece is x - 3 ; in the third piece, a; - 7 ; in the fourth piece, x- 10. Therefore, by the condition of the problem, we have a;+(a;- 10) = (a;-3) + (!c-7), or 2a; - 10 =2a; - 10. This equation is an identity, and is therefore satisfied by any finite value of X. If it be solved in the usual way, we obtain (2 - 2)a; = 10 - 10, or x = lij^i? - 2. 2-2 That is, the conditions of the problem will be satisfied by any number of yards in the first piece. Infinite Solutions. 9. Pr. A cistern has three pipes. Through the first it can be filled in 24 minutes ; through the second in 36 minutes ; through the third it can be emptied in 14| minutes. In what time will the cisteru be filled if all the pipes be opened at the same time ? Let X stand for the number of minutes after which the cistern will be filled. In one minute ^j of its capacity enters through the first pipe, and hence in x minutes — of its capacity enters. Por a similar reason, — of 24 '36 Its capacity enters through the second pipe in x minutes ; and in the same time — of its capacity is discharged through the third pipe. Therefore, after x minutes there is in the cistern ^ + 36~72' =(A + ^.-A)»:- of its capacity. But by the condition of the problem, that the cistern is then filled, we have 1 1 whence ^ + -h-^ 200 ALGEBRA. [Ch. XII This result means that the cistern will never be filled. This is also evi- dent from the data of the problem, since the third pipe in a given time discharges from the cistern as much as enters it through the other pipes. The Problem of the Couriers. 10. Pr. Two couriers are traveling along a road in the direction from Mto N; one courier at the rate of wii miles an hour, the other at the rate of ma miles an hour. The former is seen at the station A at noon, and the other is seen h hours later at the station B, which is d miles from A in the direction in which the couriers are traveling. Where do the couriers meet ? A B -o- M Ci Ca Ci N Fig. 3. Assume that they meet to the right of JS at a point Ci, and let x stand for the number of miles from B to the place of meeting C\ (Fig. 3). The first courier, moving at the rate of mi miles an hour, travels d + x miles, from A to Ci, in hours ; the second courier, moving at the mi rate of m^ miles an hour, travels x miles, from B to Ci, in — hours. By m d and 7»2 > »»i, or when hmi < d and m^ < mi. A positive result means that the problem is possible with the assumption made ; i.e., that the couriers meet at a point to the right of B. (ii.) A Negative Result. — The result will be negative either when Ami > d and m2 < mi, or when hmi < d and m2 > mi. Such a result shows that the assumption that the couriers meet to the right of B is untenable, since, as we have seen, in that case the result is positive. That under the assumed conditions the couriers can meet only at some point to the left of B can also be inferred from the following considera- tions, which are independent of the negative result : If hmi > (I, the first courier has passed B when the second courier is seen at that station ; that INTERPRETATION OF SOLUTIONS. 201 is, the second courier is behind the first at that time. And since also m2 < TOi, the first courier is traveling the faster, and must therefore have overtaken the second, and at some point to the left of B. On the other hand, if Jimi < d, the first courier has not yet reached B when the second is seen at that station ; that is, the first courier is behind the second at that time. And since also m^ > mi, the second courier is traveling the faster, and must therefore have overtaken the first, at some point to the left of B. Similar reasoning could have been applied in (i.). (iii.) A Zero Result. — A zero result is obtained when hmi = d, and m2 =jfc mi ; that is, the meeting takes place at B. This is also evident from the assumed conditions. For the first courier reaches B h hours after he was seen at A ; and since the second courier is seen at B, h hours after the first was seen at A, the meeting must take place at B. (iv.) Indeterminate Result. — An indeterminate result is obtained if hmi = d, and toz = mi. In this case every point of the road can be regarded as their place of meeting. For the first courier evidently reaches B at the time at which the second courier is seen at that station ; and since they are traveling at the same rate, they must be together all the time. The problem under these conditions becomes indeterminate. (v.) An Infinite Result. — An infinite result is obtained when h'mi=^d, and rrii = mi. In this case a meeting of the couriers is impossible, since both travel at the same rate, and when the second is seen at B the first either has not yet reached B or has already passed that station. An infinite result also means that the more nearly equal mi and wa are, the further removed is the place of meeting. EXERCISES. Solve the foUovring problems, and interpret the results. Modify those problems which have negative solutions so that they will be satisfied by positive solutions. 1. A and B together have $ 100. If A spend one-third of his share, and B spend one-fourth of his share, they will then have $ 80 left. What are their respective shares ? 2. In a number of two digits, the digit in the tens' place exceeds the digit in the units' place by 5. If the digits be interchanged, the resulting number will be less than the original number by 45. What is the number ? 3. A father is 40 years old, and his son is 13 years old ; after how many years will the father be four times as old as his son ? 4. The sum of the first and third of three consecutive even numbers is equal to twice the second. What are the numbers ? 202 ALGEBRA. [Ch. XH 6. In a number of two digits, the tens' digit is two-thirds of the units' digit. If the digits be interchanged, the resulting number will exceed the original number by 36. What is the number ? 6. A father is 26 years older than his son, and the sum of their ages is 26 years less than twice the father's age. How old is the son ? 7. A teacher proposes 30 problems to a pupil. The latter is to receive 8 marks in his favor for each problem solved, and 12 marks against him for each problem not solved. If the number of marks against him exceed those in his favor by 420, how many problems will he have solved ? 8. In a number of two digits the tens' digit is twice the units' digit. If the digits be interchanged, the resulting number will exceed the original number by 18. What is the number ? 9. In a number of two digits, the digit in the units' place exceeds the digit in the tens' place by 4. If the sum of the digits be divided by 2, the quotient will be less than the first digit by 2. What is the number ? 10. A has $ 100, and B has $ 30. A spends twice as much money as B, and then has left three times as much as B. How much does each one spend ? Discuss the solutions of the following general problems. State under what conditions each solution is positive, negative, zero, indeterminate, or infinite, Also, in each problem, assign a set of particular values to the general numbers which will give an admissible solution. 11. In a number of two digits, the tens' digit is m times the units' digit. If the digits be interchanged, the resulting number will exceed the original number by n. What is the number 1 12. A father is a years old, and his son is 6 years old. After how many years will the father be n times as old as his son 1 13. What number, added to the denominators of the fractions - and -, will make the resulting fractions equal ? 14. Having two kinds of wine worth a and 6 dollars a gallon, respec- tively, how many gallons of each kind must be taken to make a mixture of n gallons worth c dollars a gallon ? 15. Two couriers, A and B, start at the same time from two stations, distant d miles from each other, and travel in the same direction. A travels n times as fast as B. Where will A overtake B ? CHAPTER XIII. SIMULTANEOUS LINEAR EQUATIONS. § 1. SYSTEMS OF EQUATIONS. 1. If the linear equation in two unknown numbers x-^y = 5 (1) . be solved for y, we obtain y = 5 —X. This value of y is not definitely determined. We may sub- stitute in it any particular numerical value for x, and obtain a corresponding value for y. Thus, when a; = 1, 2/ = 4; when a- = 2, / = 3 ; when x=3, y = 2; etc. In like manner the equation could have been solved for x in terms of y, and corresponding sets of values obtained. Any set of corresponding values of x and y satisfies the given equation, and is therefore a solution. An equation which, like the above, has an indefinite number of solutions, is called an Indeterminate Equation. 2. The equation y — x = l (2) also has an unlimited number of solutions. Solving this equar tion for y, we have y = 1 -\-x. Then, when x = l, y = 2; when x = 2,y = 3; when a; = 3, y = 4 ; etc. Now, observe that equations (1) and (2) have the common solution, a; = 2, 2/ = 3. It seems evident, and we shall later prove, that these equations have only this solution in common. Equations (1) and (2) express different relations between the unknown numbers, and are called Independent Equations. Also, since they are satisfied by a common set of values of the imknown numbers, they are called Consistent Equations. 203 204 ALGEBRA. [Ch. XIII 3. The equations x + y = 5 and 3x + 3y = 16 are not satis- fied by any common set of values of x and y. For any set of values which reduces x + y to 5 must reduce 3x + 3y, OT 3 (x -{- y), to 15, and not to 16. These two equa- tions express inconsistent relations between the unknown numbers, and are called Inconsistent Equations. 4. The three equations x + y = 5{l), y-x = l{2), 2x + y = 9(3), are not satisfied by any common set of values of x and y. For, by Art. 2, equations (1) and (2) are satisfied by the values x = 2, y = 3. But equation (3) is evidently not satisfied by this set of values. The three equations express three inde- pendent relations between x and y. 5. A System of Simultaneous Equations is a group of equa- tions which are to be satisfied by the same set, or sets, of values of the unknown numbers. A Solution of a system of simultaneous equations is a set of values of the unknown numbers which converts all of the equations into identities; that is, which satisfies all of the equations. The examples of Arts. 1-4 are illustrations of the following general principles, which will be proved later : A system of linear equations has a definite number of solutions, (i.) When the number of equations is the same as the number of unknown numbers. (ii.) When the equations are independent and consistent. EXERCISES I. 1. Are equivalent equations consistent ? Are they independent ? Whicli of the following systems have inconsistent equations ? Which have equations not independent ? Which have equations consistent and independent ? (2x + Zy = ^, _ f3a;4-10{/ = 42, l4 = '3x-f62/ = ll, f2ii;-F32/ = 4, .4x + 7v=15. ■ \4tx-Qy = %. is- 9!/ = 4, r 18 a; -15 2/ = 51, 4x-62/ = 9. 'I 6a;- 52/ = 17. 6 a; 4- 20 2/ = 84. 5 a; -1-4 2/= 6, 7 a: -)- 6 2/ = 10. §2] EQUIVALENT SYSTEMS. 205 §2. EQUIVALENT SYSTEMS. 1. Two systems of equations are equivalent when every solution of either system is a solution of the other. E.g., the systems (I.) and (II.) : 3x + 2y = 8,) ,j, 3x + 2y = 8, x-y = l,j ^ '^ 2x-2y=z2,j (II.) are equivalent. For they are both satisfied by the solution, x = 2, y = l, and, as we shall see later, by no other solution. 2. The solution of a system of equations depends also upon the following principles of the equivalence of systems : (i.) If any equation of a system be replaced by an equivalent equation, the resulting system will be equivalent to the given one. Thus, the systems (I.) and (II.) above are equivalent. The equation x — y = 1 of (I.) is replaced by the equivalent equa- tion 2 a; - 2 3/ = 2. (ii.) If any equation of a system be replaced by an equation obtained by adding or subtracting corresponding members of two or more of the equations of the system, the resulting system will be equivalent to the given one. Thus, the system (II.) is equivalent to the system 5x + 2y= 8, (3 a; + 2 2/) + (2 a; -2 2/)= 10, , Sx + 2y= 8, 1 or ' i) ' 5a! =10 (III.) (iii.) If one equation of a system be solved for one of the un- known members, and the resulting value be substituted for this unknown number in each of the other equations, the derived system will be equivalent to the given one. E.g., the systems ^-y= ^>\ (IV-) and x = 2 + y,- 5x-3y = 12,j ^ -^ 6(2 + 2/) -32/ ==12, / are equivalent. (V.) 206 ALGEBRA. [Ch. Xlll The proofs of the principles enunciated are as follows : (i.) Let A = 3, and C=D, (I.) be two. equations in two unknown numbers, say x and y ; and let C = D be equivalent to G = D. Then the system A = B, and 0' = D', (11.) is equivalent to the system (I.). For, by definition of equivalent equa^ tions, the same sets of values which satisfy C = D also satisfy C = D', and vice versa. .Therefore, any one of these sets of values, which also satisfies A = B, is a. solution of both systems. Consequently, every solu- tion of either system is a solution of the other. In like manner, the principle can be proved for a system of any num ber of equations. (ii.) The proof is very similar to that of Ch. IV., § 3, Art. 5 (i.). (iii.) Let A= B, (1) and C = D, (2) (L) be two equations in two unknown numbers, say x and y ; and let a; = P be the equation derived by solving (1) for x, and C = D' be the equation obtained by substituting P for x in (2). Then the system a;=P, (3) and G' = JD', (i) (IL) is equivalent to the system (I.). Since equation (3) is equivalent to equation (1), any solution of the system (1) must satisfy equation (8); that is, must give to x and P one and the same value. But (4) differs from (2) only in having P where (2) has X. Therefore, since x and P have the same value, any value of x, with the corresponding value of y, which makes C and Z> equal must make G' and D' equal. Therefore, every solution of the system (I.) is a solution of the system (II.). Since equation (1) is equivalent to equation (3), any solution of the system (II.) must satisfy equation (1); that is, must make A and B equal. But (2) differs from (4) only in having x where (4) has P. Therefore, since any solution of (II.) makes x and P equal and C and D' equal, it must also make G and D equal. Therefore, every solution of the system (II.) is a solution of the system (I.). Consequently, the two systems are equivalent. In like manner, the principle can be proved for a system of any num- ber of equations. 3. Elimination is the process of deriving from two or more equations of a system an equation with one less unknown num- ber than the equations from which it is derived. The unknown number which does not appear in the derived equation is said to have been eliminated. §3] SYSTEMS OF LINEAR EQUATIONS. 207 E.g., if the equations x -\-y = 1, CB - 2/ = 1, be added, we obtain 2 a; = 8, in which the unknown number y does not appear. We say that y has been eliminated from the given equations. §3. SYSTEMS OP LINEAR EQUATIONS. Linear Equations in Two Unknown Numbers. 1. There are several methods for solving two simultaneous equations in two unknown numbers. The object in all of them is to obtain from the given system an equivalent system of which one equation contains only one unknown number. (I-) (11.) Elimination by Addition and Subtraction. 2. Ex. 1. Solve the system 3 a; + 4 2/ = 24, (1) 5 SB -62/ = 2. (2). To eliminate x, we multiply both members of equation (1) by 0, and both members of equation (2) by 3, thereby making the coefficients of x in the two equations equal. We then have 15 a; + 20 2/ = 120, (3) 15 a; -18 2/ = 6. (4).' The system (II.) is equivalent to the system (I.), by § 2, Art. 2 (i.). The system (II.) is, by § 2, Art. 2 (i.) and (ii.), equivalent to the system 3a; + 42/ = 24, (1)] (15a; + 20^) -(15a!-182/) = 120 -6; (5)^ or, performing the indicated operations, to 3a! + 42/ = 24, (1)1 3a; + 42/ = 24, (1)1 38 2/ = 114; (6)1 (^^-^ °^ 2/ = 3. (7)1 ^^"^ The system (V.) gives the required solution, since equation (7) gives the value of y, and equation (1) the corresponding value 1} <"■> 208 ALGEBRA. [Ch. XIII of X, by § 2, Art. 2 (iii.). Substituting 3 for y in (1), we obtain 3 a; + 12 = 24 ; whence a; = 4. Consequently the required solution is a; = 4, j/ = 3. This solution may be written 4, 3, it being understood that the first number is the value of x, and the second the value of y. The work has been given in full to emphasize that by each step one system has been replaced by an equivalent system. In practice the work may be contracted as follows : Multiplying (1) by 5, 15 a; + 20 «/ = 120. (3) Multiplying (2) by 3, 15x-l^y = 6. (4) Subtracting (4) from (3), 38 ?/ = 114 ; (6) whence 2/ = 3. (7) Substituting 3 for 2/ in (1), 3 a; + 12 = 24 ; (8) whence a; = 4. (9) In a similar way y could have been first eliminated. Ex. 2. Solve the system I±^_2^^ = 33/-5, (1) o 4 4a;-3 ^5y-7^.^3_g^_ ^2) 6 2 Clearing (1) and (2) of fractions, 28 + 4a; -10a; + 5 2/ = 602/ -100, (3) 4a;-3 + 152/-21 = 108-30a;. (4) Transferring and uniting terms, 6a; + 562/ = 128, (5) 34 a; + 15 2/ = 132. (6) Multiplying (5) by 3, 18 a; + 165 y = 384. (7) Multiplying (6) by 11, 374 x + 165y = 1462. (8) Subtracting (7) from (8), 356 x = 1068 ; (9) whence a; = 3. * (10) §3] SYSTEMS OF LINEAR EQUATIONS. 209 Substituting 3 for x in (5), 18 + 66 ?/ = 128 ; (11) whence 2/ = 2. (12) Consequently, the required solution is 3, 2. Notice that (1) and (2), (3) and (4), (6) and (6), (7) and (8), and (10) and any preceding equation except (9), form equiva- lent systems. In forming with (10) an equivalent system, we take the simplest of the preceding equations, in this case (6). 3. The examples of the preceding article illustrate the fol- lowing method of elimination by addition and subtraction. Simplify the given equations, if necessary, and transfer the terms in x and y to the first members, and the terms free from X and y to the second members. ■ Determine the L. C. M. of the coefficients of the unknown num- ber to be eliminated, and multiply both members of each equation by the quotient of the L. G. M. divided by the coefficient of that unknown number in the equation. The coefficients of the unknown number to be eliminated being now equal, or equal and opposite, in the two equations, subtract, or add, corresponding members, and equate the results. A final equation in one unknown number will thus be derived. The Solution of the given system is then obtained by solving this derived equation, and substituting the value of the unknown num- ber thus obtained in the simplest of the preceding equations. EXERCISES II. Solve the following systems of equations by ttie method of addition and subtraction : ' x-i-y = a, x — y = h. ■3x+ 2/ = 31, 5x-2y = 16. nx — ay = 0, n^x — ay = an. 6x + iy = 4Qi, _2x + 7y = 63. 1. 4. 10. x + y = n, x — y = 7. x-12y = 3, x+ 4 2/ = 19. 10x-3y = 25, bx-9y=-26. f 12 a; -I- 15 2/ = 8, 16a! -h 92/ = 7. 8. 11. 3. 6. 9. 12. 7x + ny = 2, 7a;- 112/ = 0. 4x-7y = 19, x + 9y = 3'J. 6a; + 2/ = 6, ix + 3y = n. 5a: -32/ = 12, 19a;-52/ = 73J. 210 ALGEBKA. [Ch. XIII 13. 15. 17. \ -5x + '. ley = 5, ■28^ = 19. fl8x- 202^=1, [15x+lQy = Q. ^^-^^ + 14 = 18, 2 ^^ + ^ + 16 = 19. 3 19. X m — a X ■ + = 1. n ■ + m — b y—=i. 14. 16. 18. 20. a n — b 21 »+ 82/ =-66, 28 K - 23 2/ = 13. ' 12 a; - 14 2/ = - 4, 8a; -21 2/ = -8.5. 3x + ^ = 22, ll2/- — = 20. ^ 5 ax — by _ J a2 + 62 ' Sx + g;/ _ ^ a2 + 62 Elimination by Substitution. 4. Ex. Solve the system 5x — 2y = l, (1) 4a; + 52/ = 47. (2) J If we wish to eliminate x, we proceed as follows : Solving (1) for x, x= ^ - 5 Substituting """^ ^ for a; in (2), (3) 4(^l±2l) + 5 2, = 47. (4) (I-) (II.) (III.) The system (II.) is equivalent to the system (I.), by § 2, Art. 2 (iii.). Solving (4) for y, 2/ = 7. (5) Substituting 7 for y in (3), a; = 3. (6) . The system (III.) is, by § 2, Art. 2 (iii.), equivalent to the system (II.), and hence to the given system. Therefore the required solution is 3, 7. In a similar way y could have been first eliminated. 5. The example of the preceding article illustrates the following method of elimination by substitution: 8olve the simpler equation for the unknown number to be eliminated in terms of the other, and substitute the value thus 3] SYSTEMS OF LINEAR EQUATIONS. 211 obtained in the other equation. The derived equation mil contain hut one unknown number. The solution of the given system is then obtained by solving the derived equation, and substituting the value of the unknown num- ber thus obtained in the esspression for the other unknown number. EXERCISES III. Solve the following systems of equations by the method of substitution : 4. 7. 10. 13. 16. x = 2y-3, 2/=2x-15. 3. ix = 3y-1, y = 3x-19. 3. a!-4=K2/+6). liy-3x = 2. 5 \x + y = a, g 1x = 6y, y = 14a;. x = ny. 15?/ = 28a; -70 5x = 8y-n, 6y = 7x-21. 8. i Tx-3 = hy, g ,7 2/- 3 = 8a;. ay = bx, a + y = b + x. x-3y = 0, 11 r3a;+ 42/ = 2, ^^ ■5a; + 7 2/ = 49, ,25 a; + 48 2/ = 287. 9a; + 202/ = 8. i 7a; + 52/ = 47. r fa; = 10-12/, 14 [^x + ^y^n, ^^ (ix-iy = 2. 4|2/ = 5a;-7. 2 a; + 3 2/ = 43. 2a; + 32/ = 60. [■ 7 + x 2x — y_ 5 4 5y-7 . 4a;-3 2 6 = 32,- = 18- 5, 17. 5 a;. ^ + y -2. a + h a — b 18. (a; - a) (o + &) = (a - &) (,y - a), X _ y Elimination by Comparison. 6. Ex. Solve the system 7a! + 2j/ = 20, (1)- 13x-3y=n. (2), To eliminate y, we proceed as follows : 20 -7 a; Solving (1) for y, y ='■ Solving (2) for y, y = - 13 a; - 17 (3) (4) (I-) (II.) 212 ALGEBRA. [Ch. XIII (III.) The system (II.) is equivalent to the system (I.), by § 2, Art. 2 (i.). Substituting in (4) for y, its value given in (3), 20 - 7 » 13 a; - 17 ,^, The last equation and equation (3) form a system which is equivalent to the system (II.), and hence to the given system. Solving (6) for x, x = 2. Substituting 2 for x in (3), y = 3. . The system (III.) is equivalent to the system formed by equations (3) and (5), and therefore to the given system. Con- sequently the required solution is 2, 3. In a similar way, y could have been first eliminated. 7. The example of the preceding article illustrates the fol- lowing method of elimination by comparison : Solve the given equations for the unknmvn number to be elimi- nated, and equate the expressions thus obtained. The derived equation will contain but one unknown number. The solution of the given system is then obtained by solving this derived equation, and substituting the value of the unknown numr ber thus obtained in the simplest of the preceding equations. EXEECISBS IV. Solve the following systems of equations by the method of comparison : 1. 10, x = 3y-2, X = y — 12. 5a; = 72/ — .1, 7x = 9y+ 1.7. lix = 7y-38, liy = Tx-72. 1+72/ = 99, I + 7a; = 51. 5. [2/ = 3 a; -17, (y = 2x-10. ix = iy-l, i2/ = Jx-2. ■ 5 a; + 9 2/ = 28, 7x + 3y^20. 7 11. 2 3 3 2 :0, :0. 3. 9. 12. 52/ = 2a;+l, 82/ = 5 a;- 11. 2ix- Hy = 10, n\x- Hy = 55. 21a;- 23 2/ = 2, 7a;- 19 2/ = 12. x.y 2 "^6 = 11 X , y 5 24 _7 2 § 3] SYSTEMS OF LINEAR EQUATIONS. 213 13. 8x + Qy = 26, f 63 a; - 46 « = 29, 14. ] .. . . 15. S2x-3y = 26. [42x-69y = X + ay + 1 = 0, y+c(^x+ 1) = 0. , 5x + iy = 9a — b, 16. ] 17. 7x-6y = a-13b. ax — by = a^ + b^, (a-b)x + (a+ 6)2/ = 2(n2- 62). The General Solution of a System of Two Linear Equations in Two Unknown Numbers. 8. Any linear equation ui two unknown numbers can evidently be brought to the form ax+ by = c. 9. Let aix+biy = Cj, (1) a^x + biy = C2, (2) be any two linear equations. Multiplying (1) by 62, 0162* + 61&2I/ = 62C1. (3) Multiplying (2) by 61, 02610; + btb^y = bid. (4) Subtracting (4) from (3), (ai62 — 0261) x = 62''.i — 61C2 ; (5) whence, if 0162 - aafii ^0, x = ^2°' " ^'"a . (6) ai62 — flafti In like manner, if 0162 - 0261 =^0, y = "iCz - "zd . (7) O162 — O261 But if 0162 — 0261 = 0, we have no authority for dividing both members of equation (5) by 0162 — (Hb\. Consequently, the general solution of two linear equations in two unknown numbers is 62C1 — 61^2 C1C2 — fl2Cl 0162 — flz^i 0162 — fla^i when 0162 — 0261 =^ 0. 10. Every system of two independent and consistent equations of the first degree in two unknown numbers has one, and only one, solution. For the system aix + biy = Ci, (1) a^x + b^y = C2, (2) is, by Art. 9, equivalent to the system (ai62 — a2bi)x = 62C1 — 61C2, (3) (0162 — 0261)2/ = aiC2 — fl!2Cl. (4) But equations (3) and (4) are each linear in one unknown number, and each, therefore, has one, and only one, solution. Consequently, the given system has one, and only one, solution. 214 ALGEBRA. [Ch. XIIl 11. Three independent linear equations in two unknown numbers can- not be satisfied by any common set of values of the unknown numbers. If the values of x and y which constitute the solution of equations (1) and (2) , Art. 9, satisfy a third equation, asx + bzv = Cs, (3) , 62C1 — 61C2 . , fliCz — aaCi we have as x -^^ ~ + 6s x -V ^ = "a. 01O2 — aaOi ai02 — aa&i From this relation, we obtain „ V, "^862 — 0263 , ,, ai^s — asfti Cs = Ci X — r J- + C2 X wherein I = 0162 — O261 0162 — a^bi Ici + mc2, (i.) 0362 — ^263 -, 0163 — 0361 0162 — a-ibi ~ 0162 — 0261 We also have , ^azbi — aiba 0163 — 0361 63(0162 — 0261) 61 X — J r- + 62 X — 7 r = — ^^ 7 — = 6s- 0162 — 0261 0162 — O261 O162 — O261 That is, 63 = Ibi + mb^. (ii.) In like manner, it can be shown that os = ?oi + mo2. (iii.) Observe that (ii.) and (iii.) are identities, and hold for all values of the o's and 6's which do not reduce 0162 — O261 to ; while (i.) is not an identity, but imposes a condition upon the values of the knovm numbers in the three equations. When this condition is satisfied, Cs is obtained from ci and C2, just as Os is obtained from oi and 02, and 63 from 61 and 62. That is, when the solution of (1) and (2) is also a solution of (3), the last equation is not independent of the other two. Linear Equations in Three or More UnknoTiTU Numbers. 12. The following examples will illustrate the methods for solving systems of three linear equations in three unknown numbers : Ex.1. Solve the system 2x — 5y + 5z = ll, (1) 5a; + 42/-6« = -6, (2) -4a; + 7^-8«=-14. (3) §3] SYSTEMS OF LINEAR EQUATIONS. 215 To eliminate x, we proceed as follows : Multiplying (1) by 5, 10x-Wy + 25z = 55. (4) Multiplying (2) by 2, 10x + 8y-12z = - 10. (5) Subtracting (4) from (5), 23 ?/ - 37 z = - 65. (6) Multiplying (1) by 2, 4 a; - 6 ?/ + 10 z = 22. (7) Adding (3) and (7), y + 2z = 8. (8) Solving (6) and (8), 2/ = 2, z = 3. Substituting 2 for y and 3 for z in (1), x = l. Notice that (1), (2), (3); and (1), (6), (8) form equivalent systems. Consequently the required solution is 1, 2, 3. Ex. 2. Solve the system ay — cz = 0, (1) z-x = -b, (2) ax + by = a' + b(a + c). (3) Notice that by eliminating z from (1) and (2) we obtain an equation in x and y, which with equation (3) gives a system of two equations in the same two unknown numbers. Solving (2) for z, z = x — b. (4) Substituting a; — & for z in (1), ay — cx + cb = 0. (5) Multiplying (3) by a, o?x + aby = a' + a^& + abc. (6) Multiplying (5) by b, — 6ca; + ahy = — 6^c. (7) Subtracting (7) from (6), (a^ + 6c) a; = a' + a=6 + abc + Vc = a2(a + 6) + 6c(a + 6) = (a^ + 6c)(a + 6); (8) whence x = a-\-b. Substituting a-\-b for x in (4), z = a. Substituting a for z in (1), y = c- To solve three simultaneous equations in three unknown num- bers, eliminate one of the unknown numbers from any two of the equations; next eliminate the same unknown number from the 216 ALGEBRA. [Ch. XIIl third equation and either of the other two. Two equations in the same two unknown numbers are thus derived. Solve these equations for the two unknown numbers, and sub- stitute the values thus obtained in the simplest equation which contains the third unknown number. 13. From four equations in four unknown numbers, we can by eliminating one of the unknown numbers obtain three equations in three unknown numbers. We then solve these equations for the three unknown numbers and substitute the values thus obtained in the simplest equation which contains the fourth unknown number. Number of Solutions of a System of Linear Equations. 14. The examples of the preceding articles illustrate the following principles : (i.) A system of n independent and consistent linear equations, in n unknown numbers, has one, and only one, determinate solution. From the given system a system of ra — 1 equations in ji — 1 unknown numbers can he derived hy eliminating one of the unknown numbers. By eliminating from the latter system another unknown number, a second system of n — 2 equations in ji — 2 unknown numbers is derived ; and so on. Finally, a single equation in one unknown number is obtained. By the principles of equivalent equations the given system is equiva- lent to a second system which contains the following equations : any one of the given equations in n unknown numbers, any one of the ra — 1 derived equations in to — 1 unknown numbers, and so on, to any one of the three derived equations in three unknown numbers, either of the two derived equations in two unknown numbers, and the last derived equation in one unknown number. The last equation in one unknown number has one, and only one, definite solution. If the value of this unknown number be substituted in the next to the last equation of the second system described above, one, and only one, definite value for a second unknown number is obtained. If the values of these two unknown numbers be substituted in the equa- tion in three unknown numbers, one, and only one, definite value of a third unknown number is obtained ; and so on. Consequently the given system is satisfied by one, and only one, defi- nite set of values of the unknown numbers. §3] SYSTEMS OF LINEAR EQUATIONS. 217 (ii.) A system of n independent linear equations, in more than n unknown numbers, has an indefinite number of solutions. For, by each elimination of an unknown number, we derive a set of equations, one less in number, and containing one less unknown number. Finally, as in (i.), we obtain a single equation. But since the original system contained more unknown numbers than equations, the last derived equation will contain more than one unknown number. Since this equa- tion, therefore, has an indefinite number of solutions, we conclude that the given system has likewise an indefinite number of solutions. (iii.) A system of n independent linear equations, in less than n unknown numbers, does not have a determinate finite solution. For, if we take from the given system as many equations as there are unknown numbers, the system formed by these equations will have by (i.) one, and only one, definite solution. But since the other equations of the given system are Independent of the equations selected, that is, express independent relations between the unknown numbers, they cannot be satisfied by this solution. Therefore, the given system cannot be satisfied by any one definite set of values of the unknown numbers. BXEECISBS V, Solve the following systems of equations : - 05 + 2^ = 28, 1. 11. a: + 3 = 30, 2. 3/ + z = 32. 3 a; -2/ = 7, 3 y — 2 = 5, 5. Zz -x = 0. 3x + 2y-4:Z = l5, 5x-3y + 2z = 28, Sy + 4z-x = 24. X + y — z = c, x + z-y = b, y + z — X = a. 2x- 4y+ 9z = 28, 7x+ 3y - 52 = 3, 9x + 10y -llz = i. x + y = 2c, x + z = 2b, (y + z=:2a. 3x + 5y = 35, 3y + 6z = 21, I 3 2 + 5 a; = 34. 3. \y x-y-- x + \ 2, 2 = 3, a; + 2 = 9. a; + 2/ + 2 = 50, y = 3x-2\, 2 = 4 a; - 33. -2 = 1, 10. 12. 8a; + 3?/ -62 = 1, 3z — 4x — y = \. ix-3y + 2z = % 2x + 6y-3z = i, 5x + (iy-2z = 18. x-2y + 3z==6, 2x + 3y -iz = 2Q, 3x-2y + bz = 26. 218 ALGEBRA. [Ch. xin 13. 15. 17. 19. 21. 6 9 10 5 + l?_A = ll, 3 2 25 2 18 10 bx + cy + az = a^ + b^+ c% (xc + ay + bz = a^ + b^ + cK r (c + a) a; - (c - a) 2/ = 2 6c, I (a + 6)t/-(a-6)« = 2ac, I (6 + c) 2 - (6 - c) a; = 2 a6. x + y + z = Q, a + 2/ + M = 7, a; + z + M = 8, 2/ + 2 + M = 9. 7a;-22 + 3« = 17, 4 2/ - 2 s + J = 15, 5j/-3a;-2M = 8, 42/-3m + 2« = 17, 3 3 + 8 w = 33. 14. 16. 18. 20. 22. a + b - + 6 + c z = c + a, 6-c. c — a c + a _K z b — c a — b x + y + z = A, ax+by + cz = 0, a'x + b^ + d'z = 0. x+y+z={a+b+cy, ay+bz+cx=3(ab^+bc^+ca^), I ax+62/ + C2=(i'+6'+c'+6a6c. a;+^ + 3 — M = ll, x + y — z + u = n, x-y+z + u = 9, — x + y + z + u = \2. ■3a;-42/+3z+3B-6« = - 1, 3x-6y+2z-4:u = - 7, 102/-3z + 3m-2?)=35, 5z+iu+2v-2x= 15, 6m-3i;+4x-23/=21. §4. SYSTEMS OF FRACTIONAL EQUATIONS. 1. If some or all of the equations of a system be fractional, and lead, when cleared of fractions, to linear equations, the solu- tion of the system can be obtained by the preceding methods. Any solution of the linear system which is derived by clear- ing of fractions, is a solution of the given system, unless it is a solution of the L. C. D. (equated to 0) of one or more of the fractional equations. (See Ch. X., Art. 4.) Ex. 1. Solve the system Clearing (2) of fractions, 6x — 5y-- 6a; + l. = 0, 13 ll' Ay + S 66x + ll = 52y + 65. (1) (2) (3) §4] SYSTEMS OF FRACTIONAL EQUATIONS. 219 Transferring and uniting terms in (3), and dividing by 2, 33 »- 262/ = 27. (4) The solution of (1) and (4) is 15, 18. In clearing (2) of fractions, we multiplied by 11 (4 2/ + 5). Since x = 15,y = 18 is not a solution of 4 ^ + 5 = 0, it is a solution of the given system. Ex. 2. Solve the system x + y = !cy, (1) 2x + 2z = xz, (2) ■ (I.) 3y + 3z = yz. (3). Observe that the given equations are neither linear nor frac- tional. Yet they can be transformed so that they will contain only the reciprocals of x, y, and z. Dividing (1) by xy, (2) by xz, (3) by yz, we have : l + i = l. (4) ? + ?=l. (6) § + ? = l. (6) (II.) y X z X z y We will solve this system for -, -, — X y z Multiplying (4) by 2, ? + ? = 2. (7) y X Subtracting (5) from (7), ? - ? = 1. (8) Solving (6) and (8) for i and ^, i = :^, - = - :^- ■' y z y 12 z 12 Substituting :^ for - in (4), - = t^- Consequently, a solution of the given system is i^-, ^, — 12. It is important to notice that we cannot assume that the system (II.) is equivalent to the system (I.), since the equa- tions of (II.) are derived from the equations of (I.) by dividing by expressions which contain the unknown numbers. But if any solution of (I.) be lost by this transformation, it must be a solution of the expressions (equated to 0) by which the equations of (I.) were divided ; that is, of xy = 0, xz = 0, yz = 0. (III.) 220 ALGEBRA. [Ch. XIIl The system (III.) has the solution 0, 0, 0, and this solution evidently satisfies the system (I.). We therefore conclude that the gi-ii _ a az + ex _ b ~(a-6)(c-&) ' (4 -x) (244 -!/) = «, (7 -X) (124 -2/) = a, (13 -x) (64 -2/) = 2. f X + ^ + z = 0, 29. 31. 7 + y +- a+b b—c c+a - = 2c, X y a—b b — -+- c c ^ =2 — a a, X y a-b b~ - ^ =2 c c+a a-2c (z + x)a - -(« -x)b-- = 2yz, {x + y)b- -(X -y)c-- = 2xz, (2/ + 3)C - -(^ -z)a-- = 2xy. 32. 33. ■34. 35. (6 + c)x + (c + a)y + (a + b)z = box + cay + abz = 1. r 5x + 7y + 2 3x + 4|M-7_„ ia: + 5j/ + 7 _ 5 *■ 3 7x + 3i7 + 4 4 4x -3j/ + 19 4 5a; -4 2/ + 21 6 8 2a; -32/ 5 + 17 •2J + 9. :», 6 3a;-22/-2 ^ 9 '^^ + 5x-82/ + 39 = 0, + 16 2/ - 10 X = 88J. 36. l2.a;-3 2/ + 17 (a - 6)x + (6 - c)!/ + (c - a)z = 2(a^+ 6^+ c''- ab-ac- 6c), (a -. 6)2/ + (6 - [ X + y + z = 0. 2/z + xz + a;2/ = xyz, yu + xu + xy = xyu, zu + XU+ xz = xuz, zu + yu + yz = uyz. x + y+z + u = 16, x + y + z + v — 18, x + y + u + v = 20, x+ z + u + v = 22, y + z + u + i>=:2i. c)z + (c — a)x = a6 + ac + 6c — a'' — 6? - 38. 40. a* + a'x + a'^y + az -\- u = 0, 6* + b^x + 6^2/ + 6z + M = 0, c* + c^x -1^ 0^2/ + cz + M = 0, d* + (2Sa; + dZj/ + dz + M = 0. 2M-3'u = 2a- 7 6+:;2c, « + 2 z = 7 6, i" 3z+ 2/ = 3a + 6 6, 42/ — 2x = 8a, l-3x — 5m = o — 56 — 5 c. CHAPTER XIV. PROBLEMS. 1. As was stated in Ch. V., Art. 2 (ii.), every problem which can be solved must contain as many conditions, expressed or implied, as there are required numbers. In solving a problem by means of one equation in one unknown number, one of the required numbers was usually taken as the unknown number of the equation. All but one of the conditions of the problem were used to express the other required numbers in terms of the one selected as the unknown number. The remaining con- dition furnished the equation of the problem. But a problem which contains more than one condition can be solved by means of a system of equations in which the un- known numbers are usually the required numbers of the prob- lem. Each condition then furnishes an equation. The solution of the system of equations thus obtained gives the solution of the problem, if the conditions of the latter be consistent. 2. We will first solve by means of a system of two equations one of the problems which was solved in Ch. V. by means of one equation in one unknown number. Pr. 1. (Pr. 4, Ch. V.) At an election at which 943 votes were cast, A and B were candidates. A received a majority of 65 votes. How many votes were cast for each candidate ? Let X stand for the number of votes cast for A, and y for the number of votes cast for B. Then, by the first condition, a54-y = 943; (1) and by the second condition, a; - 2/ = 65. (2) 226 226 ALGEBRA. [Ch. XIV Whence x = 504, the number of votes cast for A, y = 439, the number of votes cast for B. Had we substituted the value of y obtained from (1), namely 943 — a;, for y in (2), we should have obtained the equation of the solution in Ch. V., X - (943 - a;) = 65. Pr. 2. A tank can be filled by two pipes. If the first be left open 6 minutes, and the second 7 minutes, the tank will be filled; or if the first be left open 3 minutes, and the second 12 minutes, the tank will be filled. In what time can each pipe fill the tank ? Let X stand for the number of minutes it takes the first pipe to fill the tank, and y for the number of minutes it takes the second pipe. Let the capacity of the tank be represented by 1. Then in 1 minute the first pipe fills - of the tank, and in f* aj Y 6 minutes - of the tank ; the second pipe fills - of the tank X y in 7 minutes. Therefore, by the conditions of the problem, ^ + 1 = 1; ^ + 15 = 1. X y X y Whence x = V)\, y = 17. BXBBCISES I. 1. Find two numbers whose sum is 19 and whose difference is 7. 2. If one number be multiplied by 3 and another by 7, the sum of the products will be 58 ; if the first be multiplied by 7 and the second by 3, the sum will be 42. What are the numbers ? 3. In a meeting of 48 persons, a motion was carried by a majority of 18. How many persons voted for the motion and how many against it ? 4. If one of two numbers be divided by 6 and the other by 5, the sum of the quotients will be 52 ; if the first be divided by 8 and the second by 12, the sum of the quotients will be 31. What are the numbers ? 5. Find two numbers, such that if 1 be subtracted from the first and added to the second the results will be equal ; while if 5 be subtracted from the first and the second be subtracted from 5, these results will also be equal. PROBLEMS. 227 6. If 45 be subtracted from a number, the remainder will be a certain multiple of 5 ; but if the number be subtracted from 135, the remainder will be the same multiple of 10. What is the number, and what multiple of 5 is the first remainder ? 7. If 1 be added to the numerator of a fraction, the resulting fraction will be equal to i ; but if 1 be added to the denominator, the resulting fraction will be equal to ^. What is the fraction ? 8. If 1 be subtracted from the numerator and denominator of a cer- tain fraction, the resulting fraction will be equal to J ; but if 1 be added to the numerator and denominator of the same fraction, the resulting fraction will be equal to ^. What is the fraction ? 9. A said to B : " Give me three-fourths of your marbles and I shall have 100 marbles." B said to A: "Give me one-half of your marbles and I shall have 100 marbles." How many marbles had A and B ? 10. A bag contains white and black balls. One-half of the number of white balls is equal to one-third of the number of black balls, and twice the number of white balls is 6 less than the total number of balls. How many balls of each color are there ? 11. The sum of two numbers is 47. If the greater be divided by the less, the quotient and the remainder will each be 5. What are the numbers ? 12. A father said to his son : " After 3 years I shall be three times as old as you will be, and 7 years ago I was seven times as old as you then were." What were the ages of father and son ? 13. A merchant received from one customer $26 for 10 yards of silk and 4 yards of cloth ; and from another customer 1 23 for 7 yards of silk and 6 yards of cloth at the same prices. What was the price of the silk and of the cloth ? 14. A merchant has two kinds of wine. If he mix 9 gallons of the poorer with 7 gallons of the better, the mixture will be worth 1 1.37 J a gallon ; but if he mix 3 gallons of the poorer with 5 gallons of the better, the mixture will be worth $ 1.45 a gallon. What is the price of each kind of wine ? 15. A man has a gold watch, a silver watch, and a chain. The gold watch and the chain cost seven times as much as the silver watch ; the cost of the chain and half the cost of the silver watch is equal to three- tenths of the cost of the gold watch. If the chain cost $ 40, what was the cost of each watch ? 16. A and B make a purchase for $ 48. A gives all of his money, and B three-fourths of his. If A had given three-fourths of his money and B all of his, they would have paid $1.50 less. How much money had A and B? 228 ALGEBRA. [Ch. XIV 17. A mechanic and an apprentice together receive $40. The me- chanic works 7 days and the apprentice 12 days ; and the mechanic earns in 3 days $7 more than the apprentice earns in 5 days. What wages does each receive ? 18. I have 7 silver halls equal in weight and 12 gold halls equal in weight. If I place 3 silver halls in one pan of a balance and 5 gold balls in the other, I must add to the gold balls 7 ounces to maintain equilibrium. If I place in one pan 4 silver balls and in the other 7 gold balls, the balance is in equilibrium. What is the weight of each gold and of each silver ball ? 19. A tank has two pumps. If the first be worked 2 hours and the second 3 hours, 1020 cubic feet of water will be discharged. But if the first be worked 1 hour and the second 2J hours, 690 cubic feet of water will be discharged. How many cubic feet of water can each pump dis- charge in one hour ? 20. It was intended to distribute $ 25 among a certain number of the poor, each adult to receive f 2.50 and each child 75 cents. But it was found that there were 3 more adults and 5 more children than was at first supposed. Each adult was therefore given 1 1.75 and each child 50 cents. How many adults and how many children were there ? 21. A man ordered a wine-merchant to fill two casks of different sizes with wine, one at 1 1.20 and the other at $ 1.50 a quart, paying 1 88.50 for both casks of wine. By mistake the casks were interchanged, so that the purchaser received more of the cheaper wine and less of the dearer. The merchant therefore returned to him $1.50. How many quarts did each cask hold ? 22. A and B jointly contribute $ 10,000 to a business. A leaves his money in the business 1 year and 3 months, and B his money 2 years and 11 months. If their profits be equal, how much does each contribute ? 23. A merchant sold 12 gallons from each of two full casks of wine, and then found that the larger contained twice as much as the smaller. After he had sold more wine from both casks, he found that each one contained one-third of its original capacity. If he had then added i gallons of wine to each cask, the contents of the smaller would have been three-fourths of the contents of the larger. What was the capacity of each cask ? 24. One boy said to another: "Give me 5 of your nuts, and I shall have three times as many as you will have left." "No," said the other, " give me 2 of your nuts, and I shall have five times as many as you will have left." How many nuts had each boy ? 25. A father has two sons, one 4 years older than the other. After 2 years the father's age will be twice the joint ages of his sons ; and 6 years ago his age was six times the joint ages of his sons. How old is the father and each of his sons ? PROBLEMS. 229 26. If a number of two digits be divided by the sum of the digits, the quotient will be 7. If the digits be interchanged, the resulting number will be less than the original number by 27. What is the number ? 27. A man walks 26 miles, first at the rate of 3 miles an hour, and later at the rate of 4 miles an hour. If he had walked 4 miles an hour when he walked 3, and 3 miles an hour when he walked 4, he would have gone 4 miles further. How far would he have gone, if he had walked 4 miles an hour the whole time ? 28. Two trains leave different cities, which are 650 miles apart, and run toward each other. If they start at the same time, they will meet after 10 hours ; but if the first start 4J hours earlier than the second, they will meet 8 hours after the second train starts. What is the speed of each train ? 29. If the base of a rectangle be increased by 2 feet, and the altitude be diminished by 3 feet, the area will be diminished by 48 square feet. But if the base be increased by 3 feet, and the altitude be diminished by 2 feet, the area will be increased by 6 square feet. Find the base and the alti- tude of the rectangle. 30. A number of three digits is in value between 400 and 500, and the sum of its digits is 9. If the digits be reversed, the resulting number will be If of the original number. What is the number ? 31. The report of a cannon travels with the wind 344.42 yards a second, and against the wind 335.94 yards a second. What is the velocity of the report in still air, and what is the velocity of the wind ? 32. Two messengers, A and B, travel toward each other, starting from two cities which are 805 miles distant from each other. If A start 5| hours earlier than B, they will meet 6| hours after B starts. But if B start 5f hours earlier than A, they will meet 5|- hours after A starts. At what rates do A and B travel ? 33. Each pi two servants was to receive 1 160, a dress, and a pair of shoes for one year's services. One servant left after 8 months, and re- ceived the dress and .f 106 ; the other servant left after 9J months, and received a pair of shoes and 1 142. What was the value of the dress, and of the pair of shoes ? 34. On the eve of a battle, one army had 5 men to every 6 men in the other. The first army lost 14,000 men, and the second lost 6000 men. The first army then had 2 men to every 3 men in the other. How many men were there originally in each army ? 35. If the sum of two numbers,'each of three digits, be increased by 1, the result will be 1000. If the greater be placed on the left of the less, and a decimal point be placed between them, the resulting number will V 230 ALGEBRA. [Ch. XIV six times tlie number obtained by placing the smaller number on the left of the greater, with a decimal point between them. What are the num- bers 1 36. A vessel sails 110 miles with the current and 70 miles against the current in 10 hours. On a second trip, it sails 88 miles with the current and 84 miles against the current in the same time. How many miles can the vessel sail in still water in one hour, and what is the speed of the current ? 37. A and B run a race of 400 yards. In the first heat A gives B a start of 20 seconds, and wins by 50 yards. In the second heat A gives B a start of 125 yards, and win&by 5 seconds. What is the speed of each runner ? 38. A merchant had two casks containing different quantities of wine. He poured from the first cask into the second as much wine as was in the second ; next he poured from the second cask into the first as much wine as was left in the first ; finally he poured from the first cask into the sec- ond as much wine as was left in the second. Each cask then contained 80 quarts. How many gallons did each cask originally contain ? 39. A and B formed a partnership. A invested 1 20,000 of his own money and $5000 which he borrowed; B invested $22,000 of his own money and $ 8000 which he borrowed at the same rate of interest as was paid by A. At the end of a year, A's share in the profits amounted to $ 1750 more than the interest on his $ 5000, and B's share to ? 2000 more than the interest on his •$ 8000. What rate per cent interest did they pay, and what rate per cent did they realize on their investments ? • 40. Two bodies move along the circumference of a circle in the same direction from two different points, the shorter distance between which, measured along the circumference, is 160 feet. One body will overtake the other in 32 seconds, if they move in one direction ; or in 40 seconds, if they move in the opposite direction. While the one goes once around the circumference, the distance passed over by the other exceeds the circum- ference by 45 feet. What is the circumference of the circle, and at what rates do the bodies move ? 41. A number of workmen, who receive the same wages, earn together a certain sum. Had there been 7 more workmen, and had each one re- ceived 25 cents more, their joint earnings would have increased by $ 18.65. Had there been 4 fewer workmen, and had each one received 16 cents less, their joint earnings would have decreased by f 9.20. How many workmen are there, and how much does each one receive ? 42. A courier rode from A toward B, which is 64 miles distant from .4. Five hours after his departure, a second courier started from B and rode PROBLEMS. 231 toward A. The couriers met 7 hours after the second courier started. It the second courier had started from E 2 hours before the iirst started from A, they would have met 8 hours after the second courier started. At what rate did each courier ride ? 43. A farmer has enough feed for his oxen to last a certain number of days. If he were to sell 75 oxen, his feed would last 20 days longer. If, however, he were to buy 100 oxen, his feed would last 15 days less. How many oxen has he, and for how many days has he enough feed ? 44. An alloy of tin and lead, weighing 40 pounds, loses 4 pounds in weight when immersed in water. Find the amount of tin and lead in the alloy, if 10 pounds of tin lose If pounds when immersed in water, and 5 pounds of lead lose .375 of a pound. 45. Two men were to receive $ 96 for a certain piece of work, which they could do together in 30 days. After half of the work was done, one of them stopped for 8 days, and thtn the other stopped for 4 days. They finally completed the work in 35J days. How many dollars should each one receive, and in what time could each one have done the work alone ? 46. Two boys, A and B, run a race from Pto Q and return. A, the faster runner, on his return meets B 90 feet from Q, and reaches P 3 minutes ahead of B. If he had run again to Q, he would have met B at a distance from P equal to one-sixth of the distance from P to Q. How far is Q from P, and how long did it take B to run from P to Q and return ? 47. It took a certain number of workmen 6 hours to carry a pile of stones from one place to another. Had there been 2 more workmen, and had each one carried 4 pounds more at each trip, it would have taken them 1 hour less to complete the work. Had there been 3 fewer workmen, and had each one carried 5 pounds less at each trip, it would have taken them 2 hours longer to complete the work. How many workmen were there, and how many pounds did each one carry at every trip ? 48. Three carriages travel from A to B. The second carriage travels every 4 hours 1 mile less than the first, and is 4 hours longer in making the journey. The third carriage travels every 3 hours If miles more than the second, and is 7 hours less in making the journey. How far is B from A, and how many hours does it take each carriage to make the journey 1 49. Water enters a basin through one pipe and is discharged through another. Through the first pipe four more gallons enter the basin every minute than is discharged through the second. When the basin is empty, both pipes are opened, the first one hour earlier than the second, and after a certain time the basin contains 1760 gallons. The pipe through which water enters is then closed, and after one hour is again opened. If 232 ALGEBRA. [Ch. XIV both pipes be then left open as long as they were open together in the former case, the basin will contain 880 gallons. In what time can the one pipe fill the basin and the other empty it, if it hold 1760 gallons ? 50. A body moves with a uniform velocity from a point ^ to a point B, which is 323 feet distant from A, and without stopping returns at the same rate from iJ to ^. A second body leaves B 13 seconds after the first leaves A, and moves toward A with a uniform but less velocity than the velocity of the first. The first body meets the second 10 seconds after the latter starts, and on returning to A overtakes the second body 45 seconds after the latter starts. What is the velocity of each body ? 51. A fox pursued by a dog is 60 of her own leaps ahead of the dog. The fox makes 9 leaps while the dog makes 6, but the dog goes as far in 3 leaps as the fox goes in 7. How many leaps does each make before the dog catches the fox ? 52. The sum of the three digits of a number is 14 ; the sum of the first and the third digit is equal to the second ; and if the digits in the units' and in the tens' place be interchanged, the resulting number will be less than the original number by 18. "What is the number ? 53. The sum of the ages of A, B, and C is 69 years. Two years ago B's age was equal to one-half of the sum of the ages of A and C, and 10 years hence the sum of the ages of B and C will exceed A's age by 31 years. What are the present ages of A, B, and C ? 64. The total capacity of three casks is 1440 quarts. Two of them are full and one is empty. To fill the empty cask it takes all the contents of the first and one-fifth of the contents of the second, or the contents of the second and one-third of the contents of the first. What is the capacity of each cask ? .55. Three brothers wished to buy a house worth 1 70,000, but none of them had enough money. If the oldest brother had given the second brother one-third of his money, or the youngest brother one-fourth of his money, each of the latter would then have had enough money to buy the house. But the oldest brother borrowed one-half of the money of the youngest and bought the house. How much money had each brother ? 56. The sum of the three digits of a number is 9. The digit in the hundreds' place is equal to one-eighth of the number composed of the two other digits, and the digit in the units' place is equal to one-eighth of the number composed of the two other digits. What is the number ? 57. Find the contents of three vessels from the following data : If the first be filled with water and the second be filled from it, the first will then contain two-thirds of its original contents ; If from the first, when full, the third be filled, the first 'will then contain five-ninths of its origi- PROBLEMS. 233 nal contents ; finally, if from the first, when full, the second and third be filled, the first will then contain 8 gallons. S8. Three hoys were playing marbles. A said to B : " Give me 5 marbles, and I shall have twice as many as you will have left." B said to C: "Give me 13 marbles, and I shall have three times as many as you will have left." C said to A ; " Give me 3 marbles, and I shall have six times as many as you will have left." How many marbles did each boy have ? 89. Three cities, A, J?, and C, are situated at the vertices of a triangle. The distance from ^ to C by way of B is 82 miles, from B to .4 by way of O is 97 miles, and from O to B by way of A is 89 miles. How far are A, B, and G from one another ? 60. A father's age is twenty-one times the difference between the ages of his two sons. Six years ago his age was six times the sum of his sons' ages, and two years hence it will be twice the sum of their ages. Find the ages of the father and his two sons. 61. A regiment of 600 soldiers is quartered in a four-story building. On the first floor are twice as many men as are on the fourth ; on the second and third are as many men as are on the first and fourth ; and to every 7 men on the second there are 5 on the third. How many men are quartered on each floor ? 62. The sum of the three digits of a number is 9. If 198 be added to the number, the digits of the resulting number are those of the given number written in reverse order. Two-thirds of the digit in the tens' place is equal to the difference between the digits in the units' and in the hundreds' place. What is the number ? 63. Four men are to do a piece of work. A and B can do the work in 10 days, A and C in 12 days, A and D in 20 days, and B, C, and Din 1\ days. In how many days can each man do the work, and in how many days can they all together do the work ? 64. The year in which printing was invented is expressed by a number of four digits, whose sum is 14. The tens' digit is one-half of the units' digit, and the hundreds' digit is equal to the sum of the thousands' and the tens' digit. If the digits be reversed, the resulting number will be equal to the original number increased by 4905. In what year was print- ing invented ? Discussion of Solutions. 3. Pr. 1. A merchant has two kinds of tea ; the first is worth a cents a pound, and the second & cents a pound. How much of each kind must be tafeen to make a mixture of one pound worth c cents ? 234 ALGEBRA. [Ch. XIV Let X stand for the part of a pound of the first kind, and y for the part of a pound of the second kind. Then, by the first condition, x + y = 1; (1) and by the second condition, ax + by = c. (2) Whence x = ^-^^, y = ?^^^- a — b a — b (i. ) If a > c > 6, the values of x and y are both positive, and the solu- tion satisfies the conditions of the problem. Thus, if o = 100, 6 = 75, and c = 85, we have a; = f , y = |. If a < c < 6, then x and y are both positive, and satisfy the conditions of the problem. That is, if the value of the pound of mixture be inter- mediate between the values of a pound of each of the two kinds, a definite solution is always possible. (li.) If c > a > 6, then a; will be posiijee and 2/ Jiegfaiiw. The solution does not satisfy the conditions of the problem. Thus, if a = 100, 6 = 75, c = 110, we obtain x — ^, j/ = — J. It is evident that a one-pound mixture of two kinds of tea which is worth more than either kind cannot be made. (iii.) If a = 6 = c, then x = %, y = %. This solution shows that the conditions of the problem may be satisfied in an indefinite number of ways. It is evident that a one-pound mixture of two kinds of tea, that are the same in price, can be made in any number of ways, if the mixture be the same in price. (iv.) If a = 6, and a=^c, then x = oo and y = co. This solution does not satisfy the conditions of the problem, since X and y must be finite proper fractions. It is also evident that a one- pound mixture of two kinds of tea which are the same in price cannot be made, if the mixture is to be of a different price. EXERCISES II. Solve the following problems, and discuss the results : 1. If an alloy of two kinds of silver be made, and a ounces of the first be taken with & ounces of the second, the mixture will be worth m dollars an ounce. If 6 ounces of the first be taken with a ounces of the second, the mixture will be worth n dollars an ounce. How much is an ounce of each kind of silver worth ? 2. Two bodies are separated by a distance of d yards. If they move toward each other with different velocities, they will meet after m seconds ; but if they both move in the same direction, the one will overtake the other after n seconds. With what velocities do the bodies move ? CHAPTER XV. INDETERMINATE LINEAR EQUATIONS. 1. An Indeterminate Equation was defined in Ch. XIII., § 1, Art. 1, as an equation which has an indefinite number of solutions ; as x -\- y = b. An Indeterminate System is a system of equations which has an in- definite number of solutions. Thus, if the system a: + «/ — z = 9, 2a;-j/ + 7a = 33, be solved for x and y, we obtain a; = 14-2z, ^=3z-5. In these values of x and y we may assign any value to z, and obtain corresponding values of x and y. Evidently the number of solutions will be more limited it only positive integral values of the unknown numbers are admitted. In this chapter we shall consider a simple method of solving in positive integers linear indeterminate equations and systems. 2. Ex. Solve 4 a; + 7 2/ = 94, in positive integers. Solving for x, which has the smaller coefficient, we obtain , = ?4-l, = 23-,+2^, ^ (1) 4 4 or a; - 23 + y = ^~^^ - 4 Since x and y are to be integers, must be an integer ; that is, 4 y must have such a value that 2 — Sy shall be divisible by 4. O Q a. Let - — ^^ = m, an integer. 