OlotttfU IniuerattH 2Itbraw| atljata, JfMB Inrk ' i f *• BOUGHT WITH THE [NCOME OF THE SAGE ENDOWMENT FUND THE GtFT OF HENRY W. SAGE 1891 Cornell University Library arV19306 A brief course in college algebra, 3 1924 031 221 538 olin.anx The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031221538 COLLEGE ALGEBRA A SERIES OF MATHEMATICAL TEXTS EDITED BT EARLE RAYMOND HEDRICK THE CALCULUS By Ellbky Williams Davis and William Chables Brbnke. ANALYTIC GEOMETRY AND ALGEBRA By Alexander Ziwet and Louis Allen Hopkins. ELEMENTS OF ANALYTIC GEOMETRY By Alexander Ziwet and Louis Allen Hopkins. PLANE AND SPHERICAL TRIGONOMETRY By Alfred Monroe Kenton and Louis Ingold. ELEMENTS OP PLANE TRIGONOMETRY By Alfred Monroe KEispyoN and Louia Ingold. ELEMENTARY MATHEMATICAL ANALYSIS By John Wesley Young and Frank Millett Morgan. PLANE TRIGONOMETRY By John Wesley Young and Frank Millett Morgan. COLLEGE ALGEBRA By Ernest Brown Skinner. MATHEMATICS FOR STUDENTS OF AGRICULTURE AND GENERAL SCIENCE By Alfred Monroe Kenyon and William Vernon Lovitt. MATHEMATICS FOR STUDENTS OF AGRICULTURE By Samuel Eugene Rasoe. THE MACMILLAN TABLES Prepared under the direction of Eable Raymond Hedrick. PLANE AND SOLID GEOMETRY By Walter Burton Ford and Charles Ammerman. CONSTRUCTIVE GEOMETRY Prepared under the direction of Eaele Raymond Hedrick. JUNIOR HIGH SCHOOL MATHEMATICS By William Ledley Vosbukgh and Frederick William Gentleman. A BRIEF COURSE IN COLLEGE ALGEBRA By Walter Burton Ford. A BRIEF COURSE IN COLLEGE ALGEBRA BY WALTER BURTON FORD PKOFESSOB OF MATHEMATICS THE UNIVEESITT OF MICHIGAN THE MACMILLAN COMPANY 1922 AU rights reserved H PEINTED IN THE UKITBD BTATieB OF AMBBIOA f\^vl'lo\ COPYBIGHT, 1922, bt the macmillan company. Set up and electrotyped. Published August, 1939. PREFACE The present book, which is intended as a text for use in the freshman year of college or technical school, has been pre- pared with the following considerations particularly in mind. (1) In view of the fact that a considerable number of pupils enter college today for whom no knowledge of quad- ratic equations can be assumed, it seems desirable to include a complete treatment of this important topic in the present- day college text. Two chapters (II and III) are therefore devoted to quadratic equations, the first of which, however, is altogether elementary and may be omitted at the discretion of the teacher. (2) In order to meet the needs and customs of different institutions, the various chapters have been made quite inde- pendent of each other, thus permitting a ready adjustment of the book to either a long course or a short one, and affording the teacher the greatest possible flexibility in the choice of topics for any course of given length. In this connection the author feels that it should be frankly recognized that today college algebra in most institutions is pursued but a few weeks. This makes it impossible to cover a wide range of topics and forces such a selection as may fit best the needs of the particular situation. Much may be gained, however, from a brief but intensive study of a few special topics in algebra at this period of the pupil's career. (3) In view of the importance in elementary physics and other applied fields of the subject of variation, this topic has been treated somewhat more fuUy than usual. On the other hand, such topics as complex numbers (vector addition, multiplication, etc.), partial fractions, limits and infinite series vi PREFACE have been omitted in the behef that, even in case there is time to include them in the course, they may be taken up to greater advantage at a later time when the pupU is face to face with their chief applications. (4) The ideal problem for a freshman text is a short one which illustrates pointedly the mathematical principle at issue. Problems long in statement and dealing with the tech- nique of the arts and sciences should have but httle place in the freshman year. At this period the essential task of both teacher and text should be to train the pupil in accuracy and conciseness of thought and expression, the mathematics itself forming, for the most part, the medium through which this may be accomplished. Mention may be made of the fact that certain sections of the book have been starred (*) to indicate that they are of minor importance and may be omitted without destroying the continuity of the whole. Also, in view of the natural over- lapping of certain parts of the college course with the more advanced parts of the usual second, or advanced course of the high school, the author has ^not hesitated in the treatment of some of the earlier topics, such as the progressions, variation, binomial theorem and logarithms, to follow closely the treat- ment of these same topics to be found in the later pages of the "Second Course in Algebra" by Ford and Ammerman (Mac- millan), the exercises, however, being changed. The author would here express his thanks to Professor E. B. Lytic, of the University of Illinois, who read the manu- script and offered valuable suggestions, and to Professor J. L. Markley and Mr. R. W. Barnard, of the University of Michigan, who assisted in reading the proofs. Walter Btjeton Fobd. University of Michigan. CONTENTS CHAPTER • PAGE I. Review Topics .... 1 II. Quadratic Equations 28. III. Properties of Quadratic Equations 43 IV. Simultaneous Quadratic Equations .... 65 V. The Progressions . . . 70 VI. Variation . 86 VII. Logarithms 98 VIII. Compound Interest and Annuities 125 IX. Mathematical Induction — Binomial Theorem . 130 X. Functions , 140 XI. The Theory of Equations . . 159 XII. Permutations and Combinations . 182 XIII. Probability 194 XIV. Determinants . 204 Answers . . 231 Tables ... . . 241 Index 263 yu COLLEGE ALGEBRA CHAPTER I REVIEW TOPICS 1. Algebraic Reductions. The process of reducing, or simplifying, a given algebraic expression makes frequent use of the following principles from elementary algebra, f Pbinciple 1. A parenthesis preceded by a minus sign may be removed from an expression if the signs of all the terms in the parenthesis are changed. Thus a — (6— c) =a — 6+c. Likewise a+b—(c — d+e)=a-\-b—c-l-d—e. A parenthesis preceded by a plus sign may be removed without changes in the signs of the terms which it includes. Thus a+(b—c)=a+b—c. Likewise a+b+(c—d+e)=a+b+c—d+e. EXERCISES Simplify each of the following expressions by removing all paren- theses and combining terms wherever possible. 1. x — (,y — z). 4. m—{n—2a). 2. x — (—y—z). 5. 5a — 26 — (a— 26). Ans. 4a. 3. -(a+6)+2. 6. a-(6-c+o)-(c-6). fNo attempt will be made in the present chapter to give a complete summary of the topics treated in high school algebra. Only a few will be considered, particularly those which are important for the study of the later chapters of this book. The student wiU do well to have at hand at all times for reference purposes a textbook in elementary algebra, preferably the one which he has used in the high school. 1 2 COLLEGE ALGEBRA [I, § 1 7. 2xy+Sifi-(x^+xy-y'^). 8. m+{3m—n) — (2n—m)+n. 9. a2b+62c+o2c2-(2a2b2_3a2c) + (4a26-5a2c2-6a262). 10. ^+(y~^)~('^~^) a+b-(2a+b-c) 11. (o+b)2-(o-6)2. j2 2ab-(a+b)2 a;2-(x-2/)2 13. o(b-c)+6(o-c)-c(o-b). Principle 2. Multiplying or dividing both the numerator and the denominator of a fraction by the same number does not change the value of the fraction. Thus Likewise Also This principle is frequently used to change, or reduce a fraction to a form having a given denominator. Thus, suppose it is desired to change the fraction a/{a+b) to a form having cfi—lfi as its denominator. To do so, we multiply both numerator and denominator by a—h, as follows: a ^ aja-b) ^ cfi—ab ^^_ a+b (o+b)(o-b) o2-b2 The principle is also used to reduce a fraction to its lowest terms. Thus, suppose we are to reduce the fraction 2la2x^y 30a3xz to its lowest terms. The process consists in dividing both numerator 2 3" 2X2 4 3X2 6 8 8^2 4 10 lO-i-2 5 a — = b aXa bXa a2 \b' mrfi-. -mn -mn I,§1] REVIEW TOPICS and denominator by all the factors which they have in common; that is, in the present case, by 3, by a2, and by x. In practice, the work is done by cancellation as shown below: 7 X ^S^/z lOaz lOo EXERCISES 1. Change 2/3 to a fraction whose denominator is 21. 2. Change 4/5 to a fraction whose denominator is 125. 3. Change 5a/7 to a fraction whose denominator is 42. 4a2 4. Change — to a fraction whose denominator is 20?/*. 52/ x-Z 6. Change to a fraction whose denominator is {x — \)K x — 1 6. Reduce ■ to a fraction whose denominator is 9— a2. d — c 7. Reduce to a fraction whose denominator is a — 6. o—a Reduce each of the following fractions to its lowest terms. R. "^^-i^ 1B 3a2+3a6 a^xy a^6-a63 g o2b2a;2 1p 3a2!,-363 63x2/2 2a6-262 1ft 16m2?ii2z2 yj a^bc—lfic 4Qaw?yzi 3a26+36s 11 77o7x5632/ 1R a(a +26)* 121a366c7 6(a2 -462)2 12 im+l2/ ^9 ^2-2x4+3^ ajj/m+l X2-X6 „ a2-fe2 jjO m—nfi—n+mn (a+6)2 m—mn+rfi—n 1^ o2-2ab+62 21 a6+oc— d6— dc o2-b2 OT6+mc 4 COLLEGE ALGEBRA [I, § 2 2. Addition and Subtraction of Fractions. In case two fractions have the same denominator, their sum will be equal to the sum of their numerators divided by this denominator. Thus 2 5_2 + 5_7 3 3 3 3 Likewise a . c _ a+c . b b b ' mn p—q _'mn+p—q x'hj x^y x^y In case two fractions do not have the same denominator, they may be added by first changing them, as in §1, so that they shall come to have equal denominators, and then pro- ceeding as mentioned above. Thus 2 3^£^ ^8+9^17^^5 • 3 4 12 12 12 12 Likewise o cad 6c_ad+6c. b d bd bd bd m+n m-n _ (m+n)^ (m—n)^ _ m?+2m,n+n^+nfi—2mn+n^ _ 2m^+2n^ m-n m+n m^—rfi T/fi-r^ nfi—n^ irfi—rfi In practice, when adding several fractions, it is best to determine first the least common multiple {L. C. M.) of the several denominators, that is, the expression of lowest degree which exactly contains each of them, then change each frac- tion so that its denominator shall be this L. C. M. and add as indicated above. Thus, in adding 2/15 and 3/10, the L. C. M. of the denominators is 30. The two fractions, when changed so as to have 30 as denomi- nator, are respectively 4/30 and 9/30. Hence the desired sum is (4+9)/30 = 13/30. Likewise, in adding a/{mM) and b/{mrfi), the L. C. M. of the denominators is rrfirfi, so that _a 6 _ an . bm an+bm . m^n mrfi m'hi^ m'W nfin^ I, §2] REVIEW TOPICS Similar statements apply whenever one fraction is to be subtracted from another, or when both addition and sub- traction are involved any number of times. Thus Sa',b „ , 1 6ob , 5b2 30b , 10 6o6+5b2-306+10 a„„ 5 2 b 10b 10b 10b 10b 10b Likewise {a-b)^ {a+b)^ 6ob a—b a+b 6ab a+b a-b a'^-b^ o2-b2 a^-b^ ' af-b^ (a-b)g-(a+b)g+6ab ^ a2-2a6+ba-o2-2ab-bii+6ab ^ 2ab ^^g_ EXERCISES Simplify each of the following expressions by performing the indi- cated additions and subtractions. g X X — 2 x-2 x+2 10. a:-4 g-6 ^ x+8 3 8 6 ' ax—bx+ab 1. Sx 7x _ 4 10 2. 6a;-l 3x-2 x-5 8 7 4 3. a—b b—c ab be 4. ax , +a. a—x a a(a—x) Hint. 0=- = -^^ - 1 a—x 5. ,,,a^+b^ a+b+ a—b 6. b—c a—c be ac 7. a+b a—b a—b a+b R .l„ '^'+y 11. 1-- l2 10 12. °+l 6__^ 02-9 a+5 a+3 m'^+n^ , 13. m Vn. m—n 14. -°' "~^ a 3 2 a+2 4- HlNT. 4—02 (fi-4: 16. ffl+& a^+b2 b-o a-b b^-a^ a+b 16. (o+b)2 ^ ^ 2ob a24-62 o2-b2 17. l+i+^£-_2. a; l+a; 6 COLLEGE ALGEBRA [I, § 3 3. Multiplication and Division of Fractions. Principle 1. In order to multiply two or more fractions together, multiply their numerators together to get the numerator of the product, and multiply their denominators together to get the denominator of the product. In performing such multiplications, it is desirable to cancel like factors from numerator and denominator wher- ever possible. 2 Thus Likewise Similarly 1 2 6_ 1XXX^ _2 2 3 7 ^X/X7 7 4x2/ 3a^ lixj/-i^ lax a—b d jfl'-^d rX~ a+b (^-b^ (a+6)j:a'-67(a+6) {a^bf Pbinciple 2. In order to divide one fraction by another, invert the divisor and proceed as in multiplication. Thus 1^21 33 2^3~2^2~4' Likewise x 5mn 10m?n pipn/ pa^ ax 2 2m Similarly a— 6 (a-bf _ c?-ab jfl-^ b b{a-b) a+b b a+b a^fl^=-Vy a(a+b) EXERCISES Perform each of the following indicated multiplications. J 5x^ 3as g 2ox 106^ ' 2ac 101/2- ■ i2by^'^' n imn I5bx . a^b" Gr' -X *■ V- Zxy mnf 4x a^-^b^" I, § 3] REVIEW TOPICS - a 6 o 4a— 6 2a 4x^—2/^ 5. N/ . o. y X ' a+b a—b 2x+y 4a^—ab 4 6. ^y' ,^ 25-lOa: ^ x^+5x+6 3^+7a:+10 {a-b)\, b (a+bf ^„ /^ 2 \ /m2+m-2N a^-ab a^-b^ \ to-1/\ mHm / 7. a+6 '"c?—ob'"c? — b Perform each of the following indicated divisions x+2y 11. 12a% 25oc 4aa; ■24c2 12. 5ab 3aV ■ 256^ 15a2 16. ?!_^+(x2-3xi/+2j/2) j,j x2^^_2 3?-x-e 7x^ 21x^2/2 13. li-^ 14. 41/3 • i4(j x2-5x+4 ■x2+x-20 - ("D^(''4)- (to+2/)2 ' m^-y^ 2a+l „. , , , . lo?—(? . o— c 15. (4a+2)^?^. 20. (a+c). (^.f^)- 5a \l+x 1-x^/ Simplify each of the following expressions. xrh^ i_^ _1 x+y_oc+y 24 '^'' - 25 ^. 26. ^. 27. — ^ — ^ ^-^ 1+5 1+- ab" x^ a; y x 8 COLLEGE ALGEBRA [l, § 4 4. Simple Equations. By a simple equation is meant one which, when cleared of fractions, contains the unknown number to no higher power than the first. The usual method of solving such equations is illustrated below. Example. Solve the equation x+1 2x-b lla+5 1-13 2 5 " 10 3 ' Solution. The L. C. M. of the denominators is 30. Hence, multi- plying both sides of the equation by 30 in order to clear of the fractions, we obtain 16(x+l)-6(2s-5)=3(llx+5)-10(x-13), or 15x+15-12a;+30 = 33x+15-10a;+130. Transposing and collecting like terms now gives -20x = 100. Therefore x= —5. Ana. Check. Placing x = — 5 in the original equation gives -5 + 1 -10-5 -55+5 -5-13 2 5 ~ 10 3 ' or -4 15 -50 18 2 5 ~ 10 3 ' or -2+3= -5+6, or 1 = 1. EXERCISES Solve each of the following equations for x, checking your answer for each of the first five. 1 2x-3 x+1 5x+2 4 ~6~~ 12 2 ^-5 2x+3 ^ 7x+3 3 6 12 3. 1 2i X 3x _ 3 4x" 19 24' 4. X 3~ 3x- -5x -7" 2 ^3' I, § 4] REVIEW TOPICS 9 5 6x+3 3a:-l 2x-9 g 2x+l _ 8 2a:-l 15 5x-25~~5 " 2x-l 4x^-1 ~2i+l' In each of the following problems, let x represent the unknown quan- tity, then form an equation and solve it: 7. Divide 38 into two parts whose quotient is 7/12. 8. Divide 96 into two parts such that 3/4 of the greater shall exceed 3/4 of the smaller by 6. 9. I have $110 in one bank and 875 in another one. If I have $45 more to deposit, how shall I divide it between the two banks in order that they may have equal amounts? 10. A motor boat traveUng at the rate of 12 miles per hour crossed a lake in 10 minutes less time than when traveling at the rate of 10 miles per hour. What was the width of the lake? [Hint. Time = Distance -=- Rate.] 11. A freight train goes 6 miles an hour less than a passenger train. If it goes 80 miles in the same time that a passenger train goes 112 miles, find the rate of each. 12. A tank can be filled by one pipe in 10 hours, or by another pipe in 15 hovu-s. How long will it take to fiU it if both pipes are used at the same time? [Hint. Let x = the number of hours. Then l/x = the part both can fill in one hour.] 13. Two pipes are connected with a tank. The larger one is an intake pipe which can fill the tank in 3 hours, while the smaller one is an outlet pipe entering at the bottom which can empty the tank in 4 hours. With both pipes open, how long before the tank will fill? 14. A can do a piece of work in 16 hours, while B can do it in 20 hours. If A works 10 hours, how many hours must B work to fimsh? 15. An aviator made a trip of 95 miles. After flying 40 miles, he increased his speed by 15 miles an hour and made the remaining distance in the same time it took him to fly the first 40 miles. What was his rate over the first 40 miles? 16. A 5-gaUon mixture of alcohol and water contains 80% alcohol. How much water must be added to make it contain only 50 % alcohol? 17. What weight of water must be added to 65 pounds of a 10 % salt solution to reduce it to an 8 % solution? 18. A train 660 feet long, running at 15 miles an hour, will pass completely through a certain tunnel in 49J^ minutes. How long is the tunnel? 10 COLLEGE ALGEBRA [I, § 5 5. Elimination. In case two simple equations (see §4) are given, each containing the two unknown values x and y, these values may usually be obtained by the process of elimination as is illustrated below. Example 1. Solve the equations (1) 2x+3y = 2, (2) 6a; -42/ = 28. SoLtJTiON. From (1) we have (3) 2x = 2-3j/. Therefore -2— 3u x = 2 SubstitutiBg this value for x in (2), we find (4) 5(^)-42/ = 28. In (4) we have an equation containing only y; that is, x has been eliminated from (1) and (2). Clearing (4) of fractions and simplifying, we obtain —23y = i6. Therefore y= —2. Substituting —2 for y in (1), we find 2a;-6 = 2, or 2x = 8, or a; = 4. Hence the required values of x and y are a; = 4 and y= —2. Check. Substituting a; = 4 and y= —2 in (1), we have 2X4+3(-2)=8-6=2, as desired. Likewise, with x = 4t and y=—2, equation (2) is satisfied, since it becomes 5X4-4X(-2) =20+8 = 28. The preceding method of solution, wherein the value of one of the letters, as x, is obtained from one of the equations and then substituted in the other equation, thus giving an equation, like (4), containing only one letter, is called elimination by substitution. Another common and very useful method of elimination is illustrated below. Example 2. Solve the equations (1) 3x+iy = 12. (2) 2a;-5j/ = 54. I, § 5] REVIEW TOPICS 11 Solution. Multiplying (1) by 2 and (2) by 3, the two equations become (3) 6a;+82/ = 24, (4) 61-152/ = 162. The coefficient of x is now the same in both (3) and (4) so that, upon subtracting (4) from (3), we obtain (5) 23!/= -138. Therefore 2/= —6. Substituting y= —6 in (1), we now have 3x-24 = 12, or 3x = 36. Therefore a; = 12. Hence the required values of x and y are x = 12,y=—&. EXERCISES Find, by any method of eUmination, the values of x and y in each of the following pairs of equations. Check your answers in the first five. ^ { x-y = i, \4y—x = 14 3x , 2y 3x-42/ = 26, /o+X=17. 2 4 x+y x—y j y+l=3x, 0)2 3 ■ \ x-8y = 22. 3. \5x+9 = 3y. ■ \ x+y ^-y _^^ , j iy = 10-x, ^^ * l-x=3y, ^°' )5y-8 , 5x-3 5 4 +^ = 18-5x. \3il-x)=40-y. (^ 2 a:+|=ll. jj )x-l x+y ' l?^Q,.^oi )-^+3 = 0. +3^ = 21. Ix-y r4 3_14 P = ll-^> 12 r ^ 5 )3 2 ]B , 5^?5 [37 ■ [Hint. Solve first for 1/x and 1/2/.] 12 COLLEGE ALGEBRA [I, § 5 S5 6 -+- = 64, X y In each of the following examples, let x and y represent the desired unknown quantities, form two equations and solve. 15. The sum of two numbers is 76 and their difference is 5. What are the numbers? 16. One-third the sum of two numbers is 10, while one-sixth of their difference is 1. Find the numbers. 17. A father's age is 1}4 that of his son. Twenty years ago his age was twice his son's. How old is each? 18. A part of $2500 is invested at 6% interest and the remainder at 5%. The yearly income from both is $141. Find the amount of each investment. 19. A and B together can do a piece of work in 12 days. After A has worked alone for 5 days, B finishes the work in 26 days. In what time could each do the work alone? [Hint. If x = the time in which A can do it alone, and j/ = the time in which B can do it alone, then the part which A can do in one day = 1/1, etc. See Ex. 12, p. 9 and Ex. 12, p. 11.] 20. An errand boy went to the bank to deposit 38 bills, some of them being $1 biUs and the rest $2 bills. If their total value was $50, how many of each were there? 21. A grocer wishes to make 50 pounds of coffee worth 32 cents a pound by mixing two other grades, which are worth 26 and 35 cents per pound, respectively. How much of each muiit he use? 22. One cask contains 18 gallons of vinegar and 12 gallons of water; another contains 4 gallons of vinegar and 12 of water. How many gallons of each must be taken so that when mixed there may be 21 gallons containing half vinegar and half water? 23. Two cities are 140 miles apart. To travel the distance between them by automobile takes 3 hours less time than by bicycle, but if the bicycle has a start of 42 miles, each takes the same time. What is the rate of the automobile, and what the rate of the bicycle? 24. The perimeter of a certain rectangle is 16 feet. If the length be increased by 3 feet and the breadth by 2 feet, the area is increased by 25 square feet. What are the original length and breadth? I, §6] REVIEW TOPICS 13 6. Graph of an Equation. In reviewing this topic, it is desirable first to recall the following fundamental ideas and definitions. Let two lines XX' and YY' be drawn on a sheet of squared (coordinate) paper, the line XX' being horizontal and YY' vertical. Two such lines constitute a pair of coordinate axes. XX' is called the x-axis, YY' is called the y-axis. The point where they intersect is called the origin. Having chosen any point, as P, in the plane of the axes, draw the perpendiculars PA and PB. Then PA, which is parallel to the x-axis, is called the abscissa of P, while PB, which is parallel to the y-axis, is called the ordinate of P. The ab- scissa and ordinate taken together are called the coordinates of the point P. Thus, in Fig. 1, the abscissa of P is 3 units, while its ordinate is 4 units. Note that the x- and y- unit scales are indicated along the x- and y-axes, respectively. All abscissas measured to the right of the y-axis are taken as positive, while all abscissas measured to the left of the same axis are taken negative. Thus the abscissa of Q in Fig. 1 is —2; that of B is —3; that of S is +3. Similarly, all ordinates above the x-axis are taken positive, while all ordinates below the same axis are taken negative. Thus the ordinate of Q is +Z; that of R is —4; that of 5 is —2. In reading the coordinates of a point, the abscissa is always read first and the ordinate second. Thus, in Fig. 1, P is the point (3, 4); Q is (-2, 3); R is (-3, -4); S is (3, -2). Y A p p (3 4) (■2 ,3)' 1 X' 3 B X — , 3 - 4 2- 1, I 3 ^ ) S (3 r ) R -4 ( ^, ■4] Y' Fig. 1 14 COLLEGE ALGEBRA [I, §6 Let us now consider the following simple equation con- taining the two unknown numbers, x and y: (1) x+y=5. Since any pair of values (x, y) whose sum is 5 will satisfy this equation, it follows that there are an unlimited number of such {x, y) pairs, or solutions. For example, the following four pairs are particular solutions: (2) (x = 6,2/=-l);(x = 2,j/ = 3);(a; = i,2/ = f);(x = 8,2/=-3). If we now regard each of these solutions as the coordinates of a point, and locate (plot) the four points thus obtained, it will be found that they all he upon one and the same straight line, as shown in Fig. 2. This line is called the graph of the equation (1). Similarly, the graph of any simple (first degree) equation con- taining two letters may be drawn. However, it may be observed that in order to draw the graph it suf- fices to plot merely two solutions, since two points completely de- termine a line. Such a line will necessarily pass through, or contain, all the other solutions. If, instead of one equation being given, there are two of them, as for example \ Y \ \^ y -) \ i^ 3) -t \ ^, ^ \ X \ (6 -I'l 1 S^ J i§i 3) ^ Fig. 2 (3) x+y = Q, dx-2y=-2, and if we draw the graph of each, as in Fig. 3, then the point where the two graphs intersect will have as its coordinates a pair of values (x, y) which satisfies hath of the equations at once; in other words, it will give their common solution. In I, §6] REVIEW TOPICS 15 the present case this is seen to be the point a; = 2, j/ = 4. This common solution is the same as would be obtained if one followed the method of eUmination described in § 5. Hence, Fig. 3 may- be regarded as giving the graphical meaning of such a solution. Note. In exceptional cases, the graphs of two simple equations may turn out to be parallel lines so that they nowhere intersect. In such a case, the two equations have no coramon solution. Fig. 3 's Y / \ >v Ai'' / Vy ' \ &M ^\ X c ^ r ^1/ fT 1 s / \ / EXERCISES 1. Plot (on coordinate paper) each of the following points. (2, 4); (-2, 3); (-2, -4); (2^, -3); (0, -5); (4, 0); (0, 0). 2. Describe (a) the location of all points whose abscissa is zero; (6) of all points whose ordinate is zero; (c) of all points whose abscissa and ordinate are both negative. Draw the graph of each of the following simple equations. 3. x-y = 5. 5. 2x-3y = l. 7. x-3j/ = 3. 9. 6x+7i/ = 2. 4. 2x+2/ = 3. 6. 2a; +3)/ = 12. 8. 2x = 3y. 10. 6x-8j/= -1. Draw the graphs of each of the following pairs of equations and thus determine the values of x and y which form their common solution, if they have one. Check your results in each by solving by elimination (§5). 11. 12. 13. fx+2y = 3, \2x+y = 3. x+y = 3, Zx-y = l. x+2y = 5, x-2y = 5. 14. 15. 16. x-2y = 3, 2x-iy = l. 4x— 2/ = 0, 3x+y = 7. Uy-2x = 5, \4x+2y = 5. 17. 18. 19. x+2y=-l, /3x-3j/=-5, \3x+2]/ = 40. f8x+4j/ = 5, \ x-y = }i. 16 COLLEGE ALGEBRA [I, § 7 7. Literal Equations and Formulas. Equations in wiiich some, or all, of the known quantities are represented by letters are called literal equations. The known quantities are generally represented by the first letters of the alphabet, as a, b, c, etc. Literal equations are solved by the same processes as numerical equations. Example. Solve the following literal equation for x: ax = hx-\-7c. Solution. Transposing, we find ax — hx=^c. Combining like terms, we have (a— 6)x = 7c. Dividing by (a — 6), we obtain x = — —■ Ans. a—b It is to be noted that a literal equation is said to be solved for the unknown number, as x, only when this number has been expressed in terms of the other (known) letters, as illus- trated in the preceding example. An important special class of literal equations are the so- called formulas that occur in geometry, physics, engineering, etc. For example, if a represents the length of the base of any triangle and h represents the altitude, then the area, A, of the triangle is given (determined) by the formula (1) A^lah. Here the area is expressed in terms of the base and the altitude. I, § 7] REVIEW TOPICS 17 Similarly, the circumference, C, of any circle expressed in terms of the radius r is given by the following formula, in which IT represents the incommensurable number, 3.1416 (approximately) , (2) C=2Trr. Again, the area, A, of a circle in terms of the radius r is given by the formula ^ (3) A=irr\ As an example of an important formula in physics, it is readily seen that if an object moves during t seconds with the constant velocity of v feet per second, then the dis- tance, s, passed over is given by the formula (4) s = vt. Again, the following is an important formula in engineering : The horse-power, represented by HP, which a steam engine is delivering when the area of the piston is A square inches, the pres- sure of the steam per square inch is P pounds, the length of the piston stroke is L feet and the number of ^^°- ^ strokes per minute is N, is given by the formula „P PLAN ^^^ 33,000 The following important formulas from plane and solid geometry afe to be especially noted: (6) h^ = a''+b\ which is the theorem of Pythagoras concerning the square of the hypotenuse of a right triangle. (7) F=3TrrS, which gives the volume of a sphere in terms of its radius. 18 COLLEGE ALGEBRA [I, §7 (8) A = iTrr^ which gives the area of a sphere in terms of its radius (9) V^Tn'h, which gives the volume of a right circular cyhnder in terms of the radius of the base, r, and the alti- tude, h. (10) A = 2Tnh, pwr which gives the area of the lateral surface of a right circular cylinder in terms of the radius of the base, r, and the altitude, fi. (11) V=^Trr'h, which gives the volume of a cone of circular base, r, and of altitude, h. (12) A=Trrl. which gives the area of the lateral surface of a cone of circular base, r, and of slant-height, I. (13) '=t1<^ :=+b^)-|-- FiG. 9 which gives the volume of a spherical seg- ment, or slice of a sphere between two par- allel cutting planes, in terms of its altitude, h, and the radii, o, h, of its bases. (14) 7=2^71, which gives the area of a zone, or portion of the surface of a sphere lying between two parallel planes, in terms of the altitude, h, of the zone and the radius, r, of the sphere. The following formulas from physics and mensuration may also be noted. If an elastic ball (like a bilUard ball) weighing PFi ounces and moving with a velocity of Fi feet per second strikes (impinges upon) a second ball of like size but weighing W^ I, § 7] REVIEW TOPICS 19 ounces, the latter standing at rest, then, after the impact, the first ball and the second ball will move with velocities vi and va which are given respectively by the formulas , , W1-W2., -, 2Wi ,, ,^ Q5) ^1= 7i ft. per sec, i;2= ki ft. per sec. ^ ' W1+W2 Wi+Wi It is understood in the experiment just described that the first ball moves directly toward the center of the second one before the impact. Both continue in this same line after the impact. EXERCISES Solve each of the following literal equations for x, checking your answer in the first five. 1. ax — \=h. 2. ax-{-hx = c. 3. 3a;+b=x-3b. 4. 2+b=5+„. a "• |-a=a;— 1. c 10 . x—b , x—c „ x+a x-S x+2 7. 9. X 7 ''ah' 49 ahx 6 ■'' 2x_ a a a'+b'' 2bx a-b b 2b3? X a-b+: c I i—a+c Solve (by the method of elimination) for x and y in each of the following pairs of equations. {3x+5y = 2a, j^ ^3ax+2by=ab, "• \2x-3y = ib. ' 1 ax-by=ab.. "• \cx+dy^3. j5_ )x y fax+by=m, j — - 13. \cx+dy=n. l^ V [Hint. Solve first for \/x and 1/y. See Ex. 12, page 11] 16. Divide o into- two parts whose quotient is m. 17. If A can do- a piece of work in a days, and B can do it in 6 days, how long will it take them if working together? (See Ex. 19, page 12.) 18. If the base of a triangle is 3 feet long, what must the altitude be in order that the area may be 30 square feet? [Hint. Use formula (1).] 20 COLLEGE ALGEBRA [I, § 7 19. If the area of a circle is 44 square inches, what is the value (approximately) of the radius? [Hint. Use formula (3), taking ir = 3^]. 20. How long will it take a person to walk 1 mile if his rate of walking is 8 feet per second? 21. An automobile traveled T hours at the rate of v miles per hour. By how much would this rate have had to be increased in order that the distance be covered in t minutes less time? P 22. The formula for the area A of a trapezoid whose bases are B and 6 and whose altitude is /i is / |!| A = \h{B+h). Using this formula, answer the following question : How long should the upper base of a trapezoid be in order that, if the lower base is 20 feet long and the altitude is 15 feet, the area may be 180 square feet? 23. The inside diameter of the piston of a steam engine is 8 inches, while the length of stroke is 1}^ feet. When the steam gauge registers a pressure of 60 pounds per square inch, how many strokes per minute must the piston make if the engine is to deliver 22 horse-power? Work by formula (5). 24. The velocity, v, of sound, measured in feet per second, is given by the formula i; = 1090-|-1.14(<-32), where t is the temperature of the air in Fahrenheit degrees. Find (a) the velocity when the temperature is 75°; (b) the tem- perature when sound travels 1120 feet per second. 