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4


ELEMENTS OFALGEBRA,
BY
S." F.' LACROIX,
TILANSLATED FRLOM~ THEraFmIaEC


FOR THlE USE OF THE STUDENTS OF T1 UNIVERSITY
AT
CAMBRIDGE, NEW ENGLAND.
BY JOHN FARRAR,
PROFNSSOR OF MATHEMATICS AND KATURAL PHILOSOPHY.
THILD EDITION,
BOSTON:
HILLIARD, GRAY~ LITTLE, AND WILKINS,.




DISTRICT OF MASSACHUSETTS, TO WIT:
District Clerk's Office.
BE IT REMEMBERED, that on the twenty-seventh day of October, A. D. 1830, in
the fifty-fifth yeavohe Independence of the United States of America, John
Farrar of the sai d crict has deposited in this office the title of a book, the
right whereof he claims as author, in the words following, viz:
" Elements of Algebra, by S. F. Lacroix. Translated from the French. for the
use of the Students of the University at Cambridge, New England. By John
Farrar, Professor of Mathematics and Natural Philosophy." Third Edition
In conformity to the Act of the Congress of the United States, entitled, " An
Act for the encouragement of learning, by securing the copies of maps, charts,
and books, to the authors and proprietors of such copies, during the times therein mentioned": and also to an Act, entitled, " An Act, supplementary to an Act,
entitled,' An Act for the encouragement of learning, by securing the copies of
maps, charts, and books to the authors and proprietors of such copies, during
the times therein mentioned'; and extending the benefits thereof to the arts of
designing, engraving, and etching historical and other prints."
JNO. W. DAVIS,
Clerk of the District of Massachusetts.


E. W. METCALF AND CO., CAMBRIDGE,
Printers to the University.




ADVERTISEMENT.
LACROIX'S ALGEBRA has been in use in the French schools for
a considerable time. It has been approved by the best judges,
and been generally preferred to the other elementary treatises,
which abound in France. The following translation is from the
eleventh edition, printed at Paris in 1815.  No alteration has
been made from the original, except to substitute English instead
of French measures in the questions, where it was thought necessary.  When there has been occasion to add a note by way
of illustration, the reference is made by a letter or an obelisk,
the author's being always distinguished by an asterisk.
In this third edition examples are introduced of the leading sections of the work, selected principally from the celebrated collection
of Meier Hirsch, and intended as an exersise for the learner.
Cambridge, JTovember, 1830.




I




CONTENTS.
Page.
Preliminary Remarks upon the Transitionfrom Arithmetic to
Algebra      -       -    -     --                      1
The nature and object of algebra     -           -ib.
Explanation and use of algebraic signs -  -  -               2
Examples of the solution of problems by means of algebraic signs  3
Explanation of algebraic formulas      -    --               8
Of Equations       --                                      11
To solve questions by the assistance of algebra  -  -  -    ib.
Explanation of the words, equation, members, and terms  -  - ib.
Resolution of Equations of the First Degree, having but one
Unknown Quantity       -     -    -                    12
Rule for transposing any term from one member of an equation
to the other    -    --.                   14
To disengage an unknown quantity from multipliers  -  -     15
Of equations, the terms of which have divisors  -   -       16
Rule for making the denominators in an equation to disappear  17
To write a question in the form of an equation  -  -  -     18
Examples                                                    l -  - - — ib..Methods forperforming, as far as is possible, the operations
indicated upon Quantities, that are represented by Letters  23
Explanation of the terms, simple quantities, binomials, trinomials,
quadrinomials, and polynomials            -    -    -  24
Addition of AZlgebraic Quantities                      -    ib.
Of coefficients  -       -      -              -    -      ib.
Rule for performing addition  -             -              25
Rule for the reduction of algebraic quantities -  -  -   - ib.
Subtraction of Algebraic Quantities        -    -          26
Rule for performing subtraction          -         -    -  27
Multiplication of dlgebraic Quantities  -  -    -    -     ib.
Manner of indicating multiplication  -  -                  28
What is to be understood by powers of a quantity     -     29
Of exponents         -    -    -      -    -30




VI


Contents.


Method of fii(ding the powers of a number       -     -    -
Rule for the multiplication of simple quantities   -    - 
What is meant by the degree of a product   -    -     -    -
Note-on the term dimensions        -    -    --
Multiplication of compound quantities              -       -
Rule respecting the signs     -    -    -    -     --
Rules for performing multiplication   -    -    -     -    -
Examples for illustration    -     ---
Of homogeneous expressions       -    --              -
Expression of the product of the sum of two quantities by their
difference    ----                                 -
Of the square and cube of a binomial       -    -     -
Manner of indicating the multiplication of compound quantities


30
ib31
ib.
32
33
ib.
34
37
ib.
ib.
38


Division of Algebraic quantities  -   -     -    -    -   ib.
Rules for the division of simple quantities  -  -  -    - 39
Value of a quantity whose exponent is zero  -    -    -   40
How an expression employing division may be simplified when
the operation cannot be performed    -    -    -   - ib.
Division of compound quantities   -    -                   42
What is understood by arranging the terms of a quantity  - 43
Rules for performing division  -  -    -                  44
Examples in division  -    -    -   -    -- 45
Method of arrangement, when the quantity has the same power
in several terms, both of the dividend aud divisor  -  47
Example     -    -    -    -   -    --             -      ib.
Examples in multiplication of compound quantities  -      49
Examples in division of compound quantities  -          - ib.
Of Algebraic Fractions    -     -    -    -    -     -    50
To abridge an expression when the algebraic division cannot be
performed           -    --            -    -ib.
The greatest common divisor of two algebraic quantities  - 51
To find the greatest common divisor  -   -    -         - ib.
Necessary precautions, when the quantity, which we take for a
divisor, contains several terms having the letter, with reference to which the arrangement is made, of the same degree 54
To obtain the divisor independent of this letter  -  -  -  56
Recapitulation of the rules for the calculus of fractions  -  57
Resolution of a literal equation of the first degree  -  -  59
Examples in division, in which the divisor is not an aliquot part of
the dividend     -    -    -         -   -    -    - 60
Examples in the reduction of fractions  -   -   -    -    ib.






Contents.


Vii


Examples in the reduction of fractional expressions to their simplest terms   -    -    -     -    -    -    -     -    -61
Examples in the multiplication of fractional expressions  -     62
Examples in the division of fractional expressions  -  -     -  ib.
Equations of the first degree with one unknown quantity   -     63
Of Questions having two Unknown Quantities, and of Negative
Quantities     -    -                                     64
Examples        -                                               65
To resolve an equation, when the two members have the sign-     66
A question in which one of the unknown quantities has the
sign          -    -     -    -    -    -     -    -    -  ib.
How this sign is to be understood    -     -    -    -    -     68
How values affected with the sign- must satisfy the conditions
of a problem       -     -    -    -    --              -  69
Examples for illustration  -    -    -     -70
Recapitulation of the above remarks     -    -    -     -    - 71
Of Negative Quantities                    -                    72
Demonstration of the rules for the calculus of insulated negative
quantities         - -   --                                ib.
The combination, with respect to their signs, of insulated simple
quantities       -    -     -    -73
How to find the true enunciation of a question involving negative
values   -    -    -     -    — 74
Examples for illustration  -    -    --                   -     ib.
What is signified by the expression     -                       79
What is signified by the expression  -     -    -               81
Note-on the use of the term identical   -    -    -    -     -  ib.
General conclusion from what precedes      -    -    -     -    82
Of changing the signs of quantities to comprehend several questions in one  -    -    -     -    -                       83
Solution of the preceding questions by employing only one unknown quantity        -    -     -    -    --              84
To resolve equations of the first degree, when there are two unknown quantities   -     -    -         -86
Of the Resolution of any given number of Equations of the
First Degree, containing an equal number of Unknown
Quantities     -    -     -                               89
General rule for deducing an equation having only one unknown
quantity by exterminating or eliminating successively all
the rest                                                  ib...ib.




VlII


Contents.


Examples to illustrate the above rule      -              89
Questions to be performed  -   -    -    -94
General Formulas for the Resolution of Equations of the First
Degree   -    -                                      96
General process for exterminating, in two equations, an unknown
quantity of the first degree  - --                    97
General value of unknown quantities, in equations of the first
degree, when there are three unknown quantities -  -  100
General rule for obtaining the value of unknown quantities  101
Application of the general formulas           -- 103
Equations of the first degree with two or more unknown quantities     -   -    -                -               104
Examples in equations involving negative results  -  -  - 106
Equations of the Second Degree having only one Unknown
Quantity    -     -    -                            107
Examples of equations of the second degree, which contain only
one unknown quantity        -                        ib.
Extraction of the square root of whole numbers  -    -   108
Of numbers which are not perfect squares          -    - 113
Method of determining whether the root found is two small  -  114
To find the square and square root of a fraction  -  -  - ib.
Every prime number, which will divide the product of two numbers, will necessarily divide one of these numbers  -  ib.
Whole numbers, except such as are perfect squares, admit of no
assignable root, either among whole numbers or fractions  115
What is meant by the term incommensurable or irrational  -  116
How to denote by a radical sign, that a root is to be extracted  ib.
The number of decimal figures in the square double the number
of those in the root  -    -- 117
Method of approximating roots  -  --                      ib.
Method of abridging, by division, the extraction of roots -  - 118
To approximate a root indefinitely, by means of vulgar fractions 119
Most simple method of obtaining the approximate root of a fraction, the terms of which are not squares  -  -   -  120
Resolution of equations, involving only the second power of the
unknown quantity  -    -                           - ib.
The square root of a quantity may have the sign + or  -  122
The square root of a negative quantity is imaginary  -  - 123
Complete equations of the second degree              -   ib.
General formula for resolving equations of the second degree,
having only one unknown quantity  -.     -     125




Contents.


ix


General rule for the above process  - --                 126
Examples showing the properties of negative solutions  -  -  ib.
In what cases problems of the second degree become absurd  129
Expressions called imaginary          -.                130
An equation of the second degree has always two roots -  - 131
Resolution of certain problems   -    -   -,132
To divide any number into two parts, the squares of which shall
be in a given ratio          - -  -                  135
Examples in the extraction of the square roots of numbers -  139
Examples in equations of the second degree, involving only the
second power of the unknown quantity, and those which may
be solved without completing the square  -   -    - 140
Examples in equations of the second degree solved by completing
the square         -    -                             ib.
Extraction of the Square Root of Jlgebraic Quantities  -  141
Transformation for simplifying radical quantities  -  -   ib.
Extraction of the square root of simple quantities  -  -  - 142
Extraction of the square root of polynomials  -  -  -    143
Examples in the abbreviation and transformation of radical quantities,-                                           146
Examples in the extraction of the square roots of algebraic expressions          -                                 147
Examples in the extraction of the square roots of incomplete
squares     -   -    -    -                           ib
The Formation af Powers of Simple Quantities, and the Extraction of their Roots         -                        148
Table of the first seven powers of numbers from I to 9  -  ib
To obtain any power whatever of a simple quantity  -  -  149
To extract the root of any power whatever of a simple quantity  ib
To simplify radical expressions containing one term  -   150
Of imaginary roots in general  -  -                      151
Of fractional exponents  -    -    -    -   -    -    -    b.
Of negative exponents  -      -.                         152
Of the Formation of Powers of Compound Quantities    -153
Manner of denoting these powers  -        -    -         154
Form of the product of any number whatever of factors of the
first degree.-  -ib.
Method of deducing from this product the developement of any
power of a binomial     -   -                        156
Theory of permutations and combinations  -  -    -       157
Rule for the developement of any power whatever of a binomial 161
b




x


Contents.


General term of the binomial formula  -    -    -    -    161
The binomial formula illustrated by examples  -   -    - 162
Transformation of the formula to facilitate its application  -  1603
Its applioation to trinomial quantities       -           164
Extraction of the Roots of Compound Quantities  -    -     ib.
To extract the cube root of whole numbers                  ib.
To extract the cube root of fractions  -   -    -    -    168
Method of approximating the cube root of numbers which are
not perfect cubes     -    -    -   -.. 169
Extraction of the roots of higher degrees  -  -  -   -    170
To extract the roots of literal quantities  -  -   -   - 172
Of Equations with two Terms      -                        173
The division of xm- am by x -a      --                 -  175
Factors of the equation xn- aam = 0, and the roots of unity  176
General law for the number of roots of an equation, and the distinction between arithmetical and algebraical determination 178
Of Equations which may be resolved in the same manner as
those of the Second Degree   -    -    -    -          b.
To determine their several roots                       -    b.
Calculus of Radical Expressions  -    -    -    -         180
Process for performing on radicals of the same degree the four
fundamental operations                                 ib.
To raise a radical quantity to any power whatever  -      183
To extract the root of any degree   -    -    -         - 184
To reduce to the same degree, any number of radical quantities
of different degrees  -  -   -    -    -    -    -    185
To place under the radical sign a factor that is without it  ib.
Multiplication and division of certain radical quantities  ib.
Remarks on peculiar cases which occur in the Calculus of Radical Quantities  -    -   -    -    -    -    -    -186
To determine the product of v  _/-a X  /-a  -    -   -      b.
m      n
Different expressions of the product /a X ^/b  -   -    - 187
Calculus of Fractional Exponents      -    -    -    -    189
How to deduce the rules given for the calculus of radical quantities      -    -    -    -                           ib.
Examples of the utility of signs, shown by the calculus of fractional exponents             - -                      191
Examples in the formation of powers of compound algebraic expressions   -    -    -    -    -   -    -    -..   -  ib




Contents.


xi


Examples in the extraction of the cube roots of numbers  "  192
Examples in the extraction of the cube roots of algebraic expressions   -   -    -- 193
Examples in the multiplication of radical expressions     ib.
Examples in the division of radical expressions  -  -  - 194
Examples in the calculus offractional exponents  -  -    195
Multiplication of fractional exponents ib.
Division of fractional exponents  196
Powers of powers     -   -    - 197
General Theory of Equations               -              ib.
The form which equations assume    -    -    -      -    198
Of the root of an equation  --                            ib.
Fundamental proposition of the theory  - --               ib.
Of the decomposition of an equation into simple factors  -  201
The number of divisors of the first degree, which any equation
can have   -    -    -    -    -   --             - 202
An equation composed of simple factors                    ib.
Formation of its coefficients -           -               ib*
How far an equation may have factors of any given degree  -  203
Of Elimination among Equations exceeding the First Degree  204
By substituting the value of one of the unknown quantities -  ib.
Rule for making the radical sign to disappear  -    - - 205
General formulas for equations having two unknown quantities
and how they may be reduced to equations having only one  206
Formula of elimination in two equations of the second degree  ib.
To determine whether the value of any one of the unknown
quantities satisfies, at the same time, the two equations proposed     -   -    -    -    -                      207
A common divisor of two equations leads to the elimination of
one of the unknown quantities  -                      ib.
How to proceed after obtaining the value of one of the unknown
quantities in the final equation in order to find that of the
other    -    -    -    -             -             208
Singular cases, in which the proposed equations are contradictory, or leave the question indeterminate  -  -   - 210
To eliminate one unknown quantity in any two equations  -  211
Euler's method of solving the above problem  -   -    - 212
Inconvenience of the successive elimination of the unknown
quantities when there are more than two equations and indication of the degree of the final equation  -  -  216




xii


Contents.


Qf Commensurable Roots, and the equal Roots of Jumerical
Equations     -    -   --                           217
Every equation, the coefficients of which are entire numbers,
that of the first term being 1, can only have for roots entire numbers or incommensurable numbers  -   -    -   ib.
Method of clearing an equation of fractions  -  -   -     ib.
Investigation of commensurable divisions of the first degree  - 220
How to obtain the equation, the roots of which are the differences between one of the roots of the proposed equation
and each of the others  -    --                      223
Of equal roots                                   -    -224
To form a general equation, which shall give all the differences
between the several roots combined two and two  '    227
Method of clearing an equation of any term whatever  -  - 229
To resolve equations into factors of the second and higher degrees..        -    -    -                  230
Of the Resolution of JNumerical Equations by Approximation  231
Principle on which the method of finding roots by approximation depends    -                                     ib.
Note-on the changes of the value of polynomials  -  -    232
To assign a number which shall render the first term greater
than the sum of all the others  -   -    -   -    - 234
Every equation denoted by an odd number has necessarily a real
root, with a sign contrary to that of its real term  -  -  237
Every equation of an even degree, the last term of which is negative, has at least two real roots, the one positive and the
other negative  -    -    -    -                      ib.
Determination of the limits of roots, example  -  -  -    ib.
Application to this example of Newton's method for approximating the roots of an equation  -  -  -- 238
How to determine the degree of the approximation obtained  239
Inconvenience of this method when the roots differ but little  241
Note-respecting equal roots      -    -    -   -    -     ib.
To prove the existence of real and unequal roots  -  -  -  ib.
Use of division of roots for facilitating the resolution of an equation when the coefficients are large  -  -           244
Method of approximation according to Lagrange  -  -   - 245
Of Proportion and Progression   -    -    -    -    -    248
Fundamental principles of proportion and equidifference  -  ib.
Of the changes which a proportion may undergo  -    -    250
Of progression by differences  -   -- 253




Contents.


I..
XlIl


To determine any term whatever of this progression  -  - 254
To determine the sum of the terms  -  -             -     ib.
Of progression by quotients -  -   --                 - 255
General term. Sum       -                           -    256
Progressions by quotients, the sum of which has a determinate
limit  -   -     -   -    -                          257
Manner of reducing all the terms of a progression by quotients
from the expression of the sum  -   -             - 258
Division of m by m - 1, continued to infinity  -  -  -   259
In what cases the quotient of this operation is converging and may
be taken for the approximate value of the fraction  m1-  260
Of diverging series    -      -       -. -ib
Examples in arithmetical progressions      -        -    262
Examples in geometrical progressions  - -- 263
Theory of Exponential Quantities and of Logarithms  -     ib.
Remarkable fact, that all numbers may be produced by means of
the powers of one    -       -    -      -   -    - 265
What is meant by the term logarithm, and the base of logarithms  266
Method of calculating a table of logarithms  -  -   -     ib.
Note-Long's method, and a table of decimal powers   - -268
The characteristic of logarithms           -             271
Of the logarithms of fractions     -    -    -    -   -   ib.
The arithmetical complement      -    -         -        273
To change a logarithm from one system to another  -      274
The logarithm of zero  -    -    -         -    -   -    275
The use of logarithms, in finding the numerical value of formulas ib.
Application of logarithms to the Rule of Three  *  -  - 276
The logarithms of numbers in progression by quotients form a
progression by differences   -    -             -     ib.
The use of logarithms in resolving equations  -   -   -   ib.
Application to finding the logarithms of literal expressions -  277
Application to finding the logarithms of numerical expressions  278
Calculation of numerical expressions by logarithms  -  -  ib.
Questions relating to the Interest of Money  -  -   -    279
Of simple interest and compound interest                  ib.
Of annuities  -                     -                    283
To compare the values of sums payable at different times  - 285
Questions relating to interest and annuities  -  -  -    285








ELEMENTS OF ALGEBRA.
Preliminary Remarks upon the Transition from alrithmetic to llgebra-Explanation and Use of Algebraic Signs.
1. IT must have been remarked in the Elementary Treatise of
Arithmetic, that there are many questions, the solution of which
is composed of two parts; the one having for its object to find
to which of the four fundamental rules the determination of the
unknown number by means of the numbers given belongs, and
the other the application of these rules.  The first part, independent of the manner of writing numbers, or of the system of
notation, consists entirely in the developement of the consequences
which result directly or indirectly from the enunciation, or from
the manner in which that which is enunciated connects the numbers given with the numbers required, that is to say, from  the
relations which it establishes between these numbers. If these
relations are not complicated, we can for the most part find by
simple reasoning the value of the unknown numbers. In order
to this it is necessary to analyze the conditions, which are involved in the relations enunciated, by reducing them to a course
of equivalent expressions, of which the last ought to be one of
the following; the unknown quantity equal to the sum or the dfference, or the product, or the quotient, of such and such magnitudes.
This will be rendered plainer by an example.
To divide a given number into two such parts, that the first shall
exceed the second by a given difference.
In order to this we would observe 1., that,
The greater part is equal to the less added to the given excess, and
that by consequence, if the less be known, by adding to it this
excess we have the greater; 2., that,
slg.                     1




2


Elements of dlgebra.


The greater added to the less forms the number to be divided.
Substituting in this last proposition, instead of the words, the
greater part, the equivalent expression given above, namely, the
less part added to the given excess, we find that
The less part, added to the given excess, added moreover to the
less part, forms the number to be divided.
But the language may be abridged thus,
Twice the less part, added to the given excess, forms the number
to be divided;
whence we infer, that,
Twice the less part is equal to the number to be divided diminished
by the given excess;
and that,
Once the less part is equal to half the difference between the number to be divided and the given excess.
Or, which is the same thing,
The less part is equal to half the number to be divided, diminished by half the given excess.
The proposed question then is resolved, since to obtain the
parts sought it is sufficient to perform operations purely arithmetical upon the given numbers.
If, for example, the number to be divided were 9, and the
excess of the greater above the less 5, the less part would be,
according to the above rule, equal to I less -, or 4, or 2; and
the greater, being composed of the less plus the excess 5, would
be equal to 7.
2. The reasoning, which is so simple in the above problem,
but which becomes very complicated in others, consists in general of a certain number of expressions, such as added to, diminished by, is equal to, &c. often repeated.  These expressions
relate to the operations by which the magnitudes, that enter into
the enunciation of the question, are connected among themselves,
and it is evident, that the expressions might be abridged by
representing each of them by a sign. This is done in the following manner.
To denote addition we use the sign +, which signifies plus.
For subtraction we use the sign -, which signifies minus.
For multiplication we use the sign X, which signifies multiplied by.
To denote that two quantities are to be divided one by the




Preliminary Remarks.


s


other, we place the second under the first with a straight line
between them; 4 signifies 5 divided by 4.
Lastly, to indicate that two quantities are equal, we place between them the sign =, which signifies equal.
These abbreviations, although very considerable, are still not
sufficient, for we are obliged often to repeat the number to be
divided, the number given, the less part, the number sought, &c. by
which the process is very much retarded.
With respect to given quantities, the expedient which first
offers itself is, to take for representing them determinate numbers, as in arithmetic; but this not being possible with respect to
the unknown quantities, the practice has been to substitute in
their stead a conventional sign, which varies as occasion requires. We have agreed to employ the letters of the alphabet,
generally using the last; as, in arithmetic, we put x for the fourth
term of a proportion, of which only the three first are known.
It is from the use of these several signs that we derive the
science of Algebra.
I now proceed by means of them   to consider the question
stated above (1). I shall represent the unknown quantity, or
the less number, by the letter x, for example, the number to be
divided and the given excess by the two numbers 9 and 5; the
greater number, which is sought, will be expressed by x + 5,
and the sum of the greater and less by x + 5 + x; we have
then
x + 5 + x=9;
but by writing 2 x for twice the quantity x, there will result
2x + 5 =9.
This expression shows that 5 must be added to the number
2 x to make 9, whence we conclude that
2 x =9 -   5,
or that                2x    4,
and that lastly         x = 4 = 2.
By comparing now the import of these abridged expressions,
which I have just given by means of the usutal signs, with the
process of simple reasoning, by which we are led to the solution,
we shall see that the one is only a translation of the other.
The number 2, the result of the preceding operations, will
answer only for the particular example which is selected, while
the course of reasoning considered by itself, by teaching us, that




4


Elements of./lgebra.