4 Then y = , an inconvenient form from which to determine in- o 2 3 w tegral values of y. But since the expression — -— ^ is to be an integer, any multiple of it will be an integer. We therefore multiply its numera- 235 236 ALGEBRA. [Ch. XV tor by the least number which will make the coefficient of y one more than a multiple of the denominator, i.e., by 3. We then have ^~^y = i_ 2 2/ + ?^=-^, an integer. 4 4 Therefore, as above, ~ ^ = m, an integer. 4 Whence y = 2 — im. (2) Then, from (1) and (2), a; = 20 + 7 m. (3) Any integral value of m will give to z and y integral values. But since y is to be positive, m < 1 ; and, since x is to be positive, in > — 3. Therefore the only admissible values of m are 0, — 1, — 2. When TO = 0, z = 20, y= 2; m=— 1, x = 13, y= 6; m = -2, X = 6, y = 10. In solving a system of two linear equations in three unknown numbers, we first eliminate one of the unknown numbers, and apply to the result- ing equation the preceding method. 3. Pr. A party of 20 people, consisting of men, women, and children, pay a hotel bill of $ 67. Each man pays 1 5, each woman 1 4, and each child $1.50. How many of the company are men, how many women, and how many children ? Let X stand for the number of men, y for the number of women, z for the number of children. Then, by the conditions of the problem, x + y + z^20, (1) 5x + 4!/ + |3 = 67. (2) Eliminating z, we obtain 7 a; + 5 ;/ = 74. Whence 74-7x ^-^^_^^ 4-2a ,g, 5 5 A _ 2 X Since x and y are to be integers, must be an integer, and there- fore Sxi^^, =12^:1^, =2-x + 2^ 5 5 6 must be an integer. Finally, let 2 — x = m, an integer. INDETERMINATE LINEAR EQUATIONS. 237 Whence x = 2 -5m. (4) Then, from (3) and (i), y = 12 + 7m; (5) and from (1), (4), and (5), « = 6 — 2 m. (6) Since x, y, and z are to be positive, we have : from (4), m < 1 ; from (5), m >- 2. Therefore the only admissible values of m are — 1, and 0. When m = 0, a; = 2, ?/ = 12, = 6 ; m=— 1, x = 7, y=: 5, = 8. Consequently the company may have consisted of 2 men, 12 women, 6 children ; or of 7 men, 5 women, and 8 children. 4. Not every linear indeterminate equation can be solved in positive integers. The general form of such an equation is evidently ax + by = c, wherein a, b, and c are integers. If a and 6 have a common factor, / say, then a = fa', 6 = fb', wherein a' and 6' are integers. The equation may then be written fa'x +fb'y = c, or a'x + b'y = — Since a' and 6' are integers, - must be an integer, if x and y are to have integral values ; that is, / must be also a factor of c. Therefore, (i. ) The linear indeterminate equation ax + by = c cannot be solved in positive integers if a and b have a common factor, which is not a factor of c. E.g.,2x — iy = 5 cannot be solved in positive integers. We can infer at once that (ii.) If a and b are positive and c negative, the equation ax+ by = c cannot be solved in positive integers. For ax + by would then be positive and could not be equal to c, a negative number. S.g., 2x + 5y = — 6 cannot be solved in positive integers. The case in which a ^nd 6 are negative and c positive is evidently included in (li.). We therefore conclude that (iii.) The linear indeterminate equation ax + by = c can be solved in positive integers only when a, b, and c are all positive, or when a and b have opposite signs ; and when, in both cases, a and b do not have a com- mon factor which is not also a factor of c. 238 ALGEBRA. [Ch. XV But even when the conditions given in (iii.) are satisfied, a solution in positive integers is not always possible. Thus, the equation 7 x + 9 ?/ = 15 cannot be thus solved. For the least possible positive integral values of x and y are 1 and 1. But these give 7 a; + 9 y = 16 ^t 15. EXERCISES. Solve in positive integers ; 1. re + 2/ = 10. 2. 2 X + 3 2^ = 25. 3. 5 x + 7 !/ = 52. 4. 5 X + 8 2/ = 29. 5. 3 X + 5 2/ = 10. 6. 12 x + 13 «/ = 175. 7. 25 X + 15 2/ = 215. 8. 5 x + 13 ^ = 229. 9. 34 x + 89 2/ = 407. '8x + 3 2/-2z = 8, x + 32/ + 53 = 44, 3x+52/ + 70 = 68. 7 X — 2 2/ — Z-. Solve in least positive integers : 12. 91 X- 221 2^ = 0. 13. 3x- 52^ = 1. 14. 17 x- 11 2/ = 86. 15. 89 X- 144 2^ = 1. 16. 14 x + 49 j^ = 133. 17. 67 x- 43 2/ = 5. 18. Divide 1000 into two parts so that one part shall be a multiple of 13, and the other a multiple of 53. 19. What positive integers when divided by 4 give a remainder 3, and when divided by 5 give a remainder 4 ? 20. Divide -^-^ into two fractions with denominators 13 and 9 respec- tively. 21. A farmer received $ 16 for a number of turkeys and chickens. If he was paid 1 2 for each turkey and $ .75 for each chicken, how many of each did he sell ? 22. A gardener has fewer than 1000 trees. If he plants them in rows of 37 each, he will have 8 left ; but if he plants them in a different num- ber of rows of 43 each, he will have 11 left. How many trees has he ? 23. A wheel with 17 teeth meshes in a wheel with 13 teeth. After how many revolutions of each wheel will each tooth occupy its original position ? 24. A said to B : " If I had eight times as much money as I now have, and you had seven times as much money as you now have, and I were to give you $1, we should have equal amounts." How many dollars had each ? CHAPTER XVI. EVOLUTION. § 1. DEFINITIONS AND PRINCIPLES. 1. A qth Root of a number or an expression is a number or an expression whose qth power is equal to the given one. E.g., since (+ 5)^ = 25 and (- 5)^ = 25, therefore +5 and — 5 are second roots of 25. The statement, + 5 and — 5, is usually written ± 5. Since (a + by = a? + 3 a% + 3 aft^ + &«, therefore a.+ & is a third root of o' + 3 a?b + 5aV + h\ A second root of a number is usually called a square root; and a third root a CM6e root. 2. It follows from the definition of a root that a ^th root of a number is one of q equal factors of the number. Thus, either +3 or — 3 is one of two equal factors of 9. 3. The Radical Sign, t/, is used to denote a root, and is placed before the number or expression whose root is to be found. The Radicand is the number or expression whose root is required. The Index of a root is the number which indicates what root of the radicand is to be found, and is written over the radical sign. The index 2 is usually omitted. E.g., -y/9, or -^9, denotes a second, or square root of 9; the radicand is 9, and the index is 2. 4. A vinculum is often used in connection with the radical sign to indicate what part of an expression following the sign is affected by it. E.g., V9 + 16 means the sum of ^9 and 16, while V9 + 16 means a square root of the sum 9 + 16. Likewise -^a? x 6* 239 240 ALGEBRA. [Ch. XVI means the product of -y/a^ and b% while s/a^ x &* means a cube root of a%*. Parentheses may be used instead of the vinculum in connec- tion with the radical sign ; as -,^(9 + 16) for V9 + 16. 5. Like and Unlike Roots. — Two roots are said to be like or unlike according as their indices are equal or unequal, whether or not their radicands are equal. E.g., -y/a, -y/b, are like roots ; -y/a, -^/a, are imlike roots. 6. In this chapter we shall consider only roots of numbers which are powers with exponents equal to or multiples of the indices of the required roots ; as -y/16, = V^^ ■V'^^> -V<^'^- An Even Root is one whose index is even; as -y/aP, -^a*, ^/a?^. An Odd Root is one whose index is odd; as -^8, -^aP, ^«t^a^+\ Number of Roots. 7. (i.) A positive number has at least two even roots, equal and opposite; i.e., one positive and one negative. E.g., since (±4)^=16, Vl6=±4; since {±ay=a*, -^a*=±a. In general, since ( ± a)^ = a% ^/a^ = ±a. (ii.) A positive or a negative number has at least one odd root of the same sign as the number itself. E.g., since (;-3f=-27, -^-27= -3; since 2==32, ^32=2. In general, since ( + a)^«+^ = + a^+^, ^■!^+ o^+^ = + a. Since ( - a)^+' = - a^+\ ='1^ - a^+i = - a. The principle enunciated in (ii.), when the radicand is negar tive, may also be stated as follows : (iii.) An odd root of a negative number is equal and opposite to a like root of a positive number which lias the same absolute value. E.g., since -^- 8 = - 2 and - ^8 = - 2, therefore ^ - 8 = - ^8 ; since ^+-^—a^+^ = — a and - ^t^a^j+i _ _ „^ therefore '^+^ - a^'+i = - ^'i^a^+i. §1] DEFINITIONS AND PRINCIPLES. 241 Consequently, to find an odd root of a negative number, find a like root of the positive number which has the same absolute value, and prefix the negative sign to this root. (iv.) Since 0" = 0, therefore ^0 = 0. (v.) Since (+ 4)^ = + 16 and (— 4)^ = + 16, there is no num- ber, with which we are as yet familiar, whose square is — 16. Consequently -^—16 cannot be expressed in terms of the numbers as yet used in this book. In general, since (± a)^ = + a% we cannot express ^J/— a** in terms of numbers hitherto used. 8. It was shown in Art. 7 that a positive number, which is the gth power of a number, has at least one qth root, and when q is even at least two; also that any negative number, which is an odd power of a negative number, has at least one odd root. It will be proved in Chapters XXI. and XXII. that any num- ber has two square roots, three cube roots, four fourth roots ; and in Part II., Text-Book of Algebra, that, in general, any number has q qth roots. Principal Roots. 9. The Principal Root of a positive number is its one positive root. ^.g., 3 is the principal square root of 9. The Principal Odd Root of a negative number is its one nega- tive root. E.g., — 2 is the principal cube root of — 8. 10. -v/4^ = -v/16 = ± 4, if other than the principal root be admitted, and (V4)2=(± 2)^ = 4; therefore, V4' = (V4)', only for the principal square root. In general -^a' = (-^o)'' is true only for the principal qth root. In subsequent work the radical sign will be understood to denote only the principal root, unless the contrary is stated. E.g., V9 = 3, -Vl6 = -4, ^-27 = -3. 242 ALGEBRA. [Ch. XVI BXEECISBS I. Write 1. Two square roots of 49. 2. Two fourth roots of 81. 3. Two sixth roots of 64. 4. Two square roots of 5^. Write one cube root of 5. 64. 6. - 125. 7. 1000. 8. - a*-. Find the value of the indicated principal root of each of the following numbers : 9. ^256. 10. .5/2I6. 11. .J/- 512. 12. ^625. 13. ^-729. 14. ^1296. 15. ^243. 16. .J/64. From the definition of a root, express a as a root of the second member of each of the following equations : 17. a' = b. 18. a* = b^. 19. a^ = &'. 20. a" = b". 21-24. In Exx. 17-20, express 6 as a root of the first member of each of the equations. Evolution. 11. Evolution is the process of finding any required root of a given number or expression. Since even roots of negative numbers are not considered in this chapter, and since, by Art. 7 (iii.), an odd root of a negar tive number can be found from the like root of a positive number, we shall give now only methods for finding principal roots of positive numbers and expressions. In the following principles the radicands are limited to positive values, and the roots to principal roots. Principles of Roots. 12. The like principal roots of equal numbers or expressions are equal. If a=b, then -^a = -^b. This principle follows directly from axioms (i.) and (iii.). 13. The process of evolution depends upon the following principles : § 1] DEFINITIONS. AND PRINCIPLES. 243 (i.) The principal root of a power of a number is equal to the same power of the like principal root of the number, and con- versely ; or, stated symbolically, i/a" ={\/a)''. In particular, ^a" = (i/a)". E.g., ^8= =(^8)' = 2^ = 32; ^32« = (^32)« = 32. (ii.) !Z%e principal root of a power is obtained by dividing the exponent of the power by the index of the root; or, stated sym- bolically, S!ja'"i = a<' =a*. E.g., -^a' = a^ = a^ (iii.) Tlie principal root of a product of two or more factors is equal to the product of the like principal roots of the factors, and conversely; or, stated symbolically, ^(ab) = -^ax^b, and ^axyb=V{ab). E.g., V(16 X 25) = V16 X V25 = 4 X 6 = 20 ; ^ (8 a^b^ = -^8x-^a^ \-^¥ = 2 xaxb^ = 2 ah\ (iv.) Tlie principal root of a quotient of two numbers is equal to the quotient of the like principal roots of the numbers, and conversely ; or, stated symbolically, ^7 Ch. II., § 6, Art. 7, or o = JJ«, since ( -^a)' = a, by definition of a root ; and a? = (S')p, by Ch. II., § 6, Art. 7, or aP =(iJP)«, since (i??)?' =(iJ'')«. "Whence .^a*" = -J/CiJ^)', by Art. 12, = ii», by Art. 10. Substituting ^a for B in the last equation, we have (ii.) Let the q root of a** be denoted by M, or iJ = ^a**. Then iJ« = (^a's)', by Ch. II., § 6, Art. 7, = «*« = (a*)«. Whence ^JJ' = -C/Ca')', or .B = a*, by Art. 10. Substituting ^a'^ for JB in the last equation, we have .j/a'« = a* = a«. (iii.) Let ^a = Ii, and ^b = Bi. Then ( .J/a)' = JJ«, and ( -^6)' = JBi«, by Ch. II. , § 6, Art. 7, or a = B^, and 6 = Bt^. Therefore ab = i?«iJi« = (i2iJi)«. Whence -^(aft) = BBu by Arts. 10 and 12. Substituting .J/a for B, and -J/6 for JK; in the last equation, we have ^{ab) = -^a^b. In like manner, the principle can be proved for the gth root of a product of any number of factors. In like manner, (iv.) and (v.) can be proved. § 2.. ROOTS OF MONOMIALS. 1. The principal {positive) root of a positive number or ex- pression can be found by applying the principles of § 1, Art. 13. The negative even root of a positive number or expression is found by prefixing the negative sign to its principal root. § 2] ROOTS OF MONOMIALS. 245 Tlie negative odd root of a negative number or expression is found by prefixing the negative sign to the principal root of the radicand taken positively. Ex. 1. V(16 «"&*) = V16 X Va' X V^' = 4 ab^, the principal square root. Therefore ' ± V(16 a'b*) = ± 4 ab\ In the following examples we shall give only the positive even roots. Ex. 2. ^(- 27 a^2/V) = ^- 27 x b* x 2a^b^). 6. .^C- 7. ^(3 «^^" X 27 aH^). 8. .J/(9 a%iV" x 3 oW?/")- 9. ^(5,35.a^!/8n-i2). 10. .«/(64 ajiV'z'^ X a^i/V)- 11. y/lSl a^a^ + x^)'}. 12. V(6ia'&*c^-*)- 13. *•+!/(_ a2n+166»+3). 14. .J/[3|ic»"-8(9;-l)S]. "■ V"Mi^' V~"3i3" "• Vel^ ,. 1 9 a664^ 4;625^V! 20 '/iM«^. 246 ALGEBRA. [Ch. XVI 5/ a6a;2» oo »la"6^c6 21. ',' °°^^ ■ 22. '1^ ^64 aiiV'^ 23. Vv^(49' X 648). 24. ^VC^^^'V)- 25. ^V(4098 a^^ft"). 26. y/y^ClGa*"-*). 27. ^^(27"a3»6«")2. 28. ^/^/C-SiSa^WS") 29. V(100 a^-ft*)'. 30. -J/(tJ)''. 31. ^(S a<^b^x^^y. § 3. SQUARE BOOTS OP MULTINOMIALS. 1. The square root of a trinomial which is the square of a binomial and the square roots of certain multinomials can be found by inspection (Ch. VIII., § 1, Art. 9). 2. Since (a + 6^ = a= + 2a6 + 6^ we have VC'*^ + 2 ab + b^)— a + b. From this identity we infer : (i.) 7%e first term of the root is the square root of the first term of the trinomial ; i.e., a=^aK (ii.) If the square of the first term of the root be subtraxsled from the trinomial, the remainder will be 2ab + b', =(2 a + 6)6. Twice the first term of the root, 2 a, is called the Trial Divisor. (iii.) The second term of the root is obtained by dividing the first term of the remainder by the trial divisor ; i.e., 6 = 2a (iv.) If twice the first term of the root plus the second term, 2a + b (the complete divisor), be multiplied by the second term, 6, and the product be subtracted from the first remainder, the second remainder will be 0. The work may be arranged as follows : a + b a' a' + 2ab + b^ 2ab 2ab + V 2 a ' trial divisor 2 a6 -7- 2 a = 6, second term of root 2 a + 6 complete divisor = (2 a + 6)6 §3] SQUARE ROOTS OF MULTINOMIALS. 247 Ex. Find the square root of 4 a;* — 12 a;^ + 9 y\ The work, arranged as above, writing only the trial and the complete divisor, is : 4 a;* - 12 iB«2/ + 9 «/= 4«« -12a?y - 12 x^y + 9 y^ 2a?~Zy 4a;2 4a^-3y 3. When the multinomial is the square of a trinomial, the process of finding the root is an extension of the method of Art. 2. The multinomial whose root is required should be arranged to powers of a letter of arrangement. Since {a + b + cf = {a + bf + 2{a + h)c + 2 2ab 2a6+&2 2ac 2ac+2'bc+c^ a + h+c 2a required root trial divisor 2ab-i-2a=b, second term of root 2a + b complete divisor of first stage 2ac-7-2a=c, tliird term of root 2 a+ 2 6 + C, complete divisor of sec- oDd stage Observe that 2 ac is the first term of the remainder after subtracting (a +by = d' + 2ab + b\ For, in finding the first two terms of the root we first subtracted a-^ and then 2ab + 61 Notice also that the complete divisor at any stage is twice the part of the root already found, plus the term last found. 248 ALGEBRA. [Ch. XVI Ex. 1. rind the square root of Ax*-12a? + 29x'-30x + 25. The work follows : 4:X*-12a? + 29x'-30x + 25 4 a;* -12a^ -12a?+ 9a^ 20 a? 2(ix'-30x + 25 2a?-3x + 5 4:0e'-3x 4 K^ — 6 a; + 5 Only the trial divisor and the complete divisor of each stage are written, the other steps being performed mentally. 4. The preceding method can be extended to find square roots which are multinomials of any number of terms. The work consists of repetitions of the following steps : After one or more terms of the root have been found, obtain each succeeding term, by dividing the first term of the remainder at that stage by twice the first term of the root. Find the next remainder by subtracting from the last remainder the expression {2a + b) b, wherein a stands for the part of the root already found, and b for the term last found. EXERCISES III. Find tlie square root of each of the following expressions : 1. K* - 4 x' + 8 a; + 4. 2. 4 m* - 4 jii^ + 5 m^ - 2 m + 1. 3. a;*-2a;= + 39;2_2x + 1. 4. 4x* + 12xS+ Sx^ - 6x+ 1. 5. 9x*+12x3-26x2-20x+25. 6. 4x* - 28x3 + 51 x^- 7 x + J. 7. x*y^ - 4 xV + 6 x2j/2 -^xy + \. 8. Jx* + ^x3i/ + 2xV-12x2/3 + 9!/«. 9. X* - 6 ax3 + 13 aV - 12 aH + 4 a*. 10. 4 a^ + 9 62 + 16 c2 - 12 aft + 16 ac - 24 be. 11. 49 x8 + 42 x6 - 19 X* - 12 x2 + 4. 12. 25 x* - 30 axs + 49 a^x^ _ 24 aH + 16 a*. 13. a2 + 4a3 + 4a2 + 2a + 4 + J-. §4] CUBE ROOTS OF MULTINOMIALS. 249 14, 5^-5^+4xV + 6x-12j/S + 9r. y* y z^ 15. x4 + ?^ + *! + 2aa; + 2 + ^. 16. 1 + 2x -x^ + 3xi -2x^ + x^. a a" x^ 17. ifi-6 asfi + 15 a^rK* - 20 aV + 15 a*x^ -6a^x + o«. 18. 1 - 4 a + 64 a6 - 64 a6 - 32 a^ + 48 a* + 12 a^ 19. 4 a^ + 17 a2 - 22 aS + 13 a< - 24 a - 4 a5 + 16. 20. 9x^ + 63fy + 43 si;*?/^ ^ 2 xV + 45 x^y* - 28 xj/S + 4 j/S. 21. a;* + 4a;8 + 6»2 + 5a; + 6 + 5 + _2-+i + I. a; 4a;2 a;' -g* 22. a2"'a;2» + 10 a^-^x^''+^ - 6 a^+^x'+i + 25 a^m-ix,^," (1) we have -^(a? + 3a''b + 3ab'+¥)=a + b. (2) Prom the identity (2), we infer : (i.) The first term of the root is the cube root of the first term of the multinomial; i.e., a= -^a'. (ii.) If the cube of the first term of the root be subtracted from the multinomial, the remainder will be 3a^b + 3aW+W, ={3a^ + 3ab + by. Three times the square of the first term of the root, 3 a^, is called the Trial Divisor. (iii.) The second term of the root is obtained by dividing the first term of the remainder by the trial divisor; i.e., 6= "■ • o a (iv.) If the sum 3 a^+ 3 a& + b", the complete divisor, be multi- plied by the second term of the root, and this product be sub- tracted from the first remainder, the second remainder will be 0. 250 ALGEBRA. [Ch. XVI The work may be arranged as follows : a' +3 0^1 +3 ab'+b^ 3a'b 3a'b+3ab''+b^ a+b 3 a" trial divisor (1) Sa'b-T-Sd'::^ b, second term of root (2) 3 a'' + 3 a& + V, complete divisor (3) = (3a2+3a6+&^)x6 (4) Ex. 1. Find the cube root of 27 0!^ + 54 a^y + 36 V 4. 8 f. The work, arranged as above, is : 27oi?+54.3?y+3&xy^+2,f 54ic^y 54a!^y+36a^^+8?/^ 3a;+2y 3 (3 a;)2= 27 «?, trial divisor (1) 54:x'y-i-21x^=2y, second term of root (2) 3(3 a;)^+3(3a;)(22/) + (2 2/^=27 a;^ +18a;2/+42/^ complete divisor (3) = (27 0^+18 a;2/+42/^)(2 2/) (4) 2. The preceding method can be extended to find cube roots which are multinomials of any number of terms, as the method of finding square roots was extended. The work consists of repetitions of the following steps : After one or more terms of the root have been found, obtain each succeeding term, by dividing the first term of the remainder at that stage by three times the square of the first term of the root. Find the next remainder by subtracting from the last remainder the expression (3 a^ + 3 a6 + 6^) b, wherein a stands for the part of the root already found, and b for the term last found. The given multinomial should be arranged to powers of a letter of arrangement. Ex. 27-27 a; + 90a;2_65x3 + 90a;*-27a;5+27x6 27 ' -27 a; -27 x+ 9a;2_ x' 81x2_54a;8 81a;2_54a;3+< )a;«-27a!5+27a;6 .S-x+3x2 3(3)2=27 3(.3)2+3(3)C-x) + (-a:)''=27-9x+x'' 3(3-x)2+.3(3-x) (3 x2) + (3 x2)2= 27-18 x+30x2-9x3+9x^ § 5] HIGHER BOOTS. 251 EXBECISES IV. Find the cube root of each of the following expressions : 1. a;6-6x6+ 15a;4-20a;=' + 15a;2-6a; + l. 2. 8a;6-36a;6 + 66a;*-63a;3 + 33a;2- 9a;+ 1. 3. 156a^- 144a5-99a3 + 64o6 + 39o2-9a + l. 4. 1 + 3a; + 6x2 + 7a;3 + 6a;* + 3a;5 + a;«. 5. 1 -6a; + 9a;2 + 4a;8-9y>-6a;5_a;6, 6. 8a;8-12a;2 + 12x-7+5--^+-i-. 7. 27 a8a;6 + 54 a^K^ + 9 a*xi - 28 aV - 3a%2 + 6 ax - 1. 8. 8a^+ 48 a'fe + 60 a*62 - 80 a^fts - 90 a'^b* + 108 afts _ 27 6». 9. a;3 + 3xT - 9x" - 27a;is - 6x5 _ 54a;i3 + 28x9. 10. 108a5_48a.4 + 8o34.54aT_i2a8 + ^9 _ 112 ^6. 11. 8 a6 _ 48 a^x + 60 a^x^ - 27 x^ - 108 ax^ - 90 a^x*' + 80 aV. 13. 1 +3x-8x3-6x* + 6x5 + 8x6-3x8-x!'. 13 125^6 150 g,6 X65y4 172^3 99^^-. 54 y ^^ X^ X^ X* x8 X^ X 14. 64x3»-144x3«-i+12x3»-2+117x»»-3-6xS"-*-3ax3"-«-8x3«-6. §5. HIGHER ROOTS. 1. Since -^N — ^y/N, -wherein N stands for any multi- nomial, the fourth root is most easily found as the square root of the square root of the given multinomial. In like manner, since -^N=-^^N, the sixth root can be found as the cube root of the square root of the given multi- nomial. And so on for any root whose index can be factored. 2. The process of finding the wth root of a multinomial is the inverse of raising a multinomial to the Mth power. The method can be derived from the expression for (a + 6)", which will be given in Ch. XXVIII. EXERCISES V. Find the fourth root of each of the following expressions : 1. x8 + 4x6-)-6x*-l-4x2-(-l. 2. aS + 4 a?h + 10 a^V^ -f- 16 a^b^ + 19 a^¥ + 16 a^b^ -I- 10 a^b<^ 4- 4 a6' -f- fts. 252 ALGEBRA. [Ch. XVI 3. 16a;8 _ leOa;' + 408 a;6 + HOifi - 2111 x* - 1320 a;3 + 3672 a;^ + 4320a; + 1296. 4. 625 xs + 5500 a;' + 17150 a;6 + 20020 a;^ + 721 a;* - 8008 x^ + 2744 a;" -352X + 16. Find the sixth roots of each of the following expressions : 5. 64x12 _ 192 xi» + 240a;8 - 160x6 + 60x4 - 12 x^ + 1. 6. ai2 + 6 a"6 +21 a^ft^ + 50 aV + 90 a'¥ + 126 a'b^ + 141 a<^b^ + 126 a^f + 90 a^b' + 50 a»b^ + 21 a^fti" + 6 aft" + b^^ §6. ROOTS OF AEITHMETICAL NUMBERS. Square Roots. 1. Since the squares of the numbers 1, 2, 3, •■•, 9, 10, are 1, 4, 9, ■ ■ -, 81, 100, respectively, the square root of an integer of .one or two digits is a number of one digit. Since the squares of the numbers 10, 11, ■••, 100, are 100, 121, •••, lOOOO, the square root of an integer of three or four digits is a number of two digits. In general, the square root of any integer of 2 n — 1 or 2 n digits is a number of n digits. Therefore, to find the number of digits in the square root of a given integer, we first mark off the digits from right to left in groups of two. The number of digits in the square root ivill be equal to the number of groups, counting any one digit remaining on the left as a group. 2. The method of finding square roots of numbers is then derived from the identity (a+by = a^+(2a + b)b, (1) wherein a denotes tens, and b denotes units, if the square root be a number of two digits. Ex. 1. Find the square root of 1296. We see that the root is a number of two digits, since the given number divides into two groups. The digit in the tens' place is 3, the square root of 9, the square next less than 12. Therefore, in the identity (1), a denotes 3 tens, or 30. 6] ROOTS OF ARITHMETICAL NUMBERS. 253 The work then proceeds as follows : a + b 12 '96 I 30 + 6 = 36 9 00 I 2a = 60. trial divisor 3 96 3 96 2 a = 60, trial divisor (1) (2a& + 62)^2a = 396-i-60 = 6+ (2) = (2a + 6)x6=(60 + 6)x6 (3) The first remainder, 396, is equal to 2ab + V, and cannot be separated into the sum of two terms, one of which is 2 ah. We cannot, therefore, determine 6 by dividing 2ab by 2 a, as in finding square roots of algebraic expressions. Consequently step (2) suggests the value of b but does not definitely determine it. As a rule we take the integral part of the quotient, 6 in the above example, and test that value by step (3). This method may be extended to find roots which contain any number of digits. At any stage of the work a stands for the part of the root already found, and 6 for the digit to be found. Ex. 2. Find the square root of 51529. The root is a number of three digits, since the given number divides into three groups. The digit in the hundred^ place is 2, the square root of 4, the square next less than 5. Therefore in the identity (1), a denotes 2 hundreds, or 200, in the first stage of the work. The work then proceeds as follows : 200 + 20 + 7 = 227 5' 15' 29 4 00 00 1 15 29 84 00 31 29 31 29 2 a = 400, trial divisor (2ab + b'^-h2a = 11529 -^ 400 = = (2a + b)b = (400 + 20) x 20 = 20 + (2 a6 + &2) H- 2 a = 3129 -r- 440 : = (2a+6)6 = (440 + 7)x7 ■7 + (1) (2) (3) (4) (5) In the second stage of the work, a stands for the part of the root already found, 220, and b for the next figure of the root. In practice the work may be arranged more compactly, omitting unnecessary ciphers, and in each remainder writing only the next group of figures. Thus : 254 ALGEBRA. [Ch. XV] 6' 16' 29 227 4 1 15 11 -- 4 = 2 + (2) 84 42 31 29 312 -- 44 = 7 + (4) 31 29 447 Observe that the trial divisor at any stage is twice the part of the roet already found, as in (2) and (4). The abbreviated vFork in the last example illustrates the following method : After one or more figures of the root have been found, obtain the next figure of the root by dividing the remainder at that stage (omitting the last figure), by the trial divisor at that stage. See lines (2) and (4). Annex this quotient to the part of the root already found, and also to the trial divisor to form the complete divisor. Find the next remainder by subtracting from the last remainder the product of the complete divisor and the figure of the root last found. 3. Since the number of decimal places in the square of a decimal fraction is twice the number of decimal places in the fraction, the number of decimal places in the square root of a decimal fraction is one-half the number of decimal places in the fraction. Consequently, in finding the square root of a decimal frac- tion, the decimal places are divided into groups of two from the decimal point to the right, and the integral places from the decimal point to the left as before. Ex. 14' 46.28' 09 38.03 9 5 46 5 44 68 2.28 09 2.28 09 76.03 § 6] ROOTS OF ARITHMETICAL NUMBERS. 255 In finding the second figure of the root, we have ^ = 9 ; but 69 X 9 = 621, which is greater than 646, from which it is to be subtracted. Hence we take the next less figure 8. Cube Roots. 4. Since the cubes of the numbers 1, 2, 3, ■ ■ •, 9, 10 are 1, 8, 27, ••-, 729, 1000, respectively, the cube root of any integer of one, two, or three digits is a number of one digit. Tlie cube roots of such numbers can be found only by inspection. Since the cubes of 10, 11, •■-, 100 are 1000, 1331, ••-, 1000000,. respectively, the cube root of any integer of four, five, or six digits is a number of two digits. In general, the cube root of any integer of 3 n — 2, 3 w — 1, or 3 n digits is a number of n digits. Therefore, to find the number of digits in the cube root of a given integer, we first mark off the digits from right to left in groups of three. The number of digits in the cube root will be equal to the number of groups, counting one or two digits remaining on the left as a group. 5. The method of finding cube roots of numbers is derived from the identity (a + 6)^ = a» + (3a2 + 3a6 + V)b, wherein a denotes tens, and 6 denotes units, if the cube root is a number of two digits. Ex. Find the cube root of 59319. The digits in the tews' place of the root is 3, the cube root of 27, the cube next less than 59. Therefore in identity (1), a denotes 3 tens or 30. The work may be arranged as follows : 59' 319 27 000 32 319 32 319 a + b 30 + 9 3 a2 = 3 (30)2 = 2700 ^^-^ (3 a?b + 3 aft^ + 6') -r- 3 a* = 32319 -i- 2700 = 9 + (2) Ba^ =3(30)" =2700 3a6 = 3(30)9 = 810 &2 = 9" = 81 = (3a^ + 3a& + 6")x& = 3591 x9 (3) 59' 319 39 27 32 319 2700 810 81 32 319 3591 256 ALGEBRA. [Ch. XVI As in finding square roots of numbers, step (2) suggests the value of b, but does not definitely determine it. If the value of b makes (3 a^ + 3 a6 + b^) X b greater than the number from which it is to be subtracted, we must try the next less number. In practice the work may be arranged more compactly, omit- ting unnecessary ciphers, and in each remainder writing only the next group of figures ; thus (1) (2) (3) 6. The preceding method may be extended to find roots that contain any number of digits. At any stage of the work a stands for the part of the root already found, and b for the digit to be found. The method consists of a repetition of the following steps : T7ie trial divisor at any stage is three times the square of the part of the root already found ; as 27 in the preceding example. After one or more figures of the root have been found obtain the next figure of the root by dividing the remainder at that stage (omitting the last two figures) by the trial divisor. In the last example, 9 + =323 --27. Annex this quotient to the part of the root already found. Add to the trial divisor (with two ciphers annexed) three times the product of the part of the root already found (with one cipher annexed) and the figure of the root just found, and also the square of the figure of the root just found. The sum is called the complete divisor. Find the next remainder by subtracting from the last remainder the product of the complete divisor and the figure of the root last found. 7. Evidently, in finding the cube root of a decimal fraction the decimal places are divided into groups of three figures from §6] ROOTS OF ARITHMETICAL NUMBERS. 25T the decimal point to the right, and the integral places from the decimal point to the left as before. Ex. 11'089.667 22.3 8 3 089 1200 120 4 2 648 1324 441.567 1452.00 19.80 .09 441.567 1471.89 EXERCISES VI. Find the square root of each of the following numbers : 1. 196. 2. 841. 8. 1296. 4. 65.61. 6. 3481. 7. 667489. 8. 170569. 10. 582169. 11. 1.737124. 12. 556.0164. Find the cube root of each of the following numbers 14. 2744. 15. 39304. 16. 110.592. 18. 1.191016. 19. 74088000. 20. 340068392. 22. 584067.412279. 23. 375601280.458951. 5. 7396. 9. 1664.64. 13. .00099225. 17. 328509. 21. 426.957777. 24. .041063625. Find the value of each of the following indicated roots : 25. .J/279841. 26. .J/3010936384. 27. 4/164204746.7776. CHAPTER XVII. INEQUALITIES. 1. One number is greater or less than a second number according as tlie remainder of subtracting the second number from the first is positive or negative. Thus, a > 6, when a — 6 is positive, i.e. , vrhen a — 6 > 0. a < 6, when a — 6 is negative, i.e., when a — 6 < 0. 2. An Inequality is a statement that two numbers or expressions are unequal ; as a^ + 6'2> a'^. The members or sides of an inequality are the numbers of expressions which are connected by one of the signs of inequality, > or <. 3. Two inequalities are of the Same or Opposite Species, or are said to subsist in the same or opposite sense, according as they have the same or opposite sign of inequality. H.g., 8 > 3 and — 5 > — 7 are inequalities of the same species ; 0> — 1 and < 1 are inequalities of opposite species. 4. Observe that a relation of inequality between two numbers can be stated in two ways ; as 7 > 3, or 3 < 7. That is, if the members of an inequality be interchanged, the sign of inequality must be reversed. Principles of Inequalities. 5. If one number be greater than a second, and this second number be greater than a third, then the first number is greater than the third; that is, If a > 6 and 6 > c, then a>c. In like manner, if a < 6 and 6 < c, then a 2, 2 > 1, and 3>1; -3<-2, -2<0, and - 3 < 0. 6. Addition and Subtraction. — The following principles of inequali- ties involve the operations of addition and subtraction : (i. ) If the same number, or equal numbers, be added to or subtracted from both members of an inequality, the resulting inequality will be of the same species ; that is. If a>b, then a ±m^b ±m. E.g., 3>2, and3 + 1>2 + 1, and3-l>2 - 1. 258 INEQUALITIES. 259 (ii.) If the corresponding members of two or more inequalities of the same species be added, the resulting inequality will be of the same species; that is, If ai>6i, ai>bi, as>b3, •••, then 01+02 + 03+ ••• >6i + 62 + 63+ ■••■ E.g., -5>-7, 3>2, 0>-4, and -5+3 + 0>-7 + 2-4 ; i.e., -2>-9. (iii.) If the members of one inequality be subtracted from the corre- sponding members of another inequality of the same species, the resulting inequality will not necessarily be of the same species ; that is, If oi > 61 and 02 > 62, then Oi — 02 may or may not > 61 — 62. E.g., 11>6, 4>3, and ll-4>6-3; 5>4, 3>1, l3ut5-3<4-l. (iv.) If the members of an inequality be subtracted from the corre- sponding members of an equality, the resulting inequality will be of the opposite species ; that is, if • a = b, and c^d, then a — c<^b — d. E.g., 4 = 4, 3>-2, and 4 - 3<4 -(- 2), or 1<6. The proof of the principle enunciated in (i.) follows ; the other princi- ples are easily proved in a similar manner. (i.) If a > 6, then a — 6 is positive ; and a — 6±m + m is positive. Therefore (a ± m) — (b±m) is positive ; and hence a ± m > 6 ± m. ■ 7. Multiplication and Division. — The following principles of ine- qualities involve the operations of multiplication and division : (i.) If both members of an inequality be multiplied or divided by the same positive number, or by equal positive numbers, the resulting ine- quality will be of the same species ; that is, if o > 6, then an > bn, and - > -, n n wherein m is a positive number. E.g., -3>-5, and -16>-25, and - 1 >- |. (ii.) If both members of an inequality he multiplied or divided by the same negative number, or by equal negative numbers, the resulting ine- quality will be of the opposite species ; that is, if o>6, then o(-n)<6(— n), and -^< , — n — n wherein — n is a negative number. E.g., 2>-l, and 2(-3)<(- l)(-3), or -6<3; and -i-<^, or - l 61, 02 > 62, 03 > bg, then aia^Os > 61626s) wherein ai, 61, 02, 621 03, 63 are all positive. E.g., 12 > 4, 3 > 2, and 12 x 3 > 4 x 2, or 36 > 8. The proof of the principle enunciated in (ii.) follows ; the other princi- ples can be easily proved in a similar way. (ii.) If a > 6, then a — & is positive. Let — m be any negative num- ber. Then -m(a-6), =-ma-(-mb), and ^"~^) , =-^ ^ — m — m — m are negative. Therefore — ma<,— mb, and _ ^<_5 — — m — m 8. Powers and Roots. — The following principles follow directly from those of the preceding article : (i.) If both members of an inequality be positive, and be raised in the same positive integral power, the resulting inequality will be of the same species ; that is, if a>6, then a">6", wherein a and 6 are positive, and n is a positive integer. E.g., 9>4, and 81>16. (ii.) If the same principal root of both members of an inequality be taken, the resulting inequality will be of the same species ; that is, if a>6, Va>V6- E.g., 9>4, and 3>2; -27<-8, and -3<-2. 9. Transformation of Inequalities. — The preceding principles enable us to make the following transformations of inequalities : (i.) Any term may be transferred from one member of an inequality to the other, if its sign be reversed. E.g., ii a — 6 > c, then a>b + c. (ii.) If the signs of both members of an inequality be reversed from + to — , or from — to +, the sign of inequality must be reversed. E.g., -3<5, and 3>-5. INEQUALITIES. 261 (iii.) An inequality may he cleared of fractions by multiplying both members by the L. G. D., taken positively. E.g., it _«L_&- b-c b + c b^-c^' then x(b + c) — ^(6 — c) > z, if 6^ — c'^ be positive, i.e., if 6 > c, while x(b + c) — y(b — c)<,z, if 6^ — c^ he negative, i.e., if 6 < c. (iv.) Common positive factors can be canceled from both members of an inequality. E.g., 8>-12, and 2>-3. If x(a^ - &2) < (a 4 6)2, then x(a — 6)< (a + 6), when a + 6 is positive ; but x(a — 6) > (a + 6), when a + 6 is negative. (v.) If the reciprocals of the members of an inequality, which are either both positive or both negative, be taken, the resulting inequality will be of the opposite species. E.g., 3>2, and 1<1; -5<-2, and -i>-^. 10. An Absolute Inequality is one which holds for all values of the literal numbers involved ; as a^ + ft'' > a^. Such inequalities are analogous to identical equations. A Conditional Inequality is one which holds only for values of the literal numbers lying between certain limits. E.g., a;2 + 1 ;> 2, only for values of x greater than 1 and less than — 1 ; that is, for values of x between 1 and + oo, and between — 1 and — oo. Absolute Inequalities. 11. Ex. 1. Prove that if a =^ 6, then a^ + ja > 2 ab. We have (a-6)2>0, (1) since the square of any positive or negative number is positive, and there- fore greater than 0. From (1), a2-2a6 + 62>0; whence 0= + &2>2 a6, by Art. 9 (i.). Ex. 2. Which is greater, "' "*" ^ or " "^ ^ ^ , in which a and 6 are .,. , a + 5& a + 36 positive 7 262 ALGEBRA. [Ch. XVII We can determine which fraction is greater by finding their difference a + 45 _ a + 26 _ ITf- a + 56 a + 36 (a + 5 6) (a + 3 6)' Since this remainder is positive, we have a + 46 -^ a + 26 a + 56 a + 36 Conditional Inequalities. 12. Ex. 1. Between what limits must x lie to satisfy the inequality a; > 5 X - 10 ? Transferring terms, — 4 x > — 10 ; whence x < f , by Art. 7 (ii.). That is, the inequality is satisfied by all values of x between f and -co. Ex. 2. What values of x satisfy the inequality x2 + 5x>-6? Transferring - 6, x^ + 5 x + 6 > ; or (x + 2) (x + 3) > 0. In order that the product (x + 2)(x + 3) may be greater than 0, i.e., positive, the two factors must be either both positive or both negative. The factors x + 2 and-x + 3 will be both positive, when x >— 2. Thus, if x = -l, then (x + 2)(x + 3) = (- 1 + 2)(- 1 + 3) = 2. The factors will be both negative, when x < — 3. Thus, if x = -4, then (x + 2)(x + 3) = (- 4 + 2)(- 4 + 3) = 2. Therefore, the given inequality will be satisfied by all values of x between —2 and + co, and between —3 and — qo. Ex. 3. What values of x and y satisfy the inequality bx + Zy>\\, (1) and the equality 3 x + 5 ?/ = 13 ? (2) Multiplying (1) by 3, 15 x + 9 j/ > 33. (3) Multiplying (2) by 5, 15 x + 25 ?/ = 65. (4) Subtracting (4) from (3), - 16s(>- 32, or y <2. Multiplying (1) by 5, 25 x + 15 j^ > 55. (6) Multiplying (2) by 3, 9 x + 15 y = 39. (6) Subtracting (6) from (5), 16x>16, or x>l. Notice that not any value of x greater than 1 taken with any value of y less than 2, will satisfy both (1) and (2). But such values of x and y as satisfy (1) and (2) simultaneously, must be greater than 1 for x, and less than 2 for y. If we assign to x any value greater than 1, we can INEQUALITIES. 263 determine from (2) the corresponding value of y, which will always be less than 2 ; these corresponding values of x and y will then satisfy (1). E.g., let a; = I ; then from (2), j/ = |, < 2 ; these values of x and y satisfy (1). EXERCISES I. Prove the following inequalities, in which the literal numbers are all positive and unequal : 1. a2 + 62 + c2>a6+aic+ &c. 2. a%^+h^c'^+a,^c'^>abc{a,+ h+e). 3. a6(a + 6)+ 6c(6 + c)+ac(a + c)>6a6c. 4. If P + m^+ >i2 = 1, and l^^ + mi^ + m^ = 1, then Ih + mmi + nn^ < 1. 5. aS + 68 > a^fi + a&2. 6. a^ + b*>a»b + ab». 7. (a + 6)(6 + c)(c + a)>8a6c. 8. 3(a2 + &2 + c^) > (a + 6 + c)2. 9. a«-b<'>3ifib -3ab^, if a>6; <3 a^fi - 3 a&2, if a<6. 10. (ab+xy)(ax+by)>'t abxy. 11. a' + ft^ + c^ > 3 a&c. 12. aH6*+c*>o6c(a+6+c). 13. (a+6 + c)8>3(a+6)(a + c)(6 + c). If a; be positive, which fraction is the greater : 14. ^ or ^±1? 15. ^+6 or ^±3, if ^>6? a: + 3 a;+l x-Q x-i Determine the' limits between which the values of x must lie to satisfy each of the following inequalities : 16. a;-8>4. 17. - 3(a; + 10)>-20. 18. i^Jri_a;<§Ij:il^ + 9. 19. ""-^ >g^i^. 4 3 ia+b b — a 20. a; «f_0. 23. a;2_a;_6>0. 24. ^ii>0. 25. 6^ZlI^±20, (\x-kx + \x>x + S, |25-4a;>0. ' \\(x + 2)>~\(x-2). ■x2- 12a; + 32>0, (x^-Zx~4,>Q, !-13a;+22>0. ' \x^-x~Q>Q. What value of x satisfies each of the following systems : f 2x2-5x4-2 = 0, rx2 + x-6 = 0, 30. 31. J x2~l>0? lx2 + 3x-4>0? 264 ALGEBRA. [Ch. XVII f2. + 3,=-4, ^^ x-y>2. Determine the limits between whicli tlie values of x and y must lie to satisfy the following systems : ■7x + !/ = 15, 3 a; - 2 !/ > 14. Determine the limits between which the values of a must lie to make each of the following values of x positive : 34. a; =-^ 35. jc = l^^l^. 36. cc = *^^^. 11 -2a 15-4(1 9-2a Problems. 13. Pr. 1. Divide 80 into two parts, such that the greater part shall exceed twice the sum of 4 and the less part. Let X stand for the greater part ; then 80 — a; will stand for the less. By the given condition, X > 2(80 - X + 4), or 3 X > 168 ; whence x > 56. Therefore any number greater than 56 (and less than 80) will satisfy the condition of the problem. E.g., if X = 60, the greater part, then 80 — x = 20, the less part ; and 60 >2 X 24. Pr. 2. A man receives from an investment an integral number of dol- lars a day. He calculates that if he were to receive $6 more a. day his investment would yield over $ 270 a week ; but that, if he were to receive $ 14 less a day, his investment would not yield as much as 1 270 in two weeks. How much does he receive a day from his investment ? Let X stand for the number of dollars which he receives a day. Then, by the first condition, 7 (x + 6) > 270 ; whence x > 32f And, by the second condition, 14 (x - 14)< 270 ; whence x < 33f Therefore he receives •§ 38 a day from his investment. EXERCISES II. 1. What integers have each the property that one-half of the integer, increased by 5, is greater than four-thirds of it, diminished by 3 ? 8. What integers have each the property that, if 9 be subtracted from three times the integer, the remainder will be less than twice the integer, increased by 12 ? 3. A has three times as much money as B. If B gives A f 10, then A will have more than seven times as much as B will have left. What are the possible amounts of money which A and B have 1 INEQUALITIES. 265 4. Find a multiple of 25, such that three-fourths of it is greater than one-half of it, increased by 15, while five times the number is less than three times the number, increased by 200. 5. "What positive numbers have each the property that, if the number be subtracted from a and be added to 6, the product of the resulting numbers will be greater than the product of the given numbers ? 6. In a class-room can be placed 6 benches, but it contains fewer. If 5 pupils be seated on each bench, then 4 pupils will be without seats. But if 6 pupils be seated on each bench, some seats will be unoccupied. How many benches are in the room ? A Property of Fractions. 14. If the denominators of the fractions — , — , ^■■■be all positive, di dz da then the fraction "i + "^ + «3 -I ^-^ „^g(j{g,. j^(j^ j/jg i^^igt and less di + d2 + ds + ■■■ than the greatest, of the given fractions. Let ^ be the greatest of the given fractions, and let di ^ = a;, or m = dix. •• (1) "1 Then —l + nd. We have a" -.&" =(a - 6)(a»-i -1- a^-^h -\ 1- a6''-2 + 6"-i), or a" = 6" + (a - 6) (a"-' + a-'-^fi + • •• + a&»-2 + 6"-!) . Let a > 6. Then fl!"-l>6»-^ a^-^fe > ft"-!, ■•-, ah''-^>h''-\ ft"-! = &»-i. Therefore a"-^ + a''-^ H (- a6"-2 + 6"-' > 6"-" + 6"~^ + --n terms > raft"-!. Consequently, since a — 6- is positive, a" > 6" + n(a — 6)6"-'. Now leta = l + d, and 6 = 1. Then (l + d)''>l + nd. CHAPTER XVIII. IRRATIONAL NUMBERS. 1. If a be the g'th power of a number, say 6, then .-^a, = -J/6«, has, as we have seen in Ch. XVI., a definite value ; as .J/16 = 2. "We shall now consider roots of positive numbers which are not powers with exponents equal to or multiples of the indices of the required roots. 2. The qth root of a positive fraction, whose terms (^either or both) are not qth powers of positive integers, cannot be expressed either as an integer or as a fraction. The proof of the general case will be first illustrated by the -y/i. The -^2 must be a number whose square is 2. But since 1^ = 1 and 2^ = 4, the ^2 cannot be an integer. Let us assume that -^2 can be expressed as a fraction, — , reduced to its lowest terms. Then from V2 = -, (1) we have 2 = ^. (2) d d? Since — is in its lowest terms, -^ is in its lowest terms [Ch. IX., d d^ Art. 29 (iii.)]. Consequently, by Ch. IX., Art. 29 (v.), ?i2 = 2, and d"^ = 1. (3) But since 2 is not the square of an integer, the first of equations (3), and therefore also (1), is untenable. Consequently, y/2 cannot be expressed as a fraction. In general, 4/—, wherein N and J) (either or both) are not gth powers of positive integers, cannot be expressed as a fraction -• The proof is identical with that for the y/2. Observe that, if d be assumed equal to 1, the preceding proof shows that i / — cannot be expressed as a positive integer ; also, if I> be assumed equal to 1, that the ^iV cannot be expressed as a positive integer, or as a positive fraction. 266 IRRATIONAL NUMBERS. 267 3. It is therefore necessary to exclude such roots from our considera- tion or to enlarge our idea of number. The latter alternative is in accordance with the generalizing spirit of Algebra. We therefore assume that y/2, and in general, -y/— i is a number, and include it in our number system. The properties of these new numbers must be consistent with the defi- nition of a root ; that is, with the relations, (V2)^ = 2,and(^fy=f 4. Before operating with or upon the numbers thus introduced into the number system, we must prove that they obey the fundamental laws of Algebra, which were proved in Chs. II. and III. only for integers and fractions. The following property will lead to another definition of the ■^2, and in general of the ^—, which is consistent with that given in Art. 3, and from which the fundamental laws can be easily deduced. iV If — be a fraction whose terms (either or both) are not qth poioers of positive integers, numbers can always be found, both greater and less than -. / — , which differ from •» / — by as little as we please ; that is, by less than any assigned number, however small. The proof of the general case will iirst be illustrated by the -^2. Since 2 lies between 1^ and 2^, the y/2 lies between 1 and 2, i.e., 1 \wl \ 10 / — is found to lie, wherein fci is or any positive integer. Then since — lies between (^Y and (bdl^Y the ij— lies between D \\0l V 10 / ' \Z> ^and*l^; i.e., 10 10 ' io ' *■ ^ wherein p = l, 2, 3, ■••, co. These two series have the following properties : (i.) The numbers h + ^-+ ... + A wherein p = 1,2, 3, ••■, «, 10 102 ^10? of the first series increase as p increases, but remain always less than the numbers *i+ii-+... + ^2-±i, wherein p = 1, 2, 3, ■•■, oo, 10 102 iop ' of the second series; and the numbers of the second series decrease as p increases, but remain always greater than the numbers of the first series. That is, the numbers of the one series more and more nearly approach the numbers of the other series, but never meet them. (ii.) The difference between a number of the one series and the corre- sponding number of the other series can be made less than any assigned number, however small, by taking p sufficiently great. 7. Two series of numbers which possess the properties (i.) and (ii.), Art. 6, are said to have a common limit, which lies between them. Two such series therefore define the number which is their common limit. This number is approached by both series and not reached by either. The two numbers between which the i— lies can be reduced to a common denominator lO^". Let us designate 10? by n. Then since these two numbers differ by — , = -, they may be represented by — , IOp n n and ^ — ^!^ respectively. In the theory which follows, we shall let n ^4--^+ ii = !2- en *i+_^4. I ^p + 1 _ w + 1 C21 10 102"^'""^10y n' ^^ 10 102 10? „ ■ ^ ■' That is, ™<'/i^<5L±i. (I.) n \D n Irrational Numbers. 8. An Irrational Number is a number which cannot be expressed either as an integer or as a fraction, but which can be inclosed between two fractions ultimately differing from each other, and therefore from the inclosed number, by less than any assigned number however small. IRRATIONAL NUMBERS. 271 An irrational number, /, is therefore defined by the relation n n wherein — and ?Ldl_L have the properties (i.) and (ii.), Art. 6; as y/i. n n 9. Whatever value p, and therefore — and '""*" , may have, there n n will always be numbers, integers or fractions, lying between any two numbers of the series — and ™ "'" ■ But no such number can be n n selected which will not be passed by numbers of one or the other series, if p be sufficiently increased. Therefore there is no number in the sys- tem defined so as to include only integers and fractions, which is greater than every number of the series (1.) and less than every number of the series (2.); that is, which is approached by both series and not reached by either. Since, however, these series cannot meet, we conclude that there was a gap between them which could not be filled by any integer or fraction. Consequently by including irrational numbers in the number system, continuity has been introduced where before it was lacking. Negative Irrational Numbers. 10. If the fractions of the series which define an irrational number be negative, the number thus defined is called a Negative Irrational Num- ber. Therefore a negative irrational number is defined by the relation _ m + 1 ^ _ j^ _ m n n wherein the two series of fractions, — "^ "^ and — — , have the properties n n (i.) and (ii.). Art. 6. 11. The positive and negative irrational numbers defined by the relations m ^ r ^ m, + \ m + 1 ^ j ^ m n n ' n n are called equal and opposite. The absolute value of an irrational number is its value without regard to quality. The Ftindameutal Operations with Irrational Numbers. 12. Addition. — Let /i and /a be two positive irrational numbers defined by the relations ^ ni ria wi n^ determine a positive number which lies between them. This number is defined as the product 7i • In. That is, mt ma mi + 1 ma + 1 Hi "ni'^ '■' ''^ ni ' mi The following definition of multiplication of irrational numbers is con- sistent with the preceding result. The product of two irrational numbers is the product of their absolute vahies, with a sign determined by the laws of signs for the product of two rational numbers. The Associative, Commutative, and Distributive Laws for Multiplication of Irrational ITumbers. 16. These fundamental laws hold also for irrational numbers. That is, hh = hh ; hhh = Iiihh) = etc. ; (Zi ± 72)/s = Us ± hh. The proofs of these principles are similar to those given in Art. 14. 17. Reciprocal of an Irrational Number. — It can easily be proved that the reciprocal of the numbers of the series which define / have the properties (i.) and (ii.), Art. 6. Therefore the two series of numbers, and — , define a positive m + 1 m '^ n n number which lies between them. This number Is defined as the reciprocal oil. That is, -<_<_. m + 1 / TO n n 18. Division. — Division by an irrational number may be defined as follows : To divide any number by an irrational number, not 0, is equivalent to multiplying it by the reciprocal of the irrational number. From this definition it follows that the fundamental laws hold also for division of irrational numbers. 19. It follows from the preceding theory that the laws governing the fundamental operations with irrational numbers are the same as those governing these operations with rational numbers. E.g., v'2-(V3-v'5) = v'2-V3 + v'5; (^2^3)' = (v'2)«(v'3)». CHAPTER XIX SURDS. 1. In Ch. XVI. we consiiered only roots whose radicands are powers with exponents equal to or multiples of the indices of the roots. In Ch. XVIII. we assumed the existence of roots of numbers which are not powers with exponents equal to or multiples of the indices of the required roots, and proved that such roots obey the fundamental laws of Algebra ; as -^2 x V3= -y/3 x V^j etc. Such roots were called Irrational Numbers. 2. A Rational Number is a number which can be expressed as 2x an integer or as a fraction; as 2, — , -1/(27 a*). A Rational Expression is an expression which involves only rational numbers ; as | a + ^ &, a& + -^a'. 3. A Radical is an indicated root of a number or expression ; as V7, V9; a/(« + b). A Radical Expression is an expression which contains radi- cals ; as 2 V7, -y/x + -yjy, ■y/(a + ^b). A Surd is an irrational root of a rational number ; as -y/l, -y/a. Observe that -^(1 + -^7) is not a surd, since 1 + -y/7 is not a rational number. Notice the difference between arithmetical and algebraical irrationality. Thus, -y/a is algebraically irrational ; but if a = 4, then -^a, = v'4, = 2, is arithmetically rational. Classification of Surds. 4. A Quadratic Surd, or a Surd of the Second Order, is one with index 2 ; as -^3, -sja. ^76 276 ALGEBRA. [Ch. XIX A Cubic Surd, or a Surd of the Third Order, is one with index 3 ; as^(a + 6), ^7. A Biquadratic Surd, or a Surd of the Fourth Order, is one with index 4; as -y/(a&), -^5. A Simple Monomial Surd Number is a single surd number, or a rational multiple of a single surd number ; as -y/3, 2y'5. ■ A Simple Binomial Surd Number is the sum of two simple surd numbers, or of a rational number and a simple surd number ; as V2 + -V^, 3 + V^- 5. The principles enunciated in Ch. XVI., and their proofs, hold also for irrational roots. Each principle will be restated as occasion for its use arises in this chapter. As in Ch. XVI., we shall limit the radicands to positive values, and the roots to principal roots. Reduction of Surds. 6. A surd is in its simplest form when the radicand is in- tegral, and does not contain a factor with an exponent equal to or a multiple of the index of the root ; as ■y/2, ~^(a^b), -y/a". A surd can be reduced to its simplest form by applying one or more of the following principles : Agr (i.) ^aKv = at = aK [Ch. XVI., § 1, Art. 13 (ii.)] (ii.) V(''6) = Va X V* [Ch. XVI., § 1, Art. 13 (iii.)] (iv.) In a root of a power (or a power of a root) the index of the root and the exponent of the power may both be multiplied or divided by one and the same number; or, stated symbolically, yw =*-^a*'', and Va" = Va*- E.g., ^a==^a*; ■^a^ = ^a\ The proof is left as an exercise for the student. 7. The following examples will illustrate the methods of re- ducing surds to their simplest forms : Ex. 1. V(18 a'b") = V(9 a*6^ x V(2 «) = 3 a'by/(2 a). SURDS. 277 Ex. 2. ^(a''+^h^+'^ = ^{a^V") x ^(ab'^ = ab^-^yiab^). Observe that the radicand is separated into two factors, one of which is a power with the highest exponent which is»equal to or a multiple of the index of the required root. The result is then obtained by multiplying the rational root of this factor by the irrational root of the second factor. Ex 3 /3 a" _ V(3 a') _ ^a? X V3 _ a^^ ■ \4 62 V(4=&') ■^/{^V) 2b When the required root of the denominator of a fraction cannot be expressed rationally, multiply both terms of the fraction by the expression of lowest degree which will make the denominator a power with an exponent equal to the index of the root. Ex.4. J2 = J6=^. \3 \9 3 E^.5. ^./^^^n/g^^-^(«p. By Art. 6 (iv.), a given surd can frequently be reduced to an equivalent surd of a lower order. Ex. 6. ^(27 a%«)= ^b^ X ^(Sa)^ = 6V(3a). EXERCISES I. Reduce each of the following surds to its simplest form : 1. ^32. 2. V75- 3- VIOS. 4. y/v?. 5. ^{oK). 6. V(a*6*)- 7. V(4 «'»")• «■ VCSOoxV). 9. y/{c?'}fl-\-a^(?'). 10. V(a&8c<-6ge8). 11. v'(6-c)(63-c8). 12. V(a2-l)(l + a). 13. V(9a;'-18a;2 + 99;). 14. V(4o^& - ^a^S'' + 4o68). 15. ^192. 16. .£/(-10i). 17. -yC-oi"). 18. S- '^/-?- 33. 278 ALGEBRA. [Ch. XIX 37. J6i«. 38. .m^. 39. Ji«^. 40. J-MoL. \81& \ 125 66 \9 6V yi6 b^sfi 41. '11 42. '/^. 43. 4/^. 44. 'Z^^. \9 \6» \27 6 \4 682^ 53. ';^^. 54. .t/25. 65. +"V). 67. '^(aS - 2 ax + a;2)8. Addition and Subtraction of Surds. 8. Similar or Like Surds are rational multiples of one and the same simple monomial surd, as ■y/12, = 2-y^3, and 5-.^3. Like surds, or such surds as can be reduced to like surds, can be united by algebraic addition into a single like surd. Ex. 1. V12+2V27-9V48=2V3+6V3-36V3=-28V3. Ex. 2. 8.^40 + 3..^135 - 2^625 = 16^5 -f 9^5 - 10^5 = 15^5. Ex. 3. V2-Vi+V-02=V2-iV2 + TVV2 = |V2- Ex. 4. ^(a'b) + 2^(a'b^ + ^(a¥) = a'^(ab) + 2 ab^(ab) + b^^{ab) = («+&) V(«&)- EXERCISES II. Simplify each of the following expressions : 1. v'24-V6 + V'150. 2. 2v8 + 5v72-7v'18. 3. V54 + 2^/24-9^96. 4. 6^3-2^48 + 5^/108. 8. 3V75 + 4iVl92 - 2}V12. 6. y/2^ + y/b^ - y/l 7. 4Vf-?VA-2v'27. 8. 2Vf + v'60-Vl5 + Vi 9 8^/48 + 3.^162-2^384. 10. 5 .J/54 + 9^/250 - .J/686, 11. 2|.J/500 + i.J/256 - 3^.J/32 - f .J/108. 18. 1.5 ^1| - 2J^12.8 - 3f .J/5f + 4.6.J/43.2. 13. .J/40-5.J/ji5 + 4.J/(-.625) + |.J/16J. SURDS. 279 14. 2y/3-y/12+*/9. 15. 4/24 + 3 {/9 - 5 .J/192. 16. V(4a») + \/(9«') + V(25«»)-V(81a'). 17. y/ (12 a^b) +^(1 5 a^b)--y/ (21 a^b). 18. .J/(64 aSfiS) + .J/(125 a^b^) - ^(a^b^). 19. o VCa'fe') + 6V(a'6') - 2 a&V(a'6») + ^(a^^b'^). 20. 3 «.J/(250 x^z') - 5 a;,.J/(128 ajgS) + 3 xe^(16 x^). 21. 3 a?b^(Z2 a?b) + 5.J'(108 a»h*) - ab^(bOO a%. 22. ^(9 a262) + ^(27 a^ft) + 5.{/(729 aSfta). 23. 2 1/(3 x^y) - .J/(9 a^?/^) + ^(125 ^^ _ ."/(ajSj^). 24. v'(9« + 2V)+3V(4o + 12). 25. V(4a' + 4a26) + v'(4«6^ + 4 6'). 26. 7xV(25a + 75)-5V(9a;^a + 27x2). 27. 2v'(2x=)-V(8a;)-v'(2a;'-4a;2 + 2a;). 28. .5/((i866 + 3 66) + a.^(32 a' + 96 6) - .J/(a8 + 3 o^B). 29. Va8 - d'b - Vab'^ - 6' - V(a + 6) (a^ - b^). »« o ia — x „ la — x „ /o — a; . la — x 30. 3 a-v — ; 3 x^ — ; 2 a^ — ; — + 4 x^ — ; — \a + x \a + x • ya + x ya + x Redaction of Surds of Different Orders to Equivalent Surds of the Same Order. 9. Surds of different orders can be reduced to equivalent surds of the same order by the principle given in Art. 6 (iv.): ^ao =''^a'"i [Art. 6 (iv.)] Ex. Keduce y'S, -^(2 a), and -^(5 6) to equivalent surds of the same order. We have V^ = ^3« = 3^729 ; ^(2 a) =^(2 ay = '^{8 a?); ^(5b) ='^i5by = ^(25b^. Observe that the L. C. M. of the given indices is taken as the common index of the equivalent surds, and that each radicand is raised to a power whose exponent is equal to the quotient of this L. C. M. divided by the index of the given root. 280 ALGEBRA. [Ch. XIX 10. Any rational number can be expressed in the form of a surd by writing under the radical sign a power of the number ■whose exponent is equal to the index. E.g., 2=V4=^8 = -=V2'. 11. Two surds, or a surd and a rational number, can be com- pared by first reducing them to equivalent surds of the same order, and then comparing the resulting radicands. Ex. Which is greater, y/2 or -^3 ? We have ^2=-^i, and ^3 =^9. Since 9 > 8, therefore ^9 > ^8, or ^3 > y/2. EXERCISES III. Reduce to equivalent surds of the same order : 1. V2, -t^S. 2. V3, -t/6. 3. ^7. -SyiO. 4. Vi> Vi- 5. 5, ^10. 6. 6, ^4. 7. -J/2, -5/3. 8. !J/15, !^10. 9. V(3a2), 6*. 10. ^a\ Vi/h\ 11. "+-J/(x3j/), "--i/C^J/'). 12. V5, 4^10, -t/15. 13. -J/2, 3, .^5. 14. V^ 68, ^c*. 16. ^"yaS, ^6^ «!!yc5. Which is the greater, 16. 2-^3 or 3-^2? 17. V5 or -J/10? 18. ^-5/25 or ^V^ ? 19. -J/o= or -^a, when a < 1 ? 20. -{/a;' or -J/x*, when a; > 1 ? Which is the greatest, 21. V3, ^5, or -J/IO? 22. VI. v^l, or -J/|? Multiplication of Surds. 12. Multiplication of Monomial Surds. — The product of two or more monomial surds is found by applying the principle Ex. 1. 5^4 X 2-^6 = 10^24 = 20^3. If the surds are of different orders, they should first be re- duced to equivalent surds of the same order. Ex. 2. V* X a/^" = -S/a' X -V^^ = \/a' = «\/«- Ex. 3. ■{/{a'b) X -^(a'b') X •{/(a^65) = 3^(a%^) x ^{aV) x ^(a^^") SURDS. 281 Ex. 4. 2^5 X 3^20 X VIO = 6^5 x -^(2' x 5) x V(2 X 5) = 6^5< x ?^(2« X 5«) X ^(2" X 68) =6^(5^^ X 2^2) = 60J2/5. When the radicands contain numerical factors it is advisable to express them as powers of the smallest possible bases, as in Ex. 4. It is frequently desirable to introduce the coefficient of a surd under the radical sign. Ex. S. 3 a^(2 «6) = ■.»/(27 a=) x ^(2 ab) = -.^(64 a'b). 13. Multiplication of Multinomial Surd Numbers. — The work may be arranged as in multiplication of rational multinomials. Ex. Multiply 2-3V2 + 5V6 by V2-V3- We have 2 - 3 V2 + 6 V6 V2-V3 2^2 - 6 + 10V3 -15v2 - 2V3 + 3V6 -13V2-6+ 8V3 + 3V6 Observe that the terms of each partial product are simplified, and that similar surds are then written in the same column. 