26. Derive formulas for the following: (o). The number N of turns made by a wagon wheel d feet in diam- eter in traveling s miles. (6). The number N of dimes in m dollars, n quarters and q cents. (c). The value of a number containing three digits if the digit in unit's place is a, the digit in ten's place is 6 and that in hundred's place is c. I, §8] REVIEW TOPICS 21 26. The centrifugal force F, measured in pounds, with which a body weighing W pounds puEs outward when moving in a circle of radius r feet with a velocity of v feet per second is determined by the formula F = 32r" Use this formula to answer the following questions. A pail of water weighing 5 pounds is swung round at arm's length at a speed of 10 feet per second. If the length of the arm is 2 feet, find (o) the pull at the shoulder when the pail is at the uppermost point of its course; (b) when at the lowest point of its course. Also find the least velocity possible without water dropping out at the uppermost point of the course. , 27. The weight W that can be raised by the device shown in Fig. 11 is given by the formula 27riflP W=- dr where P represents the pressure appUed at the handle and where R, r, d and I have the lengths indicated in the figure. Show, by pjg n means of this formula, that if d be halved and the number of teeth on the wheel be correspondingly doubled to fit the new gear, other parts remaining the same, then twice as much weight W can be rs,ised as before with a given pressure P on the handle. 8. Exponents. The laws of exponents are briefly ex- pressed in the following five formulas. I. Thus n. Thus m. Thus (3*)==3«'. (ab)"'=a'"b'". (2-3)2 = 2='-3^ 22 COLLEGE ALGEBRA [I, §8 Thus V. -—=0'"-". a" Thus - = 2^-2 = 2''. 2'' These formulas apply not only when m and n are positive integers, but in all cases. Thus 2 J • 2~i = 2t~i = 25['^, (3f)^=3txf = 3t, The use of formulas I — V in this universal way implies (as shown in elementary algebra) that the expression a"" must have the same meaning as the qth root of a"- That is, we have VI. a9 = Vaf' Thus 5* = -^ = V^25! Similarly, any quantity, as a, when raised to a negative exponent, as — n, must have the same meaning as the recip- rocal of a". That is, vn. a-"=— . a" 2 1 Thus 6-5 = -5-- 6' Again, for any value whatever of a, except 0, the expression a" has the value 1. That is, when a is not vin. axis. In this way determine graphically the values of the solutions, and check the correctness of your answers by actually solving the equation. 1. ^-x-2 = Q. 5. 2x2+5x+2 = 0. 2. ar*-10x+24 = 0. 6. :^-7x+\2 = Q. 3. a?-2x-15 = 0. 7. x2+7x+12 = 0. 4. 3x2_ga. = 3_ 8. 2^+^x = 9. 26. Determining Graphically Whether Solutions Are Real or Imaginary. In order to apply the method described in § 25 for determining graphically the solutions of a given quadratic it was essential that the graph should cut the a;-axis. However, quadratic equations may easily be found whose graphs do not cut or touch the cc-axis at all. For example, consider the equation (1) a;2-6x+15 = 0. Proceeding as in § 25 to draw the graph, we place (2) 2/ = x2-6a;+15, Ill, § 27] PROPERTIES OF QUADRATIC EQUATIONS 53 and determine various pairs of values (x, y) which satisfy this equation. The table below shows several such {x, y) pairs. When x = -1 22 15 1 10 2 7 3 4 5 6 then y = 6 7 10 15 Plotting the various points ix, y) thus obtained and draw- ing the curve through them, we obtain the graph indicated in Fig. 14. It is to be noted that this graph lies entirely above the x-axis, thus not cutting (or touching it) in any manner. The significance of such a result is that the two roots of (1) are imaginary. If they were real, the graph would cut the X-axis, as shown in § 25. In reality, we find upon solving (1) that its two solu- tions have the following unaginary values: a;=3zt-\/^- Thus, in general, we have the follow- ing result: The solutions of a quadratic equation are real or imaginary according as its graph does or does not cut or touch the x-axis. fig. 14 EXERCISES Find, by drawing the graph, whether the solutions of each of the following equations are real or imaginary. 1. 3?+2x+3 = 0. 3. a?-22+3 = 0. B. 6x2+5x-hl=0. 2. 3?+2x-3=0. 4. 33i?+4x+l = 0. 6. 2a^-33;+4 = 0. 27. The Nature of the Solutions Considered Geometri- cally. We have seen in §§ 25, 26 that whenever a quadratic has two distinct real solutions its graph will cut the x-axis in y on -1 u?2- 3 t I I t " ^ , . 1 I t i r ■0' V 7 3 -t \v/ -1- ^ n_ 5 _ 54 COLLEGE ALGEBRA [III, § 27 two points, while if the solutions are imaginary the graph fails to cut the x-axis at all. Suppose now that we have a quadratic equation whose two solutions are real and equal to each other, for example the equation (1) 4x2-12a;+9 = 0. Here the discriminant (§ 22) is equal to (-12)2-4X4X9 = 144-144 = 0, so that the roots must be equal by the rule of § 22. If we now proceed to draw the graph corresponding to (1) in the usual manner by placing y = 4:X^ — 12x+9, it appears that the resulting graph just touches the x-axis instead of actually cutting through it. This " was to be expected, since the equahty of the roots means that there is but one root, and this, when considered as in § 26, can be possible only when the graph merely touches (is tangent to) the X-axis. Thus, in general, we. have the following result. // the two roots of a quadratic equation are real and equal, the graph of the equation will be tangent to the x-axis, and conversely. 2 Kg. 15 EXERCISES Draw the graph of each of the following equations and examine whether they do or do not illustrate the statement at the end of § 27. If not, what statement is illustrated (see §§ 25-27). 1. x2-2x + l=0. 2. 3?-6x + l2 = 0. 3. x2+6x+12 = 0. 4. 4x^+ix+l=Q. 5. x2-2x-8 = 0. 6. 3i2+4x+l=0. 7. 3x*+4a;+2=0. 8. 4a?'-12x+9 = 0. CHAPTER IV SIMULTANEOUS QUADRATIC EQUATIONS I. One Equation Linear and the Other Quadratic 28. Graphical Solution. In § 6 we have seen how to determine graphically the solution of two simple (first degree) equations each of which contains the two unknown numbers x and y. The method consists in drawing the graph of each equation, then observing the x and the y of the point where the two graphs intersect. The particular pair of values (x, y) thus obtained constitutes the solution. We often meet with a pair of equations similar to those just mentioned except that one (or both) of the equations is not of the first degree. For example, consider the pair, or system, of equations (1) x-y = l, (2) a;2+2/2 = 25. In order to solve this pair of equations, that is, to find the particular pair (or pairs) of values {x, y) which will satisfy them both, we may proceed graphically in a manner precisely analogous to that employed in the study of simple equations. Thus the graph of (1) is found (as in § 6) to be the straight line shown in Fig. 16. In order to draw the graph of (2), we first solve this equation for y in terms of x, thus obtaining (3) 2/=±\/25^=^. By giving various values to x in (3), we obtain the y-values corresponding to each. The table below shows the ^/-values thus obtained corresponding to a; = 0, +1, +2, etc., to x= +5. When a; = +1 +2 +3 +4 +5 tlien y = ±-s/25 ±5 ±\/24 ±4.8 ±V21 ±4.5 ±\/l6 ±4 ±V9 ±3 ±Vo 55 56 COLLEGE ALGEBRA [IV, §28 1 Y 5 A ^ ■^ ^ / / \ / / / / / X - 1, / 1 5 \ -> \ / \ / / / ^ ^ ^ ^ / / Fig. 16 Observe that to a; =0 correspond the two values 2/= ±5; similarly to x = \ correspond the two values 2/=±4.8 (approximately), etc. Moreover, if we assign to x the negative value, a;= — 1, we find in the same way that corresponding to it y has the two values, 2/= ±4.8. Likewise, for x=—2 we find y=±4.5, etc., the values of y for any negative value of X being the same each time as for the corresponding positive value of x. Plotting all the points (a;, j/) thus found and drawing the smooth curve through them, we obtain as the graph the curved line shown in Fig. 16. This curve is a drde, as appears when we plot more and more of the points (a;, y) pertaining to the equation (3). Note. The form of (3) shows that there can be no points in the graph having x values greater than 5, for as soon as x exceeds 5 the expression 25— a^ becomes negative and hence \/25— x2 becomes imag- inary, and there is no point that we can plot corresponding to such a result. Similarly, it appears from (3) that x cannot take values less than —5. Thus the graph can contain no points lying outside the circle already drawn. Returning now to the problem of solving (1) and (2), we know (§ 6) that wherever the one graph cuts the other we shall have a point whose x and y form a solution of (1) and (2) , that is, we shall have a pair of values (x, y) that will satisfy both equations at once. From the figure it appears that there are in the present case two such points, namely (a; = 4, y = 3) and (a;= — 3, ?/= — 4). Equations (1) and (2) therefore have the two solutions (a; = 4, y = 2i) and (a;= — 3, i/= — 4). Ans. IV, § 28] SIMULTANEOUS QUADRATIC EQUATIONS 57 Check;. For the solution (a; = 4, y = 3) we have a;— j/ = 4— 3 = 1, and a^-f2/^ = 16 -1-9 = 25, as required. For the solution {x=—3, y=—i) we have x—y=—S — (—i) = l, and x'+^ = 9 + lQ = 25, as required. The following are other examples of the graphical study of non-Unear simultaneous equations. Example 1. Solve the system (4) 2x-9y+lQ = Q, (5) 422+V = 100. Solution. The straight line representing the graph of (4) is drawn readily. To obtain the graph of (5), we have V = 100-4a;2. Hence 2/2 = ^(100-4x2) =|(25-A and therefore (6) 2/=±|V25=?. Corresponding to (6), we find the following table: When x= -HI + 2 -1-3 -f4 +5 greater than +5 then y= ±ll/25 ±1(5) ±3.3 ±ll/24 ±i(4.8) ±3.2 ±W2i ±1(4.5) ±3.0 ±tl/l6 ±S(4) ±2.6 ±It/9 ±1(3) ±2 ±ll/0 ±0 imaginary imaginary imaginary For any negative value of x, the y-vahies are the same as for the corresponding positive value of x, (See the discussion of (2).) The graph thus obtained for (6), or (6), is an oval shaped curve. It belongs to the general class of curves called ellipses. The two graphs are seen to inter- sect at the points (x=4, 2/ = 2) and {x=—5,y=0). Therefore the desired solutions of (4) and (6) are (a;=4, y = 2) and (x= —5, y=0). Y ^"" "^^--' -s — ^"^ X S ^1 FiQ. 17 Arts. 58 COLLEGE ALGEBRA [IV, § 28 Example 2. Solve the system (7) 2x-2/=-2, (8) x!/ = 4. Solution. The graph of (7) is the straight line shown in Fig. 18. To obtain the graph of (8), we have (9) y=~- from which we obtain the following table: When X = 8 7 6 1 5 1 4 1 3 1 2 2 1 4 i 8 i 12 i 16 20 then y = This table concerns only positive values of x, but it appears from (9) that for any negative value of x the appropriate y-value is the negative of that for the corresponding positive value of x. The graph thus obtained for (9), or (8), consists of two open curves, each indefinitely long, situated as in Fig. 18. These taken together (that is, re- garded as one curve) form what is known as a hyperbola (pronounced hy- per'-bo-la). The part (branch) of the curve lying to the right of the 2/-axis corresponds to the table above, while the other branch corresponds to the negative x-values. The two graphs are seen to intersect in the points (x = l, ?/ = 4) and (x= —2, 2/= -2). Therefore the desired solutions of (7) and (8) are (x = (x=-2, y=-2). Arts. Note. Ellipses and hyperbolas are extensively considered in the branch of mathematics called analytic geometry. Both of these curves are of wide application in physics, astronomy, and engineering, as illustrated in the fact that the orbit of each of the planets in the solar system is an ellipse. Both curves belong to a wider class of curves known as the conic sections. Fig. 18 = 1, 2/ = 4) and IV, § 28] SIMULTANEOUS QUADRATIC EQUATIONS 59 Example 3. Consider graphically the system (10) x+y = 10, (11) x2+2/' = 25. Solution. The graph of (10) is found in the usual manner, and is represented by the straight line in Fig. 19. The graph of (11) has already been worked out (see discussion of (2)), being a circle of radius Fig. 19 5 jvith center at the origin. The peculiarity to be especially observed here is that these two graphs do not intersect. This means (as it naturally must) that there are no real solutions to the system (10) and (11); in other words, the only possible solutions are imaginary. Likewise, whenever any two graphs fail to intersect, we may be assured at once that the only solutions their equations can have are imaginary. The system (10) and (11) and other such systems will be considered further in the next article. EXERCISES Draw the graphs for the following systems and use your result to determine the solutions whenever they are real. 1 I x = 2y, • Ix^ +2/^ = 20. x+y = 7, 3xH^ = 43. o / x-2y=-l, 6. x+y = 7, xy = 10. \2x+y=7. x+y=2, y=^. _ f2x+j/ = l, ■ \i/=4a^+2x+l. g fx'+xy = 12, ■ \ x-y = 2. a \x = &-y, ■ 1x5+2/3 = 72. 60 COLLEGE ALGEBRA [IV, § 29 29. Solution by Elimination. Let us consider again the system (1) and (2) of § 28. (1) x-y=l, (2) x^+y^ = 25. Instead of solving this system graphically, we may solve it by eUmination; that is, by the process employed with two linear equations in § 28. Thus we have from (1) (3) y = x-l. Substituting this value of y in (2), thus ehminating y from (2), we obtain a;2+(a;-l)2 = 25, or a;2+x2-2a;+l = 25, or 2a;2-2a;-24 = 0. or, dividing through by 2, (4) x2-a;-12 = 0. Solving (4) by formula (§ 56), gives as the two roots -(-1)+V(-1)^-4(1)"ML2) l+Vl+48 1+7 '"' 2 ~ 2 ~ 2 ~^' and ^^ -(-!)- V(-l)'-4(l)(-"l2) _ l-Vl+48 _ l-7 _ _ 2 2 2" When X has the first of these values, namely 4, we see from (3) tjiat y must have the value y = 4:—l, or 3. Similarly, when x takes on its other value, namely —3, we see that y has the value y= —3 — 1, or —4. The solutions of the system (1) and (2) are, therefore, (a; = 4, 2/ = 3) and (a;= — 3, 1/=— 4). Ans. Observe that these results agree with those obtained graphically for (1) and (2) in § 28. Further applications of this method are made in the examples that follow. IV, § 29] SIMULTANEOUS QUADRATIC EQUATIONS 61 Example 1. Solve the system (5) 2x+2/ = 4, (6) 3^+2/2 = 12. Solution. From (5), (7) j/ = 4-2x. Substituting this expression for y in (6), we find a^ + (16-16x+4x2) = I2, or (8) 5ar*-16x+4 = 0. The two roots of (8), as determined by formula (§ 21), are -(-16)±-v/0^n;6)2-4(5)(4) _ 16±V256-80 _ 16±Vl76 ^~ 2(5) ~ 10 ~ 10 _16±4Vn _8±2Vn ~ 10 ~ 5 " . The first of these values, namely x = (8+2-\/ll)/5) when substi- tuted in (7), gives as its corresponding value of y, 16+4-^/11 4-4v'n y=^ — I — =— ^ — The second value, namely 2 = (8— 2\/lI)/5, when substituted in (7), gives as its corresponding value of y, ^ 16-4Vn 4+4\/lT « = 4 = . 6 5 Hence the desired solutions are 8+2VT1 r 8-2vTI = ' \x= ) 5 j 5 ,_ and < ^ 4-4V1I 4+4vTl To obtain the approximate values of the numbers thus obtained, we have VH = 3.31662 (tables), and hence the above solutions reduce to the forms x = 2.9266, , fx = 0.2734, 2/= -1.8533, \2/ = 3.4533. These are the solutions of the system (5), (6), correct to four "places 0/ decimals, which is sufficient for ordinary work. |. 62 COLLEGE ALGEBRA [IV, § 29 Example 2. Solve the system (9) x+y = lO, (10) x2+!/2 = 25. Solution. From (9), y = 10—x. Substituting this expression in (10), a?+(100-20a;+x2)=25, or (11) 2:!;2_20a:+75 = o. Solving (11) by formula, we find its solutions to be, after reduction, 10+ SV^ , 10-5v^^ x = and x = 2 2 Since these x-values contain the square root of the negative number —2, they are imaginary. The j/- values are also imaginary, as appears by substituting the x-values just found into (9), which gives the results 10- &\/^ , 10+5\/^ y = and y The desired solutions of the systems (9), (10) are therefore 10+5 V^ ( _10-5V^ and l- 10+5\Ar2 i" 2 r 2 This result should now be contrasted with what we saw in Example 3 of § 28 regarding this same system (9) and (10). There we found graphically that the solutions must be imaginary because the graphs failed to intersect, but we could not find the actual imaginary numbers which form the solutions. EXERCISES Solve each of the following systems by the method of elimination, and, in case surds are present, find each solution correct to two places of decimals by use of the tables. [ x-y=5. ■ \ x-2y = Z. 9. \Sy'^5x l? Il0x+y = 3xy, i:^+3xy -2^ = 43, \ y-x = 2. ' \ x+2j/ = 10. (x'+xy = 12, /x2+3x7/ = i/+23, \ x-2/ = 2. • \ x+3y = 9. 10. / x-2y = 2, /3x2-xj/-5/ = 5, 1x^+4^ = 25. \ 3x-5i/ = l. IV, § 30] SIMULTANEOUS QUADRATIC EQUATIONS 63 II. Neither Equation Linear 30. Two Quadratic Equations. In each of the systems considered in §§ 28, 29 one of the two given equations was hnear. However, the same methods of solving may often be employed in case neither equation is linear. In such cases four solutions may be present instead of two. Example 1. Solve the system (1) 9x2+16i/2 = 160, (2) x'-j/' = l5. Solution. Here only x^ and y^ appear and we begin by finding their values. Thus, multiplying (2) through by 16 and adding the result to (1), we eliminate 2/^ and find that 25x^ = 400, or (3) x^ = 16. Substituting this value of a^ in (2"), we find (4) 2/^ = 1. From (3) and (4) we now obtain (5) a: = ±4 and 2/=±l. Forming all the pairs of values (x, y) that can come from (5), we obtain as our desired solutions and (x=4, y = \); (a;=-4, j/ = l); (a;=4, y=-l); (x= — 4, 2/= — 1). Ans. Check. Each of these pairs of values of x and y is immediately seen to satisfy both (1) and (2). Let the student thus check each pair. When considered graphically, equation (1) gives rise to an ellipse (compare § 28, Ex. 1), while (2) gives a hyperbola situated as shown in Fig. 20. These two curves intersect in four points which correspond to the four solu- tions just obtained. Y \ ^/ ^ , z \^^ ^'.Z \/ \t -A \- ^ H ^ h 7\ A ^^=.- -^^^ 7 S 7 \ Fig. 20 64 COLLEGE ALGEBRA [IV, § 30 Example 2. Solve the system (7) xH2/' = 25, (8) xy=-12. Solution. Here we cannot proceed as in Example 1 because we cannot find readily the values of x^ and y^. But if we multiply (8) by 2 and add the result to (7), we obtain (9) 3?+2xy+y' = l. Taking the square root of both members of (9) gives (10) x+j/=±l. Similarly, multiplying (8) by 2 and subtracting the result from (7), a?-2xy+if = i9, and hence (11) x-y=±7. Taking account of the two choices of sign in (10) and (11), we see that they give rise to the four simple (hnear) systems: (a) x+2/ = l, x-v = 7; (6) x+y=-l,x-y=7; (c) x+y = l, x-y=-7; (d) x+2/=-l, x-i/=-7. Thus we have replaced the original system (7) and (8) by the four simple systems (a), (6), (c), and (d), each of which may be immediately solved by elimi- nation, as in § 28. Since the solutions of (a), (6), (c), (d) are respectively (x = 4, y=—3), (x=3, 2/= -4), (x=-3, 2/ = 4), and (x = — 4, 2/ = 3), we conclude that these are the desired solu- tions of (7) and (8). Ans. The graphical significance of these solutions is shown in Fig. 21, where the circle x^-l-3y^ = 25 is cut by the hyperbola xy= —12 in four points that correspond to the four solutions just found. Check. That these four solutions each satisfy (7) and (8) appears at once by trial. I Y I r 7 i- k^ -^ 7 ^ =.^? V t ^ A ^0 X 1 5 V 7 ^^ \ d^ ^^ ^2 ^----2 r 1 I Fig. 21 IV, § 31] SIMULTANEOUS QUADRATIC EQUATIONS 65 While no general rule can be stated for solving two equa- tions neither of which is linear, the following observation may be made. Unless the equations can be solved readily for x^ and 'ip- (as in Example 1), the system should first be examined with a view to making such combinations as will yield one or more new systems each of which can be solved (as in Example 2) by methods already familiar. All solutions obtained in this way should be checked in order to avoid false combinations of the x- and ^/-values thus obtained. EXERCISES Solve each of the following systems, and draw a diagram for each of the first three to show the geometric meaning of your solutions. ^- 1 2x^-2/2 = 31. *• \xy+u' = 15. [Hint to Ex. 4. First add, then subtract the two equations, thus showing that the given system is equivalent to two others each of which may be solved as in § 29. Compare Ex. 2, § 30.] fx'+xy+^^lSl, . ja^+xy = 77, °- \ x2+2/2 = 106. "• \xy-Tf=12. C^xy+2x+y = 25, (xy-Q = 0, 10. ??^1^. \x'+y'^ = xy+7. [ y x *31. Systems Having Special Forms. The systems of equations considered in §§ 29, 30 illustrate the usual and more simple types such as one commonly meets in practice. It is possible, however, to solve more complicated systems provided they are of certain prescribed forms. We shall here consider only two such type forms. I. When one (or both) of the given equations is of the form aa^+bxy+cif = 0, where the coefficients a, 6, c are such that the expression a3?-{-bxy+cy^ can be factored into two rational linear factors. 66 COLLEGE ALGEBRA [IV, § 31 Example. Solve the system (1) x2+2x-2/ = 7, (2) x^-xy-2^ = 0. SoLtmoN. Here we see that (2) is of the form mentioned above, since ^—xy—2y^ can be factored into {x—'2,y){x+y). (2) may thus be written in the form (3) {x-2y){x+y)=Q. It follows that either x—2y = Q, or x-\-y = 0. Hence the system (1), (2) may be replaced by the two following systems: U+2x-y = 7, _^, l3?+2x-y = 7, \ x-2y = 0, ^"""^ \ x+y = 0. Each of these two systems may now be solved as in § 30, and we thus find that the solutions of the first system are (x = 2, 2/ = l) and (x= -^, j/= — J), while the solutions of the second system are fx = i(-3+-v/37), \y=U3-V27), and fx=-|(-3-\/37), \?/=i(3 + V37). The desired solutions of (1) and (2) consist, therefore, of these four solutions just obtained. Ans. II. When both the given equations are of the farm ox^ +bxy -\-cy^ = d, where a, h, c and d have any given values (0 included). Example. Solve the system (4) ^-xy+if = 3, (5) a?+2xy = 5. Solution. Let v stand for the ratio x/y; that is, let us set (6) x = vy. Substituting in (4) and (5), we find, (7) v'y'-vif+y' = Z. (8) «V+2j'j/' = 5. IV, § 32] SIMULTANEOUS QUADRATIC EQUATIONS 67 Solving (7) and (8) for y^, (9) 2/2= ^ 5 ir+2v Equating the values of i^ given by (9) and (10), 5 3 Clearing of fractions, (11) 2v^-\lv+b=0. Solving (11) by formula (§ 21), _ ll±Vl21-40 ll±V81 _ llzb9 4 ~ 4 4 ' Therefore v = 5, or j;=^. Substituting 5 for v in (9), or (10), Hence Substituting -^ for w in (9) or (10), y^ = i. Hence 2/= +2 or —2. The only values that y can have are, therefore, l/V?, — l/Vv, 2, and -2. Since a; = z;2/ (see (6)), the value of x to go with 2/ = l/\/7 is a; = 5(1/ y/7) = 5/ -\/7. Similarly, when y=—l/ y/l we have x = 5(-l/V7) = -5/^/7. Likewise, when j/=2 (in which case v=\, as shown above, then by (6) we have x= \- 2 = 1. Again, when y= —2, then x=^(— 2) = — 1. Therefore the only solutions which the system (4), (5) can have are (x = 5/V7, 2/ = l/\/7); (x=-5/-v/7, 2/=-l/V7); (x = l, 2/=2); (a;= — 1, y= — 2); and it is easily seen by checking that each of these is a solution. Ans. 32. Conclusion. Every system of equations considered in this chapter has been such that we could solve it by finally solving one or more simple quadratic equations. We have examined only special types, however, and the student should not conclude that all pairs of simultaneous quadratics can be 68 COLLEGE ALGEBRA [IV, § 32 solved so simply. In fact, the solution of simultaneous quad- ratics in general involves a study of equations of higher degree than the second such as considered in Chapter XI. MISCELLANEOUS EXERCISES Solve the following simultaneous quadratics. The star (*) indicates that the exercise depends upon § 31. ^- \ x+y = 7. I xy = 7. i. 2. '•{ xy{x- *10. *11. 22/) = 10, xy = 10. lo?+xy+2)/ = n, \ 2i?+5y^ = 22. (2o?+xy-i/ = 0, \ 23?+y = l. -Sy-y'^S, 5xy—&if = 0. 6. jxy+2x = 5, \2xy-y = 3. xif+xy=24:, xi^+x='56. a^— xj/ = 6, 3?-y' = 8. 8. x^-y*=3m, x2-2/2 = 9. f 3^+2/2 = 100, \(x+2/f = 196. *9. *''■ {Ji *13. *14. 15. |x2-7x2/ + 122/' = 0, \ xy+3y-2x = 21. \xy+2y' = S, \a^+2xy = 12. f x'-xy-y' = 20, \:^-Sxy+2jf = 8. / x-2y = 2(a+b}, \xy+2yF = 2bib-a). APPLIED PROBLEMS In working the following problems, let x and y represent the two unknown quantities, then form two equations and solve them. If radicals occur, find their approximate values by use of the tables. 1. The sum of two numbers is 12, and their product is 32. What are the numbers? 2. The sum of two numbers is 82, and the sum of their square roots is 10. What are the numbers? 3. A piece of wire 48 inches long is bent into the form of a right triangle whose hypotenuse is 20 inches long. What are the lengths of the sides? 4. If it takes 52 rods of fence to inclose a rectangular garden con- taining 1 acre, what are the length and breadth of the garden? 5. If, in the adjoining figure, the combined area of the two circles is 15^ square feet and the distance CC between centers is 3 feet, what / 22 \ are the lengths of the two radii? I Take ir = — .1 V 7 / FiQ. 22 IV, § 32] SIMULTANEOUS QUADRATIC EQUATIONS 69 6. Work Ex. 5 in case the circles are situated as in Fig. 23, taking the shaded area to be 110 square feet and CC to be 5 feet. 7. The area of a triangle is 160 square feet, and its altitude is twice as long as its base. Find, cor- rect to three decimal places (using tables), the base and altitude. 8. The area of a rectangular lot is 2400 square feet, and the diagonal across it measures 100 feet. ^'°' ^^ Find, correct to three decimal places, the length and breadth. 9. The dimensions of a rectangle are 5 feet by 2 feet. Find the amounts (correct to two decimal places) by which each dimension must be changed, and how, in order that both the area and the perimeter shall become doubled. 10. Two men working together can complete a piece of work in 6 days. If it would take one man 6 days longer than the other to do the work alone, in how many days can each do it alone? (Compare Ex. 19, p. 12.) 11. The fore wheel of a carriage makes 28 revolutions more than the rear wheel in going 560 yards, but if the circumference of each wheel be increased by 2 feet, the difference would be only 20 revolutions. Find the circumference of each wheel. 12. A sum of money on interest for a certain time at a certain rate brought $7.50 interest. If the rate had been 1% less and the principal $25 more, the interest would have remained the same. Find the prin- cipal and the rate. 13. A man traveled 30 miles. If his rate had been 5 miles more per hour, he could have made the journey in 1 hour less time. Find his time and rate. (See Ex. 10, p. 9.) 14. Show that the formulas for the length I and the width w of the rectangle whose perimeter is a and whose area is h are l=\{a+^/a^-mh), a' = i((i-Vo2-166). 15. If the difference of the areas of two circles be d and the sum ot their circumferences be s, show that their radii r^ and rj, must have the following values: 47r(i4-s* s^—i'Trd CHAPTER V THE PROGRESSIONS I. Arithmetic Progression 33. Definitions. An arithmetic progression is a sequence of numbers, called terms, each of which is derived from the preceding by adding to it a fixed amount, called the common difference. An arithmetic progression is commonly denoted by the abbreviation A. P. Thus 1, 3, 5, 7, ••• is an A. P., since each term is derived from the preceding by adding 2 to it. Hence 2 is the common diiierence. The dots following the 7 indicate that the series may be extended as far as one pleases. Thus the first term after 7 would be 7+2, or 9; the next would be 9+2, or 11; etc. Again, 5, 1, —3, —7, —11, ••• is an A. P. Here the common differ- ence is —4. EXERCISES Determine which of the following progressions are arithmetic pro- gressions, and for such as are, determine the common difference. 1. 3, 6, 9, 12, -. 6. 0, 2a, 4a, 6a, -. 2. 3, 5, 6, 8, — . 7. o, a+4, a+8, a+12, ■••. 3. 6, 3, 0, -3, — . 8. a, a+d, a+2d, a+3d,---. 4. 30, 25, 20, 15, -. 9. x-iy, x-2y, x-y, •••. 5. -1, 2, 5, 8, — . 10. Zx+Zy, &x+2y, 9x+i/. ■••. 11. Write the first five terms of the A. P. in which (a) The first term is 4 and the conamon difference is 2. (5) The first term is 3a and the common difference is —6. 34. The Formula for the nth Term. From the definition (§ 33) it follows that every arithmetic progression is of the typical form a, a-\-d, a+-2d, a+3d, Here the first term is a and the common difference is d. 70 V, § 35] THE PROGRESSIONS 71 Observe that the coefficient of d in any given term is 1 less than the number of that term. Thus, in the third term the coefficient of d is 3 — 1, or 2; hkewise in the fourth term the coefficient of d is 4—1, or 3. Thus, in general, the coefficient of d in the nth term is (n— 1). Hence, if we let I stand for the entire nth term, we have the formula Z=a+(n-l)c?. Example. Find the 11th term of the A. P. 1, 3, 5, 7, •••. Solution. Here o = l, d = 2, n = ll, 1 = 1 Hence, substituting in the formula, we find ? = o+(n-l)d = l+10X2 = l+20 = 21. Am. This result may be checked by actually writing out the series so as to include the 11th term. 35. The Formula for the Sum of the First n Terms. Let a represent the first term of an A. P., d the common difference and I the nth term, as in § 34. Then the sum of the first n terms, which we will denote by S, is (1) S = a+{a+d) + (a+2d) + ia+M)-\ \-(l-d)+l. This value for S may be much simplffied, however, as we shall now show. Write the A. P. (1) in its reverse order, thus obtaining (2) S = l+{l-d) + il-2d) + il-3d)+ ■ • ■ +{a+d)+a. Now add (1) and (2), noting the cancellation of d with —d, of 2d with — 2d, etc. The result is 2S = ia+l) + ia+l) + (a+l)+'- ■ ■ +ia+l) + (.a+T), or 2S = n(a+l). Therefore S=|(a+Z). If we replace I by its value a+ (n—l)d (§ 34), this formula takes the form S=|-{2a + (n-l)d}. 72 COLLEGE ALGEBRA [V, § 35 Example. Find the sum of the first 12 terms of the A. P. 2, 6, 10, 14, -. Solution. Here a = 2, d = 4, m = 12, s = ? Substituting in the second of the formulas just obtained, we find S = — |4+ 11X41 =6|4+44| =6X48 = 288. Am. 36. Arithmetic Means. The terms of an arithmetic pro- gression that lie between any two given terms are called the arithmetic means between those terms. Thus the three arithmetic means between 1 and 9 are 3, 5, 7, since 1, 3, 5, 7, 9 form an A. P. Whenever a single term is thus inserted between two numbers, it is briefly called the arithmetic mean of those two numbers. Thus the arithmetic mean of 2 and 10 is 6 because 2, 6, 10 form an A. P. A formula for the arithmetic mean between any two num- bers a and b is easily obtained. Thus, if x is the desired mean, then a, x, b must form an A. P. Hence, if d be the common difference, we must have x — a = d and b—x = d. It foUows that we must have x—a = b — x. This equation, when solved for X, gives as the desired formula a+b Thus, it follows that the arithmetic mean of two numbers is equal to half their sum. Note. The arithmetic mean of two numbers is also called their average. Example. Insert five arithmetic means between 3 and 33. Solution. We are to have an A. P. of 7 terms in which o = 3, Z = 33, andn = 7. We begin by finding d. Thus l = a+{n-l)d (§ 34) so that 33 = 3+6d. Solving, d = 5. The progression is therefore 3, 8, 13, 18, 23, 28, 33, and hence the desired means are 8, 13, 18, 23, 28. Ans. V, § 36] THE PROGRESSIONS 73 EXERCISES Find, by the fonnulas of §§ 34, 35, the numbers called for in Exer- cises 1-6 below. 