the less part is equal to half the number to be divided, minus half
the given excess, renders it evident, that the unknown number is
composed of the numbhers given, and furnishes a rule by the aid
of which we can resolve all the particular cases comprehended
in the question.
The superiority of this method consists in its having reference
to no one number in particular; the numbers given are used
throughout, without any change in. the language by which they
are expressed; whereas, by considering the numbers as determinate, we perform upon them, as we proceed, all the operations
which are represented, and when we have come to the result,
there is nothing to show how the number 2, to which we may
arrive by any number of different operations, has been formed
from the given numbers 9 and 5.
3. These inconveniences are avoided by using characters to
represent the number to be divided and the given excess, that
are independent of every particular value, and with which we
can therefore perform any calculation. The letters of the alphabet are well adapted to this purpose, and the proposed question
by means of them may be enunciated thus,
To divide a given number represented by a into two such parts
that the greater shall have with respect to the less a given excess
represented by b.
Denoting always the less by x;
The greater will be expressed by x + b;
Their sum, or the number to be divided, will be equal to
x + x + b, or 2x +6 b;
The first condition of the question then will give
2x +b =-a.
Now it is manifest that, if it is necessary to add to double of
x, or to 2 x, the quantity b in order to make the quantity a, it
will follow from this, that it is necessary to diminish a by b to
obtain 2 x, and that consequently 2 x = a - b.
a   b
We conclude then that half of 2 x or x  2 — -,.
This last result, being translated into ordinary language, by
substituting the words and phrases denoted by the letters and
signs which it contains, gives the rule found before, according to
which, in order to obtain the less of two parts sought we subtract




Preliminary Remarks.


from half of the number to be divided, or from  half ofthe given
b
excess, or 2.
Knowing the less part, we have the greater by adding to the
less the given excess. This remark is sufficient for effecting the
solution of the question proposed; but Algebra does more; it
furnishes a rule for calculating the greater part without the aid of
the less as follows;
a-   being the value of this, augmenting it by the excess b,
a   b             a b
we have for the greater part -   +- b. Now 2  - + b shows
a
that after having subtracted from 2 the half of b, it is necessary
to add to the remainder the whole of b, or two halves of b,
a                        b
which reduces itself to augmenting 2 by the half of b, or by b.
a   b              a    b
It is evident then that --  - b becomes 2 +; and by translating this expression we learn, that of the two parts sought the
greater is equal to half of the number to be divided plus half of
the given excess.
In the particular question which I first considered, the number to be divided was 9, the excess of one part above the other
5; in order to resolve it by the rules to which we have just
arrived, it will be necessary to perform upon the numbers 9 and 5,
the operations indicated upon a and b.
The half of 9 being 9 and that of 5 being s, we have for the
less part
9-. 5_ 4= 2,
and for the greater
5 -      14 = 7.
2 T 2    2- 
4. I have denoted in the above the less of the two parts by x,
and 1 have deduced from it the greater. If it were required to
find directly this last, it should be observed, that representing it
by x, the other will be x - b, since we pass from the greater to
the less by subtracting the excess of the first above the second;
the number to be divided will then be expressed by x + x - b,
or by 2 x - b, and we have consequently 2 x - b = a.
This result makes it evident that 2 x exceeds the quantity a




6


Elements of Algebra.


by the quantity b, and that consequently 2 x = a -- b.   By
taking the half of 2 x and of the quantity which is equal to it, we
obtain for the value of x
a    b
x   2+,
which gives the same rule as the above for determining the
greater of the two parts sought. 1 will not stop to deduce from
it the expression for the smaller.
The same relation between the numbers given and the numbers required may be enunciated in many different ways. That
which has led to the preceding result is deduced also from the
following enunciation:
Knowing the sum a of two numbers and their difference b, to find
each of those numbers; since, in other words, the number to be
divided is the sum of the two numbers sought, and their difference is the excess of the greater above the less. The change in
the terms of the enunciation being applied to the rules found
above, we have
The less of two numbers sought is equal to half of the sum
minus half of the difference.
The greater is equal to half of the sum plus half of the difference.
5. The following question is similar to the preceding, but a
little more complicated.
To divide a given number into three such parts, that the excess
of the mean above the least may be a given number, and the excess
of the greatest above the mean may be another given number.
For the sake of distinctness I will first give determinate values
to the known numbers.
I will suppose that the number to be divided is 230;
that the excess of the middle part above the least is 40; and
that of the greatest above the middle one is 60.
Denoting the least part by x,
the middle one will be the least plus 40, or x -+ 40, and the
greatest will be the middle one plus 60, or x + 40 + 60.
Now the three parts taken together must make the number to
be divided; whence,
x + x + 40 + x + 40 + 60 = 230.
If the given numbers be united in one expression and the unknown ones in another, x is found three times in the result, and
for the sake of conciseness we write
3x+ 140= 230.




Preliminary Remarks.


7


But since it is necessary to add 140 to triple of x to make 230,
it follows, that by taking 140 from 230 we have exactly the
triple of x, or
3 x -230-    140,
or                  3x =90,
whence it follows that
x = 90 - 30.
By adding to 30 the excess 40 of the middle part above the
least, we have 70 for the middle part.
By adding to 70 the excess 60 of the greatest above the middle part, we have 130 for the greatest.
6. If the known numbers were different from those which I
have used in the enunciation, we should still resolve the question
by following the course pursued in the preceding article, but we
should be obliged to repeat all the reasonings and all the operations, by which we have arrived at the number 30, because there
is nothing to show how this number is composed of 230, 40,
and 60.   To render the solution independent of the particular
values of numbers, and to show how the value of the unknown
quantity is fixed by means of the known quantities, I will enunciate the problem thus;
To divide a given number a into three such parts, that the excess
of the middle one above the least shall be a given number b, and the
excess of the greatest above the middle one shall be a given number c.
Designating as above by x the unknown quantity, and making
use of the common signs and the symbols a, b, c, which represent the known quantities in the question, the reasoning already
given will be repeated.
The least part = x
the middle part =x + b,
the greatest    = x + b + c.
and the sum of these three makes the number to be divided;
hence,
x   x + b -    + b + c = a.
This expression, which is so simple, may be still further
abridged; for since it appears that x enters three times into the
number to be divided and b twice, instead of x + x +- x, I shall
write 3 x, and instead of + b + b, I shall write + 2 b, and it will
become
3 x+ 2 b + c = a.




8


Elements of Algebra.


From this last expression it is evident, that it is necessary to
add to triple the number represented by x, double the number
represented by b, and also the number c, in order to make the
number a; it follows then, that if from the number a we take
double the number b and also the number c, we shall have exactly
the triple of x, or that
3 x = a   2b -c.
Now x being one third of three times x, we thence conclude
that
a-2b- c
3
It should be carefully observed, that having assigned no particular value to the numbers represented by a, b, c, the result to
which we have come is'equally indeterminate as to the value of
x; it shows merely what operations it is necessary to perform
upon these numbers, when a value is assigned to them, in order
thence to deduce the value of the unknown quantity.
a- 2b - c
In short, the expression -  --, to which x is equal, may
be reduced to common language by writing, instead of tie letters, the numbers which they represent, and instead of the signs,
the kind of operation which they indicate; it will then become,
as follows;
From the number to be divided, subtract double the excess of the
middle part above the least, and also the excess of the greatest above
the middle part, and take a third of the remainder.
If we apply this rule, we shall determine, by the simple operations of arithmetic, the least part. The number to be divided
being for example 230, one excess 40, and the other 60, if we
subtract as in the preceding article twice 40, or 80, and 60 from
230, there will remain 90, of which the third part is 30, as we
have found already.
If the number to be divided were 520, one excess 50, and the
other 120, we should subtract twice 50, or 100, and 120 from
520, and there would remain 300, a third of which or 100 would
be the smallest part. The others are found by adding 50 to 100,
which makes 150, and 120 more to this, which makes 270, so
that the parts sought would be
100, 150, 270,
and their sum would be 520, as the question requires.




Preliminary Remarks.


9


It is because the results in algebra are for the most part only
an indication of the operations to be performed upon numbers
in order to find others, that they are called in general formulas.
This question, although more complicated than that of article
1., may still be resolved by ordinary language, as may be seen
in the following table, where against each step is placed a translation of it into algebraic characters.
PROBLEM.
To divide a number into three such parts, that the excess of
the middle one above the least shall be a given number, and the
excess of the greatest above the middle one shall be another given
number.
SOLUTION.
By common language.          By algebraic characters.
Let the number to be divided be denoted by        a.
The excess of the middle part
above the least by       b.
The excess of the greatest
above the middle one by  c.
The least part being       x.
The middle part will be thee 
least, plus the excess of the  The middle part will be x + b.
mean above the least. 
Tile greatest part will be the)
middle one, plus the excess of
the greatest above the middle Thr i          b 
one. The three parts will to- The greatest will be x + b + c.
one. The three parts will together form the number proposed.
Whence the least part, plus'
the least part, plus the excess
of the middle one above the
least, plus also the least part,  Whence
plus the excess of the middle Kx + x + b + x + b +    c = a.
one above the least, plus the
excess of the greatest above
the middle one, will be equal
to the number to be divided. J.alg.             2




10


Elements of Algebra.


Whence three times the least
part, plus twice the excess of
the middle part   above  the
least, plus also the excess of 3x 4- 2 b + c = a.
the greatest above the middle
one, will be equal to the number to be divided.
Whence three times the least
part will be equal to the number to be divided, minus twice
the excess of the middle part 3x = a -2b —c.
above the  least, and  minus
also the excess of the greatest 
above the middle one.       J
Whence in fine, the least partl
will be equal to a third of
what remains after deducting
from the number to be divid- I     a-2b- c
ed twice the excess of the mid- x           ' 3
die part above the least, and
also the excess of the greatest
above the middle one.       J
7. The signs mentioned in article 2. are not the only ones
used in algebra.  New considerations will give rise to others,
as we proceed.   It must have been observed in article 2. that
the multiplication of x by 2, and in articles 5. and 6. that of x by
3 and that of b by 2, is denoted by merely writing the figures
before the letters x and b without any sign between them, and I
shall express it in this manner hereafter; so that a number
placed before a letter is to be considered as multiplied by the
number represented by that letter, 5 x, 5 a, &c. signify five times
3;
x, five times a, &c. 4 x or —, &c. signifies 3 of x, or three times
x divided by 4, &c.
In general, multiplication will be denoted by writing the factors in order one after the other without any sign between them,
whenever it can be done without confusion.
Thus the expressions a x, b c, &c. are equivalent to a X x,
b X c, &c., but we cannot omit the sign when numbers are concerned, for then 3 X 5, the value of which is 15, becomes 35. In
this case we often substitute a point in the place of the usual
sign, thus, 3. 5.




Equations.


11


Equations.
8. If the solution of the problems in articles 3. and 6. be examined with attention, it will be found to consist of two parts entirely
distinct from  each other. In the first place, we express by
means of algebraic characters the relations established by the
nature of the question between the known and unknown quantities, from which we infer the equality of two quantities among
themselves; for instance, in article 3. the quantities 2 x + b and a,
and in article 6. the quantities 3 x +2 b + c and a.
We afterwards deduce from this equality a series of consequences, which terminate in showing the unknown quantity x to
be equal to a number of known quantities connected together by
operations, that are familiar to us; this is the second part of the
solution.
These two parts are found in almost every problem which belongs to algebra. It is not easy, however, at present to give a
rule adapted to the first part, which has for its object to reduce
the conditions of the question to algebraic expressions. To be
able to do this well, it is necessary to become familiar with the
characters used in algebra, and to acquire a habit of analyzing
a problem in all its circumstances, whether expressed or implied.
But when we have once formed the two numbers, which the question supposes equal, there are regular steps for deducing from
this expression the value of the unknown quantity, which is the
object of the second part of the solution. Before treating of
these I shall explain the use of some terms which occur in algebra.
An equation is an expression of the equality of two quantities.
The quantities which are on one side of the sign= taken
together are called a member; an equation has two members.
That which is on the left is called the first member, and the
other the second.
In the equation 2 x + b = a, 2 x + b is the first member, and
a is the second member.
The quantities, which compose a member, when they are separated by the sign + or -, are called terms.
Thus, the first member of the equation 2 x - b = a contains
two terms, namely, 2 x and + b.




12


Elements of llgebra.


The equation 43 x + 7 = 8 x -12 has two terms in each of
its members, namely,
x3 and + 7 in the first,
8 x and -- 12 in the second.
Although I have taken at random, and to serve for an example merely, the equation - x + 7 = 8 x-12, it is to be considered, as also every other of which I shall speak hereafter, as
derived from a problem, of which we can always find the enunciation by translating the proposed equation into common language. This under consideration becomes,
To find a number x such, that by adding 7 to! x, the sum shall
be equal to 8 times x minus 12.
Also the equation ax + b c - cx = ac - b x, in which the
letters a, b, c, are considered as representing known quantities,
answers to the following question;
To find a number x such, that multiplying it by a given number a,
and adding the product of two given numbers b and c, and subtracting from this sum the product of a given number c by the number x,
we shall have a result equal to the product of the numbers a and c,
diminished by that of the numbers b and x.
It is by exercising one's self frequently in translating questions
from ordinary language into that of algebra, and from algebra
into ordinary language, that one becomes acquainted with this
science, the difficulty of which consists almost entirely in the
perfect understanding of the signs and the manner of using them.
To deduce from an equation the value of the unknown quantity, or to obtain this unknown quantity by itself in one member
and all the known quantities in the other, is called resolving the
equation.
As the different questions, which are solved by algebra, lead
to equations more or less compounded, it is usual to divide them
into several kinds or degrees. I shall begin with equations of the
first degree. Under this denomination are included those equations in which the unknown quantities are neither multiplied by
themselves nor into each other.
Of the resolution of Equations of the First Degree, having but one
unknown quantity.
9. WE have already seen that to resolve an equation is to
arrive at an expression, in which the unknown quantity alone in




Equations of the First Degree.


13


one member is equal to known quantities combined together by
operations which are easily performed. It follows then, that in
order to bring an equation to this state, it is necessary to free
the unknown quantity from known quantities with which it is
connected. Now the unknown quantity may be united to known
quantities in three ways;
1. By addition and subtraction, as in the equations,
x - 5    9 -x,
a + x    b - x.
2. By addition, subtraction, and multiplication, as in the
equations,
7x - 5 -12 + 4x,
ax - b    cx +4d.
3. Lastly, by addition, subtraction, multiplication, and division,
as in the equations,
5x      -11
ax         d    mx + 9,
+ cx -d        -+ -.
The unknown quantity is freed from additions and subtractions, where it is connected with known quantities, by collecting
together into one member all the terms in which it is found; and
for this purpose it is necessary to know how to transpose a term
from one member to the other.
10. For example, in the equation
7x-5 = 12-+4x,
it is necessary to transpose 4 x from tie second member to the
first, and the term - 5 from the first member to the second.
In order to this, it is obvious, that by cancelling + 4x in the
second member, we diminish it by the quantity 4x, and we must
make the same subtraction from the first member, to preserve
the equality of the two members; we write then- 4 x in the
first member, which becomes 7 x - 5 - 4 x, and we have
7 x-5 -4x - 12.
To cancel - 5 in the first member, is to omit the subtraction
of 5 units, or in other words, to augment this member by 5 units;
to preserve the equality then we must increase the second member by '5 units, or write + 5 in this member, which will make
it 12 + 5; we have then
7x-4       =- 12+ 5.




14


Elements of   ilgebra.


By performing the operations indicated there will result the
equation                 3 x= 17.
From this mode of reasoning, which may be applied to any
example whatever, it is evident, that to cancel in a member a
term  affected with the sign +, which of course augments this
member, it is necessary to subtract the term  from  the other
member, or to write it with the sign -; that on the contrary
when the term to be effaced has the sign minus, as it diminishes
the member to which it belongs, it is necessary to augment the
other member by the same term, or to write it with the sign +whence we obtain this general rule;
To transpose any term whatever of an equation from one member to the other, it is necessary to efface it in the member where it is
found, and to write it in the other with the contrary sign.
To put this rule in practice, we must bear in mind that the
first term of each member, when it is preceded by no sign, is
supposed to have the sign plus. Thus, in transposing the term
c x of the literal equation a x -b =  x + d from the second
member to the first, we have
ax - b -c x - d;
transposing also - b from the first member to the second, it
becomes
a x - c   - d +b.
11. By means of this rule, we can unite together in one of the
members all the terms containing the unknown quantity, and in
the other all the known quantities; and under this form the
member, in which the unknown quantity is found, may always
be decomposed into two factors, one of which shall contain only
known quantities, and the other shall be the unknown quantity by
itself.
This process suggests itself immediately, whenever the proposed equation is numerical and contains no fractions, because
then all the terms involving the unknown quantity may be reduced to one. If we have, for example,
10x + 7 x-2 x-25 + 7,
by performing the operations indicated in each member, we shall
have in succession
17x -2x = 32,
15 x   32;
and 15 x is resolved into two factors 15 and x; we have then




Equations of the First Degree.


15


the unknown factor x by dividing the number 32, which is equal
to the product 15 x by the given factor 15, thus,
__ 32
X       --  5
This resolution is effected in like manner in the literal equations
of the form
ax- bc;
because the term a x signifies the product of a by x; we hence
conclude, that
bc
_ e
a
Let there be the equation
a x - bx + cx- a - b c,
which contains three terms involving the unknown quantity.
Since a x, b x, c x, represent the products respectively of x by the
quantities a, b, and c, the expression a x - b x + c x translated
into ordinary language is rendered as follows,
From x taken first, so many times as there are units in a, subtract so many times x as there are units in b, and add to the result
the same quantity x, taken so many times as there are units in c.
It follows then on the whole, that the unknown quantity x is
taken so many times as there are units in the difference of the
numbers a and b, augmented by the number c, that is to say, so
many times as is denoted by the number a - b - c; the two
factors of the first member are therefore a - b - c and x; we
have then
a c -bec
a — b_  - c
From this reasoning, which may be applied to every other example, it is evident, that after collecting together into one member
the different terms containing the unknown quantity, the factor,
by which the unknown quantity is multiplied, is composed of all those
quantities by which it is separately multiplied, arranged with their
proper signs, and the unknown quantity is found by dividing all the
terms of the known member by the factor which is thus obtained.
According to this rule, the equation a x - 3 x = b c gives
be
X =
a -3
Also the equation x +- a x= c - d is reduced to
c-d
x 1 + a'




16


Elements of allgebra.


for it is necessary to observe that the letter x, taken singly,
must be regarded as multiplied by one. It is besides manifest,
that in x + a x, the unknown quantity x is contained once more
than in a x, and is consequently multiplied by 1 + a.
12. It is evident that if there be a factor, which is common to
all the terms of an equation, it may be dropped without destroying the equality of the two expressions, since it is merely dividing
by the same number all the parts of the two quantities, which are
by supposition equal to each other.
Let there be, for example, the equation
6abx- 9 b c d = 12bdx + 15 a b c.
I observe in the first place, that the numbers 6, 9, 12 and 15 are
divisible by 3, and by suppressing this factor, I merely take a
third part of all the quantities which compose the equation.
I have after this reduction,
2ab x — 3 b c d =4bdx+ 5 a b c.
I observe, moreover, that the letter b, combined in each term as
a multiplier, is a factor common to all the terms; by cancelling it
the equation becomes
2 ax - 3cd = 4dx + 5ac.
Applying the rules given in articles 10. and 11., I deduce successively
2 a x - 4 dx =5 a c + 3 c d,
5 ac + 3 c d
2a-4d 
13. I now proceed to equations, the terms of which have divisors. These may be solved by the preceding rules whenever
the unknown quantity does not enter into the denominators;
but it is often more simple to reduce all the terms to the same
denominator which may then be cancelled.
Let there be, for example, the equation
+ 4       -        _ 12   7.
Arithmetic furnishes rules for reducing fractions to the same
denominator, and for converting whole numbers into fractions of
a given kind. (drith. 79, 69.)  Let all the terms of the proposed equation be transformed by these rules into fractions of
the same denominator, beginning with the fractions, which are
2 x  4x   5 
3' 5' -     '




Equations of the First Degree.


17


I convert them by the first of the rules cited into the following;
5X 7 X 2x;       3 X 7 X4x       3 x 5 x5x
3 X 5 X 7        3 X 5 x 7       3X 5X7 
Since, for converting the whole numbers 4 and 12 into fractions,
nothing more is necessary than to multiply them by the common
denominator of the fractions, namely, 3 X 5 X 7; we have
3X 5X7X4,          3 X 5X 7 X 12.
By placing all these terms in order in the proposed equation, it
will become
5X7X2zx       3X5X7X4
3X5X7          3X 3x5X7
3X7 X 4 x     3 X 5 X7X12       3 X 5 X 5 x
3 3X 5 X 7     3 X 5 X 7        3 X 5 X 7'
The denominator may now be cancelled, since by doing it we
only multiply all the parts of the equation by this denominator
(Arith. 54), which does not destroy the equality of the members.
It will become then
5 X 7 X 2 x + 3 X 5 X 7 X 4
-3X7 X 4x+ 3 X 5 X7 X 12-3 X 5 X 5x,
or           70 x + 420 = 84 x + 1260 - 75 x,
an equation without a denominator from which we deduce the
value of x by the preceding rules.
It is evident from inspection, as also from the mere application of the arithmetical rules referred to, that in the above operation the numerators of each fraction must be multiplied by the product of the denominators of all the others, the whole numbers by the
product of all the denominators; then no account need be taken of
the common denominators of the fractions thus obtained.
The equation 70 x + 420 == 84 x + 1260 - 75 x becomes
successively
70 x + 75 x - 84 x = 1260 - 420,
61 x = 840,
-- 8-40 - 134 7
W --    wr*
The same process is applicable to literal equations, it being
observed, that it is necessary only to indicate the multiplications,
which are actually performed when numbers are concerned.
Let there be, for example, the equation
ae deuefoi
we deduce from it
ehX ax - behxc= bhxdx-+bexfg,


SIg.


3




18


Elements of Algebra.


a result which may be more simply expressed by placing the
factors of each product one after the other without any sign between them, according to the method given in article 7.; and by
arranging the letters in alphabetical order, they are more easily
read; it then becomes
aehx - be h = b dhx + b efg,
from which is deduced
ae h x- b dh x= bef          c e h,
and                       X l = befg + bceh
and                       x:-           -_1.
aeh — b ah
14. Although no general and exact rule can be given for
forming the equation of any question whatever; there is, notwithstanding, a precept of extensive use, which cannot fail to
lead to the proposed object. It is this,
To indicate by the aid of algebraic signs upon the known quantities represented either by numbers or letters, and upon the unknown
quantities represented always by letters, the same reasonings and
the same operations, which it would have been necessary to perform
in order to verify the values of the unknown quantities, had they
been known.
In making use of this precept, it is necessary, in the first place,
to determine with care what are the operations which are contained in the enunciation of the question, either directly or by
implication; but this is the very thing which constitutes the difficulty of putting a question into an equation.
The following examples are intended to illustrate the above
precept. I have taken the two first from among the questions
which are solved by arithmetic, in order to show the advantage of
the algebraic method.
(I.) Let there be two fountains, the first of which running for 2 h.
fills a certain vessel, and the second fills the same vessel by running
43h.; what time will be employed by both the fountains running
together in filling the vessel?
If the time were given, we should verify it by calculating the
quantities of water discharged by each fountain; and adding them
together we should be certain, that they would be equal to the
whole content of the vessel.
To form the equation we denote the unknown time by x, and
we indicate upon x the operations implied by the question; but
in order to render the solution independent of the given num



Equations of the First Degree.