14. Conjugate Surds. — Two binomial quadratic surds which differ only in the "sign of a surd term are called Conjugate Surds. E.g., V3+V'2 and -V3+V?; l-V^ ^^^ l+V^- Either of two conjugate surds is the conjugate of the other. The product of two conjugate surds is a rational number. For, i-y/a + V&) ( Va - V&) = ( V«)' - ( V*)' = a - &. 15. Type-Forms. — Many products are more easily obtained by using the type-forms given in Ch. VI., § 1. Ex. (V2 + V3)' = ( V2)' + 2 V2 X V3 + ( V^)' = 2 + 2v'6 + 3 = 5 + 2 ve. BXEECISBS IV. Simplify each of the following expressions : 1. V3 X y/6. 2. V27 X 3v/18. 3. 4Vf x 5Vf. 4- VlxVrk- 6- ^4x^2. 6. 2 t^^n a=b'' + c + 2b^c. Solving the last equation for y/c, we obtain a — b'—c y/C-. 26 This equation asserts that y/c, an irrational number, is equal to , a rational number. This is a contradiction of terms, and therefore the hypothesis -y/a =b+ y/c is untenable. 31. If a+y/b^x+^y, (1) wherein -y/b and y/y are surds, and a and x are rational, then a = x and b=y. For if o :#: a;, let a = a; + m, wherein m :# 0. Then (1) becomes x+m+-yjb=x-ir^y, or m+y/b=y/y. (2) But by the preceding article (2) is untenable, unless m = 0. Therefore a = x, and hence y/b=y/y, or b = y. 32. If V(a + V*) = V + y/y, then V(a - V*) = V' - V/- From V(« + V^) = V* + V^) we obtain a + y/b = x + y + 2y/(xy). Whence, by Art. 31, a=x + y, (1) and y/b = 2y/(xy). (2) Subtracting (2) from (1), a-^b = x + y- 2y/(xy). (3) Therefore y/{a—y/b) = y/x — y/y. 290 ALGEBRA. [Ch. XIX Evolution of Surd Expressions. 33. A root of a monomial surd number is found by applying the principle ^■ ^^^ V^ = 1.4142 + ••■. Therefore -y/^ = .707, correct to three places of decimals. EXERCISES XI. Find by inspection the square root of each of the following expressions : 1. a'' + 2ay/b + b. 2. 4a + 9x-12^(ax). 3. 9 + 6.5/3+ .J/9. 4. .J/5 + 2.«/2 + 2.«/80. Find by inspection the cube root of each of the following expressions ; 5. Xy/x + Sy/x~3x-l. 6. 4 n + 12 n^n^ + 12 n^n + ln^. 7. 8 a;s + 66 a;2 + 33 a; - 36 x^y/x— 6Sxy/x- 9^x + 1. Find an approximate value of each of the following expressions, correct to four figures : 8. V8. 9. JV2.5. 10. V2- 11. IVl 12. V345.06. 13. V10862.321. 14. V54.0001. IS. 2 16. 3 . 17. 1 . 18. 5 V5 V8 2 .J/4 VV6 10 1+^3 20. 3 + 2V7 21. V17 , i-VS 5 - 4V11 V2.5+V6 SURDS. 293 Find an approximate value of each of the following expressions, to include four terms : 22. ^(1-x). 23. y/Ca' + b^-). 24. ^(.x'' - xy + i^). 25. ^(1+x'). 26. ^(a=-63). 27. ^{x^+x^+xy^+y«). EXERCISES XII. MISCELLANEOUS EXAMPLES. 1. Simplify 2'\3 +^5 -'^13 + 4a/3. In each of the following expressions make the indicated substitution and simplify the result : 2. In 1±-^ + ^—^ , let a = i^3. i+va + a) 1-va-a) I .r. * ii m V(l - a;'') X » ii m 4. In 2 [a6 - V(«^ - 1) y/(fi' - I)], let 2 a = k + i and 2 6 = «/ + i- 5. in y(a + 3') + V(a-^) ,letr- ^"^ ; X -■ ^{a + x)-y/ia-xy 62 + 1 6. In x^ + y^ + xy, let x = 4 [ V(a + 6) + VC" - 3 6)] and y = J[V(a + 6)t\/(« - 3 6)]- 7. In /^V+ f ^ Y, let X =J^- x + va + a:') V\6 A/a/ 9. In Xy/(\ - x% let x = ^ ^+ VC^^- ^ '''') . 10. Ina;8 + 3oa; + 26, leta;=.;/[-&+V(a' + 6'')]+v'[-6-\/(o'+6'')]. 11. Prove the following identity : V[2 a2 - 62 + 2 aV(o'' - ft'')]- VC"'' - 2 bV(. or -3i = -i-i-i; and |V-l=^ + ^,or|i = i + i. 5. Two or more multiples or fractions of the imaginary unit can be united by addition or subtraction into a single multiple or fraction of that unit. E.g., 6^-1-8^-1 =-2.y/-l, 01 6i-Si^-2i; aV- 1 + 1>^/- 1 =(a + b)^- 1, or ai+bi=(a + b)i. 6. Multiplication by /. — We define multiplication, when the multiplier is the imaginary unit, by assuming that the Com- mutative Law holds, that is, by the relation •y/— 1 X a = oV— 1) or ia = ai. E.g., i2 = 2i=i + i. That is, i is used like a real factor. 7. The following particular cases of Art. 6 deserve special mention : i. 1 = 1. i = i; 8.0 = 0-1 = 0. 8. It follows directly from Arts. 5 and 6 that the Distribu- tive and Associative Laws hold when the imaginary unit is a factor of the product. E.g., (a ±b)i = ai ± bi; a/'bi = abii = abi\ 296 ALQEBRA. [Ch. Xy, 9. Division by /. — It follows from the definition of division that — is a number which multiplied by i gives ai. i But a X i = ai. Therefore 1 Observe again that i is used like a real factor. 10. We now have, in addition to the double series of real numbers, the double series of imaginary numbers : 3 J, — 2 i, — I, 0, i, 2i, 3i, ••■. Between any two consecutive numbers of this series there are fractional and irrational multiples of i. Thus, between i and 2 i lie f i, -^2 i, etc. 11. Powers of /. — The following valiies of the positive integral powers of V ~ ^j o^ h follow directly from the defini- tion of i and Art. 8 : V— 1 = V— 1> or i = i, (V-l)^=-l, i^ = -l, (V- 1)^= (V- i)X V- 1) = - V- 1, i^^i^-i=-i, (V- i)^=(v- mv- ^y= + 1> i^ = i^ . ^^ = + 1, (V-iy=(V-i)*(V-i)=+V-i. i^ = i^-i=+i, (V-i)''=(V-i)*(V-i)'=-i> i^ = i*.i^ = -L The preceding results give the following properties of powers of i: (i.) All even powers of i are real. (ii.) All odd powers of i are imaginary. 12. Since (V-a)^ = -a, and {-yJax^-Vf = {-^a)\-y/-Vf = -a, we have (-^— a)" = (ya x V" 1)^- Whence y'— a = V Xyf—l. E.g., V-9=V9xV-l = V-l- IMAGINARY AND COMPLEX NUMBERS. 297 13. Addition of Imaginary Numbers. — Imaginary numbers are united by addition and subtraction just as real numbers are united. Ex. 1. V- 9 + V- 16 = 3V- 1 + 4V- 1 = 7 V- 1 = 7 i Ex. 2. 4V- 5 - lOV- 5 + 3V- 5= -3V- 5 = -3V5V-l- Ex. 3. P + i>= = i + (- 1) = 0. 14. Multiplication of Imaginary Numbers. — The following principles enable us to simplify a product of imaginary factors : V- a X V6 = V(«6) X V- 1 = V(- a*) ; V— a X V— * = — V'(<'*)- For v- « X V6=v'«V-lV6=v'«-\/&V-l = V(a6)V-l=V(-«6); and V-« X V-6=\/«v'-l x V^V-l = ^/a^/biy/-ly = -V(«6)- Ex. 1. V- 9 X V16 = SV- 1x4 = 12V- 1 = 12 i. Ex.2. v-2xV-8 = -Vl6 = -4. Ex.3. V-5xV-10xV-15=V5xVl0xVl5x(V-l)' = -5V30V-l=-5V30 -i. 15. Division of Imaginary Numbers. — The following princi- ples enable us to simplify quotients which involve imaginary numbers : For V^^^ ygy-l /g^ ^_ V6 V6 ^/s ■x/g _ y V- 6 V& X V- 1 Vft X (V- 1)' and -v^:^ = V«V-1 =A^ =K V-1 (V-1)^ -1 ^ 298 ALGEBRA. [Ch. XX EXERCISES I. Simplify each of the following expressions : 1. 2V-9-3V-25. 2. 7V- 81 + 5^-144. 3. 8^-75+7-147. 4. -5V-8-3V-32. 5. 2V-a^ + 5v'(-9a2)-3V(-16a2). 6. 2V(-a*6)-4V(-«26=)+ 2^-65. 7. ^3 • V- 3. 8. V3 ■ V- 12. 9. V- 60 - V- 2. 10. V- 3 • V- 6. 11. V- 2 • V- 6 • V- 24. 12. V- 5 • V- 20 • V8- 13. VS • V- 12 • V- 3. 14. V- 15 ■ VIO ■ V2- 15. V- «« • V- a;*- 16. y/(3x^) ■ V - 3. 17. V(- «'''6) • VC- a&') • V(- »^)- 18. V(- ™*»^) • V(- ™»*') • -v/(- ™'«') • -v/C- ™''«)- 19. (V- 3 + V- 2) (7-3 + 7-5). 20. (37-5+47-6)(2V-5-3V-6). 21. (v'-a+v'-6)(V-a-V-6)- 22. [V-(« + 6) + V-6][v'-(a + 6)-V-6]. 23. V-*^- 24. (V-a;)^- 35. V" «*• 26. (V- «)*• 27. y/(l-x)- ^{x -1). 28. VC^-ft)- V(6-«)- 29. i^. 30. ii8. 31. i". 32. ii". 83. i*+J'*. 34. i=6-isi. 35. Cy/-ayK 36. (-aV-a)*^- 37. V" 27-^-3. 38. V-8-V-2- 39. 6 V- 35 -=- 2v'7. 40. y/B ^ i-^ - 5. 41. y/-a-i-y/-a^. 42. ^(,-ab)^^-b. 43. (^_6+V-8)-=-V-2. 44. (V- 12 - v'18)- V- 3. 45. i. 46. V 47. V 48. ^— ■ 49. -J— • Complex Numbers. 16. A Complex Number is the algebraic sum of a real and an imaginary number ; as, 3 ± 2 i. The general form of a complex number is evidently a + hi, wherein a and h are real numbers. When 6 = 0, we have any real number. When a = 0, we have any imaginary number. 17. Two complex numbers which differ only in the sign of their imaginary terms are called Conjugate Complex Numbers ; as, 2 — 3 i and 2 + 3 i- IMAGINARY AND COMPLEX NUMBERS. 299 18. Two complex numbers are said to be equal when the real term of one is equal to the real term of the other, and the imagi- nary term of one is equal to the imaginary term of the other; as,2 + 3i = 2+3i. That is, if a + bi = c + di, then a = c, and bi = Square Root of a Complex Number. 30. If ■y/{a + bi)=^x-\-i^y, then V(« — ^0 = -\I^— WV- For, from V(« + ^0 = V'*' + *V2/j we have a + M = x-y + 2^(xy) • i. Therefore, by Art. l^,a=x-y, and & = 2^{xy). Consequently, a—bi:=x — y — 2^{xy) ■ i = {_y/x-i-y/y)\ Whence V(* - ^*) = V"' — WV- 302 ALGEBRA. [Ch. XX 31. The square root of a complex number can be expressed as a complex number. Assuming V(13 - 20 V - 3) = V« - iy/V, (1) we have V(13 + 20V- 3) = V^* + ^^V- (2) Multiplying (1) by (2), V(i69 + 1200)=a; + y, or a; + 2/ = 37. (3) Squaring (1), 13 - 20 V- 3 = a; - y - 2^{xy) • i ; whence x — y = 13. (4) From (3) and (4), a; = 26, y = 12. Therefore V(13 - 20V- 3)= 5 - 2V3 • i. In general, assuming -^{a + 6i) =^x + iy/y, (1) we have .y/{a — l)i) = ^x—i^y. (2) Multiplying (1) by (2), V(«' + b'^) = x + y. (3) Squaring (1), ' a-\-bi = x — y-\- 2-^{xy) ■ i ; whence a = x — y. (4) From (3) and (4), ^ = VK + 6^)+a^ ^ ^ V("^ + ^^)-'' . Therefore V(a 4.60 = -^ ^^"' + g'^)+ « + ,^ V(a' + 5')- ^ . Since a and 6 are real, therefore ^(a^ +6'') is real. 32. Assuming a = 0, in the general result of the preceding article, we have V(6i) =-y/| + 'VI = ^^^^ *^ ■ "^^ "*" '^• In particular, if 6 = 1, EXERCISES II. Simplify each of the following expressions : I. (l+V-9) + (4-V-4). 2. (6-V-16)-(-5-v/-36). 3. (2 + 4i) + (-3 + 2i). *• (7 - 5i)-(3 - 4 0- 5- (VJ-iV-2)V-8- 6. (V5-V-3)V-3. 7. (2 + 3V-l)(3-4V-l). 8. (5 + 3v/-l)(5-3v'-l)- 9. (5-2v/-6)C3-4v'-3). 10. (7 + V- 5)(7 - V" 6)- IMAGINARY AND COMPLEX NUMBERS. 303 11. (V12-3i)(V3 + 5i). 12. (2 + i^S)(2-iy/3). 13. (2v3+5v'-7)(2v/3-5V-7). 14. (^8 - V- 12)(V2 + V- 3) IS- (i-iV3-i)(3 + 3v'3.0- 16. (V5-2Jv'6)(v'5 + 2iV6)- 17. [a-6+V(-2a6)][a-6-V(-2a&)]. 18. [x+iv'(a-a;2)][«-iV(o-a;^)]. 19- V(l+ V-l)x V(l- V"!)' 20. V( V2 + V3 • «) X V(a/2 - ^3 • 0- 21. (1+^-2)2. 22. (V-2+V3)''. 23. (^^2 ± iV" 3)». 24. (3-5V-2)'. 25. [iV3(3 - i)P- 26. (V-75-3)«. 27. (1+2 at)". 28. (a + &0». 29. (V«+V6-J)*- Reduce each of the following expressions to the form of a complex number : 30 6V3-V-15 gj 8f + 6V2 -t + a -2v'-3 " ■ 4i " <"• ^ 33. 1 34. ^^ l+V-2 2-V-3 3 + 2V5-i 35. 3 + 2V-1. 3g 3 + 4V-5 . 3^ 1+i . 2 - 3v'- 1 4 - SV- 5 (1 + 0* 38. 2 39 lO + V-5 V2-V-2-2V-1 1-V-3+V-5 Factor each of the following expressions : 40. a;2-6a; + 29. 41. a;2 + 4a! + 67. 42. x^-Ux + Gl. 43. 6x^~6x + 2. 44. ix'^+^xy + Zy\ 45. 16 a;^ - 8 aj; + 5 j/». Find the square root of each of the following expressions : 46. l+V-S. 47. 5-V-n. 48. 3 + 4i. 49. -3 + 4i. 50. -15 + 3^-11. 51. ia + a^-h. Make the indicated substitution in each of the following expressions, and simplify the results : 52. In a;2 - 6a; + 14, . let a = 3 + V- 5. 53. In 3a;2-5a; + 7, let a; = 2 - 3^-2. 54. In 5a;8 + 2a;2-3a;-l, - leta; = l-2t\ Simplify each of the following expressions : 65. ^ ^ 56. V-^ + V-i ■ 4-V-14 2-^-14 6 + V-6 3-V-l 67. _3 L_ + _^. 58. a + &» j,a-&». 1+i 4 — 2 i 1 — i a — M a +6i' CHAPTER XXI. QUADRATIC EQUATIONS. 1. A Quadratic Equation is an equation of the second degree in the unknown number or numbers. E.g., a^ = 25, a;^ - 5 a; + 6 = 0, a:? + 2xy = 7. A Complete Quadratic Equation, in one unknown number, is one which contains a term (or terms) in a^, a term (or terms) in X, and a term (or terms) free from x, as oi^—2ax+b=cx—d. A Pure Quadratic Equation is an incomplete quadratic equa- tion which has no term in a;, as a;^ — 9 = 0. In this chapter we shall consider quadratic equations in only one unknown number. 2. The following example illustrates a principle of the equivalence of a quadratic equation to two derived linear equations. The equation aj^ + 6 a; + 9 = 16, or (x + Sf = 16, (1) is equivalent to the two equations a; + 3 = 4, and a; + 3 = - 4, (2) obtained by equating the positive square root of the first mem- ber in turn to the positive and to the negative square root of the second member. For (1) is equivalent to (x + 3)' - 16 = 0. (3) This equation is equivalent to a; + 3-4 = and a!-f-3-f-4 = jointly. But the latter equations are equivalent to x + 3 = A, a;-|-3 = -4. Equations (2) are usually written a; + 3 = ± 4, 304 QUADRATIC EQUATIONS. 305 In general, if the positive square root of the first member of an equation be equated in turn to the positive and to the negative square root of the second member, these two derived equations are jointly equivalent to the given equation. For, the equation M= N is equivalent to M— N=0; that is, to ( V Jf + V-ZV) ( ^/M - -^N) = 0. The last equation is equivalent to ^M-y/N=0 and ^11+^^=0 jointly ; that is, to y/M = ± ^/N. Pure Quadratic Equations. 3. Any pure quadratic equation can be reduced to the form x' — m. rrom this equation we obtain a; = ± -\/m, by Art. 2. Ex. Solve the equation (2 a; — 5) (2 « + 5) = 11. Simplifying, a^ = 9 ; whence a; = ± 3. EXERCISES I. Solve each of the following equations : 1. 3? = 289. 2. a;2_2809. 8. a;" = 3.61. 4. k2 = 53.29. 5. fa;2 = 1536. 6. |a;2 = 1479.2. 7. 9 a;2 - 36 = 5 a;2. 8. 1x^-8 = 9x^-10. 9. (3a;-4)(3a; + 4) = 65. 10. (7 + a;)2 + (7 - a!)2 =130. 11. (2a;-3)(3a;-4)-(x-13)(a;-4) = 40. 12. (5 a; - 7)(3 a; + 8) - (a; - 10)(9 - a;)= 1634. 13. (4 + a;)(3 - a!)(2 - a;)-(a; + 2)(a; + 3)(a; - 4)= 30. 14. (5 - a;)(3 - a;)(l + a;) + (5 + a;)(3 + a;)(l - a;)= 16. 15. 8(2 - a;)2 = 2(8 - a;)2. 16. (3 - a;)^ = 3(1 - a;)^. 17. aa;2 + 6 = 6a;2 + o. 18. a(,x^ + b)=b{x^+ a). Solution by Factoring. 4. The principle on which the solution of an equation by factoring depends was proved in Ch. VIII., § 4, Art. 1. The methods given in Ch. VIII., § 1, Arts. 9-13; Ch. XIX., Art. 20, and Ch. XX., Art. 29, enable us to factor any quadratic expression. The roots of the given quadratic equa- 306 ALGEBRA. [On. XXI tion are the roots of the equations obtained by equating to each of its factors. Ex. 1. Solve the equation 4(a; — f)^' = 6 a; + 20. Reducing the first member, 4 k'' - 12 a; + 9 = 6 a; + 20. Transferring and uniting terms, Factoring first member, 4(a;-f + |V5)(a;-|-|V5)=0. Equating each factor to 0, a; — | + f -y/6 = 0, a;-f-fV5 = 0. Whence a; = | — f V^j ^-^d « = f + f V^- Ex. 2. Solve the equation 4 m'a^ + 4 m'n + 1 = 4 ma;. Transferring terms, 4 mV — 4 mx + 1 + 4 m^n = Q, or (2 mx - If - (2 m V- w)^ = 0. Equating to the factors of the first member, 2 mx — 1+2 m-y/— n = 0, 2 ma; — 1 — 2 m-yj— w = 0. Whence a; = - V— »»)anda; = hV— w. 2m 2m BXEBCISES II. Solve each of the following equations : 1. a;2-7a: = 4a;. 2. x^-2x-n = Q. 3. 05^ - 6 a; + 8 = 0. 4. a;" - 4 a; + 8 = 0. 5. a;2 - 4 a; - 71 = 0. 6. a;^ + 10 a; + 24 = 0. 7. 13a:-6-6a;2 = 0. 8. (a; + 10)2 = 28. 9. 6a;-a;2 = 18. 10. 7a;2-3x = 160. 11. (%x-iy = 2. 12. a;(5a;-2) = -6. 13. a;2-2V2a;-l = 0. 14. 36 a;^ - 36^5 a; + 17 = 0. 15. (a; + 8)(a; + 3) = a;-6. 16. (x + 7)(a; - 7) = 2(a; + 50). 17. (2a; + l)(a; + 2) = 3x«-4. 18. (a; - l)(2a; + 3)= 4a;2- 22. 19. a;2-3 = KK-3). 20. a;(a; + 5) = 5(40 - a;) + 27. QUADRATIC EQUATIONS. 307 21. 7a;(a;-l)=7-4(a;-l). 22. (2x + iy = x(,x + 2). 23. x^-2ax + a^ = bK 24. x^-2mx-l= 0. 25. x^-2ax + a^ + b^ = 0. 26. x^ -ib^ = a(2x -5a). 27. n2a;2 + 2 mna; + 2 m^ = 0. 28. a;^ - (a + b}x + ab = 0. 29. (o2 + 62)a; _ a6x2 - o& = 0. 30. x^ + 2mx + m'' = n. 31. a;2 + 2a + l = 2(a:-a). 32. a^aj^ - 2 aa; + 1 = a^. 33. 4a;2-12ax + 9a2 = 462. 34. (a; + a)2 = 5 aa; - (« - a)". 35. a;2-4(a + 6)+l=2a;. 36. a;^ = 2(a + 6)a; + 2(0^ + 6^). 37. (to2 - l)a;2 - 2 (m^ + l)a; + m^ - 1 = 0. 38. mnx'^ - (m + m) (_mn + l)a; + (m^ + 1) (re^ + 1) = 0. 39. (m - 7t)2(m + n)x'^ + 2(m-n)(m+ n^x + 6 m% + 2 «» = 0. Solution by Completing the Square. 5. The following examples illustrate the solution of a quad- ratic equation by the method called Completing tJie Square. Ex. 1. Solve the equation o^ — 5a!-|-6 = 0. Transferring 6, a? — 5 x = — 6. To complete the square in the first member, we add (—f)', = ^, to this member, and therefore also to the second. We then have x^-5x + s^ = s^-6 = \. Equating square roots, a; — |- = ± ^, by Art. 2. Whence x = ^±^. Therefore the required roots are 3 and 2. Ex. 2. Solve the equation 7ce= + 5a!-|-l = 0. Transferring 1, 7a:?-|-5a; = — 1. Dividing by 7, a? -\- ^ x = — ^. Adding (^y = ^, ^ + lx + ^^^ = ^-^ = ^. Equating square roots, x -|-y\ = ± ^y^. Whence a; = — ^ ± ^. Therefore the required roots are -A + AV-3 and -^_^V-3. 308 ALGEBRA. [Ch. XXI Ex. 3. Solve the equation (a^ - b'')x' -2a'x + a' = 0. Transferring a% {a^ — h^)3? — 2d'x = — a^. Dividing by a? — W, a? — ■ d'-h^ ^^•l^^g (-^^J' = (^36i)i' *° ^°*^ members, .., 2a?x a* ^ a' a' _ a^W Equating square roots, x — — = ± Whence x = a'-b^ a" ± ab a^-b^' Therefore the required roots are and a—b a+b The preceding examples illustrate the following method of procedure: Bring the terms in x and a? to the first member, and the terms free from x to the second member, uniting like terms. If the resulting coefficient of a? be not +1, divide both members by this coefficient. Complete the square by adding to both members the square of half the coefficient of x. Equate the positive square root of the first member to the posi- tive and negative square roots of the second member. Solve the resulting equations. EXERCISES III. Solve each of the following equations : 1. a;2-4a; + 3 = 0. 2. a;2-53; = -4. 3. a;2 + 2 K + 1 = 0. 4. 2 a;^ - 7 a; + 3 = 0. 5. 3 a;2 - 53 X + 34 = 0. 6. 14 x - 49 x^ - 1 = 0. 7. x2-4x + 7=0. 8. (2x-l)(x-2) = (x + l)2. 9. x2 _ 2 X + 6 = 0. 10. x2 - 1 + x(x - 1) = x2. 11. (3x-2)(x-l)= 14. 12. 110x2-21x+ 1 =0. QUADRATIC EQUATIONS. 309 13. (2 a; - 3)2 = 8 a;. 14. (5 a; - 3)' - 7 = 40 a; - 47. 15. (2a; + l)(x + 2)=3»2-4. 16. (x + ])(2 a; + 3) = 4a;2 - 22. 17. (x-7)(a;-4) + C2a;-3)(x-5)=103. 18. 10 (2 X + 3) (x - 3) + (7 X + 3)^ = 20 (x + 3) (x -1). 19. (x-l)(x-3) + (x-3)(x-5)=32. 20. (x - l)(x - 2) + (x- 3)(x -4) = (x - 1)^ - 2. 21. (m-«)x2-(m+n)x+2M=0. 22. 4x^-4(3 a+2 6)x+24«6=0. 23. (a + 6)2x2 + 2 a6x = ^^ — (a + 6)2 24. x2 - 2(a + 6)x + (o + 6 + c)(a + 6 - c) = 0. 25. x2 - (a - l)x - a= 0. 26. x2 _ 2 ex + ac + 6c - a6 = 0. 27. x2-4mrex = (m2-M2)2. 28. ^2x2 - 4 a6dx + 4 0^62 - 9 c2 = 0. 29. (a2 - 62)x2 - 2(a2 + 62)x + a^ - 62 = 0. 30. a6ca;2-(o262 + c2)^ + a6(._0. 31. x'^^& -(,y/2-\-y/Z)x+l = Q. 32. x2-2xV(a + 6)+26 = 0. 33. (a^ - 4a)(x + 2 6 - 2) ^_ 3^ , ^ ^ (a + 6-l)(a-6 + l) g^ x2 _ 6 + C-X _ (6 + e)x2 a'^ + ah -\- ac a + 6 + c a^ + aV) -\- a'^c General Solution. 6. The most general form of the quadratic equation in one unknown number is evidently aa? + &a; + c = 0. The coefficient a is assumed to be positive and notO, but 6 and c may. either or both be positive or negative, or 0. Dividing by a, a,2 + ^a! + -^ = 0. a a Transferring -, a ^ 1 ^™ c or + -x = • a a Adding (AJ,. 6^ 4a^ a 'tar 4= a' a b^ — Aac Equating square roots, x + — - = ± ^ ^ ^• 310 ALGEBRA. Whence 2a ' 2a ' and . ^_ & V(&^-4ac) [Ch. XXI 7. The roots of any quadratic equation can be obtained by substituting in the general solution the particular values of the coefScients a, b, and c. Ex. Solve the equation 3a^ + 7a;— 10 = 0. We have a = 3, 6 = 7, c = - 10. Substituting these values in the general solution, we obtain x = -^ + -^*°-^^"'-^''" = l, and a! = -|.- -^^*°-%^^'-"'" = -J/. BXEECISES IV. Solve each of the following equations : 1. 2a;2 = 3a;+2. 2. 5*2 _ 6a; + 1 = 0. 3. 9 x{x + 1) = 28. 4. a;2 - 62 = 2 aa; - a^. h. x^ + Qax + \ = 0. 6. a;2 + 1 = 2 J K. 7. (a; -5)2+ (a; -10)2 = 37. 8. 2 x(_S n - 4: x) = n^. 9. n2(a;2 4.i)=a2 + 2re2a;. 10. a;^ + (a; + a)^ = a". Fractional Equations vrhich lead to Quadratic Equations. 8. The principles given in Ch. X. for solving fractional equations which lead to linear equations hold also for frac- tional equations which lead to quadratic equations. Ex. 1. Solve the equation — ^ = -^^ + 2. (1) x — 1 ar — 1 Multiplying by a^ - 1, 4 (a; + 1) = 3 a; + 2 (a?" - 1). (2) Transferring and uniting terms, 2 a;" — a; = 6. (3) Dividing by 2, x^-^x = 3. (4) The roots of equation (4) are 2, — |. Since neither is a root of the L. C. D. (equated to 0) of the fractions in the given equation, i.e., of ar* — 1 = 0, they are the roots of that equation. QUADRATIC EQUATIONS. 311 Ex. 2. Solve the equation 1 1 iB2_2n-»i» (1) n + x n — x x' — n" Uniting the fractions in the first member, -2a! a?-2n-w' (2) Clearing of fractions, 2x = ae'-2n-n\ (3) Transferring and uniting terms, x'-2x = n^ + 2n. W Completing the square. 3?-2x + l = n' + 2n + l. (5) Equating square roots, a; - 1 = ± (w + 1). Therefore the roots of (5) are 1 + (n + 1) = 2 + w, and 1 - (m + 1) = - w. The number 2 + m is not a root of the L. C. D. equated to 0, that is, of a?— 71? = a. Therefore 2 + w is a root of the given equation. But —n is a root of x^ — n^ = Q, or of (a; — n) (a; + n) = 0, and is therefore not a root of the given equation. This root was introduced by multiplying the given equation by the factor x + n which was not necessary to clear it of fractions. For, transferring and uniting the fractions in equation (2), we obtain :>? -2x-2n-n-^ ^ ^ x^ — v? Factoring the numerator, (a; + w) (a; — w — 2) _ ^ a? — n^ Eeducing to lowest terms, ^~''^~ — o. x — n The numerator equated to gives a; — « — 2 = 0, whence x = 2 + n, as above. 312 ALGEBRA. [Ch. XXI 9. The work of solving an equation can sometimes be sim- plified by a simple substitution. a; + 5 a; + 2 _3 x + 2 a; + 5~2 Ex. Solve the equation If we let = y, the given equation becomes y = -• The roots of this equation are 2, — ^. We now have to solve the two equations x + 5_ and x + 2 x + 5 x + 2 ■■ 2, whence x=l; = — ^, whence a; = — 4. This method can be used when the fractional equation con- tains only two expressions in the unknown number, one of which is the reciprocal of the other. EXERCISES V. Solve each of the following equations : 1. 15a: + - = ll. X . + 1 = 7 + 1. X 7 4. 7. 10. 12. 14. 16. 18. X 14 a; + 120 3a;- 10 X— 1 3x x-1 15 , 5 :2. x + 2 x+1 _ 5. 8. = 0. X x-6 x + i + 1 = X ■ X - 9a;-3a;2 6x x+3 a; + 24 1 , 7x! x-% X 21 -7a;" 106 a; + 4 11. 60 a; + 4' 3' 3 a;-6 a; + 30 3 + a;-6 3 1 + a; 1 — a; a; + l a: + 3 _ll a;-3 2' X 1 a; + 3 ' 21a; ' 1 8a;2-72 a;-7 1 5a;2 — 5 a;+l 2a;-2 a:3 + 3gg-|-3a;+l _9 a;3 + 2 a;2 + 2 a; + 1 ?" 2 a; - 2 z^ - 1 13. 15. 17. 19. 1) 4(x + 1) , 1 , 1 : + - = 0. 2(a;2 1 X — ' x-2 a; + 2 _ 2Ca; + 3) a; + 2 a;-2~ a;-3 4a; + 67 . a; _2 40a;2-36 Sx 16 2a; -12 6x + 5 X- 1 a;2 + 1 6 'x-\-l 21. 24. x-«=^- 6 a 2a; -12 3a; I 22 30 a;2 _ 27 5 2' 16 x~ a X —1 X + a b + X 25. X x-1 x+1 QUADRATIC EQUATIONS. 313 26. 28. an x+ in x — in (a — b^x a — b 1 _ ix gg x' + l 1__ ^ X {a^-xy 4 a;3 + 5 a6 ' n^-2n 2 - nx n 30 a-^b _ 1 _ J__ 31 1 _ 2a _ a^ + b^ 8x2-262 2x+6 2a' ' x-a a^' + x^^-2ax qg, a a — 1 1 53 n + x , n — x _ n^ nx — X x^ — 2 nsi^ + nV n — x n + x n'^ — x^ g. X— a+ b _ a — b — X „, ax 1 — g ' x + a — b a + b + X ' az+ 1 a^^ — a — aH + ax Wa-x 121 a2 - x'^ 36. ('£±^Y+I.«±^+3=0. 37. \a — xj 2 a — a: 9 a2 - a;2 38. x + l.a — 6_a;-la + 6 jg X — 1 a + 6 x + 1 a — 6 a+b+x a b x 40. K— 6,a; — a_a.6 .. x—ax—bba (a-xy+(x-by_ J (a - a;)2 + (a; - 6)2 40 (a-xy+(,x~by_a^-b' ^3 (ia-xy+(x-by a' -6' (a - a;)2 + (a; - 6)= a^ + 6^ (a + 6-2a;)2 (0 + 6)2 44. x^-2nx + 2ax-n^ . x + 2n x3 _ aS x^ + ax + a^ 1 X — a Theory of Quadratic Equations. 10. A quadratic equation has two, and only two, roots. For, by Art. 6, the equations j.^ 6 I VC6'-4ac) 2o 2a and ^^A- V(6'-4«c) 2a 2a are jointly equivalent to the equation aa;2 + 6a; + c = 0. Therefore ax^ + bx + c has two, and only two, roots. 11. Relations between the roots of a quadratic equation and the coefGlcients of its terms. — If the roots of the quadratic equation aa;2 + 6a; + c = 0, or a;2 + - a; + - = a a be designated by ri and ra, we have ^ _ 6 , v/C62-4gc) r,=-A_ VC&'-4«c) . 2a 2a 314 ALGEBRA. [Ch. XXI The sum of the roots is n 4- r2 = — • (1) a The product of the roots is ^^ L 2a 2a J L 2a 2a J L 2aJ L 2 a J ia^ 4a'' a ^ The relations (1) aud (2) may be expressed thus : (i.) The sum of the roots of a quadratic equation is equal to the quotient obtained by dividing the coefficient of the first power of the un- known number, with sign reversed, by the coefficient of the second power of the unknown number. In particular, if the coefficient of the second power of the unknown number be 1, the sum of the roots is equal to the coefficient of the first power of the unknown number, with sign reversed. (ii.) The product of the roots of a quadratic equation is equal to the quotient obtained by dividing the term free from the unknown number by the coefficient of the second power of the unknown number. In particular, if the coefficient of the second power of the unknown number be 1, the product of the roots is equal to the term free from the unknown number. E.g., the roots of the equation 6a;'' — k — 2 = are f and — ^ ; their sum is \ (the coefficient of x, with sign reversed, divided by the coefficient of a;^), and their product is — J (the term free from x divided by the coefficient of a;'). The roots of the equation a;^ — 5a; + 6 = are 2 and 3 ; their sum is 5, and their product is 6. 12. Formation of an Equation from its Roots. — The relations of the last article enable us to form an equation if its roots be given. We may always assume that the coefficient of the second power of the unknown number is 1. Ex. 1. Form the equation whose roots are — 1, 2. We have n + r2 = — 1 + 2 = 1, the coefficient of x, with sign reversed ; and riTi =— 1x2= — 2, the term free from x. Therefore the required equation is a;^ — a; — 2 = 0. Ex. 2. Form the equation whose roots are 1 + 2-^3, 1 — 2^*3. We have n + r2 =(1 + 2V3) + (1 - 2^3) = 2 ; and rir2=(l + 2V3)(l-2v3) = l-12=-ll. Therefore the required equation is x" — 2x — 11 =0. QUADRATIC EQUATIONS. 315 Ex. 3. Form the equation whose roots are 2 + 5-^/— 3, 2 — Sy"^ 3. We have n + ra = (2 + 5 V- 3) + (2 - 5 V- 3) = 4 ; and ri>-2 = (2 + 6 V- 3) (2 - 6 V- 3) = 4 + 75 = 79. Therefore the required equation is x^ — ix+ 19 = 0. 13. The roots of a quadratic equation, all of whose terms are in the first member, are the roots of the two linear factors into which this mem- ber can be resolved. Consequently a quadratic equation whose roots are given can be formed by multiplying together the two linear factors which (equated to 0) have as roots the given roots. Ex. Form the equation whose roots are —1,2. Since — 1 is the root of a; + 1 = 0, and 2 is the root of a; — 2 = 0, the required quadratic is (a; + 1) (a; — 2) = 0, or x^ — x — 2 = 0. When the roots are irrational or imaginary, the method of the pre- ceding article is to be preferred. 14. One Root Known. — When one root of a given quadratic equation is known, the other root can be found without solving the equation. Ex. 1. One root of the equation a;'^ — 5a; + 6 = is 3; what is the other root ? Since 5 is the sum of the roots, the required root is 5 — 3, =2. Or, since a;^ — 5 a; -F 6 is the product of two linear factors, one of which is a; — 3, the other factor is x^-5x + 6 ^j._2 a;-3 ' The required root is therefore the root of a; — 2 = 0, or a; = 2. Ex. 2. One root of the equation a;2-|-2a; — 1 = 0, is— 1-|-y'2; what is the other root ? The required root is - 2 - ( - 1 -h V^) , = - 1 - V^- 15. Expressions Symmetrical in the Roots. — The value of an ex- pression which is symmetrical in the roots of a given quadratic equation can be found without solving the equation. Ex. If n and ra be the roots of x'' +px + q = 0, find the value of ri^ + r^. We have r-? -h r^ = (n + rj)^ - 2 nra = (-p)^ - 2 g =p2 _ 2 g-. 16. We can sometimes form an equation whose roots have definite relations to the roots of a given qualdratic equation without solving the latter. 316 ALGEBRA. [Ch. XXI Ex. Form an equation whose roots are the reciprocals of the roots of the equation a;^ — 8 x + 15 = 0. Let ri and ra be the roots of the given equation. Then — and — are the roots of the required equation. ^^ '^ We have — H — = ^^ ^^ = — , the coefficient of x (with sign reversed) ri J-2 J'ir2 15 in the required equation ; and — x — = = — , the term free from x in the required equation. ''^ '"^ ''^'^ Consequently the required equation is x^-^x + ^ = 0, or 15a;2-8a;+ 1 =0. Nature of the Roots of a Quadratic Equation. 17. In many applications it is Important to know, without having to solve an equation, the nature of its roots, i. e. , whether they are both real and unequal, whether they are both real and equal, whether they are imaginary, etc. In the general solution 2a 2a ^ 2a 2a of the equation aa;^ + 6a; + c = 0, > a, 6, and c are limited to real, rational values. (i. ) The two roots are real and unequal when Ifl — ^ac is positive, i.e., lohen 6^ — 4 ac > 0. E.g., in x'' + 'ix-12 = 0, a =1, 6 = 4, c = — 12 ; and since 6^ _ 4 g.c, = 16 + 48, is positive, the roots of this equation are real and unequal. (ii. ) The two roots are real and equal when V^ — iac is equal to ; i.e., when V = 4oc. ^.^., in a;2-4x+4 = 0, a = l, 6= — 4, c = 4; and since 6^ = 4 ac, the roots of this equation are real and equal. (iii.) The two roots are conjugate complex numbers when b^ — iac is negative ; i.e., when 6^ _ 4 3c < 0. E.g., in a;^ - 2 a; + 3 = 0, a = l, 6= — 2, c = 3; and since 6^ — 4 ac, =4 — 12, = — 8, is negative, the roots of this equation are complex numbers. (iv. ) The two roots are real and rational when 6^ — 4 ac is positive and the square of a rational number. QUADRATIC EQUATIONS. 317 Eg-, in x^-5x + i = 0, a = l, 6=— 5, c = 4; and since V — i ac, = 9, is tlie square of a rational number, the roots of this equation are real and rational. (v.) The two roots are real, but conjugate irrationals, when b'^ — 4^ac is positive and not the square of a rational number. JS.g., in x^-6x + 2 = 0, a = l, 6=-5, c = 2; and since 6^ _ 4 ac, =25-8, = 17, is positive and not the square of a rational number, the roots of this equation are conjugate irrational numbers. (vi.) The two roots are equal and opposite when 6 = 0. We then have n = V^"^""), and r^ = - V{-^a<0 . 2a 2a Notice that in this case the roots are real or imaginary according as ac is negative or positive. E.g., from 2 a;^ + 5 = we have x = V— f = ± i^/- 10. (vii.) One root is zero when c = 0. We then have n = — -+ ^ = -^ + -^ = ; 2a 2a 2a 2a r, = — L_A^' = _A__L = _6. 2a 2 a 2a 2a a E.g., from 2 a;^ _ 3 a; = 0, we have a; (2 a; - 3) = ; whence a; = 0, and a; = |. (viii.) Both roots are zero when 6 = and c = 0. We then have n = + 0, ?-2 = - 0. E.g., from 3 a;'' = we obtain a; = and a; = 0. 18. As long as a is not equal to 0, however near its value may be to 0, the values of r\ and ra given above constitute the solution of the equa- tion. If, then, we assume that these values still give the roots when a = 0, we must determine the nature of the roots under this assumption. (i.) One root is infinite and the other — '-, when a = 0, 6 =7^ 0, c =^ 0. 6 For n - -^+ VC6^-^ °"=) = r-6+ VC5^-4 fflc)1 [-6- V(6'-4 ae)^ 2a 2a[-6-V(6^-4ac)] ^ 62 _ (62 _ 4 ac) 2c 2a[-6-v/(6''-4ac)] -j) -^(pi -iac) If in we let o = 0, we obtain n = — — — = - -. -6-^(6^— 4 ac) , -6-6 b 318 ALGEBRA. [Ch. XXI Also, r2 = -i^— = —^ = oo. In this case we are apparently dealing with, the linear equation bx + c = 0, and not with a quadratic. But in applications of Algebra it is frequently necessary to consider the coefficient of a;^ as growing smaller and smaller without limit, i.e., as approaching 0. The meaning of the results given ahove are that as a grows smaller and smaller without limit, one root grows larger and larger without limit, and the other root becomes more and more nearly equal to 6 E.g., the equation O.K2 + 2a; + 3 = has one root oo and one root — |. (ii.) Both roots are infinite when a = 0, 6 = 0, c =^ 0. 2 c We have n = , as above. -6-V(6°-4ac) And r _ -6-V(5^-4ac) ^ 62_(-62_4ac) 2a 2a[-6 + V(6''-4ac)] 2c -6 + V(&'^-4«c) If we now let a = 0, 6 = 0, c T^ 0, we obtain 2c J 2c ri = — = 00, and r2 = — = co • Attention is called to the remarks at the end of (i.). H.g., the equation O.a;2 + 0.a; + 2 = has two infinite roots. 19. If, in simplifying a quadratic equation, the terms of the second degree in the unknown number be canceled, an infinite root is lost. E.g., solve the equation (1 + 2 a;) (2 — 3 a) = 5 — 6 a;^. Performing indicated operations, 2 + x — 6x^ = 5 — 6x^. Transferring and uniting terms, k = 3. In canceling — 6 a;^ an Infinite root was lost. It was for this reason that in the principle for adding or subtracting the same number or expression to or from both members of an equation the roots were limited to finite values. Maxiiaa and Minima. 20. Pr. 1. What is the least value of a;^ + 6 a; + 11 for real values of a; ? What is the value of x which gives this least value ? Let x'^ + ex + ll = y. Then we are to find the least value of y for aU possible real values of x. QUADRATIC EQUATIONS. 319 We then have »^ + 6x + 9 = y — 2; or a; + 3=v(^-2). Now X -will be real only for values of ?/ = or > 2. That is, 2 is the least value ot y, =x^ + 6x+ 11, for real values of x. When 2^ = 2, x=-3. Ft. 2. Between what bounds do the values of the fraction ^ + ag — 8 lie for real values of a; ? x — 2 Let t±iE^ = y. (1) Whence a;^ + (4 - y)x + (2y -8) = 0. Then, by Art. 17 (i.), the values of x will be real when (4-2/)2-4(2 2/-8)>0; that is, when y^-l6y + i8>0, or (i/ -i)(y - 12)> 0. This expression will be greater than 0, for all values of J/ < 4, and for all values of !/ > 12. That is, the given fraction can have all values between — oo and 4, and between 12 and + oo. The values of x corresponding to the bounds of the fraction are found by solving the equations obtained by equating the given fraction to 4 and 12 respectively. We then have : When y = i, jc = ; when ^ = 12, x = i. Pr. 3. Under what condition can aa;2 + 2hxy + by^ + 2gx + 2fy + G be resolved into linear rational factors ? Solving the given expression as a quadratic in x we obtain ax + hy + g = ±^l(ih^- ab)y^ + 2(hg - af)y + (g^ - ac)-]. (1) Equation (1) gives two linear rational factors when the expression under the radical is a perfect square ; that is, when (hg - a/y - (K> - ah) (g^ - ac) = ; or abc + 2fgh - aP - bg^ - ch^ = 0. EXERCISES VI. Form the equations whose roots are : 1. 8, 2. 2. - e, - 3. 3. 10, 10. 4. 7, - 3. 5. 4,-10. 6. 2^, If. 7. -l-lh 8. -J, 8. 9. 2, 0. 10. a, b. 11. - a, - 1. 12. a^, - 4 a^. 13. ^, ^. 14. ^+-^, 1. 15. « * b a a-b 2a-2b 2b-2a 320 ALGEBRA. [Ch. XXI 16. V2. -V2- 17. 5V7, -5V7. 18. W- 3, - iV- 3. 19. 1+V7, 1-V7- 20. ^-iVlia + iVll- 21. 3-V-5,3+V-5- 22. l-iv-l, l + iV-1- Find the second root of each of the following equations, without solving the equation: 23. a;2 - 9 a; + 20 = 0, when n = 4. 84. 6 a;2 _ X - 1 = 0, when n = - f 25. a;2 _ (a ^. j'ja _ o, when n = a + 6. 26. x^ - (a2 - 62)a; = o, when n = 0. 27. 62a;2 + 2 afta; + o^ = o, when n = - -• 6 28. (a^ - 62)a;2 + 4 afta - a^ + 6^ = o, when n a + b a;2 - 2 a; - 1 = 0, when n = 1 - y/2. 30. x2 + a; V5 + 1 = 0, when n = - i( ^5 + 1). 31. a;2 - 6 X + 13 = 0, when n = 3 + 2 V- 1. 32. (a + 6)2a;2 - (a + 6)ca; - ac = 0, when ri = " ~y/^' "^.f ""•' • 2(a + 6) If ri and r2 stand for the roots of the equation x^ +pz + q = 0, express each of the following symmetrical expressions in terms oip and q : 33. n^ + r^'. 34. n^ + r^. 36. ri* + j-2*. 36. 1+1. 37. '^+^^. 38. !i±i:?+rL±i^. 39-44. Express each of the relations given in Exx. 33-38 in terms of the roots of the equation a;^ + a; — 6 = 0. Without solving the equations x^ +px + q = and a;'' — 10 a; + 40 = 0, form the equations whose roots are 45. -The opposites of the roots of the given equations. 46. The reciprocals of the roots of the given equations. 47. Twice the roots of the given equations. 48. The roots of the given equations increased by 2. 49. The roots of the given equations diminished by 5. 50. The squares of the roots of the given equations. "Without solving the following equations, determine the nature of the roots of each one : 51. a;2 + 17 X + 70 = 0. 52. x^ - 6 x = 27. 53. x2 + 12x = -40. 54. x2-6x + 9 = 0. 55. x2 + 5x-14 = 0. 56. x2 + 20x = -100. QUADRATIC EQUATIONS. 321 57. a;2-x = 12. 58. a;' -8a; + 25 = 0. 59. a;2 -13a; + 22 = 0. 60. a;2-8a: = 16. 61. 9a;2-12a; + 4=0. 62. 8a;2 - 2a; - 25 = 0. 63. 16a;2 + 8x + 49 = 0. 64. lOa;^ -21x - 10 = 0. 65. 16S62 + 40a; + 26 = 0. 66. 25a;2 + 4a;- 77 =0. 67. 209;2 + 19a; = -3. 68. 4x2 + 52 a; = 87. For what values of m are the roots of each of the following equations equal ? For what values of m are the roots real and unequal 1 And for what values of m are the roots complex numbers ? 69. mx^ + 4 X + 1 = 0. 70. 2 x^ + mx + 1 = 0. 71. 3 x2 + 6 X + m = 0. 72. mx^ + mx + 1 = 0. Find the greatest, or the least, value of each of the following expres- sions for real values of x, and the corresponding value of x : 73. x2-ex + 7. 74. 10 + 3x-x2. 75. 2 x2 - 5 X + 7. 76. 3 + 8 X - 5 x2. Find the bounds of each of the following fractions for real values of X, and the corresponding values of x : ^_ x^ + 3x-3 ,^g x^ + X + 2 ,^g 4x2 + 24x + 27 x-l' ■ x + 2' ■ 13 -8x' 80 *^^ + 3 81. ^^^ + '^ ■ 82. ^ + 3 4x2 + 8x + 3 4x2 + 8x + 2 x2 + 6x-3 Which of the following expressions can be separated into rational linear factors? 83. 3x2 + 13xy-10j/2- 11X + 132/-4. 84. 5x2-14x!/-3!/2-8x + 8j? + 3. 85. 2x2 -4x2^-62/2- 3a; + 17^-4. 86. ix^-lxy + Zy^+Qx-By + S. 87. Divide 100 into two parts so that their product shall be a maximum. 88. Divide 20 into two parts so that the sum of their squares shall be a minimum. 89. Two points, A and B, move along two perpendicular lines. A is 13 feet and B is 16 feet from the point of intersection, P. A moves at the rate of 4 feet a second toward P, and B at the rate of 3 feet a second away from P. After how many seconds will A and B be the least dis- tance from each other ? 322 ALGEBRA. [Ch. XXI Problems. 21. Fr. 1. The sum of two numbers is 15 and their product is 56 ; what are the numbers ? Let X stand for one of the numbers ; then, by the first con- dition, 15 — a; stands for the other number. By the second condition a;(15 — a;)=56; whence x = 7, and 8. Therefore x = 7, one of the numbers, and 15 — a; = 8, the other number. Observe that if we take a; = 8, then 15 — a; = 7. That is, the two required numbers are the two roots of the quadratic equation. Pr. 2. Divide 100 into two parts whose product is 2600. Let X stand for tlie less part, and 100 — x for the greater. By the second condition, a; (100 — a;) = 2600. The roots of this equation are 50 + 10-^— 1 and 60 — 10-^/— 1- An imaginary result always indicates inconsistent conditions in the problem. The inconsistency of these conditions may be shown as follows : Let d stand for the difference between the two parti of 100. Then &0 + ^d stands for the greater part, and 50 — |- a for the less. The product of the two parts is (50 + 1 d)(50 - i d), = 2500 - {\ df = 2500 - J d^. Since d^ is positive for all reed values of d, the product 2500 — \d' must be less than 2500. Consequently 100 cannot be divided into two parts whose product is greater than 2500. 22. When the solution of a problem leads to a quadratic equation, it is necessary to determine whether either or both of the roots of the equation satisfy the conditions expressed and implied in the problem. Positive results, in general, satisfy all the conditions of the problem. A negative result, as a rule, satisfies the conditions of the problem, when they refer to abstract numbers. When the QUADRATIC EQUATIONS. 323 required numbers refer to quantities which can be understood in opposite senses, as opposite directions, etc., an intelligible meaning can usually be given to a negative result. An imaginary result always implies inconsistent conditions. 23. The interpretation of a negative result is often facilitated by the following principle : If a given quadratic equation have a negative root., then the equation obtained from the given one by changing the sign of x has a positive root of the same absolute value. Let — r be a root of ax^ + 6a; + c = 0. (1) Then, since — r must satisfy the equation, we have a(-r)2 + 6(-j-) + c = 0, or ar^-br + c = 0. (2) But equation (2) shows that r satisfies the equation ax^ — bx + c = 0, which is obtained from (1) by changing the sign of x. Pr. A man bought muslin for $3.00. If he had bought three yardi more for the same money, each yard would have cost him 5 cents less. How many yards did he buy ? Let X stand for the number of yards the man bought. Then 1 yard cost — cents. If he had bought a; + 3 yards for the same money, each "^ 300 yard would have cost cents. Therefore — - ^'^ = 5 ; whence a; = 12 and - 15. X x + S The root 12 satisfies the equation and also the' conditions of the prob- lem ; the root — 16 has no meaning. But if x be replaced by — x in the equation, we obtain a new equation 300 300 _ g ^j. J00__300^5^ ^2) — X -a; + 3 ' a;-3 x whose roots are — 12 and + 15. Equation (2) evidently corresponds to the problem: A man bought muslin for $3.00. If he had bought 3 yards less for the same money, each yard would have cost him 5 cents more. Notice that the intelligible result, 12, of the first statement has become — 12 and is meaningless in the second statement. Attention is called to the remarks in Ch. XII., Art. 6. 324 ALGEBRA. [Ch. XXI EXERCISES VII. 1. If 1 be added to the square of a number, the sum will be 50. What is the number? 2. If 5 be subtracted from a number, and 1 be added to the square of the remainder, the sum will be 10. What is the number 1 3. One of two numbers exceeds 50 by as much as the other is less than 50, and their product is 2400. What are the numbers 1 4. The product of two consecutive integers exceeds the smaller by 17,424. What are the numbers ? 5. If 27 be divided by a certain number, and the same number be divided by 3, the results will be equal. What is the number ? 6. What number, added to its reciprocal, gives 2.9 ? 7. What number, subtracted from its reciprocal, gives ?i? Let ra = 6.09. 8. If n be divided by a certain number, the result will be the same as if the number were subtracted from re. What is the number 1 Let m=4. 9. If the product of two numbers be 176, and their difference be 5, what are the numbers ? 10. A certain number was to be added to \, but by mistake \ was divided by the number. Nevertheless the correct result was obtained. What was the number? 11. If 100 marbles be so divided among a certain number of boys that each boy shall receive four times as many marbles as there are boys, how many boys are there ? 12. The area of a rectangle, one of whose sides is 7 inches longer than the other, is 494 square inches. How long is each side ? 13. The difference between the squares of two consecutive numbers is equal to three times the square of the less number. What are the numbers ? 14. A merchant received 1 48 for a number of yards of cloth. If the number of dollars a yard be equal to three-sixteenths of the number of yards, how many yards did he sell ? 16. In a company of 14 persons, men and women, the men spent |24 and the women $24. If each man spent |1 more than each woman, how many men and how many women were in the company ? 16. A pupil was to add a certain number to 4, then to subtract the same number from 9, and finally to multiply the results. But he added the number to 9, then subtracted 4 from the number, and multiplied these results. Nevertheless he obtained the correct product. What was the number ? QUADRATIC EQUATIONS. 325 17. A man paid $ 80 for wine. If he had received 4 gallons less for the same money, he would have paid f 1 more a gallon. How many gallons did he buy ? 18. A man left $31,500 to be divided equally among his children. But since 3 of the children died, each remaining child received $3375 more. How many children survived ? 19. Two bodies move from the vertex of a right angle along its sides at the rate of 12 feet and 16 feet a second respectively. After how many seconds will they be 90 feet apart ? 20. A tank can be filled by two pipes, by the one in two hours less time than by the other. If both pipes be open 1| hours, the tank will be filled. How long does it take each pipe to fill the tank ? 21. From a thread, whose length is equal to the perimeter of a square, 36 inches are cut off, and the remainder is equal in length to the perime- ter of another square whose area is four-ninths of that of the first. What is the length of the thread ? 22. A number of coins can be arranged in a square, each side contain- ing 51 coins. If the same number of coins be arranged in two squares, the side of one square will contain 21 more coins than the side of the other. How many coins does the side of each of the latter squares con- tain? 23. A farmer wished to receive $2.88 for a certain number of eggs. But he broke 6 eggs, and in order to receive the desired amount he increased the price of the remaining eggs by 2§ cents a dozen. How many eggs had he originally ? 24. Two bodies move toward each other from A and B respectively, and meet after 35 seconds. If it takes the one 24 seconds longer than the other to move from A to B, how long does it take each one to move that distance ? 25. It takes a boat's crew 4 hours and 12 minutes to row 12 miles down a river with the current, and back again against the current. If the speed of the current be 3 miles an hour, at what rate can the crew row in still water ? 26. A man paid $ 300 for a drove of sheep. By selling all but 10 of them at a profit of $2.50 each, he received the amount he paid for all the sheep. How many sheep did he buy ? 27. Two men start at the same time to go from A to B, a distance of 36 miles. One goes 3 miles more an hour than the other, and arrives at B 1 hour earlier. At what rate does each man travel 1 326 ALGEBKA. [Ch. XXI 28. It took a number of men as many days to dig a ditch as there were men. If there had been 6 more men, the work would have been done in 8 days. How many men were there ? 29. The front wheel of a carriage makes 6 revolutions more than the hind wheel in running 36 yards ; if the circumference of each wheel were 1 yard longer, the front wheel would make but 3 revolutions more than the hind wheel in running the same distance. What is the circumfer- ence of each wheel ? 30. Two men formed a partnership with a joint capital of $ 500. The first left his money in the business 5 months, and the second his money 2 months. Each realized f 450, including invested capital. How much did each invest ? 31. Two trains run toward each other from A and B respectively, and meet at a point which is 15 miles further from A than it is from B. After the trains meet, it takes the first train 2f hours to run to B, and the second train 3f hours to run to A. How far is A from B ? 32. The perimeter of a rectangular lawn having around it- a path of uniform width is 420 feet. The area of the lawn and path together exceeds twice the difference of their areas by 1200 square yards, and the width of the path is one-sixth of the shorter side of the lawn. Find the dimen- sions of lawn and path. 33. Water enters a forty-gallon cask through one pipe and is discharged through another. In 4 minutes one gallon more is discharged through the second pipe than enters through the first. The first pipe can fill the cask in 3 minutes less time than it takes the second to discharge 66 gal- lons. How long does it take the first pipe to fill the cask ? 34. In a rectangle, whose sides are a and 6 inches respectively, a second rectangle is constmcted. The sides of the inner rectangle are equally distant from the sides of the outer, and the area of the inner rectangle is one-nth of the remaining part of the outer. What are the lengths of the sides of the inner rectangle ? Let a = 70, 6 = 52J, re = 1. 35. It has been found by experiment that when an object is removed to a point 2, 3, 4, •■■ times its original distance from the source of light, its illumination is 2^, 3^, 4^, ;•■ times as feeble. A lamp and a candle are 4 feet apart. At what point oa the line joining them will the illumination from the candle be equal to that from the lamp, if the light of the lamp be 9 times as intense as that of the candle ? CHAPTER XXII. EQUATIONS OF HIGHER DEGREE THAN THE SECOND. We shall consider in this chapter a few higher equations which can be solved by means of quadratic equations. 1. A Binomial Equation is an equation of the form x" = a, wherein 71 is a positive integer. Certain binomial equations can be factored into linear and quadratic factors or factors which can be brought to quadratic form by proper substitutions. Ex. 1. Solve the equation k' — 1 = 0. (1) Factoring, (a; -l}(x^ + x + l) = 0. (2) This equation is equivalent to the two equations a; — 1 = 0, whence sc = 1 ; and x^ + a; + 1 = 0, whence x=—i± iy/— 3. This example gives the three cube roots of 1, since x' — 1 = is equiva.- lent to x^ = l, or a; = ^1. Therefore the three cube roots of 1 are 1, — J + i\/~ 3, — i — iV~ ^• In general, the three cube roots of any number can be found by multi- plying the principal cube root of the number in turn by the three algebraic cube roots of 1. E.g., ^8 = 2^1 = 2, -l±V-3; the three cube roots of a are -J/a, v^a(— J ± ^y/— 3), wherein ^a denotes the principal cube root of o. Ex. 3. Solve the equation a^ + 1 = 0. Factoring, (a;^ + 1 + Xy/2) (a;^ + 1 - x V^) = 0. This equation is equivalent to the two equations x^ + 1 + Xy/2 = 0, whence x = iV2(- 1 ± V- 1); and x^ + 1- Xy/2 = 0, whence x = iV^Cl ± V- !)■ Since the given equation is equivalent to x* = — 1, or x =: -J/— 1, we conclude that the four fourth roots of — 1 are iV2(-l±V-i). W2(i±V-l) 327 328 ALGEBRA. [Ch. XXII The four fourth roots of any negative number can be found by multi- plying the principal fourth root of the radicand taken positively in turn by the four fourth roots of — 1. E.g., ^-16 = 2^-l=v'2(-l±V-l). VaaiV-l)- 2. Ex. 1. Solve the equation a* — 9 = 2 a;^ — 1. Since a;^ = (x'^'Y, we may take x"^ as the unknown number and solve this equation as a quadratic in x^. "We then have {x^Y - 2 a;2 _ 8, =0. Factoring, {x^ - 4) (x^ + 2) = 0. Whence a;^ _ 4 = o, or a; = ± 2 ; and a;^ + 2 = 0, or a; = ± y/- 2. In general, any equation containing only two powers of the unknown number, one of which is the square of the other, can be solved as a quad- ratic equation. Ex. 2. Solve the equation a;^ ~ 3 a;' = 40. Since x^ =(afiy, we take x^ as the unknown number. We then have {x^y - 3 a;' = 40. Solving this equation for a;', we obtain a;' = 8, whence x = -^8; and x^ = — 5, whence x = — ^5. Therefore, by Art. 1, Ex. 1, the six roots of the given equation are 2, - 1 ± V- 3, - -J/S, ^^5(1 T V- 3), wherein y/5 denotes the principal cube root of 5. Ex. 3. Solve the equation (a;2 - 3x + l)^ = 6 + 6(x^ - 3x + 1). In this example x^ — 3x + X is regarded as the unknown number, and may temporarily be represented by the letter y. The equation then becomes 2/2 = 6 + 5 J/ ; whence t/ = 6, and — 1. We therefore have the two equations x" _ 3 X + 1 = 6, whence x = f ± ^^29 ; x^ — 3x-|-l=— 1, whence x = 2, x = 1. Therefore the roots of the given equation are f ± ii/^S, 2, 1. Attention is called to the fact that, in each example, we have obtained as many roots as there are units in the degree of the equation. Frequently equations which do not at first appear to come under this case can, by a proper arrangement of terms, be made to do so. Ex. 4. Solve the equation x* + 2 x^ - 7 x'' - 8 x + 12 = 0. The given eqxiation can be written X* + 2 x3 + x2 - 8 x2 - 8 X + 12 = 0, or (X2 + 05)2 - 8(x2 + X) + 12 = 0. EQUATIONS OF HIGHER DEGREE. 329 If we now let a;^ + a; = ^, we have y^ — 8y + 12 = ; whence y = 2, and 6. We then have to solve the equations x^ + x = 2 (1), and a;2 + a; = 6 (2). The roots of (1) are 1, — 2 ; and the roots of (2) are 2, — 3. 3. Ex. Solve the equation '^^ + a: + 1 ^ a?" - a + 2 ^ ^. _ x^-x + 2 x2 + a;+l ' If we let '^ +'^+ = y^ the given equation becomes y +- = 2^. The roots of this equation are |, |. We now have to solve the two equations a;" + a: + 1 ^3 ,^. ^^^ a' + a: + 1 ^2 -^s a;2 - a; + 2 2 ^ ^' a;" - a; + 2 3 ^ ^' The roots of (1) are found to be 4, 1 ; and the roots of (2) are found to be - 1 ± ^^29- An equation can be solved by this method when it contains only two expressions in the unknown number, one of which is the reciprocal of the other, and when the numerators and denominators of these expressions are of degree not higher than the second. EXERCISES. Solve each of the following equations : 1. x2 + l = 0. 2. (a;- 1)8 = 8. 3. (a; + 2)' + 4 = 0. 4. (a;+l)3=(3-a;)3. 5. x^ = (2 a- xy. 6. a^ - 1 = 0. 7. (a: + l)*=16. 8. a;*+625(x+l)*=0. 9. a;^ + 1 = 0. 10. a;6-l=0. 11. a^ + 9 = 10a;a. 12. a^-6a;2 = -l. 13. (a;2-9)(a;2-16)=15a;2. 14. (a;2 - 10) (a;2 - 18) = 13 x^ 18. x0-65x« = -6i. 16. a;8 + 5a^ = 6. 17. (a + a;)3 + (a-a!)8 , g a;* + 6 a;2 + 1 _ 3 ■ a^-6a:2 + l 2 19. (x - 2)6 - 19(a; - 2)3 = 216. 20. (3a;2-5a;+l)2-9a:2+15x=7. 21. 15 a;2 - 35 a; - 3(7 a; - 3 a;2 + 8)2 + 310 = 0. 22. (a;2-a; + l)2 = 3a;(a;-l)+l. 23. (x + ^] +a; = 42-°. \ xJ X 24. x^-a^ x^ + a^_ 34 x^ + a^ 3^- a^ 15 „ a;2-5a; + 3 a;2 + 5a;-3_8 ■ a;2 + 5a;-3 x^-bx + Z 3 26- a;S + a;2 - a; - 1 = 0. 27. 3a;3-13a;2+13a;-3 = 0. 28. a?'-2a;8 + 2a;2-2a;+l = 0. 29. 2a;*-5a;3 + 4a;2-5a; + 2 = CHAPTER XXIII. IRRATIONAL EQUATIONS. 1. An Irrational Equation is an equation whose members are irra^ tional in the unknown number or numbers ; as, y/(x + l)=3. Notice that we cannot speak of the degree of an irrational equation. 2. The solution of an irrational equation depends upon the principle : If both members of an equation be raised to the same positive integral power, the resulting equation will have as roots the roots of the given equation, and, in general, additional roots. Let M=N be the given equation. Squaring both members, M^ = NK Whence M^-N^ = 0, or {M- N)(M+ iV)= 0. This equation is equivalent to the two equations ilf — iV= 0, or M= N, the given equation ; and M+ iV"= 0, or M= — N, an additional equation. That is, the equation obtained by squaring both members of the given equation is equivalent to the given equation and an additional equation which differs from the given one in the sign of one of its members. In like manner the principle can be proved for any positive integral powers of the members of the given equation. JE.g., if both members of the equation a;+ 1 = 2 be squared, we have (a; + 1)^ rr 4 ; whence a; = 1, and — 3. The root 1 satisfies the given equation ; the root — 3 is a root of the equation a; + l=-2, which was introduced by squaring, and does not satisfy the given equation. 3. To solve an irrational equation, we must first derive from it a rational, integral equation. This step, which is usually effected by rais- ing both members of the equation to the same positive integral power one or more times, is called rationalizing the equation. 330 IRRATIONAL EQUATIONS. 331 In the following examples the roots will be limited to principal values •. Ex. 1. Solve the equation z + V(25 — a^^) = 7. Before squaring, it is better to have the radical by itself in one member. Transferring x, y/(ib -x^)=T -x. (1) Squaring, 26 - a;2 = 49 - 14 a; + x"^. (2) The roots of this equation are 3, 4. Both roots of (2) satisfy the given equation, since 3 + V(25 — 9) = 7, and 4 + VC^S — 16) = 7. Therefore no root was introduced by squaring both members of the given equation. This is also evident from the following considerations : Any root of the additional equation, V'(25 - a;2) = - (7 - x), or - V(25 - a;=) = 7 - a;, (3) obtained by changing the sign of one of the members of the given equa- tion when prepared for squaring, must be a root of tlie rational integral equation (2). But both, roots of this equation, 3 and 4, make the first member of (3) negative, and the second member positive. That is, equa- tion (3) is an impossible equation. Ex. 2. Solve the equation x - V(25 -x^)=l. Transferring x, - ^(25 -(ifi)-l-x. (1) Squaring 25 - a;^ = 1 - 2 a; + x". (2) The roots of this equation are 4 and — 3. The number 4 is a root of the given equation, since 4-^(25-16)=!; but the number — 3 is not a root of the given equation, since -3-^(25 -9) = - 7, notl. Therefore, the root — 3 is a root of the additional equation -V(25-a;2) = _(l_a;), or y/(25-x'^)=l -x, introduced by squaring. That — 3 is not a root of the given equation is also evident from the form of the equation. For any real value of x which makes a; - V (25 - x'^) equal to 1 must be greater than 1, and therefore cannot be equal to — 3. The preceding examples illustrate the following method of solving irfar tional equations : Transform the given equation so that one radical stands by itself in one member of the equation. Equate equal powers of the two members when so transformed. Bepeat this process until a rational equation is obtained. 332 ALGEBRA. [Ch. XXllI 4. In the preceding article the indicated roots in the equations were limited to principal values. At the same time an irrational equation, if written arbitrarily, may be inconsistent with the laws governing the relations between numbers. In such a case the equation is impossible, that is, it cannot be satisfied by either real or imaginary values of the unknown numbers. E.g-i y/(x + 6) + .^'(a; rl- 1) = 1 is an impossible equation. Tor it cannot be satisfied by any complex value of x, since by Ch. XX., Arts. 31 and 22, ^(x + 6) + ^(x + 1) must be complex if x be complex, and hence cannot be equal to 1. It cannot be satisfied by any real positive value of x, since, in that case, either .^(x + 1) or ^(x + 6) is greater than 1. It cannot be satisfied by any real negative value of x, since, if x be negative and Its absolute value be less than 1, y'(a; + 6) will be greater than 1, and if x be negative and its absolute value be greater than 1, y/{x + 1) will be imaginary. 5. But if the restriction to principal roots be removed, any irrational equation contains in itself the statements of two or more equations. E.g., if both positive and negative square roots be admitted, the equa- <^'on ^(X + 6) + y/(X + l)=l is equivalent to the four equations V(K + 6) + v'(«+l)=l (1). V(« + 6)-V(« + l)=l (2), -V(» + 6) + V(a:+l)=l (3), -V(» + 6)-v(a; + l)=l (4), in which the roots are limited to principal values. The same rational integral equation will evidently be derived by ration- alizing any one of these equations. Therefore the roots of this rational equation must comprise the roots of these four irrational equations. Con- sequently, in solving an Irrational equation, we must expect to obtain not only its roots, but also the roots of the other three equations obtained by changing the signs of the radicals in all possible ways. Some of these equations can be rejected at once as impossible. The roots of the other irrational equations will be the roots of the rational equation. Thus, of the above equations, (1), (3), and (4) can be rejected at once as impossible. The rational equation derived from any one of the four equations Is X + 1 =i; whence x = 3. The number 3 is a root of the one equation not rejected, since V(3 + 6)-V(3 + l)=l. The same conclusions could have been reached by substituting the roots of the integral equation successively in the irrational equations, rejecting those which are not satisfied by any root. IRRATIONAL EQUATIONS. 333 Special Devices. 6. Ex. 1. Solve the equation VCS a;2 _ 2 a; + 4) - 3 x3 + 2 a; = - 16. Since -3x2 + 2a; = -(3a;2 - 2a; + 4) + 4, we may take ^(_S a;^ - 2 a; + 4) as the unknown number, replacing it temporarily by y. We then have 2/ - 2/2 + 4 = - 16. The roots of this equation are 5, and — 4. Equating y'(3 a;2 - 2 a; + 4) to each of these roots, We have V(3x2 - 2a; + 4)= 5, whence a; = 3, - |. V(3 a;2 - 2 a; + 4) = - 4, whence a; = ^(1 ± V37). The numbers 3, — ^ satisfy the given equation, and are therefore roots of that equation. The numbers iy/(l ± y/31) do not satisfy the given equation. But if the value of the radical be not restricted to the principal root, the given equation comprises the two equations v/(3a:2-2a; + 4)-3a;2 + 2a; = -16, (1) -V(3a;2-2a; + 4)-3a;2 + 2a;=-16. (2) Then i(,l± y/S7) are roots of (2). Ex. 2. Solve the equation */(3 x^ + 13) + V(3 x^ + 13) = 6. Assuming .^(3 a;2 + 13) as the unknown number, and representing it by y, we have y + y^ = 6. The roots of this equation are 2 and — 3. Equating y/(3 x^ + 13) to each of these roots, we have .J/(3 a;2 + 13) = 2, whence a; = ± 1, .J/(3 a;2 + 13) = - 3, whence x=± V^ = ± 1^/51- The numbers ± 1 are roots of the given equation, since ■y/lQ+ y/16=:6. The numbers ± f V^l are evidently not roots of the given equation, but are found to be roots of the equation - .t/(3 a;2 + 13) + V(3 a;2 + 13) = 6. The preceding examples illustrate the following principle : If a radical equation contain one radical, and an expression which is equal to the radicand or which can be made to differ from the radicand (or a multiple of the radicand) by a constant term, it can be solved as a quadratic equation. The same is true if the equation contain two radi- cals, one the square of the other, and in addition only constant terms. In both cases, the radicand must, in general, be a linear or a quadratic expression. 