1. The 12th term of 3, 6, 9, 12, ■••. 2. The 21st term of 4, 2, 0, -2, -4, ■••. 3. The nth term of x-y, 2x-2y, Zx-^y, ■■■. 4. The sxmi of the first ten terms of 3, 6, 9, 12, •••. 6. The sum of the first thirteen terms of 1, SJ, 6, •■•. 6. The sum of the A. P. of eleven terms, the first of which is —5 and the last of which is 20. 7. When a small heavy body (as a bullet) drops vertically downward it passes over 16.1 feet during the first second, three times as far during the second second, five times as far during the third second, etc. Hence answer the following questions. (a) How far does it go during the 12th second? (6) How far does it go during the first twelve seconds? 8. If you save 5 cents during the first week in January, 10 cents the second week, 15 cents the third week and so on, how much will you save during the last week of the year. Also, what will be the total of the year's savings? 9. Find the sum of all odd integers less than 100. 10. The first term of an A. P. is \ and the 12th term is 11^. What is the sum of the 12 terms? 11. In Fig. 24 the sixteen dotted Hues are equally spaced, and hence their lengths form an arithmetic progression. If the highest one is 6 inches long and the lowest one is 3 feet long, what is the sum of all their lengths? Fig. 24 12. The rungs of a ladder diminish uniformly from 2 feet 4 inches in length at the base to 1 foot, 3 inches at the top. If there are 24 rungs altogether, what is the total length of wood they contain? 13. A piece of rope, when coiled in the usual manner shown in Fig. 25, is found to have 12 com- plete turns, or layers. If the innermost turn is 4 inches long and the outermost is 37 inches long, es- timate the total length of the rope. Fia. 25 74 COLLEGE ALGEBRA [V, § 36 14. Fifty-five logs are to be piled so that the top layer shall contain 1 log, the next layer 2 logs, the next layer 3 logs, etc. How many logs will lie on the bottom layer? 15. A row of numbers in arithmetic progression is written down and afterwards all erased except the 7th and the 12th, which are fomid to be —10 and 15 respectively. What was the 20th niimber? 16. A small rope is wound tightly round a cone, as shown in Fig. 26, the number of complete turns being 24. Upon unwinding from the top, the first and second turns are found to measure respectively 2\ inches and 3^ inches. Estimate the length of the rope. 17. Prove that equal multiples of the terms of an arithmetic progression form another arithmetic progres- sion. 18. Prove that the sum of n consecutive odd inte- , gers, beginning with 1, is n^. 19. Show that the first formula for S obtained in § 35 may be translated into words as follows: "The sum of n terms of an arithmetic progression is equal to n multiplied by the arithmetic mean of the first and the last terms." 20. In the figure below is shown the fnistum of a cone with its "mid- section," or section midway between the bases. Similarly, the frustum of a pyramid and its "mid- section" are shown. It is proved in solid geometry that in all such cases the perimeter of the mid-section is the arithmetic mean of the perimeter of the two bases. Hence, answer the following questions: (a) If the perimeters of the bases are 30 inches and 10 inches respectively, what will be the perimeter of the mid-section? (6) If the radius of the upper base is 2 inches and that of the lower base 8 inches, what will be the perimeter of the mid-section? 21. If d = 2, n = 21 and 5 = 147, find o and I. 22. Show that if any three of the quantities a, d, I, n, S are given, it is always possible to find the other two. In particular, prove that the value of a in terms of d, I and S is given by the formula V, § 38] THE PROGRESSIONS 75 II. Geometric Progression 37. Definitions. A geometric progression is a sequence of numbers, called terms, each of which is derived from the preceding by multiplying it by a fixed amount, called the common ratio. A geometric progression is commonly denoted by the abbreviation G. P. Titus 2, 4, 8, 16, 32, •■■ is a G. P., since each term is derived from the preceding by multiplying it by 2, which is therefore the common ratio. Likewise, 10, —5, 5/2, —5/4, ■•• is a G. P. whose common ratio is -1/2. The next two terms would be 5/8, -5/16. EXERCISES Determine which of the following are geometric progressions, and for such as are, determine the common ratio. 1. 3, 6, 12, 24, 48, •••. 2. i'i'i'T^'-- 3. -1, 2, -4, 8, - -16, -. 4. a, (f, a?, . a^ -. 6. (a+6), (a+6f, {a+bf, (a+bf, •■ •, 6. re' n^ 7. Write the first five terms of the G. P. in which (o) The first term is 4 and the common ratio 4. (b) The first term is —3 and the common ratio —2. (c) The first term is a and the conamon ratio r. 38. The Fonnula for the nth Term. From the definition in § 37 it follows that every geometric progression is of the type form a, ar, ar^, ar^, ar^, . . ., where a is the first term and r is the common ratio. Observe that the exponent of r m any one term is 1 less than the number of that term. Thus 2 is the exponent of r in the third term; 3 is the exponent of r in the fourth term, etc. 76 COLLEGE ALGEBRA [V, § 38 Therefore the exponent of r in the nth term must be (n— 1), so that if we let I stand for the nth term we have the formula Example. Find the 7th term of the G. P. 6, 4, f, •••. SoLTJTioN. We have a = 6, r = |^, n = 7, 1 = 1 The formula gives I = ar'^-^ = Qx(A =2X3 X- IJIJ^. An.. a* 3^ 243 39. The Formula for the Sum of the First n Terms. Let a be the first term of a geometric progression, r the common ratio and I the nth term. Then the sum of the first n terms, which we will caU S, is (1) S = a-\-ar+ar'+a7^^ . . . +af^+ar'^K This value for S may, however, be written in a very much more condensed form, as we shall now show. Multiply both members of (1) by r, thus obtaining (2) rS = ar+ar^+ar^+ar*+ . . .ar'^-^+ar''. Now subtract equation (2) from equation (1), noting the cancellation of terms. This gives S—rS=a—ar". Solving this equation for S, we find (3) S = °-^". 1— r This is the condensed form for S mentioned above. It is to be observed also that since Z=ar''~i (§ 38), we may write rl = ar^. Placing this value of ar" into the formula just found for S, we obtain as a second expression for S 1 — r Example. Find the sum of the first six terms of the G. P. 3 6, 12, 24, -. Solution. a = 3, r = 2, n = 6, iS = ? „ a-ar" 3-3-2" 3-3-64 3-192 -189 V, § 40] THE PROGRESSIONS 77 40. Geometric Means. The terms of a geometric pro- gression that he between any two given terms are called the geometric means between those two terms. Thiis, if we wish to insert three geometric means between 2 and 32, they would be 4, 8, 16, since 2, 4, 8, 16, 32 forms a G. P. Whenever a single term is inserted in this way between two numbers, it is briefly called the geometric mean of those two numbers. Thus the geometric mean of 2 and 32 is 8, since 2, 8, 32 forms a G. P. A formula for the geometric mean of any two numbers, as a and b, is easily obtained. Thus, if x denote the mean, then a, x, b forms a G. P. so that x/a = b/x, each of these fractions being equal to the common ratio of the G. P. Clearing this equation of fractions, and solving for x we find x=\/ab. Thus it follows that the geometric mean of two numbers is equal to the square root of their 'product. Example. Insert four geometric means between 3 and 96. SoLTTTiON. We are to have a G. P. in which a = Z, Z = 96 and w = 6. We begin by finding r. Thus l=ar'^~^ (§ 38), so that 96 = 3 -r*, or r* = 32. Hence r = 2. The progression is therefore 3, 6, 12, 24, 48, 96, and hence the four desired means are 6, 12, 24, 48. Atis. Historical Note. It is related that when Sessa, the inventor of chess, presented his game to Scheran, an Indian prince, the latter asked him to name his reward. Sessa begged that the prince would give him 1 grain of wheat for the first square of the chess board, 2 for the second, 4 for the third, 8 for the fourth, and so on to the sixty- fourth. The number of grains of wheat thus called for was (see (3), § 39) 1 _i .9^ 2^ — 1 = =2^-1 = 18,446,744,073,709,551,615. 1-2 1 This amount is greater than the world's annual supply at present. History does not relate how the claim was settled. (From Godfrey and Siddons' Elementary Algebra, Vol. II, pp. 336, 337.) 78 COLLEGE ALGEBRA [V, § 40 EXERCISES Find, by the formulas of §§ 38, 39, the following numbers. 1. The ninth term of 2, 4, 8, 16, •••. 2. The eighth term of ^i ^i 1, ■•■. 3. The tenth term of 4, 2, 1, -|^, •■■. 4. The eleventh term of ax, aV, oV, a*x*, •■•. 5. The tenth term of 2, \/2, 1, ••-. 6. The sum of eight terms of 2, 4, 8, •••. 7. The sum of six terms of 1, 6, 25, ••■. 8. The sum of ten terms of — |-i -|-j —^i •■■. 9. The sum of ten terms of 1, a^, a^, •■•. 10. What is the sum of the series 3, 6, 12, •••, 384? 11. What is the sum of the series 8, 4, 2, •••, -^^t 12. Find the sum of the first ten powers of 2. 13. Find the sum of the first seven powers of 3. 14. For every person there has lived two parents, four grandparents, eight great grandparents, etc. How many ancestors does a person have belonging to the 7th generation before himself (assuming no dupli- cation)? Answer also for the 10th generation. 16. From a grain of corn there grew a stalk which produced an ear of 100 grains. These grains were planted and each produced an ear of 100 grains. This was repeated until there were 5 harvests. If 75 ears make a bushel, how many bushels were there the fifth year? 16. A series of five squares is drawn such that a side of the second is twice as long as a side of the first, a side of the third twice as long as a side of the second, etc. If a side of the first is 2 inches long, find (by § 39) the sum of the areas of all the' squares. 17. Half the air in a certain sealed receptacle is removed by each stroke of an air pump. What fraction of the original amount of air has been removed by the end of the 7th stroke? 18. A wheel is making 32 revolutions per second when the steam is turned off and the wheel begins to slow down, making half as many revolutions each second as it did during the preceding second. How long before it will be making only 2 revolutions per second? 19. It is found that the number of bacteria in milk doubles every 3 hours. By how much will it be multiplied by the end of one day? V, § 40] THE PROGRESSIONS 79 Fig. 28 20. Show that if a principal of $p be invested at r% compound interest, the sum of money accumulating at the ends of successive years will form a geometric progression, but if the investment be made at simple interest, the sums similarly accumulating will form an arith- metic progression. 21. From a cask of vinegar ^ the contents is drawn off and the cask then filled by pouring in water. Show that if this is done 6 times, the cask will then contain more than 90% water. [Hint. Call the original amount of vinegar 1, then express (as a proper fraction) the amount of water in the cask after the first refiUing, second refilling, etc.] 22. In Fig. 28 a series of ordinates equally spaced from each other has been drawn, the first one being laid off 1 unit long, the second one being laid off equal to the first one increased by ^ its length, etc. Show that these ordinates represent the successive terms of the G. P. whose first term is 1 and whose common ratio is l\. In this sense, the figure may be called the diagram corresponding to the G. P. in which a = l, r = l^. 23. Draw the diagram for the G. P. in which (a) a = l,r = li, (b) a = 2, r = l|, (c) a = 4, j- = |. 24. Prove that the reciprocals of the terms of a geometric progression form another such progression. 25. If a series of numbers are in geometric progression, are their squares likewise in geometric progression? Answer the same question for the cubes of the given numbers; also for their square roots and their cube roots. Answer the same questions for an arithmetic progression. [Hint. See that your reasoning is general; that is, do not base it merely upon the examination of special cases.] 26. Find, correct to four decimal places, the geometric mean of 6 and 27. (Use the tables.) 27. In Fig. 29 a square is placed (in any manner) within another square whose side is twice as long. Show that the area between the squares is equal to three halves of the geometric mean of the areas of the two squares, Pig. 29 80 COLLEGE ALGEBRA [V, § 41 41. Infinite Geometric Progression. Consider the geo- metric progression (1) 1, h h h ^' - ■ Here a=l, r = ^, and hence, by § 39, the sum of n terms is a-ar'' 1-1 • (if l-iiT S = - ^-r 1-i i Now, if the value selected for n is very large, the expres- sion (1/2)" which here appears is very small, being the frac- tion I multiplied into itself n times. In fact, as n is selected larger and larger, this expression (1/2)"' comes to be as small as we please, so that the value for *S, as given above, comes as near as we please to 1-0 — 1— ' 2 which is the same as 2. So we say that 2 is the sum to in- finity of the geometric progression above, meaning thereby simply that as we sum up the terms, taking more and more of them, we come and remain as near as we please to 2. The meaning gf this result is illustrated in Fig. 30. i \ — ^^ Fig. 30 Here, beginning at the point marked 0, we first measure off 1 unit of length, then, continuing to the right, we measure off § unit, then J unit, then | unit, etc., each time going to the right just one-half the amount we went the time before. As this is kept up indefinitely, we evidently come as near as we please to the point marked 2, which is 2 units from 0. This corresponds exactly to what we are doing when we add more and more of the terms of the given progression ^' h h h TS' '" ' A progression Hke the one just considered, in which the V, § 41] THE PROGRESSIONS 81 value of n is not stated but may be taken as large as one pleases, is called an infinite geometric progression. Having thus considered the sum to infinity of the special infinite geometric progression (1), let us now suppose that we have any infinite geometric progression, as a, ar, ar^, ar^, •••, and (as before) that r has some value numerically less than 1. Then the sum of the first n terms is, by § 39 and, as n is taken larger and larger, the expression r" which appears here becomes as small as we please, since we have supposed r to be less than 1. Hence, as n increases indefi- nitely, the value of S comes as near as we please to a — a ■ 1-r or 1-r We have therefore the following theorem: The sum to infinity of any geometric progression whose common ratio r is numerically less than 1 is given by the formula 1— r Example. Find the sum to infinity of the progression 3, 1, ^, -1, .jL, •••. Solution. o = 3, r = |. Since r is numerically less than 1, we have by the formula of § 41, S = -^ = A = |=?=4i. Ans. 1-r 1-i I 2 ^ 82 COLLEGE ALGEBRA [V, § 41 EXERCISES Find the sum to infinity of each of the following progressions, and state in each case what your answer means, drawing a diagram similar to Fig. 30 to illustrate. 1. 1, |, |, ^, ... . 2. ^,|, |, .... 3. 1, _J, 1, _^, ... . [Hint. r= —J- and hence is numerically less than 1. The formula of § 41 therefore applies.] 4. 4, .4, .04, .004, ••■. 6- i, -1*8. ¥\) •■• • 6. 1— x+rc^— x'+ •■• when a;=f-. 8. ?, _ 2V^ 4 sVs' 9 7. x^, 1,-L, 1, \/3 3 9.1, 6 5V3 15 10. A pendulum starts at A and swings to B, then it swings back as far as C, then forward as far as D, etc If the first swing (that is, the cir- cular arc from A to B) is 6 inches long and each succeeding swing is five-sixths as long as the one just preceding it, how far will the pendulum bob travel before coming to rest? 11. At what time after 3 o'clock do the hands of a watch pass each other? [Hint. We may look at this as follows: The large (minute) hand first moves down to where the small (hour) hand is at the beginning, that is, through 15 of the minute spaces along the dial. Meanwhile the small hand advances ^ as far, or -j-f of a minute space. This brings the small hand to the position indicated by the dotted line in the figure. The large hand next passes over this -J-f of a minute space. Meanwhile the small hand again advances -^ as far, which is -j^ of a minute space. The large hand next covers this ^'^^ of a minute space, but the small hand meanwhile ad- vances T^ as far, or i\^-g of a minute space, etc. Thus, the successive moves of the large hand. Fig- 32 counting from the first one, form the G. P. 15, 14, J5 -I4.5., ... .1 V, § 42] THE PROGRESSIONS 83 42. Variable. Limit. We have seen (§ 41) in connection with the geometric progression 1, i, i, i, ••■, that the sum of its first n terms is a quantity which, as n increases indefinitely, comes and remains as near as we please to the exact value 2. The usual way of stating this is to say that as n increases, the sum of the first n terms approaches 2 as a limit. The sum of the first n terms is here called a variable since it varies, or changes, in the discussion. A similar remark apphes to all the infinite geometric progressions which we have consid- ered. In every case the sum to infinity is the hmit which the sum of the first n terms, considered as a variable quantity, is approaching. Note. It may be asked whether the sum of the first n terms of the G. P. 1, -J, \, ^, ■■■ could ever actually reach its limit 2. The answer is that it may or it may not, depending upon circumstances. Thus, if we think of the terms, beginning with the second, as being added on at the rate of one a minute we could never reach the end of the adding process, since the number of the terms is inexhaustible and hence the minutes required would have no end. In other words, the sum of the first n terms could never reach its hmit on this plan. But suppose that instead of this we were to add on the terms with increasing speed as we went forward. For example, suppose we added on the -|^ in ^ a minute, then the -^ in :^ of a minute, then the ^ in ^ of a minute, etc. On this plan we would actually reach the limit 2 in 2 minutes of time. Here the constantly increasing speed of the adding process exactly counterbalances the fact that we have an indefinitely large number of terms to add, with the result that we reach the end of the process in the definite time of 2 minutes. This idea is practically illustrated in Ex. 11, p. 82, where the hands of the watch would never pass each other at all except for the fact that the successive moves of the large hand, which constitute the terms of the progression 15, |-|' i*A' T"7 fr' ' ' ' are added on in less and less time as the process goes on, each being added on in -^-^ the time occupied by the one just before it. The question of whether a variable can reach its limit is intimately connected with the famous problem considered by the Schoolmen of antiquity and known as the problem of Achilles and the tortoise. In this problem, Achilles, who is a celebrated runner and athlete, starts out from some point, as A, to overtake a tortoise which is at some point, as T, the tortoise being famous for the slow rate at which it crawls 84 COLLEGE ALGEBRA [V, § 42 along. Both start at the same instant and go in the same direction, as indicated in the figure. Achilles soon arrives at the point T, from which the tortoise started, but in the meantime the tortoise has gone Fig. 33 some distance ahead. Achilles now covers this last distance, but this leaves the tortoise still ahead, having again gained some additional distance. This continues indefinitely. How, therefore, can Achilles ever overtake the tortoise? The Schoolmen never quite answered this question satisfactorily to themselves. The secret of the diflBculty lies in the fact that, as in the other problems mentioned above, the successive moves which Achilles makes are done in shorter and shorter intervals of time, with the result that, although the number of moves necessary is indefinitely great, they can aU be accomplished in a definite time. 43. Repeating Decimals. If we express the fraction ^| decimally by dividing 12 by 33 in the usual way, we find that the quotient is .363636 • • • , the dots indicating that the divi- sion process never stops (or is never exact) but leads to a never-ending decimal. However, the digits appearing in this decimal are seen to repeat themselves in a regular order, since they are made up of 36 repeated again and again. Such a decimal is called a repeating decimal. More generally, a repeating decimal is one in which the figures repeat them- selves after a certain point. Thus, .12343434 • • , and 1.653653653 • ■ •, are repeating decimals. Let us now turn the question around. Thus, suppose that a certain repeating decimal is given, as for example .272727 • • -, and let us ask what fraction when divided out gives this decimal. This kind of question is usually too difficult to answer in arithmetic, but it can be easily answered as follows by use of the formula in § 41. Thus the decimal .272727 ■ ■ • may be written in the form V, § 43] THE PROGRESSIONS 85 This is an infinite geometric progression in which a = -^, ^ — jiTS- The sum of this progression to infinity must be the value of the given decimal. Hence, the desired value is 1- iVir ^27 100_27_ 3 l-ris 100^99 99 11 ■ Ans. This answer may be checked by dividing 3 by 11, the result being .272727 • • •, which is the given decimal. Note. It is shown in higher mathematics that every rational frac- tion in its lowest terms (that is, every number of the form a/b, where o and b are integers prime to each other) gives rise when divided out to a never-ending repealing decimal (including the cases ia which all the digits after a certain point are zero), while every irrational number (such as -\/2) gives rise when expressed decimally to a never-ending non-repeating decimal. EXERCISES Find the values of the following repeating decimals and check your answer for each of the first six. 1. 0.153153 — . 4. 0.3414141 — . Solution. 0.3414141 2. 0.135135 3. 0.543543543 5. 0.17272 — 6. 1.212121 •• 7. 3.2151515 •=.3 -I- .0414141 ••• =-^+i^(i3;^) _.i 1 41 100 ~ 10 10^100^ 99 3 41 338 169 ~10 990~990~495' Ans. 8. 5.032032032 9. 6.008008008 10. 34.5767676 •• CHAPTER VI VARIATION 44. Direct Variation. One quantity is said to vary directly as another when the two are so related that, though the quantities themselves may change, their ratio never changes. Thus the amount of work a man does varies directly as the number of hours he works. For example, if it takes him 4 hours to draw 10 loads of sand, we can say it will take him 8 hours to draw 20 loads. Here the first ratio is x% and the second is -^ and the two are equal, though the numbers in the second have been changed from what they were in the first. In general, if the man works twice as long, he will draw twice as much ; if he works three times as long, he will draw three times as much, etc.; all of which implies that the ratio of the time he works to the amount he draws in that time never changes. EXERCISES Determine which of the following statements are true and which are false, giving your reason in each instance. 1. The amount of electricity used in lighting a room varies directly as the number of lights turned on. 2. The amount of water in a cyUndrical pail varies directly as the height to which the water stands in the pail. 3. The amount of gasoline used by an automobile in any given time (one week, say) varies directly as the amount of driving done. 4. The time it takes to walk from one place to another at any given rate (3 miles an hour, say) varies directly as the distance between the two places. 5. The time it takes to walk any given distance (5 miles, say) varies directly as the rate of walking. 6. The perimeter of a square varies directly as the length of one side. 7. The circumference of a circle varies directly as the length of the radius. 8. The area of a square varies directly as the length of one side. 9. X varies directly as Wx. 10. X varies directly as lOx^. 86 VI, § 46] VARIATION 87 45. Inverse Variation. One quantity, or number, is said to vary inversely as another when the two are so related that, though the quantities themselves may change, their product never changes. Thus the time occupied in doing any given piece of work varies inversely as the number of men employed to do it. For example, if it takes 2 men 6 days, it will take 4 men only 3 days. The point to be observed here is that the first product, 2X6, equals the second product, 4X3. In general, if twice as many men are employed it will take half as long; if three times as many men are employed, it will take one-third as long, etc. In all these cases, the number of men employed multiplied by the corresponding time required to do the work remains the same. Note. The term varies inversely as is due to the fact that in case xy never changes (as required by the above definition), it follows that x-r-(l/y) never changes, since xy = x-i-(,l/y). That is, x varies directly as the reciprocal, or inverse, of 2/ (§ 44). EXERCISES Determine which of the following statements are true and which are false, giving your reason in each instance. 1. The time it takes water to drain off a roof varies inversely as the number of (equal sized) conductor pipes. 2. The time it takes to walk any given distance (5 miles, say) varies inversely as the rate of walking. 3. The weight of a pail of water varies inversely as the amount of water that has been poured out of it. 4. X varies inversely as 10/x. 5. X varies inversely as 10/a^. 46. Joint Variation. One quantity, or number, is said to vary jointly as two others when it varies directly as their product. Thus the area of a triangle varies jointly as its base and altitude, for if A be the area of any triangle and b its base and h its altitude, we have A =\bh, which may be written A/bh =^. Hence A varies directly as the product bh (§ 44); that is, the ratio of A to bh is always the same, namely -j in this instance. 88 COLLEGE ALGEBRA [VI, § 46 EXERCISES Determine whether the following statements are true, giving your reason in each instance. 1. The area of a rectangle varies jointly as its two dimensions; that is, as its length and breadth. 2. The pay received by a workman varies jointly as his daily wage and the number of days he works. 3. The amount of reading matter in a book varies jointly as the thickness of the book and the distance between the lines of print on the page. 4. The interest received in one year from an investment varies jointly as the principal and rate. 5. The volume of a rectangular parallelepiped (such as an ordinary rectangular shaped box) varies jointly as its length, breadth, and height. [Hint. Here we have one quantity varying jointly as three others. First make a definition of what such variation means.] 47. Variables and Constants. When we say that the amount of vpork a man does varies directly as the number of hours he works, we are dealing with two quantities, namely the amount of work done and the time used in doing it. But it is to be observed that these are not being regarded as fixed quantities, but rather as changeable ones, the only essential idea being that their ratio never changes. In gen- eral, quantities which are thus changeable throughout any discussion or problem are called variables, whUe quantities which do not change are called constants. (Compare § 42.) 48. The Different Types of Variation Stated as Equa- tions. We may now state very briefly and concisely what is meant by the different types of variation described in §§ 44r-46 and certain other important types also. To do this, let us think of x, y, and z as being certain variables and k as being some constant. Then we may state the fol- lowing facts : VI, § 48] VARIATION 89 I that X varies directly as y means (§ - = k, or X = ky, where k is a constant. (1) To say that x varies directly as y means (§ 44) that X y' (2) To say that x varies inversely as y means (§ 45) that xy = k, or a; = -, where kis a constant. (3) To say that x varies jointly as y and z means (§ 46) that — =k, or a; = kyz, where k is a constant, yz Two other important types of variation are described below: (4) To say that x varies directly as the square of y means that X -j = fc, or x = ky^, where k is a constant. (5) To say that x varies inversely as the square of y means that k xy^ = k, or 0; = -;' where k is a constant. In all these types of variation it is important to observe that the value which must be given to the constant k depends upon the particular statement or problem in hand. For example, consider the statement that "The area of a rec- tangle varies jointly as its two dimensions." This means (see [3]) that if we let A be the variable area and a and h the variable dimensions, then A = kab. But in this case we know by arithmetic that A = ah, so the value of k here must bel. On the other hand, consider the statement that "The area of a triangle varies jointly as its base and altitude." Letting A be the variable area and h and h the variable base and altitude, respectively, this means that A = kbh. But here, as we know from geometry, k = ^. 90 COLLEGE ALGEBRA [VI, § 48 EXERCISES Convert each of the following statements into equations, supplying for each the proper value for the constant k mentioned in § 48. 1. The circumference of a circle varies directly as the radius. [Hint. Let C stand for circumference and r for radius.] 2. The circumference of a circle varies directly as the diameter. 3. The area of a circle varies directly as the square of the radius. 4. The area of a circle varies directly as the square of the diameter. 5. The area of a sphere varies directly as the square of the radius. 6. The volume of a rectangular parallelopiped varies jointly as its length, breadth, and height. 7. Interest varies jointly as the principal, rate, and time. 8. The volume of a sphere varies directly as the cube of the radius. [Hint. First supply for yourself the definition of what this type of variation means.] 9. The volume of a circular cone varies jointly as the altitude and the square of the radius of the base. (See formula (11), § 7). 10. The distance, measured in feet, through which a body falls if dropped vertically downward from a position of rest (as from a window ledge) varies directly as the square of the number of seconds it has been falling. [Hint. It is found by experiments in physics that the value of the constant k is in this case 32 (approximately).] 