19


bers, and at the same time to abridge the expression where fractions are concerned, we will represent them also by letters, a
being written instead of 2~h. and b instead of 34h.
This being supposed, by putting the capacity of the vessel
equal to unity, it is evident, that
The first fountain, which will fill it in a number of hours denoted  by a, will discharge into it in one hour a quantity of
water expressed by the fraction -, and that consequently, in a
1    x
number x of hours it will fuirnish the quantity x X a, or 
(Arith. 53).
The second fountain, which will fill the same vessel in a number of hours described by b, will discharge into it in one hour a
quantity of water expressed by the fraction L, and consequently in a number x of hours, it will furnish the quantity x X ),
or.
The whole quantity of water then furnished by the two fountains,
will be
+b;
and this quantity must be equal to the content of the vessel,
which was considered as unity; we have then the equation
a+~-1.
a   b
This equation reduced by the foregoing rules, becomes
b x - ax - ab,
ab
-b+a
The last formula gives this simple rule for resolving every
case of the proposed question.
Divide the product of the numbers, which denote the times employed by each fountain in filling the vessel, by the sum of these
numbers; the quotient expresses the time required by both the fountains running together to fill the vessel.
Applying this rule to the particular case under consideration,
we have
2 X3 33      x 1 =,7
2    3 = 5        + 145  20 ++ 30  50
2  4  2   4, f




20


Elements of algebra.


whence                x =   5=.
(2.) Let a be a number to be divided into three parts, having
among themselves the same ratios as the given numbers m, n, and p.
It is evident that the verification of the question would be as
follows;
denoting the 1st part by x, we have
m   n:: x: the 2d part =, (3rith. 116.)
m: p:: x: the 3d part — P;
m ~
the three parts added together must make the number to be
divided. We have then the equation
n x   px
x +     -+ - = a.
m     m
By reducing all the terms to the denominator m, it becomes
mx +- nx +- px =am;
and we deduce from this
a ma
x  -- _.
m +- n —p'
This result is nothing more nor less than an algebraic expression of the rule of Fellowship (.Irith. 124); for by regarding
the numbers m, n, p, as denoting the stocks of several persons
trading in company, m + n +-p is the whole stock, a the gain
to be divided, and the equation
m a
x =        n --- p
m ~ n + p
shows that a share is obtained by multiplying the corresponding
stock into the whole gain, and dividing the product by the sum of
the stocks; which reduced to a proportion, becomes
the whole stock: a particular stock:: the whole gain: the particular gain.
15. To form an equation from the following question, requires an attention to some things, which have not yet been considered.
A fisherman, to encourage his son, promises him 5 cents for each
throw of the net in which he shall take any fish, but the son, on the
other hand, is to remit to the father 3 cents for each unsuccessful
throw. After 12 throws thefather and the son settle their account,
and the former is found to owe the latter 28 cents. What was the
number of successful throws of the net.




Equations of the First Degree.


21


If we represent this number by x, the number of unsuccessful
ones will be 12 - x; and if these numbers were given, we should
verify them by multiplying 5 cents by the first, to obtain what
the father was bound to pay the son, and 3 cents by the second,
to find what the son engaged to return to the father. The first
number ought to exceed the second by 28 cents, which the
father owed the son.
We have for the first number x times 5 cents, or 5 x. With
respect to the second, there is some difficulty. How are we to
obtain tie product of 3 by 12 - x?  If instead of x we had a
given number, we should first perform the subtraction indicated,
and then multiply 3 by the remainder; but this cannot be done
at present, and we must endeavour to perform the multiplication
before the subtraction, or at least, to give the expression an
entire algebraic form, similar to that of equations that are readily
solved.
With a little attention we shall see, that by taking 12 times
the number three, we repeat the number 3 so many times too
much, as there are units in the number x, by which we ought
first to have diminished the multiplier 12, so that the true product will be 36 diminished by 3 taken x times, or 3 x,
or more simply            36 - 3 x.
This conclusion may be verified by giving to x a numerical
value. If, for example, x were equal to S, we should have 3 to
be taken 12 times —  8 times, and, if we neglect -  times, we
should make the result 8 times the number 3 too much; the
true product then will be
3 X 12 -    3 X 8 -  36 -  24 =  12.
This result agrees with that which would arise from first subtracting 8 from 12; for then
12 —8 =    4, and 3 X 4 = 12.
This being admitted, since the money due from the father to the
son is expressed by 5 x, and that which the son owes the father
by 36 - 3 x, the second number must be subtracted from the
first in order to obtain the remainder 28; but here is another
difficulty; how shall we subtract 36 - 3 x from 5 x, without
having first subtracted 3 x from 36?
We shall avoid this difficulty by observing, that if we neglect
the term — 3 x, and subtract from 5 x the entire number 36, we
shall have taken necessarily 3 x too much, since it is only what




22


Elements of Algebra.


remains after having diminished 36 by 3 x that is to be subtracted from 5 x; so that the difference 5 x - 36 ought to be augmented by 3 x in order to form the quantity that should remain
after having taken from 5 x the number denoted by 36 -3 x.
This quantity will then be
5 x - 36 + 3 x,
and we have the equation
5 x - 36 + 3 x = 28,
which becomes successively
8 x - 36 = 28,
8 x   28 + 36,
8 x   64,
x - 64    8.
There have been then 8 successful throws of the net and 4
unsuccessful ones.
Indeed 8 throws at 5 cents a throw give 40 cents,
4 throws at 3 cents a throw give  12
difference                28
as required by the conditions of the question.
To render the solution general, let a represent the sum given
by the father to the son for each successful throw of the net, and
b the sum returned by the son for each unsuccessful one, and c
the total number of throws, and d the sum received on the whole
by the son. If x be put equal to the number of successful throws,
c-x will express the number of unsuccessful ones; each throw
of the former kind being worth to the son a sum a, x throws
would be worth a X x, or a a, and the unsuccessful throws would
be worth to the father the sum b multiplied by the number c - x.
The reasoning by which we have found the parts of the product of 3 by 12- x, applies equally to the general case. If we
neglect in the first place - x in forming the product b c of b by
the whole of c, the sum b will be repeated x times too much, and
consequently the true product will be b c- b x.
In order to subtract this product from the sum a x, it is necessary to observe, as in the numerical example, that if we subtract
the whole of the quantity b c, we take the quantity b x too much,
by which the former ought to have been first diminished, and
that consequently the true remainder is not merely a x - b c, but
ax-b c + b x.




Quantities represented by Letters.


23


As this sum is equal to d, we have the equation
a x - b c + b x = d,
which gives
ax + 6 x =d+ bc,
ad+bc
xa=4- b.
As this general formula indicates what operations are to be
performed upon the numbers a, b, c, d, in order to obtain the
unknown quantity x, we may reduce it to a rule, or carefully
write instead of the letters a, b, c, d, the numbers given. This
last process is called substituting the values of the given quantities, or putting the formula into numbers. Applying here those of
the foregoing example, as we have
28 +3 X 12
x_      5+3; 
by performting the operations indicated, it becomes
28 + 36    64
x= S -8 8 —.
Methods for performing, as far as is possible, the Operations indicated upon Quantities that are represented by Letters.
16. FROM the preceding question it is evident, that in certain
cases a multiplication indicated upon the sum or difference of
several quantities is made to consist of several partial multiplications; and in art. 11. we have exactly the reverse, by resolving the quantity ax - b x + c x, which represents the result of
several multiplications, followed by additions and subtractions,
into the two factors a - b + c and x, which indicate only a single multiplication preceded by addition and subtraction.  The
reasoning pursued in these two circumstances will suggest rules
for performing, upon quantities represented by letters, operations
which are called algebraic multiplication and division, from the
analogy which they have with the corresponding operations of
arithmetic.
We have also by the same analogy two algebraic operations,
which bear the names of addition and subtraction, in which the
object is to unite several algebraic expressions in one, or to take
one expression from  another.  But these operations, like the
preceding, differ from  those of arithmetic in this, that their
results are, for the most part, only indications of the operations




24


Elements of Algebra.


to be performed; they present only a transformation of the
operations originally indicated into others, which produce the
same effect. All that is done, is either to simplify the expressions, or to give them a proper form for exhibiting the conditions
that are to be fulfilled.
In order to explain these operations, we give the name of
simple quantities to those which consist only of one term, as
- 2 a, -3 a b, &c.; binomials to those which consist of two, as
a + b, a —b, 5 a - 2x, &c.; trinomials to those which consist
of three terms; quadrinomials to those which consist of four terms,
and polynomials to those which consist of more than four terms.
It may be observed also, that we call polynomials campound quantities.
Of the Addition of Allgebraic Quantities.
17. THE addition of simple quantities is performed by writing
them one after the other, with the sign + between them; thus, a
added to b is expressed by a + b. But when it is proposed to
add together several algebraic expressions, we aim at the same
time to simplify the result by reducing it to as small a number
of terms as possible by uniting several of the terms in one.
This is done in articles 2. and 5. by reducing the quantity x + x
to 2x, and the quantity x + x + x to 3 x. It can take place
only with respect to quantities expressed by the same letters,
and which are for this reason called similar quantities. A literal
quantity that is repeated any number of times is regarded as a
unit; it is thus, that the quantities 2 a and 3 a considered as two
and three units of a particular kind, form when added 5 a or 5
units of the same kind. Also 4 a b and 5 a b make 9 a b.
In this case, the addition is performed with respect to the
figures which precede the literal quantity, and which show how
many times it is repeated. These figures are called coeficients.
The coefficient then is the multiplier of the quantity before which
it is placed, and it must be recollected, that when there is none
expressed, unity is understood; for 1 a is the same as a.
18. When it is proposed to unite any quantities whatever, as
4 a + 5 b and 2 c + 3 d,
the sum total ought evidently to be composed of all the parts
joined together; we must write then
4 a + 5 b + 2 c + 3 d.




Addition of Algebraic Quantities.


25


If we have on the contrary
4a + 5 b and    2c - 3d,
the sign -must be retained in the sum, to mark as subtractive
the quantity 3 d, which, as it is to be taken from 2 c, must necessarily diminish by so much the sum formed by uniting 2 c with the
first of the quantities proposed; we have then,
4a+ 5 b +2c- 3 d.
From these two examples it is evident, that in algebra the addition of polynomials is performed by writing in order, one after the
other, the quantities to be added with their proper signs, it being
observed that the terms which have no signs before them are considered as having the sign +.
The above operation is, properly speaking, only an indication
by which the union of two compound quantities is made to consist in the addition and subtraction of a certain number of simple
quantities; but, if the quantities to be added contained similar
terms, these terms might be united by performing the operation
upon their coefficients.
Let there be, for example, the quantities
4 a + 9 b -2 c,
2 a -3 c + 4 d,
7b+    c -     e;
the sum indicated would be, according to the rule just given,
4 a + 9 b -2 c + 2 a -3 c+4 d + 7 b + c -e.
But the terms 4 a + 2 a, being formed of similar quantities,
may be united in one sum equal to 6 a.
Also the terms + 9 b + 7 b give + 16 b.
The terms-2c and - 3c, being both subtractive, produce
on the whole the same effect as the subtraction of a quantity
equal to their sum, that is to say, as the subtraction of 5 c; and
as by virtue of the term + c, we have another part c to be added,
there will remain therefore to be subtracted only 4 c.
The sum of the expressions proposed, then, will be reduced to
6a + 16b-4c + 4d-e.
The last operation exhibited above, by which all similar
terms are united in one, whatever signs they have, is called
reduction. It is performed by taking the sum of similar quantities
having the sign +, that of similar quantities having the sign -,
and subtracting the less of the two sums from. the greater, and
giving to the remainder the sign of the greater..lg.                    4




26


Elements of dlgebra.


It is to be remarked, that reduction is applicable to all algebraic
operations.
The following examples of addition, with their answers, are
intended as an exercise for the learner.
(1.) To add the quantities
7m + 3 — 14p+ 17r
3a + 9n —11m +        2r
5p -   4m -- 8n
11n - 2b -m —r+-s.
Answer, 7m+3n-14p+17r+3a+9n-1           lm+2r+5p-4m+8n
+ 11 n- 2b —m-r +.s.
By making the reduction, this quantity becomes
- 9m- +31 n-9p + 18r + 3 a — 2 b +s,
or       31n-      9m — 9p+18r+-3 a —2b + s,
by beginning with the term having the sign +.
(2.) To add the quantities
11bc+ 4-ad-8 ac+ 5cd
8ac+- 7bc -2ad- 4mn
2 cd -  3 ab+5 ac + an
9 an -2bc-2ad + 5 cd.
11b c + 4ad- 8ac+ 5 c d- 8 ac+ 7 bc —2ad             4 m n
2cd-3ab - 5ac+ an + 9an              2bc —2ad +5cd.
By reducing this quantity it becomes
16 b c   5 a c  12 c d +4mn —3 a b + 10 an.
Of the Subtraction of Algebraic Quantities.
20. THE subtraction of single quantities, according to established usage, is represented by placing the sign - between the
quantity to be subtracted, and that from which it is to be taken;
b subtracted from a is written a - b.
When the quantities are similar, the subtraction is performed
directly by means of the coefficients.
If 3 a be subtracted from 5 a, we have for a remainder 2 a.
With regard to the subtraction of polynomials, it is necessary
to distinguish two cases.
(1.) If the terms of the quantity to be subtracted have each the
sign +, we must clearly give to each the sign -, since it is
required to deduct successively all the parts of the quantity to be
subtracted.


i
i
I
i
i
i
i
I
i
I
I




Multiplication of Algebraic Quantities.


27


If for example, from 5 a - 9 b + 2 c we would take
2 d + 3 e + 4f,
we must write 5 a - 9  + 2 c - 2 d- 3 e -4f.
(2.) If any of the terms of the quantity to be subtracted have
the sign -, we must give them  the sign plus. Indeed, if from
the quantity a we would take b- c, and should first write a- b,
we should thus diminish a by the whole quantity b; but the subtraction ought to have been performed after having first diminished b by the quantity c; we have taken therefore this last
quantity too much, and it is necessary to restore it with the sign 4-,
which gives for the true result a - b + c.
This reasoning, which may be applied to all similar cases, shows
that the sign - of c must be changed into the sign +-; and by
connecting this result with the preceding, we conclude that the
subtraction of algebraic quantities is performed by writing them in
order after the quantities, from which they are to be taken, having
first changed the signs + into - and the signs - into -.
After this rule has been applied, the quantities are to be reduced when they will admit of it, according to the precept given in article 19., as may be seen in the following examples;
(1.) To subtract from  17 a+ 2 m -   9b - 4c+23d
the quantity            51 a- 27 b -- 11 c - 4d.
Result              17 a + 2 m-   9 b -4 c + 23d
-51 a + 27b-        l c+4d.
When reduced it becomes
- 34a + 2 m + 18b-       15 c + 27 d,
or rather        2 m -34 a    18 b -  15 c + 27 d.
(2.) To subtract from    5 a c - 8ab + 9 b c - 4 a m
the quantity              8 am - 2 ab + 11 a c- 7 cd.
Result                 5 a c - 8 a b - 9 b c - 4a m
-8 am+ 2 a b-11ac + 7cd.
Reduced it becomes
-6 a c - 6a b + 9 b c- 12 a m + 7 cd,
or                 9b c- 6 ac -6 a b- 12 am - 7 cd.
Of the Multiplication of Algebraic Quantities.
21. So far as letters are considered as expressing the numerical values of the quantities for which they stand, multiplication
in algebra is to be regarded like multiplication in arithmetic.




28


Elements of Algebra.


(1rith. 21, 66.)  Thus, to multiply a by b is to compound with
the quantity represented by a another quantity, in the same manner
as the quantity represented by b is with unity.
We have already explained in articles 2. and 7. the signs used
to indicate multiplication; and the product of a by b is expressed by a X b, or by a. b, or lastly by a b.
We have often occasion to express several successive multiplications, as that of a by b, and that of the product a b by c, also
that of this last product by d, and so on. In this case, it is evident, that the last result is a number having for factors the numbers a, b, c, d, (arith. 22); and to give a general expression of
this method we indicate the product by writing the factors composing it in order, one after the other, without any sign between them;
we have accordingly the expression a b c d.
Reciprocally every expression, such as a b c d formed of several
letters written in order one after the other, designates always the
product of the numbers represented by these letters.
I have already availed myself of this method, in which the numerical coefficients are also included, since they are evidently factors of the quantity proposed. Indeed 15 a b c d, designating the
quantity a b c d taken fifteen times, expresses likewise the product
of the five factors 15, a, b, c, d.
It follows from this, that in order to indicate the multiplication
of several simple quantities, such as 4 a b c, 5 d ef, 3 m n, it is
necessary to write the quantities in order, one after the other, without any sign between them, and it becomes
4ab c 5 d ef3 mn;
but since, as is shown in arithmetic, (art. 82.) the order of the factors of a product may be changed at pleasure without altering the
value of this product, we may avail ourselves of this principle, to
bring together the numerical factors, the multiplication of which is
performed by the rules of arithmetic; to express then this product,
as indicated in the order 4. 5. 3 a b c d efm n, we multiply together the numbers 4, 5, 3, which give simply
60 a b c d efm n.*
* As the use of algebraic symbols abridges very much the demonstration of this proposition, I have thought it proper to suggest here
a method by these symbols.
If we write the product a b c d ef as follows, a b c X de X f, and




Multiplication of Algebraic Quantities.


29


23. The expression of the product may be much abridged
when it contains equal factors. Instead of writing several times
in order, the letter which represents one of the factors, it need
be written only once with a number annexed, showing how
many times it ought to have been written as a factor; but as
this number indicates successive multiplications, it ought to be
carefully distinguished from a coefficient, which indicates only
additions. For this reason, it is placed on the right of the letter
and a little higher up, while a coefficient is always placed on the
left and on the same line.
Agreeably to this method, the product of a by a, which would
be indicated, according to article 21., by a a, becomes a2. The
2 raised, denotes that the number, designated by the letter a, is
twice a factor in the expression, to which it belongs. It ought
not to be confounded with 2 a, which is only an abbreviation of
a + a. To render evident the error, which would arise from
mistaking one for the other, it is sufficient to substitute numbers
instead of the letters. If we have for example a = 5, 2 a would
become 2. 5= 10, and a2    a X a = 5. 5   25.
Extending this method we should denote a product in which a
is three times a factor by writing a3 instead of a a a; also a5
represents a product in which a is five times a facter, and is
equivalent to a a a a a.
24. The products formed in this manner by the successive
multiplications of a quantity, are called in general powers of that
quantity.
The quantity itself, as a, is called the first power.
The quantity multiplied by itself, as a a, or a2, is the second
power. It is called also the square.
The quantity multiplied by itself twice in succession, as a a a,
or a3, is the third power, and is called also the cube.*
change the order of the factors of the product to e d instead of de,
(Arith. 22,) it becomes a b c X e d Xf, or a b ce df. It is evident
that we may, by analyzing the product differently, produce any
change which we wish in the order of the factors of the product in
question.
* The denominations square and cube refer to geometrical considerations. They interrupt the uniformity in the nomenclature of
products formed by equal factors, and are very improper in algebra.
But they are frequently used for the sake of conciseness.




30


Elements of.lgebra.


In general, any power whatever is designated by the number
of equal factors from which it is formed; a5 or a a a a a is the
'fifth power of a.
I take the number 3 to illustrate these denominations, and I
have
1st. power                      3
2d.                          3.3=     9
3d.                       3. 3. 3    9. 3 =27
4th.                   3..3.3.=27.3= 81
5th.                3. 3.3.3.3 = 81.3 - 243
&c.
The number which denotes the power of any quantity is called
the exponent of this quantity.
When the exponent is equal to unity, it is not written; thus a
is the same as a1.
It is evident then, that to find the power of any number, it is
necessary to multiply this number by itself as many times less one,
as there are units in the exponent of the power.
25. As the exponent denotes the number of equal factors,
which form the expression of which it is a part, and as the product of two quantities must have each of these quantities as factors; it follows that the expression a5, in which a is five times a
factor, multiplied by a3, in which a is three times a factor, ought
to give a product in which a is eight times a factor, and consequently expressed by a8, and that in general the product of two
powers of the same number ought to have for an exponent the sum
of those of the multiplicand and multiplier.
26. It follows from this, that when two simple quantities have
common letters, we may abridge the expression of the product of
these quantities by adding together the exponents of such letters of
the multiplicand and multiplier.
For example, the expression of the product of the quantities
a2 63 c and a4 b5 c2 d, which would be a2 b3 c a4 b5 c2 d, by the
foregoing rule, art. 21., is abridged by collecting together the
factors designated by the same letter, and
a2 a4 63 5 c C2 d,
becomes                a6 b8 C3 d,
by writing             a6 instead of a2 a4
b8 instead of b3 b5
C3 instead of c c2 or of cl c2.




.Multiplication of algebraic Quantities.


31


27. As we distinguish powers by the number of equal factors
from which they are formed, so also we denote any products by
the number of simple factors or firsts which produce them; and
I shall give to these expressions the name of degrees. The product a2 b3 c,,for example, will be of the sixth degree, because it
contains six simple factors, viz; 2 factors a, 3 factors b, and 1
factor c. It is evident that the factors a, b, and c, here regarded
as firsts, are not so, except with respect to algebra, which does
not permit us to decompose them; they may, notwithstanding,
represent compound numbers, but we here speak of them only
with respect to their general import.*
The coefficients expressed in numbers are not considered in
estimating the degree of algebraic quantities; we have regard only
to the letters.
It is evident (21, 25,) that when we multiply two simple quantities the one by the other, the number which marks the degree
of the product is the sum of those which mark tile degree of each
of the simple quantities.
28. The multiplication of compound quantities consists in that
of simple quantities, each term of the multiplicand and multiplier
being considered by itself; as in arithmetic we perform the
operation upon each figure of the numbers which we propose
to multiply.  (.rith. 33.)  The particular products added together make up the whole product.     But algebra presents a
circumstance which is not found in, numbers. These have no
negative terms or parts to be subtracted, the units, tens, hundreds, &c. of which they consist, are always considered as
added together, and it is very evident, that the whole product
must be composed of the sum of the products of each part of
the multiplicand by each part of the multiplier.
* We apply the term dimensions, generally, to what I have here
called degrees, in conformity to the analogy already pointed out in
the note to page 29. This example sufficiently proves the absurdity
of the ancient nomenclature, borrowed from the circumstance, that
the products of 2 and 3 factors, measure respectively the areas of the
surfaces and the bulks of bodies, the former of which have two and
the latter three dimensions; but beyond this limit the correspondence between the algebraic expressions and geometrical figures fails,
as extension can have only three dimensions.




32


Elements of llgebra.


The same is true of literal expressions when all the terms are
connected together by the sign +.
The product of            a + b
multiplied by                c
is                           ac + b c,
and is obtained by multiplying each part of the multiplicand by
the multiplier, and adding together the two particular products
a c and b c. The operation is the same when the multiplicand
contains more than two parts.
If the multiplier is composed of several terms, it is manifest
that the product is made up of the sum of the products of the
multiplicand by each term of the multiplier.
The product of          a + b
multiplied by              c + d.i                         a c + bc
ls                        - +-ad+b d
for by multiplying first a + b by c, we obtain a c - b c, then by
multiplying a -b by the second term d of the multiplier, we
have a d + b d, and the sum of the two results gives
c - b c + ad - bd
for the whole.
29. When the multiplicand contains parts to be subtracted,
the products of these parts by the multiplier must be taken from
the others, or in other words, have the sign - prefixed to them.
For example,
the product of           a -   b
multiplied by             c
is                      ac-bc;
for each time that we take the entire quantity a, which was to
have been diminished by b before the multiplication, we take the
quantity b too much; the product a c therefore, in which the
whole of a is taken as many times as is denoted by the number
c, exceeds the product sought by the quantity b, taken as many
times as is denoted by the number c, that is by the product b c;
we ought then to subtract b c from a c, which gives, as above,
a c - b c.
The same reasoning will apply to each of the parts of the multiplicand, that are to be subtracted, whatever may be the number and whatever may be that of the terms of the multiplier, pro



Multiplication of Algebraic Quantities.