334 ALGEBRA. [Ch. XXIII 7. Irrational equations containing cube and higher roots in general lead to rational, integral equations of a higher degree than the second, and therefore cannot be solved by means of quadratic equations. But in some cases their solutions can be effected by special devices. Ex. Solve the equation .J/(8 x + 4)- ^(8 a; - 4) = 2. Cubing, 8a; + 4 - 3[^(8a; + 4)]2^(8a; - 4) + 3^(8a; + 4)[4/(8!B - 4)]2 -8x + 4 = 8. (1) Transferring and uniting terms, and dividing by — 3, [ P=0,1 (b), (c), Q = 0,] s = o, w, Q = o,] E = 0, obtained by taking each factor of one equation with each factor of the other. 340 ALGEBRA. [Ch. XXIV For, every solution of the given system must reduce either P or Q, or both P and §, to 0, and at the same time must reduce either B or 8, or both B and S, to 0. Now any solution of (I.) which reduces P to and iJ to 0, is a solution of (a) ; any solution which reduces P to and 8 to 0, is a solution of (6) ; and so on. Therefore, every solution of the given system is a solution of at least one of the derived systems. And any solution of (a) reduces P to and B to 0, and therefore reduces P x Q to and P x iS to 0. Therefore, every solution of (a) is a solution of (I.). In like manner it can be shown that every solution of the three other derived systems is a solution of the given system. Ex. 1. Solve the system {x — 2y)(x — ^y)=0, (a! + 3/-4)(a;-2/ + 2)=0. The given system is equivalent to the four systems x--2y = 0,- a; + 2/ - 4 = 0, (a), -22/ = 0,l Q>), a; - 3 2/ = 0, a; + 2/ - 4 = 0, . t.\ a; - 3 2/ = 0, ■ ^^^' .-2. + 2 = 0,.K'^- The solution of (a) is |, | ; the solution of (6) is — 4, — 2 ; the solution of (c) is 3, 1 ; the solution of (d) is — 3, — 1. These are therefore the solutions of the given equations. Ex. 2. Solve the system 2«? -7 xy + &y^ = (i, (1) a;2 + 2/2 = 13. (2) The first member of (1) is (a; — 2 2/) (2 a; — 3 y), and the first member of (2), when 13 is transferred to that member, cannot be resolved into rational factors. The given system is there- fore equivalent to the two systems x-2y = 0, 1 2a;-32/==0, | a;= + 2/^ = 13jW' a? + f = lB,]^^> The solutions of (a) and (6), and therefore of the given system, are respectively 2V¥» V¥; -2V¥» -V¥; 3, 2; -3,-2. § 1] SIMULTANEOUS QUADRATIC EQUATIONS. 341 4. When all the terms which contain the unknown numbers in both equations of the system are of the second degree, a system can always be derived whose solution is obtained by the method of the preceding article. Ex. Solve the system a? + xy + 2y^=74:, (1) 2a^ + 2a;y + i/2 = 73. (2) Multiplying (1) by 73, 73 3^ + 73xy + 146 2/^ = 74 x 73. (3) Multiplying (2) by 74, 148 x' + 148 a^ + 74 f=74:X 73. (4) Subtracting (3) from (4), 75 x^ + 75 xy - 72 y^ = 0, or 25x' + 25xy-24:y^ = 0, or (5x — 3y)(5x + 8y) = 0. Therefore the given system is equivalent to 5x — 3y = 0, 1 5x + 8y=:0, a^ + !B2/ + 2y2=74j W' a^-\-xy + 2y^ = 7i,. (b). The solutions of these systems, and hence of the given system, are respectively 3, 5 ; — 3, — 5 ; 8, — 6 ; — 8, 5. In applying this method to such systems, we must first derive from the given equations a homogeneous equation in which there is no term free from the unknown numbers. 5. Such examples can also be solved by a special device. Ex. Solve the system x^ + i'f= 13, (1) acy + 2 2/2 = 5. (2) In both equations, let y = tx. (3) Then from (1), x^ + 4: xH^ = 13, whence a;2 = ^^ ^ ; (4) and from (2), xH + 2 xH^ = 5, whence x^ = — ^—-- (5) , t -\- ^ IT Equating values of xK —r = -—,• (6) Whence < = i, and J = - f . When t = i, ■3fi = — 5i_^ = 9 whence a; = ±3. ' 1 + 4 «2 When « = — I, x^ = \, whence x = ± Vi- Whena;=±3, y = tx = \{-kZ^ = ^\. Whena; = ±Vi. 2/ =- f (±v'i) = =F f Vi- 342 ALGEBRA. [Ch. XXIV After assuming y = to, we have a system of three equations in k, y, and t. Then the system (1), (2), (3) is equivalent to (3), (4), (5), which is in turn equivalent to (3), (4), and (6). From (6) we obtain the values of t, from (4) the corresponding values of x, and from (3) the correspond- ing values of y. The solutions of the given system therefore are 3, 1; -3, -1; ^\, -fv'i; - Vi Wi BXEBCISES III. Solve each of the following systems : (a;-8)(2/-6) = 0, a; + y = 13. x^ + xy = 78, ' — xy^l. x^ + xy + y^ = 52, 3. 5. 6. 8. 10. 11. xy — a;-' = ». x^ + xy + 4y^ = G, 3x^ + 8y^ = 14. x'' + xy + y^ = 13 x, x^ — xy + y^ = T X. ■ X + iy/{xy) + 4 s/ + Va; + 2 V2/ = 12, (a;-5)(j^-3)=0, (a:-4)(2/-7)=0. a;2 + 4 2/2 = 13, xy + 2 2/2 = 5. x^ — xy + y^ = 21, 2/2 - 2 KJ/ + 15 = 0. k2 - 2 X2/ + 3 8/2 = 9, a;2 _ 4 scj/ + 5 j/2 = 5. r a;2 + 2/2 = 61 - 3 a;2/, I x2 - 2/2 = 31 - 2 xy. 12 (a;-3)(j/-2)=0, 7 3 13- • - + - = 2. X J/ I x2 + 2 X2/ = 12. 6. i/"(Ae members of one equation of a system of two equations PB=QS, (1) P=Q, (2) (I.) contain as factors the corresponding members of the second equa- tion, then the system is equivalent to the following two systems : B = S, P=Q,(a), and P=0, Q = 0,(b). The first equation of the system (a) is obtained by dividing the members of equation (1) of the given system by the cor- responding members of equation (2), and the second equation is equation (2) of the given system. § 1] SIMULTANEOUS QUADRATIC EQUATIONS. 343 The system (6) is obtained by equating to the two members of equation (2) of the given system. The given, system is equivalent to the system PJi-QS=0, P-Q = 0, (II.) or, replacing Q by P, to the system P(iJ -S)=0, P-Q = 0. (III.) The system (III.) is, by Art. 3, equivalent to the two systems M- 8 = 0, P-Q = 0, and P=0, P-Q = 0. That is, to the systems (a) and (6). Ex. 1. Solve the system {x — T)(x — y + 2) = (y + l)(x + y), x — y + 2 = x^y. The given system is equivalent to the following two systems : x-l = y + l,] x-y + 2 = 0,] \ (a), and ^ \ (b). x-y + 2 = x + y,r ^' x + y = 0,j^'' The solution of (a) is 3, 1 ; the solution of (b) is — 1, 1. Ex. 2. Solve the system a^ — y^ = 8, X +y =3. The given system is equivalent to the two systems The solution of system (a) is ^, ^. Since the equation 3=0 is impossible for finite values of x, the system (6) is impossible. EXERCISES IV. Solve each of the following systems : f a;2 - 4 2/2 = 21, ( x^ - 'f + (x + yy = 2i, ' \x-2y = 3. ■ ' \x + y. = 4:. (\bx^ + 2xy-y^ = 15, f (x^ - 1) (y^ - 1) = 2800, ^' [&x-y = 3. *■ |(a;-l)(^-l)=40. r (3a; -1)2 -(4^ + 2)2 = 60, [3x + iy = 6. 1. Sjrmmetrical Equations. — A Symmetrical Equation is one which remains the same when the unknown numbers are interchanged. 344 ALGEBRA. [Ch. XXIV A system, of two symmetrical equations can be solved by first finding the values ofx + y and x — y. Ex. 1. Solve the system a^ + 2/^ = 13, (1) xy = 6. (2) Multiplying (2) by 2, 2xy = 12. (3) Adding (3) to (1), x' + 2xy + y^ = 25. (4) Subtracting (3) from (1), x^-2xy + y'' = l. (5) Equating square roots of (4), x + y = ±5. (6) Equating square roots of (5), x — y = ±l. (7) The given system is equivalent to the system (4), (6), which is equivalent to the systems (6), (7). The latter systems are <«+y=5, ] x+y=5, I x+y= -5, 1 , , x+y^ -5, x-y=l,\^''^'x-y=-l,r^'x-y=+lM'K-y=-l,l (d). The solutions of these four systems are respectively 3, 2 ; ^,05 — Zj — o J "~ Oj — Z. The solutions of (6) and (7) should be obtained mentally, without writing the equivalent systems (a), (&), (c), (d). Each sign of the second member of (6) should be taken in turn with each sign of the second member of (7). Notice that these solutions differ only in having the values or X and y interchanged. This we should expect from the definition of symmetrical equations. When the equations are symmetrical, except for sign, the solution can be obtained by a similar method. Ex. 2. Solve the system " x — y = l, (1) xy=2. (2) Squaring (1), x'-2xy + y^ = l. (3) Adding four times (2) to (3), 0? + 2xy + y'' = 9. (4) Equating square roots of (4), x + y = ± 3. (5) The solutions of (5) and (1) are 2, 1, and — 1, — 2. Notice that the solutions in this case differ not only in having the values of x and y interchanged, but also in sign. § 1] SIMULTANEOUS QUADRATIC EQUATIONS. 345 Observe that the system (2), (3), and therefore the equivalent system (3), (4), is equivalent to the two systems : x~y=-l, x-y=. ■h] y the given system, and xy = 2,) ^ [xy=:2. Consequently, if in the system (8), (4), equation (3) be replaced by (1), the resulting system x-y=:l, (1) x^ + 2xy + y2 = 9, (4) is equivalent to the given system. 8. Many systems which are not symmetrical can be solved by the method of the preceding article. Ex. 1. Solve the system 2x + 3y = 8, (1) xy = 2. (2) We should first obtain the value of 2x — 3y. Squaring (1), 4 a^ + 12 a^ + 9 j/^ = 64. (3) Subtracting 24 times (2) from (3), 4:X'-12xy+9y'=16. (4) Equating square roots of (4), 2 a; — 3 y = ± 4. (5) The solutions of (1) and (5) are 3, f ; 1, 2. BXEBCISBS V. Solve each of the following systems : 1. 7. 10. 12. 15. ■x + y = 12, xy = 32. ' x — y = %, xy =— 15. f a;2 + 2/2 = 40, 1 xy = 12. 9x^ + y^ = 31a^, xy=~2 aK {x^-^y^ = 137, a; + 2/ = 15. a;2 _ 2,2 == 28, xy — 48. 13. 16. X + 2/ = a, xy = h. X — y = m, xy = n. a;2 + 2/2 = 181, xy=- 90. 11. ■x^ + y^ = 61, x + y = ll. a;2-42/2 = -3, xy = -l. 3. 6. 9. 14. 17. ix + 5y = 31, xy = 28. 6x-Ty = 5S, 3xy=-60. 25x2+92/2=148, 6xy = 8. 5x2 4.22/2 = 5c8M 8 62, xy = 2 ab. ' 6x + 3y = ll, 25 a;2 + 9 j,2 = 73. ■x^ + y^ = 58, x-y = 5. 346 fa52 + y2 = 74, 18. i. \x-y = 2. 19. 21. 24. 27. 1 + ^ X y = 3, 1 = 2. » _y _16 !/ a; 15' a; - !/ = 2. Vx + y/y = 5, a;y = 36. x + xy + y = x^ + xy + 2/2 32. ^»=^ + ^^-(^- xy + x-y = 28. ALGEBRA. •9x2 + 2/2 = 82, 3a; -2/ = 10. xy = 80, 1 _ 1 _ 1_ X y 5 i+i = io, a; 2/ i + i = 58. a;2 y^ ' y/x + y/y = 1, x + y = ZT. 26. [Ch. XXIV 16a;2+492/2=113, 4a; + 72/=l. p + 2/ = 16, 1 + 1 = 1. X y Z a;2 + 2/2 = 2f a;2/, l + l = li. x^y ^ fa;2+a;+2/=18-2/2, [a;y = 6. 30, 34. I x2 + 2/2 + a; - I a;!/ + X - 2/ = 29, = 61. -2/) =20, 1. 2/ = a, 6. „, f a;2 + 2/2 + 7 a:2/ = 171, 31. \ { xy = 2(a; + y). gg ra;2 + 2/2-a;-2/ = 22, la; + 2/ + a;2/=-l- 36. 1^ + ^ = '' [ a;2 + 2/2 + X2/ = 3. 36. 38. a; + s? = 9, . a;2 + 2/2 — a;^ 4 a;y = 96 — a;2! x + y = Q. 40. 37. x^ + xy + 2/2 = 2 TO, ■ = 21. " I, a;2 - a;2/ + 2/^ = 2 ». 39 fV[(2 + a!)(l + 2/)] = 2, t V(2 4-a;)-va+2')=i• a; + -v/* + 2/ + V2/ = 18, (x + v'a;)(2/ + v'2/)='?2. Simultaneous Quadratic Equations in Three Unknown Numbers. 9. No definite methods can be given for solving simultaneoufi quad- ratic equations in three unknown numbers. Ex. 1. Solve the system xy = 2, (1) X3 = 3, (2) 2/0 = 6. (3) Multiplying corresponding members of (1), (2), and (3), xV^'^ = 36. (4) Equating square roots of (4), xyz = ± 6. (5) Dividing (5) in turn by (1), (2), and (3), z=±3, 2/=±2, a; = ±l. The required solutions are therefore, 1, 2, 3, and — 1, — 2, — 3. § 1] SIMULTANEOUS QUADRATIC EQUATIONS. 34T Ex. 2. Solve the system x(^y + z)=5, (1) y{x + z) = S, (2) z(^x + y)=9. (3) Adding corresponding members of (1), (2), and (3), and dividing by 2, xy + xz + ye = 11. (4) Subtracting in turn (1), (2), and (3) from (4), 3/s = 6, (5) xz = 3, (6) xy = 2. (7) The solutions of equations (5), (6), and (7) are 1, 2, 3, and — 1, — 2, -8. EXERCISES VI. Solve eaoli of the following systems : 1. 9. 11. 13. IS. 17. xy = 30, xy/y = a. a;2 + 2/2 = 13, yz=- 60, 2. XyJZ = 6, 3. a;2 + z2 = 34, .xz = - 50. Vy/Z = c. . 2/2 + z2 = 29. x^ + yz = 5, a;2 + ?/2 = a ■ x{y + 2) = 5, y^-\-xz = 5, 5. 2/^ + ?2 = 6i 6. 2/(a; + z) = 8, g2 + jcj; = 5. z^ + k2 = c2 .z(a; + 2/)=9. ■ x^ -V xy -\- y^ = 61, a;(a; + 2/ + «) = 6, x^ + xz + z'' = 21, 8. y(x, + y + z)=\2. .y2 + yz + z^ = 13. z{x + y + z)=\%. r xyz ^2 x + y \^ + y= (a 6)^ xyz xyz _3 x + z 2 10. ^+'- (a c)^ iCJ/a xyz _6 y + z 5 xyz 3x = 5y, ■x + y + z = a. x(^z+-2) = yz + S2, 12. x{x + y)=b\ x(z-l) = (jy+l)z-l .z{_z + y)=cK ■ xz = y^, y/(jK!^ + y-^ + z^)=lZ, x + y + z = li , 14. a; + 2/ + z = 19, ■ X^ + 2/2 + «2 = 133. a;(2/ + z) = 48. XZ = 360, ■y = l{x + z). S/(z-10) = 40 » 16. ^ a;2 + 2/2 = 458, .a;(« + 8) = 400 + K«- 2). . 2/2 + z" = 730. fx^ + y^ + z^ = 29, ■ (jl + z)(x + y-\-z) = 6, xy + xz + yz- = -10, 18. (z + a:)(a; + 2/ + z) = 8. .x+y-z=- 5. . (a; + 2/)(a; + 2/ + z) = -6 348 ALGEBRA. [Ch. XXIV §2. SIMULTANEOUS HIGHER EQUATIONS. 1. The solutions of certain equations of higher degree than the second can he made to depend upon the solutions of quadratic equations. Ex. 1. Solve the system x^ + y^ = 9, (1) x + y = 3. (2) Dividing (1) hy (2), x^ -xy + y^ = 3. (3) Subtracting (3) from the square of (2), Zxy = G, or xy = 2. (4) The solutions of (2) and (4), and therefore of the given system, are 1, 2, and 2, 1. Ex. 2. Solve the system a;* + j/* = 17, (1) x + y = Z. (2) We first find the value of xy. Let xy = z. (3) Squaring (2) , x^ + 2xy + y'^ = 9, (4) or x2 + j/2 = 9 - 2 z . (5) Squaring (5) , x* + 2 x'^y^ + !/* = 81 - 36 « + 4 ««, (6) or X* + ^4 = 81 - 36 s + 2 ^. (7) Since x* + ^ = 17, we have from (7), 2 «2 _ 36 3 + 81 = 17. (8) Whence z = 16, and 2. (9) Therefore, from (3) and (9), xy = 16, (10) and xy = 2. (11) The solutions of (2) and (10) and of (2) and (11) are readily found, and should he checked hy substitution. Ex. 3. Solve the system (x2 + 2/2)(a;3 + 2,8)=45, (1) x + y = 3. (2) Erom (2) x^ + j/2 = 9 - 2 xy, (3) and x8 + j/8 = 27 - 3 xy{x + y) = 27 — 9 xy, since x + j^ = 3. (4) Substituting in (1) for x^+y^ and x^+y^ their values from (3) and (4), (9 - 2 x?/) (27 -9x8^) = 45, or 2 xV. -15xy = - 22. (6) Equation (5) can be solved as a quadratic in xy, and the results com- bined with equation (2). The results should be checked by substitution. §2] SIMULTANEOUS HIGHER EQUATIONS. 349 1 3 5. 9. 11. 13. 15 17. 2. 4. 6. 8. 10. 13. 14. 16. 18. EXERCISES VII, Solve each of the following systems : x + y = b, x^ + y^- 35. 2(x + y-)=5, 32(x3 + 3^8) = 2285. (a;-7)s+(5-y)8 = 9, x~y = 5. a;* + 2/* = 82, xy = S. x* + y* = 257, x-y = 3. (ix^-y2)(x + y)=9, xy(x + y)=6. x~y = 342, i/x-^y = 6. 'x + y + V(a; + y)= 12, a;8 + 2/8 = 189. ^(90 -a;) -^(9-2,) =2, x + y = n. x-y = l, lfi-y« = 7. (a:-l)8 + (y-2)8 = 28, x + y = l. K* — ^ = 554, a;2 + 2/2 = 34. a;4 + j/4 = 97, a; + 2/ = 5. (a; -7)1 + (2/ -3)4 =257, a; - 2/ + 1 = 0. (a; + 2/)(a;2 + 3^2)=175, (a;-2/)(a;2-2/2)=7. f a; + 2/ = 30, ^(a; + 7) + ^(2/-9) = 4. ^(a;+7) + ^(!/-5)=3, a; + 2/ = 15. a:* + 2/* = 4097, x + y + ^{x+y)=l2. BXBBCISBS VIII. MISCELLANEOUS EXAMPLES. Solve each of the following systems by the methods given in this chap' 1. ter, or by special devices : ' a; - 2 2/ = 2, 2. xy = 12. •a;2_2,2 = 8(a;-s/), a;2 + t/2 = 50. 3. ■ a; + 2/ = a;2, 3 2/ — a; = 2/2. 6. 8. a; + y ^ a; - j ^5^, a; — y a; + 2/ 2 a;2 - 3 2/2 = 24. (5a;-3)(3 2/ + 2)=0, (4a;+5)(2j/-3) = 0. ■ a;(7 a; - 8 J/) = 159, 5 a; + 2 ?/ = 7. a;2-2/2 = 2(a; + y), a;2 + 2/^ = 100. a;2 - 2 a;2/ + 3 2/^^ _ 1^ Zx^-ixy + y^ Z a;2 - 3 2/ = 1. (2a;+32/-7)(a;-42/+2)=0, a;2 + y2 ^ 2 a; - 7 J/ = 2. 9. 350 ALGEBRA. [Ch. XXIV 10. 18. 14. 16. 18. 23. 26. 28. 30. 34. 36. 38. S x'' + 5y^ + 2x - iy = li. ' x!^ + y'^ + X - y = 62, (x^ + y^)(x-y) = 61. I ^(5x+Sy-5) + 16-6y=6x, I 2 K^j/Z + 2 = 5 K!/. x3 - 2/3 = 26, a; — 2/ = 2. a; + 2/ = 19, ' -^x + -^y = 4:. ( «-y _ \x^y-'^' I a;2 + j/2 x" + y"= a", xy = b. •3s;(a;2 + 2/2)=10x, 2 a;(a;2 _ y2) = 48 y. xi + (kY 4- 2,4 = 133^ x^ — xy + y^ = 1. xi-x^ + y*-y^ = 84, x^ + xy + y^ = 19. k' + y^ + K2/(a: + 2/) = 65, x^y^(x^ + y^)=i(>8. »; + «/+ veals') = 14, a;2 + y" + xt/ = 84. J/2 = xz, x + y + z = 2S, xyz = 512. 11. 13. 15. 17. 19. 21. 25. 27. 31. 33. 35. 37. 39. - a;2 + 2/2 + 5 a; - 9 J/ = 84, a;2 _ 2,2 + 5 a; + 9 2/ = 84. a;2 + s/2 = 485, aV = 57834 - 5 a;?/. 3(a; + 2/)2 = -f(x + j/) + i^^, li«V-Ax2/ = 43. f 2a; + 3sf + 6a;2/ = ll, [ 4a;2 + 92/2 + 12a;y = aiY - 11. 40 a;2 + 2^2 = X + ] xy = - 12 x-\-y a;Y - ^V = 1152, ai'^j' — <'^y^ = 48. » - 2/ - V(» - 2') = 2, a;S - 2/8 = 2044. 2/ 3 xj/ X 3 X* + y* = 14 xYi X + 2/ = o. (x2 + 2/2) (x8 + y') = 455, X + 2/ = 5. X* + xV + 3,4 = 84 x2, x'^ + xy + y'' = 14 x. x2 - xV + 2/2 + 23 = 0, x-X2/ + 2/ + l = 0. (x + 2/) (x!/ + 1) = 18 X2/, (x2+2'^) (xY+ 1) =320 xY- 'x + y--y/(xy)=T, a;2 + 2/2 + xy = 133. x + y + z = 5, x2 + 2/2 = ^2, a;s 4. 2/S + s8 _ 8. 3] PROBLEMS. 351 40. xu = ye, x + u = lS, y + z = U, a;2 + 2,2 + g2 ^. „2 _ 340. 41. xu — yz, x + u = 1, y + z = 6, K* + 2^ + s* + M* = 1394. §3. PROBLEMS. 1. Pe. The front wheel of a carriage makes 6 more revolu- tions than the hind wheel in traveling 360 feet. But if the circumference of each wheel were 3 feet greater, the front wheel would make only 4 revolutions more than the hind wheel in traveling the same distance as before. What are the circumferences of the two wheels ? Let X stand for the number of feet in the circumference of front wheel, and y for the number of feet in the circumference of hind wheel. Then in traveling 360 feet the front wheel makes ■ — ■ revolutions, and the hind wheel makes revolutions. By the first condition, = \- 6. (1) •' ' X y ^ ' If 3 feet were added to the circumference of each wheel, the front wheel would make revolutions, and the hind wheel revolutions. By the second condition, = + 4. (2) Whence x = 12, the circumference of the front wheel, and y = 15, the circumference of the hind wheel. EXERCISES IX. 1. The square of one number increased by ten times a second num- ber is 84, and is equal to the square of the second number increased by ten times the first. 2. The sum of two numbers is 20, and the sum of the square of the one diminished by 13 and the square of the other increased by 13 is 272. What are the numbers ? 3. Find two numbers such that their difference added to the differ- ence of their squares shall be 150, and their sum added to the sum of their squares shall be 330. 352 ALGEBRA. [Ch. XXIV 4. Find two numbers whose sum is equal to their product and also to the difference of their squares. 5. The sum of the fourth powers of two numbers is 1921, and the sum of their squares is 61. What are the numbers 1 6. If a number of two digits be multiplied by its tens' digit, the prod- uct will be 390. If the digits be interchanged and the resulting number be multiplied by its tens' digit, the product will be 280. What is the number ? 7. If a number of two digits be divided by the product of its digits, the quotient will be 2. If 27 be added to the number, the sum will be equal to the number obtained by interchanging the digits. What is the number ? 8. The product of the two digits of a number is equal to one-half of the number. If the number be subtracted from the number obtained by interchanging the digits, the remainder will be equal to three-halves of the product of the digits of the number. What is the number ? 9. If the difference of the squares of two numbers be divided by the first number, the quotient and the remainder will each be 5. If the differ- ence of the squares be divided by the second number, the quotient will be 13 and the remainder 1. What are the numbers ? 10. The sum of the three digits of a number is 9. If the digits be written in reverse order, the resulting number will exceed the original number by 396. The square of the middle digit exceeds the product of the first and the third digit by 4. What is the number ? 11. A rectangular field is 119 yards long and 19 yards wide. How many yards must be added to its width and how many yards must be taken from its length, in order that its area may remain the same while its perimeter is increased by 24 yards ? 12. The floor of a room contains 30J square yards, one wall contains 21 square yards, and an adjacent wall contains 13 square yards. What are the dimensions of the room ? 13. A merchant bought a number of pieces of cloth of two difierent kinds. He bought of each kind as many pieces and paid for each yard half as many dollars as that kind contained yards. He bought altogether 19 pieces and paid for them $921.50. How many pieces of each kind did he buy ? 14. The diagonal of a rectangle is 20f feet. If the length of one side be increased by 14 feet and the length of the other side be diminished by 2 1 feet, the diagonal will be increased by 12| feet. What are the lengths of the sides of the rectangle ? § 3] PROBLEMS. 353 15. A certain number of coins can be arranged in the form of one square, and also in the form of two squares. In the first arrangement each side of the square contains 29 coins, and in the second arrangement one square contains 41 more coins than the other. How many coins are there in a side of each square of the second arrangement ? 16. A piece of cloth after being wet shrinks in length by one-eighth and in breadth by one-sixteenth. The piece contains after shrinking 3.68 fewer square yards than before shrinking, and the length and breadth together shrink. 1.7 yards. What was the length and breadth of the piece ? 17. A merchant paid f 125 for two kinds of goods. He sold the one kind for $91 and the other for $ 36. He thereby gained as much per cent on the first kind as he lost on the second. How much did he pay for each kind? 18. Two workmen can do a piece of work in 6 days. How long will it take each of them to do the work, if it takes one 5 days longer than the other ? 19. Two men, A and B, receive different wages. A earns $ 42, and B $40. If A had received B's wages a day, and B had received A's wages, they would have earned together !| 4 more. How many days does each work, if A works 8 days more than B, and what wages does each receive ? 20. In 8 hours workmen remove a pile of stones from one place to another. Had there been 8 more workmen, and had each one carried 5 pounds less at each trip, they would have completed the work in 7 hours. Had there been 8 fewer workmen and had each one carried 11 pounds more at each trip, they would have completed the work in 9 hours. How many workmen were there and how many pounds did each one carry at every trip ? 31. A man has two square fields in which he wishes to plant trees, the outer rows to be distant from the edges by half the distance between the rows. If he plants the trees in the first field 2i yards apart, and in the second field 2J yards apart, he will need 11,412 trees. But if he plants the trees in the first field 2| yards apart, and in the second field 3 yards apart, he will need only 569 trees. How long is a side of each of the fields ? 22. A tank can be filled by one pipe and emptied by another. If, when the tank is half full of water, both pipes be left open 12 hours, the tank will be emptied. If the pipes be made smaller, so that it will take the one pipe one hour longer to fill the tank and the other one hour longer to empty it, the tank, when half full of water, will then be emptied in 15f hours. In what time will the empty tank be filled by the one pipe, and the full tank be emptied by the other ? CHAPTER XXV. RATIO, PROPORTION, AND VARIATION. § 1. RATIO. 1. The Ratio of one number to another is the relation be- tween the numbers which is expressed by the quotient of the first divided by the second. E.g., the ratio of 6 to 4 is expressed by f , = f. The ratio of one number to another is frequently expressed by placing a colon between them ; as 5 : 7. The first number in a ratio is called the First Term, or the Antecedent of the ratio, and the second number the Second Term, or the Consequent of the ratio. Thus, in the ratio a:h, a is the first term, and 6 the second. 2. Since, by definition, a ratio is a fraction, all the proper- ties of fractions are true of ratios ; as a : 6 = ma : mb. 3. The definition given in Art. 1 has reference to the ratio of one number to another. But it is frequently necessary to compare concrete quantities, as the length of one liae with the length of another line, etc. If two concrete quantities of the same Mnd can be eaypressed by two rational numbers in terms of the same unit, then the ratio of the one quantity to the other is defined as the ratio of the one number to the other. Kg., the ratio of 2J yards to 1| yards is 2^ : li, =?i = ^. Observe that by this definition the ratio of two concrete quantities, is a number. Also that the quantities to be com- pared must be of the same kind. Dollars cannot be compared with pounds, etc. 354 § 1] RATIO. 355 4. If two concrete quantities cannot be expressed by two rational numbers in terms of the same unit, they are said to be Incommensurable one to the other. Thus, if the lengths of the two sides of a right triangle be equal, the length of the hypothenuse cannot be expressed by a rational number in terms of a side as a unit, or any fraction of a side as a unit. If a side be taken as the unit, the hypothenuse is expressed by ^2, an irrational number. And the ratio of the hypothe- nuse to a side is ^'2 : 1, = ^2, a number comprised in the number system. In the following article we will prove that the ratio of any two incommensurable quantities can be expressed as a number comprised in the number system. 5. Let P and § be two incommensurable quantities. We assume that the ratio P : Q is greater than the ratio P' : §, wherein P' is less than P and is commensurable with §, and that the ratio P : § is less than the ratio P" : Q, wherein P" is greater than P and is commensurable with Q. Let us take -Q as the unit. Then we can find two consecutive n , integral multiples of - §, which are therefore commensurable with §, between which P lies. Let — Q and ™Ji_ a be these multiples. The ratios of these multiples to Q are respectively — and "^ "*" . Then by the hypothesis n n The two rational numbers — and '""*" , between which the ratio P : Q n n lies, have the properties (i.) and (ii.), Art. 6, Ch. XVIII. They therefore define an irrational number. EXERCISES I. What is the ratio of 1. 6ato96? a. |a26to^a62? 3. ^x^yio^xy'^'> 4. -toi? 5. ^to^? 6. " *" ^ » ah b d x-3 (x-S)^ 7. Which is the greater ratio, a + 26:a + 6ora + 36:o + 26? 8. If ^'^ + ^y = 10, what is the value of the ratio x-.y? Zx — y 356 ALGEBRA. [Ch. XXV § 2. PEOPORTION. 1. A Proportion is an equation whose members are two equal ratios. E.g., 4:3 = 8:6, read the ratio of 4, to 3 is equal to the ratio of 8 to 6, or 4 is to 3 as 8 is to 6. Instead of the equality sign a double colon is frequently- used ; as 4 : 3 : : 8 : 6. 2. Four numbers are said to be in proportion, or to be pro- portional, when the first is to the second as the third is to the fourth. E.g., the numbers 4, 3, 8, 6 are proportional, since 4:3 = 8:6. The individual numbers are called the Proportionals, or Terms of the proportion. The Extremes of a proportion are its first and last terms ; as 4 and 6 above. The Means of a proportion are its second and third terms ; as 3 and 8 above. The Antecedents and Consequents of a proportion are the antecedents and consequents of its two ratios. E.g., 4 and 8 are the antecedents, and 3 and 6 the conse- quents of the proportion 4:3 = 8:6. Principles of Proportions. 3. In any proportion the product of the extremes is equal to the product of the means. li a:b = c:d, we are to prove ad = be. By§l,Art.l, f = |. Clearing of fractions, ad = 6c. 4. If the product of two numbers be equal to the product of two other numbers, the four numbers are in proportion. Let ad = be. 2] PROPORTION. Dividing by bd, - = -, 01 a:b = c:d; d hy cd, a b , , - = -, ov a:c = b:d: G d hy ab, d c J , - = -, 01 d:b = c:a; a hy ac, - = -, ov d:c = b:a. c a Interchanging the ratios in (1), (2), (3), (4), c:d = a:b; b: d = a: c; c: a = d: b; b:a = d:c. 357 (!)• (2) (3) (4) (5) (6) (7) (8) Notice that the two numbers of either product may be taken as the extremes, the other two as the means. In (1) to (4), a and d are the extremes,' c and b the means; in (5) to (8), d and a are the means, c and b the extremes. 5. In Art. 4, we may regard the proportions (2) to (8) as being derived from (1), and thus obtain the following proper- ties of a proportion : (i.) The means may be interchanged; as in (2). (ii.) The extremes may be interchanged; as in (3). (iii.) The means may be interchanged, and at the same time the extremes; as in (4). (iv.) The means may be taken as the extremes, and the ex- tremes as the means; as (8) from (1), (7) from (2), etc. 6. If any three terms of a proportion be given, the remaining term can be found. Ex. What is the second term of a proportion, whose first, third, and fourth terms are 10, 16, and 8 respectively ? Letting x stand for the second term, we have 10 : aj = 16 : 8, or 16 a; = 80 ; whence x = 5. 358 ALGEBRA. [Ch. XXV 7. The products, or the quotients, of the corresponding terms of two proportions form again a proportion. If a : 6 = c : cJ, or - = -) (1) b d and a! : 2/ = I? : M, or - = -) (2) y u •we have, multiplying corresponding members of (1) and (2), — = — ; wlience ax:by = cz: du. by du Dividing the members of (1) by the corresponding members of (2), we have a c X z , abed T- = - : whence -:- = -: — b d X y z u y u 8. In any proportion, the sum of the first two terms is to the first {or the second) term as the sum of the last two terms is to the third {or the fourth) term. Let a:b = c:d. Then ^ = ^- b d Adding 1 to both members, - + ! = -+!, b d or a + b ^ c + d b d Whence a + b:b = c + d:d. In like manner it can be proved that a + b : a = c + d: c. These two proportions are said to be derived from the given proportion by Composition. 9. In any proportion, the difference of the first two terms is to the first {or the second) term as the difference of the last two terms is to the third {or the fourth) term. a:b = c:d. a+b:b = c + d: d; a — b:b = c — d: ■.d. a + b _^_c + d _ a — b'^ c — d' 1, a + b _c + d § 2] PROPORTION. 359 If a:b = c:d, then a — b:a = c — d:c, and a — b : b = c — d : d. The proof is similar to that of Art. 8. These two proportions are said to be derived from the given proportion by Division. 10. In any proportion, the sum of the first two terms is to their difference as the sum of the last two terms is to their difference. Let By Art. 8, and by Art. 9, Then by Art. 7, u. - or a—b c—d Whence a + b:a — b = c + d:c — d. This proportion is said to be derived from the given one by Composition and Division. 11. A Continued Proportion is one in which the consequent of each ratio is the antecedent of the following ratio ; as, a:b = b:c = c:d = etc. 12. In the continued proportion a:b = b: c, b is called a Mean Proportional between a and c, and c is called the Third Proportional to a and b. 13. The mean proportional between any two numbers is equal to the square root of their product. From a:b = b:c, we have, by Art. 3, b' = ac; whence b = ^(ac). 14. The following examples are applications of the preced- ing theory : Ex. 1. Find a mean proportional between 5 and 20. Let X stand for the required proportional. Then, by Art. 13, x = V(5 X 20) = 10. 360 ALGEBRA. [Ch. XXV Ex. 2. If a:b = c:d, then ab + cd:ab-cd = b' + d?:b'- d\ Let - = -=£». b d Then a=bx and c = dx. Therefore ab + cd = b^x + d% and a6 — cd = 6^a; — d^x. We then have ab + cd ^ b^x + d^a> ^b^ + d\ ab-cd" b'^x - d^x b^ - d^' Whence ab + cd: ah - cd = b'' + d^ -.y^ - d\ Ex. 3. Solve the equation V(2 + a;) + V(2-a;) _ ^2 V(2 + a;)-V(2-a;) ' 1 By composition and division, V(2 + a;) _3 V(2-a!) 1' Squaring and clearing of fractions, 2+a; = 18 — 9a!; whence a; = f. EXERCISES II. Verify each of the following proportions : 1. 2J:lJ = lJ:f 2. 14| : 4| = 200 : 60. 3 4a6 . aJJ + ft' ,. 2a6 . 1 a^-Tfi' a-b a^-b*' 2a-2b Form proportions from each of the following products, in eight different ways : 4. 2x = Sy. 5. m^ = rfi. 6. a^ -b« = x^ - y^. Find a fourth proportional to 7. 1, 2, and 8. 8. |, |, and ^. 9. ab, ac, and 6. Find a third proportional to 10. 2 and 6. 11. J and J. 12. a and 6. Find a mean proportional between 13. 2 and 18. 14. | and |. 15. a^fi and afi^. 16. «±6 and «i:^'. 17. ^+i and Ka^ - !)• § 2] PROPORTION. 361 Find a value of a; to satisfy each of" the following proportions : 18. a;:2 = 12:3. 19. i +y/a : i - a = x: ^a - 2 a. 20. a + b+^^:(SL±^-l = ,c:a-b. a — b 2ab If a : 6 = c : d, prove that 21. a + c:b + d = aM: b^e. 32. a^ + &2 : a^ - 6^ = c^ + d' -.d^-cP. 23. (a±by:ab=(G±dy:cd. 24. 2a + 36:4a; + 56 = 2c + 3d:4c + 5(?. 25. a + 6 : c + (^ = V(a^ + V^) ■ y/(.c^ + cP). 26. ^(a^ 4- 62) : V(c2 + (^^) = ^(a^ + &») : ^/(ca + (28)= a : c. If & be a mean proportional between a and c, prove that 27. ffi2 J. 62 . 63 J. c2 = a : c. 28. aP'b'^c^— + i + 1\ = a' + 6' + c'. If a:6 = &:c = c:<2, prove that 29. (6 + c)(& + d) = (a + c)(c + (2). 30. (a= + 62 + c2) (62 + c2 + d^) = (fl!6 + 6c + C(2)2. If a:6 = c:d = e:/, prove that 31. a^:b^ = ce:df. 32. c^ + d^ le^ +/2 = ciJ: e/ 33. a2 + c2 + e2 : 62 + ^ _^ p = c'^:d^. 34. n a2^^ito^c^-ag^6g-cy then a; = g = y : 6 = g :c. c 6 a Solve each of the following equations : 35 VCl +x) + y/(l -a') _3 36 -v/Ca + a;) - VCa - a;) _ 1 V(1 + «)-a/(1-») ' ' V(« + 'e)+V(«-a;) V6' 37. Find two numbers whose ratio is 7 : 5, and the difference of whose squares is 96. 38. A works 6 days vyith 2 horses, and B works 5 days with 3 horses. What is the ratio of A's work to B's work ? 39. The ratio of a father's age to his son's age is 9 : 5. If the father is 28 years older than the son, how old is each ? 40. Find three numbers in a continued proportion whose sum is 39, and whose product is 721. 41. Find two numbers such that if 1 be added to the first and 8 to the second, the sums will be in the ratio 1 : 2, and if 1 be subtracted from each number, the remainders will be in the ratio 2 : 3. 42. What is the ratio of the numerator of a fraction to its denominator, ii the fraction be unchanged when a is added to its numerator and 6 to its denominator ? 43. The sum of the means of a proportion is 7, the sum of the extremes is 8, and the sum of the squares of all the terms is 65. What is the proportion ? 362 ALGEBRA. [Ch. XXV § 3. VARIATION. 1. Frequently two numbers or quantities are so related to each other that a change in the ralue of one produces a corre- sponding change in the value of the other. Thus, the distance a train runs in one hour depends upon its speed, and increases or decreases when its speed increases or decreases. The illumination made" by a light depends upon the intensity of the light, and varies when the intensity varies. The value of y given by the equation y = 2x — 3 depends upon the value of x and varies when the value of x varies. Thus, if K = 1, 2/ = — 1 ; if a; = 2, 2/ = 1, etc. We shall in this chapter consider only the simplest kinds of variation. 2. Direct Variation. — Two quantities are said to vary directly, one as the other, when their ratio is constant. Thus, if X varies directly as y, then - = fc, a constant. For example, if a train runs at a uniform speed, the number of miles it runs varies directly as the number of hours. If it runs at the rate of 30 miles an hoiir, in 1 hour it will run 30 miles, in 2 hours 60 miles, in 3 hours 90 miles, and so on ; and the ratios 1 : 30, 2 : 60, 3 ; 90, etc., are equal. The symbol of direct variation, oc, is read varies directly as. The word directly is frequently omitted. V If y = 3x, then yozx (read y varies as x), since - = 3, a constant. 3. Inverse Variation. — One quantity is said to vary inversely as another when the first varies as the reciprocal of the second. Thus, if X varies inversely as y, then xix — Therefore, :j- = &, a constant ; whence oey = k. ;3] VARIATION. 363 That is, if one quantity varies inversely as another, the product of the quantities is constant. If 6 men can do a piece of work in 12 hours, 3 men can do the same work in 24 hours, and 1 man in 72 hours, and the products 6 X 12, 3 X 24, 1 X 72 are equal. That is, the num- ber of hours varies inversely as the number of men working. 3 If y=:-,y varies inversely as x, since xy = 3. 4i Joint Variation. — One quantity is said to vary as two others jointly, when it varies as the product of the others. Thus, if X varies as y and z jointly, then — = A;, a constant. For example, the number of miles a train runs varies as the number of hours and the number of miles it runs an hour jointly. It will run 40 miles in 2 hours at a rate of 20 miles an hour, 90 miles in 3 hours at the rate of 30 miles an hour, 40 90 120 and 2 X 20 3 X 30 5 X 24 5. One quantity is said to vary directly as a second and in- versely as a third, when it varies as the second and the recip- rocal of the third jointly. Thus, if X varies directly as y and inversely as z, then — 5- = Tc,a, constant ; or — = ft. 1 V y •- 6. In all the preceding cases of variation, the constant can be determined when any set of corresponding values of the quantities is known. Ex. 1. If a; oc y, and a; = 3 when y = 5, what is the value of the constant ? We have - = A;, or a; = ky. y Therefore, when a; = 3 and y = 5, 3 = 5 &, whence Jc = ^. Con sequently x = ^y. 364 ALGEBRA. [Ch. XXV Ex. 2. If X varies inversely as y, and if y = 4 when x = 7, find the value of x when y = 12. Prom xy = Jc,we obtain k = 28. Therefore xy = 28. Consequently, when y = 12, 12 a; = 28 ; whence x = 2^. Ex. 3. The volume of a gas varies inversely as the pressure when the temperature is constant. When the pressure is 15, the volume is 20 ; what is the volume when the pressure is 20 ? Let V stand for the volume and p for the pressure. Then from pv = k we obtain k = 300. Therefore pv = 300. Consequently, when p = 20, 20 v = 300 ; whence v = 15. EXERCISES III. Ji xazy, what is the expression for x in terms of y, 1. If a; = 10 when 2/ = |? 2. Ifx = a when y = 2a? 3. If xccy^, and x = 6 when y = 1, what is the expression for x in terms of «/? 4. If a; « V2/1 and a; = 3(o8 + b«) when ?/ = 25(a2 + 2 a6 + &2), what is the expression for y in terms of k ? If XX-, what is the expression for x in terms of y, 6. If a; = 10 when y = i? 6. If a; = 3J when J/ = M ? 7. If a; a; — , and a; = 4| when 2/ = f , what is the expression for y in terms of a; ? ^ 8. If a;oc — , and a; = 4 when y = 25, what is the expression for x in terms oiy? ^^ 9. If xxy, and a; = 10 when 2/ = 5, what is the value of x when 2/ = 12i? 10. If xxy, and a; = a when ?/ = f a^, what is the value of y when a;=a26? 11. If X X 2/2, and x = 5 when 2/ = — 3, what is the value of x when 2/ = 15? 12. If X Qc -^2/. and X = a + m when y ={a — m)^, what is the value of X when j/ = (as + m)* ? 13. If X oc -, and x = 3 when 2/ = f 1 what is the value of x when j/ = 4^ ? § 3] VARIATION. 365 14. The circumference of a circle whose radius is 6 feet is 37.7 feet. What is the circumference of a circle whose radius is 9.5 feet, if it be known that the circumference varies as the radius ? 15. An ox is tied by a rope 20 yards long in the center of a field, and eats all the grass within his reach in 2J days. How many days would it have taken the ox to eat all the grass within his reach if the rope had been 10 yards longer ? 16. The volume of a sphere whose radius is 7 inches is 1437.3 cubic inches. What is the volume of a sphere whose radius is 10 inches, if it be known that the volume varies as the cube of the radius ? It has been found by experimeHt that the distance a body falls from rest varies as the square of the time. 17. If a body falls 256 feet in 4 seconds, how far will it fall in 10 seconds ? 18. From what height must a body fall to reach the earth after 15 It has been found by experiment that the velocity acquired by a body falling from rest varies as the time. 19. If the velocity of a falling body is 160 feet after 5 seconds, what will be the velocity after 8 seconds ? 20. How long must a body have been falliug to have acquired a velocity of 256 feet ? 21. The surface of a cube whose edge is 5 inches is 150 square inches. What is the surface of a cube whose edge is 9 inches, if it be known that the surface varies as the square of its edge ? 22. It has been found by experiment that the weight of a body varies inversely as the square of its distance from the center of the earth. If a body weighs 30 pounds on the surface of the earth (approximately 4000 miles from the center), what would be its weight at a distance of 24,000 miles from the surface of the earth ? It has been found by experiment that the illumination of an object varies inversely as the square of its distance from the source of light. 23. If the illumination of an object at a distance of 10 feet from a source of light is 2, what is the illumination at a distance of 40 feet ? 24. To what distance must an object which is now 10 feet from a source of light be removed in order that it shall receive only one-half as much light ? 25. At what distance will a light of intensity 10 give the same illumi- natioij as a light of intensity 8 gives at a distance of 50 feet ? CHAPTER XXVI. DOCTRINE OP EXPONENTS. 1. We have already abbreviated such products as aa, aaa, aaaa, ■■■, aaa---n factors, by a?, a^, a\ •••, a", respectively, and called them the second, third, fourth, ■■-, nth, powers of a. This definition of the sym- bol a" requires the exponent n to be a positive integer. Thus 2' means the product of 5 factors, each equal to 2. But 2° has, as yet, no meaning, since 2 cannot be taken times as a factor. For a similar reason, 2~', 2^, 2^% and 2'^~\ are, as yet, meaningless. 2. Nevertheless, having introduced into Algebra the symbol a", it is natural to inquire what it may mean when w is 0, a rational negative or fractional number, an irrational number, etc. We shall find that, by enlarging our conception of powers, quite clear and definite meanings can be given to such expres- sions as 2°, 3-^ 4^, 6V2, 6^"^ 3. The discussion of powers, in general, therefore naturally divides itself into six cases. (1) Powers with positive integral exponents. (2) Powers with zero^ exponents. (3) Powers with negative integral exponents. (4) Powers with fractional {positive or negative) exponents. (5) Powers with irrational exponents. (6) Powers with imaginary exponents. The consideration of powers with imaginary exponents will be given in Part II., Text-Book of Algebra. 366 DOCTRINE OF EXPONENTS. 367 Positive Integral Powers. 4. The principles upon which operations with positive inte- gral powers depend have been proved in the preceding chapters. For the sake of emphasis, and for convenience of reference in enlarging our conceptions of powers, we restate them here : (i.) a'"a" = a"'+". (ii.) — = a"-", when m > w ; — =1, whenm = »i: ^ a" a" — = , when m<.n. Qti a."-'" (iii.) (a™)" = a""". (iv.) (ab)"' = aH". ($) ay" a"' b" Zeroth Powers. 5. The meaning of a symbol may be defined by assuming that it stands for the result of a definite operation, as was done in letting oT' z= a • a- a- •■• n factors ; or by enlarging the meaning of some operation or law which was previously restricted in its application. In the latter way, negative numbers were introduced by extending the meaning of subtraction. 6. We now enlarge the meaning of powers by assuming that the principle a" holds also when m = n. We then have — = a™ a™ But since -i = 1> a™ it follows that a° = l. That is, the zeroth power of any base, except 0, is equal to 1. E.g., 1" = 1, 5» = 1, 99» = 1, (a 4- &)" = 1, etc. 368 ALGEBRA. [Ch. XXVI 7. Thus, by the assumption that the stated law holds when m = m, a definite value of the zeroth power of a number is obtained. Nevertheless, it will doubtless seem strange to the student that all numbers to the zeroth power have one and the same value, namely 1. But it should be distinctly noted that aP is by definition a symbol for — ; i.e., for the quotient of two like powers of the same base. Thus, 2'= — = — = — = 1 23 2^ 2" Negative Integral Po'v^rers. 8. We now still further enlarge the meaning of powers by assuming that the principle a" holds not only when m^n and m = w, but also when m < n. In this case, m — w is a negative number. Since m+!-" = a^-' = x-^ = -■ 6^ Ex. 2. aM X a-^b* = aH^+* = a-^&V = ^ a* (i.) m positive and n negative, and the absolute value of m less than the absolute value of re. Let re = — rei, so that ni is positive. Then /7"* 1 1 , . gmfin ~ o^a-"! = -— = = — = = a'^+(-ni) = a^+n, a"l a"l~"* (J— ["*+(— ni)] In a similar way the principle can be proved for other cases in which the exponents are or negative. That the principle holds when the exponents, either or both, are frac- tions, follows from the definition of a fractional power. EXERCISES III. Simplify each of the following expressions : 1. xV. 2. x-'A 3. a-SflS. 4. m-^m,-^. 5. a?a^. • 6. a^a^. 7. b~h^. 8. c'^c'i. 9. 5 a-8 X 3 a«. 10. - ^ 6-2 x If 6"'. 11. 3f x^ x 2f x~k 12. a86-2 X aM. 13. Sx^y^x^x-^y"^. 14. 2 a~ V^c' x 5 aSc"^. m p 1 i* 1 _p_ 15. o^ft-" X a-rft'. 16. xyf x a;»2/~». 17. a'»-»6' x a'»+''& '. 18. li«lfx-^. 19 7 e-« . 35 ar^ ^q a-^ft-" . c-i re-2 9ra» ■ 3a3 ■ 6c2 Jc o-2»6-2n 31. (a^ + a;-2)(a^-a;-2). 22. (a^ + a"^)(a^- a"^. 23. {J-ah^ + h^){a^ + h^). 24. (a;^!/"* + aj^"^ + l)(a;2/"^ - 1). 25. (o^+fl!i6i+&2)Ca^-ai6i+6^). 26. (a^+6^ + a~^6)(a6"^-a^+6*). 27. (a-' + a-6 - a-s) (a' + a^ + a^). 28. (a;8 _ x-^ - 2 a;-s + 5) (10 a;"' + sc"! - 5 a;-^) . 29. (i a-6a;-» + f x-^" - a-^x-^") (| a^^a^ - 4 a-''a;P+2» + a-ia^+»). 30. (a;^ - xy^ + x^y - y^) {x + x^y^ + y). 374 ALGEBRA. [Ch. XXVI 31. (a^ + 0^ -a^- a"^) (o^ + a~^ + 1). 32. (x^ + 2 E-S + .3 K^ + 2 x^ + 1) (x^ - 2 X* + ] ). 33. (a-is + 6-1-6 _ a-756-8) (a-^s + 6" -8). 34. (If a V^ + 2 a^ + I xi-8 + 6 ax=) (a^ - 3 x^ + f x^ Vi). Quotients of Powers. (II.) 5=""""' for all rational values of m and n. Ex. 1. — = cc^-t-') = a^+' = ar>. Ex.2. ^=a-i-i&W = «-i&^=^. "We have — = a'^a-" = a^+t-"' = o"-". a" EXERCISES IV. Simplify each of the following expressions : 1 ^. 2 ^. 3 ^. i ^. 5. ^. 6. ^■ 5-a ^i x-5 ^-i 7. ^. 8. «i. 9. ^. 10. '^^. 11. ^- 12. ^• -J a-2 X-" x-" x»-i x"^ 13. (lJ6-s)-4-(3&2). 14. l-:-Qa6-i). 15. (3| a"6-4)- (I «"&-') • 16. 12 a-i6->x-^ ^ i«^. 17. |^ x ^-^. x-t 3a^6-* 7^3^,* 18. (ai-6^)-=-(ai + 6^). 19. (x - j^)h-(x^ - 2/*). 20. (x-i + 2/-1) -^ (x~* + y~i) . 21. (x"* - «/-*) h- (a;-i + 2/-1) . 11 1111 »!_?!! r"-! 22. (x+xV+2/)-(«*+a;V+2/^)- 23- (a^ - a 2)-4-(a2 - a =). 24. [(a-l)-2-l]-=-[(a-l)-i-l]. 25. (3 o-w + a6 - 4 a"') ^ (2 a'^ + ra^ + 3 «-6). 26. (2 x-8 - 8 x-2 - 2 x-i + 2 - x) H- (x-» + 1). 27. (x-i - 3 x~^ + 3 - 3 x^ + 2 x) H- (x~^ - 2 x-i + x~^ - 2). DOCTRINE OF EXPONENTS. 375 28. (2 o' - 3 o' - 23 a-i + 15 a-^ + 9 a-^) h- (a* + 2 - 8 o"*). 29. (6 a;^ + 9 x~^ - 2 a:-i - 13) ^ (3 a;^ + 2 a;~i - 5). 30. (x^ -xy^ + x^y -y^)^(^x^ -y^). 31. («* + a^x^ + a^) -i- (x^ + a'a;^ + a^). 82. (o^ - a2&* - ah^ + 6^) -=- (a^ - o6* + o^& - 6^). 33. (6xi-7x-l9x^ + 2xi + 8xi)^(2x^-Sxi-4xh. Powers of Powers. (III.) (o™)" = a"", for all rational values of m and n. Ex. 1. (x^-» = a;2(-3) ^ j„-6 ^ 1. Ex. 2. (1024^)-^ = 1024-iSr = ^^„ }^^^^^ = 1. ^ '' (?»/1024)s 8 Ex. 3. (l + a;^)2 = l + 2a;~^ + (a;"^)2=l+-^ + -. (i.) m and ra both negative integers. Let m = — mi and re = — mi, so that mi and m are positive. We have (a"')»=(a-'»i)-"i =( -s" ) ' = (ai°H)"i = a^'^ = at-™!"-"!) = a"". In a similar manner the principle can he proved for other cases in which the exponents are or negative integers. (ii.) m a fraction, and n a positive or a negative integer, or 0. Let m = —, wherein g is a positive integer and p is a positive or a negative integer. We then have («'»)»= (a«)»= [(0')"]"= (0')""= a^ = af = a"". In a similar manner the principle can he proved when m is an integer and re is a fraction. (iii.) m and re both fractions. Let m = — and n = -, where q s' r are positive or negative integers. Let m = — and n = -, wherein q and s are positive integers, andp and q s 376 ALGEBRA. [Ch. XXVI If (a')' be raised to the g'sth, = sgth power, we have p r p r p p [(o«)»]«» = {[(a« )"]»}«= [(a* )•■]« = [(««)«]'• = (aJ')'' = a^. p r Consequently {afly is the qs root of a^*"; or, by definition of a frac- tional power, p r pr ■p r EXERCISES V. Simplify each of the following expressions : 1. (a;2)-2. 2. (a3)i 3. [(-a;)*]^. 4. {x'^y. 5. (a;"*)i5. 6. (a-3)i 7. (ftS)"^. 8. (a;-2)-6. 9. («"*)"*. 10. (a»)-2. 11. (a-'»)-3. 12. (a «) ". 13. (.J/a-2)4. 14. (Va)"^- 15- (v/x^)"^- 16. (.ya-"')-s. 17. (a^ - a"^)2. 18. (1 - x^f. 19. (a^ + a"*)'. Powers of Products. (IV.) (a6)'" = a'"6'", for all rational values o.f m. Ex. 1. (2 a;)-' = 2-«a!-' = ^• Ex. 2. (3 a;"i2/^-' = Z'^^y-^ = a^ (i.) ma negative integer. Let m =— nii, so that mi is positive. Then (a6)"» = Cab)-'^ = — - — = — - — = a-'^b-'^ = a'^b'". (aft)"! a™i6™i P (11.) m a fraction. Let m = -, where p is a positive or negative inte- ger, and g is a positiije integer. If (a6)« he raised to the gth power, we have p [(a6)«]' =(aby, since q is an Integer, = aPb'', by (1.). p p p p But (a«6^)« =(a«)«(6«)' = apftp. p p p p p p Therefore [(a6)»]' = (a«6«)« ; whence (a6)« = a«6«. DOCTRINE OF EXPONENTS. 377 Powers of Quotients. (V.) I T 1 = T— > for all rational values of m. \bj b'" Ex.1, f— 1 =^=-?-. Ex.2.' ^ * ** We have (")"* = (a6"^)" = a"6-'" = ^• BXBBCISBS VI. Simplify each of the following expressions : 1. (aVi)-2. 2. (ia)~^. 3. (8 «-6)*. 4. (a-ifi-s)-*. 5. (2a^a;)^. 6. (x^aT^)-^. 7- (i^a;~")~^- 8. (0-26*0"*) -6. 9. (2a-^a;)-2. 10. (5a;-».J/a2)-6. 11. {Zy/x^x-^)-^. 12. [3.y(a-'»&!')]-"'i'. 13. d,4. 'M^)-*- "-(sii)-' "■(s-w^)-'- 19 /i«:My. 20. fa^I^^V. 21. f v«V'. -[(i^)T' -[(S^)1' -[(^p)"*]" 28. (aV2-l)2. 39. (fl!~Vi + a6^)2. 30. {a-%^-c^\ EXBECISBS VII. MISCELLANEOUS EXAMPLES. Simplify each of the following expressions : 2 378 ALGEBRA. [Ch. XXVI 2x^ 1 (1 - x'^y (1 -,x^)^ a -6 a^- -b^ a^ -bi "- -b a^x^ + oi^i a — x a^ + x^ a^ + K^ ' + 1 6. 8. 2ai aH a^^-x^ aH + ,i a — a; a + x «*-=.* a^ + ^ «*+! 1 ,+^i+l ^^-^-l ai + a^ + 1 a^ - a^ + 1 a^ - a^ + 1 JFlnd the square root of each of the following expressions : 10. a;^ + x~^ + 2. 11. «-*» + 2 a"^a;"^ + ax"*. 12. 4a;-*-12a;-8+13a;-2-6K-i + l. 13. 9a;H10a;-2-4a;-4+a;-6-12. 14. a2-|o^-foi + 4Ja + l. 15. %x^-b x^y^ + \^ x'^y - f k^j;^ + ^ xi/^. Find the cube root of each of the following expressions : 16. x-^ - 6 a;-6 + 12 x'* - 8 x'K 17. %x- 36 x^ - 27 a;^ + 54 xt 18. 8 a;-8 + 12 a;-2 - 30 a;-i - 35 + 45 a; + 27 a;^ _ 27 a;8. 19. a;^ - 3 a;^ + 3 a;^ + 2 a; + 3 a;^ - 3 a;^ - 6 a;i ^ + 3 ai'T* + xi 20. 8 x^!/"^ + 13 x^ + ^^ + 12 x^2/-i + 18 xV* + 9 xjr^ + 3 xV Solve each of the following equations : 21. 2x-2-x-i-l = 0. 22. x-2 + 5x-i-J^ = 0. 23. 3 x-6 - 2 x-8 - 1 = 0. 24. 5 X - 2 x^ - 16 = 0. 25. 3 x^ - 2 x^ - 4 = 0. 26. 5 x~^ - 2 x"^ - 3 = 0. 27. x^ + 3x^-10 = 0. 28. 7x"^ - 3x~* - 4 = 0. 29. 5 x^ - 3 x* - 14 = 0. 30. 2 x"^ + 5 x"^ - 7 = 0. CHAPTER XXVII. PROGRESSIONS. §1- 1. A Series is a succession of numbers, each formed accord- ing to some definite law. The single numbers are called the Terms of the series. The law may specify that each term shall be formed from the immediately preceding term in a prescribed way. E.g., in the series 1 + 3 + 5 + 7 + 9 + - (1) each term after the first is formed by adding 2 to the preced- ing term. In the series l + 2 + 4-f8+.-- (2) each term after the first is formed by multiplying the preced- ing term by 2. Or the law may state a definite relation between each term and the number of its place in the series. ^.g'., in the series 1+1+1-^^+... (3) each term is the reciprocal of its number. 2. The number of terms in a series may be either limited ot unlimited. A Finite series is one of a limited number of terms. An Infinite series is one of an unlimited number of terms. In this chapter a few simple and yet very important series will be discussed. §2. ARITHMETICAL PEOGRESSION. 1. An Arithmetical Series, or as it is more commonly called an Arithmetical Progression (A. P.), is a series in which each 379 380 ALGEBRA. [Ch. XXVH term, after the first, is formed by adding a constant number to the preceding term. See § 1, Art. 1, (1). Evidently this definition is equivalent to the statement, that the difference between any two consecutive terms is constant. E.g., in the series 1 + 3 + 5 + 7+- we have 3-1 = 5-3 = 7-5 = - For this reason the constant number of the first definition is called the Common Difference of the series. 2. Let ai stand for the first term of the series, a„ for the nth (any) term of the series, d for the common difference, and S^ for the sum of n terms of the series. The five numbers Oi, a„, d, n, S„ are called the Elements of the progression. 3. The common difference may be either positive or negative. If d be positive, each term is greater than the preceding, and the series is called a rising, or an increasing progression. E.g., 1 + 2 + 3 + 4+---, wherein d = 1. If d be negative, each term is less than the preceding, and the series is called a falling, or a decreasing progression. E.g., 1 — 1—3 — 5 , wherein d = — 2. 4. In an arithmetical progression any term is equal to the first term plus the product of the common difference and a number one less than the number of the required term, i.e., a„ = a^ + (n-l)d. (I.) By the definition of an arithmetical progression ttji = 0.], aj = «! + d, as= a2 + d = ai + 2 d, etc. The law expressed by the formulae for these first three terms is evidently general, and since the coefficient of d in each is one less than the number of the corresponding term, we have «« = «! + (« — ^)d. §2] ARITHMETICAL PROGRESSION. 381 Ex. 1. rind the 15th term of the progression 1+3+5+7+- we have o^i = 1> d = 2, w = 15 ; therefore a^ = 1 + (15 - 1) 2 = 1 + 28 = 29. This formula may be used not only to find a„, -when %, d, and n are given, but also to find any one of the four numbers involved when the other three are given. Ex. 2. If Oj = 3 (n = 6), and Ui = 1, we have 3 = 1 + 4 d ; whence d = ^. 5. In any arithmetical progression, the sum of n terms is equal to one-half the product of the number of terms and the sum of the first and the nth term, i.e., S„=l(a, + a„). (II.) Since the successive terms in an arithmetical progression, from the first to the nth inclusive, may be obtained either by repeated additions of the common difference beginning with the first term, or by repeated subtractions of the common dif- ference beginning with the nth term, we may express the sum of n terms in two equivalent ways : S^=ai+(ai+c[) + {ai+2 d)+ — +(a,+n-2 ■d)-\-(a,+n-l ■ d), S„=a„+{a„-d) + (a„-2d) + - + (a„-n^-d) + (a„-n^-d). Whence, by addition, 2S„= («! + a„) + (Ol + ctn) H h («! +»n) + {f=?-lt — 2 That is, the arithmetical mean between two numbers is half their sum. 11. Arithmetical Means between two numbers are numbers, in value between the two, which form with them an arithmetical progression. E.g., 2, 3, and 4 are three arithmetical means between 1 and 5. Ex. Insert four arithmetical means between — 2 and 9. We have n = 6, ai = — 2, aj = 9. From (I.), 9 = - 2 + 5 d, whence d = -i^. The required means are \, J^, ^, \*-. 12. If n aritliinetical means be inserted between o and 6, we have an arithmetical progression of m + 2 terms, the first term being a and the last 6. Therefore, from (I.), 6 = a + (m + 2 — 1) d, whence d = —^ — 11+1 The resulting series is therefore a,a-\ -, a + 2 — — -, a + 3 — — ■, ■••, 6; «+l ra + 1 n + \ na + b Cn-l)a + 2b ()i - 2) a + 3 6 . or o, -—r-, — , -— , •■", "• n+1 n+1 n+1 EXERCISES II. Insert an arithmetical mean between 1. 45 and 31. 2. 17^ and 14J. 3. 2 a and - 2 6. 4 «^ and ^±^. 5. ^+^ and -^+1. a+b a — b x — l a;* — 1 6. Insert 6 arithmetical means between 7 and 35. 7. Insert 12 arithmetical means between 37 and — 28. 386 ALGEBRA. [Ch. XXVIl 8. Insert 9 arithmetical means lietween ^ and 12. 9. Insert 5 aritlimetioal means between 17| and IJ. 10. Insert 20 aritlimetical means between — 16 and 26. 11. Insert 6 arithmetical means between a + b and 8 a — 13 6. Problems. 13. Pr. 1. Tlie sum of four numbers in arithmetical pro- gression is 16, and their product is 105. What are the numbers ? We can express the four required numbers in terms of two unknown numbers. Let X — 3d, X— d, x + d, x + 3d be the four required num- bers. Then, by the first condition, (x-3d) + (x-d) + {x + c[) + (x + 3d) = 16; whence a? = 4. By the second condition, (x - 3 d)(x - d) (x ^- d?){x + 3 d) = 105, or (a^-9 d") {x^ -dF) = 105. Substituting 4 for x and reducing, 9 d* — 160 d^ = — 151. From this equation we obtain d = ±l, and ± ^^151. The corresponding numbers are, when d = l: 1, 3, 5, 7 ; when d = — l: 7, 6, 3, 1 ; when d = ^ V^^^ ■ 4 - V151, 4 - iVlSl, 4 + 1 V151, 4 + V151.; _ when d = — -J-^lSl : 4+V151, 4+^Vl51, 4-IV151, 4-V151. Notice the advantage of assuming the required numbers as in the above example. Had we assumed x, x + d, x + 2d, x + 3d as the required numbers, the solution would have involved an equation of the fourth degree which could not have been solved as a qiiadratic. Pr. 2. Find the sum of all the numbers of three digits which are multiples of 7. The numbers of three digits which are multiples of 7 are 7 X 15, 7 X 16, 7 X 17, -, 7 x 142. §2] ARITHMETICAL PROGRESSION. 387 Their sum is 7 (15 + 16 H h 142). The series within the parentheses is an arithmetical progres- sion, in which Oj = 15, d = l, n = 128, and a^ss = 142. Therefore Sj^s = 10,048. The required sum is therefore 7 x 10,048, = 70,336. y EXERCISES III. 1. Find the sixth term, and the sum of eleven terms, of an A. P. whose eighth term is 11 and whose fourth term is — 1. V 2. The sixteenth term of an A. P. is — 5, and the forty-first term is 45. What is the first term, and the sum of twenty terms ? ' 3. Find the sum of all the even numbers from 2 to 50 inclusive. 4. Find the sum of thirty consecutive odd numbers, of which the last is 127. 5. The sum of the eighth and fourth terms of an A. P. of twenty terms is 24, and the sum of the fifteenth and nineteenth terms is 68. What are the elements of the progression ? 6. The product of the first and fifth terms of an A. P. of ten terms is 85, and the product of the first and third terms is 55. What are the elements of the progression ? r 7. The first, term of an A. P. of thirty terms is 100, and the sum of the first six terms is five times the sum of the six following terms. What are the elements of the progression ? 8. The sum of the second and twentieth terms of an A. P. is 10, and their product is 23|J. What is the sum of sixteen terms ? 9. The sixth term of an A. P. is 30, and the sum of the first thirteen terms is 455. What is the sum of the first thirty terms ? I 10. What value of x will make the arithmetical mean between k? and k' equal to 6 ? ^ 11. Find the sxun of all even numbers of two digits. 12. How many consecutive odd numbers beginning with 7 must be taken to give a sum 775 ? 13. Insert between and 6 a number of arithmetical means so that the sum of the terms of the resulting A. P. shall be 39. 14. Find the number of arithmetical means between 1 and 19, if the first mean is to the last mean as 1 to 7. V 15. The sum of the terms of an A. P. of six terms is 66, and the sum of the squares of the terms is 1006. What are the elements of the progression ? 388 ALGEBRA. [Ch. XXVH 16. The sum of the terms of an A. P. of twelve terms is 354, and the sum of the even terms is to the sum of the odd terms as 32 to 27. What is the common difference ? \ 17. How many positive integers of three digits are there which are divisible by 9 ? Find their sum. 18. If the sum of m terms of an A. P. is n, and the sum of n terms Is m, what is the sum oim + n terms ? Of m — re terms ? /19. Show that the sum of 2 ra + 1 consecutive integers is divisible by 2re + l. 20. Prove that if the same number be added to each term of an A. P., the resulting series will be an A. P. 21. Prove that if each term of an A. P. be multiplied by the same number, the resulting series will be an A. P. 22. Prove that if in the equation y = ax + b, yfe substitute c, c + d, c + 2d, ■■■, in turn for x, the resulting values of y will form an A. P. 23. Prove that if a\ b^, 1. V 5. Ex. 1. Given Oj = 3, r = 2, m = 6, to find Sg. From (II.), S, = ^f~^^ = 189. §3] GEOMETRICAL PROGRESSION. 391 Formulae (II.) and (III.) may be used not only to find >S'„ ■when Ou r, and n, or Oj, a„, and r are given, but also to find the value of any one of the four elements when the other three are given. Ex. 2. Given S^ = - &^, Oi = - ^, a„ = - 32, to find r. By (III.), - 63^ = -^ + S2r^ whence r = 2. 1 — r 6. Formulae (I.) and (II.), or (I.) and (III.), may be used simultaneously to determine any two of the five elements, ctj, a„, r, S„, n, when the three other elements are given. Ex. 1. Given r = - i m = 5, aj = — -i, to find % and 8^. From (I.), — i = «i(— i)^ whence Oi = — 4 ; and from (II.), using the value of % just found, -4[l-(-i)T 11 ""'- l-(-i) - 4- Ex. 2. Given r = 2, a„ = 16, 8^ — 31|, to find % and n. From (III.), 3H = ^^^^7"^ , whence a^ = \. From (I.), 16 = J ■ 2»-i, whence n = &. Ex. 3. Given ri = 7, a? = 16, /S*? = 31f , to find r and %. From (I.), 16 = Ojr^ and from (II.), 31| = '^^^ ~ '"') • ^ ' * 1— r Eliminating ai between these two equations, we obtain 63r'-127?-« = -64. Thus this example leads to an equation of the seventh degree, which cannot be solved. The value of r in such equations can often be found by inspection. In the above equation r = 2. We then have % = \. 7. In many examples the elements necessary for determin- ing the element or elements directly from (I.)-(III.) are not given, but in their place equivalent data. 392 ALGEBKA. [Ch. XXVII Ex. 1. Given ag = 48, Og = 384, to find % and r. Erom (I.), 48 = a^r*, and 384 = a^f ; whence r* = 8, or r = 2. Therefore % = 3. Or, we could have regarded 48 as the first term and 384 as the last term of a progression of four terms. Then by (I.), 384 = 48 1^, whence r = 2, as before. Ex. 2. Given S^ = 63, S^ = 511, to find % and r. Erom (II.), 63 = "-^(f-^) , and 511 = ^^'^'''~/) ; (1) ^ ■' r — 1 r — 1 ^ , r»-l 511 r« + ?