11. The following, Uke Ex. 10, are statements of well-known phys- ical laws. Convert each into an equation without, however, attempting to supply the proper value of k, since to do so requires a study of physics and experiments in laboratories. Fig. 34 (a) When an elastic string is stretched out, as represented in Fig. 34, the tension (force tending to pull it apart at any point) varies directly as the length to which the string has been stretched. (This fact is known as liooke's Law). VI, § 49] VARIATION 91 (6) If a body is tied to a string and swung round and round in a circle (as in swinging a paU of water at arm's length from the shoulder), the force, F, with which it pulls outward from the center (called cen- trifugal force) varies directly as the square of the velocity of the motion. (c) The intensity of the illumination due to any small source of light (such as a candle) varies inversely as the square of the distance of the object Uluminated from the source of hght. (d) The pressure per square inch which a given amount of gas (such as air, or hydrogen, or oxygen, or illuminating gas) exerts upon the sides of the containing receptacle (such as a bag) varies inversely as the volume of the receptacle {Boyle's Law). For example, whenever air is confined in a rubber balloon, as in the first drawing in Fig. 35, it exerts a certain pressure upon each square Fig. 35 inch of the interior surface. If the balloon be squeezed, as in the second drawing (no air being allowed to escape), until its volume is half of what it was before, this pressure wiU be exactly doubled. (e) The square of the mean distance of any planet in the solar system from the sun varies directly as the cube of the time it takes the planet to make one complete revolution around the sun (Kepler's third law of planetary motion). In the case of the earth, its mean distance from the sun is about 93,000,000 miles and its time of complete revolution is 1 year, or 365^ days. 49. Problems in Variation. The problems naturally- arising in the study of variation fall into two general classes as follows : (1) Those in which the value of the constant k mentioned in § 48 can be determined from the statement of the problem 92 COLLEGE ALGEBRA [VI, § 49 and forms an essential part in the solution. This kind of problem is illustrated by Exs. 1-10 below. The solution given for Ex. 1 should be well understood before the student undertakes Exs. 2-10. (2) Those in which it is not necessary to know the value of k. Such problems are illustrated in Exs. 11-20 below. The pupil is advised to work several problems from each group rather than to confine his attention to either. EXERCISES I. Illustrations op Case (1) 1. In a fleet of ships all made from the same model (that is, of the same shape, but of different sizes) the area of the deck varies directly as the square of the length of the ship. If the ship whose length is 200 feet has 5000 square feet of deck, how many square feet in the deck of the ship which is 300 feet long? Solution. Let A represent the area of deck on the ship whose length is I. Then the given law of variation, expressed as an equation (§ 48), is (1) A=hl?. (fc = some constant) Since the ship which is 200 feet long has 5000 square feet of deck, it follows from (1) that we must have 5000 = A; (200)2. This equation tells us that the value of k in the present problem must be 5000 _ 5000 _ 1 ~ (200)2 ~ 200X200 "8 Placing this value of k in (1), gives us an equation which deter- mines completely the relation between A and I in the present problem; that is, (2) A=iP. Now the problem asks how many square feet of deck there are in the ship whose length is 300 feet. This can be found by simply placing 1 = 300 in (2) and solving for A. Thus 1 300 X 300 A = I X (300)2 ^ = 1 1 ,250 square feet. Am. VI, § 49] VARIATION 93 Note. Observe that the first step in the above solution is to express as an equation the law of variation belonging to the problem. Next, the constant k is determined. After this, the first equation is rewritten in its more exact form obtained by assigning to k its value. The answer is then readily obtained. These steps should be followed in working each of the Exs. 2-10 which follow. 2. In a fleet of ships all of the same model, the ship whose length is 200 feet contains 6000 square feet in its deck. How long must a similar ship be made if its deck is to contain 13,500 square feet? 3. To make a suit of clothes for a man who is 6 feet 8 inches high requires 6 square yards of cloth. How much cloth will be required to make a suit for a man of similar build, whose height is 6 feet 2 inches? [Hint. The areas of any two similar figures vary directly as the squares of their heights.] 4. If 10 men can do a piece of work in 20 days, how long will it take 26 men to do it? [Hint. The time required varies inversely as the number of men employed.] 5. The horse-power required to propel a ship varies directly as the cube of the speed. If the horse-power is 2000 at a speed of 10 knots, what will it be at a speed of 15 knots? 6. A silver loving-cup (such as is sometimes given as a prize in athletic contests) is to be made, and a model is first prepared out of wood. The model is 8 inches high and weighs 12 ounces. What will the loving-cup cost if made 10 inches high, it being given that silver is 17 times as heavy as wood and costs $2.20 an ounce? [Hint. The volumes and hence the weights of any two similar figures of like material vary directly as the cubes of their heights.] 7. When electricity flows through a wire, the wire offers a certain resistance to its passage. The imit of this resistance is called the ohm, and for a given length of wire the resistance varies inversely as the square of the diameter. If a certain length of wire whose diameter is \ inch offers a resistance of 3 ohms, what wiU be the resistance of a similar wire (same length and material) ^ of an inch in diameter? 8. Three spheres of lead whose radii are 6 inches, 8 inches, and 10 inches respectively are melted and made into one. What is the radius of the resulting sphere? 94 COLLEGE ALGEBRA [VI, § 49 9. On board a ship at sea the distance of the horizon varies directly as the square root of one's height above the water. If, at a height of 20 feet, the horizon is 5.5 nules distant, what is its distance as seen from a hghthouse 80 feet above sea-level? 10. The horse-power that a shaft can safely transmit varies jointly as its speed in revolutions per minute and the cube of its diameter. A 3-inoh steel shaft making 100 revolutions per minute can transmit 85 horse-power. How many horse-power can a 4-inch shaft transmit at a speed of 150 revolutions per minute? II. Illusteations of Case (2) 11. Knowing that the force of gravitation due to the earth varies inversely as the square of the distance from the earth's center {Newton's Law of Gravitation), find how far above the earth's surface a body must be taken in order to lose half its weight. Solution. Letting W represent the weight of a given body at the distance d from the earth's center, the law stated above, when expressed as an equation, becomes k (1) W = -^- (fc= some constant) a Now let Wi represent the weight of the body when on the surface. Remembering that the earth's radius is 4000 miles (approximately), equation (1) gives Next, let X represent the desired distance, namely the distance above the surface at which the same body loses half its weight. At this distance its weight will consequently be ^Wi, while its distance from the earth's center is now 4000-|-x. So (1) gives ^ ' 2 (4000 +x)2 Dividing equation (3) by equation (2), noting the cancelation of Wi on the left and of the (unknown) k on the right, we obtain 1_ 4000^ 2~(4000+x)2' It remains only to solve this equation for x. Clearing of fractions, (4000-|-a;)2 = 2 ■ 4000^=4000^ • 2. Extracting the square root of both members, 4000 -f-a; = 4000 -\/2- VI, § 49] VARIATION 95 Solving, a; = 4000-s/2-4000 = 4000(V2-l) miles. Ans. To find the approximate value of this answer, we have ,(see tables) V2 = 1.41421 60 that a; = 4000(1.41421-l) =4000X.41421 = 1656.84 miles. Ans. 12. Show that the earth's attraction at a point on the surface is over 5000 times as strong as the distance of the moon; that is, at the (approximate) distance of 280,000 miles. [Hint. Call Wi the weight of a given body on the surface, and let Wi represent the weight of the same body at the distance of the moon from the earth's center. Then use the law expressed in (1) of the solution of Ex. 11.] 13. A book is being held at a distance of 2 feet from an incandescent lamp. How much nearer must it be brought in order that the illumi- nation on the page shall be doubled? (See Ex. 11 (6), p. 91.) 14. If two like coins (such as quarter dollars) were melted and made into a single coin of the same thickness as the original, show that its diameter would be -v/2 times as great. [Hint. Call D the diameter of the given coins and A the area of each. Note that the area of the new coin will then be 2A. Use the result stated in Hint to Ex. 3, p. 93.] 15. Find the result in Ex. 14 when four equal-sized coins are used. 16. Show that a faUing body will pass over the second 3 feet of its descent in about .4 of the time it takes it to pass over the first 3 feet. (See Ex. 10, p. 90.) 17. The time required for a pendulum to make a complete oscillation (swing forward and back) varies directly as the square root of its length. By how much must a 2-foot pendulum be shortened in order that its time of complete oscillation may be halved? 18. If the diameter of a sphere be increased by 10%, by what per cent will the volume be increased? 19. Show that if a city is receiving its water supply by means of a main from a reservoir, the supply can be increased 25% by increasing the diameter of the main by about 12%. 20. It is desired to buUd a ship similar in shape to one already in use but having a 40% greater cargo space (or hold). By what per cent must the beam (width of the ship) be increased? [Hint. See the Hint to Ex. 6, p. 93.] 96 COLLEGE ALGEBRA [VI, § 50 50. Variation Geometrically Considered. If a variable y varies directly as another variable x, we know (§ 48) that this is equivalent to having the equation y = kx, where h is some constant. If the value of k is 1, this equation takes the defi- nite form y = x, and we may now draw its graph, the result being a certain straight hne. If, on the other hand, fc = 2, we have y = 2x, and this again is an equation whose graph may be drawn, lead- ing to a straight line, but a differ- ent one. In general, whatever the value of h, the corresponding equation has a straight-Hne graph. The fact that in all cases the graph is a straight line characterizes this type of variation; that is, characterizes the type in which one variable varies directly as another. Figure 36 shows the lines corresponding to several different values of h. In case a variable y varies in- versely as another variable x, we know (§ 48) that there exists an equation of the form y = h/x, where h is some constant. If we let fc = l, this becomes y=\/x. By letting x take a series of values and determining the cor- responding values of y from this equation (thus forming a table as in § 25) we obtain the graph. Similarly, corresponding to the value fc = 2 we have y = 2/x, and this equation has a definite graph which is different from the one just mentioned. In general, whatever the value of k, m ' if '""1 1 rrHJIN JJ 2 X FiQ. 37 VI, § 50] VARIATION 97 the corresponding equation has a graph, but it is now to be noted that these graphs are not straight lines; they are hyperbolas. (See Ex. 2, § 28.) Figure 37 shows the curves corresponding to several different values of k. Note. Though these curves differ in form, they have the foUowing feature in common: Through the origin draw any two straight hnes (dotted in figure). Then the intercepted arcs AB, CD, EF, GH, etc., are similar; that is, the smallest arc when simply magnified by the proper amount produces one of the others. EXERCISES Draw diagrams to represent the geometric meaning of each of the following statements. 1. y varies directly as the square of x. 2. y varies inversely as the square of x. 3. y varies as the cube of x. 4. y varies directly as x, and y = Q when s = 2. [Hint. The diagram here consists of a single line.] 5. y varies inversely as x, and j/ = 6 when x = 2. 6. The cost of n pounds of butter at 40c per pound is C = 40n. 7. The amount of the extension, e, of a stretched string is propor- tional to the tension, t, and e = 2 in. when i = 10 lb. (See Ex. 11 (c), p. 91.) 8. The pressure, p, of a gas on the walls of a retaining vessel varies inversely as the volume, v; and p = 40 lb. per square foot when 11 = 10 cu. ft. 9. The length, L, of any object in centimeters is proportional to its length, I, expressed in inches; and L = 2.54 cm. when l = lin. CHAPTER VII LOGARITHMS I. General CoNsiDEKATioNsf 51. Definition of Logarithms. If we ask what power of 10 must be used to give a result of 100, the answer is 2 because 10^ = 100. Another common way of stating this is to say that "the logarithm of 100 is 2." In the same way, the power of 10 needed to give 1000 is 3 because 10^=1000, and this is briefly stated by saying that "the logarithm of 1000 is 3." Similarly, the power of 10 that gives .1 is —1 because 10~'-=^\, or .1 by (B), § 8, and this is equivalent to saying that "the logarithm of .1 is — 1." Likewise, the loga- rithm of .01 is —2. From these illustrations we readily see what is meant by the logarithm of a number. It may be defined as follows: The logarithm of a number is the power of 10 required to give that number. Note. A more general definition will be given in § 67, but this is the one commonly used in practice. We write log 100 = 2 to indicate that the logarithm of 100 is 2. Similarly, log 1000 = 3, log .1=-1, log .01= -2, etc. EXERCISES 1. What is the meaning of log 10000? What is its valiie! 2. What is the value of log .001? Why? 3. What is the value of log .00001 ? Why? 4. What is the value of log 10? 6. What is the value of log 1? (See VIII, § 8.) 6. As a number increases from 100 to 1000, how does its logarithm change? fParts I and II give definitions and essential theorems which should be well understood before Part III, which describes the important applications, is taken up. 98 VII, § 52J LOGARITHMS 99 7. As a number decreases from .1 to .01 how does its logarithm change? Answer the same as the number goes from .01 to .001 ; from 1 to 10; from 1 to 1000. 8. Explain why the following are true statements: (o) log 100000 = 5. (b) log .0001 = -4. (c) logVl0 = i. [Hint. Remember VlO = lOX] (d) log AyiO_=i. (e) log -^100 =|. [Hint. Remember •v/I00= •v^I^ = 10». (§8.).] (/) logVl=-h 52. Logarithm of Any Number. Suppose we ask what the value is of log 236. What we are asking for (see defini- tion in § 51) is that value which, when used as an exponent to 10, will give 236; that is, we wish the value of x which will satisfy the equation 10^ = 236. This question resembles those in § 51, but is different because we cannot immediately arrive at the desired value of x by mere inspection. AU we can say here at the beginning is that x must he somewhere between 2 and 3, because 10^=100 and 10^=1000, and 236 lies between these two numbers. In order to find a; to a finer degree of accuracy, it is now natural to try for it such values as 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, and 2.9, all of which he between 2 and 3. The result (which for brevity we shall here state without proof) is that when x = 2.3 the value of 10"^ is slightly less than our given number, 236, while if we take x = 2.4 the value of 10* is slightly greater than 236. Thus x hes somewhere between 2.3 and 2.4. In other words, the value of log 236 correct to the first decimal place is 2.3. It is now natural, if we wish to obtain x to stiU greater accuracy, to try for it such values as 2.31, 2.32, 2.33, 2.34, 2.35, 2.36, 2.37, 2.38, and 2.39, all of which lie between 2.3 and 2.4 The result (which again is here stated without proof) is that when x = 2.37 the value of 10"^ is shghtly less than our 100 COLLEGE ALGEBRA [VII, § 52 number 236, while if we take a; = 2.38 the value of 10^ is slightly greater than 236. This means that the second figure of the decimal is 7, after which we may say that the value of log 236 correct to two places of decimals is 2.37 Proceeding further in the same maimer, it can be shown that when a; = 2.372 the value of 10^ is slightly less than 236, while for x = 2.373 the value of lO"" is slightly greater than 236. Thus the value of log 236 correct to three places of decimals is 2.372 Similarly, it can be shown that the number in the fourth decimal place is 9, and this is as far as it is necessary to carry out the process, since the result is then sufficiently accurate for all ordinary purposes. Hence log 236 = 2.3729, correct to four places of decimals. Note. It thus appears that logarithms do not in general come out exact, though they do so for such exceptional numbers as 100, 1000, 10,000, .1, .01, etc. They can be expressed only approximately, yet as accurately as one pleases by carrying out the decimal far enough. In this respect they resemble such numbers as \/2, \/2, v^j etc. Other examples of logarithms are given below. Note especially the decimal part of each, which is correct to four places. log 283 = 2.4518 log 196 = 2.2923 log 17=1.2304 log 6 = 0.7782 log 3.410 = 0.5328 log 5.75 = 0.7597 53. Characteristic. Mantissa. We have seen that the logarithm of a number consists (in general) of an integral part and a decimal part. These two parts of every logarithm are given special names as follows: The integral part of a logarithm is called the characteristic of the logarithm. The decimal part of a logarithm is called the mantissa of the logarithm. Thus, since log 236 = 2.3729, the characteristic of log 236 is 2, while its mantissa is .3729 Similarly, the characteristic of log 6 is 0, while its mantissa is .7782 VII, §54] LOGARITHMS 101 EXERCISES 1. What is the characteristic of log 100? What the mantissa? Answer the same questions for log 1000, log 10, and log 1. 2. What is the characteristic of log 185? [Hint. Note that 185 lies between 10^ and 10'.] 3. What is the characteristic of log 310? of log 1287? of log 85? of log 21? of log 4? of log 12? of log 13987? 4. For what kind of number can one tell hy inspection both the characteristic and the mantissa of its logarithm? (See § 51.) 54. Further Study of Characteristic and Mantissa. We have seen (§ 53) that log 236 = 2.3729, which is the same as saying that (1) 102-3™ = 236. Let us now multiply both members of (1) by 10. The left side becomes 102-3729+1 or 103-3729 (§ g, Formula I) while the right side becomes 2360. That is, we have 103-3729 = 2360, which is the same as saying that log 2360 = 3.3729 If, instead of multiplying both sides of (1) by 10, we divide both by 10, we obtain in like manner 102-3729-1 = 23.6 (§8, Formula V). That is, we have 101-3729 = 23.6, which is the same as saying that log 23.6 = 1.3729 Finally, if we divide both sides of (1) by IO2, or 100, we obtam 102-3729-2 = 2.36 That is, we have 100-3729 = 2.36 which is the same as saying that log 2.36 = 0.3729 What we now wish to do is to compare the results which we have just been obtaining, and for this purpose they are ar- ranged side by side in a column below. (Zofl- 2360 = 3.3729 . . Jlog 236 = 2.3729 ^''' Slog 23.6 = 1.3729 [log 2.36 = 0.3729 Note that the mantissas here appearing on the right are all the same, namely .3729, while the numbers appearing on 102 COLLEGE ALGEBRA [VII, § 54 the left (that is, 2360, 236, 23.6, and 2.36) are alike except for the position of the decimal point; that is, they contain the same significant figures. This illustrates the following important rule. Rule I. If two or more numbers have the same significant figures (that is, differ only in the location of the decimal point), their logarithms will have the same mantissas; that is, their logarithms can differ only in their characteristics. Thus, log 243, log 2430, log 24.3, log 2.43, log .243, and log .0243 all have the same mantissas. It is only their characteristics that can be different. EXERCISE Apply Rule I, § 54, to tell which of the following logarithms have the same mantissas. log .167 log 8100 log 16.7 log 81 log .0072 log .081 log 7.2 log 720 log 1670 log 16700 II. To Determine the Logarithm of Any Number 55. Purpose of This Part. When we wish to determine the value of a logarithm, as for example, to find log 236, we can work out the characteristic and mantissa as explained in § 52, but this requires considerable time. What we do in practice is to use certain simple rules for determining the characteristic, and we determine the mantissa directly from certain tables which have been carefully prepared for the purpose. We shall now state these rules (§§ 56-58) and explain the tables and how to use them (§§ 59-61). 56. Characteristics for Numbers Greater Than 1. If we look again at the results in (2) of § 54, we see that the char- acteristic of log 2360 is 3. Thus the characteristic is 1 less than the number of figures to the left of the decimal point. Note. 2360 is the same as 2360., so that there are four figures here to the left of the decimal point. VII, § 56] LOGARITHMS 103 Again, we see from (2) of § 54 that the characteristic of log 236 is 2 and this, as in the case already examined, is 1 less than the number of figures to the left of the decimal point. Note. 236 is the same as 236., so there are three figures here to the left of the decimal point. Similarly, since the characteristic of log 23.6 is 1 (see (2) of § 54) this again obeys the same law as just observed in the other two cases; that is, the characteristic is 1 less than the mmiber of figures to the left of the decimal point. Finally, since the characteristic of log 2.36 is 0, the same law is again present here. The law which we have just observed can be shown in like manner to hold good for the characteristic of the logarithm of any number greater than 1; hence we may state the following general rule. Rule II. The characteristic of the logarithm of a number greater than 1 is one less than the number of figures to the left of the decimal point. Thus, the characteristic of log 385.9 is 2; that of log 8.679 is 0. EXERCISES State, by Rule II, § 56, the characteristic of the logarithm of each of the following numbers. 1. 385.4 7. 18.831 2. 461. 8. 3.1568 3. 7962. 9. 401.005 4. 2.7 10. 2967.6 5. 75.54 11. 85. 6. 165,781 12. 2.46879 State how many figures precede the decimal point of a number if the characteristic of its logarithm is 13. 2. 15. 1. 17. 5. 14. 3. 16. 0. 18. 4. 104 COLLEGE ALGEBRA [VII, § 57 57. Characteristics for Positive Numbers Less Than 1. We have seen (see (2) in § 54) that log 2.36 = 0.3729, which is the same as saying that (1) 10''-3™ = 2.36 Let us now divide both members' of this relation by 10. We thus obtain (§ 8, Formula V) 100.3729-1 = .236 or 10-i+o-3'29 = .236, which means (by § 51) Zoff .236= -1+0.3729 Observe that — 1 +0.3729 is really a negative quantity, being equal to —(1—0.3729) which reduces to —0.6271 However, it is more con- venient for our present purposes to keep the longer form —1+0.3729 Note that this cannot be written as —1.3729 because the latter is equal to -1-0.3729 instead of -1+0.3729 If, instead of dividing both members of (1) by 10, we divide both by 10^, or 100, we obtain 100.3729-2 = _0236 (or IQ-^+o-sns = .0236), which means that Zofl- .0236 =-2+0.3729 Similarly, by dividing (1) by 10', or 1000, we find that Zofl' .00236= -3+0.3729 Finally, if we divide (1) by lOS or 10000, we find that log .000236= -4+0,3729 Let us now compare the four results just obtained. Be- ginning with the last result, we see that in the number .000236 there are three zeros immediately to the right of the decimal point; that is, between the decimal pomt and the first significant figure. Corresponding to this, the charac- teristic on the right is minus four. Hence the characteristic is negative and 1 more numerically than the number of zeros between the decimal point and the first significant figure. VII, § 58] LOGARITHMS 105 Similarly, in the number .00236 there are two zeros between the decimal point and the first significant figure, and corre- sponding to this there is a characteristic on the right of minus three. Hence, as before, the characteristic here is negative and numerically 1 more than the number of zeros between the decimal point and the first significant figure. This statement, which is true in all cases mentioned above, can be proved for the characteristic of the logarithm of any positive number less than 1. Hence we have the following rule. Rule III. The characteristic of the logarithm of a (positive) number less than 1, is negative, and is numerically 1 greater than the number of zeros between the decimal point and the first significant figure. Thus, the characteristic of log .0076 is —3; that of log .28 is —1. Note. The logarithm of a negative number is an imaginary quan- tity (as shown in higher mathematics), and hence we shall consider here the logarithms of positive numbers only. 58. Usual Method of Writing a Negative Characteristic. In § 57 we saw that log .236= -1+0.3729 If we add 10 to this quantity and at the same time sub- tract 10 from it we do not change its value, but we give it the new form 9-f0.3729-10, which is the same as 9.3729 — 10. That is, we may write log .236 = 9.3729-10. This is the form used in practice. Likewise, instead of writing log .0236= -2+0.3729 (see § 57) we write in practice log .0236 = 8.3729-10, 106 COLLEGE ALGEBRA [VII, § 58 and similarly we write log .00236 = 7.3729-10. Thus, the usual method of expressing the characteristic whose value is —1 is to write 9—10 for it; if it is —2, we write 8 — 10 for it; if it is —3, we write 7 — 10 for it, etc. For example, log .0076 has the characteristic 7-10. EXERCISES State, by Rule III, § 57, the value of the characteristic of the loga- rithm of each of the following; state how it would be written if expressed in the usual form described in § 58. 1. .06 -2, or 8-10. Ans. 6. .0835 2. .0071 7. .4578 3. .81 8. .00875 4. .00053 9. .15681 5. .835 10. .00005 How many zeros lie between the decimal point and the first sig- nificant figure of a number when the characteristic of its logarithm is 11. -3 13. -5 15. 7-10 12. 9-10 14. 8-10 16. 6-10 59. Determination of Mantissas. Use of Tables. Sup- pose we wish to determine completely the value of log 187. By Rule II, § 56, we know that the characteristic is 2. To find the mantissa, we turn to the tables (p. 108) and look in the column headed N for the first two figures of the given number, that is, for 18. The desired mantissa is then to be found on the horizontal line with these two figures and in the column headed by the third figure of the given number; that is, in the column headed by 7. Thus in the present case the mantissa is found to be .2718 Note. For brevity, the decimal point preceding each mantissa is omitted from the tables. It must be supplied as soon as the mantissa is used. The complete value (correct to four decimal places) of log 187 is therefore 2.2718 VII, § 59] LOGARITHMS 107 Again, suppose we wish to determine log 27.6. The char- acteristic (by § 56) is 1. The mantissa, by Rule I, § 54, is the same as that of log 276; and the latter, as given in the tables, is .4409 Therefore, log 27.6 = 1.4409 Ans. As a last example, suppose we wish to determine log .0173 The characteristic (by § 57) is —2, or 8-10. The mantissa, by the rule in § 54, is the same as that of log 173 and the latter, as obtained from the tables, is .2380 Therefore, log .0173 = 8.2380 - 10. Ans. These examples illustrate how the tables together with Rules II and III, §§ 56, 57, enable us to determine completely the logarithm of any number provided it contains no more than three significant figures. We may now summarize our results in the following rule. Rule IV. To find the logarithm of a number of three signifi- cant figures: 1. Look in the column headed N for the first two figures of the given number. The mantissa vnll then be found on the hori- zontal line, opposite these two figures and in the column headed by the third figure of the given number. 2. Prefix the characteristic according to Rules II and III, §§ 56, 57. EXERCISES Determine the logarithm of each of the following numbers, expressing all negative characteristics as explained in § 58. 1. 561 2. 217 3. 280 4. 800 5. 72.5 [Hint to Ex. 5. Note how log 27.6 was obtained in § 69.] 6. 7.25 7. 93. 8. 9. 0. .0136 10. .936 11. .0036 [Hint. Write as .00360] 12. 7550. 15. .35 18. .000831 13. .071 16. 55.7 19. i- 14. .7 17. 