33


vided they all have the sign  -.  Recollecting that the terms
which have no sign are considered as having the sign —, we see
by the examples, that the terms of the multiplicand affected by
the sign + give a product affected by the sign +, while those
which have the sign -  give one having the sign -.  It follows
from this, that when the multiplier has the sign +, the product has
the same sign as the corresponding part of the multiplicand.
30. The contrary takes place when the multiplier contains
parts to be subtracted; the products arising from these parts
must be put down with a sign, contrary to that which they
would have had by the above rule. This may be shown by the
following example.
Let the multiplicand be    a - b
and the multiplier         c -d
the product will be       a acd b 
for the product of the multiplicand, by the first term of the multiplier, will be by the last example a c - b c; but by taking the
whole of c for the multiplier instead of c diminished by d, we
take the quantity a -b so many times too much as is denoted
by the number d; so that the product a c - b c exceeds that
sought by the product of a - b by d. Now this last is, by what
has been said, ad —bd, and in order to subtract it fiom the
first it is necessary to change the signs (20). We have then
a c - b c- a d + b d for the result required.
31. Agreeably to the above examples, we conclude, that the
multiplication of polynomials is performed by multiplying successively, according to the rules given for simple quantities (21-26),
all the terms of the multiplicand by each term of the multiplier,
and by observing tiat each particular product must have the same
sign, as the corresponding part of the multiplicand, when the multiplier has the sign +, and the contrary sign when the individual
multiplier has the sign -.
If we develope the different cases of this last rule, we shall find,
(1.) That a term having the sign -+, multiplied by a term having the sign -, gives a product having the sign +;
(2.) That a term having the sign -, multiplied by a term having the sign —, gives a product which has the sign-;
(3.) That a term  having the sign +, multiplied by a term having the sign -, gives a product which has the sign -;
Jllg.                 5




34


Elements of Algebra.


(4.) That a term having the sign -, multiplied by a term having the sign -, gives a product which has the sign —.
It is evident from this table, that when the multiplicand and
multiplier have the same sign, the product has the sign -+, and
that uhen they have different signs, the product has the sign -.
To facilitate the practice of the multiplication of polynomials
I have subjoined a recapitulation of the rules to be observed.
(1.) To determine the sign of each particular product according
to the rule just given; this is the rule for the signs.
(2.) Toform the coefficients by taking the product of those of
each multiplicand and multiplier (22); this is the rule for the coefficients.
(3.) To write in order, one after another, the different letters
contained in each multiplicand and multiplier (21); this is the
rule for the letters.
(4.) To give to the letters, common to the multiplicand and multiplier, an exponent equal to the sum of the exponents of these letters in the multiplicand and multiplier (25); this is the rule for
the exponents.
32. The example below will illustrate all these rules
Multiplicand   5 a4  -  2 a3 b + 4 a2 62
Multiplier       a3  -  4 a2 b + 2 b3
5a7     -2a6b +      4a5b2
Several   _7
pSeveral  i   20 a6 b + 8 a5 b2 — 16 a4 63
products. ( + 10a4b 3- 4 a3 b4 +  8 a2 65
Result reduced       5a7-22a6b+ 12a5b2-6a463 —44a34+8ab5.
The first line of the several products contains those of all the
terms of the multiplicand by the first term a3 of the multiplier;
this term  being considered as having the sign +, the products
which it gives have the same signs as the corresponding terms
of the multiplicand (31).
The first term 5 a4 of the multiplicand having the sign plus,
we do not write that of the first term of the product, which would
be +; the coefficient 5 of a4 being multiplied by the coefficient
1 of a3, gives 5 for the coefficient of this product; the sum  of
the two exponents of the letter a is 4 + 3, or 7, the first term of
the product then is 5 a7.
The second term -   2 a3 b of the multiplicand having the sign
-, the product has the sign minus; the coefficient 2 of a3 b mul



AMultiplication of Algebraic Quantities.


35


tiplied by the coefficient 1 of a3, gives 2 for the coefficient of the
product; the exponent of the letter a, common to the two terms
which we multiply, is 3 + 3, or 6, and we write after it the
letter b, which is found only in the multiplicand. The second
term of the product then is - 2 a6 b.
The third term -+4 a2 2 gives a product affected with the
sign —, and by the rules applied to the two preceding terms, we
find it to be + 4 a5 62.
The second line contains the products of all the terms of the
multiplicand by the second term- 4 a2 b of the multiplier. This
last having the sign -, all the products which it gives must
have the signs contrary to those of the corresponding terms of
the multiplicand; the coefficients, the letters, and the exponents
are determined as in the preceding line.
The third line contains the products of all the terms of the
multiplicand by the third tern  + — 2 b of the multiplier. This
term having the sign  -, all the products which it gives have the
same sign as the corresponding terms of the multiplicand.
After having formed all the several products which compose
the whole product, we examine carefully this last, to see whether
it does not contain similar terms; if it does, we reduce them
according to the rule (19), observing that two terms are similar,
which consist of the same letters under the same exponents. In
this example there are three reductions, viz;
-   2a6b and -20a6, which give - 22 a6b;
+   4a562and+    8 a5b2, which give + 12 a5 62;
-      16 a4 b3 and + 10 a4 b3, which give -  6 a4 b3.
These reductions being made, we have for the result the last
line of the example.
See another example to exercise the learner, which is easily
performed after what has been said.




Multiplicand  5a46+7a%- 5       +23b2d4-176c'd2-9abcdrn2
Multiplier 1 I163-8c3+,5abc-2bdm
Several -40a0b9c3-56a36b3C3+ 1200ct-1I 8462c3d4+1I36bc6d2+72abc~d~rn2
products   +25ab3C + 3 5a4b4c-75 AbC2+ I115a63cd 4-85ab2C4 d2 —45a262c'd n2
-1 Oa46dm14a b4dm+3Oa bcdm-46b d m34b cd~m            1 Sab2cdm
Result    55a 65+77a3b6-I 40abc+2531Aa4- 8764cOd2-9ab cdm2-40a462c-56a~bc
reduced + 120a 5C41_ 84b2 C3d4 +1 36b6c d2+72abC4dM2+ 35a4 b4C-7 5a6bc2 + I11 5ab3cd4-S 5al2C4 d2
(eucd -45a2b 2cdm2-i1oa4b 3dm     1 4a3b4dm+3Oa5bcdmn-46b3d5rn+34b 2C3d3Mn+ l8ab2cd2 M3.




Multiplication of algebraic Quantities.


37


33. From the manner of proceeding in multiplication, it is
evident that if all the terms of the multiplicand are of the same
degree (27), and those of the multiplier are also of the same
degree, all the terms of the product will be of a degree denoted
by the sum of the numbers, which mark the degree of the terms
of each of the factors.
In the first example, the multiplicand is of the fourth degree,
the multiplier of the third; and the product is of the seventh.
In the second example, the multiplicand is of the sixth degree,
the multiplier of the third; and the product is of the ninth.
Expressions of the kind just referred to, the terms of which
are all of the same degree, are called homogeneous expressions.
The above remark, with respect to their products, may serve to
prevent occasional errors, which one may commit by forgetting
soine of the factors in the several parts of the multiplication.
34. Algebraic operations performed upon literal quantities, as
they permit us to see how the several parts of the quantities
concur to form the results, often make known some general properties of numbers independent of every system of notation.
The multiplications that follow, lead to conclusions of the greatest importance, and of frequent use in the subsequent parts of
this work.
a+b                      a +b
a -b                     a + b
a  -+ ab                 a2 + ab
-ab — b2               -+ab+ b2
aa2  b2                  a3  - 2 ab + b2
a   - 2 ab  +  ba
a+b
as + 2 aa b + a b2
+ a2 b + 2 ab + b3
a3 + 3a2 b + 3 ab2 + b3
It appears from the first of these products, that the quantity
a + b, multiplied by a-b, gives a2 -b2; whence it is evident
that, if we multiply the sum of two numbers by their difference, the
product will be the difference of the squares of these numbers.
If we take, for example, the sum 11 of the numbers 7 and 4,




38                  Elements of dAgebra.
and multiply it by the difference 3 of these numbers, the product
3 X 11, or 33, will be equal to the difference between 49, the
square of 7, and 16, the square of 4.
By the second example, in which a + b is twice a factor, we
learn; that the second power, or the square of a quantity composed
of two parts a and b contains the square of the first part, plus
double the product of the first part by the second, plus the square
of the second.
The third example, in which we have multiplied the second
power of a + b by the first, shows; that, the third power or cube
of a quantity composed of two parts contains the cube of the first,
plus three times the square of the first multiplied by the second,
plus three times the first multiplied by the square of the second
plus the cube of the second.
35. As we have often occasion to decompose a quantity into
its factors, and as the algebraic operations are dispensed with,
when it can be done, in order to exhibit the formation of the
quantities to be considered, as distinctly as possible, it is necessary to fix upon some signs proper to indicate multiplication
between complex quantities.
We use indeed the marks of a parenthesis to comprehend the
factors of a product. The expression
(5 a4 — 3 a2   + b4) (4 a b2 -a c2 + d) (b2 - c2),
for example, indicates the product of the compound quantities
5a4 -3 a2 b2 +b4, 4ab2 -ac2 + d, and b2 -c2.
Bars were used formerly by some authors placed over the factors thus,
5 a4 -3 a2 b2 +b4 X 4 a b2 - a c      d3 X b2 - c2;
but as these may happen to be too long or too short, they are
liable to more uncertainty than the marks of a parenthesis,
which can never admit of any doubt with respect to the quantity
belonging to each factor. They have accordingly been preferred.
Of the Division of.lgebraic Quantities.
36. ALGEBRAIC division, like division in arithmetic, is to be
regarded as an operation designed to discover one of the factors
of a given product, when the other is known. According to this
definition, the quotient multiplied by the divisor must produce
anew the dividend.




Division of Jlgebraic Quantities.


39


By applying what is here said to simple quantities we shall
see by art. 21., that the dividend is formed from the factors of the
divisor and those of the quotient; whence, by suppressing in the
dividend all the factors which compose the divisor, the result will be
the quotient sought.
Let there be, for example, the simple quantity 72 a5 b3 c2 d to
be divided by the simple quantity 9 a3 b c2; according to the
rule above given, we must suppress in the first of these quantities
the factors of the second, which are respectively
9, a3, b, and c2.
It is necessary then, in order that the division may be performed, that these factors should be in the dividend. Taking them
in order, we see in the first place that the coefficient 9 of the divisor, ought to be a factor of the coefficient 72 of the dividend, or
that 9 ought to divide 72 without a remainder. This is in fact the
case, since 72 = 9 X 8. By suppressing then the factor 9, there
will remain the factor 8 for the coefficient of the quotient.
It follows moreover, from  the rules of multiplication (25), that
the exponent 5 of the letter a in the dividend, is the sum of the
exponents belonging to the divisors and quotient; this last exponent therefore will be the difference between the two others, or
5 - 3 -= 2. Thus the letter a has in the quotient the exponent
2. For the same reason, the letter b has in the quotient an exponent equal to 3 -  1, or 2. The factor c2 being common to the
dividend and divisor is to be suppressed, and we have
8 a2 b2 d
for the quotient required.
The same will apply to every other case; we conclude then,
that, in order to effect the division of simple quantities, the course
to be pursued is,
To divide the coefficient of the dividend by that of the divisor;
To suppress in the dividend the letters which are common to it
and the divisor, when they have the same exponent; and when the
exponent is not the same, to subtract the exponent of the divisor
from that of the dividend, the remainder being the exponent to be
affixed to the letter in the quotient;
To write in the quotient the letters of the dividend which are not
in the divisor.
37. If we apply the rule now given for obtaining the exponent of the letters of the quotient, to a letter which has the same




40


Elements of 3lgebra.


exponent in the dividend and divisor, we shall find zero to be
the exponent which it ought to have in the quotient; a3 divided
by a3, for example, gives a~. To understand what is the import of such an expression, it is necessary to go back to its origin and to consider, that if we represent the quotient arising
from the division of a quantity by itself, it ought to answer to
unity, which expresses how many times any quantity is contained in itself. It follows from this, that the expression a~ is a symbol equivalent to unity, and may consequently be represented by 1.
We may then omit writing the letters which have zero for their
exponent, since each of them signifies nothing but unity. Thus
a3 b c2 divided by a2 b c2, gives a' b~ co, which becomes a as is
very evident by suppressing the common factors of the dividend
and divisor.
We see by this, that the proposition, every quantity which has
zero for its exponent, is equal to 1, is nothing, properly speaking,
but the explanation of a conclusion to which we are brought by
the common manner of writing the powers of quantities by exponents.
In order that the division may be performed, it is necessary,
1. that the divisor should have no letter which is not found in
the dividend; 2. that the exponent of any letter in the divisor
should not exceed that of the same letter in the dividend; 3.
that the coefficient of the divisor should exactly divide that of
the dividend.
38. When these conditions do not exist, the division can only
be indicated in the manner pointed out in the 2d article.  Still
we should endeavour to simplify the fraction  by suppressing
such factors, as are common to the dividend and divisor, if there
are any such; for (Arith. 57) it is manifest, that the theory of
arithmetical fractions rests upon principles which are independent of every particular value of their terms, and which would
apply to fractions represented by letters, as well as to those which
are represented by numbers.
According to these principles, we in the first place suppress the
numerical factors common to the dividend and divisor, and then
the letters which are common to the dividend and divisor, and which
have the same exponent in each.  When the exponent is [not the
same in each, we subtract the less from  the greater, and affix the
remainder, as the exponent to the letter, which is written only in
that term of the fraction which has the highest exponent.




Division of Algebraic Quantities.


41


The following example will illustrate this rule.
Let 48 a3 b5 c2 d be divided by 64 a3 b3 c4 e; the quotient can
only be indicated in the form of a fraction
48 a3 b5 c2 d
64 a3 b3 c4 e
But the coefficients 48 and 64 being divisible by 16, by suppressing this common factor, the coefficient of the numerator becomes 3, and that of the denominator 4. The letter a having
the same exponent 3 in the two terms of the fraction, it follows
that a3 is a factor common to the dividend and divisor, and may
consequently be suppressed.
To find the number of factors b common to the two terms of
the fraction, we must divide the higher b5 by the lower bV according to the rule above given, and the quotient b2 shows, that
b = b3 X b2. Suppressing then the common factor b3, there will
remain in the numerator the factor b2.
With respect to the letter c, the higher factor being c4 of the
denominator, if we divide it by c2 we shall decompose it into
c2 X c2; and by suppressing the factor c2 common to the two
terms, this letter disappears from the numerator, but will remain
in the denominator with the exponent 2.
Finally, the letters d and e will remain in their respective
places, since, in the state in which they are, they indicate no
factor common to both.
By these several operations the proposed fraction is reduced to
3 b2 d
4 c2 e 
and it is the most simple expression of the quotient, except we
give numerical values to the letters; in which case it might be
further reduced by cancelling the common factors as before.
39. It ought to be remarked, that, if all the factors of the dividend enter into the divisor, which besides contains others peculiar to it, it is necessary after suppressing the former to put unity
in the place of the dividend, as the numerator of the fraction. In
this case indeed we may suppress all the terms of the 'numerator, or, in other words, divide the two terms of the fraction by
the numerator; but this being divided by itself must give unity for
the quotient, which becomes the new numerator.
Suppose, for example, the fraction
4 a2 b c
12 a2 b3 cd;
ilg'.          6




42


Elements of J.lgebra.


the factors 12, a,2 b3, and c may be divided respectively by the
factors 4, a2, b, and c, or we may divide the two terms of the
fraction by the numerator 4 a2 b c. Now the quantity 4 a2 b c,
divided by itself, gives 1 for the quotient, and the quantity
12 a2 b3 c d, divided by the first, gives by the above rules 3 b2 d;
the new fraction then is
1
3 b2 d'
40. It follows from the rules of multiplication, that when a
compound quantity is multiplied by a simple quantity, this last becomes a factor common to all the terms of the former. We may
make use of this observation to simplify fractions of which the
numerator and denominator are polynomials, having factors that are
common to all their terms.
Let there be the expression
6 a4  3a2 bc+ 12 a2 c2
9 a2 b - 15 a2 c+24a3 
by examining the quantity 6 a4-   3 a2 bc + 12a2c2, we see
that the factor a2 is common to all the terms, since a4 = a2 X a2,
and that, besides, 6, 3, and 12 are divisible by 3; so that,
6a4- 3a2bc+ 12a2c2 = 2a2X 3a2-bc X 3a2+ 4c2X 3a2.
Also the denominator has for a common factor 3 a2; for the
factors a2 and 3 enter into all the terms, and we have
9a2b-15a2c+24a3 =           3bx3 a2 -5c X 3 a2+ Sa X3a2.
Suppressing therefore the 3 a2 as often in the numerator as in
the denominator, the proposed fraction will become
2a2- bc + 4 c2
3b6-5c+8a
41. I pass now to the case where the numerator and denominator are both compound, and in which one cannot perceive at first
whether the divisor is or is not a factor of the dividend.
As the divisor multiplied by the quotient must produce the
dividend, it is necessary that this last should contain all the several products of each term of the divisor by each term of the
quotients and, if we could find the products arising from each
particular term of the divisor, by dividing them by this term,
which is known, we should obtain those of the quotient, after the
same manner as in arithmetic we discover all the figures of the
quotient by dividing successively by the divisor the numbers,
which we regard as the several products of this divisor by the




Division of  Ilgebraic Quantities.


43


different figures of the quotient. But in numbers the several
products present themselves in order, beginning with the units at
the last place on the left, on account of the subordination established between the units of each figure of the dividend according
to the rank which they hold. But as this is not the case in algebra, we supply the want of such an arrangement by disposing all
the terms of the dividend and divisor in the order of the exponents of the power of the same letter, beginning with the highest
and proceeding from left to right, as may be seen with reference to
the letter a in the quantities
5 a7 -22 a6 b + 12 a5 b2 - 6 a4 b3 -   4 a3 b4+ 8 a265,
5 a4 -  2 a3 b + 4 a2 b2,
of which one is the product and the other the multiplicand in the
example of art. 32. This is called arranging the proposed quantities.
When they are thus disposed, it is evident, that whatever be
the factor by which it is necessary to multiply the second to obtain the first, the term  5 a7, with which this begins, results from
the multiplication of 5 a4, with which the other begins, by the
term in the factor sought, in which a has the highest exponent,
and which takes the first place in this factor when the terms of it
are arranged with reference to the letter a. By dividing then
the simple quantity 5 a7 by the simple quantity 5 a4, the quotient
a3 will be the first term of the factor sought. Now as the entire
product ought by the rules of multiplication to contain the several
particular products arising from the multiplication of the whole
multiplicand by each term of the multiplier, it follows that the
quantity here taken for the dividend, ought to contain the products of all the terms of the divisor, 5 a4 - 2 a3 b + 4 a2 b2, by
the first term of the quotient a3; and consequently, if we subtract
from the dividend these products, which are 5 a7 - 2 a6 b + 4 a5 b2,
the remainder -  20 a6 b + 8 a5 2-  6 a4 b3 -  4 a3 b4 + 8 a2 b5
will contain only those, which result from the multiplication of
the divisor by the second, third, &c. terms of the quotient.
The remainder then n'ay be considered as a part of the dividend, and its first term, in which a has the highest exponent, cannot be obtained, otherwise than by the multiplication of the first
term of the divisor by the second term of the quotient. But the
first term of this part of the dividend having the sign -, it is
necessary to assign that which is to be prefixed to the corresponding term of the quotient. This is easily done by the first




44


Elements of Algebra.


rule art. 31., for the quantity - 20 a6 b, being regarded as a part
of the product, having a sign contrary to that of the multiplicand
5 a4, it follows that the multiplier must have the sign -. Division then being performed upon the simple quantities,-20 a6 b
and 5 a4, gives -  4 a2 b for the second term of tie quotient.
If now we multiply this by all the terms of the divisor, and subtract the  product from  the partial dividend, the remainder
+ 10 a4 b3 —4 a3 b4 + 8 a2 b5 will contain only the products of
the third &c. terms of the quotient.
Regarding this remainder as a new dividend, its first term
10 a4b3 must be the product of the first term of the divisor by
the third of the quotient, and consequently this last is obtained
by dividing the simple quantities, 10a4 b3 and 5 a4, the one by
the other.  The quotient 2b3 being multiplied by the whole of
the divisor furnishes products, the subtraction of which, exhausting
the remaining dividend, proves that the quotient has only three
terms.
If the question had been such as to require a greater number
of terms, they might evidently have been found like the preceding,
and if, as we have supposed, the dividend has the divisor for a
factor, the subtraction of the product of this divisor by the last
term of the quotient ought always to exhaust the corresponding
dividend.
42. To facilitate the practice of the above rules;
(1.) [le dispose the dividend and divisor, as for the division of
numbers, by arranging them with reference to some letter, that is,
by writing the terms in the order of the exponents of this letter, beginning with the highest;
(2.) We divide the first term of the dividend by the first term
of the divisor, and write the result in the place of the quotient;
(3.) We multiply the whole divisor by the term of the quotient
just found, subtract it from the dividend, and reduce similar terms;
(4.) We regard this remainder as a new dividend, the first term
of which we divide by the first term of the divisor, and write the
result as the second term of the quotient, and continue the operation
till all the terms of the dividend are exhausted.
Recollecting that when a product has the same sign as the
multiplicand, the multiplier has the sign +, and, that when a
product has the contrary sign to that of the multiplicand, the
multiplier has the sign —(31), we infer that, when the term of
the dividend and the first term of the divisor have the same sign, the




Division of Algebraic Quantities.