-8 + l 611 whence -^ = -^, or -^^^ = — • Therefore 63 r= - 448 r^ = 448 ; whence r' = 8, and — f ; and ?• = 2, and — -fv^S. We then have from (1) : % = 1, when r = 2 ; and % = IOOtV (2^3 + 3), when r = - 1^3. Such examples as the last in general lead to equations of higher degree than the second. EXERCISES IV. Find the last term and the sum of the terms of each of the following geometrical progressions': 1. 3 + 6H to 6 terms. 2. 2 — 4h to 10 terms. 3. 32-16 + ...to 7 terms. 4. 1| + 2f + ... to 6 terms. 5. 2-22+ ...to 11 terms. g. — + iH to n terms. 7. 1 + (1 + o) + ••• to 4 terms, to n terms. 8. 02" + csJ'+« + •■• to 7 terms, to n terms. In each of the following geometrical progressions find the values of the elements not given : 9. ai = 1, J- = 4, n = 5. 10. ai = 2J, ?• = — 2, n = 6. 11. a„ = 10, r = 2, n = 4. 12. a„ = 1.2, r = - .2, n = 5. 13. »• = 2, n = 5, Sn = 62. 14. r = 10, re = 7, & = 3,333,333. 15. ai = 5, re = 9, a„ = 327,680. 16. ai = 74f, re = 6, a„ = 2f 17. a„ = 96, re = 4, «„ = 127.5. 18. a„=7, re = 9, (?„ = 68,887. §3] GEOMETRICAL PROGRESSION. 393 19. Oi = 1, »• = 2, a„ = 64. 20. ai = 7, r= 10, «„ = 700. . 21. oi = 74|, a„ = 2^, & = 147. 22. ai = 1, o„ = 512, & = 1023. 23. a„ = 44, r = 4, & = 55. 24. a„ = 3125, r = 5, ;?„ = 3905. 25. ai = 40, r = \, S„ = 75. 26. ai = 4, r = 3, Sn = 118,096. 27. ai = 3, ra = 3, Sn = 1953. 28. ai = 100, n = 3, S„ = 700. Sum of an Infinite Geometrical Progression. 8. When r is less than 1, the term air" in the formula ^_ ai-air" 1 — r decreases as n increases. As n grows indefinitely large, oi)"" becomes indefinitely small. Tor, as in Ch. XVII., Art. 15, we have an+i > Jn+l + (m + 1) (a - &) 6". Now let a = 1, 6 = 1 — d, wherein d is a positive proper fraction. Then 1 >(1 - (?)"+i +(« + l)d(l - d)" ; or, 1 >(1 - d)"(l + nd). Therefore (1 _ d)n <- — I — ^ ^ 1+nd Evidently, — - — can be made less than any assigned number, however 1 + nd small, by increasing n indefinitely. Therefore, (1 — d)", which may rep- resent any proper fraction, say r, can be made less than any assigned number, however small. That is, when m = oo, ajr" = 0. We therefore have, when r < 1 and re = oo, S„ ai l-r Ex. 1. Find the sum of the infinite series l + i + i + i+ ■■•- in which r = i- We have S„ = —^=2. The meaning of this result is that the sum of the given series approaches the finite value 2 more and more nearly as more and more terms are included in the sum, and that the sum can be made to differ from 2 by as little as we please, by taking a sufficient number of terms. The exact sum 2, however, can never be obtained. - This theory can be applied to find the value of a repeating (recurring) decimal. 394 ALGEBRA. [Ch. XXVII Ex. a. Verify that .6 = |. ■We have .666- = A + t^ + t^+ ..., a geometrical progression whose first term is ^ and whose ratio is -j^. Consequently EXBBOISBS V. Find the sum of the following Infinite geometrical progressions : M. 6 + 4 + .... 3. 60 + 15 4 -. 3. 10 - 6 + .... 4. i + i+-- V.V 1-J+.... 6. 5-i + .... Vt. -|-f+.... Vs. vf + Vf+-- 9- V-2 + VTi7+-- 10. l + a;H-a;2+ •••, when ,^<1. 11. 1 + i + i+ ••., when a > 1. X x^ Find the value of each of the follpwing repeating decimals : '-' 12. .44-. ■ 13. .99-., M4. .2727-. 15. .015015-. 16. .199-. 17. 1.0909 •■•. 18. .122323..-. 19. .201475475-. Verify each of the following identities : 20. V-44- = -66-. 21. V.6944- = .833-. Geometrical Means. 9. A Geometrical Mean between two numbers is a number, in value between the two, which forms with them a geometrical progression. E.g., + 2, or — 2, is a geometrical mean between 1 and 4. Let O be the geometrical mean between a and h. Then by definition of a geometrical progression, -=^; whence G = ± y/{ab). O, Gr That is, the geometrical mean between two number's is the square root of their product. Ex. Find the geometrical mean between 1 and |. We have (? = ±V(lx|)=±|. 10. Geometrical Means between two numbers are numbers, in value between the two, which form with them a geometrical progression. E.g., 4 and 16 are two geometrical means between §3] GEOMETRICAL PROGRESSION. 395 1 and 64 ; and 2, 4, 8, 16, 32 are five geometrical means between 1 and 64. Ex. Insert five geometrical means between 2 and 11. We have 11 = 2 r^, whence r = -^^■ The required means are 2^^, 2-^{^r, 2^(Jf)», 2^(^)^ 2^(^)». 11. If n geometrical means be inserted between a and 6, we have a geometrical progression of n +.2 terms. Consequently, by (I.), ^. n+l 16 6=ar'^S or r = -\\— The progression is therefore, or a, "+^(a"&), ''+^(a'-i62), »ti/(a"-268), .-6. BXBRCISBS VI. Insert a geometrical mean between ^. 2 and 8. . 3. 12 and 3. 3. J and ^k- 4. VandVC^a). S. 75 m^ and 3 mra*. 6. - and -■ 7. (a - &)2 and (a + 6)2. 8. (a^ + l)(a2 - l)-i and iCa* - 1). ^9. Insert 5 geometrical means between 2 and 1468. 10. Insert 7 geometrical means between 2 and 512. 11. Insert 6 geometrical means between 3 and — 384. 12. Insert 6 geometrical means between 5 and — 640. 13. Insert 8 geometrical means between 4 and — ifff'- 14. Insert 9 geometrical means between 1 and -5^^. Problems. 12. Pr. The sum of the terms of a geometrical progression of five terms is 484 ; the sum of the second and fourth terms is 120. What is the progression ? Let — , -, X, xr, xr^ be the required terms. 396 ALGEBRA. [Ch. XXVII By the first condition, E + '^ + x + xr + xr^ = 484. (1) r r By the second condition, -+xr — 120. (2) Sxibtracting (2) from (1), ^ + x + xr' = 364. (3) Dividing by x, and adding 1 to both members, l + 2 + r^ = ^ + l. (4) r X Squaring (2), a^fl + 2 + A = 14400, (6) 1 , „ , , 14400 ,n. or _ + 2 + r^ = — -^— (6) r or Equating second members of (4) and (6), 364 . H 14400 ,„. -^ + ^ = ^' ^^) whence x = 36, and — 400. Substituting 36 for as in (2), we obtain r = 3, and \. The value, — 400, of x gives imaginary values of r, and there- fore must be rejected. When a = 36 and r = 3, the series is 4, 12, 36, 108, 324- when a; = 36 and r = \, the series is 324, 108, 36, 12, 4. EXERCISES VII. 1. The first term of a G. P. of six terms is 768, and the last term is one-sixteenth of the fourth term. What is the sum of the six terms of the progression ? S. The first term of a G. P. of ten terms is 3, and the sum of the first three terms is one-eighth of the sum of the next three terms. Find the elements of the progression. 3. The twelfth term of a G. P. is 1536, and the fourth term is 6. What is the ratio, and the sum of the first eleven terms ? 4. The sum of the first and fourth terms of a G. P. of ten terms is 455, and the sum of the second and third terms is 140. What are the elements of the progression 1 §3] GEOMETRICAL PROGRESSION. 397 5. In a G. P. of eight terms, the sum of the first seven terms is 444 J, and is to the sum of the last seven terms as 1 to 2. Pind the elements of the progression. Y 6. The sum of the first four terms of a 6. P. is 15, and the sum of the terms from the second to the fifth inclusive is 30. What is the first term, and the ratio ? 7. Find the elements of a G. P. of six terms vrhose first term is 1, and the sum of whose first six terms is twenty-eight times the sum of the first three terms. 8. The sum of the first three terms of a G. P. is 21, and the sum of their squares is 189. What is the first term ? 9. The product of the first and third terms of a G. P. of seven terms is 64, and the sum of the fourth and sixth terms is 6. What are the ele- mejits of the progression ? V^IO. The product of the first three terms of a G. P. is 216, and the sum of their cubes is 1971. What is the first term, and the ratio ? 11. Find the mth and the nth terms of a geometrical progression whose (m + w)th term is h, and (m — w)th term is I. ^12. If the numbers 1, 1, 3, 9 be added to the first four terms of an A. P., respectively, the resulting terms will form a G. P. What is the first term, and the common difference of the A. P. ? 13. A 6. P. and an A. P. have a common first term 3, the difference between their second terms is 6, and their third terms are equal. What is the ratio of the G. P., and the common difference of the A. P. ? 14. The sum of the eight terms of an A. P., whose first term is 1, is 3,294,176. The first and last terms of a G. P. of eight terms are equal to the corresponding terms of the A. P. Find the fifth term of the G. P. 15. The first and last terms of an A. P. of fifteen terms are equal to the corresponding terms of a G. P. of fifteen terms, and the ninth term of the A. P. is equal to the eighth term of the G. P. What is the ratio of the G. P. ? 16. The ratios of two geometrical progressions having the same number of terms are ^ and \ respectively, and their first terms are equal. The last term of the first progression is 243, and the last term of the second is 32. Find the elements of each progression. 17. Show that, if all the terms of a G. P. be multiplied by the same number, the resulting series will form a G. P. 18. Show that the series whose terms are the reciprocals of the terms of a G. P. is a G. P. 19. Show that the product of the first and last terms of a G. P. is equal to the product of any two terms which are equally distant from the first and last terms respectively. S98 ALGEBRA. [Ch. XXVIl 20. If the numbers a, b, c, d form a G. P., show that (a - dy = (6 - c)2 + (c - a)2 + (d - &)=. 21. A merchant's investment yields him each year after the first, three times as much as the preceding year. If his investment paid him | 9720 in four years, how much did he realize the first year and the fourth year ? 23. On one of the sides of an acute angle a point is taken a feet from the vertex ; from this point a perpendicular is let fall on the second side, cutting off 6 feet from the vertex. From the foot of this perpendicular a perpendicular is let fall on the first side, and from the foot of this per- pendicular a third perpendicular is let fall on the second side, and so on indefinitely. Eind the sum of all the perpendiculars. 23. Given a square whose side is 2 a. The middle points of its adja- cent sides are joined hy lines forming a second square inscribed in the first. In the same manner a third square is inscribed in the second, a fourth in the third, and so on indefinitely. Find the sum of the perimeters of all the squares. §4. HAEMONICAL PROGRESSION. 1. A Harmonical Progression (H. P.) is a series the recipro- cals of whose terms form an arithmetical progression. E.g., 1 + 1+1 + !_ + .. . is a harmonical progression, since H-2 + 3 + 4 + -.. is an arithmetical progression. Consequently to every harmonical progression there corre- sponds an arithmetical progression, and vice versa. 2. If three numbers be in harmonical progression, the ratio of the difference between the first and second terms to the difference between the second and third terms is equal to the ratio of the first term to the third term. Let the three numbers a, b, c be in harmonical progression. By the definition of a harmonical progression, 111 a b c are in arithmetical progression. Consequently 1111 a— 6 b—c , a—b a — or — ;— = , whence b a c b ab be b — c c §4] HARMONICAL PROGRESSION. 399 3. Any term of a harmonical progression is obtained by finding the same term of the corresponding arithmetical pro- gression and taking its reciprocal. Ex. Find the eleventh term of the harmonical progression 4 2 4 ... ^) ^J S! • The corresponding arithmetical progression is 4 4 4 ••• ^) ^> ^> > and its eleventh term is ^. Therefore the eleventh term of the given progression is ^. 4. No formula has been derived for the sum of n terms of a harmonical progression. 5. A Harmonical Mean between two numbers is a number, in value between the two, which forms with them a harmonical progression. E.g., f is a harmonical mean between ^ and — f. Let H stand for the harmonical mean between a and b, then — is an arithmetical mean between - and -. Consequently i+i 1 a b „ 2 ab H 2 a + b Ex. Insert a harmonical mean between 2 and 5. 2x2x5 20 We have H = - 2 + 5 7 6. Harmonica! Means between two numbers are numbers, in value between the two, which form with them a harmonical progression. E.g., f , 1, f, f, •!■ are five harmonical means between 3 and |. Ex. Insert four harmonical means between 1 and 10. We have first to insert four arithmetical means between 1 and ^, and obtain 41 32 23 14 3Tr> rjT) TiT) TS- The required harmonical means are therefore 50 50 50 50 ¥T' T^i TS! Tt- 400 ALGEBRA. [Ch. XXVII 7. To insert n harmonioal means between a and 6, we insert n aritli- metical means between - and -, .and take their reciprocals. Tlie n a b aritlimetical means are «l + i (n-l)i + 2l {n-2)- + z\ a b a b a b n+ I n + l n + 1 a + nb 2a+(n-l)b 3a+(n-2)b ... or ■ 1 ^ ^ — » ^ — 5 (m+l)o6 (n + l)a6 (n+l)ab Consequently tlie required harmonical means are (n+ l)ab (n + l)ab (n + l)ab a + nb ^ 2a + in-\)b 3a + (n-2)b' Problems. 8. Pr. 1. The geometrical mean between two numbers is |, and the harmonical mean is -f. What are the numbers ? Let X and y represent the two numbers. Then V(*2/) = h or »2/ = i ! (1) and ^^L=-, OT 5xy = x + y. (2) x + y 5 Solving (1) and (2), we obtain x = l, y = \, and x = \,y = l. Pr. 2. If to each of three numbers in geometrical progression the second number be added, the resulting series will form a harmonical progression. What are the numbers ? Let -, a, ar represent the three numbers. r Then we are to prove that - + a, 2 a, ar+a is a harmonical progression ; that is, that r 1 1 o(l + r) 2a a(l+r) is an arithmetical progression. *• +- 1 We have ajl +r) ajl + r) 1 2 2a §4] HARMONICAL PROGRESSION. 401 That is, -— is an arithmetical mean between ~ and 1 2a ^ a(l + r) /-I I — r- Consequently, - + a, 2 a, ar + a is a, harmonical pro- gression. EXERCISES VIII. Find the last term of each of the following harmonical progressions : 1- 1 + 1 + i + - to 8 terms, t/z. i + i + A + - to 15 terms. 3. 2-2-1 to 11 terms. >1, by Ch. XVII., Art. 8 (ii.). We are then to prove that d can be made less than any assigned num- ber, however small, by increasing n indefinitely. From (1), we have 6» = 1 + d, or & =(1 + dy. By Ch. XVII., Art. 15, 1 + 7id< (1 + d)". Therefore l + nd6''2 and 6 "1 6 "2 determine a positive number which lies between them. This number is defined as the product b'lb's. mi m2 mi+1 m2+l That is, b'^b'^ 6 "^ . In like fnanner it can be shown that the two series 6"! "2 and 6 "> "^ determine a positive number which lies between them. This number is defined as 6^1+^2. »ni m2 flii+l nig+l That is, 6"i "^ < b'^+^' < 6 "' '^ ' ffli 7712 ^4.^ mi+l m2+l mi+1 mj+l But since 6"' 6"= = b"'- "^ and 6 "' & "2 = 6 "^ "2 ' the two numbers fe'i+^z and 6^'6^2 are determined by the same relation, and are therefore equal. That is, 6^'6^2 _ j/i+Jj. In a similar manner the principle can be proved when the exponents, either or both, are negative irrational numbers. (ii.) We have ^ = &^i6-'2 = 6i+(-« = 6A--f2 wherein Ii and I2, either or both, are negative. Principles (III.)-(V.), Ch. XXVI., can be proved in a similar manner. LOGARITHMS. 411 7. In the proofs of the preceding principles, the base was assumed to be greater than 1. Similar reasoning will, however, apply when the base is less than 1 and positive. For, if 6 < 1 and positive, it can be shown that the irrational power is defined by the relation m+l m 6 " < 6' < &" . Solution of the Equation b' = a. 8. (i.) First, let 6 > 1. The powers -, 6-2, 6-1, 6°, 61, 62, ..., increase toward the right beyond any positive number, however great, as the exponents increase without limit. For, by Ch. XVII., Art. 15, (1 +^)»>l + 7id, wherein n and d are positive. We can make 1 + nd greater than any assigned number, however great, by increasing n without limit. But 1 + d represents any number greater than 1. The same series decreases toward the left below any assigned positive number, however small, as the absolute values of the exponents increase without limit, by Ch. XXVII., § 3, Art. 8. For 6-" = (-Y, and - < 1. Then a will either be equal to one of these powers or lie between two consecutive powers. In the former case, x is equal to a positive or a negative integer. In the latter case, x lies between two consecutive num- bers of the series ...-2, -1, 0, +1, +2, .... Let 6' and 6'+' be the two powers between which a is found to lie ; i.e., b''<.a. Then a, which lies between 6* and 6'+i, must either be equal to one of these powers, or lie between two consecutive powers. In the former case, x is equal to a fraction ft + ^, wherein fti is one of the numbers 1, •■■, 9. In the latter case, let 6 i" and 6 i° be the two powers between which a is found to lie ; that is, 6 w 1 and a is positive, there is a definite value of x which satisfies the equation 6»= = a. (ii.) When 6<1 and positive. Then -> 1 and positive. Therefore, by (i.) there is a value of x such that ©•=- But, {-] =6-». (!)•=' Therefore, there is always a value of x such that 6"= = a, when 6 < 1 and positive. 9. The value of a; which satisfies the equation 6'= a is called the logarithm of a to the base b. The Logarithm of a given number a to a given base 6 is, there- fore, the exponent of the power to which the base b must be raised to produce the number a. E.g., since 2' = 8, 3 is the logarithm of 8 to the base 2 ; since 10^ = 100, 2 is the logarithm of 100 to the base 10. 10. The relation b' = a is also written x = logj a, read x is the logarithm of a to the base b. Thus, 2« = 8 and 3 = logs 8, 10^=100 and 2=logiol00, are equivalent ways of expressing one and the same relation. 11. The theory of logarithms is based upon the idea of representing . all positive numbers, in their natural order, as powers of one and the same base. 414 ALGEBRA. [Ch. XXIX Thus, 4, 8, 16, 32, 64, etc., can all be expressed as powers of a common base 2 ; as 4 = 2^, 8 = 2', 16 = 2'', etc. Since, also, all the numbers intermediate between those given above can be expressed as powers of 2, the exponents of these powers are the logarithms of the corresponding numbers. The logarithms of all positive numbers to a given base form what is called a System of Logarithms. The base is then called the base of the system. 12. Neither the number 1, nor any negative number, can be taken as the base of a system of logarithms. For, since any power of 1 is 1, it is evident that any other number than 1 can- not be represented as a power of 1. It is also impossible to represent all positive numbers as real powers of a given negative number. It follows from Art. 8 that any positive number except 1 may be taken as the base of a system of logarithms. 13. The following properties of logarithms evidently follow from the properties of powers. (i.) 77*6 logarithm of 1 to any base is 0. For b" = 1, or log,l=0. (ii.) The logarithm of the base itself is 1. For 6* = b, or logj 6 = 1. The following properties of logarithms hold when the base is greater than 1. (iii.) The logarithm of an infinite is an infinite. For 6" = oo , or logs 00 = CO . (iv.) The logarithm of is a negative infinite. For 6~" = 0, or logs = — 00 . (v.) The logarithm is positive or negative, according as the number is > or <1. For any positive number > 1 lies between 1 and + oo . Therefore its logarithm lies between and + oo , and is positive. Any positive number < 1 lies between and 1 ; therefore its logarithm lies between — oq and 0, and is negative. LOGARITHMS. 416 14. The following relation will be found useful : b^o&b" = a. For let logs a = x, or 6"^ = a. Substituting in the last relation logj a for x, we have ft'oSj a=:a. The truth of this relation is self-evident. It asserts that the logarithm of a, to the base 6, is the exponent logs a ; but the italicized words are just the words for which the expression logj a stands. EXERCISES I. Express the following relations in the language of logarithms : 1. 52 = 25. 2. 26 = 32. 3. 73 = 343. 4. 3' = 2187. 5. 4* = 256. 6. 33 = 27. 7. 58 = 125. 8. 83 = 612. Express the following relations in terms of powers : 9. logs 81 =4. 10. log9 81 = 2. 11. log4 64 = 3. 12. log264 = 6. 13. log8512 = 3. 14. logs 729=6. 15. logil6=-4. 16. logio.001 = -3. Determine the values of the following logarithms : 17. log2 32. 18. log, 128. 19. log2 .5. 20. log2.25. 21. log4 64. 22. logeiS. 23. log2.125. 24. logs. 04. 25. log729 3. 26. logsmS. 27. logjwiT. 28. log2 .03125. 29. log4. 15625. SO. log27iEh- 81. logei.S. 32. log82.125. To the base 16, what numbers have the following logarithms ? 33. 0. 34. J. 35. - 2. 36. f. 37. -J. Solve the following equations : 38. logs a; = 3. 39. log2a; = .5. 40. log2 a; = j^s- 41. log^ 9 = 2. 42. log, 27 = - 3. 43. log, 8 = J. Principles of Logarithms. 15. The logarithm of a product is equal to the sum of the loga- rithms of its factors; or, logs (m xn) = logfi m + logi,n. Let logj m = x and logj nr=y; then b' = m and ft* = n, and therefore, mn = b'b» —b'+i. Translated into the language of logarithms, this result reads logj(mn)=a; + y. 416 ALGEBRA. [Ch. XX : : But X = logjm and y = logs n, and consequently logj(mw) = logs m + logj n, for all positive values of 6. This result may be readily extended to a product of any number of factors. For, logiimnp) = \ogb{mn) + logsp = logs m + logs n + logsp. And, in like manner, for any number of factors. E.g. Given logj 32 = 5, and logj 64 = 6 ; what is tlie logar rithm of 2048 to the base 2 ? Since 2048 = 32 • 64, we have loga 2048 = log2 32 + log2 64 = 6 + 6 = 11. 16. The logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor; or, logb {m-^n) = logs m — logs n. Let logs m = x and logs n = y; then b" = in and by = n, and therefore m -i- n = b" -i- by = b^-n. In the language of logarithms the last equation is logs (m^n)=x — y = logs m — log jre, for all positive values of 6. E.g. Given logj 3 = 1 and logj 2187 = 7, what is the loga- rithm of 729 to the base 3 ? Since 729 = i^, we have logs 729 = logs 2187 - logs 3 = 7 - 1 = 6. 17. Both m and « may be products, or the quotient of two numbers. E.g., log,„ 1^ = logio (4 X 5) - logio (9 x 8) = logio 4 + logio 5 - logio 9 - logio 8. 18. The logarithm of the reciprocal of any number is the opposite of the logarithm of the number. LOGARITHMS. 417 For, logjl = log6 1-logjn n = - logs n, since logj 1=0. E.g., logs 4 = 2, and logj i = - 2. 19. Tfte logarithm of any power, integral or fractional, of a number is equal to the logarithm of the number multiplied by the exponent of the power; or, log(/n'') = yBlogm. Let logj m = x, then 6* = m. Raising both sides of the last equation to the pth power, we have ftJ" = m", or logi (m") =px=p logt m. E.g., if logs 25 = 2, what is logs (25)" ? We have log, (25)3 = 3 log^ 25 = 3 x 2 = 6. 20. When the exponent is a positive fraction whose numera- tor is 1, this principle may be conveniently stated thus : The logarithm of a root of a number is the logarithm of the number divided by the index of the root. For, logj (m?) = i log m = l2iL™. E.g., If log, 2401 = 4, what is log, V2401 ? We have log, V2401 = |log,2401 = I • 4 = 2. 21. It can readily be seen from the preceding principles and examples that if the logarithms of all numbers to any one base are given, certain numerical calculations can be greatly simplified by replacing the operations of multiplication and division by those of addition and subtraction, and the opera- tions of involution and evolution by those of multiplication and division. 418 ALGEBRA. [Ch. XXIX EXERCISES II. Express the following logarithms in terms of log a, log h, log c, and logd: ^•'°^f- '-'^^ ^■'<^- '^•'°^0- 8. logoV^V^Vc. 6. log;^^- 7. log. "-'^^ Express the following sums of logarithms as logarithms of products and quotients. 8. log a + log 6 - log c. 9. log a — (log b + log c). 10. 31oga-^log(6 + c). 11. |log(l-j;)+|log(l+x). 12. 21ogf+31og-- 13. 21oga-|log6 + ilogc. o a Given logio2 = .30103, logio3=.47712, logio5=.69897, logio? =.84510, find the values of the following logarithms, to the base 10 : 14. log 25. 15. log 6. 16. logs.- 17. log 9. 18. log 12. 19. log 36. 20. log 108. 21. log 4^. 32. log2f 33. logSf. 34. logSf. 28. log 360. 36. log 3072. 37. log 3500. 38. log 5880. 39. logV72. 30. logVlSO. 31. logVl716. - -^- 33. log-£^9fxVl05 ^ ^72 X ^8f 34. log iaill. (llf)^ Find the values of the following fractions : 35. l°g^2. log 9 36. ^°S^2^- log 125 37 log 243 log 3 38. Prove that the ratio of the logarithms of two numbers is the same for all bases. Systems ol Logarithma. 22. The \mo most important systems of logaritlims are : (i.) The system whose base is 10. This system was intro- duced, in 1615, by the Englishman, Henry Briggs. Logarithms to the base 10 are called Common, or Briggs's Logarithms. LOGARITHMS, 419 (ii.) The system whose base is the sum of the following infinite series, 1 1 1-2 1.2.3 1-2-3.4 + ' The value of this sum, which to seven places of decimals is 2.7182818, is denoted by the letter e. Logarithms to the base e are called Natural Logarithms; sometimes also Napierian Logarithms, in honor of the inventor of logarithms, the Scotch Baron Napier, a contemporary of Briggs. Napier himself did not, however, introduce this sys- tem of logarithms. These two systems are the only ones which have been gen- erally adopted ; the common system is used in practical calcu- lations, the natural system in theoretical investigations. The reason that in all practical calculations the common system of logarithms is superior to other systems is because its base 10 is also the base of our decimal system of numeration. The logarithms of most numbers are irrational, and thus approximate values are used. Properties of Common Logarithms. ^c/>>v 23. In the following articles the subscript denoting the base 10 will be omitted. We now have •10° = (a) (6) 1, or logl =0, 10'= 10, or log 10 =1 10=^= 100, or log 100 =2 [10^=1000, or log 1000 = 3 rlO-i = ■1, or log .1 — -1 10-^ = .01, or log 01 = -2 10-' = .001, or log 001 = -3 iio-^= .0001, or log .0001 = -4 Evidently the logarithms of all positive numbers, except positive and negative integral powers of 10, consist of an inte- 420 ALGEBRA. [Ch. XXIX gral and a decimal part. Thus, since 10' < 85 < 10^, we have 1 < log 85 < 2, or log 85 = 1 + a decimal. 24. The integral part of a logarithm is called its Character- istic. The decimal part of a logarithm is called its Mantissa. 25. Since a number having one digit in its integral part, as 7.3, lies between 10° and 10', it follows from table (a) that its logarithm lies between and 1, i.e., is 0-)-a decimal. Since any number having two digits in its integral part, as 76.4, lies between 10' and 10^, its logarithm lies between 1 and 2, that is, is 1 + a decimal. In general, since any number having n digits in its integral part lies between 10""' and 10", its logarithm lies between m — 1 and n, i.e., is re — 1 -|- a decimal. We there- fore have : (i.) The characteristic of the logarithm of a number greater than unity is positive, and is one less than the number of digits in its integral part. E.g., log 2756.3 = 3 -f- a decimal. Since a number less than 1 having no cipher immediately following the decimal point lies between 10" and 10~', it follows from table (&), that its logarithm lies between and — 1, i.e., is — 1 + a positive decimal. Since a number less than 1 having one cipher immediately following the decimal point lies between 10~' and 10~^, its logarithm lies between — 1 and —2, i.e., is — 2 + a positive decimal. In general, since a number less than 1 having n ciphers immediately following the deci- mal point lies between 10~" and 10~'"+'', its logarithm lies between — n and — (n -|- 1), i.e., is — {n + l) + a positive deci- mal. We therefore have : (ii.) Tlie characteristic of the logarithm of a number less than 1 is negative, and is numerically one greater than the number of ciphers immediately following the decimal point. E.g., log .00035 = - 4 + a decimal fraction. It follows conversely from (i.) and (ii.) : LOGARITHMS. . 421 (iii.) If the characteristic of a logarithm he +n, there are n+1 digits in the integral part of the corresponding number. (iv.) If the characteristic of a logarithm be —n, there are n—1 ciphers immediately following the decimal point of the correspond- ing number. 26. It has been found that 638 = lO^'^ws to five decimal places, or log 538 = 2.73078. We also have log .0538 = log tMI^ = log 538 - log 10000 = 2.73078 - 4 = .73078-2; log 53800 = log (538 x 100) = log 538 + log 100 = 2.73078 + 2 = 4.73078. These examples illustrate the following principle: If two numbers differ only in the position of their decimal points, their logarithms have different characteristics but the xame positive mantissa. If n denote the number of places through which the decimal point has buen moved in a given number a, we have log (a 10") = log a + re log 10 = log a + n, and log (a -h 10») = log a — m log 10 = log a — re, since moving the decimal point a given number of places to the right or left is equivalent to multiplying or dividing by a power of 10. 27. The characteristic and the mantissa of a number less than 1 may be connected by the decimal point, if the sign (— ) be written over the characteristic to indicate that the character- istic only is negative, and not the entire number. Thus instead of log .00709 = .85065 - 3 = - 3 + .85065, we may write 3.85065 ; this must be distinguished from the expres- sion — 3.85065, in which the integer and the decimal are both negative. Similarly, log .082=2.91381, while log 820=2.91381. Five-Place Table of Logarithms. 28. The logarithms, to the base 10, of a set of consecutive integers have been computed. In tabulating these logarithms, compactness is important. 422 ALGEBRA. [Ch. XXIX Por this reason, all unnecessary detail is omitted. Since the characteristic of the logarithm of any number can, as we have seen, be determined by inspection, it is unnecessary to write it with the mantissa in the table. Consequently, only the man- tissas, witliout the decimal points, are there given. Neither is it necessary to give the logarithms of decimal fractions, since their mantissas are the same as the mantissas of the numbers obtained by omitting the decimal point. The logarithms may be carried to any number of decimal places, and the extent to which they are carried depends upon the degree of accuracy required in their use. 29. The accompanying five-place table gives the mantissas of the logarithms of all consecutive integers from 1 to 9999 inclusive. In this table the first three figures of each number are given in the column headed N, and the fourth figure in the horizontal line over the table. The first figure, which is the same for all numbers in a given column, is printed in every tenth number only. The columns headed 0, 1, 2, 3, etc., contain the mantissas, with decimal points omitted. In the column headed 0, when the first two figures are not printed, they are to be taken from the last mantissa above which is printed in full. In the columns headed 1, 2, 3, etc., the last three figures only are printed ; the first two are to be taken from the column headed in the same horizontal line. When a star is prefixed to the last three figures of a man- tissa, the first two figures are to be taken from the line below. To Find the Logarithm of a Given Number. 30. When the Number consists of Four or Fewer Figures. — Take the mantissa that is in the horizontal line with the first three figures and in the column under the fourth figure of the given number Determine the characteristic by Art. 26. LOGARITHMS. 423 E.g., log 2583 = 3.41212, log 46.32 = 1.66677. In writing logarithms with negative characteristics it is customary to modify the characteristics so that 10 is uniformly subtracted from the logarithms. Thus, 2.45926 = .45926 - 2 = 8.45926 - 10 ; 4.37062 = .37062 - 4 = 6.37062 - 10. That is, we add 10 to the negative characteristic, and write — 10 after the logarithm. log .5757 = 9.76020 - 10, log .02768 = 8.44217 - 10. Observe that the first two figures of the mantissa of log .5757 are taken from the line below, in accordance with the directions in Art. 29. If the given number consists of fewer than four figures, annex ciphers until it has four figures, in taking the mantissa from the table. E.g., mantissa of log 78 = mantissa of log 7800 = .89209, and log 78 = 1.89209. In like manner, log 583 = 2.76567, log .02 = 8.30103 - 10. 31. When the Number consists of more than Four Significant Figures. — The method used is called interpolation, and depends upon the following property of logarithms : The difference between two logarithms is very nearly propor- tional to the difference between the corresponding numbers when this difference is smaU. The error made by assuming that these differences are exactly proportional will be so small that it may be neglected. Ex. 1. Find log 27845. Omitting, for the moment, the decimal points from the mantissas, we have mantissa of log 27850 = 44483, mantissa of log 27840 = 44467, difference of mantissas = 16. ^ 424 ALGEBRA. [Ce. XXIX Let a; stand for the difference between the mantissas of log 27845 and log 27840 ; that is, for the correction to be added to the smaller mantissa to give the required mantissa. Then, by the above property, X ^ 27845-27840 ^ 5 ^ g 16 27850-27840 10 " ' Whence a; = .5 x 16 = 8. Therefore, mantissa of log 27845 = 44467 + 8 = 44475, and log 27845 = 4.44475. Observe that, by Art. 26, the mantissa of log 27850 is the same as the mantissa of log 2785. In subsequent work such ciphers will be omitted. The method can now be stated more concisely for practical work: Subtract the mantissa corresponding to the first four figures of the given number from the next mantissa in the table ; multiply this difference by the remaining figure or figures of the given number, treated as a decimal ; add the product to the first (and smaller) mantissa. Prefix finally the proper characteristic. In thus finding the mantissa, a decimal point in the given number is ignored, in accordance with Art. 26. The difference between two consecutive mantissas in the table is called the Tabular Difference. Ex. 2. Find log 78.1283. We have mantissa of log 7813 = 89282, mantissa ^ + 243 = 36 . 3». 22. 3"'>g» = 9. 23. 6iog2» - 625. 24. le'oBS' = 32i''8»^. 25. 5" = 10. 26. 16« = 45. 27. 11"= = 310. 28. 25' := 10. 29. T = 300. 30. 3.594« = 359600. 81. V9.8926 = 1.29. 32. 5x = 73.14. 33. a;V2 = .J/3. 34. 6'+^ = 1000. 38. 7»+i = 5. 36. l.SS'^-s = 9.847. 37. 5"^+! = ll"-!. 38. 31+7 _ 71+3. 39. 31«+3 = 25"^+*. 40. 35X+2 = 40X-1. 41. r 3» • sv = 75, 42. 2» . 7* = 98. 14» ■ 8!' = 896, 6''. when the n things are all different. Evidently „Pi = n. (1) From each permutation of n things one at a time we obtain, by annex- ing to it each of the m — 1 remaining things in turn, m — 1 permutations two at a time. Therefore nPi =(n- l)„Pi = n(ra - 1). _ (2) Again, from each permutation of n things two at a time we obtain, by annexing to it each of the m — 2 remaining things in turn, n — 2 permuta- tions three at a time. Therefore „P8 = («-2)„P2 =m(«-l)(w -2). (3) In like manner, „P4 = (M-3)„Ps=n(m-l)(ra-2)(n-3). (4) The method is evidently general. The number subtracted from n in the last factor in (l)-(4) is one less than the number of things taken at a time. Therefore, „P, = n(»i-l)(>t-2)..- [n-(r-l)] = re(n-l)(m-2).-(i>i-r + l). § 2] PERMUTATIONS. 443 4. Observe that the number of factors in the formula for „Pr is equal to the number of things taken at a time. E.g., 8^5 = 8x7x6x6x4 = 6720. 5. If all the things are taken at a time, i.e., if r = n, we have „P„ = n(w-l)(n-2)-(n-n+l) = w(w-l)(n-2>-3x2xl. E.g., 5P5 = 5x4x3x2x1= 120. 6. The continued product »i(ji-l)(n-2)-3x2xl is called Factorial-/?, and is denoted by the symbol [w or w ! Therefore the formula of the preceding article may be written E.g., 7P7 = [7, or 7 !, = 7 x 6 x 5 x 4 x 3 x 2 x 1. 7. In many applications the things considered are not all different. We will now derive a formula for the number of permutations of n things, taken all at a time, when some of them are alike. Let p of the n things be alike, and suppose the permutations m at a time to be formed. In any one of these permutations, let the p like things be replaced by p unlike things, different from all the rest. Then by changing the order of these p ne w things only, we can form |p permutations from the one permutation. In like manner, \p permutations can be formed from each of the given permutations. Therefore „P„ (all different) = |p «P„ {p alike), or aPn {p alike) = - — = i= ■ In like manner, it can be proved that nPnXp alike, q alike, •••) = - [^x|9 X- 18 E.g., »Pt (3 alike) = — = 6720. ALGEBRA. [Ch. XXX EXERCISES I. 3. loJ'io- 4. 12P9. 5. soft. 8. „+lPn-l- 9. n+kPk- 10. m+nPm—n- 444 Find the values of I. 18P4. 2. isPs. 6. n+lft- 7. 2n+lJ'6- Find the value of n, II. "When„P4 = 3„P3. 12. „P6 = 20 „P4. 13. n+sPi = 15 „P3. 14. ■When„+iP4=30„_iP2. 15. „+4P8=8„+3P2. 16. 2»+iP4=140„P8. Find the value of k, 17. WhenioP*+6=3ioP4+5. 18. 7P*+i=12 7P4-i. 19. i2Pi=20i2Pj_2. 20. How many numbers of 4 figures can be formed with 1, 2, 3, 4, 5, 6, 7 ? 21. How many numbers of 4 figures can be formed with 0, 1, 2, 3, 4, 5, 6, 7 ? 22. How many even numbers of 4 figures can be formed with 4,5,3,2? 23. In how many ways can 6 pupils be seated in 10 seats ? 24. In how many ways can 4 tickets be placed in 6 different boxes, so that no box shall contain more than 1 ticket ? 25. How many numbers of 5 figures can be formed with 1, 2, 3, 4, 5, 6, 7, 8, 9, if the figure 7 be in the middle of each number ? 26. If the permutations of 1, 2, 3, 4, taken all at a time, be arranged in a column, how many times is each figure found in each column ? 27. How many permutations can be formed with the letters in the word Philippine? 28. How many permutations can be formed with the letters in the word lloilo 9 29. In how many ways can 7 men be seated at a round table ? 30. In how many ways can a bracelet be made by stringing together 7 pearls of different shades ? §3. COMBINATIONS. 1. The formula for the number of combinations of n things, r at a time, which is denoted by „C^, is most readily obtained by deriving a relation between „P, and „(7,. The method will be illustrated by a particular example. § 3] COMBINATIONS. 445 The combinations of the four letters a, b, c, d, taken three at a time, evidently are : abc, abd, acd, bed. From the combina- tion abc we obtain, by changing the order of the letters in all possible ways, [3 permutations. In like manner, each of the combinations gives [3 permutations. Therefore ^3 = 13.C3, or ,C3 = i§ = i><^. In general, „C, = n(n-l)(n-2)...(n-r + l) ^ /n\ wherein the n things are all different. For from each combination that contains r things can he formed \r permutations, hy changing the order of the things in all possible ways. Therefore '— \r_ [r ^8x7x6 Kft E.g., gCa = 13 = 56. 2. The formula for „C, can be put in a more convenient form for purposes of theory. We have ^ n(n-l) — (n - r + 1) X (n -r)(n - r - 1) — 3 x 2 x 1 "^'~ \rx{n-r)(n-r-l)— 3x2x1 \r \n — r 3. We have C "^ : " ' \r\n-r and c L- . [^ "^''-' in-r\n-(n- -r) [n — r\r therefore, fir — nCn-V 446 ALGEBRA. [Cii. XXX That is, the number of combinations of n dissimilar things r at a time is equal to the number of combinations of the n things n — r at a time. This relation is also evident from the definition of a com- bination. For, every time that r things are taken from the n things to form a combination, there is left a combination of n — r things. E.g., looC'os = 100^2 = -Tj K~ ~ 4950. This relation is thus useful in computing the number of com- binations when the number of things taken at a time is large. 4. The Greatest Value of nCr. — We have p _ n(n - l)(n -2) •■■ (n - r + 2)(n - r + \) 1-2-3.-. (r-l)r _ n(n ^ l)(n - 2) •■■ {n - r + 2) n- r+ I 1 .2-3..-(r- 1) r = „a-. x?l:^^=„a_,(^-l). (1) Also, „C.+i = n{n-l){n-2).:(n-r + l)in-r) ' 1-2-3. ..»•(?• + 1) _ K( ra-l)(m — 2)--.()2 — r-f-1) n - r " l-2-3"-r r + l r + 1 Whence „a = „C.+i x ^^^- (2) n - r From (1), „Cr > nCr-i, when ^^ii - 1 > 1, or j- < 2ii. (3) r 2 That is, the number of combinations of n things, taken any number less than at a time, is greater than the number of combmations taken one less at a time, and therefore greater than the number of combinations taken any number less at a time. From (2), „C,>„a+i, when ''-±^>\, or )'>^-=^- (4) n — r 2 That is, the number of combinations of n things, taken any number n — 1 greater than at a time, is greater than the number of combinations taken one more at a time, and therefore greater than the number of § 3] COMBINATIONS. 447 combinations taken any number more at a time. Consequently „Cr is greatest when r, an Integer, lies between " ~ and " "*" . (1) n Even. Let re = 2 m. Then r is an integer in value between ^'"~ \ = m - i, and 2nL±i, = m + i. That is, r = m, = -. There- fore, when re is even, the greatest number of combinations is „C„. 2 (ii) re OM. Let n = 2 m + 1. Then r should have an integral value 2 7n 2 wi -I- 2 between —^, =to, and — ^ , =?» + !. This is evidently impossible, since m and m -)- 1 are consecutive integers. re 4- 1 But, when r = m, < m + 1, = T , then by (3), 2 n — 1 and, when r = m + 1, > i«, = , then by (4), 2m+l C/m+l > 2m+l Om+2' Also, by § 3, Art. 3, Im+lCm = 2m+lC'2m+l-,n, = 2m+lCm+l. Consequently, when re is odd, the greatest number of combinations is 2m+lC'm, = 2m+l^m+l, Or „CV-1, =«C/n-fI. 2 2 Ex. 1. When re = 4, the greatest number of combinations is 4C2. We have tCi = 4, 4C2 = 6, 4C3 = 4, 4O4 = 1. Ex. 2. When re = 5, the greatest number of combinations is 5C2, = s C3. We have 5C1 = 5, 6C2 = 10, 5C3 = 10, 5O4 = 5, 5C5 = 1. EXERCISES II. Find the values of ■ 1. iiCi. 2. 16C7. 3. 2SC20. 4. gsCgs- S. „C„_5. Find the value of re, 6. When2„C6 = 9„-206. 7. When S^Os = 10„_2C2. 8. When4„+iC4 = 15„_i03. 9. When „+iP4 = 112 „_iC2. 10. When„+iP4 = 84„_iC8. 11. When „P2 = 24„C„_i. Find the value of k : 12. 8Ps = 24 8<7s. 13. 6Ps+i = 486Ci. 14. loPj = 144ioCj-i. 15. Prove that „C, + „Cr-i = „+iG,. 16. In how many ways can a committee of 4 men be appointed from 25 men ? 448 ALGEBRA. [Ch. XXX 17. In how many ways can 3 books be selected from 15 books ? 18. In a plane are 20 points, no 3 of which are in the same straight line. How many triangles can be formed with 3 of the points as vertices ? How many quadrilaterals, with 4 of the points as vertices ? How many hexagons, with 6 of the points as vertices ? 19. How many triangles are formed by 7 straight lines in a plane, if no 2 are parallel and no 3 intersect in a common point? How many by 10 lines? How many by n lines? Find the values of r and „C„ when „Cr is greatest, in 20. 7 a. 21. sC,.. 22. 11 a. 23. 14^. 24. i,0,. § 4. TWO IMPORTANT PRINCIPLES. 1. The following example illustrates an important principle. Pr. Between two cities A and B there are five railroad lines. In how many ways can a man go from A to B and return by a different road ? He can go to B in either of five ways. With each of these five ways he has a choice of four ways of returning. Hence he can make the round trip in 6 x 4, = 20, ways. Evidently, if he were not required to return by a different road, he could make the trip in 5 x 6, = 25, ways. The general principle is : If one thing can he done in a ways, and another thing can he done in b ways, and the doing of the first thing does not interfere with the doing of the second, the two things can be done in ab ways. The truth of the principle is evident. 2. The following relation will be useful in subsequent work : ni+nCr = mCr + mCi'-\nC\ + mCr-2nCi+ ••• + mCinCr-2-\- mC\nCr-\-\- nCr, (1) in which m > or =»•,»> or = r. The number of combinations of the m + n things »• at a time is evidently the sum of : The number of combinations of m things taken »■ at a time, or ,„(7,. The number of combinations of m things taken r — 1 at a time, multi- plied by the number of combinations of n things taken one at a time, or And so on. §§ 4-5] PROBLEMS. 449 This relation may tie written r:")=(T)+(.^)B-HT)UOU:)- ^^^ 3. Tlie relation of the preceding article requires m, n, and r to be integers. Evidently, however, the second member of (2) could be made identical with the first member by ordinary reduction. We, there- fore, conclude that this relation holds for all rational values of m and n, provided r is a positive integer. §5. PROBLEMS. 1. Pr. 1. In how many ways can a committee of 3 Repub- licans and 4 Democrats be appointed from 18 Republicans and 12 Democrats ? The 3 Republicans can be chosen in ^^Cg, = 816, ways, and the 4 Democrats in 12C4, = 495, ways. Since any 3 Republicans can be associated with any 4 Democrats to form the committee, the required number of ways is 816 x 495, = 403,920. Pr. 2, A box contains 20 balls numbered 1 to 20. In how many ways can 7 balls be selected, if 1 be included, and 2, 3 be excluded? We first set aside 1 to be included, and 2, 3 to be excluded, and from the remaining 17 balls select 6 balls. Then 1 may be combined with each of the latter in one way, giving a com- bination of 7 balls. Therefore the problem is equivalent to determining the number of combinations of 17 things, 6 at a time. „ ri 17. 16 -15 -14 -IS- 12 .„q7« Hence .jd = 6.5.4.3.2.1 ' ^ ' is the required number of ways. EXERCISES III. 1. A man has 3 coats, 4 vests, and 5 pairs of trousers. In how many ways can he dress ? 2. In how many ways can 4 white balls, 3 black balls, and 2 red balls be selected from 8 white balls, 7 black balls, and 5 red balls ? 3. In how many ways can permutations be formed, with 10 conso- nants and 4 vowels, each one to contain 5 consonants and 2 vowels ? 450 ALGEBRA. [Ch. XXX 4. In how many ways can 4 hearts, 3 diamonds, 2 clubs, and 1 spade be drawn from a pack containing 13 cards of each kind ? 5. In how many ways can 7 pears, 5 apples, and 4 oranges be given to 16 children, each child to receive a piece of fruit ? 6. How many numbers of 7 figures can be formed with 1, 2, 3, 4, 6, 6, 7, if the figures 4, 5, 6 be kept together? t. In a company are 10 men, 12 women, and 15 children. In how many ways can a party be formed, consisting of 4 men, 6 women, and 10 children ? 8. How many permutations of 10 letters can be formed from 5 con- sonants and 5 vowels, if no two consonants be adjacent ? 9. How many permutations of 9 letters can be formed from 5 conso- nants and 4 vowels, if each vowel be placed between two consonants ? 10. In how many ways can a committee of 5 men be appointed from 20 men, if there be no restriction in the choice ? In how many ways, if a particular man be always chosen ? In how many ways, if a particular man be always excluded ? 11. In school are 96 pupils. In how many ways can a teacher divide them into sections of 12 ? 13. In how many ways can 4 ladies and 4 gentlemen be seated at a Square table, so that a gentleman and a lady shall be seated at each side ? 13. How many throws can be made with 2 dice, if such throws as 1, 2 and 2, 1 be regarded as the same ? How many with 3 dice ? 14. In how many ways can the sum 10 be thrown with 2 dice ? With 3 dice ? 15. In how many ways can 52 cards be divided into 4 sets of 13 cards each ? 16. A box contains 15 balls, numbered 1 to 15. In how many ways can 5 balls be selected, if 1, 2, 3 be included ? In how many ways, if 1, 2 be included, and 3 excluded ? In how many ways, if any two of the numbers 1, 2, 3 be included, the other excluded ? 17. In how many ways can 10 different coins be arranged in a row, if the faces of each coin are distinct ? In how many ways can they be arranged in a circle ? 18. In how many ways can a number of 6 figures be formed with 1, 1, 1, 2, 2, 3, the first and last figure of each number to be an even digit ? 19. A man can go to his ofBce in 3 ways. In how many ways can he arrange to go to his ofBoe for 6 days ? 20. In how many ways can 7 gentlemen and 10 ladies arrange a game of lawn tennis, each side to consist of 1 lady and 1 gentleman ? CHAPTER XXXI. PROBABILITY. 1. In this chapter we shall consider the likelihood that an event, about whose happening there is uncertainty, will happen, or fail to happen. Thus, if a coin be tossed once, it may fall heads up, but it is not certain so to fall. It may fall tails up. One way of falling is as likely to happen as the other. Now, ^ of the whole number of ways in which a coin can fall is heads up. It seems natural, therefore, to take i as the mathematical expression of the likelihood, or probability, that the coin will fall heads up. Then, ^ is also the probability that the coin will fall tails up. Again, let 4 white balls and 6 red balls be placed in a box, and one ball be drawn at random. If the balls cannot be distinguished by the sense of touch, one ball is as likely to be drawn as any other. Now, one ball can be drawn in 10 dif- ferent cases, in 4 of which a white ball can be drawn. That is, the number of cases in which a white ball can be drawn is ^, = |., of the whole number of cases. We therefore take f as the mathematical expression of the probability of drawing at random a white ball. The probability of not drawing a white ball, which is the same as the probability of drawing a red ball, is evidently f. If data relating to the number of times an event has hap- pened in a large number of cases be collected, these data will indicate quite surely how often the same event will happen in the same number of cases under similar conditions. Thus, from tables used by life insurance companies, we find that of 9'5,966 healthy persons of sixteen, 96,293 have lived to 451 452 ALGEBRA. [Ch. XXXI be seventeen. We therefore take |||g | as the probability that a person of sixteen, in good health, will live to be seven- teen. 2. The considerations of the preceding article naturally lead to the following definitions : The Favorable Cases are those in which an event can happen, or has happened in an extended number of cases. The Unfavorable Cases are those in which the event can fail to happen, or has failed to happen in an extended number of cases. The Probability that an event will happen is the ratio of the number of favorable cases to the whole number of cases. Evidently the probability that an event will not happen is the ratio of the number of unfavorable cases to the whole number of cases. If a be the number of cases in which an event can happen, and b be the number of cases in which it can fail to happen, and each case be equally likely to happen, we have : is the probability that the event will happen; a + b b a + b is the probability that the event will not happen. The Odds in favor of an event is defined as the ratio of the number of favorable cases to the number of unfavorable cases. That is, - are the odds in favor of the event; b in like manner, - are the odds against the event. 3. Since an event is certain to happen or fail to happen, the number of ways favorable to its happening-or-failing is a-\-b. Therefore the probability of the event's happening-or-failing, that is, certainty, is a + h _ a , b a+b a+b a+b ■ 1. PROBABILITY. 453 4. If P be the probability that an event will happen, it follows from the preceding article that 1 — P is the probability that the event will not happen. Ex. What is the probability of throwing at least 4 in a single throw with two dice ? The number of cases favorable to throwing at least 4 is the number of cases in which 4, 6, 6, •••, 12 can be thrown. The number of unfavorable cases is the number of cases in which 2 and 3 can be thrown. The required probability can be obtained most readily by first finding the probability of the event's not happening. The sum 2 can be thrown in one case, 1, 1. The sum 3 can be thrown in two cases, 1, 2 and 2, 1. The two dice can be thrown in 6 x 6, = 36, different cases, counting 4, 5 and 5, 4, say, as different throws. Therefore the probability of not throwing a sum at least 4 is -^, = yV) 3^ •••• Hence, AP, + P,P,+ P,P, + P,Pi+:. = :^ + ^ + l+^+.... But, by Ch. XXVII., § 3, Art. 7, the sum of the series on the right approaches 1 as a limit. That is, limit of {APi + P,P,+ P^Pt + PjP, Jr-) = AB. Again, 1 + - becomes more and more nearly equal to 1, as ** / 1\ 1 n increases indefinitely, and ( 1 -f - ) — 1, = -, will become and V nj n remain less than any assigned positive number, however small. 2. It follows from the definition of a limit that the variable may be always greater, or always less, or sometimes greater and sometimes less than its limit. Thus, by Ch. XXVII., § 3, Art. 8, we have limit(l-^-i-^-...)=0, (1) limit (l+| + J + J+...) = 2, (2) limit (1-i + i- J + ...) = !. (3) Andin(l), S, = l, S,= \, S, = \, S,= I,...; (4) in (2), S,= l, S,= l ^3=1, S,= ^^, ■■■; (5) in (3), S, = l, S, = i, Ss = h S,^ f, '••. (6) 3. The symbol, =, read approaches as a limit, or simply approaches, is placed between a variable and its limit. The word limit may be abbreviated to lim. Thus, 2: i 1 (1 — ^) = 0, read the limit of 1 —x,asx approaches 1, is 0. § 2] LIMITS. 459 Infinites and Infinitesimals. 4. The fractions given in Ch. Ill, § 4, Art. 14 are particular values of the fraction -, in which the denominator x is as- X sumed to be a variable. It is evident that the value of this fraction can be made greater than any assigned number, how- ever great, by taking its denominator sufficiently small. A variable which can become and remain numerically greater than any assigned positive number, however great, is called an Infinite Number, or simply an Infinite. 5. The fractions given in Ch. Ill, § 4, Art. 19, are also particular values of the fraction -, in which, as above, the X denominator x is assumed to be a variable. It is evident that the value of this fraction can be made less than any assigned number, however small, by taking the denominator sufficiently great. A variable which can become and remain numerically less than any assigned positive number, however small, is called an Infinitesimal. No symbol by which to denote an infinitesimal variable has been generally adopted. It follows from the definition that the limit of an infinitesi- mal is 0. 6. It is important to keep in mind that both infinites and infinitesimals are variables. Their definitions imply that fixed values cannot be assigned to them. An infinitesimal should therefore not be confused with 0, which is the constant difference between any two equal numbers. 7. The statement, x becomes infinite, or x increases numeri- cally beyond any assigned positive number, however great, is fre- quently abbreviated by the expression, a; = oo. 8. The conclusions reached in Ch. Ill, § 4, Arts. 14 and 19, can now be restated thus : 460 ALGEBRA. [Ch. XXXII (i.) If the numerator of a fraction remain finite and not 0, and the denominator approach zero, the value of the fraction will become infinite; or stated symbolically, n . . „ - = 00, as X = 0, wherein n is finite and not 0. (ii.) If the numerator of a fraction remain finite and not 0, and the denominator become infinite, the value of the fraction will approach ; or stated symbolically, - = 0, as X = , X wherein n is finite and not 0. Observe that these principles hold not only when w is a constant, not 0, but also when w is a variable, provided it does not become infinite. 9. 27*6 difference between a variable and its limit is evidently an infinitesimal ; that is, if lim X = a, then lim {x — a) = 0. Consequently, if lim x = a, we have X ~ a = x', or a; = a + a;', wherein a/ is a variable whose limit is 0. Conversely, if a; = a + a;', and a;' be a variable whose limit is 0, then lim x=a. 10. If the limit of a variable be 0, the limit of the product of the variable and any finite number is 0. That is, if lim X = 0, and a be any finite number, lim ax = 0. Let k be any number, however small. Then x can be made less numerically than _, and, therefore, ax less than h. Hence, lim ax = 0. a Fundamental Principles of Iiixuits. 11. (i.) If two variables be always equal, and one of them approach a limit, the other approaches the same limit. That is, if X =y, and x = a, then y = a. § 2] LIMITS. 461 (ii.) If two variables he always equal as they approach their limits, their limits are equal. That is, if lim X = a, lim/ = b, and x =/, then a = b. (iii.) TJie limit of the algebraical sum of a finite number of variables is the sum of their limits. That is, if limJif = a, lim/ = b, ••■, then lim {x +y + ••■) = a + b-\ . (iv.) The limit of the product of a' finite number of variables is the product of their limits, if none of the limits be oo. That is, if limJr=a, lim/ = A, •••, then lim(jrf •••) = a6 ■■•. (v.) The limit of the quotient of two variables is the quotient of their limits, if the limit of the divisor be not 0. That is, if iimx = a, lim/=6, then limf — )=^, when limy^O. The proofs follow : (1.) We have x = a + x', wherein, by Art. 9, x is a variable whose limit is 0. Then, since y = x always, we have y = a + x'. Hence lim y = a. (ii.) This principle follows directly from (i.). (iii.) We have x = a + x', y = b + y', ••■, wherein, by Art. 9, x', y', ••• are variables whose limits are 0. Then x + y + - =(,a + b + ■■■) + (,x' + y' + ■■■). Let k be any assigned number, however small. Then each of the z. variables x', y', •■■ can be made less than-, wherein n is the number of n variables. Therefore, x' + y' + ■■■ can be made less than k. Conse- quently lim (x + y ■■■')= a + 6 + ■■■. (iv.) We have x = a + x', y =:h -\-y', •••, wherein x', y', ■■■ are vari- ables whose limits are 0, and a, b, ••• are finite. Then xy = ab + bx' + ay' + x'y'. Therefore, by (iii.), lim xy = lim ab + lim bx' + lim ay' + lim x'y' = ab, since lim bx' =0, ■••, by Art. 10. In like manner the principle can be proved for any finite number of factors. (v.) Let- = q OT X = yq. Then, by (iv.), lim x — lim y lim q. Therefore, lim g = '4^, or lim? - ""^ * lim y y lim y 462 ALGEBRA. [Ch. XXXII Indeterminate Fractions. 12. It follows from the definition of a fraction that t: is a number which multiplied by gives 0. But any finite number multiplied by gives 0, or w = 0. Consequently - may denote any number whatever. Eor this reason, such a fraction is called an Indeterminate Fraction. a^ — 9 13. The fraction becomes - when a; = 3, and has no a; — 3 definite value. But as long as xi=3, however little it may differ from 3, we may perform the indicated division. We therefore have x-S a; + 3, when x^3. Now since the limit of the fraction depends upon values of X which differ from 3, however little, we have lim a;^ — 9 lim / , r,\ a Although the given fraction is indeterminate, it is clearly desirable that it shall have a definite value. We therefore 3,2 _ 9 assign to the value 6, when x = 3. X — 3 That is, we define an indeterminate fraction to be the limit of the fraction as the variable approaches that value which renders it indeterminate. In this way we may obtain a definite value when the fraction involves but one variable. 14. The fraction ^ is a number which multiplied by oo gives CO. But any finite number multiplied by oo gives oo. There- fore ^ is also an indeterminate fraction. 15. The fraction —^ — becomes ^, as n = oo. Dividing nu- n + 1 merator and denominator by n, we have •2] LIMITS. 463 Since 1 - 1 w-1 n w + 1 1+i n n as n. = cc, lim n-1 lim 1- 1 n , = GO n + 1 n = 00 1 + 1 n 1. BXBBCISBS II. Find the limiting values of the following fractions : 1 ^^-6^ + 5 ^i^g„ ^ ^5 2 5^^3^+2 ^^^^ ^ ^ 2. a;2 - 8 a; + 15 a;" + x - 6 „ 3 a2 - a6 - 2 62 3. , when a = — * 6. 9 a2 + 12 a6 + 4 &2' ^ - 9 a;2 — 30 m; + 25 2/2 , . . 4. ^^-^ " ^ ^, when x^iy. 3x^-2xy-5y^ ' 5. ?i±l^i^^^,whenx = l. a;2 + a! - 2 e. ('^ ~ ^^^ , when X = 1. 7. ^i^^l^Jl^ ^hen a: = 1. a;-l a;2-7a; + 5 8. ^'-^^ + ^ when a: = 1. 9. ^5^1=l-\ when a; = 0. a;3 - 8 X + 2' a'^ - 1 Find the limiting values of the following fractions when n = co : 11 IL+1. 12 ^^-^ 13 "C"-!) . 1. ■ to2 - 1 ■ m + 3 ■ |2 «2 ,- ?i2 _ 3 „ _[_ 2 ^ wx(rea: — 1) (wx — 2) J^_ ■ 2 »i2 _ 3 re + 4' ■ [3 ' ras' CHAPTER XXXIII. INFINITE SERIES. 1. The infinite series 1+1 + 1 + ^+- is a decreasing geometrical progression, whose ratio is f. Let >S'„ = l + | + | + -/7 + ---tomternis. Then, by Ch. XXVII., § 3, Art. 8, as n increases indefinitely. By actual computation, we obtain S,= l, S, = ll, ^3 = 2i, ;S, = 2^,etc. These sums approach 3 more and more nearly, as more and more terms are included. This infinite series may therefore be regarded as having the finite sum 3. But the sum of the series 1 + 2+4 + 8 + ..- increases beyond any finite number, however great, as the number of terms increases indefinitely. 