25,300 20. |- 108 COLLEGE ALGEBRA [VII, § 59 N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 0000 0414 0792 1139 1461 0043 0463 0828 1173 1492 0086 0492 0864 1206 1523 0128 0531 0899 1239 1553 0170 0569 0934 1271 1584 0212 0607 0969 1303 1614 0253 0645 1004 1335 1644 0294 0682 1038 1367 1673 0334 0719 1072 1399 1703 0374 0755 1106 1430 1732 15 16 17 18 19 1761 2041 2304 2553 2788 1790 2068 2330 2677 2810 1818 2095 2355 2601 2833 1847 2122 2380 2625 2856 1876 2148 2405 2648 2878 1903 2175 2430 2672 2900 1931 2201 2465 2695 2923 1959 2227 2480 2718 2945 1987 2253 2504 2742 2967 2014 2279 2529 2765 2989 20 21 22 23 3010 8222 3424 3617 3802 3032 3243 3444 3636 3820 3064 3263 3464 3655 3838 3075 3284 3483 3674 3856 3096 3304 3502 3692 3874 3118 3324 3622 3711 3892 3139 3345 3641 3729 3909 3160 3365 3560 3747 3927 3181 3385 3579 3766 3945 3201 3404 3508 3784 3962 25 26 27 28 29 3979 4150 4314 4472 4624 3997 4166 4330 4487 4639 4014 4183 4346 4502 4654 4031 4200 4362 4518 4669 4048 4216 4378 4533 4683 4065 4232 4393 4548 4698 4082 4249 4409 4664 4713 4099 4265 4425 4579 4728 4116 4281 4440 4594 4742 4133 4298 4456 4609 4757 30 31 32 33 34 4771 4914 5051 5185 5315 4786 4928 5065 5] 98 5328 4800 4942 5079 5211 5340 4814 4955 5092 5224 5353 4829 4969 5105 5237 5366 4843 4983 5119 6260 5378 4867 4997 6132 5263 5391 4871 5011 5145 5276 5403 4886 5024 5169 5289 .6416 4900 5038 5172 5302 5428 35 36 37 38 39 5441 5563 5682 6798 5911 5453 5575 5694 5809 5922 5465 6587 5705 5821 6933 5478 6599 5717 6832 5944 6490 5611 5729 5843 5955 5602 5623 5740 5855 5966 6514 5635 6752 5866 697J 5627 6647 5763 5877 6988 5539 5658 5775 6888 5999 5551 5670 5786 5899 6010 40 41 42 43 44 6021 6128 6232 6335 6435 6031 6138 6243 6345 6444 6042 6149 6253 6355 6464 6053 6160 6263 6365 6464 6064 6170 6274 6375 6474 6075 6180 6284 6385 6484 6085 6191 6294 6396 6493 6096 6201 6304 6405 6603 6107 6212 6314 6415 6513 6117 6222 6325 6425 6522 45 46 47 48 49 6632 6628 6721 6812 6902 6542 6637 6730 6821 6911 6551 6646 6739 6830 6920 6561 6656 6749 6839 6928 6571 6665 6758 6848 6937 6580 6675 6767 6857 6946 6590 6684 6776 6866 6955 6599 6693 6785 6875 6964 6609 6702 6794 6884 6972 6618 6712 6803 6893 6981 50 51 52 53 54 6990 7076 7160 7243 7324 6998 7084 7168 7251 7332 7007 7093 7177 7259 7340 7016 7101 7185 7267 7348 7024 7110 7193 7275 7356 7033 7118 7202 7284 7364 7042 7126 7210 7292 7372 7060 7135 7218 7300 7380 7059 7143 7226 7308 7388 7067 7152 7235 7316 7396 VII, § 59] LOGARITHMS 109 N O 1 2 3 4: 5 6 7 8 9 55 66 67 58 89 7404 7482 7559 7634 7709 7412 7490 7566 7642 7716 7419 7497 7574 7649 7723 7427 7605 7582 7657 7731 7435 7513 7589 7664 7738 7443 7520 7597 7672 7745 7451 7528 7604 7679 7752 7459 7536 7612 7686 7760 7466 7543 7619 7694 7767 7474 7551 7627 7701 7774 60 61 62 63 64 7782 7853 7924 7993 8062 7789 7860 7931 8000 8069 7796 7868 7938 8007 8075 7803 7875 7945 8014 8082 7810 7882 7952 8021 8089 7818 7889 7959 8028 8096 7825 7896 7966 8035 8102 7832 7903 7973 8041 8109 7839 7910 7980 8048 8116 7846 7917 7987 8055 8122 65 66 67 68 69 8129 8195 8261 8325 8388 8136 8202 8267 8331 8395 8142 8209 8274 8338 8401 8149 8215 8280 8344 8407 8156 8222 8287 8351 8414 8162 8228 8293 8357 8420 8169 8235 8299 8363 8426 8176 8241 8306 8370 8432 8182 8248 8312 8376 8439 8189 8254 8319 8382 8445 70 71 72 73 74 8451 8513 8573 8633 8692 8457 8519 8579 8639 8698 8463 8525 8585 8645 8704 8470 8531 8591 8651 8710 8476 8537 8597 8657 8716 8482 8543 8603 8663 8722 8488 8549 8609 8669 8727 8494 8555 8615 8675 8733 8500 8561 8621 8681 8739 8506 8567 8627 8686 8745 75 76 77 78 79 8751 8808 ■ 8865 8921 8976 8756 8814 8871 8927 8982 8762 8820 8876 8932 8987 8768 8825 8882 8938 8993 8774 8831 8887 8943 8998 8779 8837 8893 8949 9004 8785 8842 8899 8954 9009 8791 8848 8904 8960 9015 8797 8854 8910 8965 9020 8802 8859 8915 8971 9025 80 81 82 83 84 9031 9085 9138 9191 9243 9036 9090 9143 9196 9248 9042 9096 9149 9201 9253 9047 9101 9154 9206 9258 9053 9106 9159 9212 9263 9058 9112 9165 9217 9269 9063 9117 9170 9222 9274 9069 9122 9175 9227 9279 9074 9128 9180 9232 9284 9079 9133 9186 9238 9289 85 86 87 88 89 9294 9345 9395 9445 9494 9299 9350 9400 9450 9499 9304 9355 9405 9455 9504 9309 9360 9410 9460 9509 9315 9365 9415 9465 9513 9320 9370 9420 9469 9518 9325 9375 9425 9474 9523 9330 9380 9430 9479 9528 9335 9385 9435 9484 9533 9340 9390 9440 9489 9538 90 91 92 93 94 9542 9590 9638 9685 9T31 9547 9595 9643 9689 9736 9553 9600 9647 9694 9741 9557 9605 9652 9699 9745 9562 9609 9657 9703 9750 9566 9614 9661 9708 9754 9571 9619 9666 9713 9759 9576 9624 9671 9717 9763 9581 9628 9675 9722 9768 9586 9633 9680 9727 9773 95 96 97 98 99 9777 9823 9868 9912 9956 9782 9827 9872 9917 9961 9786 9832 9877 9921 9965 9791 9836 9881 9926 9969 9795 9841 9886 9930 9974 9800 9845 9890 9934 9978 9805 9850 9894 9939 9983 9809 9854 9899 9943 9987 9814 9859 9903 9948 9991 9818 9863 9908 9952 9996 110 COLLEGE ALGEBRA [VII, § 60 60. To Find the Logarithm of a Number of More Than Three Significant Figures. Suppose we wish to determine log 286.7 Here we have jour significant figures, while our tables tell us the mantissas of numbers having three (or less) significant figures (as in § 59 and in the preceding ex- ercises). In such cases we proceed as follows. From the tables on pp. 108-109 we have log 286 = 2.4564 ] log 286.7 = ? [Difference = 2.4579-2.4564 = .0015 log 287 = 2.4579 J Since 286.7 lies between 286 and 287, its logarithm must lie between their logarithms. Now, an increase of one unit in the number (in going from 286 to 287) produces an increase of .0033 in the mantissa. It is therefore assumed that an increase of .7 in the number (in going from 286 to 286.7) pro- duces an increase of .7 of .0015, or .00105, in the mantissa. Therefore, log 286.7 = 2.4564-1- .7 of .0015 = 2.4564-f .00105 = 2.45745, so that log 286.7 = 2.4574 (approximately). Ans. In practice the answer is quickly obtained as follows: The difference between any mantissa and the next higher one in the table (neglecting the decimal point) is called the tabular difference. The tabular difference in this example is 4579-4564, or 15. Taking .7 of this, we obtain 10.5, which (keeping only the first two figures) we call 10, and adding this to 4564 we find 4574. This, therefore, is the required mantissa of log 286.7, so that log 286.7 = 2.4574 (approximately). VII, §60] LOGARITHMS 111 Similarly, in finding log 286.75 the tabular difference (as before) is 15. Taking .75 of 15 gives 11.25, which (keeping only two figures) has the approximate value 11. Hence the mantissa of log 286.75 is 4564+11=4575. Therefore log 286.75 = 2.4575 Ans. Below are two examples further illustrating how the above processes are quickly carried out in practice. The student should form the habit of writing the work in this form. Example 1. Determine the value of log 48.731 Solution. Mantissa of log 487 = 6875 , Tabular difference = 9 Mantissa of log 488 = 6884 , .31X9 = 2.79 = 3 (approximately). Hence mantissa of log 48.731=6875+3 = 6878. Therefore log 48.731 = 1.6878. Ans. Example 2. Determine the value of log .013403 Solution. Mantissa of 134 = 1271 1 „, , ,.^ > Tabular difference =32. Mantissa of 135 = 1303 J .03 X 32 = .96 = 1 (approximately) . Hence mantissa of log .013403 = 1271+1 = 1272. Therefore log .013403= -2 + . 1272 = 8.1272-10. Ans. Note. The process which we have employed for determining a mantissa when it does not actually occur in the tables is called inter- polation. When examined carefully, it will be seen that the process is based upon the assumption that if a number is increased by any frac- tional amount of itself, the logarithm of the number will hkewise be increased by the same fractional amount of itself. Thus, in finding the mantissa of log 286.7 at the middle of p. 110, we assumed that the increase of .7 in going from 286 to 286.7 would be accompanied by Mke increase of .7 in the logarithm. Such an assumption, though not exactly correct, is very nearly so in most cases and is therefore suffi- ciently accurate for all ordinary purposes. 112 COLLEGE ALGEBRA [VII, § 60 Tables of logarithms much more extensive than those on pages 108, 109 have been prepared and are commonly used. See, for example. The Macmillan Tables. By means of these, any desired mantissa may usually be obtained as accurately as is necessary directly, that is, without interpolation. EXERCISES Obtain the logarithm of each of the following numbers. 1. 678.3 12. .07235 2. 332.2 13. 745.23 3. 675.3 14. 132.36 4. 481.6 15. 51.745 6. 956.7 16. 430.07 6. 22.17 17. 5.2178 7. 8.467 18. 4.2316 8. 3.706 19. 1.6086 9. 2.408 20. .14653 10. 2.767 21. .074568 11. ,3456 22. .00738 61. To Find the Number Corresponding to a Given Loga- rithm. Thus far we have considered how to determine the logarithm of a given number, but frequently the problem is reversed, that is, it is the logarithm that is given and we wish to find the number having that logarithm. The method of doing this is the reverse of the method of §§ 59, 60, and is illustrated in the following examples. Example 1. Find the number whose logarithm is 1.9547 Solution. Locate 9547 among the mantissas in the table. Having done so, we find in the column N on the line with 9547 the figures 90. These form the first two figures of the desired number. At the head of the column containing 9547 is 1, which is therefore the third figure of the desired number. Hence the number sought is made up of the digits 901. The given characteristic being 1, the number just found must be pointed off so as to have two figures to the left of its decimal point (Rule II, § 56). Therefore the number is 90.1 Ans. Vn, §61] LOGARITHMS 113 Example 2. Find the number whose logarithm is 0.6341 Solution. As in Example 1, we look among the mantissas of the table to find 6341. In this case we do not find exactly this mantissa, but we see that the next less mantissa appearing is 6336, while the one next greater is 6345. The numbers corresponding to these last two mantissas are seen to be 430 and 431 respectively. Whence, if x represents the number sought, we have Mantissa of log 430 = 6335 \ j-^.^ ^ "j Mantissa of log x = 6341 / ' >• Tabular difference = 10. Mantissa of log 431 = 6345 ) Since an increase of 10 in the mantissa produces an increase of 1 in the number, we assume that an increase of 6 in the mantissa will pro- duce an increase of t%, or .6, in the number. Hence the number sought has the digits 4306. Since the given characteristic is 0, it is evident that the number must be 4.306 (§ 56). Ans. Note l. The student will observe that in Example 1 the given man- tissa actually occurs in the tables, while in Example 2 it does not, thus making it necessary in this last case to interpolate. (See the Note in § 60.) Note 2. The number whose logarithm is a given quantity is called the antilogarithm of that quantity. Thus 100 is the antUogarithm of 2; 1000 is the antilogarithm of 3, etc. EXERCISES Find the numbers whose logarithms are given below. 1. 2.6656 11. 3.7430 2. 1.8351 12. 0.5240 3. 0.2742 13. 0.6970 4. 2.5855 14. 9.7400-10 5. 9.6830-10 15. 8.3090-10 6. 8.8028-10 16. 7.5308-10 7. 7.6425-10 17. 9.0046-10 8. 6.8842-10 18. 8.0012-10 9. 1.2517 19. 3.4968-10 10. 2.8583 20. 5.9654-10 114 COLLEGE ALGEBRA [VII, §62 III. The Use of Logarithms in Computation 62. To Find the Product of Several Numbers. The pro- cesses of multiplication, division, raising to powers, and ex- traction of roots, as carried out in arithmetic, may be greatly shortened by the use of logarithms, as we shall now show. Let us take any two numbers, for example 25 and 37, and determine their logarithms. We find that log 25 = 1.3979 and log 37 = 1.5682 This means (§ 136) that 25 = 101-3979 and 37=10^-5682 Multiplying, we thus have 25X37 = 101-39^3 +1-5682 (§ 8, Formula I) The last equality means (§ 51) that log (25X37) = 1.3979+1.5682, or log (25X37) =log 25+log 37. Similarly, if we start with the three numbers 25, 37, and 18 we can show that log (25X37X18)= log 25-Mog 37+log 18. Thus we arrive at the following important rule. Rule V. The logarithm of a 'product is equal to the sum of the logarithms of its factors. Thus log (13X.0156X99.8)=log 13-Hog .0156+log 99.8 The way in which this rule is used to find the value of the product of several numbers is shown below. Example 1. To find the value of 13X.0156X99.8 Solution, log 13 =1.1139 log .0156 = 8.1931-10 log 99.8 =1.9991 Adding, 11.3061 - 10, or 1.3061 Hence, by Rule V, the logarithm of the desired product is 1.3061 It follows that the product itself is the number whose logarithm is 1.3061 When we look up this number (as in § 61) we find it to be 20.23 Hence 13 X. 0156X99.8 = 20.23 (approximately). Am. VII, § 62] LOGARITHMS 115 Example 2. To find the value of 8.45 X. 678 X. 0015 X956X. Ill Solution, log 8.45= 0.9269 log .678= 9.8312-10 log .0015= 7.1761-10 log 956= 2.9805 log .111= 9.0453-10 Adding, 29.9600 - 30 = 9.9600 - 10. Hence, by Rule V, the logarithm of the desired product is 9.9600 — 10. Therefore the product itseH is found (as in § 61) to be .912 (approxi- mately). Ans. These examples illustrate the following rule. Rule VI. To multiply several numbers: 1. Add the logarithms of the several factors. 2. The sum thus obtained is the logarithm of the product. 3. The product itself can then be determined as in § 61. EXERCISES Find, by Rule V, § 62, the value of each of the following logarithms. 1. log (38.2X6.31). 3. log (167 X7.31X. 00456). 2. log (6X4.21X.0015). . 4. log (3.81 X. 00175X1.87). Find, by Rule VI, § 62, the value of the following products. Check your answer in Ex. 5 by multiplying out the long way as in arithmetic. Compare the two results and see how great was the error committed by following the short (logarithmic) method. Compare also the time required for the two methods. 5. 56.8X3.47X.735 8. 34.56X18.16X.0157 6. .975X42.8X3.72 [Hint. See § 60.] 7. 896X40.8X3.75X.00489 9. 576.8X43.25X3.576X.0576 10. 60.573X8.087X.008915X1.2387 11. 23X23X23X23X23X23X23, (or 23'). 12. 1.2X2.3X3.4X4.5X5.6X6.7X7.8 13. .31 X5.198X6.831X2.584X. 00312 X. 07568 14. Since 25X15 =375 we know by Rule V, § 62, that the logarithm of 25 added to the logarithm of 15 is equal to the logarithm of 375. Show that the values given in the tables for log 25, log 15, and log 375 confirm this result. Invent and try out several other similar problems. 116 COLLEGE ALGEBRA [VII, § 63 63. to Find the Quotient of Two Numbers. Let us take any two numbers, for example 41 and 29, and write their logarithms. We find log 41 = 1.6128 log 29 = 1.4624 These mean that 41 = 10i«i28 and 29 = 10'«24 Whence, dividing the first of these equahties by the sec- ond, we obtain 101.6128 414-29= = 101.6128-1.4624 (§ §, Fonuula V) The last equality means that log (41 -^ 29) = 1,6128 -1.4624 = log 41 -log 29. This result illustrates the following general rule. Rule VII. Tine logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor. Thus log (467.3 -=-.00149)= log 467.3 -log .00149 The way in which this rule is used is shown below. Example 1. To find the value of 236 -=-4.15 Solution. log 236 = 2.3729 log 4.15 = 0.6180 Subtracting 1.7549 Hence the logarithm of the desired quotient is 1.7549 (Rule VII.) The number whose logarithm is 1.7549 is found (as in § 61) to be 56.875 Therefore 236-7-4.15 = 66.875 (approximately). Ans. Example 2. To find the value of 1.46 -h .00578 Solution. log 1.46=0.1644 = 10.1644-10 (See Note p. 117.) log .00578 = 7.7619-1 Subtracting, 2.4025 The number whose logarithm is 2.4025 is found to be 252.64 Therefore 1.46-^.00578=252.64 (approximately). Ans. VII, § 63] LOGARITHMS 117 Thus we have the following rule. Rule VIII. To find the quotient of two numbers: 1. Subtract the logarithm of the divisor from the logarithm of the dividend. 2. The difference thus obtained is the logarithm of the quo- tient. 3. The quotient itself can then be determined as in § 61. Note. To subtract a negative logarithm from a positive one, or to subtract a greater logarithm from a less, increase the characteristic of the minuend by 10, writing —10 after the mantissa to compensate. Thus, in Example 2, we wished to subtract the negative logarithm 7.7619 — 10 from the positive one 0.1644 Therefore 0.1644 was written in the form 10.1644 — 10, after which the subtraction was easily per- formed. EXERCISES Find, by Rule VII, § 63, the value of each of the following logarithms. i. log (17-:- 8). 3. log (37.5 -5- .0018). 2. log (218^7.15). 4. log (8.69-M13). Find, by Rule VIII, § 63, the value of each of the following quo- tients. . Check your answer in Ex. 5 by dividing out the long way as in arithmetic. Compare the two results and see how great was the error committed by following the short (logarithmic) method. 5. 246-^15.7 6. 34.7-^5.34 9. 3.25-^.00876 7. 389.7 H-4.353 [Hint. See Note in § 63.] [Hint. See § 60.] 10. 49.6^87.3 8. 45.67+38.01 40.3X6.35 3.72 [Hint. Find the logarithm of the numerator by Rule V, § 62.] .0036X2.36 ,„ 24.3 X. 695 X. 0831 12. 13. .0084 8.40 X. 216 14. Since 27-^9 = 3 we know, by Rule VII, § 62, that the logarithm of 9 subtracted from the logarithm of 27 is equal to the logarithm of 3. Show that the values given in the tables for log 9, log 27, and log 3 confirm this result. Invent and try out several other similar problems for yourself. 118 COLLEGE ALGEBRA [VII, § 64 64. To Raise a Number to a Power. Let us take any number, for example 25, and raise it to any power, say the fourth. We then have 25^ which means 25X25X25X25. Hence, by Rule V, § 62, we have log 25* = log 25+log 25+log 25+log 25, or log 25* = 4 log 25. This illustrates the following rule. Rule IX. The logarithm of any power of a number is equal to the logarithm of the number multiplied by the ex- ponent indicating the power. Thus log 3.17i'' = 101og 3.17; similarly, log .00174« = 6Iog .00174 The way in which this principle is used to raise a number to a power is shown below. Example 1. To find the value of 2.37* Solution. log 2.37 =0.3747 4 Multiplying, 1.4988 Hence log 2.37^ = 1.4988 (Rule IX) The number whose logarithm is 1.4988 is found to be 31.535 Therefore 2.37^ = 31.535 (approximately). Ans. Example 2. To find the value of .856^ Solution. log .856= 9.9325-10 5 Multiplying, 49.6625 - 50 = 9.6625 - 10 The number whose logarithm is 9.6625 — 10 is .4597 Therefore .856* = .4597 (approximately). Ans. Thus we have the following rule Rule X. To raise a number to a power: 1. Multiply the logarithm of the number by the exponent indicating the power. 2. The result thus obtained is the logarithm of the answer. 3. The answer itself can then be determined as in § 61. VII, §65] LOGARITHMS 119 EXERCISES Find, by Rule IX, § 64, the value of each of the following logarithms. 1. log 16^ 2. log 3.12^ 3. log .01762 4. log 36.64* Find, by Rule X, § 64, the value of each of the following expressions. 6. 8.82' Check your answer by raising 8.82 to the third power as in arith- metic. Compare the two results and see how great was the error committed by following the short (logarithmic) method. 6. 4.12* 7. 4.123* 8. .175^ [Hint. See Ex. 2 in § 64.] 9. 81^X.0152 [Hint. Combine the rules of §§ 62 and 64.] 10. 43X8.92X.075' 8.76X53.9X4.5' ■ 2.32X3.15X5.14'' [Hint. Use Rules VI, VIII, X.] 12. Since 9^ = 729 we know, by Rule IX, § 64, that three times the logarithm of 9 is equal to the logarithm of 729. Show that the values given in the tables for log 9 and log 729 confirm this result. Invent and try out several other similar problems for yourself. 65. To Extract Any Root of a Number. Let us take any number, for example 36, and consider any root of it, say the fifth; that is, let us consider -v'^. Supposing X to be the value of the desired root, we have Now the logarithm of the first member of this equality is equal to 5 log x by Rule IX. Hence 5 log a; = log 36, or log x = ^ log 36. This illustrates the following rule. Rule XL The logarithm of the root of a number is equal to the logarithm of the radicand divided by the index of the root. Thus log \/Tn=\ log 2.73; similarly, log ■{/. 01685=1 log .01685 The way in which this principle is used to extract the roots of numbers in arithmetic will now be shown. 120 COLLEGE ALGEBRA [VII, § 65 Example 1. To find the value of v^86.2 SoLTJTioN. log 85.2 = 1 .9304, so that i of log 85.2 = 0.4826 Hence log -v'^5i2 = 0.4826 (Rule XI) The number whose logarithm is 0.4826 is 3.038 (§ 61) Therefore •v^85.2 = 3.038 (approximately). Ans. Example 2. To find the value of -v^.0875 Solution. log .0875 = 8.9420-10, so that ^ of log .0875 = -^(8.9420 -10) =^(48.9420 -50) = 9.7884 - 10. (See Note below.) The number whose logarithm is 9.7884-10 is .6143 (§ 61) Therefore ■\^ .0875 = .6143 (approximately). Ans. These examples illustrate the following rule. Rule XII. To find any root of any number. 1. Divide the logarithm of the number by the index of the root. 2. The quotient obtained is the logarithm of the desired root. 3. The root itself can then be determined as in § 61. Note. To divide a negative logarithm, write it in a form where the negative part of the characteristic may be divided exactly by the divisor giving —10 as quotient as in Example 2. EXERCISES Find, by Rule XI, § 65, the value of each of the following logarithms. 1. log \/T6. 2. log -(7312 3. log -i^m75 4. log ^^38^ Find, by Rule XII, § 65, the value of each of the following expres- sions. Check your answer in Ex. 5 by extracting the square root of 315 (correct to three decimal places) as in arithmetic. Compare the two results and see how great was the error committed by following the short (logarithmic) method. 6. VsTE 9. -\/8.76X.0153 6. 1^4.32 [Hint. Use Rules IX and XL] 7. v'T325 10. -^/STeX v^8?76 8. -v/^0957 11 / 576X9.132 [Hint. See Example 2 in § 65.] '^3.8X5.32^' VII, § 65] LOGARITHMS 121 APPLIED PROBLEMS Solve the following exercises by logarithms. 1. How many cubic feet of air are there in a schoolroom whose dimensions are 50.5 ft. by 25.3 ft. by 10.4 ft.? 2. How many gallons will a rectangular tank hold whose dimensions are 8 ft. 10 in. by 9 ft. 3 in. by 10 ft. 1 in.? 3. How much wheat will a cyhndrical bin hold if the diameter of the base is 9 ft. 5 in. and the height is 40 ft. 4 in.? 4. How much would a sphere of soHd cork weigh if its diameter was 4 ft. 3 in., it being known that the specifio gravity of cork is .24? [Hint. To say that the specific gravity of cork is .24 means that any volume of cork weighs .24 times as much as an equal volume of water. Water weighs 62.5 pounds per cubic foot.] 6. The diameter d in inches of a wrought-iron shaft required to transmit h horse-power at a speed of n revolutions per minute is given by the formula <^ = 'V~5f Find the diameter required when 135 horse- power is to be transmitted at a speed of 130 revolutions per minute. 6. A wire 135 feet long is suspended from two poles of equal height placed 130 feet apart. Compute the sag, using the formula of Ex. 20, page 42. 7. If the three sides of a triangle are of lengths a, b, c respectively, and we place s=^((i+6+c), then the area is expressed by the formula s= ■\/s{s — a){s — b)(s — c). Determine the area of the triangle whose sides are 3.15 inches, 4.87 inches and 2.68 inches. 8. The height ff of a mountain in feet is given by the formula F = 49,000f^Vl+— V ' \R+rJ \ 900 / where B, r are the observed heights of the barometer in inches at the foot and at the summit of the mountain, and where T, t are the observed Fahrenheit temperatures at the foot and summit. Find the height of a mountain if the height of the barometer at the foot is 29.6 inches and at the summit 25.35 inches, while the temv perature at the foot is 67° and at the summit 32°. 122 COLLEGE ALGEBRA [VII, § 66 66. Solution of Exponential Equations. The equation (1) 2^ = 32, wherein the unknown number, x, appears in the exponent, is an example of an exponential equation. In the present instance, the equation may be solved immediately by inspec- tion, X being equal to 5, since 2* = 32. But if, instead of (1), we start with the foUowiag equally simple exponential equa- tion (2) 2" = 48 the value of x can be obtained only approximatively, and its determination involves the use of logarithms in the manner shown below: Solution. Taking the logarithm of each member in (2), X log 2 = log 48. (Rule IX) Therefore .=!^ = 1:^^5.58+ Ar.. log 2 0.3010 EXERCISES Solve each of the following exponential equations, using logarithms. 1. 4^ = 10. 6. 32^-20-3^+99 = 0. 2 2^ = 80 f^"^- 32^-20-3^+99 = (3^-9) (3^-11). 3. 3P = 23. ^ /3- = 2y, 4. .2^=3. {"^^-y- '2^+» = 6, 5. 13^ = .281 ®' \2^+i = 3i'. IV. General Logarithms *67. Logarithms to Any Base. In § 51 we defined the logarithm of a number as the power to which 10 must be raised to obtain that number. Thus, from such equalities as 10^ = 100, 10^ = 1000, etc., we had log 100 = 2, log 1000 = 3, etc. Strictly speaking, this defines the logarithm of a number to Ike base 10, or, as it is usually called, a common logarithm. We may and frequently do use some other base than 10. For example, since 3^ = 9, 3^ = 27, 3^ = 81, etc., we can say that the loga- VII, § 70] LOGARITHMS 123 rithm of 9 to the base 3 is 2, the logarithm of 27 to the base 3 is 3,the logarithm of 81 to the base 3 is 4, etc. The usual way of denoting this is to write log39 = 2, log327 = 3, logaSl =4, etc. The number being used as the base is placed to the right and just below the symbol log. Similarly, we have log2l6 = 4, log864 = 2, log5l25 = 3, etc. Thus we have the following general definition. The logarithm oj any number x to a given base a is the power of a required to give x. It is written logaX. Any positive number except 1 may be used as the base. Note. When the base a is taken equal to 10 (that is, in the usual case) we write simply log x instead of logio^; ♦EXERCISES State first the meaning and then the valiie of 1. log24. 2. logaS. 3. legale. 4. logg^. 6. logal. 6. log4^. 7. log5.2 8. log832. *68. Logarithm of a Product. We can now show that Rule V, § 62, holds true whatever the base. That is, if M and TV are any two numbers, and a the base, then logd JW iV = logo Af +IogaiV. Pkoop. Let a; = logaM and 2/ = logaiV. Then a'' = M and a^ = N (§ 67). Hence a^ • aV = MN, or a''+y = MN. But the last equahty means that logoMiV=a;+2/=log^M+logJV. (§67) *69. Logarithm of a Quotient. Rule VII, § 63, holds true whatever the base. That is, if M and N are any two numbers, then l0ga{M-i-N) =logaM-logaN. Proof. Let x=logaM and j/=logoiV. Then a^ = M and aV=N. (§ 67). Hence, d'-T-aV = M^N, or a'^« = M-^N. But the last equality means that loga(M-i-N)=x-y=\ogaM-logaN. *70. Logarithm of a Power of a Number. Rule IX, § 64, holds true whatever the base. That is, if ikf is any number and n any (positive integral) power, then logaM" = wlogailf. Proof. Let x=\ogaM. Then a^ = M (§ 67) and hence o"^ = Af". But the last equality means that logaM"-=nx=n logail/. 124 COLLEGE ALGEBRA [VII, § 71 *71. Logarithm of a Root of a Number. Rule XI, § 65, holds trae whatever the base. That is, ii M is any number and n any (positive integral) root, then lo^\/M = -logcLM. Proof. Let x=logailf. Then a^ = Af (§67) and hence (0^)'^" = Af '/", or a"^'^ = v^Af . But the last equality means that 1 loga\/M-- - logaAf . *72. Summary. From the results estabhshed in §§ 67-71 it appears that Rules V-XII, §§ 62-65, are not only true when the base is 10 (as was there taken) but they are true for any base. Tables exist for various bases other than 10, but we shall not consider them. Note. The reason why 1 cannot be used as a base is that 1 to any power is equal to 1, that is, we cannot get different numbers by raising 1 to different powers. *73. Historical Note. Logarithms were first introduced and em- ployed for shortening computation by John Napiek (1550-1617), a Scotchman. However, he did not use the base 10, this being first done by the English mathematician Briggs (1556-1631), who computed the first table of common logarithms. *74. Calculating Machines. The Slide-Rule. Machines have been invented and are now coming into very general use, especially by engi- neers, by which the processes of multiphcation, division, involution, and evolution can be immediately performed. The construction of Fig. 38. The Slide Rule these machines depends upon the principles of logarithms, but to describe the machines and their methods of working would take us beyond the scope of this text. The simplest machine of this kind is the slide rule, the use of which is easily understood. A simple shde rule with directions is inexpensive and may ordinarily be secured from booksellers. A full description of the instrument and its use may be found in the Macmillan Tables (The Macmillan Co., New York). CHAPTER VIII COMPOUND INTEREST AND ANNUITIES 75. Compound Interest. The interest which P dollars will bring at the end of one year if placed at the rate of interest i is evidently PXi, ov Pi. If the interest Pi thus received be added to the principal, or P, the new principal at the end of the ^rsi year is P+Pi, or (1) P{l+i). If the principal (1) be again allowed to draw interest for one year at the same rate i, the interest received will be P(l+i)Xi, or P(l+i)i, and if this be added (compounded) to the former principal (1), the amoimt of the principal at the end of the second year becomes P(l+i)+P(l+i)i, which may be written P{l+i){l+i), or (2) P(l+iy- Similarly, the amount at the end of the third year is P(l+^)^ and, in general, we have the following formula for the amount An which will be realized from a principal P by compounding the interest upon it annually for n years at the rate i: (3) A„=pa+i)". Example 1. What will be the amount of $225 loaned for 5 years at 8% compound interest? SoLTTTioN. Here P = 225, i = .08 and n=5. Hence, using the formula, we find A5 = 225(1 +.08)5= 225 Xl.08^ The actual computation of A is now best carried out by logarithms. Thus, taking the logarithm of each member of the last equation, we have, by Rules V and IX, §§ 62, 64, log ^5= log 226+5 log 1.08=2.3522+5X0.0334 = 2.3522 +0.1670 = 2.5192 Therefore, by § 61, ^5 = 1330.50 Am. 125 126 COLLEGE ALGEBRA [VIIl, § 75 Example 2. What principal will amount to $1000 in 10 years at 5% compoxind interest? Solution. Here Aio = 1000, P = ?, t = .05, n = 10 so that the formula gives 1000 = P(l + .05)i° = P(1.05)^''- The problem thus resolves itself into solving this equation for P, and this is most readily done by use of logarithms as follows: log 1000 =log P+10 log 1.05 Hence log P=log 1000-10 log 1.05 = 3-0.2120 = 2.788 Therefore, by § 61, P = $613.70 Ans. EXERCISES 1. Find the amount of $400 for 10 years at 3% compound interest. 2. Find the amount of $100 for 20 years at 6% compound interest. 3. What principal loaned at 4% compound interest will amount to $1500 in 10 years? 4. What sum of money invested at 4% compound interest from a child's birth until he is 21 years old will yield $1000? 6. In what time will $800 amount to $1834.50 if put at compound interest at 5%? [Hint. Note that the unknown time becomes determined by an exponential equation which can be solved as in § 66.] 6. How long will it take a sum of money to double itself at 5% compound interest ? 7. What is the rate per cent when $300 loaned at compound interest for 6 years will yield $402? 8. Solve the formula for n in terms of A, P, and i. 9. Construct a graph to show the compound amount of 1 dollar at 6% as the time varies. 10. If, instead of the interest being compounded annually as in the formula of § 67, it is compounded m times a year, show that the formula becomes 11. In how many years will $300 amount to $400 at 6% compound interest, the interest being compounded quarterly? 12. What sum should be deposited in a bank paying 4% compounded semi-annually in order to discharge a debt of $7430 due ten years later. VIII, § 76] COMPOUND INTEREST 127 76. Annuities. An annuity is a series of equal payments made at equal intervals during a fixed period of time. For convenience, the first payment will here be regarded as made at the end of the first year, the second payment at the end of the second year, etc. Thus, if A has a life insurance policy in the form of an annuity in case of death to B of $1000 a year for 10 years, then at the end of the first year after A's death the company issuing the policy is to pay B $1000, and a like payment is to be made at the end of the second year, third year, etc., up to the end of the tenth year. Evidently, if interest be taken into account, such a policy will be worth more to B than the mere total of $10,000 thus received, since he may during the 10 years be reinvesting the various pajrments so as to receive additional returns. The following fundamental general problem thus arises. If we represent the amount of each payment by a, the num- ber of yearly payments by n and the interest rate by i, what will be the accumulated value F„ of the annuity at the end of the n years? The answer, expressed as a formula for y„ in terms of a, n and i, is readily obtained as follows. Using the formula of § 75, we see that the acciunulated values of the first, second, ■■■nth. payments will be: a(l+i)"-S a{l+i)"-\ ••• , ail+iy, a{l+i), a. The desired value, Vn, is therefore the sum of these n expressions. But they are seen to form a geometric progres- sion whose first term is a and whose common ratio is (l+i). The sum is therefore readily expressed by use of the first formula in § 39, which gives (4) F„ = a^^±f^. By the present value of an annuity of a dollars per annum is meant the amount in cash that one could afford to pay for the privilege of receiving the payments in their regular order. A second fundamental problem thus arises: What is the present value P of an annuity of a, payable in n yearly 128 COLLEGE ALGEBRA [VIII, § 76 installmeiits when the interest rate is if This again may be answered by simple considerations based on the properties of a geometric progression. Thus, the present value of the first payment can be obtained from the formula of § 75 by placing in it A=a, n=l and solving for P, thus giving a(l +i)~^. Similarly, the present value of the second payment is a{\+i)~^, that of the nth payment being (l+i)~". The desired value of P is therefore the sum of these, or a(l+i)-'+a(l+i)-='+ . . . +a(l+i)-". This being a sum of terms forming a geometric progression, its value can be readily expressed as before by the first formula of § 39, which gives as the desired formula (5) /'„ = ai^:^^. EXERCISES 1. What will be the accumulated value of an annuity of $100 for 10 years at 6%. Or s 1 100 , By logarithms, (1.06)'" is found to be 1.7904, hence (1.06)i''-l = 0.7904 Therefore log 7,0= log 100+log 0.7904 -log .06 = 2 + (9.8978-10)- (8.7782-10) =3.1196 Hence Vio=$1317, the accumulated value of the annuity. Ans. 2. What is the present value of an annuity of S300 for 10 years at 6%? 