45


quotient ought to have the sign +, and, if they have contrary signs,
the quotient ought to have the sign -; this is the rule for the signs.
The individual parts of the operation are performed by the
rule for the division of simple quantities.
W/e divide the coefficient of the dividend by that of the divisor;
this is the rule for the coefficients.
We write in the quotient the letters common to the dividend and
divisor with an exponent equal to the difference of the exponents of
these letters in the two terms, and the letters which belong only to
the dividend; these are the rules for the letters and exponents.
43. To apply these rules to the quantities,
5 a7 — 22 a6 b + 12 a5 b -- 6 a4 63- 4 a3 b4 + 8 a2 b5,
5 a4- 2 a3 b + 4 a2 b2,
which have been employed as an example above, we place them
as we place the dividend and divisor in arithmetic.
Dividend.                  Divisor.
5a7 —22a6b+ 12a5b2 —6a4b 3 —4a 3b4+8a 2b55a4 —2a3b+4a2b2
— 5a7+- 2a6b- 4a5b"                           Quotient.
_a3-4a b+2b3
Rem.-20a6b+Sa5b -6a 4b3-4a 3b4+Sa b5
+20Oa6 —b8ab 2 + 16a4b 3
rem.          +10a4b3-4a3b4+8a2 b5
-10a4-+-b 34a 3b4Ea 2b5
0.
The sign of the first term  5 a7 of the dividend being the same
as that of 5 a4, the first term of the divisor, the sign of the quotient must be +-, but, as it is the first term, the sign is omitted.
By dividing 5 a7 by 5 a4, we have for the quotient a3, which we
write under the divisor.
Multiplying successively the three terms of the divisor by the
first term a3 of the quotient, and writing the products under the
corresponding terms of the dividend, the signs being changed to
denote their subtraction (20), we have the quantity
-     5 a7 +  2 a6b-4 a5 b2,
which with the dividend being reduced, we obtain for a remainder
-20 a6 b +8 a5 b2 -6 a4b3 - 4a3 b4+ 8 a2 b5.
By continuing the division with this remainder, the first term
- 20 a6 b, divided by 5 a4 will give for a quotient 4 a2 b, this
quotient having the sign -, as the dividend and divisor have




46


Elements of dlgebra.


different signs. Multiplying it by all the terms of the divisor
and changing the signs, we obtain the quantity
20a6b- 8a5b +- 16 a4 b,
which taken with the dividend and reduced, gives for a remainder
+ 10a4 b- 4a3sb4+       8 a2 65.
Dividing the first term of this new dividend, 10a4 b3, by the
first term, 5 a4, of the divisor, and multiplying the whole divisor
by the result +2 b3, writing the products under the dividend,
the signs being changed, and making the reduction, we find that
nothing remains, which shows that + 2 b3 is the last term of the
quotient sought.  The quotient therefore has for its expression
a3-4a2 b + 2 b3
44. It is proper to remark here, that in division, the multiplication of the different terms of the quotient by the divisor often
produces terms that are not to be found in the dividend, and
which it is necessary to divide by the first term of the divisor.
These terms are such as destroy themselves, since the dividend
has been formed by the multiplication of the two factors, the
quotient and the divisor. See a remarkable example of these reductions;
Let a3   b3 be divided by a - b.
Division.                  JMultiplication.
a3     b3  a-b                         a -b
-a3 + a2 b a+a b - b                      a2   a b   b
+ a2 b-b3                            a3  -a2b
- a2b + a                         + a b - a b
+ a b2 -b3 
-ab2 - b3                Result a3-b3.
0       0
The first term a3 of the dividend, divided by the first term a
of the divisor, gives for the quotient a2; multiplying this quotient
by the divisor, and changing the signs of the products, we have
- a3 + a b; the first term - a3 destroys the first term of the
dividend, but there remains the term a2 b, which is not found at
first in the dividend. As it contains the letter a, we can divide
it by the first term of the divisor, and obtain + a b. Multiplying
this quotient by the divisor, and changing the signs of the products, we have- a2 b + a b; the term - a2 b cancels the one
above it, but there remains the term + a b2, which is not in the




Division of Algebraic Quantities.         47
dividend. This being divided by a gives for the quotient + b2;
multiplying this quotient by the divisor and changing the signs, we
have -a b2 + b3; the first term - a b2 destroys the first term
of the dividend, and the second + b3 destroys the other - b3.
Tile mechanical part of the operation will be better understood,
if we look for a moment at the multiplication of the quotient
a" + a b - 62 by the divisor a - b. We see that all the terms
reproduced in the process of dividing are those which destroy each
other in the result of the multiplication.
45. It sometimes happens that the quantity, with reference to
which the arrangement is made, has the same power in several
terms both of the dividend and divisor. In this case, the terms
should be written in the same column, one under the other, the
remaining ones being disposed with reference to another letter.
Let there be
-a4b2 +b2- b2 c4  - a2c4 a6 + 2 a4 23 +- b6 +  4 2 + a2 b4,
to be divided by a2 -b2 - c2.
Arranging the first of these quantities with reference to the
letter a, we place in the same column the terms -a4 b2 and
+ 2 a4 c2; in another, the terms + a2 b4 and - a2 c4; and in the
last column, the three terms + b6, + 2 b4 c 2, + b2 c4, disposing them with reference to the letter b, as may be seen in the next
page.
The first term a6 of the dividend being divided by the first
term  a2 of the divisor, gives for the first term of the quotient
— a4; forming the products of this quotient by all the terms of
the divisor, changing the signs of the products in order to subtract them from the dividend, and placing in the same column
the terms containing the same power of a, we have, after the
reduction of similar terms, the first remainder, which we take for
the second dividend.
The first term — 2a4b2 of this new dividend, being divided
by a2, gives for the second term of the quotient - 2 a2 b2; forming the products of this quotient by all the terms of the divisor,
changing the signs of the products to indicate their subtraction
from the dividend, and placing in the same column the terms containing the same power of a, we have, after the reduction of similar terms, the second remainder, which we take for the third
dividend.
The operation being continued in the same manner with the




48


Elements of A./lgebra.


second remainder and the following ones, we shall have three
terms in the quotient. The last being multiplied by all the terms
of the divisor, furnishes products which, being subtracted from
the fourth remainder, exhaust it entirely. As the division admits
of being exactly performed, it follows, that the divisor is a factor
of the dividend.
-  a6 —   a4 b2     a2 b4 +- b6     a2 -- " - -c2
-    2ac2 -  a C4 - 2 64 C2   a4    2 a2 b2   64
+    b2C4 c         a c2- b c2
+ a6 -a4 b2
-a4 c2
1st. rem. - 2 a4 b2 + a2 b4 + b6
+    a4 C2 -  a2 c4 +2 b4 c2
+    b' C4
+ 2 a4b2 -2 a2 b4
— 2 a2 b2 c2
2d rem. +     a4 c2a   a b4 4     b6
-- 2 a2 b2 c2 + 2 b4 2
a2 c4       -6 b2 C4
a4 c2 --  a2 b2 c2
+    a2 c4
3d rem.           -  a2 b4    + b6
-  a2 b2 c2 q- 2b4 c2
+4 b2 c4
+  aa b4   -b6
b4 C2
4th rem.          -  a2 b2 c2a  b4 c2
+ b2 c4
+- a2 b2 c2 -.. 4 c2
-b2 c4
0       0
46. The form under which a quantity appears, will sometimes
immediately suggest the factors into which it may be decomposed. If we have, for example,
8 a6-4a 4    a      3+b2+ 4 a  a3 -a-b2 + 1,
to be divided by 2 a3 - ba2 + 1; as the divisor forms the three
last terms of the dividend, it is only necessary to see if it is a factor of the three first; but these have obviously for a common factor 4 a3, for 8 a6 --- 4 aa b 2- 4 a3 =4 a3 (2 a3 -  b2  1).




slgebraic Fractions.


49


The dividend then may be represented by
4 a3 (2 a3 -b2 + 1+ 2 a3 -b +1,
or            (2 a3 - b2 + 1) (4 a3 + 1).
The division is performed at once by suppressing the factor
2 a3 -b2 + 1, equal to the divisor, and the quotient will be
4a3 +1.
After a little practice, methods of this kind will readily occur,
by which algebraic operations are abridged.
By frequent exercise in examples of this kind, the resolution of
a quantity into its factors is at length easily performed; and it is
often rendered very conspicuous, when, instead of performing the
operations represented, they are only indicated.
Examples in Multiplication of Compound Quantities.
1. (3 a +3b + 4c) X (3a +3b- 4c)
9 a2 + 18 a b + 9b2-16c2.
2. (4a+4b -3c-6d) X (4a+-4b+-3c+6d)
- 16 a2+32ab+ 16b2- 9 C -36cd-36 d2.
3. (5a2 - 3ab+ 7 b2) X (3 a - b)
15 a3- 14a2b +24ab2-7b3.
4. (5ab +3 ac-4bc) X (7ab- 18ac +2bc +d)
=35 a2b2- 69a2b c-18 a b2c+ 5 abd-54a2c2
+ 78 a b c2  3 a c d -8 b2 c2 -4 b c d.
5. (a -+b+c+d) X (a-b-c- d)
= a2 -b2-2bc-2bd-c2            2cd- -.
6. (   2a + 3b -c2) X (- 3f- -7 a + c2)
= 6af- 9bf+ 3c2f -          14a2 - 21ab + 5 a c2
+ 3 b C2 - C4.
7. (3a+4c- 5d)+(6b-7n-6m) X (3a+4c
- 5d)-(6b -7n- -6m)
9 a2 +24ac- 30ad+ 16c2- 40cd +25 d
36 b2 + 84 b n - 49 n2 + 72 mb- 84n m- 36 m.
Examples in Division of Compound Quantities.
1. (4ac- 2ade)-+-2a =2c-2de.
2. (8 a2- 6a b) — 2 a= -4 a + 3 b.
3. (ab-ac)+-(b-c) =a.
4. (ac-bc +ad-b d) -. (a-b)=c +d.
Alg.                   7




50


Elements of Algebra.


5. 4a2+6ab-4ax+9bx-15x'2) —(2a+3x)
= 2 a + 3 b-5 x.
6. (4x3 + 4x2 -29x+- 21) -        (2x-3)
=2x2+ 5x-7.
7. (36a-12ab + 4b2- 3bac+ 12bc+9c2) --- (6a
-2b -3c)
=6a-2b-3c.
8. (a- 4 b4) — (a4 + 2 62)   a4   22.
9. (a4 -92    6     a bc2-  4) -  (a2 - 3 a b -c2)
= a2 + 3 a b + c2.
10. (64a2"+64ab+16b2- 9d2-48d-64)                 (8 a
+4b      3d+ 8)
=8a +4b-3d-8.
11. (32a5+b5) — (2a+b)
16 a- 8 a3 b+ 4a2b- 2 a          b3 + b4.
12. (18a2 +33ab +42ac-        12ad- 30 b2+ 124bc
+ 8 b d-  16 c2- 32 cd) - (6 a + 15 b - 2c - 4d)
=3a -2b + 8c.
Of Algebraic Fractions.
47. WHEN we apply the rules of algebraic division to quantities, of which the one is not a factor of the other, we perceive
the impossibility of performing it, since in the course of the operation we arrive at a remainder, the first term of which is not
divisible by that of the divisor.  See an example;
a3  + a    +2b       a 2  a2  b2
-a3 — ab                a -+b
istrem.      a b - a b +2 b3
-a b -b3
— a b + b3.
The first term, - a b, of the second remainder cannot be divided by a2, the first term of the divisor; so that the process is
arrested at this point. We can however, as in arithmetic, annex
- a2 --- b3
to the quotient a + b the fraction  a b,  having the remainder for the numerator, and the divisor for the denominator;
and the quotient will be
b3 — ab
a + b +a2 + V ~




Algebraic Fractions.


51


It is evident, that the division must cease, when we come to a
remainder, the first term of which does not contain the letter with
reference to which the terms are arranged, or to a power inferior to
that of the same letter in the first term of the divisor.
48. When the algebraic division of the two quantities cannot
be performed, the expression of the quotient remains indicated
under the form of a fraction, having the dividend for the numerator, and the divisor for the denominator; and to abridge it as
much as possible, we should see if the dividend and divisor have
not common factors, which may be cancelled (38). But when
the terms of the fraction are polynomials, the common factors are
not so easily found, as when they are simple quantities. They
are in general to be sought by a method analogous to that, which
is given in arithmetic for finding the greatest common divisor of
two numbers.
We cannot assign the relative magnitudes of algebraic expressions, as we do not give values to the letters which they contain;
the denomination of greatest common divisor therefore, applied to
these expressions, ought not to be taken altogether in the same
sense as in arithmetic.
In algebra, we are to understand by the greatest common divisor
of two expressions, that which contains the most factors in all
its terms, or which is of the highest degree (27). Its determination rests, as in arithmetic, upon this principle; Every common divisor to two quantities must divide the remainder after their division.
The demonstration given in arithmetic (art. 61.) is rendered
clearer by employing algebraic symbols. If we represent the
common divisor by D, the two quantities proposed might be
expressed by the products jD and BD, formed from the common divisor and the factor by which it is multiplied in each of
the quantities. This being supposed, if Q stands for the entire
quotient, and R for the remainder resulting from the division of.D by BD, we have SD =BD X Q + R (./rith. 61); dividing now the two members of the equation by D, we obtain
=BQ+R
and since the first member, which in this case must be composed
of the same terms, as the second, is entire, it must follow, that
R
D is reduced to an expression without a divisor, that is to say,
that R is divisible by D.




52


2Elements of Algebra.


According to this principle, we begin, as in arithmetic, by
inquiring whether one of the quantities is not itself the divisor of the
other; if the division cannot be exactly performed, we divide the
first divisor by the remainder, and so on; and that remainder,
which will exactly divide the preceding, will be the greatest common
divisor of the two quantities proposed. But it will be necessary, in
the divisions indicated, to have regard to what belongs to the nature of algebraic quantities.
We are not, in the first place, to seek a common divisor of two
algebraic quantities, except when they have common letters; and
we must select from them a letter, with reference to which the
proposed expressions are to be arranged, and that is to be taken
for the dividend in which this letter has the highest exponent, the
other being the divisor.
Let there be the two quantities
3 a3 - 3 a2 b + a b - b3,
4a2b-5ab 2+b3,
which are already arranged with reference to the letter a; we
take the first for the dividend, and the second for the divisor.
A difficulty immediately presents itself, which we do not meet
with in numbers, and this is, that the first term of the divisor
will not exactly divide the first term of the dividend, on account
of the factors 4 and b in the one, which are not in the other. But
the letter b being common to all the terms of the divisor and not
to those of the dividend, it follows (40) that b is a factor of the
divisor, and that it is not of the dividend. Now every divisor
common to two quantities, can consist only of factors which are
common to the one and to the other; if then there be such a divisor with respect to the two quantities proposed, it is to be looked
for among the factors of the quantity 4 a - 5 a b - b, which
remains of the quantity 4 a2 b- 5 a b3 + b3, after suppressing
b; so that the question reduces itself to finding the greatest common divisor of the two quantities
3 a3 -- 3a   +ab2 -b3,
4a2-5ab      +b2.
For the same reason that we may cancel in one of the proposed quantities the factor b which is not in the other, we may
likewise introduce into this a new factor, provided it is not a
factor of the first. By this step, the greatest common divisor,
which can consist only of terms common to both, will not be




flgebraic Fractions.               53
affected. Availing myself of this principle, I multiply the quantity 3 a   3 -3a b + ab2 _-b3 by 4, which is not a factor of
the quantity 4 a2 - 5 a b - b2, in order to render the first term of
the one divisible by the first term of the other.
I shall thus have for the dividend, the quantity
12a3 -12a2 b+4ab2-4b3,
for the divisor the quantity
4a2 -5 ab +b2,
and the quotient will be 3 a.
Multiplying the divisor by this quotient, and subtracting the product from the dividend, I have for a remainder
3a2 b +ab2 -4b3;
a quantity which, according to the principle stated at the commencement of this article, must have with 4 a2 - 5 a b + b2, the
same greatest common divisor as the first.
Profiting by the remarks made above, I suppress the factor b,
common to all the terms of this remainder, and multiply it by 4,
in order to render the first term divisible by that of the divisor; I
have then for a dividend, the quantity
12a2 +-4ab-16b2,
and for a divisor, the quantity
4 a2 -5 a b + b;
and the quotient thence arising is 3.
Multiplying the divisor by the quotient, and subtracting the product from the dividend, we obtain the remainder
19 a b - 19 b4,
and the question is reduced to finding the greatest common divisor
to this quantity, and
4a2 -5ab+b2.
But the letter a, with reference to wich the division is made, not
being in the remainder, except of the first degree, while it is of
the second degree in the divisor, it is this which must be taken for
the dividend, and the remainder must be made the divisor.
Before beginning this new division, I expunge from the divisor
19 ab- 19 b2, the factor 19 b, common to both the terms, and
which is not a factor of the dividend; I have then for a dividend,
the quantity
4 a2 -5ab +b2,
and for a divisor
a- b.




54


Elements of algebra.


The division leaves no remainder; so that a - b is the greatest
common divisor required.
By retracing these steps, we may prove a posteriori, that the
quantity a - b must exactly divide the two quantities proposed, and
that it is the most compounded of those which will do it. In dividing by a- b the two quantities proposed,
3a3      3 a2 b + ab2- b2, 4 a2 b-5ab2 +b3,
we resolve them as follows;
(3 a 2+ b2) (a- b), (4 a b -b2) (a - b).
49. When the quantity, which we take for a divisor, contains
several terms having the letter, with reference to which the
arrangement is made, of the same degree, there are precautions
to be used, without which the operation would not terminate.
See an example of this.
Let there be the quantities
a2 b   ac2 -d3, ab - ac       d2
if we make the preparation as for common division,
a2 b + ac2- d3      ab - ac+ d2
- a2 b + a2 c-   ad3   a - 
Rem. a2 c -- a C2  a d2 - d3
by dividing, first, a2 b by a b, we have for the quotient a; multiplying the divisor by this quotient, and subtracting the products
from the dividend, the remainder will contain a new term, in
which a will be of the "second degree, namely, a2 c, arising from
the product of - ac by a. Thus no progress has been made;
for by taking the remainder
a2 c+ ac2- ad2 — d3
for a dividend, and multiplying by b, to render the division possible
by a b, we have
a2 bc +abc2 -abd2- bd3                  ab   a c  d
-a2 b c+   a2    - a c d                  ac
Rem. a2 c + a 6 2 - a c   d2 -- a b d - b d3
and the term - ac produces still a term a2 c2, in which a is of
the second degree.
To avoid this inconvenience, it must be observed, that the
divisor a b - a c + d3 - a (b - c) - d2, by uniting the terms
a b - a c in one; and, for the sake of shortening the operation,
making b - c = m, we have for the divisor a m + d2; but then
the whole dividend must be multiplied by the factor m, to make
a new dividend, the first term of which may be divided by a m,
the first term of the divisor; the operation then becomes




/llgebraic Fractions.


55


a2 bm +-ac2 m-d3 m       am+- d2
- a2 b m —abd       a b- +c2
1st rem. - a 6d2 --   2 m- d3 m
-ac 2 m -    2 d2
2d rem. -a bd2 - c2 d2 - d3 m.
The terms involving a2 n'ow disappear from  the dividend, and
there remain only the terms which have the first power of a.
To make these disappear, we first divide the term a c2 m by a m,
and it gives for a quotient c2; multiplying the divisor by this
quotient, and subtracting the products from the dividend, we obtain
the second remainder. Taking this second remainder for a new
dividend, and suppressing the factor d2, which is not a factor of the
divisor, we have
-ab —2 -dm,
which being multiplied anew by m, becomes
-ab nz -     m- dm2       am + d2
+ a b m + b d2               b
Rem. + b d2 — c2 m-dm2.
The remainder b d2 - c2 m - d m2 of this last division, not involving a, it follows, that if the proposed quantities have a common
divisor, it is independent of the letter a.
Having arrived at this point, we can continue the division no
longer with reference to the letter a; but it will be observed,
that if there be a common divisor, independent of a, to the quantities b d2 - c2 m - d mi and a m + d2, it must divide separately
the two parts a m and d2 of the divisor; for if a quantity is arranged with reference to the powers of the letter a, every divisor of
this quantity, independent of a, must divide separately the quantities multiplied by the different powers of this letter.
To be convinced of this, we need only observe, that, in this
case, each of the quantities proposed must be the product of a
quantity depending on a, and of the common divisor, which does
not depend upon it. Now if we have, for example, the expression
J a 4 + Ba3 + C2 a2 +Da +E,
in which the letters.3, B, C, D, E, designate any quantities
whatever, independent of a, and it be multiplied by a quantity JM,
also independent of a, the product
JMA a 4 + MB a 3 + MC a + MD a + ME,
arranged with reference to a, will contain still the same powers




56


Elements of algebra.


of a as before; but the coefficient of each of these powers will be
a multiple of JM.
This being supposed, if we restore the quantity (b -c) in the
place of m, we have the quantities
bd- c (b - c) -d(- (b -    )2,
a(b- c) +d2;
and it is evident, that b-c and d2 have no common factor;
the two quantities then under consideration have not a common
divisor.
If it were not evident by mere inspection, that there is no
common divisor between b- c and d2, it would be necessary to
seek their greatest common divisor by arranging them with reference to the same letter, and then to see if it would not also divide
the quantity
b d2 -c3 (b - c) - d(b —c)2.
50. Instead of putting off to the end of the operation, the investigation of the greatest common divisor independent of the letter
with reference to which the quantities are arranged, it is less
trouble to seek it at first, because, for the most part, the operation
becomes more complicated at each step as we advance, and the
process is rendered more difficult.
Let there be, for example, the quantities
a4 b2 + a3 b3+ b4 c2 -a4 c2 -a3 b 2 -b2 c,
aa b + ab2 +b3 -a2 c —    abc —b2 c;
having arranged them with reference to the letter a, we have
(b2 _-c) a4 + (b3 — b C2) a3 + b 4 c2-b2 C4
(b -c) a' + (b - bc) a +b3- b c;
I observe, in the first place, that if they have a common divisor
which is independent of a, it must divide each of the quantities,
multiplied by the different powers of a (49), as well as the
quantities b4 c - b2 c4 and b3 - b2 c, which do not contain this
letter.
The question is reduced then to finding the common divisors of
the two quantities b2 - c2 and b - c, and determining whether
among these divisors there is to be found one which will divide at
the same time
ba -bc2 and b2 —    c, b4 C2 -b2C4 and b3-b2 c.
Dividing b2 - c2 by b   c, we find an exact quotient b - c;
b-c then is a common divisor of the quantities b2 -c2 and
b- c, which evidently admit of no other, since the quantity b - c




llgebraic Factions.                57
is divisible only by itself and by unity. We must now see whetherb - c will divide the other quantities referred to above, or
whether it will divide the two quantities proposed; it is found that
it will, and it gives
(b + c) a4 + (b2 + b c) a3  b3 c2 + b2 c3,
a2 + ba+b2.
To bring these last expressions to the greatest degree of simplicity, we should see if the first is not divisible by b + c; it appears upon trial that it is, and we have only to find a common
divisor to the quantities
a4 +b a2 + b2 c2
a2+ba +b2.
By proceeding with these as the rule prescribes, we comes
after the second division, to a remainder containing the letter a of
the first power only; and as this remainder is not the common
divisor, we conclude that the letter a does not make a part of the
common divisor sought, which therefore can consist only of the
factor b - c.
If, beside this common divisor, another had been found, involving the quantity a, it would have been necessary to multiply these
two divisors together to obtain the common divisor sought.
These remarks will enable the learner, after a little practice in
analysis, to find in every case the greatest common divisor. He
will determine without difficulty that the quantities
6 a5 + 15a4b -4 a3c2- 10 a2 b c,
9 a3 b-27a2 bc — Oabca + 18 bc3,
have for their greatest common divisor the quantity 3 a - 2 c2.
51. The four fundamental operations, addition, subtractions
multiplication, and division, we perform in algebra as in arithmetic,
observing merely to proceed, in the operations prescribed by the
rules of arithmetic, according to the methods given for algebraic
quantities.  I shall, therefore, merely suggest these methods,
giving an example of the application of each. I shall begin as I
did in arithmetic, with the multiplication and division of fractions,
as they require no preparatory transformations.
(1.) For multiplication, we have
bX c= a    (.rith. 53),
X    = a   (srit. 70).
Rx6=d


Alg.