2. The examples of the preceding article illustrate the fol- lowing definitions : An infinite series is said to be Convergent, when the sum of the first n terms approaches a definite finite limit, as n increases indefinitely. An infinite series is said to be Divergent, when the sum of the first n terms increases numerically beyond any assigned num- ber, however great, as n increases indefinitely. 464 INFINITE SERIES. 466 3. It was shown in Ch. XXVII., § 3, Art. 8, that the sum of n terms of the geometrical progression a + ar + ar^ + ■■•, when r < 1, approaches the definite finite limit , as n increases indefinitely. Therefore, any decreasing geometrical progression is a conver- gent series. 4. Infinite series arise in connection with many mathemati- cal operations. Thus, for example, if the division of 1 by 1 — a; be continued indefinitely, we obtain as a quotient the infinite series 1+x + a? -\-di?-\ . When X is numerically less than 1, that is, lies between — 1 and 1, this series is a decreasing geometrical progression, as in Art. 1. Therefore, by the preceding article, it is convergent. Thus, when a; = f, as in Art. 1, the sum of n terms of the series approaches 3, as «. increases indefinitely ; and -3_ = ^_ = 3. \-x 1-f Consequently, we may take the series as the expansion of the fraction, for all values of x between — 1 and 1, and write -J— = 1 + a; + a^ + a;^ + a;* + -, 1 —a; for such values of x. When a; = 1, the series becomes 1 + 1+1+..., and is evidently divergent. If we assign as the value of the fraction, when a; = 1, the limit of the fraction as x approaches 1, as in Ch. XXXII., § 2, Art. 13, we have 1 = 00, 1 —a; when a; = 1. Since both, the fraction and the sum of the series are infinite when a; = 1, in this sense we may assume that they are equivalent. 466 ALGEBRA. [Ch. XXXIII When a; = — 1, we have 1-1 + 1-1 + -. The sum of n terms of this series is 1 or 0, according as n is odd or even. The series is said to oscillate, and is neither con- vergent nor divergent. But, when a; = — 1, 1 ^ 1 ^1 1-a; 1+1 2 Consequently, we cannot assume that the series is the expan- sion of the fraction when x = — 1. When X is greater than 1, numerically, we have l-g;" 1 —X J /(.» By taking n sufficiently great a;", and therefore , can X — cc be made to exceed numerically any number, however great. Therefore, the series is divergent. Thus, when a; = 2, the series becomes 1 + 2 + 4 + 8 + .... But, when a; = 2, = = — 1. 1 — X 1 — 2 Therefore, we cannot assume that the series is the expansion of the fraction when x is numerically greater than 1. In general, an infinite series, no matter how obtained from a given expression, can be regarded as the expansion of the expression, for values of x which make the latter finite, only when the series is convergent for such values of x. 5. In the preceding article the convergency or divergency of the series was determined by an examination of the formula for the sum of n terms. There are, however, many infinite series for which such formulae have not been obtained. In such cases it is necessary to determine the convergency or divergency of the series by other methods. Even when a formula for the sum of n terms is known, methods now to be given are often to be preferred. INFINITE SERIES. 467 6. In the theory which follows, we shall let S stand for the limit of the sum of n terms of the series Mi + Mj + MaH Hm„H , as n increases indefinitely. Also let /S„ = Ml + M2 + Ms + ■•• + M„, the sum of n terms, and „i?„ = u„+i + M„+2 H 1- M»+™, the sum of m terms after the first n terms. ' Then, evidently, S„ + ^R„ = \, k — l is positive, integral or fractional. Therefore (5)*"-^ is less than 1, and the series in the second member is convergent, by Ai't. 3. Consequently, by Art. 8, the given series is con- vergent, when k>\. E. jr. , the series 1 + — + -| is convergent. When k = \, the series is that which was proved to be divergent in (i.). When *< 1, 2»<2, and therefore ->-■ 2' 2 In like manner, — > - 1 —>!,... 3' 3 4* 4 Therefore, 1+1 + 1+1+... >i+1 + 1_l1+... 2* 3* 4' 2 3 4 ■ Since the series in the second member is divergent, the given series is divergent, when A; < 1. is a divergent series. INFINITE SERIES. 471 We therefore conclude : Tlie series 1 +i 4-i. + A4. ... 2' 3' 4* is convergent when fc > 1, and divergent when Jc = l, or k — ; and so for all subsequent terms. Had we given o • o ^ attention only to the first four terms, we should have inferred that the first series is convergent from a comparison with the second series, whereas the question of its convergency or diver- gency is not settled. Now compare the given series with the known convergent series l+i-L-i + i-l =1-L^_-L_J_ + ^_H . 22^32 42^ ' ^2-2 3.3 4-4 It is evident, from the forms of the denominators in the two series^ that each term of the given series, after the first, is less than the corresponding term of the last series. Hence the given series is convergent. BXBECISBS I. Determine the convergency or divergency of the series : 3. ^ + ^+-A-+.... i. ^ + J- + .J-+.... 472 ALGEBRA. [Ch. XXXIII 5. 2 + 2^ + 2_,3^ ._ 6. ^ + -3_ + ^+.... 3 3-5 3.5.7 1.3 2.4 3-5 7 2,3,4, 8 1 , 3 , .f , ... '• l+2"2 + 3'2+ • ^- ^|2+|3+ • 1 1 1 1 + 2 1 + 22 ' 1 + 23 10 <'' + ^ 4- C« + a;)(2a + a:) , ( a + g) (2 a + a) (3 a + z) , ■ 6 + a; (6 + x)(2 6+a;)^"(6 + x)(2 6 + a;)(3 6 + a;) '"' wherein a, b, and a; are positive, and 6 > a. The General Term of a Series. 15. If the general term, the nth say, be given, the entire series is known. Ex. Write the series whose nth term is 9n-l ™4n+l (2w-l)(2n + l) 4m + l Giving to n the series of values 1, 2, 3, ■••, we obtain 1 ^,_2_ ^,_^ x^. I-S'S 3-6" 9 5-7'l3 16. It is frequently necessary to write the ??,th term of a series, when only the first few ternis are given. Ex. Writ6 the nth term of the series 1 + 1 a;2 + 1_^ a;* + 1^^^ a;" + -. 2 2-4 2.4-6 The exponent of x in the second term is 2x2 — 2, = 2 ; in the third term, 2x3 — 2, = 4 ; and in the nth, 2 n — 2. The first factor in the numerator of each term after the iirst is 1, and the last is one less than the exponent of x. Hence the numerator of the nth term is 1 • 3 • 5 ••■ (2 n — 3). In the denominators, after the first term, the first factor is 2, and the last is the same as the exponent of x. Hence the denominator of the nth term is 2 • 4 • 6 •■• (2 n — 2). Therefore, the nth term is — — '- — '-^^-~ ^ x^""l ' 2.4-6-"(2n-2) Observe that only the terms after the first are obtained from the nth term as thus written. INFINITE SERIES. 473 17. If the ratio of each term of a given infinite series to the corresponding term of another infinite series he finite, the given series is convergent when the second series is convergent, and is divergent when the second series is divergent. Let ui + U2 + ui+ ■•• + M„ + ••• be the given series, and vi + U2 + 1)3 + ••• + v„+ •■■ be a series known to be convergent, or divergent. First, let tlie second series be convergent. Let ^ be a finite number greater than the greatest of the finite ratios Ml Ma ... «m Then y:^<_]c,^k(vi + Vi + vs + ■■■ + v„+ ...). Since the second series is divergent, vi + V2 + vs + •■■ + v„, and with greater reason, mi + M2 + Ms + ... + u„, increases beyond any assigned number, however great, as n increases indefinitely. Therefore the given series is divergent. Ex. 1. Examine the series 1.2-3 2-3-4 3-4.5 n(n + l)(n + 2) Compare this series with the known convergent series [Art. 13 (ii.)] Ill 1 1^2' 33 n^ 474 ALGEBRA. [Cii. XXXIII It is sufficient to examine the ratio of the nth term of the given series to the nth term of the second series. This ratio is 1 . 1 n^ ^j «(n+l)(w+2) ■ ?i«' («+l)(n+2)' as n increases indefinitely. Since, therefore, the ratio is always finite, the given series is convergent. Ex. 2. Examine the series _^+_±+_6_ + ...+ 2n _ 1-3 3-5 6-7 ^(2n-l)(2n + l) Compare with the known divergent series The ratio of the nth terms is 2n 1 2n^ .1 (2n-l)(27i + l) ■ n' (2n - l)(2n + 1)' 2' as n increases indefinitely. Since this ratio is finite for all values of n, the given series is divergent. BXBECISES II. Determine the convergenoy or divergency of the series whose nth terms are : J 2to - 5 2 1 + ra a ra + 2 w2+(n + l)2' ■ n(n + 1)(™ +2)" Determine the convergenoy or divergency of the series : 6. ^ + -J_+ -!_ + .... 7. l + _2_ + _§_+.... 1. 33-55. 7 2 3V2 4V3 8. -§- + ^- + ^_ + .... 9. ^ + J«- + ^ + .... 1-222.332. 4 2.33-44-5 10. 1 + 1 + - ^ a(o + 6) (a + 2&)(a + 3 6) (a + 4 6) (a + 5 6) INFINITE SERIES. 475 Series having Negative Terms. 18. If a series be convergent when its terms are all positive, it will remain convergent when some or all of its terms are made negative. Since iS„ remains finite and „7?„ = 0, when all the terms are positive, with greater reason S„ will remain finite and ,„ii;„ = 0, when some or all of the terms are made negative. Ex. The series 1 — i + i — i + rj-— ■••is convergent, since the series l + i + i + i+iT+"" '^ convergent. 19. A series which is convergent when all its negative terms are made positive is said to be Absolutely Convergent. Evidently every convergent series whose terms are all posi- tive is absolutely convergent. 20. If the terms of an infinite series be alternately positive and negative, and the nth term approach 0, as n increases indefinitely, the series is convergent. Let the given series be Ml - M2 + Ms - -■ + (-!)"-'«« + •••. Then Sn = uy — u^ + ua — Ui + — I- (— 1)"-'m„ = (Ml - Ma) + (M3 - Ui) + — (1) = Mi— (M2 — Ms) — (M4— Ms) — ■■•. (2) Since the terms decrease numerically, it is evident that in (1) and (2) the sums inclosed in the parentheses are positive. Therefore from (1) we infer that & is positive, and from (2) that it is less than the first term ui. Therefore 8n. is finite. Also, ™2J„ = (- 1)»[«„+1 - «n+2 + M„+8 - Mn+4 +■■■ + (- l)'»-'Mn+m] = (- 1)"[(M»+1 - "n+a) + (m„+3 - M„+4) + "•] (3) = (- 1)"[m„+i - (m„+2 - M„+s) ]. (4) Prom (3) we infer that the part of „iJ„ in the brackets is positive, and from (4) that it is less than m„+i. Since m„+i = 0, it follows that ^Rn = 0. Hence the. given series is convergent. Ex. The series 1-J + | — J+"- is convergent; but not absolutely convergent, since l-l-| + } + i+---is divergent. 476 ALGEBRA. [Ch. XXXIII The Ratio of Convergenoy. 21. In the following principle the terms of the series are not necessarily all positive : An infinite series is convergeyit, if, after some particular term, the ratio of each term to the preceding be numerically less than some fixed positive number, which is itself less than unity. Let the given series be Ml + MZ + MS + ••• + Mn+ •••, and let the ratio of each term after the A:th to the preceding be less than r, which is itself less than 1. First, assume that the terms are all positive. Then, from ?^'< r, "^K r, "^Kr, ••., we obtain ut+i 3 n — 1 2 4 Since, therefore, after the third term, the ratio of each term to the preceding is less than |, which is less than 1, the given series is convergent. Ex. 2. Examine the series" k3 + 3^^ + 5^^ + ...(2n-l)(2n + l)^„_. . 2 2^ 2' 2" The ratio of convergency is (2>i-l)(2n + l) ^„_. . (2rt-3)(2n-l) ^„_, 2" 2""' _ (2 n + l)x . X ~ (2 n - 3)2' ^ 2 478 ALGEBRA. [Ch. XXXIII By taking n sufficiently great, we can make this ratio differ from i X by as little as we please. If, therefore, x have a definite value less than 2, the ratio can be made less than some number, say k, which is itself less than 1. Hence the series is conver- gent when X <2. The term after which this ratio becomes and remains less than k is determined from 2n + l K^j , ^ &k-\-x ■ < fc, whence n > ; ' 2n-3 2 ' 2(2A;-a;) Thus, let X = |, or l^x = |, and fe = f. We find n > 19|. That is, when a; = f , the ratio of each term, after the 19th, to the preceding is less than |, which is less than 1. Evidently, when a; = 2, or a; > 2, the ratio is greater than 1 for all values of n. Therefore the series is then divergent. 25. The significance of the words, less than some number vihich is itself less than unity, is shown by an examination of the series which is known to be divergent. The ratio of convergency is , = 1 n n This ratio is less than 1, but by taking n large enough, it can be made to differ from 1 by as little as we please. The value of this ratio will therefore not remain less than some definite number, which is itself less than 1. This condi- tion of the principle of Art. 21 is not satisfied, and the test fails Also, since neither condition of Art. 22 is satisfied, the test fails to prove the series divergent. In such cases, it is necessary to try other tests,- just as the above series was by other means proved to be divergent. 26. It is not, in general, necessary to determine the number of the term after which the ratio of any term to the preceding is less than some definite number which is itself less than 1, in the case of a convergent series. The following method, illus- trated by the examples of the preceding articles, is sufficient. INFINITE SERIES. 479 Determine the limit of the ratio of convergency as n increases indefinitely. (i.) If this limit < 1, the series is convergent. (ii.) If this limit > 1, the series is divergent. (iii.) If this limit = 1, the convergency or divergency of the series is, as a rule, not settled, and some other test must be applied. But, if the ratio he always greater than 1, as it approaches the limit 1, the series is, by Art. 22, divergent. Ex. Examine the series ■5x 5-6a^ (w + 3)(w+4)a!° 1.2.32.3.4 n{n + l)(n + 2) The ratio of convergency is (n + 3) (w + 4)a;" . (n + 2) (w + 3)a;"-' n{n + l){n + 2) ' (n-l)n(n + l) ' ^(n + 4)(n-l). (n + 2)(n + 2) ' as n increases indefinitely. Hence, for values of x < 1, the series i* convergent ; for values of x>l, the series is divergent ; while, for x = l, the series is in doubt. When x = l, we have 4-5 5-6 (n+3)(n + i) . 1.2-3 2.3-4 m(n + l)(n + 2) We will try the method of Art. 17, comparing with the known divergent series i +1 + 1 + 1+ - + - + ■■■■ 2 3 4 n The ratio of the wth term of the given series to the nth term of the auxiliary series is (« + 3)(n + 4) . 1 ^ (n + 3)(n + 4) ^^ n{n + l)(n + 2) ' n' (n+l)(n + 2)' This ratio is evidently finite for all values of n. Therefore the series is divergent, when x = l. 27. The following application of the principle of Art. 21 will be required in Ch. XXXIV. 480 ALGEBRA. [Ch. XXXIII The series 1 + f ^jx + f 2)3;'+ (g)*" + is absolutely convergent, when < 1 numerically. In the above series n is finite. We will therefore take the ratio of the (fe + l)th term to the preceding. The ratio of convergence is »(w-l)--(ra-7c + l) ^4 ^ n(^n-l)-{n-k + 2) ^^.j 1^ ■ |fc- 1 as k increases indefinitely. Therefore, the series is absolutely convergent, when < 1 numerically. EXERCISES III. Determine the convergency or divergency of the series : 1. i+L' + 3_%..., 2. ? + 2i3 + 2_£:i+.... |^[3 11., 3 1-3-5 3. l + llJ+ l-3-5 +..., 4. l + l_2^1-2.3, 4 4.74-7.10 3 3-63.69 5. -^— +— ^•+ ^"^ + -. «4l a + A a + 2A; Determine for what values of x the following series are convergent or divergent : 6. 12 + 22x + 32x2+ ••■. 7. a;-Ja;2+Ja;3 . 8. 1 4 E _|-5.4. .... 9. _L + _^ + ^_ + .... |_1|2 1.22-33-4 w.Z-\ + A . 11. _L + l^ + (2£l%.... \ X Zx^ 1-33.55-7 12. l-i? + L^-.... 13. 1 +i? + i2.% .... 14. i+-^j.^;^%.... 15. i+2^.+3;.2+-. 16. a + (a + d)a: + (a + 2 d)x'' + •••. 17. ill 18 ' 1 ^ 1 *^ 1 + a: l + a;2 l+a;8 ' 1 + a; 1 + a;2 1 + »» 19. 2a:+ 1 3(2 a; -f 1)3 ' 5(2 a: + 1)6 an . X . x-^ 1 + /<; 1 + 2 & CHAPTER XXXI V. THE BINOMIAL THEOREM. § 1. THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPOMENTS. 1. In Ch. XXVIII., it was proved by induction that, when n is a positive integer, (a + 6)"= a» + /'"V-'6 + /'" V--6^ + • ■ • + ^ ** V-»+i6»-i + .... We will here give a briefer proof, based upon the theory of combinations. Consider the following continued product of n factors : + 6 a + b n factors ^ ■ • • a + 6 The first term of the product is formed by taking an a from each factor, giving a". The second term is formed by taking an a from re — 1 factors and a 6 from the remaining factor, giving a"-'6. But such a term can be formed in as many ways as one 6 can be taken from n 6's, i.e., in „(7i ways. Therefore the product so far is a" + „0ia"-'6. A third term is formed by taking an a from n — 2 factors and a & from the remaining two factors, giving a"--^b^. But such a term can be formed in as many ways as two 6's can be taken from n 6's, i.e., in „C'2 ways. Consequently, the product to this point is a" + „CiO"-'6 + nCia^~^b^. In general, an a can be taken from each of n — k + \ factors and a 6 from each of the remaining k — \ factors, giving a»-*+i6*-i. But such a term can evidently be formed in „C*-i ways. We thus obtain (a + 6)» = a" + „Cia»-i6 + „Caa»-262 + ... + „C*-ia"-*+'6*-i + .■■. 481 482 ALGEBRA. [Ch. XXXIV But „c.=(';),„c.=('^).,A=(!;),..c._,=(^^;^^). Therefore, (o + 6) " = a" -f ( " ) a"-ii + ( " ] a'^'^W' + [ M a-'-^fts Properties of Binomial Coefficients. 2. The fcth term, counting from the beginning of the expan- sion, contains 6*~\ and is „C4_ia''~*+'&*~'. The /cth term, count- ing from the end, contains a'~', and therefore 6""'+^, and is ri „k-ll..n-k+l n^n—k+V^ " But, by Ch. XXX., § 3, Art. 3, „C,^, = „C„_j+i. We therefore conclude : In the expansion of (a + b)", wherein n is a positive integer, the coefficients of terms eqtially. distant from, the beginning and end of the expansion are equal. 3. By Art. 1, the coefficient of the (A;-|-l)th term is „Ci. Therefore, by Ch. XXX., § 3, Art. 4, we have : The greatest binomial coefficient, when n is even, is „C'^ ; and when n IS oaci, IS „0„_i, ^^n^n+i- 4. In (1 4- a;)" = 1 -f „CiX + „C^ -f ■ ■ • + „C„x'\ let x = l. Then 2" = 1+„G, + „C, + ---+„C„. That is, the sum of the binomial coefficients is 2". 5. Prom Art. 4, we have That is, the total number of combinations ofn things, taken one at a time, two at a time, and so on, to n at q time, is 2" — 1. 6. In (l + xY=l + „CiX + ^0^+ — +„C„x% let x = -L §2] THE BINOMIAL THEOREM. 483 Then i_^Ci + „(7,-„(73+-=0, or l + „C2 + „C4+-=„Ci + „C3+-. That is, in the binomial expansion, the sum of the coefficients of the odd terms is equal to the sum of the coefficients of the even terms. § 2. BINOMIAL THEOREM FOE ANY RATIONAL EXPONENT. 1. From Ch. XXVIII., Art. 4, we have (l + a,)" = l + (^!j)x + Qx^+..., (1) when w is a positive integer. In this case the expansion ends with the (n + l)th term, since the coefEicients of the (n + 2)th and all succeeding terms contain n — n,ov 0, as a factor. But if n be not a positive integer, the expression on the right of (1) will continue without end, since no factor of the form n—k+l can reduce to 0. Therefore this series will have no meaning unless it be convergent. 2. In Ch. XXXIII., Art. 27, it was proved that the series is convergent when x lies between — 1 and + 1. It remains to be proved, therefore, that in this case the above series rep- resents (1 + »)", when n is a fraction or negative. 3. Since the reasoning wiU turn upon the value of n, we shall call the expression a function of n, and abbreviate it by /(n), for all rational values of n. To understand the following reasoning, the student should notice that for all positive integral values of n, (l + x)" =/(»), as, (1 + a;)' = /(•'>) ; and that it remains to prove that (1 + x)" = /(n), when re is a fraction or negative, as, for example, that (1 + x)^ =/(§)■ 484 ALGEBRA. [Ch. XXXIV 4. We now have /(„o=i.(»)..(-)x^+..-f(;:^)x-^-... for real values of x between — .1 and + 1. We will assume that the product /(m) x /(ra) is a convergent series, when the two series are convergent. The proof of this principle is beyond the scope of this book. We then have /(m)x/W=[l+(™)x + (™)xH- + (/^j)x-'+...] -+[(\Xi)>-[(;V(T)G>(")?- But (in\ fn\_/m + n\ /m\ /m\fn\,fn\_fm+ n\ ilj + llj-i 1 j' i2J + lljilj+UJ"( ^ I' and by Ch. XXX., § 4, Art. 2, therefore f(m)xf{n) = f^m + n), (1) for all rational values of m and re. Then /(m) x/(n) x/(p)=/(to + n) x/(p) = /(m + u +i)). In general, /(m)x/(re)x/(j5)x - ■X.f{r) = f(m + n+p+ — +r), (2) for all rational values of m, n, p, ••-, r. §2] THE BINOMIAL THEOREM. 485 5. Fractional Exponents. — Let u TO = n = p =••■ = »• = -5 V wherein u and v are positive integers. Talking v factors, we now have ■^\i) ""^{l) "^Kv) '^ ■■■ " ^'"''°''^ =-^{i + 1 + I + ■•■ " summands), Now, since a is a positive integer, (H-a:)»=/(M). Therefore (1 + a:)" =r/[-]T, or (l + xy=f(^\ That is, (1 + X)" = 1 + I J J a; + a:2+ ■ 6. Negative Exponents, Integral or Fractional. — In (1), Art. 4, let m = — 71. We then have /( - n) x /(n) =/(n - n) =/(0) = 1, since /(0)= 1 + -x + ... = 1. 1 Therefore /(«) =/(-n). Since re is a positive integer or fraction, (1 + x)" =f(n), and therefore 1 ;=/(-«)> or (1+ a:)-" =/(-n). That (1 + a;)" is, (l + a:)-=l + (-")x+^-"Ja;3 + 7. Expansion of (o + b)". — We have and =..,1+^; b" 1 + a\" (1) (2) 486 ALGEBRA. [Cii. XXXIV When 6 is numerically less than a, fty . . fn\h . fn\h^ and, by (1) above, = a" + (^" j a'-'6 + Qj a"-^?)^ + • • •. (3) In a similar way it can be shown that, when a is numerically less than b, (a + &)" = 6" + Q 6»-ia + ("^ j 6»-V + ■ • -. (4) Notice that when n is a fraction or negative, formula (3) or (4) must be used according as a is numerically greater or less than 6. 8. Ex. Expand -— — — to four terms. If we assume a > 4 6^ we have, by (3), Art, 7, 1 , 4&2 , 32&^ ^ 8966^ , -^a Za-^a 9a^-y/a 81 a"^a If a < 4 &^, we should have used (4), Art. 7. Any particular term can be written as in Ch. XXVIII., Art. 6. §2] THE BINOMIAL THEOREM. 487 9. Extraction of Roots of Numbers. — Ex. Find -yjll to four decimal places. We have Vl7=V(16 + l) = 4(l + ^)i = 4 1 + 1x1 + i(ni)f 1 Y+ i(-^)(-f) f 1 Y+ 2 16 1-2 Vie; 1-2-3 \\QJ = 4 (1 + .03125 - .00049 + .00002 ) = 4 X 1.03078 = 4.12312. Therefore -y/17 = 4.1231, to four decimal places. EXERCISES. Expand to four terms 1. (l+a)i 2. (!-»;)->. 3. (a;' + !/)"*. 4. {x-y'^)-*. 5. (27 + 5x)S. 6. (8aS-36)*- 7. (3 + 2a;)i 8. (Sa^-SfiS)"^- 9. (2a;^ + xy-'^yK 10. [-J/Caft) - v'(«'')]^- H' («^«~* - a~%^)-i. 12. -^-— -• 13. . , J . • 14. ^ Find the 15. 4th terra of (1 - 2 x)^. 16. 6th term of (1 + a^'h'^. 17. 5th term of (s? - a-V)~*- 18. 8th term of (aV^ - 2 6 ^a)"^ 19. *-5thtennof (1 + a;*!/^)-''. 20. 2 Mh term of [a;^ - VC^:?/)]^. 5 21. Find the term in (3 a;S - a;^^)^ oontammg a;^ 22. Find the term in f a + -^V^ containing «-". V 2Va/ Find to four places of decimals the values of 23. VB. 84. V27. 26, ^35. 26. .{/700. 27. 4/258. CHAPTER XXXV. UNDETERMINED COEFFICIENTS. § 1. METHOD OF UNDETERMINED COEFFICIENTS. 1. Upon the following principles is based an important method of changing an algebraical expression from one form to another. 2. If an infinite series a^ -\- a-^: -'r a:fi? + a^v? -\- • • ■ he con- vergent for values of x greater than 0, the sum of the series approaches a^, as x approaches 0. Let ao + oix + a^x^ + a^x^ + ... = ao + xSi, wherein Si = ai + a^x + a^x"^ + ■■.. Evidently, if the given series be convergent, that is, if ao + xSi be finite, then Si is finite. Therefore, by Ch. XXXII., § 2, Art. 10, xSi = 0, when a; = 0. Consequently ao + a-iX + UiX^ + •••, =: ao + xS, = ao, when a: i 0. 3. If two integral series, arranged to ascending powers of x, be equal for all values of x which make them both convergent, the coefficients of like powers of x are equal. Let ao + aiX + a^i? + ••• = 6o + bix + ftaS^ + ■•• for all values of x which make the two series convergent. Then the sums of the two series approach equal limits when a; = 0. But, by the preceding article, the sum of the one series approaches ao, that of the other 6o ; consequently ao = 6o, and aix + a^x'^ + -• = h^x + ftaX^ + ■••. Since by Ch. XXXIII., Art. 21, these two series are convergent for all values of x for which the original series are convergent, they are equal for values of x other than zero, and the last equation may be divided by x. 4 as §§ 1-2] EXPANSION OF CERTAIN FUNCTIONS. 489 Hence at + aix + aspc^ + ■■■ = bi-i- hx + hx^ + •■■ ; and as before, ai = 61, and a2a; + agx!^ + ■■■ = b^x + b^^ + ■■•. In like manner, we can prove as = 62, as = 63, etc. 4. The principle of Art. 3 holds with greater reason if either or both of the series be finite. The series must be equal for all values of x, if they be both finite; or, if one be infinite, for all values of x which make that series convergent. 5. The condition that the roots of the equation ax^ + 6a; + c = are equal, given in Ch. XXI., Art. 17 (ii.), can be obtained also by applying the principle of Art. 3. If the two roots be equal, aaf + bx+ c is the square of a binomial. We therefore assume ax'' + bx + c = (Ax + By ~ AV + 2 ABx + ff. By Art. 3, A^^a (1), 2 AB=h (2), B'' = c (3). From (1) and (3), A = -y/a, B = ^c. Whence, by (2), 2 V(ac)= 6, or 6^ = 4 ac. EXERCISES I. Assuming Ax- + Bx + C to be the quotient in each of the following divisions, find the values of A, B, C, and the value of m so that the division will be exact : 1. (a;3-2x2 + 6a; + 3m)H-(a;-2). 2. [(to + 5) k3 + 3 a;2 + (m - 5) a; + 3] -f- (a; + 3). 3. [a;«-9a;2- (wi + 2)x + m]H-(a;2-4a; + 2). §2. EXPANSION OF FUNCTIONS IN INFINITE SERIES. 1. We shall now give a method of expanding certain func- tions in infinite series by the principle of § 1, Art. 3. 490 ALGEBRA. [Ch. XXXV Rational Fractions. 2. In assuming d,s the expansion of a rational fraction an infinite series of ascending powers of x, we first determine with, what power the series should commence. This is done by division, when both numerator and denominator are arranged to ascending powers of x. In fact, this step also determines completely the first term of the series. Ex. 1. Expand 2-x 1 +x — or in a series, to ascending powers of x. Since the first term of the expansion is evidently 2, we assume 2-a; 2 + Bx + Co? + Do^ + Ex" -, 1 +x — x^ wherein B, C, D,--- are constants to be determined. Clearing the equation of fractions, we obtain 2-x=2 +B 2 x+C x'+D 3?+ E + B + c + D -2 -B -C x" + The series on the right is infinite; that on the left may be regarded as an infinite series with zero coefficients of all powers of x higher than the first. By § 1, Art. 3, we have 2 = 2; B+2 = ~l, C+B -2 = 0, D + C-B=0, E + D- C=0, etc., Hence, substituting these values of B,C,D,-- in the assumed series, we have 2- whence B = — 3 ; whence C = 5 ; whence Z) = — 8 ; whence S = 13 ; etc. l-\-x = 2 - 3 a; + 5 a^ - 8 ^5 + 13a;* + § 2] EXPANSION OF CERTAIN FUNCTIONS. 491 Let the student compare this result with that obtained by division. In fact, the latter method of expanding a fraction is to be preferred when only a few terms are wanted. But the successive coefficients, after a certain stage, may be computed with great facility by the method of undetermined coefficients. A moment's inspection of the preceding work will convince the student that the coefficient D, and all which follow it, are each connected with the two immediately preceding coefficients by a definite relation. Thus, £>+C-J3=0, E + D-G=0, F + E-D = 0, etc. Ex. 2. Expand Az^, in a series, to ascending powers of x. The first term in the expansion is evidently ^x~K We therefore assume -IsL^ = A a!-2 + Bx-^ + C+DX + E3?-\- Fa? + -. Clearing of fractions, we obtain l-ce = l + 3B X + 3G -B x'^ + SD -0 .r'+ By § 1, Art. 3, we have 1 = 1. 3J3-^= -1, whence 5= -|; 3C-B= 0, whence 0= -j-\; 3D-C= 0, whence Z)=-^; etc., etc. Hence -kl^= i a;-^-|a;- -^ - ^a;- .... EXERCISES II. Expand the following fractions in series, to ascending powers of x, to four terms : -[ +2x „ 2 + a: - 3 a:" 7. 1+x 1 -X 1 1 + x + x? X* - 3 a;2 + 1 2. i +x + x' S-s + Sj;! 5. — !—■ 6. ^+^- \ -x^ x + 2 a l-X a 1 l + x-x^ 5 i2 + 2 x3 2x^-6x^ + x* 492 ALGEBRA. [Ch. XXXV Surds. 3. Ex. Expand -^(l-a^ + 2 x^), in a series, to ascending powers of x. Assume y (1 ~.a^ + 2x?) = l + Bx+Cx' + Da^ + Ex*-{- Squaring both sides of the equation, we have 1 ' + 23^ = 1 + 2B X + 2C a?+2D + 2BG Equating coefficients, 1 = 1. 2 5 = 0, 2C+B^ = -1, 2D-\-2BC=2, 2E + 2BD+C^ = Q, a?+2E + 2BD ■0; x* + 1 . 2 ' whence B -- whence C ■■ whence Z) = + 1 ; whence E = — I; etc. Hence ^{1 -a? + 2a?) = l ~ ^ x"" + ^ - lx*+ ■■: EXERCISES III. Expand the following expressions in series, to ascending powers of x, to four terms : 1. V(l + «)• 4. v(4 - 2 a: + x'^). 2. v'(a^-2a;2). 5. ^(5 + .3a: + 9a;2). 3. sy(l-a:2)- 6. ^(1 - a; + x^). § 3. REVERSION OF SERIES. 1. If one variable be equal to a series of positive integral ascending powers of a second variable, the second variable can be expressed in a series of positive integral ascending powers of the first. This process is called reversion of series. Ex. 1. Revert the series y = x-\-2x^-\-^i?+ ■■■. Assume x = Ay + By'' + Cy' + ■■-, (1) and substitute in the second member of the last equation the value of y given by the first. Then §§3-4] REVERSION OF SERIES. 493 x = A(x + 2x^ + 30^ +-)+ B(x + 2 x^ + 3x'+ ■■•)' + C{x + 23y' + 3a?+:.y+... = Ax + 2A a^+3A a^+-- + B +4J3 + C Hence ^ = 1, 2A-\-B=Q, whence J3 = -2; 3^ + 45+ C=0, whence 0= 5; etc., etc. Substituting these values of A, B, C, ••■, in (1), we have x = y-2y^ + 5if + ■■■. If the series for y in terms of x contain a term free from x, we must find a value of a; in a series of powers of y minus that term. Ex. 2. Eevert the series y= 1 +x + a^ + a^+ ••■, or y-l = x-\- ay' + x'+—. (2) Assuming x = A(y — 1) + B(y — ly + ■■•, and proceeding as in Ex. 1, we obtain A — 1, 5 = — 1, C = 1. Therefore x = {y-l)-(y- If + {y - If . EXERCISES IV. Revert each of the following series to four terms : 1. y = x + x^-\-ifi+-: 2. y = x + 3x' + 6x''+—. 3. y = x-lx^+ix^---. 4. y = l-x + 2x'' . 5. y = l +- + — + ■■: 6. y = ax + bx^ + cx' + --. [1 \1 §4. PARTIAL FRACTIONS. 1. It is frequently desirable to separate a rational algebraical fraction into the simpler (partial) fractions of which it is the algebraical sum. 494 ALGEBKA. [Ch. XXXV 2a; 1 1 E.g., 1 — 3? 1—X 1 + X The process of separating a given fraction into its partial fractions is, therefore, the converse of addition (including sub- traction) of fractions ; and this fact must guide us in assuming the forms of the partial fractions. We shall separate the given fractions into the simplest partial fractions, that is, frac- tions which cannot themselves be further decomposed. We shall also assume that the degree of the numerator is at least one less than that of the denominator. A fraction whose numerator is of a degree equal to or greater than that of its denominator can be first reduced by division to the sum of an integral expression and a fraction satisfying the above condi- tion. The latter fraction will then be decomposed. The denominators of the partial fractions can be definitely assumed. For they are evidently those factors whose lowest common multiple is the denominator of the given fraction. But there is one case of doubt, namely, when a prime factor is repeated in the denominator of the given fraction. E.g., 6-2a^ ^ 3 2 1 . {l-xy(l + x) l-x'^(l-xf 1+x' 3+3? _ 2 1 (i-xy{i+x) (i-xf ' i + x We could not have decided, in advance, whether either of the two given fractions is the sum of two or of three partial fractions. There must necessarily be a partial fraction having (1 — xy as a denominator, since, otherwise, the L. C. M. of the denominators would not contain the prime factor 1 — a; to the second power. But it cannot be determined, in advance, whether there is a partial fraction having 1 — a; as a denomi- nator. In such cases, therefore, it is advisable to make provision for all possible partial fractions by assuming as denominators all repeated factors to the first power, second power, etc. §4] PARTIAL FRACTIONS. 495 The numerators of partial fractions thereby assumed, which should not have been included, will acquire the value zero from the subsequent work, so that those fractions drop out of the result. The numeroAors of the partial fractions must be assumed with undetermined coefficients. Since the numerator of the given fraction is, by the hypothesis, of degree at least one less than the denominator, the same must be true of each partial fraction. We therefore assume, for each numerator, a complete rational integral expression with undetermined coefficients of degree one lower than the corresponding denominator. If any term in the assumed form of the numerator should not have been included, its coefficient will prove to be zero. An exception to this principle occurs when the denominator of the partial fraction is the second or higher power of a prime factor, as, (1 — xy. In that case the numerator is assumed as it would be according to the above principle if the prime factor occurred to the first power only. We may briefly restate the above principles : Separate the denominator of the given fraction into its prime factors. Assume as the denominator of a partial fraction each prime factor ; in particular, when a prime factor enters to the nth power, assume that factor to the first power, second power, and so on, to the nth power, as a denominator. Assume for each numerator a rational integral expression, with undetermined coefficients, of degree one lower than the prime factor in the corresponding denominator. Let us first decompose the two fractions which we have used to illustrate the theory. Ex.1. .. ^-Jf . =A-+- ^ ■ ^ (1 - xy{l + x) l-x (l-xf 1+x Since the prime factor in the denominator of each partial fraction is of the first degree, each numerator is assumed to be of the zeroth degree. 496 ALGEBRA. [Ch. XXXV Clearing the equation of fractions, we have 6-2!^ = A{l-x){l+x)+B(l+x) + C(l-xy = {-A+C)x'+{B-2C)x + A+B+a Since this equation must be true for all values of x, we have -A + C=-2, B-2C= 0,\WheneeA = 3,B = 2,C = l. A + B + C= 6. J Consequently 6-2a^ 3 2 1 (1 - a;)=^(l + a:) 1 - a; (1 - a;)= 1 + a; Ex. 2. S + x" ^ A B G (1 - x)Xl +x) 1 - a; (1 - a;)'^ 1 + a;' The forms of the partial fractions are assumed the same as in Ex. 1. We have 3 + x^=(~A + C)x'+{B-20)x + A + B+G, and then —A + C'=l,] J5 - 2 G= 0, [ Whence A = 0, B = 2,C=1. A + B + C=3.) Therefore S + x" _ 2 1 (1 - xfil + x) (1 - a;)2 1 + a; When the factors of the denominator of the given fraction are of the first degree, as in Exx. 1 and 2, the work may be shortened. Begin with the equation 6-2x' = A{l-x){l+x)+B(l + x) + 0(1 -xf, of Ex. 1. Since this equation is true for all values of x, we may substitute in it for x any value we please. Let us take such a value as will make one of the prime factors zero. Substituting 1 for x, we obtain i = 2B, whence B = 2. §4] PARTIAL FRACTIONS. 497 Next, letting a!= — 1, we have 4 = 4 C, whence C=l. There is no other value of x which will make a prime factor zero, but any other value, the smaller the better, will give an equation in which we may substitute the values of B and C already obtained. Letting a; = 0, we obtain 6=A + B+C, whence A = 3. The same method can be applied to Ex. 2. a^-x + 3 x'-x + S A , Bx+C Ex. 3. 5 — z — = -, ^^ . , , r^rr = r + " x'-l {x-l){scF + x + l) x-1 xF + x + l In this example the one prime factor being of the second degree we assume the corresponding numerator to be a com- plete linear expression. Clearing of fractions, we have se^-x + 3 = A{x^ + x + l) + {Bx+C){x-l) = (A + B)x'' + (A - B + C)x + A-C. Equating coefficients of like powers of x, we obtain A + B=l, A-B+C=-l, A-C=3; whence, A = l, B = 0, C= -2. Or, we might have used the second method, beginning with x^-x + 3 = A{x' + x + l) + {Bx+C) (x-1). Letting x=l, we obtain 3 = 3 A, whence A=l. Since no other value of x will make a factor vanish, we take any simple values. When a; = 0, we have . 3 = A-C, whence C = - 2. Finally, letting x = — l, we have 5 = ^+2B-2C, whence B = 0. Ex. 4. 2-2x + 4a? _Ax + B Cx + D E (l+3?f{l-x) 1+3? {I + X'f 1-X 498 ALGEBRA. [Ch. XXXV The prime factors in the denominator of the first two partial fractions being of the second degree, expressions of the first degree are assumed as numerators. Clearing of fractions, we have 2-29; + 4c(^ = {Ax + B){l+x')(l-x) + (Cx+ D)il-x) + E{l+x'y = {-A + E)x' + {A - B)x' + {- A + B - C + 2 E)3^ + (A~ B+C~D)x + iB + D + E). Equating coefficients of like powers of x, we obtain -^ + £■ = 0,^1-5 = 0, -A + B-C+2E = 4:, A-B+C-D=-2,B+D + E = 2; whence A = l, B = 1, C = - 2, D = 0, E = l. rp, ,. 2-2 a; + 4 a;' x+1 2x , 1 Therefore ,, , ^,,,T „s = T^ " 7T^-^j + - (1 + a^)2(l -x) 1 + a^ (1 + a;^)''' 1 - a; Ex. 5 1 A ^ B ^ G (x + n)(x + n + l)(x + n + 2) x + n x + n + 1 a,' + )i + 2 Clearing of fractions, we have 1 = ^ (a; + ?i + l)(a; + n + 2) + 5(a; + n)(a; + M + 2) + C(a; + n)(a;+ n + 1). Letting x = — n, we have 1 = 2 A, or A = \; x = — n — 1, 1 = — B, OF B = — 1; x = -n-2, 1 = 2 C, or C= 4. Therefore 1 1 1,1 (a;+m)(a;+M + l)(a;+n + 2) 2(x+n) x + n + 1 2(a;+n+2) 2. The General Term. — The following examples illustrate the method of finding the general term of the expansion of a rational fraction in a series, to ascending powers of x. 2 A- 7 X Ex. 1. Find the general term of the expansion of — ■ • We have 2 + 7a; 3 1 l + a;-2a!2 1 — x l+2a; = 3(1 + x + x' -\- ■■■ +a;»+ •••) -\l + {- 2 x) + {- 2xY + ... +{-2 xf + ■■■■]. §4] PARTIAL FRACTIONiS. 499 The expansions of the above partial fractions, and similar ones, are readily obtained by the formula l~x The required general term is the sum of the (w + l)th terms of the above expansions. We have 3 a;" - ( - 2 a;)" = a;" [3 + (- 1)"+' 2"]. The expansion of the given fraction can be obtained from this general term. Giving to n the values 0, 1, 2, 3, •••, we obtain 2 + 7a; _ 2 j^ q^; ^jS^ HjS _... + ^S + {- iy+'2''']x"+-: l+X-23^ Ex. 2. Find the general term of the expansion of 10 -7 a; + 6 a' (2-a;)(l+a;2)' We have 10-7a; + 6a' ^ 4 3-2a; ^ 2 3-2a; (2 - a;) (1 + a;^) 2 - a; 1 + a;^ l-^a; ! + »' =2[l + ix + {lxy+- + ilx)'" + i^xy''^'+...] + (3-2a;)[l + (-aT^) + (-a;=)^+... + (-a;^)"+-] = 2[1 + ^ a;+ 1 a;^ ■•■ + (i)'"a!^+(i)'"+'a*+'+ ••■] + [3-3x^+3 a;^ + (-l)»3a;2" + -] + l-2x+2x>-2 3f+-- + (-l)''+'2oe"'+^+---]. Observe that it is necessary to distinguish between even and odd powers of x. Terms containing even powers of x are obtained from (i)*-^" + (- 1)»3 a^, = x^" [(I)'"-' + 3 ( - 1)"] ; and terms containing odd powers from (|)2V'+' + (- l)"+'2a;^''+', = a;2"+'[(^)-" + 2(- !)»+']. The expansion is readily obtained from these general terms. 500 ALGEBRA. [Ch. XXXV EXERCISES VI. Separate the following fractions into partial fractions : 7 1. (x-2)(1-2k) 3. 3x- 1 (x + 3)(x-2) 5. 6 1 -X2 8. 1 7 X - a;2 _ 12 10. 31 + 2 (x-^-l)(x-2) la 312+ 1 (a: + l)(a;-l)2 J4 5 a:(a: + 3) ■ (2a;+ l)(2a:- l)(a; + 1) 16. 17 {X - \y 19. J_±l. 20. 2. (6 + 3a;)(a: + 4) 4 1 -X (3x + 2)(x + l) 6. 6^ 7. ^+^ a:2-4 9-x-^ g a;2 + 2 a; - 1 9x2_ 16 11 a;2 + 90 X - 9 6(x2-9)(x-3) 13 x2 + 5 X + 10 (j: + 1)(x + 2)(x + 3) 15. 3 -X (2x + l)(2x + 3)(x-l) 1 18 2 ' a;3- 1 X3 + 1 , 1 21. 1 . x-^- 1 '"" x<- 1 " x2(x2+ 1) 22-28. Find the general terms of the expansions, to ascending powers of X, of the fractions in Exx. 5-11. Find the general term of the expansions of the following fractions, to ascending powers of x : 89. \ 30. 5x'-6x-13 . g^ 6 x + 26 2x(x2+l) 10(x + 3)(1 + x2) 3(x-4)(2 + 3x2) 32. 1 33. "* + * (x + »)(x + re + l) {x — m){x — n) (h — c)(c — a)(a - b) 34. ^" •^jy- "vv" 'V 35. (x - a)(x - 6)(x - c) (1 + fl!"-'x)(l +a"x) CHAPTER XXXVI. CONTINUED FRACTIONS. 1. If the numerator and denominator of f ^ be divided by the numerator, we have 30 30 -- 30 1 43 43 -=-30 1+U TS Eeducing H, and subsequent fractions, in a similar way, we obtain 30 1 '1 1 43 1 + 13^ i + ^_ 1+ 1 30^13 2 + A 2+ 1 13 3 + i 4 The complex fraction thus obtained is usually written more compactly thus : 1111 1+2+3+4 Observe that in the last form the signs + are written on a line with the denominators to distinguish the complex fraction from the sum of common fractions. It is important to keep in mind that in both forms the numerator at any stage is the numerator of a fraction whose denominator is the entire com- plex fraction which is written below and to the right of that ■particular numerator. 2. A Continued Fraction is a fraction whose numerator is an integer, and whose denominator is an integer plus a fraction 501 502 ALGEBRA. [Ch. XXXVI whose numerator is an integer, and whose denominator is an integer plus a fraction, etc. A continued fraction frequently occurs in connection with an integral term. 2+ ' -^\ In such cases it is customary to call the entire mixed number the continued fraction. The general form of a continued fraction, therefore, is: n \ i =n -\ ' 3. We shall confine ourselves in this chapter to continued fractions in which the numerators are all 1, and the denomi- nators all positive integers ; of the general form, therefore, ,111 in which the d's are all positive integers, and n is a positive integer or 0. The n and the d's are called Partial Quotients. 4. A Tenninating, or Finite Continued Fraction, is one in which the number of partial quotients is limited, as in the example given above. A Non-terminating, or Infinite Continued Fraction, is one in which the number of partial quotients is unlimited or infinite. CONTINUED FRACTIONS. 503 To Convert a Common Fraction into a Terminating Continued Fraction. 5. Compare the work in Art. 1 of reducing |§ to a continued fraction, with the work of finding the G. C. M. of 30 and 43 : 30)43(1 30 13)30(2 26 4)13(3 12 1)4(4 4 Observe that the successive quotients in the latter process are the partial quotients of the continued fraction. This is as it should be, since a comparison of the two processes shows that the successive steps of division in getting the partial quotients are identical with those in finding the G. C. M. The method is evidently perfectly general and may be ap- plied to any common fraction. If the fraction be improper, the first quotient will be the integral part of the continued fraction, and the remaining quotients the successive partial quotients of the continued fraction proper. 161 Ex. Reduce — to a continued fraction. 45 By the method of G. C. M., we have 46)151(3 135 16)46(2 32 13)16(1 13 3)13(4 12 1)3(3 3 mi, * 151 Q , 1 1 1 1 Therefore, -^ = ^+2TlT^3- 504 ALGEBRA. [Cii. XXXVI To Reduce a Terminating Continued Fraction to a Common Fraction. 6. We have only to retrace the steps taken in the preceding article in forming a continued fraction from a common frac- tion. Thus, 1 1 11^ 1 1 4^1 13 ^30 1 + 2 + 3+4 1 + 2+13 1+30 43' Evidently this method is also perfectly general. We therefore conclude that any common fraction can be converted into a terminating continued fraction, and, con- versely, that any terminating continued fraction can be reduced to a common fraction. The latter reduction becomes laborious in the case of a con- tinued fraction with many partial quotients, and a simpler method will now be given. 7. A Convergent of a continued fraction is that part of it ob- tained by stopping with a definite partial quotient. Thus, in 111 1 1 + 2 + 3 + 4' 1 112 the first convergent is -; the second is > = - : 1 1 + 2 3' the third is the fourth is 1 1 1__ ^7_, 1 + 2 + 3+' 10' 1_1^1 1 ^30 1+2 + 3 + 4' 43 For convenience, we will call the integral term, when there is one, the zeroth convergent, so that the nth convergent will always end with the nth partial quotient. W^e will denote the successive convergents by — -, —^, — , etc. Do Di D^ CONTINUED FRACTIONS. 505 The above convergents may then be -written thus : A i' A 3' Ns^ 7 ^ 3x2 + 1 ^ 3xJy2 + iVi . A 10 3x3 + 1 3xA+A' -^4^30_ 4x 7 + 2 ^ 4x-ZV3 + -?V2 A 43 4x10 + 3 4xA + A' That is, to form the numerator of the third convergent, multiply the numerator of the second by the third partial qaiotient, and to the product add the numerator of the first convergent. To form the numerator of the fourth convergent, multiply the numerator of the third convergent by the fourth partial quotient, and to the product add the numerator of the second convergent. In like manner, form the denominators from the denominators of the two preceding convergents. In general, The numerator of any convergent after the second (after the first if there be a zeroth convergent) is formed by multiplying the numerator of the immediately preceding convergent by that partial quotient with which the convergent to he computed ends, and to the product adding the numerator of the second preceding conver- gent; the denominator of the same convergent is formed in like manner from the denominators of the two convergents immediately preceding. The principle holds for the second convergent when there is a zeroth convergent ; and in all cases for the third convergent. For ^ = ?», ^=„+l = ™^lJlI, Do 1 Vi di dx N2_^ . 1 1 - „ I d% _ ndidj + n + ck Di di ■+ di di(?2 + 1 didi + 1 _ d'!(ndi + l)+n _ d^Nx + Np ^ did-i + 1 d-iDi + Do 506 ALGEBRA. [Cii. XXXVl D3 (^1 + ^2 + 0,3 di + d^dz + 1 dididi + di + (ig _ ndidjdi + nd\ + mda + (^-jcfe + ^ ~ did2d3 + di + ds _ dtindid -i + » + da) + {ndi + 1) __ dsiV^a + Ni di{d-idi+^)-\- di. diDi-hDi If the principle holds up to and including any convergent, it holds for the next convergent. Suppose it holds up to and including the Ath convergent. We then have dk diiDt-i + Dk-i Now ^^ = d + -^ 1 (^2+...- 4 + - ' dj+i differs from the preceding convergent only in having d, + -^ "4+1 as a denominator where the preceding has df Therefore, if we substitute d, + J_ ford, in 4iV^+^-' dt+i duDk-i + Dk-i we obtain an expression for — ^, without assuming that the principle holds beyond the kVa convergent. Consequently, ldi, + -^ Wj_i + Nu-i Nk+\ _ \ dt+\ I dtdk+iNk-i + Nk-i + dt+iNt-i -0*+i /-J, I L.\ n, 1 4- n, . dA+iDk-i. + Di-i + dk+iD/i-s ldk + ~\Dk-i + Dk-. _ dt+i(dtJVt_i + Nk-i)+ Nt-i _ dt+iiVt + ^k--i dk+\{dkDk-\ + Di,-i)+ Dk-, dk+iDk + Dk-i a result in accordance with the principle. Therefore, since the principle holds to and including the third con- vergent, it holds for the fourth ; then, since it holds for the fourth, it holds for the fifth ; and so on. This method of proof is called Proof by Mathematical Induction. CONTINUED FRACTIONS. 607 Properties of Convergenta. 8. (i.) 27te successive convergents, beginning with the zeroth, are alternately less and greater than the continued fraction. „, „ 151 o , 1 1 1 1 Thus, from -^ = ^ + ^^ ^^ l^-^' we have ^0^3 ^1^7 -^2^10 Ns^^ :^4^151 Do 1' A 2' A 3' A 14' A 45' 1 3^151 7_ 151 10 .151 47^151 ^"""^ 1^45' 2^45' 3^45' 14>45' The symbol ~, read difference between, is placed between two numbers to indicate that the less is to be subtracted from thegreater. .E.g-., 3~4 = 4~3 = 4-3 = 1. (ii.) The difference between any two consecutive convergents is 1 divided by the product of their denominators. ■ Thus, 3 7 1 7 10 1 10 47 1 2 1x2' 2 3 2x3' 3 14 3x14' etc. (iii.) Each convergent is nearer in value to the continued fraction than any preceding convergent. Thus, 151 3_16. 151 7^13. 151 10^_3_. 45'^1~45' 45 '^2 90' 45 '" 3 135' A 16 13 _3_ ^^^ 45 90 135 (iv.) The convergents of even order continually increase, but are always less than the continued fraction ; while the convergents of odd order continually decrease, but are always greater than the continued fraction. 3/10 .161. 7^47 E.g., i^Y'^lF' 2^14' 508 ALGEBRA. [Ch. XXXVI In a terminating continued fraction, the last convergent will, of course, be the continued fraction, and therefore neither greater nor less than itself. The proofs follow: Let V = di+ d2+ ds + (i.) The zeroth convergent is too small by J^ 1 In the first convergent, the partial quotient di is too small by -i----; 1 1 di+ hence — is too great, and therefore n-\ — is too great. di di In the second convergent, the second partial quotient (fe is too small by - — ; hence — is too great, and therefore c^i + — is also too great ; ds + di di finally is too small, and n -\ — — ~ is too small. di+ di di+ di And so on. (ii.) Since ■^*- . -^t+i _ N'tD„+\ -^ DtN;,^i we have only to prove The law holds for the first two oonvergenls. For i^^^ — 1^ ndi + 1 _ ndi -^(ndi + Ti _ _1_ Do Di 1 di I -di ~ di If it holds for any two consecutive convergents, it holds for the second of these two and the next convergent. We have N^Di+i ~ DiNi+i = Niidi+iDt + Dt-i)-^ Dt{Nidi+i + iVj-i) = NtDt-i - DiNi-i. Therefore, If the principle holds for it holds for A-1 D,' D, Dt+i CONTINUED FRACTIONS. 509 (iii.) ^^^±1 differs from Fonly in having dj+i where V has ■D*+i d*+i + T h — , = K, say. Then ^ = t^t-^t-i + ^^-2 y ^ KN^ + N^.-^ But N^-.y-^^- KJ^x + ^Vt-i ^ iVtJt-i - iVt-i A . 1 Z)4(^Z»t + i)*_0 and K But K>\ and Dk-i (^*+iA- Therefore, the error of — * is less than - — — • 510 ALGEBRA. [Ch. XXXVl Hence, to find a convergent whicli differs from the continued fraction by less than — , we have only to compute successive convergents up to — ^, wherein d^+iD^ < m. Ex. Find an approximation to 3.14159, correct to five decimal places. We have 3.14169 = 3 + J- ^- J- J- A_ J_ 1. 7+15+1 + 25+1+7 + 4 m, • , 3 22 333 355 The successive convergents are -, — , — , — , •••. 1 7 106 113 ii '' ^''' *^^^^ 25(113)^ The error of :pj-^ is less than ^^^^^^ and with greater reason less than , = .000004. 355 Therefore -— is the required approximation. To Reduce a Quadratic Surd to a Continued Fraction. 10. The general method may be illustrated by particular examples. Ex. Eeduce -^14 to a continued fraction. Since the greatest integer contained in y/1^ is 3, we assume V14 = 3 + i. di Then d, = _J_ = Vli±3 = ^Ai+3. ' V14-3 14-9 5 Since the greatest integer in this value of di is 1, we assume 5 di Then d,= ^ ^5(vl4 + 2)^Vl4+2. V14-2 10 2 CONTINUED FRACTIONS. 511 In like manner, we assume Then d,= ^ =21_Vl4 + 2)^ Vl4 + 2 ^ 1. Similarly, 1 -, etc. V14 - 3' Since this process may be continued indefinitely, we obtain an infinite continued fraction by substituting, in succession, the values obtained for di, d^, d^, •••. We then have ^ ^d, ^1+d, ^1+2+^3 ^1+2+1+6+ Observe that the value obtained for 1 d« = - V14-3 is the same as that for c?i, so that Therefore the partial quotients 1, 2, 1, 6, are repeated indefinitely. 11. A Periodic Continued Fraction is an infinite continued fraction in which the partial quotients are repeated in sets of one or more. Ex. Eeduce ~v to a periodic continued fraction. 5-V3 ^i 5-V3 1 Since -^^^< 1, we assume — -i^— = — . 512 ALGEBRA. [Ch. XXXVI And d,= 11 =lii^|V3)_3_,Z3^^,^l 4 + 3V3 -11 1 c?3 Likewise, d,= -—^—- = 3^3 + 5=10 +^. 6^6 — 5 fflj Finally, d, = ^ ^ = d^. Therefore, only the third and fourth partial quotients are repeated, and the required fraction is J_ J_ J_ J^ 1 1 1+1+6+10+5+ 10+*"' Application of Convergents. 12. It is often convenient to substitute for a fraction with large terms, or for a quadratic surd, a convergent with com- paratively small terms, provided that convergent approximates closely enough to the true value. 1 Ex. 1. T%VV 2 + 1+^ 6+^ 60 By Art. 9, we should expect the third convergent to be a close approximation, since the following partial quotient, 50, is large. We have _■ = -, ^ = -, -3=- Therefore, by Art. 9, the error of the third convergent is less than 1 50 x-20^' = .00005. Consequently, -^^ represents the true value of ^^^ correctly to four decimal places. CONTINUED FRACTIONS. 513 Ex. 2. Given ^U = 3 + :p- ^i- t^ ^ • • •, find the error of the seventh convergent. 1 + ^+1 + + The student may satisfy himself that ^ = — . ■^ ^ B, 120 The error of — ' < — - — < — - — < .OOOOll--. A 6(120)^ 86400 Therefore 4.|9 jg correct to four decimal places. To Reduce a Periodic Continued Fraction to an Irrational Number. 13. We will take as an example the result of Ex. Art. 10. Assume x = 3 + — -- —— , 1+2+1+6+1+ ' then x-3 = ~ ~ ~ — . 1+2+1+6+ Since the partial quotients 1, 2, 1, 6 are repeated in- definitely in that order, the continued fraction whose first partial quotient is the first periodic number (i.e., 1) at any stage, and which is continued indefinitely, differs in no respect from the given periodic continued fraction. For example, the periodic continued fraction which follows the heavy plus sign (+), in the value of a; — 3 above, is the same as the entire con- tinued fraction, which is the value of x — 3. We may therefore substitute x — 3 for the part of the con- tinued fraction which follows that particular plus sign. We thus have ^3^ 111 1 _111 1 ^ 11 + 3a; l+2+l+6 + a;-3 1+2+1+3 + a! 15 + 4a;' From this equation we obtain 4 a^ = 56, or a; = -^14. 14. If the continued fraction be not periodic from the beginning, we first reduce the periodic part by itself as above, and substitute its value in the given continued fraction. The latter is then a terminating continued fraction and can be reduced to a simple fraction, whose numerator and denominator will not, however, be rational. .5U ALGEBRA. [Ch. XXXVI 111111 the periodic part commencing with the third partial quotient. A 1111 5 + y Assume , y = tl — z = 7. — z = t7, — S~ ' ^ 3+5+ 3+5+2/ 16 + 32/' hence 3 2/^ + 16 2/ = 5 + ?/, and y^-15+V285 ^^^ ^ ^ 1 1 ^ 77-^. ^ 6 2+1+2/ 166 BXBRCISBS. Compute the successive convergents to 1 J_J_J LI. 2 2 + -i-iiA-i '1+2+2+1+3 ' 5+3+2+1+4" Reduce each of the following fractions to a continued fraction, find its convergents, and determine a limit to the error of the third convergent. 3. i\. 4. ^. 5. J|. 6. \\K 7. W- 8- t¥A- 9- VtV- 10- 27||. 11. .4751. 12. 5.0372. Reduce each of the following surds to continued fractions, find the first five convergents, and determine a limit to the error of the fourth convergent. 13. y/l. 14. V23. 15. v'2.5. 16. V29- 17- 2v'45. 18. I+A^. 19. ^^-V^ 20. 2+^. 21 li±^. 5 4 2 - v/3 5 Reduce each of the following periodic continued fractions to a surd -. 22 111 111 '1+2+3+1+2+3 + 23. 3 + ^- — ^ — i — - — •■■. 5+ 1 + 5+ 1 + 24. L_ 1_ 1_ 1_ 1_ 1_ .... 1+2+7+3+7+3+ 25. Express the decimal .43429 as a continued fraction, find its fifth convergent, and determine the limit to the error of this convergent. 26. Express the decimal 2.71828 as a continued fraction, find its seventh convergent, and determine a limit to the error of this convergent. 27. The true length of the equinoctial year is Ses* S*" 48"" 408. Reduce the ratio S"" 48"" 46' : 24'', to a continued fraction, and hence show how often leap year should come. CHAPTER XXXVII. SUMMATION OP SERIES. By Undetermined Coefficients. 1. When the nth term of a series is a rational, integral function of n, the sum of rt term& can be found by means of undetermined coefficients. The form which the sum of n terms of an arithmetical progression assumes will suggest a method of procedure. By Ch. XXVII., § 2, Art. 5, the sum of n terms of the A. P. 3 + 5 + 7 + 9 H ^ [3 + (n - 1)2] is 2 n + n\ Now observe that the sum of n terms is an integral function of n, of degree one higher than the nth term. In applying the method of undetermined coefficients, we start with this assumption. Ex. Find the sum of n terms of the series 1.2 + 2.3 + 3-4 + ... + n(ji + l) + .-.. Since the nth term, n{n + 1), is of the second degree, we assume 1.2 + 2.3 + 3.4 + -+ n(w+l)= A + Bn +' Cw^ + Dn^ (1) The validity of this assumption will be proved by mathe- matical induction. The method of proof will at the same time determine the values of A, B, C, D. We have now to prove that if this relation hold for the sum of n terms, it holds for the sum of n + 1 terms. Evidently the latter sum will involve n + 1 just as the sum of n terms involves n. That is, 1 . 2 + 2 . 3 + 3 . 4 + ■ • • + n (« + 1) + (» + 1) (n + 2) ^ A + B(n + l) + C(n + iy + D{n + 1)\ (2) 515 516 ALGEBRA. [Ch. XXXVII Subtracting (1) from (2), we have (n + l){n+2) = B{n + l)-Bn + C{n + lf-Cn'+D(n + lf-Dn', or 2 + 3n + n' = {B + C + D) + (2 C + 3 D)n + 3 Dn". This relation will hold, if 3D=l, 2C + 3D = 3, B+C+D = 2; that is, if -6=1, 0=1, D = l. The formula in the second number of (1) now stands A + ^7i + nP + ^nK This will hold for the first term, if l-2 = 4 + f + l+i; that is, if J. = 0. Therefore, 1 . 2 + 2 ■ 3 + — + n{n + 1) = ^n + n' + ln^= ^n(n+l){n+2). Since the formula ^n{n + l)(n + 2) holds for the first term, it holds for the sum of two terms ; then, since it holds for the sum of two terms, it holds for the sum of three terms ; and so on, to the sum of any number of terms. By separating Terms into Partial Fractions. 2. Ex. Find the sum of n terms of the series ' + . ...'. +■ ' (x+l){x+2)(x+3) (a;+2)(a;+3)(a;+4) (x+3)(a;+4)(a!+5) + ...+ 1 +-. (x+n){x+n+l)(x+n+2) We separate the nth term into its partial fractions and use the result as a formula. We have, by Ch. XXXV., § 4, Art. 1, Ex. 5, (x + n)(x + n + l){x + n + 2) 2 (x+n) x+n + 1 2 (a!+n + 2) SUMMATION OF SERIES. 517 Giving to n the values 1, 2, 3, •••, we obtain 1 ' +: 1 (a;+l)(a;+2)(a;+3) 2 (a; + l) a;+2 2(a;+3)' (a;+2)(a;+3)(a!+4) 2 (x+2) x+3 2 (x+i)' 1 +_ 1 1 1 +- 1 (a'+3)(a;+4)(a;+6) 2 (a; +3) x+i 2 (a; +5) 1 1 1 +- ' (x+n){x+n + l)(x+n+2) 2(x+n) x+n+1 2 (a;+n+2) We now have o^ 1 1 1 I 1 ' 2(a;+l) 2(a;+2) 2 (x+3) "^2 (a!+4)' c^ 1 L_ 1 I 1 ' 2(a;+l) 2(a;+2) 2 (a;+4) "^ 2 (a;+5)' and, in general, 1 1 1,1 S„ = + o " 2(a;+l) 2(a;+2) 2(a;+7i + l) 2(a;+n4-2) 1 As (a; + l)(a;+2) {x+n + l)(x+n+2)_ 1 m = OC, iSn = £ (a;+l)(x+2) Hence the given series is convergent. It is sometimes possible to determine the partial fractions by inspection. EXBBCISBS I. Find the sum of n terms of each of the following series : 1. 1.3 + 3.5 + 5-7 + -. 2. 12 + 22 + 33 + .... 3. 18 4. 3s ^_ 58 4. .... 4. 1 .3-5 + 3-5-7 + 5-7 -9+ ■••. 5. 1 . 22 + 2 . 3-i + 3 • 42 + .... 6. 1 . 32 + 3 . 52 + 5 . 7^ + .... 518 ALGEBRA. [Ch. XXXVII Find the sum of n terms, and the limit of this sum as re = co, of each of the following series : 7 I4.I4.I_L 8 1 I 1 I 1 I , '■ 17^ + 273 + ^4 +■■■• ^- rTi + 2. 5 + 3.6+ ■ 9 1 I 1 I ' I 10 1 + 1 + 1 +■... 1-3 -53. 5-75. 7-9 ' '1.3 -42. 4-53. 5-6 11, I + ^ + 1 +.... (x + l)(x + 2) (x + 2)(a; + 3) (x + 3)(x + 4) 12 '-^ + Krr-^ + ' - 13. x(x + 2)(x + 3) (x + l)(x+3)(x+4) (x + 2)(x + 4)(x + 5) 1 + « + «! 4 (1 + x) (1 + ax) (1 + ax) (1 + a^x) (1 + a^x) (1 + a^x) 14. A_.l + J_. 1+^.1+...+ »+i ^.1+.... 1-3 3 3.5 3-2 5.7 33 (2re-l)(2ra + l) 3" 15. J- + _2_+ 3 . n 1.31-3-51.3-5.7 ^1.3.5--(2re + l) Recurring Series. 3. In a geometrical progression any term after the first is formed by multiplying the preceding term by a constant multi- plier, the ratio of the series. To form such a series it is suffi- cient to know the first term and the ratio. Thus, given a =: 2, r — 3 x, the ser jes is 2-t-6a; + 18a;2 + 54af' + ---. 4. A geometrical progression. is a particular instance of a more general class of series. To form such series the first two or more terms, and as many ratios, must be given, as ex- plained in the following examples. Ex. 1. Given the first two terms, 3 -f 2 x, and the two ratios, a^ and —2x; form the series. To form the third term we multiply the first term by the first ratio, the second term by the second ratio, and add the resulting products. We thus have 3 • of + 2 x(—2 x), = — x^, the third term. SUMMATION OF SERIES. 519 To form the fourth term we multiply the second term by the first ratio, the third term by the second ratio, and add the resulting products. Then, 2x-a^+{-ae^{-2x), =4:0^, the fourth term. And so on. Therefore, the required series is 3+2a;-a;^ + 4a;3_9a^+.... Ex. 2. Given the first three terms, 1 — 4 « + 3 a^, and the three ratios, — 2 a;', a^, 4 a; ; form the series. To form the fourth term we multiply the first term by the first ratio, the second term by the second ratio, the third term by the third ratio, and add the resulting products. Therefore, l-(-2a^)-4a;.a^+3a,-2-4a;, =6a^, the fourth term. In like manner, -4a;(-2a^)+3a;2.a;2 + 6a^-4a;, =35 a^, the fifth term; and so on. Therefore, the required series is l-4a; + 3a,-2 + 6a^ + 35a^+".. 5. A Recurring Series is a series in which, from and after a definite term, each term is formed by multiplying each of two or more preceding terms by a constant multiplier, and adding the resulting products. Such a series is also called a Compound Geometrical Progression. A recurring series is said to be of the first order, if one ratio be used, as in an ordinary geometrical progression ; of the sec- ond order, if two ratios be used ; and so on. 6. If Ml, U2, Us be any three consecutive terms of Ex. 1, Art. 4, then u^ = x^u^ — 2 xu^, or Mj + 2 xMj — ^u-^ = 0. That is, any three consecutive terms of a recurring series of the second order are connected by a definite relation. 520 ALGEBRA. [Ch. XXXVII The expression 1 + 2 x — a;^, formed by taking the coefficients of Ug, u^, ih in the above relation, is called the Scale of Relation of the series. In like manner, it follows that any four consecutive terms of the series in Ex. 2, Art. 4, are connected by the relation Ui — 4 XMj — a^Mj + 2 a?Ui = 0. Therefore the scale of relation of the series is Observe that the scale of relation is obtained by subtracting the ratios in reverse order from unity. 