3. How much must a man save annually and deposit in a savings and loan company paying 5%, compounded annually, in order to pay off a mortgage of $2000 after 5 years? 4. A man buys a house and lot, paying $1500 down and agreeing to pay $1000 annually for the next 4 years. What is the equivalent cash price if money is worth 6% per year? [Hint. Note that the $1500 payment is not a part of the annuity.] SoLTjnoN. 7io=-[(H-i)''-l]=— [(1.06)i»-l]. VIII, § 76] COMPOUND INTEREST 129 5. It is estimated that a certain mine will be exhausted in 10 years. If the mine jdelds a net annual income of $10,000, what would be a fair purchase price, money being worth 5%7 6. Show that if, instead of the installments being made annually, they are made m times a year and the interest compounded at each payment, then the two formulas of § 76, remain the same except that ilm is to be substituted for i and mn for n. 1. Using the results of Ex. 6, answer the following question: A piano is sold for $100 cash and $50 to be paid semi-annually for 3 years. What is the equivalent cash price, if money is worth 6%, compovmded semi-annually ? 8. A city is to issue 20-year bonds to the amount of $100,000 for the erection of public schools and it is desired to establish a "sinking fund" to provide for the extinction of the debt when due. How much must be deposited in the sinking fund at the end of each year, money being worth 4% and compounded annually? CHAPTER IX MATHEMATICAL INDUCTION— BINOMIAL THEOREM 77. Mathematical Induction. The three following purely arithmetic relations are easily seen to be true: 1+2 = 1(2+1), 1+2+3=1(3+1), 1+2+3+4 = 1(4+1). We might at once infer from these that if n be any positive integer, there exists the algebraic relation (1) l+2+3+4+-+n = ^(n+l), the dots indicating that the addition of the terms on the left continues up to and including the number n. For example, if n = 8, this would mean that 1+2+3+4+5+6+7+8 = 1(8+1). Again, if n = 10, it would mean that 1+2+3+4+5+6+7+8+9+10 = ^(10+1). That these are indeed true relations is discovered as soon as we simplify them. Let the pupil convince himself on this point. It is to be carefully observed, however, that the inference just made, namely that (1) is true for any n, is not yet justi- fied, for we have only shown that (1) holds good for certain special values of n, and we could never hope to do more than this however long we continued to try out the formula in this way. Something more than a knowledge of special cases must always be known before any perfectly certain general inference can be made. For example, the fact that Saturday W£is cloudy for 38 weeks in suc- cession gives no certain information that it will be so on the 39th week. We shall now show how the general formula (1) may be 130 IX, § 77] MATHEMATICAL INDUCTION 131 established free from all objection; that is, in a way that leaves no possible question as to its truth in all cases. Let r represent any one of the special values of n for which we know (1) to be true. Then (2) 1+2+3+4+.. .+r=^(?-+l). Let us add (r+1) to both sides. The result is 1+2+3+4+.. .+r+(7-+l)=^(r+l) + (r+l). In the second member of the last equation we may write ^(/•+l) + (r-+l) = (r+l)(^^+l^ = (r+l)^^U^(r+2). while the first member has the same meaning as l+2+3+- + (r+l). Thus, (2) being given us, it follows that we may write (3) l+2+3+4+... + (r+l)=^(j-+2). But (3) is seen to be precisely the same as (2) except that r+1 now replaces r throughout. This means that if (1) is true when n = r, as we have supposed, then it holds true necessarily for the next greater value of n, which is r+1. The original fact which we wished to establish (namely, that (1) is true for any n) now follows without difficulty. In fact, we know (see beginning of this section) that (1) is true when n = 4, from which it now follows that it must be true also when n = 5. Being true when n = 5, the same reasoning shows that it must be true also when n = 6. Thus, we may reach any given integer n, however large it may be. Hence (1) is true for any such value of n. This method of reasoning illustrates what is termed mathematical induction. Another example of the process will now be given, in a more condensed form. 132 COLLEGE ALGEBRA [IX, § 77 Example. Prove by mathematical induction that (1) l+3^-54-7^ \-{2n — l)=n^. {n = any positive integer) Solution. When n = l, the formula gives 1 = 1*; when n = 2, it gives 1+3 = 2^; when n = 3, it gives 1+3+5 = 3*, all of which arith- metical relations are seen to be correct. Let r represent any value of n for which the formula has been proved. Then (2) l+3+5+7 + -+(2!— I)=r2. Adding (2r+l) to each member gives (3) l+3+5+7 + - + (2r+l) =j^+(27-+l) =r'+2r+l = {r+lf. But (3) is the same as (2) except that r has been replaced throughout by J-+1. Hence, if (1) is true for any value of n, such as r, it is neces- sarily true also for that value of n increased by 1. Now, we know (1) to be true when n = 3. (See above.) Hence it must be true when n = 4. Being true when n = 4, it must be true when n = 5, etc., and in this way we now know that (1) is true for any value (positive integral) of n whatever. EXERCISES Prove the correctness of each of the following formulas by mathe- 'matical induction, n being understood to be any positive integer. 1. 2+4+6+8 H |-2n = n(n+l). [Hint. First try out for n = l, n = 2, and « = 3. Let r represent a number for which the formula holds. Add 2(r + l) to both members of the resulting equation and compare results.] 2. 3+6+9 + 12H [■3n=^{n+l). 3. l*+22+32+4*+-+»i^ = in(n+l)(2ra+l). 4. 2H4*+6*H h(2w)* =|n(n+l)(2n+l). 6. l^+2?+3^+i^+-+n^ = ln\n + lf. 1 1 _1_ 1 n ■ r^ 2^ 3^"^ n(n+l)~n+l' 7. 2+2^+2^2^ H |-2" = 2(2''-l). 8. Prove that if n is any positive integer, o" — 6" is divisible by o — 6. [Hint. Since ar+^-b'^+'^=a(a'■-b'^)+¥ {a-b), it follows that a'' +^ — ¥ +^ will be divisible by a — b whenever a^ — b' is divisible by a — b.] 9. Prove that o*'*— b*" is divisible by a+b. IX, § 78] MATHEMATICAL INDUCTION 133 78. The Binomial Theorem. If we raise the binomial (a+x) to the second power, that is, find (a+xY, the result is d'+2ax+x^. Similarly, by repeated multipHcation of (a+x) into itself, we can find the expanded forms for (a+xY, (a+x)*, (a+xy, etc. The results which we find in this way have been placed for reference in a table below: {a+xy = a^+2ax+x\ {a+xy = a^+3a'x+3ax^+:(?. (a+xy = a^+5a*x+10a^x'^+10aV+5ax*+x\ {a+xy = a^+Qa^x+15a^iy'+20a'x'+15a'x*+6a3^+x',etc. Upon comparing these, it appears that the expansion of (a+cc)", where n is any positive integer, has the following properties: 1. The exponent of a in the first term is n, and it decreases by 1 in each succeeding term. The last term, or a;", may be regarded as a" x" (See § 8). 2. The first term does not contain x. The exponent of x in the second term is 1 and it increases by 1 in each succeeding term until it becomes n in the last term. 3. The coeffiderd of the first term is 1; that of the second term is n. 4. If the coefficient of any term he multiplied by the exponent of a in that term, and the product be divided by the number of the term, the quotient is the coefficient of the next term. For example, the term 6a^a^, which is the third term in the expansion of (a+x)*, has a coefficient, namely 6, which may be derived by mul- tiplying the coefficient of the preceding term (which is 4) by the exponent of a in that term (which is 3) and dividing the product thus obtained by the number of that term (which is 2). 5. The total number of terms in the expansion is n+1. 134 COLLEGE ALGEBRA [IX, § 78 The results just observed regarding the expansion of (a+z)'', where n is any positive integer, may be summarized and condensed into a single formula as follows: {a+x)" = a"+na"-ix+ "^" ^\ "-^x^ n(n-l)(n-2) ^ 1-2 -3 ^ ^^ * the dots indicating that the terms are to be supplied in the manner indicated up to the last one, or (n+l)st. This formula is called the binomial theorem. By means of it, one may write down at once the expansion of any binomial raised to any positive integral power. That the formula is true in all cases, when n is a positive integer, will be proved in detail in § 80. We assume its truth here for those small values of n for which its correctness is easily tested. Example 1. Expand (o+x)*. Solution. Here to = 6, so the formula gives 1-2-3-4-5 1-2-3-4-5-6 Simplifjdng the various coefficients by performing the possible cancelations in each, we obtain (o+a;)^ = a«+6a5x+15aV+20o^x'+15aV+6oa^+x^ Ans. Note. It may be observed that the coefficients of the first and last terms turn out to be the same; likewise the coefficients of the second and next to the last terms are the same, and so on symmetrically as we read the expansion from its two ends. Example 2. Expand (2— m)^. SoLiTTioN. Here a = 2, x= —m, and n = 5. The formula thus gives (2-m)5 = 2H5 ■ 2^-ot)+^ • 2^(-m)'+ ^ ' ^ ' ^ • 2\-mf ,5-4-3-2„^ sd ,5-4-3-2-l +rTT:¥T-4 • 2(-)^ + 1.2.3.4.5 ^-^- IX, § 79] MATHEMATICAL INDUCTION 135 Simplifying the coefficients (as in Example 1), this becomes (2-to)S=2H5 • 2^(-m)+10 • ^(-m)''+lQ ■ 2\-mf +5 • 2(-my+(-m)K Making further simplifications, we obtain (2-TO)5=32-80m+80?n2_40TOHl0m*-m^ Ans. Note. The result for {2-xf is the same as that for {2+xf except that the signs of the terms are alternately positive and negative instead of all positive. A similar remark apphes to the expansion of every binomial of the form (o— a;)" as compared to that of (a+x)". 17. (1+1)' V y/ EXERCISES Expand each of the following powers. 1. ix+yf. 9. (a^-x^)\ 2. {a+b)\ 10. (2a+l)^. 3. {x-yf. 11. {x-3yf. 5 4. {a -by. 12. (1+:d2)6. 18. ("--j • 6. (2+r)5. 13. (l-x)«. _ 6. {a+x?. 14. (x-if. "■ «^o^+„=n(n-l)-2-l = n! Thus, the five letters a, b, c, d, e may be permuted among themselves in 6! = 120 ways. 186 COLLEGE ALGEBRA [XII, § 103 EXERCISES 1. In how many ways can the letters a, 6, c, d, e be arranged if taken 3 at a time? 2. How many numbers can be made out of the digits 1, 2, 3, 4, 5, 6 using four of them at a time, no digit being repeated? 3. In how many ways can 10 trees be planted in a row? 4. In how many ways can the letters A, B, C, a, b, c be arranged so that the three capital letters shall stand first, and the three small letters shall stand last? [Hint. First find how many ways the capital letters can be ar- ranged among themselves, then similarly as regards the small letters, then use the Principle of § 101.] 5. Work Ex. 4 in case either the three capital letters or the three small letters may stand first. 6. In how many ways can 5 French boolcs, 3 Latin books and 2 Spanish books be arranged on a shelf so that the French books shall stand together, the Latin books together, and the Spanish books to- gether? 7. Work Ex. 6 when it is required that the French books shall stand first as a group, but the remaining 5 books may be arranged in any manner thereafter. 8. In how many ways can a program of 3 speeches and 3 musical numbers be arranged so that speeches and music shall alternate through- out? 9. In how many ways can the knives, forks and spoons be distrib- uted at a table where there is to be a dinner party of 6 people, each of whom is to have a dinner knife, a bread- and butter knife, a dinner fork, a salad fork, a soup spoon, a teaspoon and a coffee spoon? 10. In how many ways can the colors red, green, blue, indigo, violet be arranged so that red and green do not stand together? [Hint. The answer may be regarded as the difference between the number of arrangements when no restrictions whatever are made and the number when red and green stand together in either order.] 11. In how many ways can 4 different coins be stacked one upon the other provided that at least one must be left with its face side up? 12. Show that formula (1) of § 102 may be written in the form r -^. n' r = • r XII, § 104] PERMUTATIONS AND COMBINATIONS 187 104. Combinations. A set of things regarded without reference to the order in which they are arranged, is called a combination of them. Thus, abc, acb, bac, bca, cab and cba are the same combination because each is made out of the same letters a, 6, c and in this respect there is no difference between them. It is only when the arrangement of the letters is taken into account that any such distinctions are possible. Let US ask how many combinations, in the sense defined above, are possible out of the four letters a, b, c, d when taken 3 at a time. The answer is 4; namely, abc, abd, acd, bed. Note that each of these is different from the three others in that it is made up of different letters. Similarly, if we ask how many combinations of the letters a, b, c, d are possible when taken 2 at a time, the answer is 6; namely ab, ac, ad, be, bd, cd. Finally, if taken 4 at a time, the answer is 1; namely abed. If we ask in a more general sense how many combinations are possible out of n different things taken r at a time, we may arrive at 'a formula for it as follows. Consider any one combination. It contains r letters, which, if arranged in all possible ways would give rise to r! permutations. (See formula (2), § 103.) Since this is true of every different com- bination, it follows that if we let „Cr represent the total number of such combinations, we shaU have the equation C •r'= P where „Pr is the total number of permutations of the n things taken r at a time. From this equation we have p n* r ■- >■ 7~' r\ which, when we substitute for „Pr its value as given by (1), § 102, becomes (1) ^g^^ n(n-l)---(n-r-H) , 188 COLLEGE ALGEBRA [XII, § 104 This, therefore, is the formula desired. By multiplying both its numerator and denominator by (n — r)!, observing that the numerator then becomes n(n — 1) ••• (n—r+1) • (n—r)\ = n{n — l) ••• {n—r-}-l)(n — r)(n—r — l) ■•■ l=n\, the formula takes the more condensed form ^2) "^' = rl(n-r)\ Note. It may be noted that formula (1) for ^Cr is the same as is obtained if, in the formula as given in § 79 for the coefficient of the rth term of the binomial expansion for (a+x)", one uses (r+1) in place of r. The binomial theorem for positive integral exponents may therefore be written in the form Example 1. How many committees of 3 men each can be formed from 8 men? SoLtTTioN. Since the personnel of a committee is in nowise changed by a different arrangement of the men in it, the question resolves itself into finding the number of combinations of 8 men when taken 3 at a time. Hence, using the first of the formulas above, we obtain the answer 8-7 -6^ Example 2. How many selections each consisting of 3 oranges and 2 apples may be made from a basket containing 6 oranges and 4 apples? Solution. The number of ways in which the 3 oranges may be selected is <• 5 • 4 The number of ways in which the 2 apples may be selected is 2 Hence, by the fundamental principle of § 101, the 3 oranges and 2 apples may together be selected in 20X6 = 120 waj's. Ans. 6g3 < ^ ^ =20. XII, § 104] PERMUTATIONS AND COMBINATIONS 189 EXERCISES 1. A captain having under his command 20 men wishes to form a guard of 3 men. In how many ways may the guard be formed? 2. How many peals may be rung with 7 different bells by striking them 4 at a time? 3. How many hands of cards, each made up of 6 hearts, are there in a pack of cards? [Hint. The pack contains 52 cards, there being 13 -each of hearts, diamonds, spades and clubs.] 4. A chandeher contains 10 lights. In how many ways may the room be hghted if only 8 lights are used? 5. How many straight lines may be drawn through 8 points no three of which lie in the same straight line? 6. Out of 8 different English books and 7 different French books, how many selections of 6 books may be made each containing 3 Enghsh and 3 French books? 7. Work Ex. 6- in case each selection of 6 books must contain at least 2 English and 2 French books? [Hint. Consider separately the various possibihties, as in Ex. 7, page 183, and add results.] 8. A candidate for a certain office is to be elected in case he receives a majority of the votes cast by 10 people. In how many ways could the majority be secured? 9. Out of 15 men how many selections of 4 men each can be made each of which will contain a certain particular man? [Hint. Take out the particular man and then consider'the remaining 14 men.] 10. A whist-hand contains 13 cards. How many such hands each made up of 4 spades, 4 hearts, 4 diamonds and 1 club is it possible to form? 11. Out of a basket containing 6 oranges, 8 apples and 3 peaches, how many selections of 5 each may be made that shall contain at least one orange? [Hint. The answer may be regarded as the difference between the number of selections of 5 indiscriminately and the number when no oranges are taken.] 12. Show that the dumber of combinations of n things taken r at a time is the same a^ when taken n—r at a time. 190 COLLEGE ALGEBRA [XII, § 105 *105. Distribution into Groups. If it be asked in how many ways 10 different things may be distributed among 3 'persons A, B, C so that A shall receive 5, B shall receive 3, and C shall receive 2, the answer may be determined as follows. Starting with A, he may receive his 5 things in 10! 10C5 =rrr; ways. (See formula (2), § 104) o!o! B may now be given his 3 things out of the remaining 5 things in Finally, C may be given his 2 things out of the remaining 2 things in 2C2 = — =^ ways. (See Note, § 103) Applying the Theorem of § 101, the 10 things may therefore be distributed in the manner specified in 10! 5! 2! ^^i!y!^2l^^y'- Noting cancellations, we may reduce this product to the form 4 10! '/■^- //• /• ,6^ 7 • ,8< 9 • 10 H!^!>XV4'.X-X-X-X =2520 ways. An.. The same method of reasoning when applied more generally leads to the following result. The number of ways in which n different things may be distributed into a specified number of groups such that the first group shall contain p things, the second shall contain q things, the third shall contain r things, etc. is given by the formula (1) N= , ."•. Example. In how many ways may 14 apples be distributed among four children so that the oldest shall receive 5, the next younger 4, the pext younger 3 and the youngest 2. Solution. By means of the above general formula, the answer is 14! 5! 4! 3! 2! ^2.522,520 ways. Ans. XII, § 106] PERMUTATIONS AND COMBINATIONS 191 Note. It is to be observed that if the number of things to be put in each group is the same that is, p = q=r = "-, and if there is.no dis- tinction made between the groups (such as first, second, etc.), then the formula above must be sUghtly changed, becoming (2) N= ^ g[ pi qlrl ■ • • where g is the number of the groups. The reason for this may be immediately imphed from the following example. Example. In how many ways may 12 men be divided into three groups of 4 each? SoLTTTioN. Formula (1) would give 12! 4! 4! 4!' But to take this as the answer implies that any way of dividing the men into the three equal groups gives rise to another way by redis- tributing the same three groups among themselves, which can be done in 3! ways. Since the question is merely as to the number of possible groups without reference to their order, the result above must therefore be divided by 3!, giving as the correct answer 12! g^^^^j =5775 ways and thus agreeing with the result given by (2) for this example. *106. Permutations of Things not all Different. In the previous discussions of this chapter all the things dealt with have been regarded as different, or distinguishable, from each other. In distinction from this, consider the following example. Example. How many permutations are possible of the letters of the word infinite when taken all together? Solution. Since no new permutation arises by interchanging the three i's among themselves, or the two n's among themselves, let us suppose at first that the i's are made dissimilar by calling them respec7 tively ii, i^, ii, and likewise the n's by calling them respectively ni, n^ . Under such a supposition, the answer, by formula (2) of § 103, would be 8!, since there would then be a total of 8 dissimilar letters. If the three i's be now regarded as the same, each of these 8! peimutations gives rise (by permuting the 3 i's among themselves) to 3! permutations that are identically the same. Hence, if the i's alone be regarded as 192 COLLEGE ALGEBRA [XII, § 106 the same, the answer would be 8!/3!. But if the two n's be now re- garded as the same, each of these 8!/3! permutations gives rise by similar reasoning to 2! permutations that are the same. Hence, the correct answer is 8! The same method of reasoning when applied more generally leads to the following result: The number of permutations among themselves of n things of which ni are alike of one kind, n^ are alike of another kind, ng are alike of another kind, etc., is given by the formula P = ^ Til! 02! Hs! • • • MISCELLANEOUS EXERCISES Success in working an example in permutations and combinations depends chiefly upon one's abiUty to determine to what extent the order of the things considered must be taken into account. Examples in the following list accompanied by the star (*) depend upon §§ 105-106. 1. On a railroad there are 20 stations. How many tickets are required to connect every station with every other one? ■ ?. The Greek alphabet contains 24 letters. How many Greek letter fraternity names -can be formed, each containing 3 letters, a repetition of letters being allowed? 3. In how many ways can 6 ladies and 6 gentlemen form couples for a dance? *4. Eight persons are to play cards. In how many ways can partners be formed? 5. Show that the number of ways in which n persons may be distri- buted among themselves at a round table is (w — 1)! 6. In how many ways can a selection of at least 4 oranges be made from a basket of 8 oranges? 7. A box contains 6 red cards, 5 white cards and 4 blue cards. In how many ways can a selection of three cards be made such that (a) alli 'three are red? (b) none are red? (c) at least one is red? XII, § 106] PERMUTATIONS AND COMBINATIONS 193 *8. How many arrangements of the letters of the word Mississippi are possible? *9. How many signals can be made with 7 flags, of which 2 are red, 1 white, 3 blue and 1 yellow, displayed altogether one above the other? 10. How many dominoes are there in a set numbered from double blank to double ten? *11. A collection of 12 books is to be distributed equally among 4 people. In how many ways can it be done, no regard being had for the order in which they are given out? *12. A collection of 12 books is to be divided into 4 equal piles. In how many ways can it be done, no regard being had for the order in which they appear in each pUe? 13. Answer Ex. 12 in case regard is taken of the order of the books in each pile. 14. How many committees, each containing 4 men, can be formed from 5 Republicans and 5 Democrats, it being understood that at least one RepubUcan and one Democrat must be on the committee. 15. From a basket of 8 apples, in how many ways can a selection be made, it being understood that any or all of the apples can be taken? CHAPTER XIII PROBABILITY 107. Introduction. If a letter be chosen at random from the alphabet the chance, or probability, that it will be a is naturally regarded as 1/26 since, out of the 26 ways in which a letter may be drawn, only 1 gives a. Similarly, the proba- bility, or chance, of drawing any single letter, as m, would be 1/26. However, if we ask the probability of drawing a vowel, the answer would be 5/26, since a vowel may be drawn in any one of 5 ways; namely, either a, e, i, o or u. As another example, suppose that a bag contains 4 red balls and 5 white balls, and that a ball is drawn at random. The probability that it will be red is then 4/9, since out of the total of 9 ways of drawing a ball, 4 give red ones. Similarly, the probability of drawing a white ball is 5/9. These and other illustrations which may be readily suppKed lead to the following definition. Definition. The probability of an event is the ratio of the number of ways in which it can happen {all regarded as equally likely) to the total number of ways in which it can either happen or fail. Thus the probability of drawing an ace from a pack of cards is 4/52, or 1/13, since there are 4 ways in which the event can happen out of a total of 52 ways in which it can either happen or fail, the latter being the total number of cards in the pack. This definition, when stated in algebraic language, is as follows. Let a be the number of ways in which an event can happen, and let b be the number of ways in which it can fail (all ways being regarded as equally likely). Then the proba- bility, p, that the event will happen is (1) . = ^- a+b 194 XIII, § 108] PROBABILITY 195 CoROLLABT 1. If an event is certain to happen, its proba- bility is 1. For in (1) we then have 6 = 0, giving p — a/a = l. CoROLLABY 2. The probability that an event will happen and the probability that it mil fail, when added together, give 1. For, just as the fraction (1) is the probability that the event will happen, so the fraction (2) q = - "^ a+b is the probability that the event will fail, and it is evident that the sum of the expressions (1) and (2) is 1. 108. Value of an Expectation. If a person is to receive $100 in case a certain event happens, and the probability of the event is 3/5, then the value of his expectation is naturally 3/5X100 = $60. This amount, in other words, is what he should pay for the privilege of beiug the possible recipient of the $100. In general, we thus adopt the f oUowiag definition. Definition. If a person is to receive the sum of money M in case an event occurs whose probability is p, then the value of his expectation is pM. EXERCISES 1. A bag contains 6 red balls, 4 white balls and 3 blue balls. If a ball be drawn at random, what is the probabihty that it wiU be (a) red, (6) white, (c) blue? 2. From a suit of 13 hearts, 3 cards are drawn. What is the chance that they will be the ace, king and queen? [Hint. Three cards may be drawn in 13C3 ways.] 3. The four capital letters A, B, C, D and the four small letters o, 6, c, d are shaken together in a hat after which three letters are drawn out at random. What is the probabUity that they wiU all be capitals? Solution. Since there are 8 letters in all, the total number of ways of drawing 3 letters of any kind is ^C3= ^11-6 = 56. * ^ 12-3 196 COLLEGE ALGEBRA [XIII, § 108 Similarly, the number of ways of drawing 3 capital letters is 4- 3 • 2 ^ ^^'"l-2.3" 4 1 Hence the desired probabUity is — > or — ■ Ans. 56 14 4. Find the probability in Ex. 3 that the three letters drawn shall consist of two capitals and 1 small letter. [Hint. The number of ways of drawing 2 capitals and 1 small letter is4C3X4Ci. (§§104,101).] 5. A portfoUo contains 15 bills, 6 of which are $5 bills, 4 are $2 bills and 5 are $1 bills. If 4 bills be taken at random find the chance that (a) all are $5 bills, (6) 3 are $2 bills and 1 is a S5 bill, (c) all are $1 bills. 6. A history of Rome in four volumes is placed on a library shelf at random. What is the probabiUty that the volumes will be in their correct order: I, II, III, IV? 7. If 4 cards be drawn from an ordinary pack, what is the proba- bility that (a) they will all be hearts? (6) that there will be 1 card of each suit? 8. If two tickets be drawn from a package of 20 tickets marked 1, 2, 3, ••■, 20, what is the probabihty that both will be marked with odd numbers? 9. If three coins be tossed, what is the probability that (a) all will be heads? (b) there wUl be exactly two heads? (c) there will be at least two heads? 10. If three cards be drawn from a pack, what is the probability that they will be an ace, king and queen of different suits? 11. A person is to receive $5 in case he tosses two coins and they both come up heads. What is the value of his expectation? 12. What can a person properly afford to pay for the privilege of receiving $7.50 in case that he draws 2 tickets from a box containing tickets marked from 1 to 15 inclusive and finds that the one is odd and the other even? XIII, § no] PROBABILITY 197 109. Definitions. The preceding discussions and illus- trations of the theory of probability are the immediate conse- quences of the definition of the term "probability," as given in § 107. If one is to proceed farther into the subject, it is desirable to make certain fundamental distinctions between the possible kinds of events, as indicated below. Two or more events are called dependent or independent according as the happening of one of them does or does not affect the happening of the others. Thus, if a drawing be made at random of one letter from a box containing the letters a, b, c, d, e and this be followed by another similar drawing, the two events would be independent in case the letter first drawn was replaced in the box before the second drawing, while the events would be dependent in case this was not done. 110. Theorem Concerning Independent Events. In deter- mining the probability that two or more independent events will all happen, one may employ the following theorem. Theorem. The probability that two or more independent events will all happen is equal to the product of their respective probabilities ^ > ■ Thus, suppose that two coins are tossed. The probability that the one will come up heads is evidently 1/2, and the probability that the other will come up heads is likewise 1/2. Therefore, the probability that both will come up heads is, by the above Theorem, 1/2X1/2 = 1/4. This result may be verified by noting that the total number of ways in which the two coins may fall is 2X2=4, and of these only 1 gives two heads. Hence, the answer, as before, is 1/4. Similarly, the probability that three coins will all come up heads is 1/2X1/2X1/2 = 1/8. Proof of Theorem. Suppose that the probabilities of the separate events are respectively pr, p^, pz, •■■, Pr, and let Ui be the number of ways in which the event corresponding to pi can happen and 6i the number of ways in which this event can fail; similarly let Oa be the number of ways in which the event corresponding to p^ can happen, and 62 the 198 COLLEGE ALGEBRA [XIII, § 110 number of ways in which this event can fail, etc. Then, by the definition stated in § 107, we shall have (1) pi = — — -) P2= — -r' •••, Vr = ai+6l CI2+&2 OSr+&r Moreover, by the principle of § 101, all the separate events can happen together in ai ■ 02 ■ as ••• ctr ways out of (ai+6i)(a2+&2) ••• {cLr+h) possible ways of either happening or failing. Hence, if P be the probabihty that all the events will happen, we have by the definition in § 107, (2\ P = "^^•••^'- ^ ' (ai+&i) (02+62) - {a,+hr) But, upon using (1), the expression (2) may be written in the form which was to be proved. 