8




~~~58    ~     Elements of Algebra.
(2.) For division,
~~a   ~        a    a 
divided by c, gives b- or  X    rith54 70),
a           c       a    d.
divided by d gives  X -           th. 73).
a c
(3.) The fractions a, d' being reduced to the same denominator, become respectively
a d  b  (ctrith. 79).
-d' b —d
The fractions,
a  c  e   g
b  d' d'f  ' h
by the same reduction, becone respectively
adfh    cbfh   ebdh    gbdf
bdfh' b dfh' b dfh' bc dfh
52. I have given, in art. 79. of arithmetic, a process for obtaining, in certain cases, a denominator more simple, than that which
results from the general rule; it may be much simplified by means
of algebraic symbols, as we shall see.
a   d
If, for example, we have the two fractions,-, ' ) it is easy to
see that the two denominators would be the same, iff were a
factor of the first, and c a factor of the second; we multiply then
the two terms of the first fraction by f, and the two terms of the
af      cd                      abf
second by c, which gives  f - and -f  more simple tllan b  cf
bcf     6cf'                    b b cf
bcd
and b    f' obtained by multiplying by the original denominators.
In general, toform the common denominator, we collect into one
product all the different factors raised to the highest power found in
the denominators of the proposed fractions; and it remains only to
multiply the numerator of each fraction by the factors of this product, which are wanting in the denominator of the fraction.
a    ld      e
Having, for example, the fractions -a     and, Iform the
product bs cfg; I multiply the numerator of the first fraction
byfg, that of the second by b cg, that of third by b2f, and I
obtain
afg         b c dg       b2 ef
b cfg'      b62cfg' bCcfg




algebraic Fractions.


59


53. The sum of the fractions
a       b      c
which have the same denominator, or
a + b    c   a+b+ c (.rith. 80)
a+d+           -    -  (Ard80).
The difference of the fractions
a     b
and,
which have the same denominator, or
a   b    a-b
d d        d
The whole of a added to the fraction, or the expression
b   a c   b    a_ +/_ b
a + - = -   +             (.rith. 81).
Also, the expression
b   a c   b    ac- b
a               -
c    c    c ~
Reciprocally,
ahc +- b  ac    b       b
the expression  c q-           = a + 
ac- b     ac   b        b
the expression       -           a -.
C       c    c        c
The terms of the preceding fractions were simple quantities; but
if we had fractions, the terms of which were polynomials, we
should have to perform, by the rules given for compound quantities, the operations indicated upon simple quantities; it is thus that
we have
a2 + b    a-b      (a2 - b2) (a —b)   a3  a b2 - a b -b3
c + d    c —d      (c + d) (c —d) -        ca —d2
The quotient of the fraction
a2 + b2 a. a-b
+ d,divided by --  d
is a2 +b    c —d     (a2 +b2) (c —d) _ a2c+b2c —a2d —bd
c   d    a —b     (c+d) (a-b) -ac+ad-bc-bd'
and so of other operations.
54. Understanding what precedes, we can resolve an equation
of the first degree, however complicated.
If we have, for example, the equation
(a -- b) (x      4       2 )   a c
a - b      '             3 a -b'




60


Elements of /lgebra.


we begin by making the denominators to disappear, indicating only
the operations; it becomes then
(a+b)(x-c)(3a+b)+4b(a-b)(3a+b)=2x(a —b)(3a+b)-ac(a-b)
performing the multiplications, we have
3a2ax + 4abx + b  -3a c - 4abc-b2c+12a2b-8ab2-4b3
=6 a2 x- 4 a x-2b2 xa- a2 c + a b c;
transposing to one member all the terms involving x, it becomes
-3a2x+abx-+3b2 - =x 2a2 c + 5abc + b2 c-12a2 b8ab2 +4b3,
from which we deduce
2a c+5 a bc+ b2 c-12a2 b + 8ab2 + 4b3
x  -       _-3 a2 + S a b + 3 b2
Examples in Division in which the Divisor is not an Aliquot Part
of the Dividend.
1. 1.. (1 - b) = 1 + b + b +  + b3 + b4 +......
2. 1 -(1   b)   1-b+b2- b3+b4-.....
3. /   i\   c   bc, b2 c   b3c
4. c - (a + b) =b               - 
a 'a     a3 a4.
2z
Examples in the Reduction of Fractions.
3a    c        12ad+ 5 b c +-20bdA
* 5b' + 43d1 + —      20 b d
a    c    e   g    k
2. b    a+d    f      -
adfh+bcfh- bde h -b dfg-b df hk
-  bdf h
1   1   1   bc-a c+ab
3.-  - +  - _bs -'
a   b   c       abc
d    c    (a- b)efg- dg-cf
4. a   b
ef   eg           efg
6b c —5ab c+a3
5. c+2a —    3ac —      b2-  bc
2ab3 — b c2 + 3 abc2 - a3
b2 -bc
a 5a+6c  5a-6c
-2.      + -    -=6a.
f~~~~~ 5~~~~~~




.algebraic Fractions.      61
7. 4cg~n 4_ 4cg - n
2        2
3a-4b __2a-b —c +15a-4c _5 Sa -20 b
8.          3    +    12       84
3 a42 b  5bd-2a —3d   1 2ad+3bd +2a +3d
9 c         'c d             4 'c d
10.  a     z  _______
a +z + a -za2F-Z2'
3 a +2x  5a-x   a   a3 -4a2 X _I1IaX2 2 X3
11  a+YX a-x+ 2X         2 x a2 -xl)
1.az  -a-z     3az-a2- z2
a2 -z2  a-j.-z  a2 -z2
1. a3     ab      b    a3 +ab + b3
(a +b)3 (a +b)2+a +b     (a +b)3~
14.   3kh       2 h+        5
(h-2Z)2 +(hAX ) (Ah- 2x)  h- X
2Oh x- 22 X2
-(h-X) (k2...4hx+4Xv2)'
Examples in the Reduction of Fractional Expressions to their Simplest Terms.
axz+z2 a_ ax
1.3bx- cx - 3b6C
14a2 -7ab _7a
2. l0ac -5bC c
3.12a3X4 + 2a2 X5  2a2zX3
3. lab2 x +362X2  3b2
5a2 +Sa    5 a
a2   2   a a-x
a3 -X3 __a2 +ax+X2
(a-x)2  a-x
6. n3-2n-2n2    2
n2 -4n+4 n-2'
7. z2 + 2xz3 _  - 
z~2 + 5 x -6  x + 2'
8. 2X3 +3X2 +X=2x+ 1
2X3-..X2.-2z - z2
9.ab3V+c3 X3  a2 b2-ab cx + C2 S
a2bSc2..       ab -c x




62             Elements of.Ilgebra.
1.2 3 - (3 c + d+ 2)X2 + (3 c + d) x  2x-3c-d
1.a'2 ~bfC2 +2 a b+2a c+ 2b c  a a+b + c,
a2 -b42 -C2-~2 b c   a- ab-c'
a2 - 3a b + a c +2b2- 200bc  a-2b
12.     a2-~b2 +f2bC C2 -aIb2 a~b2 +f '2 a2- u + 2 b2 c2- a4 - O- C4
=(a+b) (a-fb ~c) (a+ b-c)
=(c-fa-b) (b-a+c)'
Examples in the Multiplication of Fractional Expressions.
1  3 ac  c  3 co
a    %   x- ~2 
2. ( 5 13 c  6k h 7d3a
6   w dbgm2 7X      ~d
3a2  39 ac  IS ah  21 a
-3. 
35f 49 2 +21 g + 'of 2  7f2+6f2.
4 2   ab +b2     ( 3a 2  2 ab  b2
\-22xy    y y2   x2    xy   yP
3a4  19 a3b  21 a2b2  9 a3  b 
=X4   1-0z3y-+ %y    ~y+y
Examples in the Division of Fractional Expressions.
C  C~~ab
2. ab +3a  a b2-f- bx3
b +c    b   b+c
4. a CZ4-cz3+X d4-dz3    a2 X4-   c +d
a c-a d +cz- dx      a-f-x  c-d
24 a2 c- 6PC- 4 a 2d+bVd  4 az-2 b32 a b+Sa n +4b2 +b n  8a +b   46+b




JIlgebratic Fractions.   63
6. (3a2~Z7 a b  2lac  56b2 83 bc  3C2)
3 a5 +4      a +   2 2
5i~ +  15'?+ 01 42 Itx
7.     55fh  29   ___   15/t x 92)
Ia2 52a ( C, b C 2    /ac\    a b c
Equations of the First Degree wit/h One Unknown Quantity.
7x  3z
4.    7 x  3x7 ~15. x =662.
+ 10 + 4~
16               3
5.    x    x + 4125;7X  316~- x -808 
2-519-s    l-7     5 
7. 4.15=-3.
6x~ +8  5 x-f-3 _27 -4x  3 X-+9~
S 1.2            3   -      x 6.
9. 4 +  3zx- 13  12 4-7z
I0 16  -   9
~7x -33 -9 + o- l 1x-7 x= 15
10    ~~8
10. 3-5  5-007i — x= 0.2 -- 0.34 x. x = 2-010424.,
11. ~a+bc c fx.  (a d b c)ce
W  d e   dc-cf
f  _j  T      ~b (d e +cf) +ad
X  d x            a c(I-3ab
a      c               c -a d
4.a2 x               c-xa.x"(b -c) (d -a)
14 ~ —d   b-c. x   a2-b2+bc




64


Elements of algebra.


15.             a bb —
15. x- b c 4- d + -. x   ==.
x              x       bc+d'
16.        ( - a       +  -  a(m-3c+3a)
16. c3a+x                   c-a+m
17. a(d2 +  2) _ac + ax      _ d
dx          d           c
C Xm      f xm       cd-af        af-cd
18.                   x_
a + b z x d-e e '      bf —ce     c e -bf'
19. 3-5zx      2a-x     a+f      dx.
a-c         d      a- c
_ d(f-2a)-2a (a-c)
(a-c)(d2 -1) -5d ~
a 2c           ac
20. (a + x) (b+x) -a (b +c) =      b  +x2.' x- b=
Of Questions having two Unknown Quantities, and of Negative
Quantities.
55. THE questions, which we have hitherto considered, involve
only one unknown quantity, by means of which, with the known
quantities, are expressed all the conditions of the question. It is
often more convenient, in some questions, to employ two unknown
quantities; but then there must be, either expressed or implied, two
conditions, in order to form two equations, without which the two
unknown quantities cannot be determined at the same time.
The question in art. 3., especially as it is enunciated in art. 4.,
presents itself naturally with two unknown quantities, that is, with
both the numbers sought. Indeed, if we denote
the least by x,
the greatest by y,
their sum by a,
their difference by b,
we have, by the enunciation of the question,
x + y = a,
y-x =b.
Each of these two equations being considered by itself, we can
determine one of the unknown quantities. If we take the second,
for example, we deduce the value of y, which is
y = b + x,
a value, which seems at first to teach us nothing with regard to'
what we are seeking, since it contains the quantity x, which is




Equations with Two Unknown Quantities.


not given; but if, instead of the unknown quantity y in the first
equation, we put this, its equivalent; the equation, containing now
only one unknown quantity x, will give the value of x by the process already taught.
We have in fact by this substitution,
x + b + x - a,
or                    2 x + b = a,
a -b
or lastly,                  x=     2 
and putting this value of x in the expression for y,
y=b +x=b         a-b a_ a     +b
2   -   2
we have then for the two unknown numbers the same expressions
as in art. 3.
It is easy to see indeed, that the above solution does not differ
essentially from that of art. 3.; only I have supposed and resolved
the second equation y - x = 6, which I contented myself with
enunciating in common language in the article cited; and from it
I deduced, without algebraic calculation, that the greater number
was x + b.
56. 1 take another question.
1 laborer having worked for a person 12 days, and having
with him, during the first 7 days, his wife and son, received 74
francs; he worked afterward with the same person 8 days more,
during 5 of which, he had with him his wife and son, and he received at this time 50 francs; how much did he earn per day himself, and how much did his wife and son earn?
Let x be the daily wages of the man,
y that of his wife and son;
12 days' work of the man will amount to 12 x,
7 days' work of his wife and son,    7 y;
we have then by the first statement of the question,
12 x+ 7y= 74;
8 days' work of the man will give  8 x,
and 5 days' work of his wife and son  5 y;
we have then by the second statement
8x + 5y = 50.
Proceeding as in the preceding question, we take the value of
y in the first equation, which is
aLg.                 9




66


Elements of Algebra.


74- 12 
y= —   -7,
and substitute this value in the second, multiplying it by 5, the coefficient, and it becomes
370-60 x
8 x q+ --          50,
an equation, which contains only the unknown quantity x. By reducing it we have
56 x + 370 - 60 x = 350,
370- 4x= 350;
transposing -4 4x to the second member, and 350 to the first, we
obtain
370 - 350 = 4 x
20 = 4x
2 0    X
5=    x.
Knowing x, which we have just found equal to 5, if we place
this value in the formula
74 — 12
Y= -- 7
the second member will be determined, for we have
74- 12 X 5 _ 74- 60      14
7         -     7       7
thus                    y = 2.
The man then earned 5 francs per day, while his wife and son
earned only 2.
57. The reader has perhaps observed, that in resolving the
above equation 370 - 4 x- =350, I have transposed 4 x to the
second member. I have proceeded thus to avoid a slight difficulty, that would otherwise have occurred, and which I will now
explain.
By leaving 4 x in the first member, and transposing 370 to the
second, we have
-4 x = 350 - 370;
and reducing the second according to the rule in art. 19., there
will result from it
- 4 x   -   20.
But as we have avoided, in the preceding article, the sign -,
which affects the quantity 4 x, by transposing this quantity to the
other member; and as in like manner the quantity 350 - 370 becomes by transposition 370 - 350; and since a quantity, by being




Equations with Two Unknown Quantities.


67


thus transferred from one member to the other, changes the sign
(10), it is evident that we may come to the same result by simply
changing the sign of each of the quantities -- 4x, + 350 -370,
which gives
4 x  -   350+ 370,
or                  4 x      370-350,
which is the same as
370    350 - 4 x.
We might also change the signs after reduction, and the equation
-4x = - 20
becomes, as above,
4x-= 20.
It follows from this, that we may transpose indifferently, to one
member or to the other, all the terms involving the unknown quantity, observing merely to change the signs of the two members in
the result, when the unknown quantity has the sign -.
58. Having undertaken, by means of letters, a general solution
of the problem of art. 56., I will now examine a particular case.
I suppose the first sum received by the laborer to be 46 francs,
and the second 30, the other circumstances remaining as before;
the equations of the question will then be
12 x+ 7 y =46,
8 x+ 5 y -30.
The first gives
46- 12 x
y -- 7 
multiplying this value by 5, in order to substitute it in the place of
5 y, in the second, we have
230 -60 z    3
8 x +       -    = 30;
the denominator being made to disappear, it becomes
56 x + 230- 60 x = 210,
or             56 x- 60 x =210 -230,
or                 -4 x= -     20,
and the signs being changed agreeably to what has just been remarked,
4x = 20,
-'- ~ = 5.
If we substitute this value instead of x in the expression for y, it
will become




68


Elements of Algebra.


46 -60
y=      7-,
-14
or                  Y     7 
Now how are we to interpret the sign -, which affects the insulated quantity 14? We understand its import, when there are
two quantities separated from each other by the sign -, and
when the quantity to be subtracted is less than that from which
it is to be taken; but how can we subtract a quantity when it is
not connected with another in the member where it is found?
To clear up this difficulty, it is best to go back to the equations,
which express the conditions of the question; for the nearer we
approach to the enunciation, the closer shall we bring together
the circumstances which have given rise to the present uncertainty.
I resume the equation
12x + 7y= 46;
I put in the place of x its value 5, and it becomes
60 + 7 y- 46.
This equation, by mere inspection, presents an absurdity. It is
impossible to make the number 46 by adding any thing to the
number 60, which exceeds it already.
I take also the second equation,
8  + 5y- 30,
and putting 5 in the place of x, I find
40 + 5y =30;
the same absurdity as before, since the number 30 is to be formed
by adding something to the number 40.
Now the quantities 12 x or 60 in the first equation, 8 x or 40
in the second, represent what the laborer earned by his own
work; the quantities 7 y and 5 y stand for the earnings of his wife
and son, while the numbers 46 and 30 express the sum given as
the common wages of the three; we must see then at once in
what consists the absurdity.
According to the question, the laborer earned more by himself,
than he did by the assistance of his wife and son; it is impossible
then to consider what is allowed to the woman and son, as augmenting the pay of the laborer.
But if, instead of counting the allowance made to the two latter
persons as positive, we regard it as a charge placed to the account




Equations with Two Unknown Quantities.


69


of the laborer, then it would be necessary to deduct it from his
wages; and the equations would no longer involve a contradiction,
as they would become
60- 7 y = 46,
40-5y -- 30;
we deduce from the one as well as from the other
y=2;
and we conclude from it, that if the laborer earned 5 francs per
day, his wife and son were the occasion of an expense of 2 francs,
which may otherwise be proved thus.
For 12 days' labor he received
5 X 12 or 60 francs;
the expense of his wife and son for 7 days is
2 X 7or 14 francs;
there remain then 46 francs.
For 8 days' labor he receives
5 X 8 or 40 francs;
the expense of his wife and son for 5 days is
2 X 5 or 10 francs;
there remain 30 francs.
It is very clear then, that in order to render the proposed probblem with the first conditions possible, instead of the enunciation
in article 56., we must substitute this;
Jl laborer worked for a person 12 days, having had with him
the first 7 days, his wife and son at a certain expense, and he received 46 francs; he worked afterwards 8 days, during 5 of which
he had with him his wife and son at an expense as before, and he
received 30 francs. It is required to find how much he earned per
day, and what was the sum charged him per day on account of his
wife and son.
Calling x the daily wages of the laborer, and y the daily
expense of wife and son, the equations of the problem will evidently be
12x —7y      46
8x- 5 y = 30;
and being resolved after the manner of those in art. 56., they
will give
x = 5 francs, y = 2 francs.
59. In every case, where we find, for the value of the unknown
quantity, a number affected with the sign -, we can rectify the




70


Elements of Algebra.


enunciation in a manner analogous to the preceding, by examining, with care, what that quantity is, among those which are
additive in the first equation, which ought to be subtractive in
the second; hut algebra supersedes the use of every inquiry of
this kind, when we have learnt to make a proper use of expressions affected with the sign -; for these expressions, being
deduced from the equations of the problem, must satisfy those
equations; that is to say, by subjecting them to the operations
indicated in the equation, we ought to find for the first member
— 14
a value equal to that of the second. Thus the expression -  4
drawn from the equations
12 x+ 7y= 46,
8 x + 5 y -30,
must, consistently with the value of x  5, as deduced from these
same equations, verify them both.
The substitution of the value of x gives, in the first place,
60 + 7 y _ 46,
40 + 5 y   30.
It remains to make the substitution of - 14 in the place of y;
and for this purpose we must multiply by 7 and by 5, having
regard to the sign -, with which the numerator of the fraction is
affected.
If we apply the rule relative to the signs given in art. 42. for
division, we have
-14        2
7  -  -
besides, by the rule for the signs in multiplication, we find
7 X -2 --     14,
5 X -2=    -10.
Hence the equations
60 + 7y=46, and      40 + 5y =30,
become respectively
60-14 = 46, and      40 - 10 =30,
and are Verified, not by adding the two parts of the first member,
but in reality by subtracting the second from the first, as was done
above, after considering the proper import of the equations.
60. The problem in art. 58. does not admit of a solution in
the sense in which it is first enunciated; that is to say, by addition, or regarding as an accession the sum considered with




Equations with Two Unknown Quantities.


71


reference to the wife and son of the laborer; neither does the
second enunciation consist with the data of the problem in art.
56.
If we were to consider in this case y, as expressing a deduction,
the equations thus obtained
12 x - 7 y    74,
x- 5y _ 50,
would give
— 14
x = 5, and    y =-  7
and the substitution of the value of x would immediately change
the equations to
60-7y =74,
40-5 y      50.
The absurdity of these results is precisely contrary to that of
the results in art. 58., since it relates to remainders greater than
the numbers 60 and 40, from which the quantities 7 y and 5 y are
to be subtracted.
The sign minus, which belongs to the expression of y, implies
an absurdity; but this is not all, it does it away also; for, according to the rule for the signs,
— 14
-7    — 2,
and               -7X -2-=+         14,
-5 X -2 =      +10.
Thus the equations
60 - 7 y =- 74,    40 - 5 y =50,
become
60 + 14 = 74,     40 + 10 = 50,
and are verified by addition; consequently, the quantities - 7 y
and - 5 y, transformed into + 14, + 10, instead of expressing
expenses incurred by the laborer, are regarded as a real gain.
We are brought back then in this case, also to the true enunciation
of the question.
61. We perceive by the preceding examples, that there may be,
in the enunciations of a problem of the first degree, certain contradictions, which algebra not only makes known, but points out also
how they may be reconciled, by rendering subtractive certain quantities which had been regarded as additive, or additive certain
quantities which had been regarded as subtractive, or by giving to
the unknown quantities values affected with the sign -.




72


Elements of Algebra.


See then what is to be understood, when we speak of values
affected by the sign -, and of what are called negative solutions,
resolving, in a sense opposite to the enunciation, the question in
which they occur.
It follows from this, that we may regard, as but one single question, those, the enunciations of which are connected together in
such a manner, that the solutions, which satisfy one of the enunciations, will, by a mere change of sign, satisfy the other also.
62. Since negative quantities resolve in a certain sense the
problems, which give rise to them, it is proper to inquire a little
more particularly into the use of these quantities, and to settle
once for all the manner of performing operations in which they are
concerned.
We have already made use of the rule for the signs, which had
been previously determined for each of the fundamental operations; but the rules have not been demonstrated with reference
to insulated quantities. In the case of subtraction, for example,
we supposed that there was to be taken from a, the expression
b - c, in which the negative quantity c was preceded by a positive quantity b.  Strictly speaking, the reasoning does not depend upon the value of b; it would still apply when b = 0, which
reduces the expression b - c to - c. But the theory of negative quantities being at the same time one of the most important
and most difficult in algebra, it should be established upon a sure
basis. To effect this, it is necessary to go back to the origin of
negative quantities.
The greatest subtraction, that can be made from a quantity, is
to take away the quantity itself, and in this case we have zero
for a remainder; thus a -a = 0. But when the quantity to be
subtracted exceeds that from which it is to be taken, we cannot
subtract it entirely; we can only make a reduction of the quantity
to be subtracted, equal to the quantity from which it was to be
taken. When, for example, it is required to subtract 5 from 3,
or when we have the quantity 3 - 5; to take, in the first place, 3
from 5, we decompose 5 into two parts 3 and 2, the successive
subtraction of which will amount to that of 5, and thus, instead
of 3 - 5, we have the equivalent expression 3 - 3 - 2, which is
reduced to -2. The sign -, which precedes 2, shows what is
necessary to complete the subtraction; so that, if we had added
2 to the first of the quantities, we should have had 3 + 2- 5, or




Equations with Two UnAnown Quantities.


73


zero. We express then, with the help of algebraic signs, the idea
that is to be attached to a negative quantity -a, by forming
the equation a - a = 0, or by regarding the symbols a - a,
b-b, &c., as equivalent to zero.
This being supposed, it will be understood, that if we add to
any quantity whatever the symbol b - b, which in reality is only
zero, we do not change the value of this quantity, and that, consequently, the expression a + b - b, is nothing else but a different
manner of writing the quantity a, which is also evident from the
consideration, that + b and -b destroy each other.
But having by this change of form introduced + b and -b into the same expression with a, we see, that in order to subtract
any one of these quantities, it is sufficient to efface it. If it were
+ b that we would subtract, we efface it, and there remains a - b,
which accords with the rule laid down in art. 2.; if on the other
hand it were - b, we efface this quantity, and there would remain
a + b, as might be inferred from art. 20.
With respect to multiplication, it will be observed, that the product of a- a by +-b must be a b - a b, because the multiplicand being equal to zero, the product must be zero; and the first
term being a b, the second must necessarily be -ab to destroy
the first.
We infer from this, that - a, multiplied by + b, must give
-a b.
By multiplying a by b - b, we have still a b - a b, because the
multiplier being equal to zero, the product will also be equal to
zero; it is therefore necessary that the second term  should be
- a b, to destroy the first + a b.
Whence + a, multiplied by - b, must give - a b.
Lastly, if we multiply -a by b -b, the first term of the product being, according to what has just been proved, — a b, it is
necessary that the second term should be + a b, as the product
must be nothing when the multiplier is nothing.
Whence - a, multiplied by - b, gives + a b.
By collecting these results together we may deduce from them
the same rules as those in art. 31.
As the sign of the quotient, combined with that of the divisor,
according to the rules proper for multiplication, must produce
the sign of the dividend, we infer from what has just been said,
that the rule for the signs given in art. 42. corresponds with that,
Alg.                    10




74


Elements of Algebra.


which it is necessary to observe in fact, and that consequently,
simple quantities, when they are insulated, are combined with respect to their signs, in the same manner, as when they make a part
of polynomials.
63. According to these remarks, we may always, when we meet
with negative values, go back to the true enunciation of the question resolved, by seeking in what manner these values will satisfy
the equations of the proposed problem; this will be confirmed by
the following example, which relates to numbers of a different kind
from those of the question in art. 56.
64. Two couriers set out to meet each other at the same time
from two cities, the distance of which is given; we know how many
miles (a) each travels per hour, and we inquire at what point of the
route between the two cities they will meet.
To render the circumstances of the question more evident, I
have subjoined a figure, in which the points A and B represent the
places of departure of the couriers.
A                          R                B
I denote the things given, and those required, in the usual way,
by small letters.
a, the distance in miles of the points of departure. and B,
b, the number of miles per hour, which the courier from Jq
travels.
c, the number of miles per hour, which]' the courier from B
travels.
The letter R being placed at the point of meeting of the two
couriers, I shall call x the distance JR passed over by the first, y
the distance BR passed over by the second; and as
AR +BR = lB,
I have the equation,
x - y= a.
Considering that the distances x and y are passed over in the same
time, we remark that the first courier, who travels a number b of