7. The follovi^ing examples will illustrate the method of finding the ratios, and the scale of relation of a recurring series, when a sufficient number of terms are given. Ex. 1. Find the ratios and the scale of relation of the series, 2 + 5 a; - a;2 + 11 a;s - 13 a;^ + 35 a;' - 61 a" + •••. It is evident that this is not a recurring series of the first order. If it be a series of the second order, then 2h+5x-k = -r', (1) Sx-h-x'-k^lla^, (2) wherein h and k are the required ratios. Solving these equa- tions for h and k, we obtain h = 2 x', k = — x. We find that these ratios give the fifth and following terms of the series. Therefore the series is of the second order, and the scale of relation is 1 _ (_ a) - 2 a;^ = 1 + X - 2 x'. Ex. 2. Find the ratios and the scale of relation of the series l-3a;-2a^ + 3a^-|-10x*-|-6a!«-17a;'= . This series is evidently not of the first order. Assuming tentatively that the series is of the second order, and proceeding SUMMATION OF SERIES. 521 as in Ex. 1, we find that the two ratios thus obtained do not give the fifth and following terms. We therefore try three ratios. Then h-3x-k-2x'-l = 3a^, -3x.h-2x'-k + 3a^-l = 10oii', -2x''-h + 3a^.k + 10a^-l = ^a:^. Whence, h = — a^, k=:~2ijir', l = x. These ratios give the seventh term. Therefore the series is of the third order, and the scale of relation is 1 — X + 2 ay' + a^. 8. When the scale of relation involves two ratios, two equa- tions are necessary in order to determine these ratios, as in Ex. 1, Art. 7. To form the first equation, three terms must be given, and to form the second equation, a fourth term must be given. Therefore, when the scale of relation involves two ratios, at least four terms of the series must be known. In like manner, we infer that, when the scale of relation involves three ratios, at least six terms of the series must be given. In general, when the scale of relation involves n ratios, at least 2 n terms of the series must be given. The Sum of a Recurring Series. 9. Let Ml + M2 + «3 + l-Mn-2 + Mn- I + Mn + ••• be a reouri'ing series of the second order, and let h and k be the ratios. Then, us = uih + uik, m = Uih, + uzk, (1) Un = Un-ih, + Un-\k. Let 8n stand for the sum of re terms ; that is, ;S>, = Ml + M2 + Ms + «4 + ■•■ + Mn-2 +M,i-1 + Un- (2) Now, substituting for each term, after the second, in (2) its value given in (1), we have 522 ALGEBRA. [Ch. XXXVII Sn = Ml + M2 + (Ulh + Uik) + (u^h + usk) + — + (M„-2ft -t- «»-!*), = Ml + tt2 + K'"'! + «2 + ••■ + ''n-z) + k(U2 + Ms + •■• + M„_l). The coefBcient of h is evidently Sn - Wn-i — «« ; and the coeflBcient of k is Sn— mi — ««■ Therefore, (S„ = Mi + a2 + ^CfSn - m»-i - ««) + !i:(Sn - mi - M„), and (1 - ft - fc)^ = Ml H- M2 — A;Mi — {hUn-l + ftM„ + ^M,,). Whence *5 = "i + "2 ~ ^''i _ ^M/.-i + hun + ku„ ' " 1 — 7i — A ]— A — A; 10. If the given series be convergent, m„ = 0, and «„_i = 0, as n = 00. Therefore, — " ! " ' = 0- 1 — h — k Conversely, if this fraction approach zero, as n increases indefinitely, „ ^ Ml + 1/2 — ^«i /1^ a definite finite number, and the given series is convergent. Consequently, the fraction in (1) may be taken as the sum of an infinite recurring series of the second order, when the series is convergent. In like manner, for an infinite recurring series of the third order, we obtain 1-h-k-l h (m„„2 + M„-l + Mn) + Ti (u„--i + U„) + IU„ 1 _ 7t _ A; - Z Then, as above, if the infinite recurring series be convergent, we have e ^ »i + t<2 + 1/3 — A-Wi - /(t/| + Ui) i-oN l-h-k-l ^' as n increases indefinitely. SUMMATION OF SERIES. 523 11. Ex. 1. Find the sum of flie infinite recurring series 2 + 5 X - x' + U a^ - 13 x' + ■•: From Ex., Art. 7, we have h = 2 x-, k = — x. Therefore, by Art. 10 (1), g ^ 2 + 5a;-(-a;)2 ^ 2 + 7a " i_2a^-(-a;) 1 + x - 2 a!^' for such values of x as make the given series convergent. 12. It is evident that the recurring series in the example of the preceding article can be obtained from its sum either by division or by the method of undetermined coefficients. On this account the fraction obtained as the sum of an infinite recurring series is called the Generating Fraction or the Generating Function of the series. Convergency of a Recurring Series. 13. The convergency or divergency of an infinite recurring series may be determined by the methods given in Ch. XXXIII., "or by Art. 10. It is therefore important to obtain the general term of such a series. When the denominator of the generating fraction can be resolved into real binomial factors, the general term can be found as in Ch. XXXV., § 4, Art. 2. Ex. Examine the series 2 + 6a; — a;^ + llar'— 13a;^H . The generating fraction of the series was found to be (Art. 11) 24-7a; l+a;-2a;2 The general term of the expansion of this fraction is, by Ch. XXXV., § 4, Art. 2, Ex. 1, a;»[3 + (- l)"+i2»]. The ratio of convergence is 1 + (_ l)»+i a;"[3 + (-ir'2"] „ 2"^^ ^_ a^-[3 + (-l)»2»-r-^ 3_ .^yV- ^^• 524 ALGEBRA. [Ch. XXXVII Therefore the given series is convergent when x is numeri- cally less than ^. BXBBCISBS II. Find the ratios and the sum of n tei'ms of the following recurring series : 1. 1 + 2 + 5 + 12 + 29+ •••. 2. 1-3-5+1 + 11 + •••. 3. 3 - 7 + 15 - 31 + 63 - 127 + 255 . 4. 2-3 + 5-21 + 60-188 + 577 . Find the ratios and the generating function of the following recurring series : 5. '2-x-3x^ + 5x^ + x* . 6. 1 + 3x-9x3- 9!c4 + •••. 7. 1 -4:x + 6x^ -Ux^ + 26x* . 8. 1 + 4a; - 5x2 + llj-x^ -22J:a;*+ .... 9. 3 - 6x-9x2-10x3- lOJx* ■•■. 10. 5-2X+ Ilx2+'l6xs + 65x*+ 178x6 + 551x6+ .... 11. 1 -2x + 3x2 + 10x3 + 13x4 + 22x^ + 51x6+ .... 12. 2-3x-2x2+ 16x3-24x4 - 16x5+ i28x6 . 13. 3 + 5x- 6x2 -15x3 + 8x4 + 37x5- lOx* . 14. 1 -3x + 7x2 + 2x3 -25x4 +42x5+ 14x6 , 15-19, Find the general term of each series in Exx. 7-11, and hence determine for what values of x each series is convergent. Method of Finite Differences. 14. In an arithmetical progression, any term after the first is formed by adding a constant difference to the preceding term. If, therefore, each term of an A. P. be subtracted from the following term, we obtain a series of differences each equal to the common difference. If each term of this series be sub- tracted from the following term, we obtain a series of dif- ferences each equal to zero. Thus, 1, 3, 5, 7, 9, .••, 2, 2, 2, 2, ..., 0, 0, 0, .... SUMMATION OF SERIES. 525 15. An arithmetical progression is a particular instance of a more general class of series, from which, by continuing to form series of differences as in Art. 14, a series of differences each equal to zero is finally obtained. Thus, Given series, 1, 4, 9, 16, 25, •••, 1st differences, 3, 6, 7, 9, •••, 2d differences, 2, 2, 2, •••, 3d differences, 0, 0, •••. 16. The first series of differences is sometimes called the First Order of Differences ; the second series of differences, the Second Order of Differences ; and so on. The nth Term of the Series. 17. To find any term of an arithmetical progression, it is suflcient to have the first term of the given series and the first term of the first order of differences (the common difference). To find any term of the more general series, it will prove to be suflScient to have the first term of the given series and the first term of each order of differences. . Observe that any term of the given series is found by adding to the preceding term the corresponding term of the first order of differences. Thus, in the example of Art. 15, 4 = 1 + 3, 9 = 4 + 5, 16=9 + 7, -. In like manner, any term of any order of differences is formed by adding to the preceding term the corresponding term of the next order of differences. Thus, in the first order of differences, 5 = 3 + 2, 7 = 5 + 2, 9=7 + 2, -. Now, let Oj be the first term of the given series, di be the first term of the first order of differences, ds be the first term of the second order of differences, and so on. 526 ALGEBRA. [Ch. XXXVII First, forming the second terms of the series as indicated above, then the third terms from the second terms, and so on, we have : Given series, Oi, fti + di, ai + 2di+d2, aj+3di+3 d^+ds, •■■, 1st order of dif., d^, dj + dj, di + 2d2 + d3, •••, 2d order of dif., dj, dj + dg, • • •, 3d order of dif., dj, ••■. We now have ai = Oi, a2 = Oi + di, ag = ttj + 2 dj + d^ etc. Observe that the numerical coefficients in the expression for ttj are the same as in a; + y, those in the expression for a^, the same as in the expansion of (x+yf, etc. That is, in the formula for each of the first four terms, the numerical coefficients are the same as in the expansion of a power of a binomial whose degree is one less than the nuviber of the term. We will now prove that, if a similar formula hold to any term, it holds to the next term. Let us assume that the formula holds to the mth term. Then, o„ = ai + (71 - l)di + ("-l)(«-2) t^ ^ ... + (n-l)(n-2)...(n-r) ^^ ^ _ \z Now, the first order of differences is a series of the same character as the given series. If, therefore, the above formula hold for the reth term of the given series, it holds for the nth term of the first order of differ- ences. Let „di stand for this term. Then, „di = di + (TO - 1)^2 + ^" ~ ^^„^" " ^^ ^8 + - + (n-\)(n-2)...(n-r + l) g^ ^ _.__ ■ |r — 1 But o„+i = a„ + „di. Therefore, [2 [r SUMMATION OF SERIES. 627 Hence, if the formula hold to the «th term, it holds to the (re + l)tL. But, as was first proved, it holds to the fourth term ; therefore it holds to the fifth ; and so on. We therefore have [* 11 Ex. Find the 15th term of the series 1+4 + 9 + 17 + 29+ •••. We have ai = l, 4, 9, 17, 29, •••, c?, = 3, 5, 8, 12, -, d, = 2, 3, 4, ..., C?3=l, 1, •••, d, = 0, -. Then, 1_L1( Q . 14-13 o , 14.13.12 ^j,Q ai5 = l +li-3 + — — - • 2-] — = 6»9. The Sum of n Terms of the Series. 18. Let it be required to find the sum of n terms of the series «! + ^2 + 03 + a4+ •■■«»+ •"• We form a new series of which the first term is 0, the second aj, the third Oi + as, the fourth ai + aa + a^, and so on. The {n + l)th term of the assumed' series i« evidently the required sum of n terms of the given series. We then have 0, tti, ai + tts, ai + fflo + a,, a^+az + as + ai, ■■•, «!, Clit (hi (^iJ '"> "d ; d-i, , Since the first order of differences of the new series is the same as the given series, the second order of differences of the new series is the same as the first order of differences of the given series, and so on. 528 ALGEBRA. [Ch. XXX VII Observe that is taken as the first term of the assumed series in order that the first order of differences may begin with ffli and therefore the first terms of the orders of differences following be in order d^, di, d^, ■■•. We now find the (n + l)th term of the assumed series, and take the result as the sum of n terms of the given series. In applying the formula of the preceding article, we must let Oj = 0, di — tti, di = di, •••■ If S„ stand for the required sum, we have wherein %, di, d^, ••■, have the same meanings, with reference to the given series, as in the preceding article. Ex. Find the sum of the squares of the first n natural numbers 12 + 22 + 32 + 4^+ -+w'- We have i-l)+2 w(« - 1) (n -2)] = ^n(n+l) (2 71 + 1). Piles of Cannon-balls. 19. Cannon-balls, oranges, and other spherical objects are usually piled in the form of pyramids, consisting of horizontal layers. (i.) Square Pyramids. — When the base of the pyramid is a square, the top layer evidently contains 1 ball, the second layer 2^ balls, the third layer 3^ balls, and so on. Therefore the number of balls in a square pyramid of n layers is the sum of n terms of the series 12 + 22+32+ •■■ +n2+ -. SUMMATION OF SERIES. 529 Hence, by Ex., Art. 18, S„ = ^n(n + l){2n + 1). Observe that n is also the number of balls in a side of the bottom layer. (ii.) Triangular Pyramids. — When the base of the pyramid is an equilateral triangle, the top layer contains 1 ball, the second layer 3 balls, the third layer 6 balls, and so on. Hence, the number of balls in a triangular pyramid of n layers is the sum of n terms of the series 1+3 + 6 + 10 + -. By Art. 18, we have wherein n is also the number of balls in a side of the bottom layer. (iii.) Rectangular Pyramids. — In a rectangular pyramid, the first layer consists of a row of balls, the second layer of two rows, the third layer of three rows, and so on. If the first layer contain Ic balls, the second layer contains 2 (fc + 1), the third 3 (fc + 2), and so on. Hence the number of balls in n layers is the sum of n terms of the series k + 2(Jc + l)+3{k + 2)+4.(k + 3) + -. We have ai = fc, di = k + 2, 6,^ = 2, c?3 = 0, •••. T liGrcforG S^ = nlc+'^^^Qc + 2)+ ^^^-^^^-'^^ .2 = ^«(2n^ + 3w& + 3A;- 2) = |m(« + 1)(2« + 3A; - 2). If m be the number of balls in the length of the bottom layer, then m = k + n — l, whence k = m — n + l, wherein n is also the number of balls in the breadth of the bottom layer. The preceding formula may now be written S„ = ^/7(/j + l)(3/n-/j + l). 530 ALGEBRA. [Ch. XXXVII Ex. 1. Find the number of balls in a rectangular pyramid, the sides of the bottom layer containing 14 and 20 balls re- spectively. Since the number of layers is the same as the number of balls in the breadth of the bottom layer, we have Su = i-U- 15- (60 - 14 + 1) = 1645. Ex. 2. Eind the number of oranges in an incomplete tri- angular pyramid, if the number in a side of the top layer be 5 and the number in a side of the bottom layer be 10. The pyramid, if complete, would consist of ten layers, and hence would contain | • 10 • 11 12, = 220, oranges. The part which is wanting would be a pyramid of four layers, and hence would contain |^ • 4 • 5 • 6, =20, oranges. Therefore the required number of oranges is 220 - 20, = 200. EXERCISES III. 1. Find the ninth term and sum of the first nine terms of the series 1, 4, 8, 13, .... 2. Find the twelfth term and the sum of the first fifteen terms of the series, 3, 2, 3, 6, •••. 3. Find the fourteenth term and the sum of the first twenty terms of the series, 5, 3, 4, 7, 11, .... 4. Find the eighth term and the sum of the first nine terms of the series 2a- Bb, 3 a -2b, 4a-2b, 5a-3b, •••. Find the nth term and the sum of the first n terms each of the series : 5. 2, 6, 16, 32, .... 6. 5, 3, 13, 35, .... 7. 5, 1, -6, - 14, -21, .... 8. 6, 3, -4, -13, -22, .... 9. a, 2a + 1, 3a -I- 3, 4a -I- 6, .... 10. Find the sura of the cubes of the first n natural numbers. Find the number of balls in 11. A square pyramid, having 12 balls in each side of the base. 12. A triangular pyramid, having 15 balls in each side of the base. 13. A rectangular pyramid, having 12 and 5 balls respectively in the length and breadth of the base. SUMMATION OP SERIES. 531 Find the number of balls required to complete 14. A square pyramid, having 144 balls in the top layer. 15. A triangular pyramid, having 5 balls in a side of the top layer. 16. A rectangular pyramid, having 22 and 8 balls respectively in the length and breadth of the top layer. 17. How many balls in an incomplete rectangular pyramid, having 16 and 12 balls respectively in the length and breadth of the top layer, and 25 in the breadth of the bottom layer ? 18. How many layers in a square pyramid, containing 91, = 7 . 13, balls? 19. A rectangular pyramid of 8 layers contains 1860 balls. How many balls are in the top row ? 20. The nuniber of balls in a square pyramid, increased by 91, is equal to twice the number in a triangular pyramid of the same number of layers. How many layers in each pyramid ? 21. Show that the number of balls in any square pyramid is one-fourth of the number in a triangular pyramid having twice as many layers. Interpolation. 20. The following example will indicate the nature of an important application of the method of finite differences. Ex. Given 1= = 1, 2^ = 4, 3^ = 9, -, find the value of (2^y. We have Oj = 1, di = 3, dj = 2, dg=0, •••. Since the first term is the square of 1, the second term the square of 2, and so on, we may call the square of 2| the 2^th term. Then, a^ = l + !-3+y-2 = -?35, which agrees with the result obtained by squaring. 21. Interpolation is the process of inserting between the terms of a given series other terms which conform to the law of the series. In thus interpolating terms in a given series it is necessary, as we have seen, to give to m a fractional value in the formula for the nth term. Interpolation is extensively applied in astronomy, and is also used in computing numbers intermediate between those given in mathematical tables. 532 ALGEBRA. [Ch. XXXVII Ex. 1. From a table of square roots we obtain y'4 = 2, V5 = 2.2361, V6 = 2.4496, V^ = 2.6458, V8 = 2.8284; find V5.25. We have ai = 2.0000, 2.2361, 2.4495, 2.6458, 2.8284, ••., rii = ;2361, .2134, .1963, .1826, d2 = -.0227, -.0171, -.0137, •••, d3=.0056, .0034, ^^ = -.0022, It should be kept in mind that in forming differences we always subtract a number from the number on its right, thereby sometimes obtaining negative remainders. Since -^5 is the second term and -y/6 is the third term, for V5.26 we take n = 2.25, = f . Then, a„ = 2 + I (.2361) + i^ (-.0227) * if. + J ■ i(- f) (.0056)+ ^ • H~ t) (- l) -(.0022) [3 |4 = 2 +.2951 - .0036 - .0002 - .00003 = 2.2914. In such examples it is not possible to obtain a series of dif- ferences whose terms are zero. But, by taking more terms in the given series, more orders of differences, with terms nearer to zero, can be obtained. The results are approximate, and the work is to be carried only so far as it will affect the last decimal place in the values of the given terms. EXERCISES IV. 1. Given log 30 = 1.47712, log 31 = 1.49136, log 32 = 1.50515, log 33 = 1.51851, log 34 = 1.53148 ; find log 31.8. 2. Given ^9 = 2.0801, ^10 = 2.1544, .J/11 = 2.2240, .J/12 = 2.2894, .J/13 = 2.3513 ; find ^11.25. 3. The latitude of a place on the earth's surface is obtained by observing the altitude of the sun at noon, and adding to the complement of the SUMMATION OF SERIES. 533 altitude of the sun's declination. On Oct. 31, 1896, tlie altitude of the sun at Philadelphia was found to be 35° 36' 37".3. The Nautical Almanac gives the following values of the sun's declination : Oct. 30, 1896, at IS-" 29' past 7 a.m., —14° 3' 3".8 ; " 31, " " " " " —14° 22' 28". 7; ■ Nov. 1, " " " " " — 14° 41' 89".8; " 2, " " " " '• —15° 0' 36".6. Find the latitude of the place of observation. 4. The length of a degree of longitude on the equator is 69.17 miles ; at 10° N. latitude, 68.13 miles ; at 20°, 65.03 miles ; at 30°, 59.96 miles ; at 40°, 53.06 miles ; at 50°, 44.55 miles. What is the length of a degree of longitude at 36°15' N. latitude ? 8. The altitude of a star as seen with the eye is greater than it really is, on account of the refraction of the rays of light from the star by the earth's atmosphere. The altitude of the star Fomalhaut was observed to be 19° 15' 37".3. Find its correct altitude, using the following table : Altitude, Kefraction. Altitude. Refraction 16° 3' 18".4 22° 2' 21 ".9 18° 2'55".8 24° 2' 8".9 20° 2'37".3 26° 1'57".9 6. The temperature of a litre of water is 23°.4 C. Find its weight, using the following table of densities, the unit being the density of water at the temperature of maximum density : Temperature. Density. Temperature. Density. 10° c. .99974 25° C. .99713 15° C. .99915 30° C. .99577 20° C. .99827 7. Nitric acid is diluted with water so that the solution contains 22.4% acid. Find the weight of a litre of the solution, using the following table of specific gravities : % of acid. Sp, gr. % of acid. Sp. gr. .999 20 1.121 10 1.058 30 1.187 CHAPTER XXXVIII. THE EXPONENTIAL AND LOGARITHMIC SERIES. 1. In this chapter we shall give two important series, and from them derive formulae for computing naperian and common logarithms. The Exponential Series. 2. The expression a", in which the variable enters as an exponent, is called an Exponential Function. 3. Let y = a", then log^ y = x log^ a, (1) wherein e is the base of the naperian system of logarithms. From (1), we obtain y = e"''°ee«, or, a" = e'"'"^^". (2) If, therefore, we expand e'^'^^e" in a series, to ascending powers of x, the result will be also the expansion of a'. 4. If m > 1, the expansion of (1 + -) by the Binomial Theorem will be a convergent series, by Ch. XXXIV., § 2, Art. 2. Therefore, nj n \2 n' [3 n" . . w(n — 1) ••• (m — r + 1) 1 , + ■ ■ • T ^ — T^ ■ — + ■■•) -1 , 1 , ^ , \ VV "V I + ^ -V V "V V ^ + 534 EXPONENTIAL AND LOGARITHMIC SERIES. 535 however great n may be. If, then, we let n increase indefinitely, 1=0, ?=0, etc. n n Hence, when w = 00, /^l4.^y= 1 + 1 + 1 + 1 + ... + -^+ •••• Denoting the series 1 4- 1 +i-f.-- + ••■ by the letter e, as in Ch. XXIX., Art. 22, we have 5. It follows from the result of the preceding article that the expansion of e" can be obtained from the expansion of K 1 +- ) , by letting n increase indefinitely. Wehave[(l+lJJ = (l+^)" . 1 , nx(nx—V) 1 , nx(nx—l)(nx—2) 1 n [2 w' [3 w' , nx{ nx — 1) ■•• {nx — r + 1) 1 [r n' x\ = 1+^+ ^ ,„ •"' + (-^), <-i)e-s + |3 xix )"-f * ~—\ Letting n increase indefinitely, we have This series is found, by Ch. XXXIII., Art. 21, to be con- vergent for all finite values of x. 536 ALGEBRA. [Ch. XXXVIII Therefore, replacing a; by a; log, a, and remembering that e'^^ie'^ — a", we have a'=l+xXog.a + '^Q^ + ^^^^^-^.... (2) This Exponential Series is convergent for all finite values of x. The Logarithmic Series. 6. From the expansion of a' in Art. 5, we have a' - 1 = a; log, a + ^!fl^ + ^S^ + .... Dividing by a., ^Izill = log.a+^fl2|^+^!M^+ .... The series in the second member is convergent for all finite values of x. Therefore, by Ch. XXXV., § 1, Art. 3, its sum approaches log^a, as aj = 0. Whence, l™^lizl = log, « ; a; = X or, replacing a; by a, ^^ ° ~ =logea. (1) a = a 7. Substituting 1 + a; for a in the relation of the preceding article, we have log.(l + a,)= lim (l + «^)--l . (1) a = \J « If a; < 1 numerically, the expansion of (1 + a;)" by the Bino- mial Theorem is a convergent series. Therefore, (l+a;)'- = l + «a; + «("-^) a^+ «("-l)(«-^) a^+..., [2 [3 and from (1), iog.(i + .)=i\«; whence, iog,(l + .r) = jr-^' + ^'- X+ •••• (2) 2 3 4 EXPONENTIAL AND LOGARITHMIC SERIES. 537 It is important to keep in mind that this expansion holds only when x lies between — 1 and + 1. It is therefore of little practical value in computing logarithms. A series which can be used for this purpose will be derived in the next article. 8. Eeplacing a; by — a; in (2), Art. 7, we have log.(l -a;)=- a;-|-|-|*- .... (1) Subtracting (1) from (2), Art. 7, log.(l+cc)-log(l-a;), =log,^, =2(^a; + | + | + ...| (2) In (2), let 1+:^=1±J;^ whence x = Then log. 1 — X n 2n + 1 1 + n n = log, (1 + n) — log. w, = 2 +^7K—r^.+^777zrr7Vs+ 2n+l 3(2 w+1)' 5{2n+iy + ■ .(3) or %,(l + «) = %,«+2[2^+3^g^3+g^2^ The series (2) is convergent when as < 1, and positive. Therefore the series (3) is convergent when ,- — — r < 1 ; or?i>o. Hence, this series is equal to log,(l + n) for all positive values of n, however great. Computation of IiOgarithms. 9. Naperian Logarithms. — By means of the formula derived in the preceding article, the naperian logarithms of all positive numbers can be computed. Letw = l. Then, log.2 = log,l+2(|+3_L + ^-l3,+ . = + 2 (.3333333 + .0123467 + .0008231 -f .0000663 + .0000057 + .0000005) = .69315, to five places of decimals. 538 ALGEBRA. [Ch. XXXVIII Letting w = 2, 4, 6, •••, we have log. 3 = log.2 + 2(| + ^ + ^ + ...) =1.09861 log, 4 = 21og,2 =1.38629 log, 5 = log,4 + 2(|+^, + ^+...) =1.60944 log, 6 = log, 2 + log, 3 =1.79176 log, 7 = log,6 + 2^1 + ^,+^+...j=1.94591 log, 8 = 3 log, 2 = 2.07944 log, 9 = 21og,3 =2.19722 log, 10 = log, 2 + log, 5 = 2.30259 and so on. It is necessary to compute only logarithms of prime num- bers, by the formula. Logarithms of composite numbers are obtained by adding together the logarithms of their factors. As the numbers increase, fewer terms in the formula are required to compute their logarithms to a given decimal place. 10. Common Logarithms. — Let n = logio N, or 10" = N. Taking logarithms, to base e, n log, 10 = log, N. Whence, n = :; — - • log, N, log, 10 °^ '°^'» '^= i^ • '''' ^' = 2:35259 • '°^-^' = -^^'^^ ^''' ""■ The number 1 1 log, 10 2.30259 = .43429 is called the Modulus of the common system with respect to the Naperian system. Hence, to compute the common logarithm of any positive number, multiply the naperian logarithm by the modulus of the common system, .43429. EXPONENTIAL AND LOGARITHMIC SERIES. 539 E.g., logio 2 = .43429 x log. 2, = .43429 x .69315, = .30103. Only common logarithms of prime numbers should be thus obtained. Common logarithms of composite numbers are ob- tained by adding together the logarithms of their factors. 11. Evidently, the naperian logarithm of a number can be obtained from the common logarithm by dividing by .43429, or by multiplying by 2.30259. The number 2.30259 is called the Modulus of the naperian system with respect to the common system. EXERCISES. 1-10. Compute the common logarithms of all integers from 11 to 20 inclusive. 11. Find the sum of I-- + 3-J+ •■■• 12. Find the sum of i-i-;i + s-;^ • 3 2 32 3 3» Expand, to ascending powers of x : 18. J(e"+e-''«)- 14. ^.(e" - e-"). X m 15. Prove Ji™ (r+'^Y = e-. 16. Show that -i- + „^ '^ . ■ ■ + „, \ ■ ■ - + - n + l 2(?j + 1)2 3(ji + 1)8 = 1_J_ + _L-.... B 2n» 3n8 17. Show that «^ + l('«^y + U«^y + - = Iog.a-log.6. 18. Show that, when a; > 1, logv'K2-l = logK-(272 + 4^ + 6ye + "T CHAPTER XXXIX. DBTERMINANTS. 1. The equations a^x + a^y = 0, (1) b,x + b^ = 0, (2) have evidently the solution 0, 0. Let us inquire if they can be simultaneously satisfied by values of x and y other than 0, 0. Multiplying (1) by bz, and (2) by a^, OibiX + Osb^y = 0, aj)ix + a^biy = 0. Subtracting, {a^bi — a^b^x = 0. This equation will be satisfied by a value of x other than 0, if Oib^ — ajfii = 0. The same result would have been obtained, had we first eliminated y. m 2. The expression Oi&g — ois&i (!•) is an example of a form which occurs frequently in mathe- matics, and for which a special notation has been devised. Such an expression is called a Determinant. The determinant (I.) is usually written in a square form : 1 2 The positive term of the determinant, obtained from the square form, is the cross-product from upper left to lower right ; the negative term is the cross-product from upper right to lower left. E.g., = 4.5-2.3 = 14; 540 3-1 2 7 = 3-7-(-l)2 = 23. DETERMINANTS. 541 3. We shall frequently call the symbolic form in which the determinant is written the determinant. The advantage of writing determinants in this form will be made evident in subsequent work. 4. The Elements of a determinant are the unconnected sym- bols in its square form ; as cii, a^, b^, b^ in (I.). A Row of a determinant is a horizontal line of elements ; as Oi, tta- A Column of a determinant is a vertical line of elements ; as The rows are numbered first, second, etc., counting from top to bottom ; and the columns from left to right. The square form has two diagonals. The Principal Diagonal is composed of the elements from the upper left-hand corner to the lower right-hand corner; as in (I.), The Order of a determinant is the number of rows or columns in its square form, ai as Thus, h h is a determinant of the second order. EXBROISBS I. Find the values of the determinants : 1. 5 3 8 5 2. _ 9 8 7 6 ■ 3. 11 -5 7 4. 4 8 5 10 5. X y 2 3 6. b 2a 2c 6 7. a + b a — b a — b a + b • Verify the following identities : 8. ai 6i = 0. 9. 6i 62 = ai 61 02 62 ■ 10. bi 62 =- 61 62 ai 02 11. ni an ai Os = 0. 12 ka\ 62 = kai kh 02 62 = *:: 02 62 13. Oi -f ai O2 61 -I- /3l 62 Oi O2 ai CEs — 61 62 + iSl &2 542 ALGEBRA. [Ch. XXXIX 5. Next, let us inquire -when the equations OiW + tta?/ + a^% = 0, h-^x + h^y + 638 = 0, CiCB + C22/ + CsK = 0, are satisfied by values of x, y, «, other than 0, 0, 0. Multiplying (2) by Cg, and (3) by 63, hjC^x + 62C32/ + fegCja = 0, hiC^x + &3C2?/ + 63C32 = 0. Subtracting, (61C3 — h^c-^x + (62C3 — b^c^y = 0. (1) (2) (3) Whence &2C3 — &3C2 In like manner, from (2) and (3), we obtain „ — blCj — 62C1 ^^ Substituting these values of y and z in (1), a,x - a, . ^l^::iM . a, + a, . ^^iZlM . a; = 0, &2C3 — &8''2 &2C3 — &3C2 or, [oi (feaCg — 63C2) — a2(&iC3 — &3C1) + as(&iC2 — &2C,)] a; = 0. This equation will be satisfied by a value of x other than 0, if the expression s — ^sh), etc. (VI.) 544 ALGEBRA. [Ch. XXXIX In (V.) and (VI.), — {a^Pi — a^^ is the minor with respect to 61 and is denoted by B^, etc. The above expansions can now be written \Bi + bA + bsB^, a^Ay + h^By + Cid, etc. In general, the minor with respect to any element is, except for sign, the determinant obtained by deleting the row and column in which the element stands. The signs to be prefixed to the determinants thns obtained alternate + and — , beginning with Oj, and going either hori- zontally or vertically, but not diagonally. Thus, the sign of the minor of a^ is +• of bj is — , of 62 is +, of 63 Is — , of c^ is + , etc. 4-1 2 3 5- 16 7 Ex. Find the value of Since the second row contains a zero element, the work is simplified by expanding in terms of the elements of this row : -1 2 6 7 +0 4 21 1 7 = 57 + 0-125= -68. 9. The terms of a determinant of the third order can also be obtained directly from the square form. Removing parentheses in (II.), Art. 5, and writing positive terms first, we have • ajb^s + fhbsCi + afi-fii — a-Jj^c^ — ajiiC^ — afi^. The terms are obtained by multiplying together the elements connected by continuous lines in the following schemes : POSITIVE TERMS NEGATIVE TEEMS DETERMINANTS. 545 Ex. 1 2 3 4 5 6 7 8 9 = 1 5-9 + 2 - 2 6-7 + 3-4-8-1.6 4-9- 3-. :0. Principles of Determinants. 10. The form in (VI.), Art. 8, can be regarded as the expansion either of a, a, C, C2 , or of a; 61 Ca That is, the value of a determinant remains the same when the rows are changed into columns, and the columns into rows. 11, If all the elements of any row, or column, of a deter- minant be 0, the determinant vanishes. For, expanding in terms of the zero elements, we have = ■0 t6, b, -0 61 62 = 0. 12. If any two rows, or columns, of a determinant be inter- changed, the sign of the determinant is changed, but its numerical value remains the same. For, ai(62C3 - &3C2)- asibiCg - b^Ci) + ag(biC2 — b^c^) - [b\{a^ — asPi)— h{aiCs — a^Ci) + b^ia^c^ — a^c^), bi 62 h Oi aj as Ch ag as h b. &3 = — Cl C2 C3 or That is, the principle enunciated holds when two adjacent rows are interchanged. To interchange the first and third rows is equivalent to interchanging the first and second rows, then in the resulting determinant the second and third rows, and finally in the last determinant the second and first rows. t6 ALGEBRA. [Ch. XXXIX These transformations give a, flj «3 6, 6a 63 b, b. &3 Ci C2 C3 &, &2 63 = - a, tta aj = Cj C2 C3 = — &1 62 &3 Cj Cj C3 Ci C2 C3 Oi fflj «3 tti aj 013 13. If two rows, or two columns, of a determinant be identical, the determinant vanishes. Let D be the value of the determinant. If the two identical rows, or columns, be interchanged, the value of the resulting determinant will be —D, by the preceding article. But this interchange of identical rows, or columns, does not alter the given determinant. Hence D = — D, and therefore Z) = 0. 14. Let D = «! as as &1 &2 h Cl C2 C3 — ai-4i + a2-42 + 03.43, (1) = a^A, + b,B, + c,Ci, etc. (2) Then b^Ai + b^A^ + 63-43 differs from (1) only in having 6's where (1) has a's. That is, 61 62 h biA^+b2A2+bsA3= 61 62 63 =0. • Cl C2 C3 In like manner, 02^, + 62£i + CjCi = 0, etc. 15. If each element of a row, or column, of a determinant be multiplied by the same number, the determinant is multiplied by that number. KCt/^ rC(l2 iCCtiQ Tor, 61 = k (uiAi + 02.^2 + 03.43). Hence the truth of the principle enunciated. It follows that, if the elements of any row, or column, of a determinant be the same multiples of the corresponding ele- ments of any other row, or column, the determinant vanishes. DETERMINANTS. 547 16. If each element of any row, or column, of a determinant be the sum of two terms, the determinant can he expressed as the sum of two determinants. For, a, + «i a2 + etj flj + eta 6, 62 h Ci C2 C3 = (a, + «]) Ai + (aj + Ui) Ai + (0.3 + a^) A3 1 2 ^S &i 62 63 Ci C2 C3 + «1 Ka «3 &1 62 &3 Cl C2 C3 In like manner, if each element of any row, or column, be the sum of m terms, the determinant can be expressed as the sum of m determinants. By a proof similar to that above, we have Oi + «! a2 &1 + /81 62 + ttj tts + «3 + 182 63+ A C2 C3 tti ttj «3 h 62 h Cl C2 C3 + «i az «3 &1 62 &3 Cl C2 C3 + Oi a2 a3 A ft A Cl C2 C3 + «! "2 «3 A ft ft Cl Cj C3 If the elements of the first two rows be sums of m and n terms respectively, the determinant can be expressed as the sum of mre terms. In like manner, the principle can be ex- tended to a determinant in which the elements of all the rows are sums of any numbers of terms. 17. If like multiples of the elements of any row, or column, be added to, or subtracted from, the corresponding elements of any other row, or column, the value of the determinant is not changed. 548 For, ALGEBRA. 61 &2 bg Ci Ci C3 [Ch. XXXIX = (Ml +0.2^^ +a3^s) + k {b,Ai +M2 + Ma), by Art. 16, = ai^i + a2^2 + Ms, since &iA + &2-42 + Ms = 0, by Art. 14. The purpose of this principle is to reduce the elements of a row, or column, to smaller numbers, as many as possible to zero, before expanding the determinant. 1 -4 2 Ex. 1. Find the value of 6 3 9 -6 8 -7 We have 1 -4 2 1 2 6 3 9 = 5 23 9 -6 8 -7 -6 -16 -7 = 1 5. -6 23 -16 -1 5 = - 23 -16 -1 6 = 99. To obtain the second determinant we multiply the elements of the first column by 4, and add the products to the corre- sponding elements of the second column ; to obtain the third determinant we multiply the elements of the first column of the second determinant by 2, and subtract the products from the corresponding elements of the third column. The third determinant is then expanded in terms of the elements of the first row. With a little practice we can replace two or more rows, or columns, at the same time. Thus, the third determinant could have been obtained directly from the first, by writing the one result of the two consecutive steps. DETERMINANTS. 549. - 4 8 15 Ex. 2. Find the value of 6-5-4 . We have 14 11 9 - 4 8 15 -2 8 15 -2 8 15 6-5 - 4 = 2 3-5-4 = 2 1 3 11 14 11 9 7 11 9 7 11 9 = 2 14 37 1 3 11 -10 -68 = -2 14 -10 37 -68 = - 2- 4 -31 -10 -68 = 1164. To obtain the second determinant, the factor 2 is removed from the elements of the first column. The elements of the first row of the second determinant are added to the corre- sponding elements of the second row, thus introducing an element 1 in the first column of the third determinant. To obtain the fourth determinant, the elements of the second row are multiplied by 2 and added to the corresponding elements of the first row, and by 7 and subtracted from the correspond- ing elements of the third row. In the fifth determinant the elements of the second row are added to the corresponding elements of the first row, to introduce smaller numbers before expanding. It is important to notice that the row, or column, which is replaced is the row, or column, that we add to, or subtract from. Thus, multiplying the elements of the first column of the determinant by 2, and adding to the corresponding elements of the second column, we have 1 3 10 = 10. Had we multiplied and added as above, but replaced the ele- ments of the first column by the sums, we should have obtained -2 10 4 = 20. 550 ALGEBRA. [Ch. XXXIX 18. The work of evalu9,ting a determinant can often be simplified by some special device. 1 a a?- Ex. 1. Find the value of 1 6 6^ 1 c c2 If we let a = 6, the resulting determinant will have two rows identical, and will therefore vanish. Hence, a — 6 is a factor of the determinant. For similar reasons 6 — c and c — a are factors. The product of these three factors is of the same degree as the determinant. Therefore this product can differ from the value of the determinant only in a numerical factor. Assume the value of the determinant to be Ti[a — 6) (6 — c)(c — a). If we identify one term of the determinant with a term of this product, all terms will be identical. The term in the principal diagonal .is fic^ ; the corresponding term in the product is 'kb(?. Hence k = \. \ a a?' h 62 =(a_6)(6_c)(c-a). Therefore 2. Find the value of a 6 + c (fi b a+o 62 c a + b c2 Adding the elements of the second column to the corresponding elements of the first column, we have a+6+c 6+c = (a + 6 + c) = (a + 6 + c) a + b + c a + c a + b + c a + b 1 b + c a^ a-b b'^-a^ 6 + c a + c a + b a-c c^ ■ a' Ex. 3. Solve the equation Factoring out k, we have = (a — 6) (a — c) (a + 6 + c) = (o-6)(a-c)(6 = 0. -(a+b) -(o+c) c)(o + 6 + c). a a X k k k b X b a a X 1 1 1 = 6 X 6 a 1 b X - X — a b (x — a)(x — 6). Therefore (x — a)(x — b) =0 ; whence x = a, x = b. DETERMINANTS. 551 EXERCISES II. Find the values of the determinants : 8 11 14 9 12 15 10 13 16 a h g 5. h b f Q f c 1 o2 a^ 1 C2 C3 x-y - 1 2 - 3 1. 5 4-10 -7 10 11 2. 19 -12 7 4. 9 18 - 21 . - 11 - 24 49 1 a a" 7. 1 b 68 1 c c3 • 8. 1 + x 1 1 10. 1 1 + y 1 • 1 1 1 1 +« a a^ — 6c a^ 12. 6 62 - ca 62 . c c" - a6 C2 Solve the follov?ing equations : 10 -3a; 4 3 14. 2-2x -2 4 = 11 - X 9 -5 3-x 7+2x ll-3a; 16. 6 + X 3-2a; 17 + 3a; 1 2 3 1 11 -9 2 6. 12 4 . -7 a b 6 c e a a2 62 11. 2j^ 2z 13. y a + b 6 a - 6 2x - z — X 2z z- a + c a a — e 2x 2y -x-y b + c c 6-c 15. = 0. 17. a — bx b — ex c - ax a + b b + c a + c 1 1 1 2a;- 1 3*4-2 - 7 3x4-2 6x-4 - 9 4a; 4- 8 7 a; 4- 1 -20 = 0. Solution of Linear Simultaneous Equations. 19. Solve the equations bix + b.^y -\-b^ = 64, CiX + c^y + c^z = C4. 1 2 ^3 Let D = bi 62 &3 (1) (2) (3) the determinant vyhose elements are, in order, the coefficients of the unknown numbers in the given equations. 552 ALGEBRA. [Ch. XXXIX Multiplying (1) by Ai, (2) by Bi, and (3) by Cj, we have UjAiX +a.iAiy + a^AjZ = a^A^, (4) b,B,x + b^B,y + b^B.z = b^B,, (6) CiCix + C2C12/ + C3C12 = C4CV (6) Adding (4)- (6), we obtain (tti^i + ftiA + CiG^x + {a^Ai + &2-B1 + CaCi)^ + (ttg^i + &3Si + CgCi)^ = 0.4^1 + 64^1 + CiCl. Now, the coefficient of x is Z), the coefficients of y and 2 vanish, by Art. 14, and the second member is the determinant b, C2 Therefore, x = ttl Ui «3 &1 64 &3 Cl C4 C3 04 ttj as = 64 &2 h C4 C2 Cg £> «! a2 a4 bi b^ 64 Cl C2 C4 In like manner, we can obtain y- Z) £> Notice that •) = 0. Therefore, by the preceding article, iJ = 0. Also, if iJ = 0, then / ( r) = 0, and r is a root of / ( k) = 0. EXERCISES I. By the synthetic method, divide 1. a:8 - 3 x^ _ 10 X + 24 by a; - 2 ; by a; - 3 ; by a: + 3. 2. 4 xs - 24 a;2 + 13 a; + 36 by a; + 4 ; by X - 6 ; by a: + 7. 3. a;* + 6 xs - 22 a; + 15 by a; - 8 ; by a: + 5 ; by a; - 4. 4. 32 a;4 - 46 a;2 + 9 a; + 5 by a; + 2 ; by a; - i ; by a; + |. Substitute for x, by the synthetic method, in x' — 8 a;^ + 17 a; — 4 the following numbers : 5. 2. 6. - 2. 7. 3. 8. - 3. 9. 4. In 2 s* — 9 a;' — 17 a;2 - 6 substitute the following numbers : 10. 4. 11. -4. 12. 5. 13. -5. 14. 6. Number of Roots. 6. As we have seen, every equation of the first degree in one unknown number has one and only one root ; every equation of the second degree in one unknown number has two and only two roots. If we assume that every equation has a root, real or imaginary, we can prove that a rational integral equation of the wth degree has n, and only n, roots. The proof that every equation has a root is too difficult for the scope of this book. Let ri be a root of /(a;) = 0. Then, by Art. 5, /(x) = (x - n) Qu wherein Qi is a rational integral function of the (n — l)th degree in x. Since every equation has a root, let ra be a root of Qi = 0. Then, ^i = (a; - ra) §2, wherein Q2 is of the (re — 2)th degree. Continuing in this way, we finally obtain a quotient of degree 1, = jj — (re — 1), in X. This quotient is therefore Q„_i. Let r„ be a root of Qn-i = 0. 560 • ALGEBRA. [Ch. XL Then, Qn-\ = (x - r„) §„, wherein §„ is free of x. Evidently the first term in Qi is aox"^^ in Q2 is aox"-'^, and so on. The first term in §„_i is aoX, and hence §„ = ao. Therefore §„-i = (sc — r„) ao. We now have f W = {x ~ n) Qi, = ix- n) {X - Ti) Q2, = {x- n) (x - j-a) (x - rs)---ix - r„)ao. We know that n is a root of /(x) = 0. Now, let x = r^. Then, /(ra) = ao (r2 - n) (j-2 - rs) ■■• (»-2 - »'„) = 0. Therefore, r2 is a root of /(x) = 0. In like manner it can be shown that rs, r4, ■■•, r„, are roots. Let k be any number different from ri, r^, •••, r„. Then, /(A;) = (ft - n) (A - ra) - (A; - r„). Since fc is not equal to any r, the factors on the right are all different from 0, and therefore their product does not reduce to ; hence k is not a root of/(x) = 0. Therefore, if an equation of the nth degree have one root, it has n, and only n, roots. Equal Roots. 7. If two or more of the r's be equal, the equation has equal roots. Ex. a;' - 5 K^ + 8 a; - 4 = (a; - 1) (a; - 2) (a; - 2). The roots are 1, 2, 2. In such cases the equation is said still to have n roots. Depression of Equations. 8. It follows from Art. 6 that if one root of an equation be known, the remaining roots can be obtained as follows ; Divide the given equation by x minus the known root. The quotient, equated to zero, is called the Depressed Equation. Solve the depressed equation. Ex. One root of the equation a^-6a^ + lla; — 6 = 0isl. THEORY OF EQUATIONS. 561 Dividing the given equation hjx—1, we obtain the depressed equation a^ — 5a; + 6 = 0. The roots of this equation are 2 and 3. Therefore the three roots of the given equation are 1, 2,3. BXEBCISES II. Solve the following equations : 1. x' + x^ — n X + lb = 0, one root being 3. 2. a;5 — 8 a;^ + 25 a; — 50 = 0, one root being 5. 3. 9 a;3 + 18 a:2 - 4 a; - 8 = 0, one root being - 2. 4. 4a;8 + 12a:2 - 11 a; + 20 = 0, one root being - 4. 5. 4 a;5 — 9 x^ + 14 x — 3 = 0, one root being ^. 6. 27 a:3 - 63 a; - 34 = 0, one root being - f . 7. a* + 3x5 - 19a;2 - 27 a; + 90 = o, two roots being 2, - 5. 8- X* — x' - 17 a;2 -I- 5 a; + 60 = 0, two roots being 4, — 3. 9. 12 x4 - 59 x8 + 33 x2 + 99 X + 27 = 0, two roots being 3, 3. 10. x^ - 20 x2 - 21 X - 20 = 0, two roots being 5, - 4. Relation between the Roots and Coefficients. 9. We can, without loss of generality, divide both mem- bers of an equation by the coefiRcient of of. We thus obtain an equation of the form x" + p,x"-^ + p^-" +-+p„ = 0, wherein pi, p^, •••, p„ denote any numbers, real or imaginary 10. If Vi, rj, Ti be the roots of the equation a? -f-iJiiB^ +P!^ +Pa = 0, we have a^ + Pia^ +P^ +Ps = ix- n) (« - Ti) (x - r^) = a^ — (n + r2 + rg) a^ + (n''2 + Vg + r^r^ x — r^r^r^. In general, a;" -fPi*""' +^2*""^ H hi?,,-!* +Pn = (x- n) (a; - r^ (x - r^ ■•• (a; - r„) = a!" - (ri -l-ra H f- r„)a!''-' + {riTi + J-irg + ••• + r„_ir„)a;"-2 - {nnn + nvi h h »-«-2»'»-i''0a'"~^ 562 ALGEBRA. [Ch. XL Equating coeflBcients of like powers of x, we have ri + r2 + r.i-\ = — pi; riPi + riz-a H Vrn^xfn = /»2 ; '•i''2''3 + i'\i'i''^ ^ f- r„_ir„.ii'„ = — /Js ; /•]/'2"-''« = (-!)>«■ From these equations we obtain the following relations be- tween the roots and the coeffi-cients, when the coefficient of the highest power of x is unity : The sum of the roots is equal to the coefficient of the second term with sign changed. The sum of the products of the roots in pairs is equal to the coefficient of the third term. Tlie sum of the products of the roots taken three at a time is equal to the coefficient of the fourth term with sign changed; and so on. Tlie product of the roots is equal to the last term if n be even, and to the last term with sign changed ifn be odd. Ex. Given aj^-So^-f 3a; + 4 = 0. The sum of the roots is — (— 5), =5. The sum of the products of the roots, two at a time, is 3. The product of the roots is — 4. 11. The following special cases may be noted : (i.) If the sum of the roots be 0, the second term is wanting. (ii.) If one root, or more than one root, be 0, the last term is wanting. 12. It is important to notice that the relations given in the preceding articles cannot be used to obtain the solution of an equation. Ex. From the equation a;' — 5»^ + 3a; + 4 = 0, we have ri + r2 + r3 = 5 (1), nrj + ^rg -f r^rj = 3 (2), r^r^r^^-i (3). Multiplying (1) by r-^", (2) by — n, (3) by 1, and adding the resulting products, we obtain rj^ — 5 rj^ + 3 ri + 4 = 0. This is a mere statement that rj satisfies the given equation, and in no way helps us to find its value. THEORY OF EQUATIONS. 563 13. The properties of Art. 10 can sometimes be used to solve an equation, if a relation between the roots be known. Ex. One root of the equation a;^ — 6a!^ + lla; — 6 = is twice a second root. Solve the equation. Let r2 = 2 r^. From the relations, ri + n + Vs = 6, r^r^ + nr^ + Va = 11, Vir-iTs = 6, we have 3ri + r3 = 6 (1), 2n' + 3nr, = ll (2), ,-^3=3 (3). From (1) and (2) we obtain rj = 1, i^, and r^ = 3, f . Since these values of Ti and r^ must also satisfy equation (3), 1, 3, is the only admissible solution. Finally, rg = 2, and the required roots are 1, 2, 3. Symmetrical Functions. 14. As in Ch. XXI., Art. 15, the value of an expression which is symmetrical in the roots can be found without solving the equation. Ex. If rj, rj, rg, be the roots of the equation a^ + PiX' + p^ + Pa = 0, find the value of rj^ + r/ + r^^ We have (n + »-2 + TsY = r^ + r} + rs' + 2 {r^r^ + r^r^ + rjrg), or {~p,Y = n' + ri + ri + 2p,; ' whence r^ + ri + ri =p^ — 2 p^. Formation of an Equation from its Roots. 15. The relations of Art. 10 enable us to form an equation if its roots be known. We may assume that the coefficient of the highest power of the unknown number is 1. Ex. 1. Form the equation whose roots are — 2, 3, 4. We have j-j -I- }.2 + rs = 5, the coefficient of x, with sign changed, j-jj-j -I- f jj-g + rgrj = — 2, the coefficient of x ; and r^r^r^ = —24, the term free from x, with sign changed. Therefore the required equation is a;^ — 5 a;^ — 2 a; + 24 = 0. 564 ALGEBRA. [Ch. XL Ex. 2. Form the equation whose roots are l+V-2, l-V-2, 1+V3, 1-V3- The work will be simplified by first forming the quadratic equation whose roots are the conjugate imaginaries, and the quadratic equation whose roots are the conjugate surds, and then multiplying together these two equations. The first of these quadratic equations is ^,•2 — 2a;4-3 = 0- and the second is a^ — 2x — 2 = 0. The required equation of the fourth degree is therefore a;^_4a!3 + 6a;2-2a;-6 = 0. We could also have formed the equation in Ex. 1 by multi- plying together the three linear factors which equated to have as roots the given roots. We should thus have obtained (a; + 2)(a;-3)(a5-4) = 0, or af-5!e'-2x + 2i = 0. EXERCISES III. Solve the following equations : 1. x' — 3 a;2 — 4 X + 12 = 0, the sum of two roots being 0. 2. 2 x' — 11 x^ + 12 X + 9 = 0, two roots being equal. 3. x" — 6 x^ — X = — 6, one root being 2 greater than a second. 4. 3 x^ _ 25 x'' — 19 X + 9 = 0, the product of two roots being 3. 5. x^ — 3 x^ — 6 X + 8 = 0, the roots being in A. P. 6. 2 x' - 9 x2 - 27 X + 54 = 0, the roots being in G. P. 7. 8 x^ — 30 x^ _|- 27 X — 5 = 0, one root being twice the sum of the other two. 8. X* — 7 x' + 9 x^ + 7 X — 10 = 0, one root being equal and opposite to a second root. 9. 6 X* — 13 x' — 18 x^ + 52 X — 24 = 0, one root being the reciprocal of a second root. 10. 8 X* + 30 x8 - 135 x2 - 135 x + 162 = 0, the roots being in G. P. 11. X* - 14 x^ + 51 x2 - 14 X - 80 = 0, the roots being in A. P. Given the equations x' + pix^ + pix + pj = and X* + pio? + p^x^ + psx +pi = 0, express the following symmetrical functions in terms of the coefBcients. THEORY OF EQUATIONS. 565 ^2- r + -+■■■• ^3. -i- + _l- + .... 14. ,.j3 + r2'+-. Form the equations whose roots are : 15. 2, - 3, 4. 16. - 1, 5, |. 17. - 6, 3 ± V3. 18. - 5, 1 ± V- 5. 19. 3, 3, - 2, - 4. 20. - 1, 4, J, \. 21. ±V2,3±V-2. 22. ±v-l. -2±V-3- 23. 2, 2, 1 ± V2, 2 ± V- 1- If n, r2, rs, he the roots of the equation x^ + pix^ + p^x +^3 = 0, form the equation whose roots are : 24. nW, nW, »-2W. 25. 1±^, ^1+^, ?1+^. ,•32 rji yj2 Real and Rational Roots. T 16. If -, a rational fraction in its lowest terms, be a root of the equation a^x" + ajOf-'^ + ••• + a„_ia; + a„ = 0, tvherein Uq, Oj, •••, a„ are now assumed to he integers, then Oa is divisible by s, and a„ is divisible by r. Since - is a root, we have s ao— + ai — -H h o„-i- + O„ = 0. Multiplying by s»-i, we obtain ao— + air»-i + a2r"-% + ••• + a„_irs"-2 + a„s'^-^ = 0, (1) or ^^ = - (air"-'- + a2r»-2s ^ 1. a„_irs»-2 + anS"-!). The second member of the last equation is integral. Consequently must reduce to an integer ; that is, aor" must be divisible by s. Since r, and therefore r", is not divisible by s, ao must be divisible by s. Now, multiplying (1) by s, and dividing by r, we readily infer that a„ is divisible by r. Ex. Forming the equation whose roots are ^, |, 5, we obtain or 6a^-37a!2 + 37 05-10 = 0. We see that 6 is divisible by 2 and 3, the denominators of the two fractional roots, and that 10 is divisible by 1, 2, and 5. 566 ALGEBRA. [Ch. XL 17. If the coefficient of the highest power of x in a rational integral equation with integral coefficients he unity, the equation cannot have a rational fractional root. For, by the preceding article, the denominator of any frac- tional root mast be a factor of this coefl&cient, and hence be 1. Surd Roots. 18. If the coefficients in a rational integral equation be rational, quadratic surds enter in conjugate pairs. That is, if a + ^b be a root of /(a;) = 0, then a — ^h is also a root. Divide /(x) by (x — a — y/V) (x — a + ^V) , and let Q stand for the quotient, and Itx+ 8 for tlae remainder, if any. Then, '/(x) = (x - a - ^l) (x - a + V&) Q + Bx + S. Substituting a + y/b for x, we have /(a + V&) =R{a+ y/h') + 8 = Ba + 8+ Ry/b. But since a + y/b is a root of /(x) = 0, f(a + y/b) = 0, and therefore, Ba + 8 + liy/b = 0. "Whence, by Ch. XIX., Art. 31, Ey/b = 0, (1) and Ila + 8 = 0. (2) From (1), since & # 0, iJ = ; and then, from (2), 8 = 0. Therefore, U a + y/b'be a, root of /(x) = 0, we have f(z) = (x-a- y/b) (x - a + y/b) Q, and a — y/b is also a root. The quadratic factor (x—a—~^b)(x~a+^b), =(x—ay—V, corresponding to a pair of conjugate surd roots, is rational in a and 6. Imaginary Roots. 19. If the coefficients of a rational integral equation be reed, imaginary or complex roots enter in conjugate pairs. That is, if a+bi be a root of /(a;)=0, then a—bi is also a root. The proof is similar to that given in the preceding article. The quadratic factor (a; — a — bi)(x — a + bi), ={x — a)^ + 61 corresponding to a pair of conjugate complex roots, is real. THEORY OF EQUATIONS. 567 Ex. One root of the equation a;* — Sa^' + Sa^ + iga; — 30 = is 2 + y' — 1 ; solve the equation. The quadratic factor corresponding to the roots 2+y — 1 and 2 — -y/ — 1 is (a;- 2 - V- !)(« - 2 + V- 1), = (a;- 2)2+ 1, = a^- 4a! + 5. Determining the other quadratic factor, we have ai'-5:^ + 3x' + 19x- 30 = (ay' -ix + 5)(a^ -x-6). The roots of the equation a^ — a; — 6 = are found to be —2, 3. Therefore the required roots are — 2, 3, 2 ±-y/— 1. EXERCISES IV. Find the factors of the first members of, and hence solve, the following equations. Also solve by Art. 11 : 1. a;3 + 3 x2 + 2 a; + 6 =0 ; one root: V" 2. 2. x3 _ 2 0:2 _ 3 a; + 6 = ; one root : - ^3. 3. 2 x8 + 2 a;2 - 19 z + 20 = ; one root : ^(3 - 1). 4. x^-Tx^ + 9x+ 5 = 0; one root : 1 + t/2. 5. 4 a^ - 4 a;S _ 23 a;2 + 4 a; + 19 = ; one root : J + ^5. 6. a;i - 8 a;3 + 21 x2 - 26 a; + 14 = ; one root -.3-^2. 7. 36 s* -24 a;3- 143x2 -146 a; -50 = 0; one root: -f-ij. 8. a;5 - 16 2;3 _ 4 a;2 _ 17 a; _ 4 = ; two roots : i, 2 + ^5. To transform an Equation into Another whose Roots are any Multiples of the Roots of the Given Equation. 20. Transform the equation a^x" + ttiof-'^ + asOf^^ + • • ■ + a„_ia; + a„ = into another equation whose roots are & times the roots of the given equation. Let y be the unknown number of the new equation. Then ^ = *x, ora; = |, and a„|_ + ai|^j + a2|^2+-+«'-i^ + «" = 0- Multiplying by S", aoV" + haiy^-' + Ifiaiy"-^ + ■•• + k^-'^Un-iy +k^a„ = 0. 568 ALGEBRA. [Ch. XL That is, to form an equation whose roots are k times the roots of a given equation : Multiply the coefficient of the second term by k; that of the third term by k^; that of the fourth term by It? ; and so on. Ex. Form the equation whose roots are 3 times the roots of the equation a? — 23? — x-{-& = Q. We have «/= - 3 ■ 2 j/^ - 3^ ?/ + 3^ • 6 = 0, or 2/3 - 6 2/2 - 9 ^ + 162 = 0. 21. Ex. 1. Form the equation whose roots are ten times the roots of the equation x* — 3 a^ + 8 a;^ + 2 a; — 19 = 0. The transformed equation is 2/^-10 X ^if + W x82/' + 10= X 2i/-10^ x 19 = 0, or, 2/' -30 2/' + 800 2/2 + 2000 2/ -190,000. To obtain the transformed equation, when the multiplier is 10 : Annex one cipher to the coefficient of the second term ; two ciphers to the coefficient of the third term ; three ciphers to the coefficient of the fourth term; and so on. If any term be wanting, it must be supplied mentally with a zero coefficient. Ex. 2. Form the equation whose roots are ten times the roots of the equation a;^ — 5a; + 6 = 0. We have y^ - 500 ?/ + 6000 = 0. 22. An important application of this transformation is to obtain an equation in which the coefficient of the highest power of the unknown number is 1 and the other coefiicients are integral. Ex. Transform the equation 12a^ — 5a^ — Ja;+1 = into an equation without fractional coefficients, in which the coefficient of the third power of the unknown number shall be 1. Dividing hj 12, a? - ^x' - ^^x + J^ = 0. Forming an equation whose roots are k times the roots of the given equation, we have if — yi %^ ~ 21 ^'u + rV ^' = ^- THEORY OF EQUATIONS. 569 We find by inspection that 12 is the least value of k which will make all the coefficients integral. Substituting 12 for k, we obtain ^ — 5 y^ — 42 ?/ + 144 = 0. To transform an Equation into Another whose Roots are Those of the Given Equation with Signs Changed. 23. Let f{x) = {x-r,){x-ri)-.{x-r„)=0 (1) be the given equation. Substituting — y for x, we have /(-2/) = i-y -n)(-y - ri)--{-y -r„) = (- l)''(2/ + n)(2/ + r2)-.(2/ + r„) = 0. (2) Evidently the roots of equation (2) are — n, - r^, ■••, — »•„. If the given equation in x be aoa;» + ais"-' + aiX"-^ + ••• + an-ii? + an-\x + a„ = 0, the required equation is aor - air~' + ••■ + (- l)»-'a„_ij/ + (- l)"a„ = 0. That is, to obtain the transformed equation : Change the sign of every alternate term, beginning with the second ; or the signs of the terms containing odd powers of x. If any term be wanting it must be mentally supplied with a zero coefficient. Ex. Transform the equation s^ +53? — Sx — 2 = into another whose roots are those of the given equation with signs changed. We have y' Jrby^ + Zy -2 = 0. Descartes' Rule. 24. If the signs of the terms in the equation a;<_3a^-4a^+2a;-3 be arranged in order, we have -\ 1 — . A Permanence of Sign is a succession of two like signs, as , above. A Variation of Sign is a succession of two unlike signs, as -) — or — f-) above. In the above equation there is one permanence of sign and three variations. 570 ALGEBRA. [Ch. XL 25. Iff{x) be multiplied by x — r, the number of variations of sign in the product will be at least one greater than the number of variations of sign infix). The number of variations of sign in a;8 + 2 x' - 5 x^ - a;5 - 4 X* + 7 ;r3 - 3 a;2 + 4 a; + 1 is four. Let us multiply this expression by x — 1. Grouping in parentheses all successive terms which have like signs, we have + ( + x8 + 2x')-( + 5x6+x5 + 4x4) + ( + 7x3)-(+3x2) + ( + 4x + l) + x -1 + (+xH2x8)-( + 5x' + x6 + 4x^) + ( + 7x'')-(+3x3) + (+4xHx) ( - xS)-( + 2x^-5xS-x°) + ( + 4xO-C + 7x3) + (+3x^-4x)-l + (+x9 + '-8)-( + 7x'-4x6+3x5) + ( + lla;*)-( + 10x8) + ( + 7x2-3x)-l Observe first that in the given expression, as grouped, the number of variations of sign is one less than the number of groups ; In fact, that in counting variations we need give attention only to the signs preceding the Also, that the grouping in the partial products and in the result corre- sponds to that in the given expression, although thereby variations of sign enter in some of the groups. We see that the number of variations of sign in the signs before the parentheses in the product is one more than in the given expression. That this should be so, can be inferred from the following considerations. The distribution of signs in the first partial product is evidently the same as in the given expression. In the second partial product there is no term in x'. Therefore in the result the sign of the first term in the first parentheses, and therefore the sign before the parentheses, is the same as in the given expression. The first term in the second parentheses is obtained by multiplying the last term in the first parentheses in the given expression, + (+ 2x'), by — 1, and is — (+ 2x'), of the same sign as the like term in the first partial product. Therefore, in the result the sign of the first term in the second parentheses, and therefore the sign before the parentheses, is the same as in the given expression. Similar considerations hold for the other groups. Therefore no variation of sign in the signs of the groups in the given expression is lost. But to the last term in the second partial product there is no corre- sponding term in the first partial product. Since this term is obtained by multiplying the last term in the last parentheses by — 1, its sign will be opposite to that before the last parentheses. Hence at this point one additional variation of sign in the signs of the groups is obtained. THEORY OF EQUATIONS. 571 Therefore the product contains at least this one more variation of sign than the given expression. Finally, observe that two variations of sign are gained in the second group. Therefore in multiplying the given expression by a; - 1 we gained three variations of sign. If any term be wanting in the given expression, it should be supplied with a zero coefBcient, and be included in the group with the preceding term. Evidently the reasoning in the preceding example is general, and holds for any rational integral expression in x. 26. Descartes' Rule. — The number of real positive roots of fix) = cannot he greater than the number of variations of sign in fix) ; and the number of real negative roots cannot he greater than the number of variations of sign in f{ — x). Let Fix) stand for the product formed by multiplying together the factors of fix) corresponding to imaginary and real negative roots, and let n, r2, ra, •.■, rj be the k real positive roots of /(a;)= 0. Then, fix)= -f (x)(a: - ri)ix - Vi) ... (a: - n). Now, Fix)ix — r{) has at least one more variation of sign than Fix); the product Fix) (x — r{) (k — rj) has at least one more variation of sign than Fix)ix — r{), and therefore at least two more than Fix), and so on. The last product, that is fix), will therefore have at least k more varia- tions of sign than Fix). Since there will be at least k variations of sign in fix), the number of real positive roots of /(x) = cannot be greater than the number of variations of sign. By Art. 23, the positive roots of /( - a:) = are, with their signs changed, equal to the negative roots of fix) = 0. But, by the preceding proof, the number of real positive roots of /( — a;) = cannot exceed the number of variations of sign in/(— x). Hence the number of real nega- tive roots of fix)= cannot exceed the number of variations of sign in fi-x)- 27. If f(x) be a complete rational integral expression, the variations of sign in /(—a;) are evidently permanences of sign in f{x). Thus, if f(x) = 4:a^-x' + 3af + 2x'-5x-7, then f{-x)='kaf + x* + 33^-2 3i^-5x + 7. The variations of sign, 3 a;' - 2 a:^ and — 5 a; + 7, in /(— x), correspond to the permanences of sign, 3a^+2ar' and —5 a;— 7, infix). 572 ALGEBRA. [Ch. XL The second part of Descartes' Rule may therefore be stated as follows : If f(x) = be a complete rational integral equation, the num- ber of real negative roots cannot be greater than the number of permanences of sign infix). 28. From Descartes' Rule, we infer the following: (i.) If the signs of the terms off{x) be all positive, the equation f(x) = does not have a real positive root. (ii.) If f(x) = be a complete rational integral equation, and the signs of the terms be alternately + and —, the equation does not have a real negative root. Limits to the Boots. 29. In determining the real roots of a given equation, the work is often simplified by knowing that the roots cannot be greater or less than a certain number. A Superior Limit to the real roots is a number greater than the greatest root. An Inferior Limit is a number less than the least root. Various formulae have been given for finding the limits to the roots, but we can as a rule determine closer limits by inspection than are obtained by using these formulae. The following examples will illustrate the method of procedure : Ex. 1. Find a superior limit to the real roots of the equation a;* + 3 a;' - 13 a;2 - 27 a; + 36 = 0. Grouping terms and factoring, we have x'ia^ - 13) +3x{af-9)+36 = 0. It is evident that the first member of this equation will not reduce to zero for any value of x greater than 4. Therefore 4 is a superior limit to the real roots. In the above example the terms are arranged in groups, each having a positive term first. We then determine by inspection what is the smallest value of X which will make all the groups positive. THEORY OF EQUATIONS. 