111. Dependent Events. Although the theorem of § 110 pertains only to independent events, it may frequently be applied to determine probability in the case of dependent events, since the latter may usually be separated so as to be regarded as independent. Example. One letter is drawn from a box containing the letters a, h, c, d, e and a second drawing is then made (the first letter obtained not being replaced before the second drawing). What is the probability that the letters thus drawn are first a and second b? Solution. The probability of obtaining a on the first drawing is evidently 1/5 and, a having been drawn, the probability of obtaining 6 on the second drawing is 1/4, since but 4 letters remain after the first drawing. Therefore, by the theorem of § 110, the desired proba- bility is 1/5X1/4 = 1/20. Am. This result may be verified as follows. The total number of ways of drawing 2 letters is 5X4 = 20 and of these there is but one that gives first a and then 6. Hence, the probability is 1/20, which agrees with the former result. XIII, § 112] PROBABILITY 199 112. Theorem Concerning Events That Can Happen in Several Ways. In determining the probability that an event wiU happen in case it can happen in any one of two or more different ways which are mutually exclusive, one may employ the following theorem. Theorem. If an event can happen in any one of two or more different ways which are mutually exclusive, the proba- bility that it will happen is the sum of the probabilities of its happening in these different ways. Thus, if it be asked what is the probabihty of getting either two heads or two tails when two coins are tossed, we may reason as follows. The probability of getting two heads, as shown in § 110, is 1/4, and similarly the probabihty of getting two tails is 1/4. Therefore, by the theorem above, the probability of getting either two heads or two tails is 1/4+1/4 = 1/2. This result may be verified by noting that the total number of ways in which the two coins may fall is 2X2 = 4, and of these 1 gives both heads and 1 gives both tails. Hence, the probabihty of getting either both heads or both tails is ■ =2/4 = 1/2, thus agreeing with the 4 former result. Pboof of Theorem. Suppose that the event can happen in two mutually exclusive ways, and let pi = ai/6i and p2 = 02/62 be their respective probabilities. Then, out of a total of 61 ■ fca possible cases leading to success or failure in either of the two ways, there are ai 62 in which the event can happen in the first way and 0261 in which it can happen in the second way. Hence, out of the fei&2 cases there are 0162+0261 cases in which the event can happen in the one or the other of the two ways, the probabihty of which is therefore 0162+0261 Ox , 02 — VT = 7-+7- = Pl+P2. O1O2 Oi O2 The theorem thus becomes proved in case there are but two ways in which the event can happen. Similar reasoning leads directly to the more general case. 200 COLLEGE ALGEBRA [XIII, § 113 113. Theorem Concerning Repeated Trials. If the proba- bility of the happening of an event in a single trial is known, the probability that it will happen exactly r times in n trials may be determined by use of the following theorem. Theobem. If p is the probability that an event vnll happen in any single trial, then the probability that it will happen exactly r times in n trials is nCrP'(t'^, where q is the probability that the event will fail in any one trial. Thus, if it be asked what is the probability of throwing exactly 3 aces in 5 throws with a single die, the answer is /ly /5y_,f, J_ 25 125 ' ' ' \6/ V6/ ' 216 ' 36~3888" Proof of Theorem. The probability that the event will happen in r specified trials and fail in the remaining (n—r) trials is, by § 110, p'"g""^. But the r trials can be selected out of the n trials in ^C^ ways. Hence, applying the theorem of § 112, it follows that the probability in question is the result of adding p''g"'^ to itself „Cr times; that is, it is equal to n'^r P 2 It is to be observed that if we expand (p+q)" by the Binomial Theorem (§ 104, Note), we obtain Thus, the terms of this expansion represent respectively the probabilities of the happening of the event exactly n times, (n — 1) times, (n — 2) times, ■■■ in n trials. Moreover, by combining this result with that of § 112, we obtain the following coroUary. Corollary The probability that an event will happen at least r times in n trials is p'' + nClP''-'q+nC,P^-V+ - +nCn-rPY'', where p and q have the meanings indicated above. In fact, by § 112, this expression comes to represent the probability that XIII, § 113] PROBABILITY 201 the event will happen either exactly n times, or exactly (n— 1) times, or exactly (n— 2) times, ••■or exactly r times; that is, that it will Happen at least r times. Thus the probability of obtaining at least 3 aces in 5 throws with a single die is .6 n\i /.;\ /1\3 /Rs2 1+5-5+10-25 276 23 6' ~ 6' ~648' ©•--©•©—©■©' EXERCISES 1. Find the probability of throwing an ace in the first only of two successive throws with a die. 2. If three cards be drawn from a pack, find the probability that they will be an ace, a king and a queen in the order earned. 3. Work Ex. 2 in case no regard is had for the order in which the three desired cards are obtained. [Hint. Consider each possible order and apply the theorem of § 112 to the separate results.] 4. Find, by use of the theorem of § 112, the probability of throwing doublets in a single throw with a pair of dice. 5. In three throws with a pair of dice, find the probabihty of throw- ing doublets at least once. 6. A bag contains 5 white and 3 black balls, and 4 are successively drawn out and not replaced. What is the probabihty that they are alternately of different colors? 7. A, B, C in the order named each draw a card from an ordinary pack, replacing the drawing each time. If the first one to obtain a spade is to win a prize, show that their expectations are in the ratio 16:12:9. [Hint. First find the probability that A obtains a spade; second, the probability that A fails to obtain a spade, but B obtains one; etc. It is understood that a total of only three drawings can be made.] 8. A and B throw with one die for a stake which is to be won by the player who first throws an ace. A has the first throw and the throwing is to continue alternately until either the one or the other wins. Show that their respective probabilities of winning are and hence that their respective expectations are in the ratio of 6:5. 202 COLLEGE ALGEBRA [XIII, § 114 114. Probability of Human Life. Mortality Table. If a person 19 years of age asks what the probability is that he will live to the age of 75, the question may be answered with good accuracy by consulting a so-caUed Experience Table of Mortality. Such a table is shown on the opposite page and is readily understood upon examination. It shows in par- ticular that out of 93,362 persons living at the age of 19 it may be expected that at the age of 75 there will remain 26,237. Hence, the answer to the preceding question is 26,237/93,362, or about 0.28 Otherwise stated, the chances that a person of 19 will live to be 75 are about 28 out of 100. The table on page 203 was compiled from the averaged observations of thirty American insurance companies to the end of the year 1874. Such a table is evidently of vital importance in answering the questions which ordinarily come before a hfe insurance company, or any person entrusted to work out a proper pension system for a group of employees, or the judge who wishes to determine what is a proper life interest of a client in an estate. Such questions depend upon the probable extent of life of an individual at a given age. EXERCISES 1. What is the probability that the average American boy of 10 years will live to vote at a public election. What is the probabOity that he will live to the age of 80? 2. A man is 47 and his son is 15. Show that the probabiUty that both wiU hve 10 years is about 0.55 [Hint. Apply the theorem of § 110.] 3. A bridegroom of 24 marries a bride of 21. Show that the proba- bihty that they will hve to celebrate their golden wedding is about 0.12 4. A and B are twins just 18 years old. Show that the probabihty that both will attain the age of 50 is about 0.55; also, that the proba- bihty that one, but not both, will die before the age of 50 is about 0.68. [Hint. Employ the theorem of § 112.] 6. Draw on coordinate paper the graph of the curve showing the probabihty of dying for each year from the ages of 10 to 90. XIII, § 114] PROBABILITY 203 American Expeeibncb Table of Mortality Age NtTMBER Living Ndmber Dying Age Number Living Number Dying Age Number Living Number Dying X ix dx X Ix dx X Ix dx 10 11 12 13 14 100 000 99 251 98 505 97 762 97 022 749 746 743 740 737 40 41 42 43 44 78 106 77 341 76 567 75 782 74 985 765 774 785 797 812 70 71 72 73 74 38 569 36 178 33 730 31 243 28 738 2391 2448 2487 2505 2501 16 16 17 18 19 96 285 95 550 94 818 94 089 93 362 735 732 729 727 725 45 46 47 48 49 74 173 73 345 72 497 71 627 70 731 828 848 870 896 927 76 76 77 78 79 26 237 23 761 21 330 18 961 16 670 2476 2431 2369 2291 2196 20 21 22 23 24 92 637 91 914 91 192 90 471 89 751 723 722 721 720 719 60 61 62 63 64 69 804 68 842 67 841 66 797 65 706 962 1001 1044 1091 1143 80 81 82 83 84 14 474 12 383 10 419 8 603 6 955 2091 1964 1816 1648 1470 25 26 27 28 29 89 032 88 314 87 596 86 878 86 160 718 718 718 718 719 66 56 57 58 59 64 563 63 364 62 104 60 779 59 385 1199 1260 1325 1394 1468 85 86 87 88 89 5 485 4 193 3 079 2 146 1 402 1292 1114 933 744 555 30 31 32 33 34 85 441 84 721 84 000 83 277 82 551 720 721 723 726 729 60 61 62 63 64 57 917 56 371 64 743 53 030 51 230 1546 1628 1713 1800 1889 90 91 92 93 94 847 462 216 79 21 385 246 137 58 18 35 36 37 38 39 81 822 81 090 80 353 79 611 78 862 732 737 742 749 756 65 66 67 68 69 49 341 47 361 45 291 43 133 40 890 1980 2070 2158 2243 2321 96 3 3 CHAPTER XIV DETERMINANTS 115. Definitions. The symbol a h c d is called a determinant of the second order and is defined as follows: a b Thus 8 3 2 4 7 3 -2 4 , =ad—bc. c a = 8 • 4-2 • 3=32-6 = 26, = 7 • 4-(-2) • 3 = 28+6 = 34, The numbers a, b, c, and d are called the elements of the determinant. The elements a and d (which lie along the diagonal through the upper left-hand comer of the determinant) form the principal diagonal. The letters b and c (which lie along the diagonal through the upper right-hand corner) form the minor diagonal. From these definitions, we have the following rule. To evaluate any determinant of the second order, subtract the product of the elements in the minor diagonal from the product of the elements in the principal diagonal. EXERCISES Evaluate each of the following determinants. 8 2 3 1 3. 3 -1 3 -4 -5 • 6. 1- 3a 66 1 • 5 - 7 1 3 • 4. 2 4 a 36 56. 6. 1 a2+b2 o2-62 4 4 204 XIV, § 116] DETERMINANTS 205 116. Solution of Two Linear Equations. Let us consider a system of two linear equations between two unknown letters, x and y. Any such system is of the form (1) aix+biy = ci, (2) 02a;+622/ = C2, where ai, 6i, d, etc., represent known numbers (coefficients). This system may be solved for x and y by elimination, as in § 5. Thus, multiplying (1) by 62 and (2) by 61, subtract- ing the resulting equations from each other, and solving for X, we find b^ci — biC2 (3) fli&2 — (hbi Likewise, we may eliminate x by multiplying (1) by Oa and (2) by aj. Subtracting the resulting equations from each other and solving for y, we find 0.1C2 ~iC2 — agbiCi —aibsd—a^biCa Oicfecs + (hdiCi + asdiC2 — asd^Ci — aidsc^ — a^diCs = } a^2Ci + ai)iCi + a^bid — a^biCi — aibiC2 — (hbiCa fl] Ws -\-(hbsdi-\-a3bidi—asb2di — aib3d2—(hbids dibiCs + oaf'sCi + asbid — agbiCi —aibid— a^iCz It is clear by § 117 that in these values for x, y, and z, each numerator and denominator is the expanded form of a deter- minant of the third order. In fact, it appears from the defi- nition in § 117, that we may now express these values of X, y, and z in the following condensed (determinant) forms: (3) X- rfl 61 Ci rfa &2 C2 ^3 ?>3 Cs fill b\ Ci 02 &2 C2 03 63 Ci Ol rfl Ci 02 d2 C2 03 di Ci y- Oi h Ci 02 h C2 03 h C3 ! Z = - Oi bi rfi 02 62 C^2 03 63 ds Oi 61 Ci 02 62 C2 tlz 63 C3 Note. The importance of these expressions for x, y, and z hes in the fact that they give at once the solution of any system such as (1) 210 COLLEGE ALGEBRA [XIV, § 118 in very compact and easily remembered forms. Here we note that: 1. The denominator determinant is the same in all three cases. (Compare statement 1 of § 116.) 2. The determinant for the numerator of the x-value is the same as that for the denominator determinant except that the numbers di, (k, ds replace the ai, 02, 03 which occur in the first column of the denominator determinant. 3. Similarly, the numerator of the y-yahie is formed from that of the denominator determinant by replacing the second column by the elements di, d^, dg; while the numerator of the g-value is formed from that of the denominator determinant by replacing the third column by the elements di, d^, dg. (Compare statements 2 and 3 of § 116.) The readiness with which (3) may be used in practice to solve a system of three linear equations is illustrated below. Example. Solve the system 2x-y+3z = 35, x+3y-15=-2z, 3x+4y = l. SoLtTTioN. Arranging the equations as in (1) of § 118, the given system is 2x-y+3z = 35, x+3y+2z = 15, 3x+iy+0z = l. Therefore, using (3) of § 118, we have at once 35 -1 3 15 3 2 14 2 -1 1 3 3 4 y=- 35 15 1 0+180-2-9-280-0 -111 0+12-6-27-16-0 -37 =3, (§ 117) 0+3+210-135-4-0 74 -37 -37 -37 = -2, 2 -1 35 1 3 15 3 4 1 -37 6+140-45-315-120 + 1 -333 -37 ~ -37 ' XIV, § 119] DETERMINANTS 211 The desired solution is, therefore, (x=3, y=—2, = 9), Ans. Check. With x = 3, 2/= —2, 2 = 9, it is readily seen that the three given equations are satisfied. EXERCISES Solve each of the following systems by determinants. r a; +22/ +32 = 14, 1. ] 2x+ 2/+22 = 10, (.3x+4i/-32 = 2. (2x- y+2z = 12, 2. ] x+3y+ 2 = 41, (2a;+ 2/+42 = 22. a;- 2/+2 = 30, 32/-X- 2 = 12, 7z-2/+2x = 141. r x+3y+4z = 83, 4. j x+ ^+ 2 = 29, (.6x+82/+32 = 156. 6. 7. 3X-22/+ 2 = 2, 2.r+5y+2z = 27, x+3y+3z = 25. x+y = 9, 6. ■! y+z = 7, . 2+X = 5. x+2/— 2 = 0, x-2/ =2b, x+z =3o+6. f ax+%+c2 = 3, 8. I abx+aby =o+6, {^bcy+bcz =b-\-c. 119. Determinants of Higher Order. The determinants thus far studied have been of either the second or third orders, the former containing 2^, or 4 elements, and the latter 3^, or 9 elements. In general, a determinant of the nth order is a square array of n' elements such as is typified by the expression fli 6i Ci di ■■• li 02 62 C2 d2 ■■■h D= as 63 Cs da ■■■ I3 '71 ^n ^n 0„"'t^ The method for obtaining the expanded form of any such determinant (compare (2), § 117) will be explained in detail in § 121. 212 COLLEGE ALGEBRA [XIV, § 120 120. Inversions of Order Consider the positive integers 1, 2, 3, 4. As here appearing, these are in their natural order, each number being less than all those which follow it. If the same numbers be arranged as follows: 4, 2, 3, 1, there are five departures from the natural order; namely, 4 before 2, 4 before 3, 4 before 1, 2 before 1 and 3 before 1. Each of these is called an inversion of order, Briefly stated, we say that 4, 2, 3, 1 contains five inversions. Similarly, any given arrangement of two or more positive integers contains a certain number of inversions, this nimiber being only in case the numbers occur in their natural order. Thus, in 3, 4, 1, 2, there are 4 inversions; namely, 3 before 1, 3 before 2, 4 before 1 and 4 before 2. Similarly, in 1, 3, 4, 5, 2 there are 3 inver- sions; in 1, 3, 2 there is 1 inversion, etc. 121. The Expanded Form of Any Determinant. An examination of the expanded form of the typical determinant of the third order, as given in (2) of § 117, shows that it may be written in the form ai?>2C3 + agbid + Oa&sCi — aibiCi — aibad — OabiCa. It is now to be observed that each of the six terms here appearing contains three factors, of which the first is an a, the second a b and the third a c, and the subscripts of these letters in any one term are all different, as for example in the third term O263C1. Moreover, in the case of the three terms which are preceded by the sign +, the number of inversions in the subscripts is even, while in the case of the three terms preceded by the sign — , the number of inversions in the subscripts is odd. Thus, in the term +036102, there are two inversions in the sub- scripts, this number being even, while in the term —asb^i there are three such inversions, this being odd. Similarly, the term +026301, is seen to be accompanied with an even number of inversions, while — 0261C3 has an odd number of them. XIV, § 121] DETERMINANTS 213 Taking now the type determinant of the fourth order, namely (1) fli bi Ci di (h ^2 C2 di as bs Cs di at 64 Ci di the observations made above suggest that its expanded form consists of all the terms that can be made, each consisting of four factors of which the first is an a, the second a b, the third a c and the fourth a d and in which no two subscripts are ahke, and with the further understanding that the sign to be prefixed to any one term as thus formed is to be + or — according as the number of inversions in its subscripts is even or odd. This, in fact, is what the meaning of (1) is taken to be, and we shall so understand hereafter. For example, +016203^4, —026304^1, +026361^4 are three particular terms in the expansion of (1). It may be observed that the total number of terms as thus described belonging to the expanded form of (1) is 24, or 4 !, since the a to be used in forming a term may fi^rst be chosen in any one of 4 ways, then the b may be chosen in any one of 3 ways (its subscript being necessarily different from that of the a chosen), then the c may be chosen in any one of 2 ways (its subscript being neither of those already used), and finally the d may be chosen in but 1 way and therefore, by § 101, the four elements for any one term may be selected in 4. 3. 2-1 = 24 = 4! ways. The student is advised to write out all of the 24 terms, prefixing the proper sign to each. Similarly, the expanded form for the typical fifth order determinant may now be supplied. In this case there are 5 != 120 terms each of the form a^ b, Ci d^ e^, where no two of the subscripts r, s, t, u, v are alike and where the sign of any one term is taken + or — according as the number of inver- sions among these subscripts is even or odd. 214 COLLEGE ALGEBRA [XIV, § 121 Likewise, for any given value of n, the determinant D of § 119 may be expanded, this expansion containing in all n! terms, each the product of n elements properly chosen. Note. For convenience, the typical determinant of the third order, namely (1) of § 117, is frequently written in the condensed form |ai62C3|. Likewise, the typical fourth order determinant may be represented by 1 016203(^4 1, and similarly for determinants of higher orders. EXERCISES 1. Write, with their proper signs, all the terms of the determinant 1016203^465! that contain both oi and 64; also all the terms that contain both 63 and 65. 2. By expanding the following determinant, show that its value is 19. 119, the first column contains so that we here have ai = 3, 3 2 2 5 3 1 6 6 -1 2 1 1 [Hint. Note that in the notation of the a's, the second column the b's etc 02 = 5, 03 = 6, 04 = 0, 61 = 2, 62 = 3, etc.] 3. Find, by expanding, the value of the determinant 8 2 2 1 3 2 1 -5 -1 1 9 6 1 122. Useful Properties of Determinants. The following theorems are useful in the study of determinants. Theorem I. Two determinants are equal in case the elements of the first column of the one are equal respectively to the elements of the first row of the other, the elements of the second column of the one are equal respectively to the elements of the second row of the other, and so on. Thus 1 -1 2 1 2 3 2 3 = -1 3 1 3 1 3 2 3 XIV, § 122] DETERMINANTS 215 Pboof. Let us consider the theorem first for deter- minants of the third order. What we are then to prove is that Oi 61 Ci 02 &2 C2 (1) as 03 Ca fll 02 aa h h 63 Cl Ci C3 The determinant on the left side of (1) when expanded by the method of § 121, is equal to (2) 016203+026301 + 036162 — 016362 — 026103 — 0362C1. As to the determinant on the right side of (1), if we place Ai = Oi, A2 = 6i, A3 = ci, £1 = 02, ^2 = 62, Bs = Ci, Ci = az, 02 = 63, C3 = C3, it becomes Ai Si Cl A2 B2 C2 Az B3 C3 and this, when expanded by the method of § 121, becomes (3) A1S2C3+ A2B3C1+ A351C2 - A1B3C2 -A2B1C3 -A3B2C1 If we replace Ai, A2, A3, Bx ••■by their values as defined above, (3) becomes (4) O162C3 + 61C2O3 + C1O263 — O1C263 — 61O2C3 — C162O3 which is seen to be the same in value as (1), thus proving the theorem. Similarly, the proof may be given for deter- minants of any order. Theorem II. // two rows (or columns) of a determinant are interchanged, the value of the new determinant thus obtained is the same as the original except that its sign is changed. Thus 2 3 1 3 2 12 1 = — -12 1 3 2 2 3 1 Here the first and last rows of the original determinant have been interchanged in obtaining the new determinant. Pboof. Consider first that we merely interchange two adjacent rows of any determinant. This will merely inter- if 216 COLLEGE ALGEBRA [XIV, § 122 change two adjacent subscripts in each term of its expansion. This will change the sign of every term in the expansion, by § 121, and hence will change the sign of the whole deter- minant. - , If, more generally, the two rows to be iiiterchanged are separated by m intermediate rows, we first note that the lower row may be brought just below the upper one by m successive interchanges of adjacent rows. To bring the upper row into the original position of the lower one then requires m+1 further successive interchanges. It follows that interchanging the two rows in question is equivalent to introducing m+(m+l)=2m+l interchanges of adjacent rows and therefore, from what is said above, is equivalent to multiplying the original deter- minant 2m-f-l times by —1; that is, by ( — 1)^™"'''. But 2m+l is necessarily an odd number whatever the (positive, integral) value of m. Hence ( — l)2"'+i jg equal in all cases to —1, so that the theorem becomes proved for the case of the interchange of any two rows. To prove it also for the case of the interchange of any two columns, it suffices to write the original determinant, as we may do by Theorem I, in a form where its successive rows and columns become interchanged and then apply to the result the reasoning already given concerning the interchange of two rows. Theorem III. If a determinant D has two of its rows {or columns) identical, its value is zero. For example, without expanding the determinant, we may write at once 1-13 4 2 5 3 1^ 1-13 4 ■ 5 6 8 7 the first and third rows being here identical. XIV, § 122] DETERMINANTS 217 Peoof. By interchanging the two identical rows we obtain, by Theorem II, the value —D. But, interchanging two identical rows does not alter the form of the original determinant. Hence, we have D= —D, or 2D = 0, or D = 0. Similarly, the proof for the case of the interchange of two identical columns follows directly from Theorem II. Theorem IV. If every element of a row (or column) of a determinant is multiplied by any given number m, the deter- minant is multiplied by m. Thus 2 -1 2 • 3 3 1 2 -2 = 2 2 -1 3 Here the elements of the last row, regarded as 3, 2, 4, are each multipMed by 2. Pboof. The theorem is an immediate consequence of the fact that one and only one of the elements that have been multiplied by m enters into each term of the expansion, thus multiplying the whole expansion by m. Theorem V. If each of the elements in a row (or column) is expressed as the sum of two numbers, the determinant may be expressed as the sum of two determinants. That is, (in the case of the third order determinant) ai+oi' bi Ci Oi fci Ci Oi' bi ci 02+02' bi C2 = 02 bi d + 02' 62 C2 03+03' h C3 03 h Ci 03' h Ci Proof. Consider any term of the expansion of the given determinant, as (oi+ai')62C3. This may be written O162C3+ ai'biCi. Likewise, every term in the expanded form of the first determinant consists of the sum of a term of the second determinant and a term of the third determinant. Hence, the first determinant is the sum of the other two determinants. Similarly, the proof can be supphed whatever the order of the given determinant. 218 COLLEGE ALGEBRA [XIV, § 122 Theorem VI. The value of a determinant is not changed if the elements in any row {or column) are multiplied by any number m, and added to, or subtracted from, the corresponding elements in any other row (or column). Thus, for example, ai+m6i Oi+mh aa+mh ai as 6i Proof. By Theorem V, the first determinant here ap- pearmg may be expressed as follows: (1) ai as 6i + mbi mh mba &2 But, the last determinant, by Theorem IV, may be written as (2) m bi Ci bi Oi h Cs and, applying Theorem III, this has the value m -0 = 0, with which the proof is complete for the third order deter- minant above considered. Similarly, the proof may be supplied in all other cases. 123. The Simplification of Determinants. The theorems of § 122, especially Theorem VI, are of great value in reducing given determinants to simpler forms. The manner in which this is done will be clear from an examination of the following examples. Example 1. = 6 17 19 23 17 2 6 17 1 21 13 15 16 = 13 2 3 =2-3 13 1 1 11 13 17 11 2 6 11 1 2 6 1 2 3 12 3 1 2 1 1 = 6-2 1 1 1 = 12 1 1 -1 = 12-3 I 2 1 2 1 XIV, § 124] DETERMINANTS 219 Explanation. First we subtracted the first column from the second and third columns. This is equivalent to making two apph- cations of Theorem VI, using to = — 1 in each. Next, we have taken the factor 2 out of the second column of the resulting determinant, and the factor 3 out of its third column (Theorem IV). Next, we have subtracted 11 times the second column from the first colvmm, and then taken out a factor 2 from the first colxmin. Finally, we have subtracted 2 times the second column from the third. Note that the last deter- minant obtained has three zero elements, thus making its expansion relatively easy to calculate, giving 3. In general, the theorems of § 122 are to be thus employed to obtain one or more zero elements and corre- spondingly reduce the labor incident to the final expansion of a deter- minant. It is not to be expected, of course, that all the elements can be reduced to zero, or even all those in any one row or column, for this would imply that the determinant had the value zero, which in general would not be the case. Example 2. 1 a b+c 1 6 c+a 1 c a+b ' = 1 a b+c+a 1 6 c+a+b 1 c a+b+c =( a+b+c) 1 a 1 b 1 c 1 1 1 = (a+b+c) -0 = 0. Note the application of Theorem III in the last step. EXERCISES Evaluate the following determinants, employing as may be desired the theorems of § 122. 20 15 25 1. 2 -2 4 2. 17 12 22 3. 19 20 16 8 4 6 2 -2 4 2. 2 3 4 5 2 6 3 4 2 6 3 7 5 1 4 1 3 2 5 124. Minors. If one row and one column of a determinant be erased, a new determinant of order one lower than the given determinant is obtained. This determinant is called a first minor of the given determinant. Similarly, by- erasing two rows and two columns, we obtain a second minor; and so on. 220 COLLEGE ALGEBRA [XIV, § 124 ai 6j Cl 02 62 C2 03 63 C3 Thus, in the determinant (1) by erasing the second row and third column, we obtain the first minor 03 63 This minor is said to correspond to the element cj, since the row and column erased both contain this element. We may represent it, there- fore, by Dc^. In general, to each element of (1) corresponds a first minor obtained by erasing the row and column in which that element stands. The minor of ai is represented by Dap the miaor of 02 by Dg^, etc. Similar remarks evidently apply to a determinant of any order. 125. Development According to Minors. An examination of the expanded form of the typical determinant of the third order (see (2) of § 117), shows that it may be written if desired in the form aiihcs — baCa) — a^ibiCz — 63C1) +a3(&iC2 — &2C1), or ai -(h 61 h +as which, by § 124, may be written in the form aiDa^-chDa^+asDos- Thus, we have the relation (1) as 61 Cl 62 C2 63 Cs =aiDoi— OaDoj+asDa As thus written, the determinant is said to be developed according to the minors of its first column. XIV, § 125] DETERMINANTS 221 In a similar way, we may show that the same determinant may be developed according to the minors of any given row or column, provided only that in forming the various products thus called for of elements into their minors, the following general rule be observed: Rule. The product of the element lying in the rth column and sth row multiplied by its minor is to be taken positively or negatively according as (r+s) is even or odd. Thus, the determinant (1), when developed according to the ele- ments of its second column, becomes -biDb,+biDb^-biDb^. Other illustrative forms of development for the same determinant are -(hDa2+b2Dbi-CiDc^, oaDaj- 63^)63 +C3DC3. Passing now to the typical determinant of the fourth order (see (1), § 121) it will be found, upon examining the terms of its expansion, that it may be developed according to the minors of any one of its rows or columns in the maimer just described, and in fact a like statement may be verified for a determinant of any order whatever. For brevity, the details of the proof will be omitted. Thus, the determinant (1) of § 121, when developed by minors according to the elements of its first column, becomes aiDai — oaDoj +03-D£i3 — O4D04. Here, of course, each of the minors, Doj, Da^, Da^, Da^, is a determinant of the third order. Other illustrative forms of development for the same determinant are -biDi^+b^b,-hsDb,+biDi^, -aJDa^+biDb^ -oJ)ci+d^d^, ciDci — cjDcj+csDcj— C4Dc^. It is frequently advantageous to develop a determinant according to its minors, especially in case several of the elements in some column (or row) are equal to zero. 222 COLLEGE ALGEBRA [XIV, § 125 Example. Find the value of the determinant 4 2 12 2 3 2 5 3 2 12 5 6 4 9 Solution. First subtract the third row from the first (Theorem VI, § 122), thus obtaining as an equivalent determinant, and one having a number of zeros in its first row,the following 10 2 3 2 5 3 2 12 5 6 4 9 Now develop by minors according to the elements of the first row. 3 2 5 2 2 5 2 3 5 2 3 2 2 12 -0 • 3 1 2 +0- 3 2 2 -0- 3 2 1 6 4 9 5 4 9 5 6 9 5 6 4 Of these four terms the last three vanish because of the factor in each, so the result reduces to the determinant appearing in the first term. We may evaluate this third order determinant, as follows: Multiplying the second column by 2 and subtracting it from both the first and last columns, this determinant takes the form -1 2 1 1 -2 4 1 Developing this according to the elements of the second row, we have -0 2 1 -1 1 -1 2 -1 1 4 1 +1 • -2 1 -0 • -2 4 -2 1 -1+2 = 1. Thus the value of the original determinant is 1. EXERCISES Evaluate each of the following determinants by using the method of development by minors. 