miles in an hour, will employ in passing over the. distance x, a time
denoted by g.


(a) In the original the distance is given in kilometres.  It is here
expressed in miles to avoid perplexing the learner.




Equations with Two Unknown Quantities.


75


Also the second courier, who travels c miles in an hour, will
employ, in passing over the distance y, a time denoted by y; we
c
have then
x    Y
b    c
The equations of the question therefore will be
x + y = a,
y
b    c.
Making the denominator b of the second to disappear, we have
by
C
putting this value in the place of x in the first equation, it becomes
+ y      a;
and we deduce from it
by+ cy= ac,         whence     y=     
Y      -- b-c'
Substituting this value of y in the expression for the value of x,
we obtain
b     ac                     abc
x=-X-+              or     x      a      (51),
C   b + c '            x     (b+c)( )
or lastly
x-    +   (38).
As the sign - does not enter into the values of x and y, it is
evident that whatever numbers are put for the letters a b c, we shall
always find x and y with the sign +-, and therefore the question
proposed will be resolved in the precise sense of the enunciation.
Indeed it is readily perceived, that in every case where two persons set off from different points and travel toward each other,
they must necessarily meet.
65. I will now suppose, that the two couriers proceed in the
same direction, and that the one who sets out from./ is pursuing
the one who sets out from B, and who is travelling toward the
same point C, placed beyond B, with respect to.3.
A                B1         R         C
It is evident that in this case, the courier who starts from the
point.3, cannot come up with the courier who sets off from the
point B, unless he travels faster than this last, and the point of




76


Elements of Algebra.


coming together, R, cannot be between d and B, but must be beyond B, with respect to d.
Having the same things given as before, and observing that
when
dR - BR =- B,
we have
x-y = a.
The second equation,
_ Y
b    c'
expressing only the equality of the times employed by the couriers in passing over the distances AR and BR, undergoes no
change.
The above equations, being resolved like the former ones, give
by
x- cb
b y
bf —_y = a,     b y-cy = ac,
C
ac
y- b-c'
b     ac       abc
x c X b -c     c (b - c)'
ab
and lastly          x =  -
b - c
Here the values of x and y will not be positive, except when 6
is taken greater than c, that is to say, except the courier starting
from the point d be supposed to travel faster than the other.
If, for example, we make
b = 20,    c = 10,
we have
20 a     20 a
-20         1-0  0   2a,
10 a     10 a
Y   20-10       10    a
from which it follows, that the point of their coming together is distant from the point A twice AB.
If we now suppose b smaller than c, and take, for example,
b = 10,    c - 20,
we find
10a      10a
x-  10 —20 ' ---10 —        a
20 a     20 a
Y- 10-20 — 10 ---2a.




Equations with Two Unknown Quantities.


77


These values being affected with the sign —, make it evident,
that the question cannot be resolved in the sense in which it is
enunciated; and indeed it is absurd to suppose that the courier
setting out from the point A, and proceeding only 10 miles in an
hour, should ever be able to overtake the courier setting out from
the point B, and travelling 20 miles per hour, and who is in advance of the first.
66. Nevertheless, these same values resolve the question in a
certain sense; for, by substituting them in the equations
x-y = a,
x _Y
we have by the rule for the signs
- a + 2 a = a,
a _     2a
10  2- 20
equations which are satisfied; since, by making the reductions, the
first member becomes equal to the second; and if we attend to
the signs of the terms which compose the first, we shall see
how it is necessary to modify the enunciation of the question, in
order to do away the absurdity.
Indeed, it is the distance a corresponding to x, and passed over
by the first courier, which is in reality subtracted from the distance
2 a, corresponding to y, and passed over by the second courier; it
is then just as if we had changed y into x, and x into y, and had
supposed that the courier starting from the point B, had run after
the other.
This change in the enunciation produces also a change in the
direction of the routes of the couriers; they are no longer travelling toward the point C, but in an opposite manner toward the
point C', as represented in the figure below;
C'    R'     A            B      R      C
and their coming together takes place in R'. The result from
this is
BR' -       =R B,
which gives
y- x = a;
we have besides constantly
xY
2;  3'




78


Elements of Algebra.


and we find
ab       10 a
c-b -20-10 
ac       20 a
Y - c-b = 2 -        = 2 a,
positive values, which resolve the question in the precise sense in
which it is enunciated.
67. The question we have been considering presents a case, in
which it is in every sense absurd. This occurs when we suppose
the two couriers to travel equally fast. It is evident, that in whatever direction we suppose them to move, they can never come together, since they preserve constantly the interval of their points of
departure. This absurdity, which no modification in the enunciation ca:. remcve, is very conspicuous in the equations.
We have now b = c, since the couriers, travelling equally fast,
pass over the same space in an hour; the equation
x   y
b   c
becomes                   - _
and gives               x = y.
Thus the equation      x- y = a
reduces itself to    x- x = a or 0 = a,
a result sufficiently absurd, since it supposes a quantity a, the magnitude of which is given, to be nothing.
68. This absurdity shows itself in a manner very singular in the
values of the unknown quantities
ab                 ac
x    b-c'         Y    b -c
their denominator becoming 0 when b = c, we have
ab                ac
X=   0            y
Y- 0O
We do not easily perceive what may be the quotient of a division when the divisor is zero; we see merely, that if we consider
b as nearly equal to c, the values of x and y become very great.
To be convinced of this, we need only take
b = 6 miles,      c   5,8 miles;
we then have        x = 0 -   30 a,
5,8 a   29
5Y   S - 0= 29a.
Y=O,2




Equations with Two Unknown Quantities.        79
If further we take  b  - 6      c   5,9.
we have
6a
x   -6  = 60 a,
0,1
5,9a
y,i — 59 a.
If moreover we make
b- 6,       c - 5,99,
it becomes
6a
c    6 a   600 a,
x -- 0,01   00 a,
5,99 a   599 a
Y    -01 -  599a;
and it is manifest, that as the divisor diminishes in proportion to the
smallness of the assumed difference of the numbers b and c, we
obtain values more and more increased in magnitude.
But as a quantity, however minute, can never be taken for zero,
it follows, that however small we make the difference of the numbers represented by the letters b and c, and however great may be
the consequent values of x and y, we never attain to those which
answer to the case where b = c.
Since these last cannot be represented by any number, however
great we suppose it, they are said to be infinite; and every expression of the form -, the denominator of which is zero, is regarded as the symbol of infinity.
This example shows that mathematical infinity is a negative
idea, since we at length get it only by the impossibility of assigning a quantity that can resolve the question.
We may ask here, how the values
ab              ac
x = -O         Y - O
satisfy the equations proposed; for it is an essential characteristic of algebra, that the symbols of the values of unknown quantities, whatever they may be, being subjected to the operations
indicated upon these quantities, shall satisfy the equations of the
problem.
By substituting them in the equations
X - y = a,
_ y
b   b'




80


Elements of Algebra.


which answer to the case where b = c, we have by the first,
ab    ab
-- = a,
ab- ab
or        ---   =   a, or ab- ab =a X O,
or lastly,    0=0,      since    aX 0     0.
The second equation gives, under the same condition,
ab      ab
0   b -OX 6b
the two members of each equation becoming equal, the equations
are satisfied.
It remains still to explain how the notion indicated by the exab
pression -O, removes the absurdity of the result found in art. 67.
For this purpose, let the two members of the equation
- y - a,
be divided by x, which gives
1-: -;
and as the equation
x y
b b
gives x = y, the first becomes
a                  a
1 —1- -,        or     0=-.
x                  x2;
The error here consists in the quantity a, by which the second
member exceeds the first; but this error becomes smaller and
smaller, in proportion to the assumed magnitude of x. It is then
with reason, that algebra gives for x an expression, which cannot
be represented by any number, however great, but which, as it
proceeds in the order of numbers becoming greater and greater,
points out in what manner we may reduce more and more the error of the supposition.
69. If the couriers travelling equally fast, and in the same ditection, had set out from the same point, their coming together
could not be said to take place at any particular point, since
they would be together through the whole extent of their route.
It may be worth while to see how this circumstance is represented
by the values, which the unknown quantities x and y assume in
this case.




Equations with Two Unknown Quantities.


81


B
A                                  C
The points./ and B being coincident, we have on this supposition a = 0, and constantly b = c; it follows then, that
Ob   0           Oc   0
X       = T       y =- o      - o
x —   o  -  '
In order to interpret these values, that indicate a division, in
which the dividend and divisor are each nothing, I go back to the
equations of the question. The first becoming
x -y=0,      gives  x=y;
and substituting this value in the second equation, which is
X -        it becomes  Y.- Y
b    '                  b   b
The last equation having its two members identical, that is to
say, composed of the same terms with the same sign, is verified,
whatever value is assigned to y, and this unknown quantity can
never be determined. Besides, it is evident that the equation
x=    becomes x    y,
and consequently can express nothing more than the first.*  The
only result, both from the one and from the other, is, that the two
couriers are always together, since the distances x and y from the
point A are equal; their value in other respects remains indeterminate.  The expression 0 then, is here a symbol of an indeterminate quantity. I say here, for there are cases where it is
not; but the expression has not then the same origin as the preceding.
70. To give an example, let there be
a(a — b2)
b(a-b) '
This quantity becomes o in its present form, when a - b; but if
we reduce it first to its most simple expression, by suppressing the factor a - b, common to the numerator and denominator,
we find
* For the sake of conciseness, analysts apply to the same equations
the epithet, identical.
Y =Y; is an identical equation, 5- 3 x  5 — 3 x is another,
b-b
and when two equations express only the same thing, we say that
these equations also are identical.
dlig.                  11




82


Elements of Jllgebra.


a (a + b)
b
which gives 2 a, when a = b.
It is not the same with the values of x and y, found in the preceding article, for they are not susceptible of being reduced to a
more simple expression.
It follows, from what I have just said, that when we meet
with an expression which becomes 0, it is proper, before pronouncing upon its value, to see if the numerator and denominator
have not a common factor, which becoming nothing, renders the
two terms at the same time equal to zero, and which being suppressed, the true value of the proposed expression is obtained.
There are, notwithstanding, some cases which elude this method,
but the limits of this work will only allow me to note the analytical fact.  It belongs properly to the differential calculus, to give
the general processes for finding the true value of quantities, which
become 0.
71. It is very evident, from what has been said, that algebraic
solutions either answer perfectly to the conditions of a problem, when
it is possible, or they indicate a modification to be made in the enunciation, when the things given imply contradictions that cannot be
reconciled; or lastly, they make known an absolute impossibility,
when there is no method of resolving with the same things given, a
question analogous in a particular sense to the one proposed.
72. It may be remarked, that in the different solutions of the
preceding question, the changing of the signs of the unknown
quantities x and y, corresponds to a change in the direction of the
journeys represented by the unknown quantities. When the unknown quantity y was counted from B towards X, it had in the
equation
+- y = a,
the sign +, and it takes the sign -  for the second case, when the
motion is in the opposite direction from B towards C, art. 65.,
since we had for the first equation
x-y = a.
By changing the sign in the second equation,
x_y
b    c'
we have
x --
6 — '"




Equations with Two Unknown Quantities.


83


a result which differs from that given in the article cited; but it
should be observed, that the journey y, being made up of multiples of the space c passed over in an hour by the courier from B,
and this space having the same direction as the space y, ought to
be supposed to have the same sign, and consequently to take the
sign -, when - is applied to y; we have accordingly,
z   -y          x   y
or  b —
A simple change of sign then is sufficient to comprehend the
second case of the question in the first, and it is thus that algebra
gives at the same time the solution of several analogous questions.
We have a striking example of this in the problem of art. 15.
It is here supposed that the father owed the son a sum d; if
we would resolve the question on the contrary hypothesis, that
is, by supposing that the son owed tile father the sum d, it would
be sufficient to change the sign of d, in the value of x, and we
have
bc-d
a-+- b
If we suppose neither to owe the other any thing, we must make
d - 0, and then the equation would be
bc
x   a+ b
Nothing can be easier than to verify the two solutions, by putting
anew the problem into an equation for each of the cases, which we
have enunciated.
73. It wvas only to preserve an analogy between the problems
56 and 64, that I have employed two unknown quantities in the
second. Each may be resolved with only one unknown quantity; for when we say that the laborer received 74 francs for 12
days' work performed by himself and 7 days' work by his wife
and son, it follows, that if we call y the daily wages of the woman
and son, and take 7 y from 74 francs, there will remain 74 - 7 y
for the 12 days' labor of the man; from which we infer that he
earned 74-    7 y per day.
By a similar calculation for the 8 days' service, we find that he
he earned 50 - 5   per day.
Putting the two quantities equal to each other, we form the
equation




84


Elements of dlgebra.


74- 7y     50- 5y
12         8
Also in the question of art. 64.,
A                      R 
if x represent the course AR of the courier from J, BR = a - x
would be that of the courier who set off from B towards.1. These
two distances being passed over in the same time by the couriers,
whose rate of travelling per hour in'"miles is denoted by numbers
b and c respectively, we have
x   a — x
whence
cx = a b - b x,
ab
X- b+c
The difference between the solutions, which I have now given,
and those of articles 56. and 64., consists merely in this, that we
have formed and resolved the first equation by the assistance of
ordinary language, without employing algebraic characters, and it
is manifest, that the further we carry this, the less will remain to be
effected by the other.
74. We sometimes add to the problem of art. 64. a circumstance, which does not render it more difficult.
A           R            C           B
We suppose that the courier, who starts from B, sets off a number
d of hours before the other, who goesfrom A.
It is evident, that this amounts only to a change of the point of
departure of the first; for if he travelled a number c of miles per
hour, he would pass over the space BC = cd in d hours, and
would be at the point C, when the other courier set off from a.;
so that the interval of the points of departure would be
SC =- JB- BC=- a-cd.
By writing then a-c d in the place of a in the equation of the
preceding article, we have
x   a   c cd- x
ab   b cd
cx =    — b + c
If the couriers proceeded i te same diection, te interval of
If the couriers proceeded in the same direction, tlhe interval of




Equations with Two Unknown Quantities.


85


A                  B           C          R
the points of departure would be
IC- =    B + BC = a +cd;
and the distance passed over by the courier from the point A
would be JR, while that passed over by the other courier would
be                  CR =.R-s C;
we have then
x   x-a- cd
b =      c
whence
ab b + b c d
b- c
75. Enunciated in this manner, the problem presents a case, in
which the interpretation of the-negative value found for x is attended with some difficulty; it is when the couriers being supposed to
proceed in opposite directions, we give to the number d a value
such, that the space BC represented by c d, becomes greater than
a, which represents JB.
b.......              s
C       R                               B
Now the courier from the point B arrives at C on the other side
of A1 at the moment lwhen the courier from A sets off towards B;
there is then an absurdity in supposing that the two couriers can
thus come together.
If we should take, for example,
a - 400m'13,  b - 121'ns, c = 8mls1,  d - 60h.,
there would result from it c d = 480mls.; thus the point C would
be 80"ml' on the other side of A, with respect to the point B; but
we find,
400  12 -60 -8   12   400- 3 -60 2 12
x —      8+12         --       2 +3
1200   1440      240 
- -      - 48.
5
Thus the coming together of the couriers takes place in a point
R, 48mls' on the other side of the point., but between./ and C;
although it seems that the courier from B, being supposed to
continue his journey beyond the point C, can be overtaken by the
other courier only after he has passed this point.
To understand the question resolved in this sense, we may
substitute in the place of x the negative member -  m, and the




86


Elements of slgebra.


equation becomes
m_ a-cd+-m
or by changing the signs in the two members,
m    cd- a - m
b         c
We see that the distance passed over by the courier from the
*. *..*.. *. B 
C        R       A                    B
point B, is c d - a - m, or what remains of B C after./B and.JR are subtracted, that is CR, and that 3.C = c d- a. This
is just what would take place if the second courier had started
immediately from the point C, where he is, at the departure of
the first; but as they travel in opposite directions, they must
necessarily meet between d and C. Thus, this case is similar
to the first of those of art. 74., where it is sufficient to change
a- c d into cd - a, in order to obtain the value, which m has
according to the above equation.*
76. The problem of art. 56., taken in its most enlarged sense,
may be enunciated as follows;,/ laborer having passed a number a of days in a family, and
having with him his wife and son during a number b of days, received a sum c; he lived afterward in the same family a number d
of days; he had with him this time his wife and son, during a
number e of days, and he received a sum f; we inquire what he
earned per day, and what was allowed per day to his wife and son.
Let x represent constantly the daily wages of the laborer,
and y that of his wife and son; for the number a of days, he has
a x, and for the number b of days, his wife and son have b y, so
that,
ax + b y    c;
for the number d of days, he has d x, and for the number e of
days, his wife and son have e y, thus,
d x + e y =f.
These are the general equations of the question.
We deduce from the first
c- by
a
multiplying this value by d, in order to substitute it in the place of


* See note at the end of Elements of Algebra.




Equations with two Unknown Quantities.


87


x in the second equation, we have
d   c d- bdy
dx    -
a
and consequently,
cd- b d y 
a-,-ba + ey=f.
By making the denominator to disappear, we obtain
cd —b dy — aey = af,
whence            aey -bdy = af-c d,
af -cd
Y     ae-bd'
Having the value of y, if we substitute it instead of y in the expression for x, this last will be known,
af- cd
a e - bd
x —
a
To simplify this expression, we should, in the first place, perform
the multiplication indicated upon the quantities
7,     1   af — cd
b,   and   ae-bd          (51)
a e - bd'
which gives
a b f - b cd
a e -b (
x-_           —;
and then reduce c to a fraction having the same denominator as the
fraction which accotnpanies it, and perform the subtraction of this
fraction (53); and it becomes
ace-b cd-abf - b cd
ae- b d
x                       -—,
a
or by being reduced
ace- abf
ae -b d.
a
* There might be some doubt as to the meaning of this expression; but it is obviated by attending to the bar denoting division,
which is placed in the middle of the line. Thus, in the expression
x-, A represents the dividend, whether integral or fractional, and
B the divisor, which may also be a whole number or a fraction. So also




8&                  Elements of AIlgebra.
Dividing by a (51) we have
a ce - abf
a2 e -a b d
Suppressing the factor a, common to the numerator and denominator (38), we find
ce-bf
X    --
a e-bd'
The values
ce-bf          af-cd
ae-bd' Y       ae-~b d'
are applied in the same manner as those, which we before found
for literal equations, with only one unknown quantity; we substitute in the place of the letters, the particular numbers in the example selected.
We shall obtain the results in art. 56., by making
a = 12,     b -7,     c =74,
d-   8,     e-=5,    f    50,
and those of art. 58., by making
a = 12,     b -7,     c-4 46,
d=   8,     e   5,   f =30.
77. The values of x and y are adapted not only to the proposed question; they extend also to all those, which lead to two
equations of the first degree with two unknown quantities, since it
is evident, that these equations are necessarily comprehended in
the formulas,
A
C.
the expression x = B signifies, that x is equal to the quotient of the
A                                  A
fraction A divided by B, and the expression x -  indicates for x the
~C                        _~C
B
quotient arising from A divided by the fraction C; and lastly, we deA
C
note by the expression x = B, the quotient resulting from the divisD
A               1B
ion of the fraction A by the fraction.
It will be perceived by these remarks, that it is necessary to place
the bars according to the result, which we propose to express.




Equations with Two Unknown Quantities.


89


a x + b y =c,
dx+ ey =f,
provided the letters a, b, d, e, denote the whole of the given quantities, by which the unknown quantities x and y are respectively
multiplied, and the letters c andf, the whole of the known terms,
transposed to the second member.
Of the resolution of any given number of Equations of the First
Degree, containing an equal number of Unknown Quantities.
78. WHEN a question has as many distinct conditions, as it contains
unknown quantities, each of these conditions furnishes an equation,
in which it often happens, that the unknown quantities are involved
with others, as we have seen already in the problems with two
unknown quantities; but if these unknown quantities are only
of the first degree, according to the method adopted in the
preceding articles, we take in one of the equations the value of
one of the unknown quantities, as if all the rest were known,
and substitute this value in all the other equations, which will then
contain only the other unknown quantities.
This operation, by which we exterminate one of the unknown
quantities, is called elimination. In this way, if we have three
equations with three unknown quantities, we deduce from them
two equations with only two unknown quantities, which are to be
treated as above; and having obtained the values of the two last
unknown quantities, we substitute them in the expression for the
value of the first unknown quantity.
If we have four equations with four unknown quantities, we
deduce from them, in the first place, three equations with three
unknown quantities, which are to be treated in the manner just
described; having found the value of the three unknown quantities, we substitute them in the expression for the value of the first,
and so on.
See an example of a question, which contains three unknown
quantities and three equations.
79. A. person buys separately three loads of grain; the first,
which contained 30 measures of rye, 20 of barley, and 10 of wheat,
cost 230 francs;
The second, which contained 15 measures of rye, 6 of barley,
and 12 of wheat, cost 138 francs;
aig.                 12




90


Elements of llgebra.


The third, which contained 10 measures of rye, 5 of barley, and
4 of wheat, cost 75 friancs;
It is asked, what the rye, barley and wheat cost, each per measure?
Let x be the price of a measure of rye,
y,   that of a measure of barley,
z,   that of a measure of wheat.
To fulfil the first condition, we observe, that
30 measures of rye are worth 30 x,
20 measures of barley are worth 20 y,
10 measures of wheat are worth 10 z
and as the whole must make 230 francs, we have the equation
30 x+ 20y + 10z= 230.
For the second condition, we have
15 measures of rye, worth 15 x,
6            barley     6 y,
12            wheat    12 z,
and consequently,
15 x + 6 y - 12z = 138.
For the third condition, we have
10 measures of rye worth 10 x,
5            barley    5 y,
4            wheat     4 z,
and consequently,
10 x+ 5 y + 4 z    75.
The proposed question then will be brought into three equations;
30x+ 20 y + 10z      230,
15x +   6y+ 12 z= 138,
10x+    5y+    4z=    75.
Before proceeding to the resolution, I examine the equations,
to see if it is not possible to simplify themn by dividing the two
members of some one of them by the same number (12), and I
find that the two members of the first may be divided by 10, and
those of the second by 3. Having performed these divisions, I
have only to occupy myself with the equations
3x+2y +      z-  23,
5x + 2y+4 z- 46,
10 x +- 5 y + 4  = 75.
As I can select any one of the unknown quantities in order to
deduce its value, I take that of z in the first equation, because this
unknown quantity having no coefficient, its value will be entire or
without a divisor, as follows;




Equations with Two Unknown Quantities.