573 An inferior limit is obtained by forming the equation whose roots are those of the given equation with signs changed, and finding a superior limit to the real roots of the transformed equation. This number with its sign changed will be an in- ferior limit to the real roots of the given equation. Ex. 2. Find an inferior limit to the real roots of the equation Changing the signs of the alternate terms, we have a;' _ 3a^ - 4x2 + 30a; - 36 = 0, or 2a;^-6K='-8a!2-|- 60a; -72 = 0. Whence a:» (x - 6) + x' (scr' - 8) + 60 (x ~ li) = 0. It is evident that 6 is a superior limit to the real roots of the transformed equation. Therefore — 6 is the required inferior limit to the real roots of the given equation. Where nearly all of the terms are negative, as in Ex. 2, it is necessary to distribute the term containing the highest power of X among the negative terms, multiplying the equation by some number, if the coefficient of this term be unity or less than the number of negative terms. BXEECISBS V. Transform the following equations into others whose real rational roots are all integers : x^-\x^-hx^ + i,x-i^ = Q. a;3 + 4.2 a;2 - 6.35 x - 1.804 = 0. X* + 12.6x2 -4.32 a; + .1539 = 0. 9-13. Transform the equations in Exs. 1-5 into others whose roots are those of the given equations with signs changed. Show that the roots of the following equations are all imaginary : 14. a;* + 8 a;2 + 5 = 0. 15. x^ + 5 a;* + 8 k^ + 10 = 0. Determine the least possible number of imaginary roots of the following equations : 16. a;*-9a;3-5x-4 = 0. 17. x^ _ 4a;S + Sa:^ + 1 = 0. 18. 3 a;' - a;* + 5 a;8 - 9 = 0. 19. 2 a;' - 3 a;^ - 4 a:^ - x = 0. 1. x8-f X2^. Jx-| = 0. 2. 3. 8 x3 - 14 x2 - 11 X + 3 = 0. 4. 5. 12 X* + 40 x2 _ 9 a; - 14 = 0. 6. 7. x8- 2.5x2 -4.5x + 9 = 0. 8. 574 ALGEBRA. . [Ch. XL 30. The principles thus far developed enable us to find all the rational roots of a given rational integral equation. Ex. 1. Solve the equation a;* — a^ — 16 a^ + 4 a; + 48 = 0. By Descartes' Rule this equation cannot have more than two positive roots, or more than two negative roots. We find that 6 is a superior limit, and — 4 an inferior limit, to the real roots. Also, since the coefficient of ^ is unity, the equation cannot have a fractional root. By Art. 10, the integral roots, positive or negative, are factors of 48. These factors, within the limits to the roots, are ±1, ±2, ±3, 4. We now determine by trial which of these factors, if any, satisfy the equation. Substituting 1 and — 1, we find that these numbers are not roots. Substituting 2, the equation is satisfied, and the de- pressed equation is found to be x3 + x^ - 14 a; - 24 = 0. Substituting — 2 in this depressed equation, we find that this factor is a root, and that the second depressed equation is a:2-a;-12 = 0. The roots of this equation are found to be — 3, 4. Therefore the roots of the given equation are 2, — 2, — 3, 4. Ex. 2. Solve the equation 6a^ + 23a^— 6a;-8 = 0. We find that 1 is a superior limit, and — 5 an inferior limit, to the roots. Therefore the integral roots, if any, are among the numbers — 1, — 2, — 4, and the fractional roots among ± l, ± J, ± i, ± f, ± l. We will first try for fractional roots. Substituting \, 6 23-6 -8(i 3 13 A 6 26 7 - 9 2 Therefore \ is not a root. Substituting — |, 6 23-6 -8(-* - 3 -10 8 20 -16 THEORY OF EQUATIONS. 575 Therefore — | is a root, and the depressed equation is 6a;2 + 20a;-16 = 0, or 3a^ + 10a;-8 = 0. The roots of this equation are found to be |, — 4. Therefore the roots of the given equation are — i, |, — 4. The above equation could also have been solved as follows : Letting a; = | ?/, we have, by Art. 20, ^3 ^23 2/2 -36 2/ -288=0. This equation can have no fractional roots. The integral roots are found to be —3, 4, —24. Then «, = J y, = — ^, |, — 4. Although by this method we avoid solving for fractional roots, yet the number of integral factors of the last term which it is necessary to test becomes greater. Newton's Method. 31. The factors of the last term are as a rule more easily tested by the method which follows. From Art. 3 we have a^oi? + a^x^ + ajX + ttj = hffi? + (ft, — bi^r) a^ + (62 — h^r) x + {R — hr) . Now, let r be a root of the given equation. Then i2 = 0, and the form of the equation which we shall use is \^ + (61 — ^o*") ^ + (pi — bir) x — b2r = 0. The last term is divisible by r, and the quotient is — ftj^ Adding this quotient to the coefficient of x, we obtain — bir. Dividing this sum by r, and adding the quotient, — 6„ to the coefficient of a:', we obtain — bf,r. Dividing this sum by r, and adding the quotient, — &o, to the coefficient of 3?, we obtain a sum 0. Arranged as in synthetic division, 60 61 — V h - biV - bir{r_ — bp —bi — &2 —boT— biV - b^r It is evident from the form of f(x), given in Art. 3, that the above method is general. Observe, also, that the numbers in 576 ALGEBRA. [Ch. XL the second line of the work above are, in order, the coefficients of the depressed equation with signs changed. We thus have the following method : Divide the last term by r and add the quotient to the coefficient of X. Divide this sum by r and add the quotient to the coefficient of x'. Divide this sum by r and add the quotient to the coeffi- cient of 3?, and so on. The last quotient will be — b^. If at any stage the division be not exact, this fact at once shows that the number being tested is not a root. Ex. Solve the equation a;^ — 4 a^ — 9 a; + 36 = 0. Substituting 1 and — 1, we find that they are not roots. Testing — 2 by Newton's Method, 1-4 -9 36(-2 -18 -27 36 Since — 27 is not divisible by 2, 2 is not a root. Testing 3, 1 -4 -9 36(3 -1 1 12 0-3 3 36 Since the division at each step is exact and the last quotient is — 1, 3 is a root, and the depressed equation is a;2 _ a; - 12 = 0. The roots of this equation are — 3, 4. Therefore the roots of the given equation are 3, — 3, 4. EXERCISES VI. Find a real root of each of the following equations, and solve the depressed equation : 1. x3 - 2 x2 - 13 a; - 10 = 0. 2. x^ - 19 a; + 30 = 0. 3. a:3-5a;2 + 17a;-13 = 0. 4. x^ - 7 a;^ + 36 = 0. 5. a;8 - 5 x2 + 5 re + 3 = 0. 6. x^ - 8x^ + Qx + 58 = 0. 7. x8 - 7 a;2 + 49 a: + 237 =0. 8. x^ + x^ _ 61 x - 205 = 0. 9. x« + 4 xs - 34 x2 - 76 X + 105 = 0. 10, x* - 15 x^ - 10 x + 24 = 0. 11. x*-2x»-iax^ + Ux + 2i = 0. 12. x*-37x2_24x + 180 = 0, THEORY OF EQUATIONS. 577 13. x*-8a;8 + 7a;2 + 40a;-60 = 0. 14. x* -a;3- 11 a;2- k- 12 = 0. 15. a:'-9a;3 + 15x2 + 393;_7o = o. 16. a* -3a;3-2a;2- 32 = 0. 17. xi+5xS + Ux^ + 61x+16Q = 0. 18. x4-26a;2 + 57 a;- 18=0. 19. z5 + 6 a;« - 2 a;3 - 36 a;2 + j; + 30 = o. 20. x^ - 11 x'' + 36 a;3 _ lo x^ - 148 x + 240 = 0. 21. x5 - 8 X* - 6 xs + 134 x2 - 43 X - 510 = 0. 23. 2x3 + x2_25x+ 12 = 0. 23. 5 x^ + 7 x^ - 46x + 24 = 0. 24. 6 X* - 13 x3 _ 18 x2 + 7 X + 6 = 0. 25. 12 x« - 64 x3 + 9 x2 + 53 X + 10 = 0. 26. 96x« + 128x3-74x2 -63x + 18 = 0. Derived Functions. 32. Given /(x) = aox" + aix"-! + aax"-^ + • • ■ + a„-ix + a„, to find the value of /(x + h). We have /(x + h) = ao(,z + ^)» + ai(x + ft)»-i + a2(x + A)''-2 + h an-i(x + /i) + a„ = agx" + aix«-i + agx"-' + h a!„_ix + a„ + h [naox"-' +(»-!) ajxo-^ 4 („ _ 2) a2X''-3 + 1- a„_i] + i^^ [» (» - 1) «!oa;"-2 + (n - 1) (re - 2) oix«-3 + . . . + a„_2] + . . . . The coefficient of /i is derived from /(x) as follovrs: Multiply each term of f (x) hy the exponent of x in that term and sub- tract 1 from the exponent. Thus, from agx" we derive naoX"-' ; from aix"-' we derive (re — 1 ) fflix''-== ; and so on. On this account the coefficient of h is called the First Derived Function, and is represented by /'(x). Observe that from a„-ix we derive 1 x flSn-ix", = a„-i. From a term without x we obtain 0. That is, from a„ = rt„x" we derive x a„x-^, = 0. In a similar way the coefficient of — is derived from the coefBcient h ; the coefficient of — from that of — ; and so on. \S [2' These successive coefficients are therefore called the Second Derived Function, the Third Derived Function, and so on ; and. are represented by /"(»), /'"(x), and so on, respectively. 578 ALGEBRA. [Ch. XL Notice that /'(x) is of degree one lower than f{x) ; that /"(x) is of degree two lower than /(x) ; and so on. We now have f{x + h) = f{x) + hf'{x) + ^ f"(jr) + . . . . In like manner /(x + h) could have been expanded in terms of ascend- ing powers of x : fix + h) =f(h) + xf'(h) + |V'W + .... Ex. Given /(x) = x» - 2 x^ + x + 3, find /(x + 2). We have /(x + 2) =/(2) + x/'(2) + ^/"(2) + .... Lf. In the terms of the second member, 2 is to he substituted for x in /(x), /'(x), and so on. We now have /(x) = x3-2x2 + x + 3, /'(x) = 3x2-4x+ 1, /"(x) = 6x-4, /"'(x) = 6, /i'(x) = 0, etc. Whence /(2) = 5, /'(2) = 5, /"(2) = 8, /"'(2)=6. Therefore, /(x + 2) = 5 + 5 x + 4 x^ + x'. Multiple Boots. 33. If ?• be ^ times a root of /(x) = 0, then x— r is k times a factor of fix), or (x — r)* is a factor. Therefore, /(x) = (x - r)*i?'(x), wherein ii'(x) stands for the product of the remaining factors affix), and does not contain the factor x — r. We now have /(x + A) = (x - r + h)''F(x + 7i). Expanding /(x + 7i) and i?'(x + ft) by the preceding article, and (x - r- + h)" by the binomial theorem, we have fix) + hf'(x)+'ff"ix) + --- = [(x - r)* + ^(x - r)'-i7j + • ■ ■] [F(x) + kF'ix) + ■••] = (X - ?-)*i^(x) +ft[^(x-r)'-ii?'(x)+ (x -»•)*-?" (x)] + •■•. Equating coefficients of ft, /'(x) = fc(x - r)*-ii?(x) + (X - r)'i!"(x) = (x - r)*-J[ftJ?'(x) + (X - »-)i^'(«)]- We now see that fix) contains x - r as a factor k — 1 times ; there- fore /'(x) = has k -1 roots equal to r. THEORY OF EQUATIONS. 579 That is, if f(x)= have a multiple root, then f'(z)= has the same root repeated one less time. 34. It follows also from the preceding article that (x — r)*-' is the H.C.F. of /(!c) and /'(x). We can therefore find the multiple roots of any given equation, f{x)= 0, by finding the roots of the H.C.F. of fix) andf'{x) equated to 0. If f{x) and /'(a) have no common factor, f{x) = Q does not have a multiple root. Ex. The equation a;*— 9a;^+ 23a;2_ 3x — 36 = has multiple roots. Solve the equation. Wehave /(x)= a* - 9a;8 + 23a;2-3a; - 36, and /'(a;)=4x3- 27a:2 + 46a;-3. The H.C.F. oif{x) and/'(x) is found to be x - 3. Therefore (x - 3)^ is a factor of /(x), and 3 is a root twice of the given equation. Removing the root 3 twice, 1-9 23 - 3 - 36(3 3-18 15 36 1 - 6 5 12 3 - 9 - 12 1-3-4 The final depressed equation is x'^-Zx -^=Q; whence, x = - 1, 4. The roots of the given equation are 3, 3, — 1, 4. EXERCISES VII. Find the successive derived functions of the following: 1. 2x2-3x + 6. 2. x* + 7x8-5x2-3x + ll. Given / (X) = x» - 9 x^ + 14 k + 24, find 3. /(x + 1). 4. /(x-2). 5. /(x-3). Solve the following equations, which have multiple roots : 6. x* + 6 x» - 14 x2 - 90 X + 25 = 0. 7. X* - 4 xs - 4x2 + i6x + 16 = 0. 8. x» - 6 x' + 38 j:2 - 112 x + 104 = 0. 9. x» + 2 x3 - 3 x2 + 136 X + 464 = 0. 10. a;5 _ 5i;4 _ 11 a;3 + 29x2 - 26x + 8 = 0. 11. x5 + 7x* -6x8- l62x2-459x-405 = 0. 580 ALGEBRA. [Ch. XL Oraphic Representation. 35. Two perpendicular straight lines divide the plane in which they lie into four parts, called Quadrants. Thus, the lines ^ XX', YT', in Pig. 5, divide the plane of the paper into four quadrants : XO Y called the first quadrant; YOX' the second quadrant ; X' Y' the third quadrant, and F' OX the fourth quadrant. The position of a point in a plane is known if its distance from these two fixed lines, and the quadrant in which it lies, be known. Thus, the position of a point Pi in the first quad- , rant is known, if we know the distances A^P^ and B^Pi ; that of the point Pj in the second quadrant, if we know the dis- tances A2P2 and B2P2. 36. The two lines X'OX and Y'OY are called Axes of Reference, and the point is called the Origin. The distance of a point P from the axis Y'OY, measured along or parallel to the axis X'OX, is usually designated by the letter x, and is called the Abscissa, or the x of the point. Thus, the x of the point Pj is OAy, — BiP^. The distance of a point P from the axis X'OX, measured along on: parallel to the axis Y'OY, is usually designated by the letter y, and is called the Ordinate, or the y of the point. Thus, the y of the point Pj is OB^, = A^P^. The abscissa and the ordinate of the point are together called the Coordinates of the point. The axis X'OX is called the Axis of Abscissas, or the x-axis. The axis Y'OY is called the Axis of Ordinates, or the y-axis. 37. The quadrant in which a point lies is determined as follows : THEORY OF EQUATIONS. 581 W- 9(2,3) <> (5,-a (-3,-4l6- Abscissas measured to the right of the origin are taken positively ; those to the left, negatively. Ordiuates measured upw^ard are taken positively ; those downward, negatively. The signs of the x and ?/ of a point thus determine the quad- rant in which it lies. Locate the point a; = 2, 3/ = 3. The most convenient way is to measure two units along the a^axis to the right, and from the point thus reached three units upward parallel to the y-nxis, as in Fig. 6. Instead of writing x = 2, " Y 2/ = 3, we may indicate the coordinates of the point by (2, 3), it being understood that the first number is the x and the second the y of the point. To locate the point (—2, 3), we measure two units to the left along the a>axis, and three units upward parallel to the 2/-axis. In like manner, the points (—3, — 4) and (5, — 2) are located, as in Fig. 6 38. If, in the equation y = 2x-3, we give to a; a series of numeri- cal values, we ob- tain correspond- ing values of y. In the table Bn the left, cor- responding values of x and y are written in the same hori- zontal line. Each set of val- ues of X and y may be taken as the coordinates of a point in the plane, in Fig. 7. If we give to x values between those in the table, we obtain Y' Fig. 6. X y -3 1 -1 2 1 3 3 -1 -6 -2 -7 582 ALGEBRA. [Ch. XL corresponding values of y. Thus, if we give to x the value \, between and 1, we obtain for y the value — 2, between — 3 and — 1. To each of these sets of values likewise corresponds a point. If, then, x be made to change continuously from to 1, passing through every intermediate value, y will change continuously from — 3 to — 1, and the points corresponding will form a continuous line from (0, — 3) to (1, — 1). In like manner we have a continuous line connecting all the points given in Fig. 7. 39. It is proved in analytic geometry that a simple equation in two unknown numbers represents in this way a straight line. It is on this account that it is called a linear equation. The figure is called the Graph of the equation, and the equa- tion is said to be the Equation of the Graph. 40. Every solution of the equation corresponds to a point on the graph ; and, conversely, the coordinates of every point on the graph, and of no other points, satisfy the equation. We thus have, corresponding to the infinite number of solu- tions of an indeterminate linear equation, an infinite number of points on a straight line. 41. Since a linear equation represents a straight line, it is necessary to locate only two points to determine its position. By letting a; = in the equation of the graph, we obtain the point where the straight line crosses the y-axis, and by making y = 0, the point where the line crosses the x-axis. Ex. Draw tlje graph of 3x + 2y=.6. When a; = 0, y = 3; the point (0, 3) is on the y-axis. When y = 0, x = 2; the point (2, 0) is on the avaxis. The graph is shown in Fig. 8. Fig 8. THEORY OF EQUATIONS. 583 Intersection of Graphs. 42. The solution of the equations x + y = 5, and a; — 2 2/ = — 4, is 2, 3. Since this solution satisfies both equations, it gives the coordinates of a point which is common to the graphs of these equations, as in Fig. 9. Therefore the point of intersection of two graphs is obtained by solving their equations. We thus see that corresponding to the one solution of two simultaneous linear equations we have the one point of inter- section of two straight lines. 43. As was shown in Ch. XIII., a definite solution of two linear equations in two un- y known numbers cannot be obtained if the equations be inconsistent or equivalent. Inconsistent Equations. 44. The two equations X + 2 2/ = 4, and x + 2y = 6, are inconsistent, and are not satisfied by any set of finite values of x and y. Drawing the graphs of these two equa- tions, we find that they represent two parallel straight lines. Fig. 10. 584 ALGEBKA. [Ch. XL Fig. 11, Equivalent Equations. 45. The two equations 2 a; - 3 2/ = 6, and 4 a; — 6 2/ = 12, are equivalent, and are there- fore satisfied by an indefi- nite number of sets of values of X and y. Drawing the graphs of these two equations, we see that they coincide, as in Fig. 11. ■ 46. Ex. 1. Graphic Representation oi Roots. Draw the graph of the equation 2/ = af'-5ar' + 2a; + 8. Sets of values of x and y are given in the table on the left. The corresponding points (0, 8), (1, 6), (2, 0), etc., indicate the general outline of the curve as represented in Fig. 12. Observe that at the points where the graph crosses the a;-axis, ?/ == ; that is, a;3_5a^ + 2a:-f8 = 0. The values of a;, 2, 4, — 1, at these points, are therefore the roots of the above equation. In like manner, if we place the first member of any rational integral equation, when all terms are transferred to that member, equal to y instead of 0, and draw the corresponding graph, the abscissas of the points where this graph crosses the a;-axis are the roots of the equation. In this graphic representa- tion the ordinate of any point i y >axis would have moved toward each other and coincided in the point where (2) touches the x-axis. The root 3 is a multiple root of the equation a;^— 6a; + 9 = 0. Finally, the second member of (3) is obtained from the second member of (1) by adding two units to the latter. The THEORY OF EQUATIONS. 587 graph of (3), therefore, could have been obtained by sliding the graph of (1) vertically upward two ui its, as shown in the iigure. This graph does not intersect the a^axis, and therefore indicates the presence of imaginary roots ; in fact, the roots of the equation a;^ — 6 a; + 10 = are conjugate imaginaries. BXEECISES VIII. 1. Locate the points (1,3); (-1,3); (2, -2); (-4, -1); (0, -2>; (3,0); (0,4); (-3,0); (0,0); (4, -2); (3,2); (-5, -3); (-5,4). Draw the graphs of the following equations : 2. a; = 2. 3. y = ~4:. 4. a; = 0. i. y = 0. 6. y = 2x~d. 1. 2x = y + i. S. Sx-4:y = 12. Draw, to one set of axes, the graphs of the equations in each of the following examples, and determine their point of intersection : lx + 2y = 6, '!2x + 6y=.-7, 9. -I 10. [3x-y = l. [4x + 3j/ = 7. ( Sx-2y = e, ( x + 5y = W, n. \ _ . . 12. ' * 6 X - 4 2/ = 6. t 2 a; + 10 2/ = 20. !y = x + 2, __ [y z=x-\, 13. \ 14. , yy = x^ + Zx + 2. [ 2/ = x2 - 3 X + 3. 15. -^ . . . 16. ' :3x2 + 2x-l. [t/ = x8-6x2 + 7x + 4. Draw the graph of each of the following functions, and locate each irrational root between two consecutive integers : 17. x8-5x. 18. 3x8-7x2 + 4. 19. 2V- 7x^-30. 20. xs - 3 x2 - 3 X + 1. 21. x8 - 5 x2 + 2 X + 12. 22. x3 - 7 x2 + 15 X - 9. 23. x* - 2 x' - 7 x^ + 8 x + 12. 24. X* - 9 x2 + 4 X + 12. 25. X* + 4 x' - 9 x2 - 120. Greatest and Least Terms in f{x). 48. In the function f(x), = Ooaj" H h a„_j_ia^+' + a„_^ + a„^^+,x'-'' + •••, the term a„_^^ can be made to exceed numerically the sum of all the terms that follow it, by taking x large enough, and to exceed the sum of all the terms that precede it, by taking small enough. 588 ALGEBRA. [Ch. XL First, let a^-iffH' = m(o,a-4+iX*-i H + an-ix +a„). "Whence, ' a„_t+iK*-i + ••■ + an-ix + a„ an-k+i + an-i a« x*-i a;* (2) By taking x large enough, we can make each term in the denominator of (2), and hence the denominator itself, as small as we please. When the denominator is made smaller than «„_*, then m>l, numerically. Therefore, from (1), a„_ix" > a„-*+ix'-i + h fln-ix + a„, numerically. In like manner the second part of the principle can be proved. 49. Since a^x" = oo, as x = co, it follows from the preceding article that f{x) = oo, as a; = oo. If we take x large enough to make the term a^x" exceed numerically the sum of all the terms that follow it,f(x) will be positive when aoX" is positive, and negative when aoW" is negative. Ex. 1. 3x^-5a^ + 7x^—19x-n = +oo, when a; = oo, or — oo. Ex. 2. 2a^ — 9a!^ — 6£e + 6 = + oo, when a; = oo , = — 00, when x = — oo. Principle of Continuity. 50. It was assumed in Art. 38 that the curve is continuous between any two points definitely located ; that is, that a point, which we may assume to be describing the curve, nowhere makes a jump. An instance of an interruption of continuity is shown in Fig. 16. Here the point which is describing the curve on arriving at the point P, whose abscissa is a, say, jumps to the point Q. We assume that Q is a finite distance from P, while the abscissa of Q is greater than a by less than any assigned number, however small. Now, let fi be a point on the curve, whose abscissa is as + h, h being an infinitesimal. Then, since ordi- nates represent values of the func- tion, we have AP = f{a), and Bi;=/(ffl+ h). Evidently,/ (a + ft) -/(a) repre- FiG. 16. THEORY OF EQUATIONS. 589 sents the change in the value of the function corresponding to the change h in the value of x. Since BB — AP, =f(,a + h')—f{a), remains finite as /i = 0, corre- sponding to an infinitesimal change in the value of x, there is a finite change in the value of the function. Oh the other hand, let the curve be continuous at a point a, as in Fig. 17. Then f{a + h)-f{a), =BB-AP, =GB. As A = 0, that is, as the point B approaches A, it is evident that >^^^ CB approaches 0. That is, vfhen the function is continuous at a; = a, corresponding to an infinitesimal change in the value of x, there is an infinitesimal change in the value of /W- We are thus led to the following definition of continuity : The function, f(x), is continuous atx = a, if f(a+ h)-fia)=0, ash = 0. Fig. 17. 51. A rational integral fraction of x is continuous for all finite values of x. Let f(x) = aax" + aia;"-i + •-. + a„_ix + a„. Then, by Art. 32, f{a + h)^f(a)+ V(a) + |f /"(o) + |/"'(«)+ •- Whence /(a + A)-/(a)= ^[/'(a) + |/"(«) + |f /'"(«)+ -J- Now, as /s = 0, A///(a)= 0, ^/"'(a)= 0, and so on. [2 [3 Therefore /'(a) + ,4/"(«) + ^/"'(«) + - =/'(«). [2 [S and h[f'(a) + 1/" (a) + 1/"'(«) +•••] = 0. Consequently, f(a + h)- f(a) = as h = 0, and f(x) is continuous at x = a. But a may be any finite value of x. Therefore f(x) is continuous for all finite values of x. 590 ALGEBRA. [Ch. XL 52. It follows from the principle of continuity that as f{x) changes from /(a) to /(&), it must do so by infinitesimal changes, and therefore pass through every intermediate value. 53. If f(a) and f{b) have opposite signs, at least one real root of f{x) = lies between a and b. By the principle of continuity, a number cannot pass from a positive value to a negative value, or vice versa, without passing through 0. Since f{x), which is continuous in passing from /(a) to /(6), changes its sign, it must pass through 0. This value of x between a and 6, for which f(x) becomes 0, Is therefore a root of /(x) = 0. In connection with this principle attention is called to the graphs of the functions in Exx. 1-3, Art. 46. Thus, in Ex. 3, /(2) and /(3) have opposite signs, and the graph in Fig. 14 cuts the cc-axis once between 2 and 3. In general, if f(d) and /(&) have opposite signs, the corresponding points on the graph are on opposite' sides of the a>axis. Therefore the graph must cross the a^axis at least once between a and b. 54. The following principles are derived from the preceding : (i.) Every equation of an odd degree has at least one real root whose sign is opposite to that of its last term. For, /(O) = a„ ; and, by Art. 49, /(+ co) =+■»,/(- oo) = - oo. If On be positive, /(O) and/(— oo) have opposite signs, and there is a negative root between and — oo. If a„ be negative, /(O) and /(+ oo) have opposite signs, and there is a positive root between and + co. (ii.) Every equation of an even degree, ivhose last term is nega- tive, has at least two real roots, one positive and one negative. For, /(O), = an, is negative, and /(+ oo) = + oo, and /(— c») = + oo. Since /(O) and /(+ oo) have opposite signs, there is a positive root between and + co. For a similar reason, there is a negative root between and — oo. 55. The following is a more general enunciation of the prin- ciple given in Art. 63 : If f{a) and f(b) have opposite signs, an odd number of real roots lies between a and b. If f(a) and f(b) have like signs, either no real root, or an even number of real roots, lies between a and b. THEORY OF EQUATIONS. 691 The proof is a simple extension of that given in Art. 63. The graphs of the functions in Art. 46 throw light also on this principle. Thus, in Ex. 2, /(— 4) and /(— 2) have oppo- site signs, and the graph in Fig. 13 cuts the SB-axis once between — 4 and — 2 ; /(— 4) and /(2) have opposite signs, and the graph cuts the a^axis three times between — 4 and 2. Again, /(— 4) and/(0) have like signs, and the graph cuts the aj-axis twice between —4 and 0; /(— 4) and/(4) have like signs, and the graph cuts the ic-axis four times between — 4 and 4. To find an Equation whose Roots are Less than the Roots of a Given Equation by a Definite Number. 56. Ex. Form the equation whose roots are 2 less than the roots of the equation a^ — 7 3f + 7x-\-15 = 0. (1) Let y be the unknown number in the required equation. Then y = x — 2, or x = y + 2. Substituting y + 2 for x in equation (1), we obtain (2/ + 2)« - 7(2/ + 2)^ + 7(2/ + 2) + 15 = 0. This is a cubic equation in y, and its simplified form can be found by performing the indicated operations, and uniting like powers of y. We thus obtain y^ — y^ — 9 y + 9 = 0. The following method will, however, lead to a simple way of determining the coefficients. Let us assume the transformed equation to be Aof + Ay + A,y + A, = 0. Substituting a;— 2 for y, we obtain Aoix - 2)' + A^(x - 2)2 + A,{x - 2) + ^3 = 0. (2) But this is the original equation, written in a different form. Therefore any result obtained from this form will be obtained by the same process from the original form. Dividing (2) by a; — 2, we obtain a quotient ^(a, _ 2)2 -f- Ai(x — 2) + A2, and a remainder A^. Dividing this quotient by a; — 2, we obtain a second quotient ji^(x — 2) + Ai, and a second remainder A2. Dividing the second quotient by a; — 2, we obtain a quotient Ad, and a remainder Ai. 1-7 2- 7 -10- 15(2 6 1-6- 3 2-6 9 1 -3 2 -9 692 ALGEBRA. [Ch. XL That is, the last term in the transformed equation is the re- mainder obtained by dividing the given equation by x — 2; the coefficient of y is the remainder obtained by dividing the first quotient by x — 2; the coefficient of y^ is the remainder obtained by dividing the second quotient by x — 2; and so on. The work may be arranged compactly, thus : The remainder in the first division is 9, and the coefficients of the quotient are 1, —5, —3. We therefore continue the work of the second division below, obtaining a second remainder — 9. By the third di- vision, we obtain a third remainder — 1, and a final quotient 1. The remainders and the last quotient have been cut off from the rest of the work by vertical lines. These numbers from left to right are now the coefficients of the transformed equation. We therefore have y' — y" — 9y + 9 = 0, as before. The method is evidently general. 57. If k be negative, the work will give an equation whose roots are Jc greater than the roots of the given equation. Ex. Form the equation whose roots are 2 greater than the roots of the equation a;^ — 6 a^ + 3 £c^ + 26 a; — 24 = 0. The synthetic divisor is now — 2. By the method of the preceding article, we find the required equation to be 2/1 _ 14 2/3 _|_ 63 2/2 - 90 2/ = 0. EXERCISES IX. Find, by Art. 53 and Descartes' Rule, the first figure of each real root of the equations : 1. x'-2x^-5x+7 -Q. 2. a;8 - 28 X + 56 = 0. 3. x' + Sx'^-4:y-l=0. i. a;3 + 2 K - 140 = 0. 5. a;3 - 12 a;2 + 16 sc + 80 = 0. 6. 2x^-Tx^-30 = 0. 7. x3 - 6 a;2 + 9 a; - 3 = 0. 8. x« - 5 a;S - 4 x + 19 = 0. 9. x* + 2x8 -16x2 -256 = 0. iq a;4 _ 6xS - 9x + 100 = 0. 11. x* + 4x8 -9x2 -120 = 0. 12. x5 + 4x8 -10x2 -8=0. 13. xi-4x»-35x2+38x+126=0, 14. x^ - Ox* + 6x8 + 2x - 1 = 0. THEORY OF EQUATIONS. 593 15-28. Find the equation whose roots are less than the roots of each equation in Exx. 1-14 by the first figure of the least real positive root. Horner's Method of Approximation. 58. The method, which will now be given, of finding an irrational root of an equation to any required degree of accu- racy is due to W. G. Horner, an English mathematician, who published it in 1819. 59. Ex. Find an irrational root of the equation- correct to three places of decimals. We first find between what two integers the required root lies. The smaller of these integers is evidently the first figure of the root. The principle of Art. 53 is in many examples suflBcient for this purpose, but in certain cases some other work may be necessary, as will be explained in Art. 70. Substitute in succession 0, 1, 2, and 3, in f(x). We have /(O) = - 3, /(I) = - 8, /(2) = - 5, /(3) = 12. Since /(2) and /(3) have opposite signs, there is, by Art. 53 and Descartes' Rule, one real root between 2 and 3. The first figure of this root is therefore 2, and the root may be repre- sented by 2 .abed, wherein a, 6, c, d, stand for the figures in the first, second, third, and fourth decimal places. In the vicinity of the root, the graph of f(x) is as shown in Fig. 18. It is important to notice that if any number a little less than this root be substituted for x, the result of the substitution will be negative ; and that if any number a little greater than the root be substituted for x, the result of the substitution will be positive. Fig. 18. 594 ALGEBRA. [Ch. XL 1 1 - 7 -3 (2 2 6 -2 1 3 - 1 -5 2 10 1 5 2 9 Diminishing the roots of the given equation by 2, we obtain f + 7f+9y-5 = 0. The root of this equation corre- sponding to the required root of the given equation is evidently .abed. This root lies between two consecutive tenths ; that is, be- tween and .1, or .1 and .2, or .2 and .3, and so on. Since this root is a decimal, its square and its cube will be considerably smaller than its first power, and we can get a rather close approximation to the figure in the first decimal place by neg- lecting the terms in ^ and y^. We then have 9y — 5 = 0, whence y = .5+. Before assuming that this is the correct figure, we should, at this stage, test it by substitution. Substituting .5, we obtain 1.375. Since the result of this substitution is positive, .5 is too great. We next try .4, and obtain —.216. Since this result is negar tive, .4 is less than the root, which therefore lies between .4 and .5, and may be represented by .4 bed. The corresponding root of the given equation is 2.4 6cd. In general, the figure suggested, at this stage of the work, by neglecting powers of y higher than the first will not be correct, and should be tested before continuing the work. We now diminish the roots of the last equation by .4 ; that is, the roots of the original equation by 2.4. The new equation is 1 7 .4 9 2.96 -5 4.784 1 7.4 .4 11.96 3.12 - .216 1 7.8 .4 15.08 (•4 8.2 g' -f 8.2 2^ + 15.08 «- .216 = 0. The corresponding root of this equation is .Obcd; that is, it lies between and .01, or .01 and .02, or .02 and .03, and so on. THEORY OF EQUATIONS. 595 As before, neglecting the terms in «* and 2^, we have 15.08 z — .216 = 0, whence z = .01 approximately. We may expect this value of z to give the correct figure in the second decimal place, since 2^ and z^ are now much smaller compared with z. If .01 be too great, the result of the substi- tution, which is the first step in the process of forming the next equation, will be +. If it be too small, the error will be shown by the figure suggested for the next decimal place being greater than 9. Diminishing the roots of the last equation by .01, that is, the roots of the original equation by 2.41, 1 8.2 16.08 -.216 (M .01 .0821 .151621 8.21 .01 15.1621 .0822 .064379 8.22 .01 16.2443 8.23 The next equation is u' + 8.23 w^ + 15.2443 u - .064379 = 0. Since the last term, — .064379, is negative, we infer that .01 is not greater than the root. Neglecting the terms in m' and u^, we have 15.2443 m - .064379 = 0, whence w=.004+. Since the figure 4, suggested for the third decimal place, is not greater than 9, we infer that .01 is not too small ; that is, .01 is correct. The root of the given equation obtained thus far is 2.414. Diminishing the roots of the last equation by .004, 1 8.23 15.2443 -.064379 (.004 .004 .032936 .061108944 1 8.234 .004 16.277236 .032952 - .003270056 1 8.238 .004 16.310188 1 8.242 We now have the equation 2)3 + 8.242 v^ + 15.310188 v - .003270056 = 0. 596 ALGEBRA. [Ch. XL Neglecting the terms in if and v^, we obtain as the approxi- mation to the root of this equation .0002. That is, the ap- proximation to the root of the given equation is 2.4142. Since the figure in the fourth decimal place is less than 5, the required root, correct to three decimal places, is 2.414. The different steps need not be kept separate, but may be arranged compactly as follows : 11 - 7 -3 (2.4142 2 6 -2 3 2 - 1 10 -5 5 2 9 7 .4 9 2.96 -5 4.784 7.4- .4 11.96 3.12 - .216 7.8 .4 16.08 8.2 .01 15.08 .0821 - .216 .151621 8.21 .01 16.1621 .0822 - .064379 8.22 .01 15.2443 8.23 .004 15.2443 .032936 - .064379 .061108944 8.234 .004 15.277236 .032962 - .003270056 8.238 .004 15.310188 1 8.242 It is not necessary to write out each new equation in order to find the first figure of the corresponding root. It is evident that this figure is, approximately, the first figure of the quo- tient obtained by dividing the last term of the equation by the THEOEY OF EQUATIONS. 697 coefficient of the first power of the unknown number, neglect- ing sign. If the coefficient of the first po-wer of the unknown number be at any stage, the corresponding figure of the root is obtained by dividing by the coefficient of the second power of the un- known number, and taking its square root. For, the approxi- mate equation is then of the form m'f — n — 0, or y=iA—. \m 60. The example of the preceding article illustrates the following method : Find the integral part of the root by the method of Art. 53. Form the equation whose roots are less than the roots of the given equation by this number. Find the next figure of the root by dividing the last term of the transformed equation by the coefficient of the first power of the unknown number, neglecting sign, and checking the first figure of the quotient. Form the equation whose roots are less by this number than the roots of the first transformed equation. Take as the next figure of the root the first figure of the quo- tient obtained by dividing the last term of the second transformed equation by the coefficient of the first power of the unknown num- ber; and so on. The last term of each transformed equation must have the same sign as the last term of the given equation. If, for any transformed equation, too great a figure be obtained by the required division, the sign of the last term of the next transformed equation will be opposite to that of the given equation. If, for any transformed equation, too small a figure be obtained by the required division, the number suggested for the next decimal place will be greater than 9. 61. In applying Horner's Method, the decimal point can be avoided as follows, referring to the example of Art. 59. The decimal point is about to appear in the second stage of the work, the root of the first transformed equation being .abed. 598 ALGEBRA. [Ch. XL Before finding the value of a, let us form the equation whose roots are 10 times the roots of this equation, as in Art. 21. The coefficients of the resulting equation are 1, 70, 900, — 5000. The corresponding root of this equation is a.bcd. We now have -^inf = ^- -^^ i^^ ^'^^- ^9, we find that the root is less than 5 and greater than 4; that is, is A.hcd. Diminishing the roots by 4, we obtain a transformed equa- tion whose corresponding root is .bed. As before, we multiply the roots of this equation by 10, and obtain as the next figure of the root, 1, = ^|gggg , approxi- mately. The work follows : (2.4142 1 1 - 7 -3 2 6 -2 1 3 - 1 -5 2 10 1 5 2 9 1 70 4 900 296 -5000 4784 1 74 4 1196 312 - 216 1 78 4 1508 1 820 1 150800 821 - 216000 151621 1 821 1 161621 822 - 64379 1 822 1 162443 1 8230 4 15244300 32936 - 64379000 61108944 1 8234 4 15277236 32952 - 3270066 1 8238 4 15310188 8242 THEORY OF EQUATIONS. 599 In general, when the decimal point is about to appear, aimex one cipher to the second coefficient {counting from the left), two to the third, and so on. Proceed as in Art. 59, noting that the synthetic divisor is then an integer. Proceed in like manner with each transformed equation. 62. A negative irrational root is obtained by first trans- forming the equation to one whose roots are those of the given equation with signs changed, and finding the corresponding positive root of this equation. Roots of Numbers. 63. An approximate value of any real root of any number can be obtained by Horner's Method. From X = -^n, we have a;« — n = 0. The q roots of this equation are the q gth roots of n. BXBRCISBS X. 1-14. Find the irrational roots, correct to three decimal places, of each of the equations in Exercises IX., Bxx. 1-14. The following equations have each two imaginary roots. Find the real rational roots ; remove them and find the irrational roots of the depressed equation. Remove these roots, and solve the final depressed equation for the imaginary roots : 15. a;4-lla;3 + 38a;2-51a; + 27 =0. 16. x^ - 2a;* - 16a;3 - 24 a;^ +U4x + 320 = 0. 17. x5 + 3a;« - 7 a;S - 18 a;2 - 11 a; - 60 = 0. 18. 2x6 - 8x6- 11 X*- 6x8 + 48x2 + 38x- 15 = 0. Each of the following equations has two rational roots which can be expressed as decimal fractions. Find them, by Horner's Method, and solve the depressed equation : 19. 32x3 - 80x2 - 66 x + 189 = 0. 20. 64 X* - 713 x2 + 732 x + 247 = 0. 21. 40x4 - 13x3- 328x2 -13x-. 368 = 0. Find, correct to three decimal places, by Horner's Method, the values of the following arithmetical roots : 22. V9- 23. V2-5. 34. V^- 8S- V9-2- 26. {/b.l. 600 ALGEBRA. [Ch. XL Sturm's Theorem. 64. Descartes' Rule does not give the exact number of positive and negative roots of an equation. The principle of Art. 55 does not inform us definitely whether there is one or an odd number of roots betvyeen a and b, vyhen /(a) and /(6) have opposite signs ; whether there is no root or an even number of roots between a and b, when /(a) and /(6) have the same sign. The following theorem, discovered by Sturm, a Swiss mathematician, in 1829, gives the exact number of real positive and nega- tive roots of an equation, and also the exact number which lie iu any interval. Before applying this theorem, it is assumed that all the multiple roots, if any, have been obtained and the depressed equation formed. Let f(x) = be this depressed equation. Then, /(x) and f'(x) have no common factor. In the method now to be employed, the process of finding the H. C. F. of two expressions is continued until a remainder free of x is obtained. But the sign of each remainder is changed before it is used in the next stage as a divisor. Let/i(a;), /2(x), etc., be the remainders, with signs changed. Then the functions, f(x), f'{x), fi(x), f-i(x), ■■■ are called Sturm's Functions. We now have f{x) = Qif(x)+Bi, or since JJi = -/i(a;), /(x) = ei/'W-/i(a;); (1) In like manner, f'{x) = §2/1 (x)-fi(x); (2) fi(.x) = Qsh(x)-fs(x); (3) fi(x) = Qk+2fk+l(.x) - fk+2(x). 65. The following principles are derived from the relations of Art. 64 : (i.) Two consecutive functions cannot reduce to for the same value of X. Suppose fi{x) and fs{x) reduce to for the same value of x ; that is, have a common factor of the form x — r. Then, from equation (3), fi(x) and f2(x) reduce to for this value of x, and have a common factor X — r. In like manner, from (2), we infer that /'(x) and /i(x), and finally, from (1), that/(x) and f'(x) reduce to for this value of x ; that is, have a common factor x — r. But this is contrary to the hypothe- sis that/(x) and/'(x) do not have a common factor. (ii.) Any value of x which reduces to any function except the first gives to the two adjacent functions opposite signs. THEORY OF EQUATIONS. 601 Let r be a value of x which reduces /2(x) to 0. Since this value of x does not reduce /i (a;; or/s(x) to 0, we have /i(r) = -/8(r). The following principle is also important : (iii.) Just before x passes through a root r, of f{x) = 0, f{x) and f(x) have opposite signs; and just after x has passed through a root r, of f(x)= 0, f(x) and f'{x) have the same sign. Let r -h be a value of x just before it reaches the root r, and r + h a, value of X just after it has passed the root r, wherein A is a small number which approaches as a; approaches r. Then, /(r + h) =f{r) + hfir) + ^f"{r) + ..., f{r - h) = f{r) - hf\r) + ^f\r) - .... Since r is a root of f{x) = 0, /(r) = 0, and the above equations become f{r + h)= hf>{r) + ^f"{r) + ..., (1) /(7--A)=-V'W + ,f/"W--- (2) \± By taking U small enough, the term hf{r) can be made numerically greater than the sum of all the terms which follow it. Therefore, as /j=0, the sign of the second member will be the same as the sign of the first term, + hf{r), in equation (1), and - A/'(r), in equation (2). From (1) we infer that /(r + h) and /'(»•) have like signs ; that is, the sign of f{x) just after x has passed the root r is the same as the sign of fix) at the root r. But we can take the interval, ft, so small that no root of /(a;) = is included between r + h and r. Then, /'(»•) and /(r+Zi) will have like signs. Consequently, /(»•+ ft) and f{r-\-h) have like signs. In like manner, it can be shown that f{r— ft) and f{r — h) have opposite signs. 66. Sturm's Theorem may now be stated as follows : If a be substituted for x in Sturm's functions and the number of vari- ations counted, and then 6, o number greater than a, be substituted for x and the number of variations counted, the number of real roots between a and b is equal to the number of variations lost. First, let x pass through a root r, of /(x) = 0. Then, just before x reaches r, f(x) and f(x) have opposite signs ; that is, there is here one variation of sign. Just after x has passed the root r, /(x) and /'(x) have the same sign, and this variation is lost. 602 ALGEBRA. [Ch. XL Next, let X pass through a root r, of one of the other functions, say fi{x). Then, by Art. 65 (ii.), when x = r, fi{x) and fs{x) have opposite signs. In passing from a value a little less than r, say r — h, to a value a little greater than r, say r + A, we can take the interval so small that no root of /i(x) = 0, or of /3(x)= 0, is included in it. Therefore /i(x) and faix) do not change signs in this interval. Then, just before x reaches r, and just after it has passed r, /i(x) and/3(a:) have opposite signs. These may be either + and — , or — and + . The sign of /2(x) will change except when r is a multiple root of /2(x)=0 an even number of times. The following arrangement of signs is now possible : /i /2 /s /i /2 fa f\ fi h /i h h x = r-li: + + - + - + + + x=r^-h: + + + - + - + + Observe that the effect of the change of sign in /2CX) is to shift the variation, and not to cause it to disappear. Evidently, if /2(x) do not change its sign, there is no loss of variation. Since a variation is lost when, and only when, x increases through a real root of /(x) = 0, the number of variations lost, as x passes from a to 6, is the same as the number of real roots between a and 6. 67. Find the number and location of the real roots of the equation X* - 4 x3 - 6 x2 + 20 X + 5 = 0. We have /(x) = x« - 4 x^ - 6 x^ + 20 x + 5, yCx) = 4 x3 - 12 x2 - 12 X + 20. The work of finding the other Sturm's Functions follows : ;c3_3x2_3x + 5)x«-4xS-6x2 + 20x+ 5(x-l x^-3x^-3x^+ 5x - xa-3x2 + 15x - x8+3x''+ 3x- 5 H- (-2) )-6x'+12x+10 /i(x)=3x2- 6x i)3x8-9x2- 3x3-6x'^- 9x + 15(x- 6x 1 -3x2- -3x2+ 4x 6x+ 5 ^ (-io)i^ lOx+lO X- 1)3x2 3x2. -6x- -3x -5(3x -3x -3x + 3 -8 /8(x)=8 THEORY OF EQUATIONS. 603 The first divisor is f{x) divided by 4. A positive factor may be tlius introduced or suppressed, since the sign of the result' is not thereby affected. Each remainder is divided by a negative factor in accordance with Art. 64, but otherwise a negative factor must not be introduced or suppressed. We now have /(a;)=a:*-4a;8-6a;2 + 20a;+ 5, /' (k) = 4 a;8 - 12 a:^ - 12 a; + 20, /i(x) = 3 a;2 - 6 a; - 5, fi(x) =x-\, fi(x) = 8. Then, by Art. 49, x / /' A h /8 Var. — 00 + _ + — + 4 + + — - + 2 + 00 + + + + + Since two variations are lost in passing from -co to 0, there are two real negative roots. And since two variations are lost in passing from Oto +00, there are two real positive roots. We next locate the positive roots and the negative roots by substituting in succession 0, 1, 2, •••, and 0, -1, -2,.... Observe that 1 reduces /'(») and /^{x) each to 0. This may be disre- garded in counting the number of variations. Since the adjacent func- tions have opposite signs, the number of variations is the same as if + or — were where the is. Since one variation is lost in passing from 2 to 3, one positive root lies between 2 and 3. The other positive root is found to lie between 4 and 5. The negative roots lie between and — 1, and — 2 and — 3. 68. The following graphic repre- sentation of the Sturm's Functions will throw some light on how the various signs change and how the variations are lost. Let a, 6, c, d, etc., be points on the x-axis between each pair of roots of any of the Sturm's Functions, in Fig. 19. The following table shows the signs of the different functions at these points, taken from their graphs. X / /' /i h h Var. + + — — + 2 1 + - + ^2 2 + — — + + 2 3 — — + + + 1 4 — + + + + 1 5 + + + + + + + — - + 2 -1 — + + — + 3 -2 — — + — + 3 -3 + — + - + 4 604 ALGEBRA. [Ch. XL rM( f /' /i /2 /3 Var. a + — + — + 4 b - - + - + 3 c — + + - + 3 d - + — - + 3 e + + — — + 2 f + — - + + 2 9 — — — + + 1 h — - + + + 1 i - + + + + 1 J + + ■ + + + Fig. 19. Observe, first, that a variation is always lost when x passes through a root of /(x), as from a to b, from /to g, etc. Also, that as x passes through a root of any other function, the signs of the functions are so distributed that no variation is lost, but a variation is moved to the left toward /(a;) and /' (k) , as in passing from 6 to c through a root of /'(x) =0, from g to h through a root of /i(x) = 0, etc. Observe, also, that although just after x has passed a root of /(x) a permanence is established between /(x) and/'(x), yet before another root of /(x) is reached, the distribution of signs has changed this perma- nence into a variation without the loss of a variation elsewhere. Thus the permanence established at b is changed to a variation at c, which is maintained at d and lost at e. Note how the variation at e between /^(x) and fi(x) is gradually shifted to the left, to be lost at j. 69. Sturm's Theorem gives the exact number of imaginary roots, since this number is equal to the degree of the equation less the number of real roots. Roots nearly Equal. 70. Sturm's Theorem gives the exact number and location of all the real roots of an equation, but, in application, it is very laborious. As a rule, the method given in Art. 63 is sufficient for determining the location of the real roots. But if two or more roots lie between the same pair of consecutive integers, it is necessary to apply Sturm's Theorem. The following example will illustrate this special case. Ex. 1. Solve the equation x* - 4 x' + x" + 6 x + 2 = 0. THEORY OF EQUATIONS. 605 By inspection, as in Art. 29, we find that 4 is a superior limit to the real roots, and — 2 an inferior limit. By Descartes' rule of signs the equation cannot have more than two real positive roots, or more than two real negative roots. Substituting in succession 0, 1, 2, 3, 4, -1,-2, that is, all consecutive integers within the limits to the roots, we obtain always positive results. We therefore conclude either that the roots are all imaginary, or that an even number of roots lies between two consecutive integers. To settle this question, we apply Sturm's Theorem. We have /(a;) = a;4 - 4 a' + x2 + 6 so + 2, /' (a;) = 4 xs - 12 a;2 + 2 a; + 6, /i(a;) = 5x2-10a;-7, /2(a;) = x-l, /s(x)=12. We infer from the table on the left that the equation has two real positive roots and two real negative roots. Since two variations are lost in passing from 2 to 3, the equation has two real roots between 2 and 3. For the same reason, it has two real roots between and — 1. We proceed to find the two roots between 2 and 3, the integral part of each being 2. Diminishing the roots by 2, we find the transformed equation, with its roots multiplied by 10, to be y* + iOy»+ 100 2/2 - 6000 y + 20000 = 0. This equation has two roots between and 10. Substituting succes- sively 0, 1, 2, ■••, we find that one root lies between 4 and 5, and one be- tween 7 and 8. Hence, the required roots to the first decimal place are 2.4 and 2.7. The roots being now separated, we proceed with each by itself, in the regular way. The two roots, to four decimal places, are found to be 2.4142 and 2.7321. Had the two roots of the transformed equation not been separated, it would have been necessary to apply Sturm's Theorem to it, as to the original equation. The two roots between and — 1 can be found in like manner, first transforming the equation into another whose roots are those of the given equation with signs changed. X / / /i /2 /a Var. — 00 + — -1- _ + 4 + + — . — -1- 2 -1-00 -1- + + + + 1 + - + 2 2 + — — + + 2 3 + + + + -1- -1 + - -1- - -h 4 606 ALGEBRA. [Ch. XL BXERCISBS XI. Determine the number and location of the real roots of the following equations. Also, find each real positive root correct to three decimal places : 1. a;8 - 9 X - 12 = 0. 2. k' - 6 a;2 + 8 a; - 1 = 0. 3. a;8 - 3 x2 - 4 a; + 13 = 0. 4. x' - 27 x + 246 = 0. 8. x3 + 3x2_4x + l =0. 6. x4- llx2 + 28x-6 = 0. 7. X* - 4 x3 - 8 x2 + 40 X - 20 = 0. 8. x* - 6 xs + 11 x2 - 10 X + 2 = 0. 9. X* -8x3 + 10x2 + 40x-63 = 0. 10. X* - 2 x8 - 17 x2 + 20 X + 70 = 0. 11. 2 X* - 4 x8 - 5 x2 + 12 X - 3 = 0. 12. 5x4 - 10x3 - 36x2 + 60x + 36 = 0. Reciprocal Equations. 71. To transform an equation into another wliose roots are the reciprocals of the roots of the given equation. Let aoX» + ai"-i + a2X»-2 + ••■ + a„-^^ + a„_ix + a„ = be the given equation. If 2/ = - , or X = - , then each value of y is'the reciprocal of a value of x. Substituting - for x, and multiplying by j/", a„J/" + fl!„-i^-^ + an-iV"-'^ + ■■■ + a^y'^ + axy + ao = 0. That is, the coefficients of the transformed equation are the coefficients of the given equation written in reverse order. Ex. The equation whose roots are the reciprocals of the roots of the equation 2x'^ — 3a? + lx + 5 = is &f + 7f-3y + 2 = (i. 72. A Reciprocal Equation is one whose roots, in pairs, are reciprocals one of the other. That is, if r be a root of a reciprocal equation, - is also a root. r 73. Now, if r be a root of the equation aox" + aix"-^ + ■•• + a„-ix + o„ = 0, (1) then, by Art. 71, - is a root of the equation a»r + a„-ir-' + • • • + ais^ + ao = 0. (2) THEORY OF EQUATIONS. 607 Equations (1) and (2) will in general be different ; but if equation (1) be a reciprocal equation, - will be a root of it also. That is, each root of (2) is a root of (1), and vice versa, and the two equations are equivalent. Hence, corresponding coefficients must be equal or proportional, and at) _ gi _ _ a„-i _ g^ an an-i Q!l CSo Whence, ao^ = a„2, ai^ = an-i\ ••■, or ao = ±a„, ai = ±an-h—. That is, in a reciprocal equation, the coefficients of terms equally distant from the beginning and end are equal, or equal and opposite. Thus, 2a^ + 3x' + 3x + 2 = 0, and aa^ - 6af' + 6a; - a = 0, are reciprocal equations. Observe that an equation of even degree has one middle term. When the equidistant coefficients are opposite in sign, this term must be wanting, since it cannot be equal and oppo- site to itself. Reciprocal Equation of Odd Degree. 74. Let UqX" + OiX""^ + a^x"-'-' + •■• ± ajX^ ± ajX ± aQ = be a reciprocal equation of odd degree. Grouping the equidistant terms, we have ao(x" ± 1) + aix{x^-' ± 1) + aiX^(_x''-* db 1) + • •• = 0, wherein the upper signs go together, and also the lower. When the upper signs are taken, the equation is divisible by x-\- 1, and — 1 is a root. When the lower signs are taken, the equation is divisible by x — 1, and 1 is a root. When a" + 1 is divided by a; + 1, or a;" — 1 by a; — 1, the first and last terms of the quotient have the same sign. These terms, multiplied by a„, are the first and last terms of the depressed equation. In like manner, it follows that other equidistant terms in the depressed equation have the same sign. Hence, a reciprocal equation of an odd degree has a root — 1 when equidistant coefficients have the same sign, and a root + 1 when these coefficients have opposite signs. 608 ALGEBRA. [Ch. XL The depressed equation is a reciprocal equation of even degree, in which the coefficients of equidistant terms have the same sign. Reciprocal Equations of Even Degree. 75. Let a^" + ajs;""' + Oja;""^ H a^ —a^x — ao = be a reciprocal equation of even degree, in which equidistant coefficients have opposite signs. Grouping terms, we have aoix" - l)+aix(x''--'' - 1)+ aix'^ix"-^ - 1)+ ••• = 0. Since n is even, k" — 1 is divisible by x'^ — 1, and ± 1 are roots. Therefore, a reciprocal equation of even degree, in which the coefficients of equidistant terms are opposite, has the roots ± 1. Tlie depressed equation is a reciprocal equation of even degree in which the coefficients of equidistant terms have the same sign. 76. Standard Form of Reciprocal Equations. — It follows from the preceding articles that any reciprocal equation can be re- duced to one of even degree in which the coefficients of equi- distant terms have the same sign, if it be not already in this form. This is therefore taken as the standard form of reciprocal equations. Ex. Eeduce the equation a^ + 2aj* — 3a^ — 3a^ + 2aT-f-l = to the standard form. By Art. 74, this equation has the root — 1. Dividing by x + 1, we obtain the depressed equation x*+3^-iaf + x + l = 0. 77. A reciprocal equation of the standard form can be reduced to an equation of half its degree. Let aoa:^" -I- aix''^-''- + ••• -f- Unx:^ + 1- aix -f ao = be a standard reciprocal equation, wherein a„x" is the middle term. Dividing by k", and grouping terms having equal coefficients, we obtain ^'^" + ^«)+"^K' + ^)+ ■■■ + '''' = "■ Let x + - = s. X THEORY OF EQUATIONS. 609 Then, fa;'-i + -i-Vx + l'\ = a;" + i+a;'-2 + J-. Whence, x" + 1 = z(x''-i + ^\-( x'^-'' +^—\. x" \ a;»-V \ K'-V When n = 2, x^+ — =z2-2; x^ Ji = 3, x» + ^= z(^ - 2)- s = sfi - 3z ; and so on. ETidently, k" + — is of the nth degree in e. Therefore, making the above substitutions, we obtain an equation of the reth degree in s. Ex. Solve the equation Dividing by a^ and grouping terms, Substituting z for * + -, we have X !^-3z-2(z'-2)+2z-2 = 0, or z'~2z^ — z + 2 = 0. The roots of this equation are found to be — 1, 1, 2. Solving the equations x + - = -l, x+- = l, x + - = 2, X XX we obtain - 1 + tV3 2 l + ^V3 2 ^ ^ 2 ' _l + iV3' 2 ' l + VS' ' ■ Binomial Equations. 78. An equation of the form a:» - a = (1), or a;" + a = (2) is called a Binomial Equation. Since a;" — a, or a;" + a, and the derived function na^~^ do not have a common factor, the n roots of a binomial equation are distinct. From (1) and (2), we obtain x — -^a, or a; = -y/— a. 610 ALGEBRA. [Ch. XL Since each equation has n distinct roots, we conclude that any positive or negative number has n distinct algebraical nth roots. 79. As in Ch. XXII., Art. 1, Ex. 1, the n roots of the equations a;" — a = 0, CB" + a = 0, can be obtained by multiplying the n roots of the equations a;" - 1 = 0, «" + 1 = 0, by the arithmetical nth root of a. But the latter equations are evidently reciprocal equations and can be solved by the methods of Arts. 74-77. Ex. Solve the equation a* + 1 = 0. We have (a; + 1) (a;* - a::^ + as^ — a; + 1) = 0. Whence a; = — 1, (1) and a!* - a?' + a^ - a; + 1 = 0. (2) From (2), by the method of Art. 77, we obtain a!=i(^+V5±V2V5-10), a; = ^(l-V5±V-2V5-10). EXERCISES XII. Solve the following equations : 1. x*-3x8+4x2-3a;+l=0. 2. 2xt+5x^-21xH5x+2=0. 3. 3a;i-2a;8-34a;2_2x+3=0. i. 6x^ - x« - Ux^ -x + 6 = 0. 5. x5-6a;i + 13a;»-13x2 + 6x- 1=0. 6. 3x5-x*-2x8-2x2-x + 3 = 0. 7. 4x« + 29x5 + 55x*-55x2-29x-4 = 0. 8. 12x6 + 16x5_25x* - 33x5 - 25x2 + 16x+ 12 = 0. 0. x* + 1 = 0. 10. X* + c(2 = 0. 11. x5 - 1 = 0. 12. x6 - 32 = 0. 13. x8 + 1 = 0. 14. x, =~p, and 3 (2/io2) (»ia) , = 3 sri«ioS, = - p ; but 3(s^io) («io), = 3 2/i3ia2, =^-p; and so on. Therefore the values of y + e, and hence of x, are restricted to 2/1 + 2i, 2/ia + eia", yia" + z^a. 85. Although the above solution is commonly called Cardan's Solviion, after Cardan, who first published it in 1645, it is not due to him. The solution was discovered by Tartaglia and Ferreo, independently of each other, about 1505. Cardan obtained it from Tartaglia. 86. The chief importance of Cardan's solution is in theoreti- cal rather than in practical applications. The more general methods, previously given, are as a rule to be preferred in solving numerical equations. Ex. Solve the equation as' — 3a; + 2 = 0. We have p = — 3, q = 2. Therefore x = -^-l+-^-l=-2. The depressed equation is a;^ — 2 a; + 1 ; whence a; = 1,1. These two roots could also have been obtained as follows : Since yj = Xj = — 1, these roots are — « — a^, = — cs" — « = 1, by Art. 82. 614 ALGEBRA. [Ch. XL When two of the roots are equal, or two imaginary, the third root can be obtained by Cardan's formula. When, however, the three roots are real and distinct, Cardan's formula does not give a practicable algebraical solution. General Solution of the Biquadratic. 87. One of the various solutions of the biquadratic equation follows. If the term in a? be present in the given equation, it is first transformed into one in which the term of the third degree is wanting. We therefore assume the biquadratic equation in the form a^ +p3? + qx + r = (i. A factor of this equation corresponding to two real roots is of the form x^ + hx + h. Also, by Arts. 18-19, a factor corresponding either to a pair of conju- gate surd roots, or to a pair of conjugate imaginary roots, is of a similur form. Therefore, whatever be the nature of the required roots, we may assume a* + J3x2 + qx+r= (pi? + hx + k) (oi? + mx + n) = x^ +(h + m)x^ +(k + n + hm)x^ + (ft» + km)x + kn. Equating coefficients of like powers of x, we have h + m = 0, (1) k + n + hm=p, (2) hn + km = q, (3) kn = r. (4) From (1) m =— ft. Then (2), (3), (4) become k+n = h^+p, (5) hin-k)-=q, (6) kn = r, (7) From (5) and (7), n - k = ^(h* + 2ph^ + p^ - 4 r). Substituting this value of to - fc In (6), squaring, and reducing, we obtain This cubic equation In K' has always one real positive root by Art. 64 (i.). When h is known, the values of k, m, n, are easily determined. Ex. Solve the equation a;^ - 9 a^ - 12 a; + 10 = 0. Here p =~ 9, q =-12, r = 10. THEORY OF EQUATIONS. 615 Hence the cubic in W is ^6 _ 18 A^ + 41 ^2 _ 144 = 0. A root of this equation is found to be 16. Whence Zi = ± 4. Let us take fe = 4. Tl\en m = — 4, and from (6)' and (7), A; + n = 7, fcw = 10 ; whence fc — n = ± 3. Taking A; — w = 3, we obtain k = 5,n=2. The required factors are a? +4,x + 5 and a^ — 4 aj + 2. Taking & — w = — 3, we obtain fc = 2, n = 5. But these values do not satisfy (6), since 4 (5 — 2) :#= — 12, and are therefore not admissible. From a^+4a; + 5=0, we obtain x = —2 ± -y'— 1 ; and from o^ - 4 a; + 2 = 0, a; = 2 ± ^2- Had we taken h = — 4:, we should have obtained m = 4, k=2, n = 5. These values give the same factors as before. BXEECISES XIII. Solve the following equations : 1. x« + 6x-1 =0. 2. a;8 - 27 a; + 54 = 0. 3. a;s _ 12 a; + 16 = 0. 4. k? - 9 a; + 28 = 0. 5. a;5 + 24 a; - 56 = 0. 6. a;S - 48 a; - 128 = 0. 7. 4 a;' - 3 a; + 1 = 0. ' 8. a;' + 72 x + 152 = 0. 9. a;3 + 9a;2 + 21a; + 18 = 0. 10. a;S - 6a;2 - 12 a; + 112 = 0. 11. a aS + 3 a:2 - 11 X + 21 = 0. 12. a;5 - 9 x^ + 171 a; - 755 = 0. 13. x*-6x2-3a; + 2 = 0. 14. x* - 11 x2+ lOx + 6 = 0. 15. x* + x2+ 36x+ 52 = 0. 16. x* - 15a;2 _ 42x - 40 = 0. 17. x4 - 4 x8 + 11 x2 - 8 X + 18 = 0. 18. x« - 2 x^ - 6 x^ + a; + 2 = 0. 19. x* + 3 x8 - 4 x2 + 16 X + 8 = 0. 20. 3 x« - 20 x8 - 30 x2 + 40 a; - 8 = 0. >;^*j>