10 2 13 7 5 2 7 5 2 3 7 9 2. 12 4 6 3. 6 3 14 12 14 10 2 4 2 13 5 6 3 2 2 3 5-4 6 3 2 5 XIV, § 126] DETERMINANTS 223 126. Cofactors. If the minor of an element of a deter- minant be taken with its proper sign, as determined by the Rule of § 125, the result is called the cof actor of that element. Thus, in the cofactor of 6i is that of hi is O] 61 Ci 1 02 62 C2 03 63 C3 - 02 C2 . «3 %' + as 01 h Cl 02 h C2 03 63 C3 It is customary to represent the cofactor of ai by Ai, the cofactor of Oa by A2, that of as by A3, that of 61 by Bi, etc. By use of these cofactors the development of any given determinant is readily expressible in various forms, in accordance with the results of § 125. Thus may be expressed in any of the following forms. oiAi +02^2+03^3, &1B1+62B2+63B3, aiA2+biB2+CiC2, C1C1+C2C2+C3C3, etc. In connection with cofactors, the following theorem is important. Theobem. If the elements in any column be multiplied respectively by the cofactors of the corresponding elements in another column, the sum of the 'prodvxis is equal to zero. Thus, in the typical determinant of the third order (see above) we have ojBi +02B2 +0363 = 0, oiCi +0202+0303=0, &141+62A2+63A3 = 0, C1B1+C2B2 +0383=0, etc. 224 COLLEGE ALGEBRA [XIV, § 126 Proof. Consider the third order determinant (see above). For this we may write, as shown above, tti bi C\ = 61-61+62^2+6353. (1) 02 62 63 Now, no one of the cofactors Bi, Bi, B3 contains any of the elements 61, 62, 63., Hence, these cofactors are unaffected if in (1) we change 61, 62, 63 to ai, (h, a^. This gives aiBi + 02^2 + azBi = ai Cti Cl 02 C2 as Ci But this determinant is equal to zero by Theorem III, § 122, thus estabhshing as desired that aiBi-\-a2B2+azBi = Q. The proof may evidently be extended to cover any case in any determinant. 127. Simultaneous Equations. It was shown in § 116 that a system of two simultaneous equations of the first degree between two unknown letters x, y can be readily solved by means of determinants, and in § 118 a like fact was shown regarding the value of the three unknown letters X, y, z pertaining to a similar system of three equations. The general formulas for such solutions are to be seen in (5) of § 116 and (3) of § 118, which should now be examined. We proceed to show that similar formulas exist also for the four values X, y, z, w pertaining to a system of four equations of the first degree between these unknowns, and similarly for a system of five equations, etc. Suppose, then, that the system is one of four unknowns, namely, UiX + 6iy + Ci2 + diw = fci, 023; + 62?/ + C2Z + cfciy = fe, aix+biy+csz+dzw = h, a^x+b^y+c^z+d^w = fc^. XIV, § 127] DETERMINANTS 225 Consider the determinant D= fli 61 Ci di 02 &2 C2 dz 03 ba d dz at bi d di Si, £2, • • •, etc., be its cofactors. and let Ai, A2, Multiplying the first equation by Ai, the second by A2, the third by A3 and the fourth by Ai and adding, we have (aiAi+chA2+a3A3+aiAi)x+ ,j. (Mi+Mz+Ms+MOj/H- {ciAi+ C2A2+C3A3+ CiAi)z+ {diAi+d2A2+dsAs+diAi)w=kiAi+k2A2+ ksAs+htAi. Here the coefficients of y, 2 and w each vanish by the theorem of § 126, so that (1) reduces to (2) {aiAi+a2A2+a3A3+aiAi)x = ^1^1+^2^2+^143+^4^4. The coefficient of x in (2) is D; the right side is what D becomes when the elements Oi, 02, as, cii are respectively re- placed by ki, ki, ks, ki. Solving (2) for x, we thus have (3) x=- h ki kz ki 61 bi Cl Ci C3 Ci d, d3 di D In hke manner, by multiplying the first of the given equations by5i, the second by S2, etc., and adding and apply- ing the theorem of § 126, we obtain cti ki Cl di Oi ki Pi di as fcs C3 di Ui ki Cl di (4) D 226 COLLEGE ALGEBRA [XIV, § 127 Likewise, the values of z and w are each expressible as the quotient of two determinants, the denominator in each instance being D and the numerators being the determinants obtained from D by replacing the elements of its third column and fourth column respectively by h, fe, kg, ki. Using for brevity the condensed form of notation ex- plained in the Note at the close of § 121, the formulas for X, y, z, w thus become respectively 1^16203^41 _|aifc2C3C?4| _ 10162^3^41 [aib2C3fc4| 1016203^4!' 1016203^4!' 1016203^4!' |ai&2C3ci!4| These four formulas are seen to be analogous in formation to the three formulas obtained ia § 118 where only three equations were under consideration. Similar statements and results evidently apply to any set of simultaneous equations of the first degree containing as many unknown letters as equations. Note. It is to be observed that in case the determinant D which appears above has the value zero, the formulas (3), (4), etc. can no longer be used, since division by zero is not a permissible operation in mathematics. Such cases require special investigation and are con- sidered in detail in higher algebra. Similar remarks apply in general, and in particular to the systems already considered in §§ 116, 118. 128. Elimination. In all the systems of simultaneous equations thus far considered it was essential that the num- ber of equations be the same as the number of unknown let- ters present. When this condition is not fulfilled, various possibilities may arise and, while space does not permit of their detailed study here, the single case in which the number of equations is one greater than the number of unknowns is particularly important and will therefore be briefly considered below. Suppose, then, that three unknowns, x, y, z are present XIV, § 128] DETERMINANTS 227 and that these are to satisfy /owr equations of the first degree, which we shall write in the form aix+hiy+ciz = di, a3X+b3y+C3Z=d3, aiX+biy+C4Z = di. Moreover, let us suppose that a certain three of these equa- tions, say the first three, when treated as in § 127, may be solved for x, y, z. We have left to determine when these values of x, y, z will satisfy also the fourth equation. Now, noting the form of the solutions for x,y,zm the first three equations (see (3), § 118) and placing them in the fourth equation, then clearing the latter of fractions it be- comes (using the condensed notation explained in the Note at the close of § 121) 04 1 dibiPi \+h\ aAcz I +C4, 1 ajjida \=di\ aib^Cz \ . Transposing all terms to the left side and noting that, by Theorem II, § 122, we may write |rfi62C3| = |biC2d3|, |aid2C3| = — |a:C2d3|, the last relation becomes (after multiplying through by —1) — 04 1 biCids I -|- &4 1 0102^3 1 — C4 1 ai W3 1 -|- ^4 1 aiftjCs I = 0. But this relation is the same as ai 61 Ci di Oi bi C2 di az &3 Cs da 04 bi Ci di as appears by expanding this determinant by minors ac- cording to the elements of its last row. Theokem. In order that the system (1) may have a set of values X, y, z that will satisfy it, it is necessary that condition (2), which relates only to the sixteen coefficients of the system, shall be satisfied. (2) =0, 228 COLLEGE ALGEBRA [XIV, § 128 The determinant appearing in (2) is called the eliminant of the system (1). Thus, the theorem above may be stated briefly as follows. In order that the system (1) may have a solution X, y, z it is necessary that the eliminant of the system shall be equal to zero. A similar theorem may now be supplied for any system of linear equations containing one more equation than unknown quantities. The student is advised to do this for such a system of five equations. EXERCISES Solve by determinants each of the following systems of equations. 2x+3y— z+ M) = 6, x+ y+ z-2w = i, 3x+2!/-32+ w=-\, X— y— 2+3«J=— 1. 2a;+3y-4z+ w = Q, X— 2/+ 2— w= —2, 7x+2y-3z+ w = 6, 5x+8y-l0z-3w = 3. Form the eliminant for each of the following systems of equations and use it to tell (by the theorem of § 128) whether the system may have a solution. In oases where there may be a solution, proceed to determine it (if possible) by the methods of § 127. !, [3x4-2^+32 = 17, 2x+32/ = 9, x+ 2/ = 4, 2x+ y+2z = W, 3x- y = 8, i. <2x- y = 5, 6. L^ , .„ , .-oq x+ . = 6. |3x-2, = 7. ['IX'IX IZf: 6. Find the value (or values) of k for which the following system may have a solution. !kx+3y = lS, x-7y=-8, x—hy = 2. 7. Eliminate m from the system (1) m^x— mx* = l, (2) TO+2x=2. Solution. First Method. Solve (2) for m, giving ot = 2— 2x, and place this value of m in (1), giving as the desired result {2-2xfx-{2-2x)x^ = \, XIV, § 128] DETERMINANTS 229 which upon reducing becomes 6a?-10x2+4a;-l=0. This equation in x alone is, then, the result of eliminating m from (1) and (2). It is an equation whose roots satisfy (1) and (2) whatever the value of m. Second Method. Multiply (2) through by m, giving (3) m^+2mx=2m. Now, arrange (1), (2) and (3) in the forms (1) x-m'—:t?-m = l, (2) Q-m^+l-m = (2-2x), (3) l-m'+ (2x-2)m = 0. Regarding this system as one of three linear equations between the two quantities m' and m, and applying the results of § 128, we obtain as the desired equation X —a? 1 1 (2 -2a;) =0. 1 (2a; -2) Upon expanding this determinant it readily reduces to 6a^-102;*+4x-l=0 and this is seen to be the same result as obtained above by the first method. In contrasting the two methods, it will be seen that the second does not depend upon solving either of the given equations for m, as did the first method. For this reason, the second method has a much wider range of apphcability, as will be illustrated in the examples which follow. The second method illustrates what is known as Sylvester's method of elimination^. 8. Eliminate m from the following system, using both the methods illustrated in Ex. 7 and noting that the result for either method is the same. m^x—2mi?+l=0, m+3?—3m:c = 0. t For details, see for example Burnside and Panton's Theory of Equations (Longmans, Green and Co.), Chapter on EHmination. 230 COLLEGE ALGEBRA [XIV, § 128 9. Write as a determinant the result of eliminating k from the system [Hint. Multiply each equation through by k and consider the resulting equations combined with the original ones.] 10. Find the condition (in the form of a determinant) that the two equations may have a common root. [Hint. The result of eliminating x, where x is regarded as the com- mon root, will express the desired condition.] 11. Find the condition (in the form of a determinant) that the two equations ax^+bx+c = 3?+qx+r = 0, may have a common root. 12. Determine the value (or values) of k for which the following two equations may have a common root. 2a?-73+3=0, a?+kx+15=0. ANSWERS Page 1. 1. x-y+z. 2. x+y+z. 3. 2-0-6. 4. m-n+2a. 6. 0. 7. a!2/+42/a-s!!. 8. 5m-2n. 9. 5a26+62c-4a2c2-8a262+3a2c. 10. IlgJ. 11.4a6. 12..=^. —a+c 2x2/— 2/2 13. 2o6-2oc. Page 3. 1; li. 2 ^52. 3 30a 16^_ ^ 21 125 42 202/3 „ 2/ n «^a; '''"-"5 '7 7 ^ni. a—b 14. ^:^. o+b 19. i^. 1+X2 a 20. i^. 1—n 29x 10. 30a . 16o22/2 g2-4a:+3 3a+a2 42 ■ 202/3 ■"• (x-l)2 ■ ''• 9-a2 ■ 2nx2 7a -3cd. 6. i, --• 7. to(-1±V2) 8. -mzhVm^+m. 9. a, 1. 10. a, ' a 11. a, - 14 18 6-2 6+2 0+6 c c 0+1 12.1, a+5 -S-' -TT- IB. ffl+6, ; 19. a, -= o+b "■ "' 7 Page 37. 1. -2, -3. 2. 9, -3. 16. -a, -b. 20.«+6,^ 13. a+1, a-1. 1 1 17. 062 026 1 6. I-, --• 7. ±2, ±V-2. 8. 3,|(-1±>/^)- 9- ±1, ±2. 3a 10. I g(-2±Vl9). 13. -1, ±lV2. 11. -1, 14. a, b. 12. 1, i(-4±9v^^). 234 COLLEGE ALGEBRA Page 39. 1. ±1, ±2. 2. ±2, ±1/3. 3. ±1, ±| 4. l,|l(-l±i/^), |(-l±^/^). 6. 3, -1. 6. ±2, ±\/^^ 7. 16, ^- 8. 9. 9. 12. 10. 2, -|i(3±V57). 11. l,i(l±- N/-15). 12. 25. 13. 2. 14. 8, -^■ 1^- l'^-^^- 1-1-1=*=^- "4 Page 40. 1. -1. 2. 8. , 8 . . 4 . „ „ 22a2 3. g. 4. 4, ^. 5. 2a2, 3 6. 0, 5. 7. 2, 1 8. a, 6. Page 40. 1. 8, 12. 2. 5, 6. 3. 20 rods by 8 rods. 4. 2 in. 5. 12. 6. 0.41 in. 7. 30 mi. per hr. 8. 5 mi. per hr. 9. 15. 10. 30 min., 45 min. 11. 20 in. 12. 6.828 sq. ft. 13. 5 in., 12 in. 14. 108.3 yds., 51.7 yds. 15. 2.89 hrs. after noon, 2.53 hrs. before noon. 17. 7 sec. 18. 7 sec. 19. About 238 ft. 20. L=i(3s± VSs^-gedz). 1 21. r =-^ {—irh+ \/ir%2^2irS). Page 43. 2. a = 2, 6= -5, c = l. 3. a = 3, 5=0, c = l. 4.^a = 2, 6 = 2, c=-l. 6. a = 2, 6= -(ot+w), c = ^- 7. a = l,b = q—p,c=—pq. 8. a = m^+l, h=2bm, c = b^—rK 9. a = 4A;2-;2, b= -(8fc2+2i2), c=4fc2_i2. Page 45. 1. -|and -2. 2. -|and-3. 3. |andi- 4. |and-~ 6. 5 and -| 6. -|±iVi7. 7. -i±i-v/i3. 8. Idb^Vs: 9. 3±\/ — 1- 10. m and — n. 11. —to and n. 12. 2a and 36. 13. -4m and -3n. 14. a(l±V2). 15. -|and|. 16. ^(-piVr^)- Page 48. 1. Real and unequal, rational. 2. Real and unequal, rational. 3. Imaginary. 4. Real and unequal, irrational. 5. Real and unequal, rational. 6. Real and unequal, irrational. 7. Real and unequal, irrational. 8. Imaginary. 9. Real and equal. 10. Real and unequal, rational. 11. Real and unequal, rational. 12. Real and unequal, rational. 13. Real and unequal, rational. 14. Real and equal. 16.(a) |- 16.(6) -2 and -| 16.(c) 1. 16.(d) 4 and -|. 17. ±'\/ahrfi+hf^. ANSWERS 235 Page 49. e. -1,1. . 1. -2, 1 3' vs 2 8. ■ 3 2* -P, 3. -ff- 2, 1. 4. 1 10.(0)4,3; !•»■■ lO.(b) ■ 7 -6' -1, -7. -1. 10. (c) 10, 13. 10. (s) -i. - Vs 3 10. (d) i 1 '3' 10. ie)V2,V5. 10 '.(/) 5 3 2* 2 Page 50. 1. a:2-3a;+2 = 0. 2. a;2+3x+2 = 0. 3. 3a^-10i+3 = 0. i. 6xi+5x+l=0. 6. x2-(V^+V3)x+V6=0. 6. x2- V2a;-4=0. 7. 2s2-(l+2V5)a;H-\/5=0. 8. 4a;2-2(\/5-l)a;--v/5 = 0. 9. a^—mx-&n^=0. 10. a;2-2oa; +02-62 = 0. 11. x2-4a;+2 = 0. 12. x2-4a;+l=0. 13. 4x2+12x+3 = 0. 14. 4a:2+4a;-l=0. Page 62. 1. (x=7, y=2) and (x= -2, y= -7). 2. (x=2, 2/ = 4) and (^=~3'2/ = 3)' 3. (a;=3, 2/ = l) and (x= — 2, ?/= — 4). 4. (a! = l±iV46, 2/=-i±iv/46). 6. (a;=3± V34, 2/=-3±^V34). a;= g-> 2/=— 3-)- 7- (a;=3, 2/=2) and (r=96,2r=-29). 8. (a;=2, 2/ = l) and ^x=-^, 2/=-MV 9. (x = 3,2/=2)and(x = |,2,=|). ^°- (^"'27' ^"27/ ^^ {x=5,y=-l). Page 65. 1. (a; =3,2/ = 1); (x=-3, y = l)\ (s=3, y=-l); (x=-3, y=-l). 2. (x=4, 2/ = l); (2=4, 2/=-l); (a;=-4, y = l); (aj= -4, j/= -1). 3. (a; = 3,j/=5); (x= -3, 2/= -5), 4. (a;=-2, 2/ = 5); (a;=2, v=-5). 6. (a; = 9, 2/ = 5); (x = 5, y = 9); (x=-5, y=-9); (a!=-9, y=-5). 6. (a;=7, y = 3); (a;=-3, y=-7); (x=3, 2/=7); (a;=-7) 2/= -3). 7. (1=3, 2/ = 2); (a;=-2, 2/= -3); (x=2, 2/ = 3); (a:=-3, y=-2). 8. (x=7, y=4); (x=-7, 2/= -4); {x=^V2, 2/=|v^); (x = -Y^, 2/=-|V2). 9. (s=4, <=±1); (s=-^>< = ±iv'=15). (OK Ok\ 1 1 a; = -- 9^ 2/= -g-j; (a!=^+jgV-1799, 2/ = --L_^V3i799); ix=±-^y/^vm, 2,= _^+^v^l799). 236 COLLEGE ALGEBRA Page 68. 1. (a;=4, y=S); (x=3, 2/=4). 2. (a;=7, y=l); (x = l, y=7y, (x=-l, 2/= -7); (x=-7, y=-l). 3. (a>=5, 2/=2); (x=-4,y=-|)- 4. (*=! 2/=2); (a; = 1.2,=3)- 6. 0«;=2,2/=3); (a;=54,j/=0- 6. («=3, 2/ = l); (a;= -3, j/=-l). 7. (a:=5, y=4); (a;=-6, 2/= -4). 8. (x=6,y=8); (x=8, 2/=6); (a;=-6, 2/= -8); (.x=-8, y=-6). 9. (a!=12, 2/=3); [(.x=-7,y — ^ ; (a;=|+|v'^ 2/=|+iv^); (x=|-|-v^, y=l~V2a). 10. (a;=Vn, 2/=0); (a;=— \/n,2/=0);(a; = l,2/=2);(a;=-l,2/=-2). 11. (a; = l, 2/= -1); -l-VS). 12. (a; =3, y=2y, (x=2, y=-3); (a; = 16, 2/=-24); (^=-r^=-f)- ^^•(^=^^=^)'(^=-^'^=-^) 14. (aj=6, 2/=2); (.x=-6, y=-2); (,x=8V^, y=6^^=iy, (X SV'^, 2/=-6v^^). Page 68. 1. 4 and 8. 2. 81 and 1. 3. 12 in. and 16 m. 4. 16 rods long, 10 rods wide. 6. 2 ft. and 1 ft. 6. 6 ft. and 1 ft. 7. Altitude = 2.529 in., Base = 1.264 in. 8. Length = 96.883 ft., Width = 24.772 ft. 9. Either increase the length by 7.38 ft. and diminish the width by .38 ft., or diminish the length by 3.38 ft. and increase the width by 10.38 ft. 10. 15 days for the one man and 10 days for the other. 11. Circimiference of fore wheel = 10 ft.; circmnference of rear wheel = 12 ft. 12. Principal = $125, rate = 6%. 13. Time = 3 hours, rate = 10 miles per hour. 14. Reduced length = 108ft., reduced width=28ft., or reduced length = 18 ft., reduced width = 168 ft. Page 73. 1. 36. 2. -36. 3. llx-ll?/. 4{ 165. 5. 208.. 6. 82i^. 7. (a) 370.3 ft., (6) 2318.4 ft. 8. $2.60; $68.90. 9. 2500. 10. 72. 11. 336. 12. 43 ft. 13. 246 in. 14. 10. ' 15. 55. 16. 237 in. 20. (o) 20 in., (6) 31f in. 21. a= -139, 1 = 53. Page 78. 1. 512. 2. 32. 3. ~ 4. oUxU. 5. 4 6. 510. 7. 3906. 8. -^- 9. i— ^- 10- 765. 11. ~ 12. 2046. 13. 3279. 10^4 1 — o^ Id 100* 1 97 14. (o) 128, (&) 1024. 16. i^ = 1333333}^ bu. 16. 1364. 17. ^ 18. 5 sec. 19. 128. 26. Either 15, 8, 1 or 5, 8, 11. ANSWERS 237 Page 82. 1. 3. 2. |. 3. |. 4. 4f • 6. j2_. 6. |. 7. |(v^+l). 8. Io-a/S). 9. — i^ — 10. 36 m. 11. 16A min. after 3 o'clock. <* 10V3-6 T>=^<. oe < 17 _ 5 „ 181 • 169 _ 19 „ , , _ „ , Page as. 1. ~ 2. 3^ 3.^ i.^6.j^^ 6. 1^. 7. 3^^, 8. 5^bV 9- eyf^- 10. 34Ui- Page 92. 2. 300 ft. 3. 7* sq. yd. 4. 8. 6. 6750. 6. $876.56. 7. IH ohms. 8. 12 in. 9. 11 mi. 10. 302 (approximately). 13. (2 -V2) ft., or approximately 0.586 ft. 15.2. 17. IJ^ft. 18. •33*%. 20. About 12%. Page 115. 1. 2.3821. 2. 8.5786-10. 3. 0.7456. 4. 8.0957-10. 5. 144.83n 6. 155.214*. 7. 178.88. 8. 9.852. 9. 4914. 10. 5.496*. 11. 3403077000 (approximately). 12. 1236 (approximately). 13. 0.006805. Page 117. 1. 0.3273. 2. 1.4842. 3. 4.3187. 4. 8. 8859-10. 5. 15.667*. 6. 6.50. 7. 89.52*. 8. 1.201*. 9. 371. 10. 0.56825. 11. 68.8. 12. 1.0114. 13. .7734. Page 119. 1. 6.0205. 2. 1.4826. 3. 6.4910-10. 4. 6.2560. 6. 686.29. 6. 288.1. 7. 288.9. 8. 0.0001641. 9. 189.6. 10. 1.437. 11. 19.011. Page 120. 1. 0.2408. 2. 0.1647. 3. 9.5607-10. 4. 0.3172. 6. 17.746. 6. 1.628. 7. 1.629. 8. 0.06253. 9. 0.605. 10. 14.312. 11. 9.16. Page 121. 1. 13285. 2. 6169.5. 3. 2189. 4. 603. 6. 4.072. 6. 15.61 ft. 7. 3.88 sq. in. 8. 4217.27 ft Page 122. 1. a; = 1.66*. 2. a; =6.323*. 3.0.91.3*. 4. a; = -0.682*. e. -0.494*. 6. a; =2 or 2.18*. 7. a; = 1.709* 2/= 3.270*. 8. a; =1.198*, J/ = 1.387*. Page 126. 1. $537.10. 2. $320.70. 3. $1014. 4. $439.50. 6. 17 years. 6. 14.2 years. 7. 5%. 11. 4.83 years. 12. $5000. Page 128. 2. $2206.50. 3. $362.22. 4. $4965.10. 5. $77,217.35. 7. $370.85. 8. $6,716. 238 COLLEGE ALGEBRA Page 135. 1. a^+3x2j/+3xj/2+!/3. 2. ai+'ia^+ea^h'+idl^+U. 3. 3?-Zoi^+Zx'!P~y\ 4. o*-4a36+6o2fe2-4afe3+64. 5. 32+80?-+ 80r2+40r3+10r4+r5. 6, a7+7o6a;+21a6x2+35a4a:3+35o3x4+21a2a:6-|- 7oa^+xV. 7. g5_i5^+9033_270ff2+405s-243. 8. aio+5a8x+10a6x2+ 10o4x3+5o2a^+26. 9. ai-AoH^+Ga'^ofi-ia^+x^. 10. 16o4+32a3+ 24o2+8a+l. 11. x5-15a%+90x32/2-270x22/3+405xj/4_2432/5. 12. l+6x2+15a^+20z6+15a;8+6xi''+xi2. 13. l-8x +28x2-56x3+ 70X«- 56x5 +28x6 -8x7+x8. 14. y^^K^jAfi_^2J^^-L. 16. 81a8-108a6+56o*-12a2+l. 16. aio+10a9x+45a8x2+120a7x3+ 210o6x4+252a5x5+210o4x6+120a3x7+45a2a;8+10ox9+xH). x'^x6j/^x62/2^X^3^a3j^^.-i;22^5T^2;2^T^j^7 18. g-sg+lO^-lO^+sg- J 19. a2+3a^A^+36^V6+ 52^. 20. 2v^+;^+3^+l, Page 136. 1. 70o4x4. 2. 56x32/5. 3. 672x6. 4. I0m5,i9. 5. -252ai06io. 6.-42504x19. 7. 462a;. 8. 12870a8. 9. 495xi2. 10. -960 -v/2. Page 139. 1. a^+|a~*x-^o~^x2+^a'^x3 -8. a"2-2a~'x+4o~*a!2-4o~^x3+ - 3. l+gX-gX2+— x3 . |(2a)-%2+_|(2a)-*63_ .... g. a-*+?a-Vx2+|a-'^x4+ ^a-tW-. 7.2*+1.2-tx-|.2-lx2+A.2-^^-.... 8. a*+ia-*x-|a-«.t2+A y = i, z = ~ 6. fc=2or-6. 8. 5x6-2x4-9a2+6a!-l=0. 12. fc=-8or-~ APPENDIX TABLE OF POWERS AND ROOTS Explanation 1. Square Roots. The way to find square roots from the Table is best unde rstoo d from an example. Thus, suppose we wish to find Vl.48. To do this we first locate 1.48 in the column headed by the letter n. We find it near the bottom of this column (next to the last number). Now we go across on that level until we get into the column headed hy^/n. We find at that place the number 1.21655. This is our answer. That is, Vl.48 =1.21655 (approxi- mately). If we had wanted Vl4.8 instead of v'1.48 the work would have been the same except that we would have gone over into the column headed VlO n (because 14.8=10X1.48). The number thus located is se en t o be 3.84708, which is, therefore, the desired value of Vl4.8. Again, if we had wished to find Vl48 the work would take us back again to the column headed Vn, but now instead of the answer being 1.21655 it would be 12.1655, In other words, the order of the digits in Vl48 is the same as for v'1.48, but the decimal point in the answer is one place farther to the right. Similarly, if we desired Vl480 the work would be the same as before except that we must now use the column headed v^lO n and move the decimal point there occm-ring one place farther to the right. This is seen to give 38.4708. Thus we see how to get the square root of 1.48 or any power of 10 times that number. 341 242 APPENDIX In the same way, if we wish to find V.MS, or V.0148, or V. 00148, or the square root of any number obtained by dividing 1.48 by any power of 10, we can get the answers from the column headed Vn or VlO n by merely placing the decim al poi nt properly. Thus , we find that V.148 = .384708, V^OMS = . 121655, V.00U8 = .0384708, etc. What we have seen in regard to the square root of 1.48 or of that number multiplied or divided by any power of 10 holds true in a similar way for any number that occurs in the column headed n, so that the tables thus give us the square roots of a great many numbers. 2. Cube Roots. Cube roots are located in the tables in much the same way as that just described for square roots, but we have here three columns to select from instead of two, namely the columns headed '^, "V^lOn, v^lOO n. Illustration. • ^1.48 occurs in the column headed v^ and is seen to be 1.13960. ''^14.8 occurs in the column headed "V^IO n and is seen to be 2.4552. 'V^148 occurs in the column headed v^lOO n and is seen to be 5.28957. To get vCTii we observe that .148= A/i^ = -\/3ii = -117148. ^^ 10 ^1000 10 Thus, we look up 'V^148 and divide it by 10. The result is instantly- seen to be .52 8957. Similarly, to get v^.0148 we observe that ^^^^^^^^ a/^ = •VtocI " in '^^^^^ '^^^^' "^^ ^°°^ ^P '^'^^ '^^^ divide it by 10, giving the result .24552. To get •v^.00148 we observe that v^.00148=^^^ = ivTiS, so that we must divide ■v'1.48 by 10. This gives .11396. Similarly the cube root of any number occurring in the column headed n may be found, as well as the cube root of any number obtained by multiplying or dividing such a number by any power of 10. TABLE OF POWERS AND ROOTS 243 3. Squares and Cubes. To find the square of 1.48 we naturally look at the proper level in the column headed n^. Here we find 2.1904, which is the answer. If we wished the square of 14.8 the result would be the same except that the decimal point must be moved two places to the right, giving 219.04 as the answer. Similarly the value of (148)^ is 21904.0 etc. On the other hand, the value of (.148)^ is found by moving the decimal place two places to the left, thus giving .021904. Similarly, (.0148)2 = . 00021904, etc. To find (1.48)'. we look at the proper level in the column headed n^ where we find 3.24179. The value of (14.8)^ is the same except that we must move the decimal point three places to the right, giving 3241.79. Similarly, in finding (.148)' we must move the decimal place three places to the left, giving .00324179. Further illustrations of the way to use the tables will be found in § 140. EXERCISES Read off from the tables the values of each of the following ex- pressions. 1. Vii 4. ■v''670 7. VMl 10. -vCOOiSi 2. VK9 5. V^ 8. \^W7 11. V. 000143 3. -v^ 6. Vma 9. V.00164 12. ^.000143 244 Table I — Powers and Roots [I n n^ Vn VlOw n^ ^ ■^10 n i/mon 1.00 1.0000 1.00000 3.16228 1.00000 1.00000 2.15443 4.64169 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.0201 1.0404 1.0609 1.0816 1.1026 1.1236 1.1449 1.1664 1.1881 1.00499 1.00995 1.01489 1.01980 1.02470 1.02956 1.03441 1.03923 1.04403 3.17806 3.19374 3.20936 3.22490 3.24037 3.26576 3.27109 3.28634 3.30161 1.03030 1.06121 1.09273 1.12486 1.15762 1.19102 1.22504 1.25971 1.29503 1.00332 1.00662 1.00.990 1.01316 1.01640 1.01961 1.02281 1.02599 1.02914 2.16159 2.16870 2.17577 2.18279 2.18976 2.19669 2.20358 2.21042 2.21722 4.65701 4.67233 4.68755 4.70267 4.71769 4.73262 4.74746 4.76220 4.77686 1.10 1.2100 1.04881 3.31662 1.33100 1.03228 2.22398 4.79142 1.11 1.12 1.13 1.14 1.15 1.16 I.IT 1.18 1.19 1.2321 1.2644 1.2769 1.2996 1.3225 1.3456 1.3689 1.3924 1.4161 1.06357 1.05830 1.06301 1.06771 1.07238 1.07703 1.08167 1.08628 1.09087 3.33167 3.34664 3.36155 3.37639 3.39116 3.40588 3.42053 3.43611 3.44964 1.36763 1.40493 1.44290 1.48164 1.52088 1.56090 1.60161 1.64303 1.68516 1.03640 1.03850 1.04158 1.04464 1.04769 1.05072 1.05373 1.05672 1.05970 2.23070 2.23738 2.24402 2.25062 2.26718 2.26370 2.27019 2.27664 2.28305 4.80590 4.82028 4.83459 4.84881 4.86294 4.87700 4.89097 4.90487 4.91868 1.20 1.4400 1.09545 3.46410 1.72800 1.06266 2.28943 4.93242 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 1.29 1.4641 1.4884 1.6129 1.5376 1.5626 1.6876 1.6129 1.6384 1.6641 1.10000 1.10454 1.10905 1.11355 1.11803 1.12250 1.12694 1.13137 1.1.3578 3.47851 3.49285 3.60714 3.32136 3.53553 3.64965 3.S6371 3.57771 3.69166 1.77156 1.81585 1.86087 1.90662 1.95312 2.00038 2.04838 2.09715 2,14669 1.06560 1.06853 1.07144 1.07434 1.07722 1.08008 1.08293 1.08577 1.08859 2.29577 2.30208 2.30835 2.31469 2.32079 2.32697 2.33311 2.33921 2.34529 4.94609 4.95968 4.97319 4.98663 5.00000 5.01330 6.02663 5.03968 6.06277 1.30 1.6900 1.14018 3.60655 2.19700 1.09139 2.35133 6.06580 1.31 1.33 1.33 1.34 1.35 1.36 1.37 1.38 1.39 1.7161 1.7424 1.7689 1.7956 1.8225 1.8496 1.8769 1.9044 1.9321 1.14455 1.14891 1.15326 1.15758 1.16190 1.16619 1.17047 1.17473 1.17898 3.61939 3.63318 3.64692 3.66060 3.67423 3.68782 3.70135 3.71484 3.72827 2.24809 2.29997 2.35264 2.40610 2.46038 2.51546 2.B7135 2.62807 2.68562 1.09418 1.09696 1.09972 1.10247 1.10521 1.10793 1.11064 1.11334 1.11602 2.35736 2.36333 2.36928 2.37621 2.38110 2.38697 2.39280 2.39861 2.40439 5.07876 5.09164 6.10447 6.11723 5.12993 6.14266 6.15614 6.16766 5.18010 1.40 1.9600 1.18322 3.74166 2.74400 1.11869 2.41014 6.19249 1.41 1.42 1.43 1.44 1.46 1.46 1.47 1.48 1.49 1.9881 2.0164 2.0449 2.0736 2.1025 2.1316 2.1609 2.1904 2.2201 1.18743 1.19164 1.19583 1.20000 1.20416 1.20830 1.21244 1.21655 1.22066 3.75500 3.76829 3.78153 3.79473 3.80789 3.82099 3.83406 3.84T08 3.86005 2.80322 2.86329 2.92421 2.98598 3.04862 3.11214 3.17652 3.24179 3.30795 1.12135 1.12399 1.12662 1.12924 1.13185 1.13445 1.13703 1.13960 1.14216 2.41587 2.42156 2.42724 2.43288 2.43850 2.44409 2.44966 2.45520 2.46072 6.20483 6.21710 5.22932 5.24148 5.25359 6.26564 6.27763 5.28967 6.30146 I] Powers and Roots 245 n n» y/n VlOw «8 ^n 5006 9.55371 9.55736 9.56101 9.56466 9.56830 9.57194 9.57557 9.57921 880 77.4400 2.96648 9.38083 681.472 2.06456 4.44796 9.58284 8.81 8.82 8.83 8.84 8.85 8.86 8.87 8.88 8.89 77.6161 77.7924 77.9689 78.1456 78.3225 78.4996 78.6769 78.8544 79.0321 2.96816 2.96985 2.97153 2.97321 2.97489 2.97658 2.97825 2.97993 2.98161 9.38616 9.39149 9.39681 9.40213 9.40744 9.41276 9.41807 9.42338 9.42868 683.798 686.129 688.465 690.807 693.154 695.506 697.864 700.227 702.595 2.06534 2.00612 2.06690 2.06768 2.06846 2.06924 2.07002 2.07080 2.07157 4.44964 4.45133 4.45301 4.45469 4.45637 4.45805 4.45972 4.46140 4.46307 9.58647 9.,5!K)09 9.59372 9.59734 9.00095 9.60457 9.60818 9.61179 9.6154C 8.90 79.2100 2.98329 9.43398 704.969 2.07235 4.46475 9.61900 8.91 8.92 8.93 8.94 8.95 8.96 8.97 8.98 8.99 79.3881 79.5664 79.7449 79.9236 80.1025 80.2816 80.4609 80.6404 80.8201 2.98496 2.98664 2,98831 2.98998 2.99166 2.99333 2.99500 2.99666 2.99833 9.43928 9.44458 9.44987 9.45516 9.46044 9.46573 9.47101 9.47629 9.48156 707.348 709.732 712.122 714.517 716.917 719.323 721.734 724.1.51 726.573 2.07313 2.07390 2.07468 2.07545 2.07622 2.07700 2.07777 2.078.54 2.07931 4.46642 4.46809 4.46976 4.47142 4.47309 4.47476 4.47642 4.47808 4.47974 9.62260 9.62620 9.62980 9.63339 9.63698 9.64057 9.64415 9.64774 9.65132 260' Powers and Boots [I n »2 yfn VIOto TO' ^ ^10 » ^100 TO 9.00 81.0000 3.00000 9.48683 729.000 2.08008 4.48140 9.65489 9.01 9.02 9.03 9.04 9.05 9.06 9.07 9.08 9.09 81.1801 81.3604 81.5409 81.7216 81.9025 82.0836 82.2649 82.4464 82.6281 3.00167 3.00333 3.00500 3.00666 3.00832 3.00998 3.01164 3.01330 3.01496 9.49210 9.49737 9.50263. 9.60789 9.51315 9.51840 9.52365 9.52890 9.53415 731.433 733.871 736.314 738.763 741.218 743.677 746.143 748.613 751.089 2.08085 2.08162 2.08239 2.08316 2.08393 2.08470 2.08546 2.08623 2.08699 4.48306 4.48472 4.48638 4.48803 4.48969 4.49134 4.49299 4.49464 4.49629 9.05847 9.66204 9.66561 9.66918 9.67274 9.67630 9.67986 9.68342 9.68697 9.10 82.8100 3.01662 9.53939 753.571 2.08776 4.49794 9.69052 9.11 9.12 9.13 9.14 9.15 9.16 9.17 8.18 9.19 82.9921 83.1744 83.3569 83.5396 83.7225 83.9056 84.0889 84.2724 84.4561 3.01828 3.01993 3.02159 3.02324 3.02490 3.02655 3.02820 3.02985 3.03150 9.54463 9.54987 9.55510 9.66033 9.56556 9.57079 9.67601 9.68123 9.58645 756.058 753.551 761.048 763.552 706.061 768.575 771.095 773.021 776.152 2.08852 2.08929 2.09005 2.09081 2.09158 2.09234 2.09310 2.09386 2.09462 4.49959 4.50123 4.50288 4.60452 4.50616 4.50781 4.50945 4.51108 4.51272 9.69407 . 9.69762 9.70116 9.70470 9.70824 9.71177 9.71531 9.71884 9.72236 9.20 84.6400 3.03315 9.59166 778.688 2.09538 4.51436 9.72589 9.21 9.22 9.23 9.24 9.25 9.26 9.27 9.28 9.29 84.8241 85.0084 85.1929 85.3776 85.5625 86.7476 85.9329 86.1184 86.3041 3.03480 3.03645 3.03809 3.03974 3.04138 3.04302 3.04467 3.04631 3.04795 9.59687 9.60208 9.60729 9.61249 9.61769 9.62289 9.62808 9.63328 9.63846 781.230 783.777 786.330 788.889 791.453 794.023 796.698 799.179 801.768 2.09614 2.09690 2.09765 2.09841 2.09917 2.09992 2.10068 2.10144 2.10219 4.61599 4.51763 4.51926 4.52089 4.52252 4.62415 4.52578 4.52740 4.52903 9.72941 9.73293 9.73645 9.73996 9.74348 9.74699 9.75049 9.75400 9.75760 9.30 86.4900 3.04959 9.64365 804.357 2.10294 4.63065 9.76100 9.31 9.32 9.33 9.34 9.35 9.36 9.37 9.38 9.39 86.6761 80.8624 87.0489 87.2356 87.4225 87.6096 87.7969 87.9844 88.1721 3.05123 3.05287 3.05450 3.05614 3.05778 3.05SM1 3.06105 3.06268 3.06431 9.64883 9.65401 9.66919 9.66437 9.66954 9.67471 9.67988 9.68504 9.69020 806.954 809.558 812.166 814.781 817.400 820.026 822.657 825.294 827.936 2.10370 2.10445 2.10520 2.10595 2.10671 2.10746 2.10821 2.10896 2.10971 4.53228 4.53390 4.53562 4.53714 4.53876 4.54038 4.54199 4.54361 4.54,522 9.76450 9.76799 9.77148 9.77497 9.77846 9.78195 9.78543 9.78891 9.79239 9.40 88..3600 3.06594 9.R9536 830.584 2.11045 4.54684 9.79586 9.41 9.42 9.43 9.44 9.45 9.46 9.47 9.48 9.49 88.5481 88.7364 88.9249 89.113R 89.3025 89.4916 89.6809 89.8704 90.0601 3.06757 3.06920 3.07083 3.07246 3.07409 3.07571 3.07734 3.07896 3.08058 9.70052 9.70567 9.71082 9.71597 9.72111 9.72625 9.73139 9.73653 9.74166 833.238 835.897 838.662 841.232 843.909 846.591 849.278 851.971 854.670 2.11120 2.11195 2.11270 2.11344 2.11419 2.11494 2.11568 2.11642 2.11717 4.54845 4.65006 4.55167 4.65328 4.55488 4.55649 4.55809 4.55970 4.66130 9.79933 9.80280 9.80627 9.80974 9.81320 9.81666 9.82012 9.82357 9.82703 I] Powers and Roots 261 n m" Vw. y/lOn n'