91


z =23-3x-2 y.
This value being substituted for z in the second and third equations, they become
5 x - 2 y + 92 -  12 x - 8 y = 46,
10 x + 5y + 92 -   12x-     8 y =75;
and reducing the first member of each, we find
92 - 7 x - 6 y = 46,
92 -2 x-3y= 75.
To proceed with these equations, which contain only two unknown quantities, I take in the first the value of the unknown
quantity y, and I obtain
92 -46 — 7x               46-7 x
Y= —,    or  y       6
and by substituting this value in the second equation, it becomes
92-2x-3X 46          7    75.
6
The denominator, 6, may be made to disappear by the usual
method, but observing that the denominator is divisible by 3, I can
simplify the fraction by multiplying it by 3, agreeably to article 54.
of Arithmetic. I have then
92-2 x-46             75.
2
The denominator 2 being made to disappear, it becomes
184- 4 x-46 + 7 x = 150;
the first member being reduced, gives
138 + 3 x= 150,
whence
150-138     12
x —           -   3, or x=4.
Substituting this value in the expression for that of y, I find
46-7X4       46- 28     18
y_ =                 G -- ' ---- or y-3;
Y=       6     =     66
and by substituting these values in the expression for that of z, we
obtain
x =23 - 3 X 4-2 2X 3= 23 -        12 - 6, or z-5.
It appears then, that the price of the rye per measure was 4 fr.,
that of the barley                        3,
that of the wheat                         5.
This example, while it illustrates the method given in the preceding article, ought to be attended to, on account of the abbreviations of calculation, which are performed in it.




92


Elements of   1lgebra.


80. I proceed now to resolve the following problem.
21 man who undertook to transport some porcelain vases of three
different sizes, contracted that he would pay as much for each vessel
that he broke, as he received for those which he delivered safe.
He had committed to him two small vases, four of a middle size,
and nine large ones; he broke the middle sized ones, delivered all
the others safe, and received the sum of 28 francs.
There were afterwards committed to him seven small vases, three
of the middle size and five large ones; he rendered this time the
small and the middle sized ones, but broke the five large ones, and
he received only three francs.
Lastly, he took charge of nine small vases, ten middle sized ones,
and eleven large ones; all these last he broke, and received in consequence only 4 francs.
It is asked what was paid him for carrying a vase of each size?
Let x be the sum paid for carrying a small vase,
y, that for carrying a middle sized one,
z, that for carrying a large one.
It is evident, that each sum which the porter received, is the
difference between what was due to him for the vessels delivered
safe, and what he had to pay for those which were broken; accordingly, the three conditions of the problem furnish respectively
the following equations;
2x-     4y+     9z-28,
7x+     3y-     5z=    3,
9x+    10y-    11 z    4.
The first of these equations gives
28 +4y -9 z
2 -;
and by substituting this value, the second and third equations
become
196 +28y -63z +3y               3
252 - 36y-        z +         z    4.
Making the denominators to disappear, we have
196+28y-63z+           6y-10z-=6,
252 + 36y-    81 z +20y- 22z =8;
reducing the first member of each, we obtain




Equations with Two Unknown Quantities.


93


196+34y-      73z-=6,
252 + 56y —103z= 8;
taking the value of y in the first of these equations, we find
73z- 190
Y-      34    *
By means of this value, the second equation becomes
252+ 56X      34      -103z=8;
being cleared of the denominator 34, it is changed into
34 X 252 + 56 X 73z —56 X 190-34 X 103 z= 34 X 8,
or into
8568 + 4088 z - 10640 - 3502 z = 272.
The reduction of the first member of this result, gives
586 z- 2072 =-272,
whence we deduce
2344
z   -o86     or z =4.
By going back with the value of z to that of y, we have
73 X 4-190     292 -190    102   or y= 3;
-- 34                  M   --— 34, or y-3;
34            34   -34'
and with these two values, we find
28 +4    3-9 X 4     28+12-36       4           2
2                  2     --      or x=2.
The prices then were 2 fr. for carrying a small vase,
3               one of a middle size,
4               a large one.
This example is sufficient to show how to proceed in all similar cases.
81. It sometimes happens, that all the unknown quantities do
not enter at the same time into all the equations; the method,
however, is not changed by this circumstance; it is sufficient,
carefully to examine the connexion of the unknown quantities, in
order to pass from one to the others.
Let there be, for example, the four equations,
3u-2y =      2,
2 x + 3 y = 39,
5x —7z     - 11,
4y + 3z     41,
containing the unknown quantities, u, x, y, and z.
With a little attention we see, that by taking the value of x




94


Elements of Allgebra.


in the second equation, and substituting it in the third, the result
containing only y and z, will, by being combined with the fourth
equation, give the values of these two quantities; and having the
value of y, we obtain those of u and x, by means of the first and
second equations. The following is the process;
39- 3 y
2
5X39-3         7z    11,
or                 195 - 15 y- 14 z = 22,
or                 15y+ 14z =173        (57).
The two equations
15 y+ 14z= 173,
4y+    3z=    41,
being resolved, give
y=5, z=7;
and by means of these values, we have
39-3X 5       39-15     24
x=       2 --        22 - 2 or x 12,
2+2y      2+ 10    12
— 3 -=-3 -=-3-, or u-=4.
The numbers sought then are
4, 12, 5, and 7.
82. The method now explained is applicable to literal equations, as well as to numerical ones; but the multitude of letters,
which it is necessary to employ to represent generally the things
given, when the number of equations and unknown quantities
exceeds two, has led algebraists to seek for a more simple manner of expressing them. I shall treat of this in the following
article; but in order to furnish the reader with the means of
exercising himself in putting a problem into an equation, and
resolving it, I have subjoined a number of questions, and have
placed at the end of each the answer that is required.
(1.). father, being asked the age of his son, said, iffrom double
the age that he is of now, you subtract triple of what he was six
years ago, you have his present age.
Answer. The child was 9 years old.
(2.) Diophantus, the author of the most ancient book on tlgebra,
that has come down to us, passed a sixth part of his life in infancy,
a twelfth part of it in youth; afterward he was married and passed




Equations with Two Unknown Quantities.


95


in this state a seventh part, and five years more, when he had a son,
whom he survivedfour years, and who attained only to half the age
of his father, what was the age of Diophantus when he died?
Answer, 84 years.
(3.) J. merchant drew, every year, upon the stock he had in
trade, the sum of 1000 francs for the expense of his family; still
his property increased every year, by a third part of what remained
after this deduction, and at the end of three years it was doubled;
how much had he at the beginning of the first year?
Answer, 14800 fr'ancs.
(4.) A merchant has two kinds of tea, the first at 14 francs a
pound, the second at 18 francs; how much ought he to take of each to
make up a chest of 100 pounds, which should be worth 1680 francs?
Answer, 30 pounds of thefirst and 70 of the second.
(5.).1 person filled, in 12 minutes, a vessel containing 39 gallons, with water, by means of two fountains, which were made to
run in succession, and one discharged 4 gallons per minute and the
other 3, how long did each fountain run?
Answer, the first 3-minutes, and the second 9.
(6.) dt noon the hour and minute hands of a watch are together,
at what point of the dial will they next be in conjunction?
Answer, at 1 hour 5 minutes and y5.
Obs. This problem refers itself to that of art. 65.
(7.)./ man, meeting some beggars, wishes to give them 25 cents
each, but finds upon counting his money, that he wants 10 cents in
order to do it; he then gives them only 20 cents each, and has 25 cents
left; how much money had he, and what was the number of beggars?
Answer, he had $1,65, and the number of the beggars was 7.
(8.) Three brothers purchased an estate for 50000 francs, and
the first wanted, in order to complete the whole paynment, half of the
property of the second; the second would have paid the entire sum
with the help of a third of what the first owned, and the third required, to make the same payment, in addition to what he had, afourth
part of what the first possessed; what was the amount of eachone's
property?
Answer, the first had 30000 francs, the second 40000, and the
third 42500.
(9.) Three players after a game count their money, one had lost,
the other two had gained each as much as he had brought to the play;




96


Elements of algebra.


after the second game, one of the players, who had gained before, lost
and the two others gained each a sum equal to what he had at the
beginning of this secondgame; at the third game, the player, who
had gained till now, lost with each of the others a sum equal to that,
which each of them possessed at the beginning of this last game;
they then separated, each having 120 francs; how much had they
each, when they commenced playing?
Answer, he who lost at the first game, had 195 francs,
he who lost at the second     105,
he who lost at the third       60.
Generalformulasfor the resolution of Equations of the First Degree.
83. To obviate the inconvenience referred to in the beginning
of the last article, we shall represent all the coefficients of the
same unknown quantity by the same letter, but distinguish them by
one or more accents, according to the number of equations.
General equations with two unknown quantities are written
thus;
ax + by = c,
a' x + b' y = c'.
The coefficients of the unknown quantity x are both represented
by a, those of y by b; but from the accent, which is placed over
the letters in the second equation, it may be seen, that they are
not considered as having the same value, as the corresponding ones
in the first. Thus a' is a quantity different from a, b' a quantity
different from b.
If there are three equations, they are expressed thus;
a x+b     y+c z-=d,
a' x + b y + c' z- d',
a" x + b" y + c" z = d".
All the coefficients of the unknown quantity x are designated by
the letter a, those of y by b, those of z by c; but the several letters are distinguished by different accents, which show, that they
dense different quantities.  Thus a, a' a", are three different
quantities. The same may be said of b, b', b", &c.
Following this method, if we have four unknown quantities, and
four equations, we may write them thus;
a  x+b     y- c   z+-d    u = e,
a' x + b' y +c' z + d     u=e',
a" x + b" y + c" z + d" u = e",
a"' x c  b"' y + c"' z + d"' u = e'".




Formulas for Equations of the First Degree.


' 97


84. To avoid fractions, and simplify the calculation, we may
vary the process of elimination in the following manner.
Let there be the equations
a x + b y= c,
a' x - b' y = c',
it is evident, that if one of the unknown quantities, x, for example, has the same coefficient in the two equations, we have only to
subtract one of these equations from the other, in order to make
this unknown quantity disappear. This may be seen at once in the
equations
10  + 11 y - 27,
10x+4   9y= 15,
which give
11y- 9 y =27 -15, or 2 y =12, or y =6.
It is evident, that the coefficients of x may be immediately made
equal in the equations
a x4- b y =c,
a' x + b' y = c',
by multiplying the two members of the first by a', the coefficient
of x in the second, and the two members of the second by a, the
coefficient of x in the first; we thus obtain,
a a' x + a' by = a' c,
a a' x + a 6' y = a c'.
Then subtracting the first of these from the second, the unknown
quantity x disappears; and we have
(a b'- a' b) y = a c' - a' c,
an equation, which contains only the unknown quantity y; from
this we may deduce,
a c' -ca'
y   ab- b a'.
The method we have just employed, may always be applied to
equations of the first degree, to exterminate any one of the unknown quantities.
By exterminating, in the same manner, the unknown quantity y,
we may find the value of x.
If we apply this process to three equations, containing x, y, and
z, we may first exterminate x from the first and second, then from
the first and third; we thus obtain two equations, which contain
only y and z, from which we may exterminate y.
When this calculation is performed, the equation containing z,
lg.                    13




98 '


Elements of Algebra.


to which we arrive, will have a factor common to all its terms,
and consequently will not be the most simple, which may be
obtained.
85. Bezout has given a very simple method for exterminating
at once all the unknown quantities except one, and for reducing
the question immediately to equations, which contain one unknown   quantity less than the equations proposed.  Although
this process is necessary, only when equations with three unknown quantities are employed, we shall, in order to give a complete view of the subject, begin by applying it to those, which contain only two.
Let there be the equations
a x -+b y =,
a' x +- b y = c 
multiplying the first by any indeterminate quantity n, we have
a m x   b m y = m c;
subtracting from this result the equation
a/ x + b6 y = c',
there remains
a mx - a' x-+ b m y — by = c-c',
or         (am - a') x +- (b m -b') y = cm - c'.
Since m is an indeterminate quantity, we may suppose it to be
such, that b m  b'. In this case, the term multiplied by y disappears, and we have
cm - c
x      aam-a
but since b m = b', it follows that,
bt
therefore
cb'
_- 
b        cb'-be'
x     b    -  a b' -b a'
b — a
If, instead of supposing bm = b', we make am-  a', the term,
which contains x, will vanish, and we shall have
cm -c
Y    bm — b"
The value of m will not be the same as before; for we shall have
at
m=-a;




Formulasfor Equations of the First Degree.


99


and by substituting this in the expression for y, we find
ca' -a c'
Y   b a' -a b'
If we change the signs of the numerator and denominator of this
value of y, the denominator will become the same, as that in the
expression for x, since we shall have
a c'  ca'
Y = a t-ba' 
86. Next let there be the three equations
a  x +b   y +c    z   d,
a' x 4+b' y+c' z-d',
a" x + 6/ y + c" z = d;
we shall be led, by an obvious analogy, to multiply the first of
these equations by m, and the second by n, m and n being indeterminate quantities, to add together the results, and from the
sum to subtract the third; by this means, all the equations will be
employed at the same time, and the two new quantities m and n,
which we may dispose of as we please, will admit of any determinate value, which may be necessary to make both the unknown
quantities disappear in the result. Having proceeded in this manner, and united the terms by which the same unknown quantity is
multiplied, we shall have
(a m + a' n - a") x + (b m +- b n - b") y + (c m + c' n - c') z
- dm + d' n- d".
If we would make the unknown quantities x and y disappear,
we must take the equations
a m -+ a n = a",
b m + b' n = b',
and then we obtain
dm +d'n —d"
C m + c' - c"
From  the two equations, in which m and n are the unknown
quantities, it is easy to deduce the value of these quantities, by
means of the results obtained in the preceding article; for it is
only necessary to change in these results x into m, y into n, and to
write instead of the letters
a, b, c   the letters a, a' a
a', b/, cv,          b, b', b",
which gives




100


Elements of Algebra.


a"' b' - b" a'
a b'-b a''
a b"   b a"
a /b -     a"
Substituting these values in the expression for z, and reducing all
the terms to the same denominator, we have,t
d(b' a"- a' b") + d' (a b" - b  ) -d" (a b' - b a')
c (b' a" - a' b") + c' (a b" - 6 a") - c (a a' -b a')'
If we had made the terms containing x and z to disappear, we
should have had y; the letters m and n would have depended
upon the equations
am +- a'n = a",     c m +- c' n - c",
and proceeding as before, we should have found
d (c' a" - a c") + d' (a c - c a") — d" (a c' — c a')
Y    b (c' a" - a' c") + 6' (a c" -- c a) -- b" (a c -c a')
Lastly, by assuming the equations
b m +- b' n = b",   c m -- c' n = c",
we make the terms multiplied by y and z to disappear; and we
have
d (c' b" - b' c") + d' (b c" - c b") - d" (b c' -c b')
-a (c' b" -  ' c") + a' (b c" - c b") - a" (b c' - c '
These values being developed in such a manner, as to make the
terms alternately positive and negative, if we change, at the same
time, the signs of the numerator and denominator, in the first and
third, we shall give them the following forms;
a b' d" - a d' b' + d a' b" -    ba' d + b d' a" -d  b' a"
z    ab' c" -ac' b"- +c a' b"-  a' c" + b c' a"-c b' a"'
a d' c" - a c' d" + c a' d-d a' c" + d c a" -c d' a"
Y - ab' c"-ac' b" +ca' b" -ba' c"l +-b c a" —cb' a"'
d b' c"- d c'b" + c d b"-bd' c" +b c' d"- c b d"
x  a  - b' c" -b c b +  a1" - ba1  -bc a1  cb' a'" 
87. Let there be the four equations
a  x +-b   y+c     z-d     u-e,
a' x +b' y+c' z+d' u =e',
a" x + b" y + c" z + d" u = e",
a"' x + b' y + c"' z + d" u -- e"';
t  d a6_b  a + dab" -ba"           ab'-ba'
a" b' —b a'      ab'-b a         ab'    b a'
b' --      b a+ d' a       b —ba'   a     - b a'
^  6a  b'. b —b   ' al        b' -  ba'
a   ' ---,b:ac'     a -    --
*  ~a:.ba b' — b a'        a b'- b a'




Formulas for Equations of the First Degree.


101


if we multiply the first by m, the second by n, the third by p,
and from the sum of their products subtract the fourth, we shall
have
(am + a' n + alp - a"') x + (b m + b' n + b"p - b"') y
+ (c m + c' n + c" P - C"') z + (d m + d' n + d" p - d') u
-     e m  + e' n  +- e/  - e"'.
In order to obtain u, we make
am -+- a' n -+ a" p - a"',
b m + b' n + b"l  = b/',
C? + c' n + c" p = c"/,
we then have
e m + e' n + e" p - e"'
dnz + d' t1 +- d" p- d"'
The preceding equations, which must give m, n, and p, may be
resolved by means of the formulas found for the case of three
unknown quantities. This method will appear very simple and convenient; but the nature of the results obtained above will furnish
us with a rule for finding them without any calculation.
88. To begin with the most simple case, we take an equation
with one unknown quantity, a x = b; from this we find
b
x — _
in which the numerator is the whole known term b, and the denominator the coefficient a, of the unknown quantity.
From the two equations
a x + by = c,    a' x +     b y = c',
we have already deduced
c b'- b c'          ac' -ca'
X-ab=  'baI7        Y   ab'   ba'
x   a   - b a,'     y~      b -b a".
The denominator in this case also is composed of the letters a, a',
b, b', by which the unknown quantities are multiplied. We first
write a by the side of b, which gives a b; we then change the
order of a and b, and obtain b a; prefixing to this the signwe have ab- b a; lastly, we place an accent over the last
letter in each term, and the expression becomes a b'-b a for
the denominator.
From this expression we may find the numerator. To obtain
that for x, we have only to change each a into c, and each b into c
for that of y, putting an accent over the last letter as before; in
this way we find c b' - b c' for the one, and a c' -c a' for the




102


Elements of ilgebra.


other.  The numerator may, therefore, be found from the denominator, as well in cases where there are two unknown quantities, as
when there is only one, by changing the coefficient of the unknown
quantity sought, into the known term or second member, and retaining the accents, which belonged to the coefficients.
The same rule may be applied to equations with three unknown quantities, as we shall see by merely inspecting the
values, which result from these equations. With respect to the
denominator, it is necessary further to illustrate the method by
which it is formed.  Now, since in the case of two unknown
quantities, the denominator presents all the possible transpositions
of the letters a and b, by which the unknown quantities are multiplied, it may be supposed, that when there are three unknown
quantities, their denominator will contain all the arrangements of
the three letters a, b, c. These arrangements may be formed in
the following manner.
We first make the transpositions a b - b a with the two letters a
and b, then, after the first term a b, write tle third letter c, which
gives a b c; making this letter pass through all the places, observing
each time to change the sign, and not to derange the order in
which a and b respectively stand, we obtain
a b c - ac b + c a b.
Proceeding in the same manner with respect to the second term
- b a, we find
-b ac + b c a-cb a;
connecting these products with the preceding, and placing over the
second letter one accent, and over the third two, we have
a b' c" - a c b" +- c a/ b" - b a' c" + b c' a" - c b a",
a result, which agrees with that presented by the formulas, obtained above.
From this it is obvious, that, in order to form a denominator
in the case of four unknown quantities, it is necessary to introduce
the letter d into each of the six products,
ab c - ac b + c ab —bac + bca- cb a,
and to make it occupy successively all the places. The term a b c,
for example, will give the four following;
abcd - a b d c    adbc- dab c.
If we observe the same method in regard to the five other products, the whole result will be twenty-four terms, in each of which,




Formulasfor Equations of the First Degree.


103


the second letter will have one accent, the third two, and the
fourth three. The numerators of the unknown quantities u, z, y,
and x, are found by the rule already given.*
89. We may employ these formulas for the resolution of numerical equations. In doing this, we must compare the terms of
the equations proposed with the corresponding terms of the general equations, given in the preceding articles.
To resolve, for example, tie three equations
7x +5y- +2z-        79,
8x+7y+9z=         122,
x -+4y + 5z =     55,
it is necessary to compare the terms with those of the equations
given in art 86. We have then
a  = 7, b = 5, c = 2, d     -   79,
a' -8, ' =     7, c' = 9, d' =  122,
a" -  1, b6 = 4, c"    5, d" =  55.
Substituting these values in the general expressions for the unknown quantities x, y, and z, and going through the operations,
which are indicated, we find
x =    4,  y = 9,   z = 3.
It is important to remark, that the same expressions may be
employed, even when the proposed equations are not, in all their
terms, affected with the sign +, as tie general equations, from
which these expressions are deduced, appear to require. If we
have, for example,
3 x —9y + 8z-=         41,
- 5 x + 4 y + 2z - - 20,
11 x — 7 y -   6 z      37,
in comparing the terms of these equations with the corresponding
ones in the general equations, we must attend to the signs, and the
result will be
a     +   3, b =-    9, c =   + 8, d  =     41,
a'    -   5, b'   + 4, c' - +      2, d'  -  20,
a" =     +  11, b =-  7 c// -     6, d" = + 37.
We are then to determine by the rules given in art. 31., the sign,
* M. Laplace, in the second part of the Memoires de l'Academie
des Sciences for 1772, p. 294, has demonstrated these rules d priori.
See also Annales des Mathelmatiques pures appliquees, by M. Gergonne, vol. iv, p. 148.




104


Elements of llgebra.


which each term of the general expressions for x, y, and z, ought
to have, according to the signs of the factors of which it is composed. Thus we find, for example, that the first term of the common denominator, which is a b' c", becoming + 3 X + 4 X - 6,
changes the sign of the product, and gives -72.  If we observe
the same method with respect to the other terms, both of the numerators and denominators, taking the sum of those which are
positive, and also of those which are negative, we obtain
2774   2834    -60 
592      62-2     30    - 2,
3022- 2932 _       90
Y    592 =- 622   -    30     3
3859 -3889 _-30 
592-   622    - 30- +- 
Equations of the First Degree with two or more Unknown
Quantities.
1.  3x+2y=118)         x-16
* x - +5y=191      y- 35.
2     5x-8'-7y —44          x42-, y-=8-.
5 5x — 8  Y --            88    y-7- 44
9 x y= 60y             - 6 211     39.8..
0.56 x + 13.421 y = 763.4.      y = 5.5421...
2x+10 = 3y+                               5 y  =5.
5 ax =b y          _  bC          tc
X*^  a+Y==  S     a+b' Y-a+b'
7y;  X + y  c }     +a
8 ax+by     G c   X   cg —b     y   ah —cf
*f x+gyy      h         ag- bf        ag-bfv
0.5  a  _3- b       2 b26a2 + d       3a2- b2+d
9. b+y 3a+j-               - 3a -              36
ax -- 2 by   d )! bc x == c-bb  6                             a 2b
I  Q        2 V              a       a+2b
10.   b2               - yc-)b3), y  ==1+.
6   -- be)        - 4 x.              -     c




Fuormulas for Equations of the First Degree.     10


105


311.  5 y (Sb-2f) bf
Lb2xb lt +(b+c+f )fy=f2x(b +2f)bff
___f      bf
b -f' ~b f
(x + y =10
12. <x+z= 19> x= 3, y= 7, z =1.6.
(y + z= 23)
x + y + z 24
13. x + y-z 184>  x16, y= 73, z=51.


14.
L


y+z=-ab x amp
m y =n x  mp+n~p +mq'
JM- anp+ 


= amq
mp + np + m q'
aC              ___X=ccbfY af-cd
I5 dx+ey=cixac~bd,       ae-bd~'
15 dx + e Y f Z= a(cl-fg) —d(bl-cg
~~gY~~hzl~~~z  It(a e- bd)
+ 411
16  +1  204  1= IS y = 32, z = 10.


(2x+ 5.y-7z =-288)
17. 5x-y + 3z=  227>
(7x+ 6 y+ ZI= 297)


x== 13, y =24, z =62.


(x +y + z=30)
18S. S8x +4y +2z =50~ x= 9, y = -7, z =3643.
2


19-+-=a
lu~g.


2
~'a-b-f-c'
2
Zb + c -a
14




106


Elements of algebra.


( 2      5    1, 1
3y+       -   sd27,
2o.   +1 I+: 6,       X =6, y = 9, Z.
l 5           4 = 4  12 —.
Examples in Equations involving Negative Results.
1. To find a number which added to b gives a sum a
(1.) Let =29, a=47.
(2.) Let b =31, a=24.
2. A father is a years, his son b years old. In how many years
will the son's age be one fourth of his father's?
(1.) Let a= 54, b    9.
(2.) Let a =45, b = 15.
3. A cistern, into which water was let by two pipes, will be filled
by them both in a number a of hours, and by the first alone, in a
number b of hours. In what time will it be filled by the second
alone?
(1.) Let a =12, b    20.
(2.) Let a= 12, b=10.
4. A person sent to buy oranges found that if he bought those
which cost a pence each, he should spend all his money, but if he
bought those which cost b pence each, he should have c pence left.
How many was he sent for?
(1.) Let a   5, b =4, c =24.
(2.) Let a = 4, b   6, c  24.
5. A boat, which had started from a certain place 10 days, is
pursued by another boat from the same place and by the same
way. The first goes 4 miles every day, the other 9. In how many
days will the second overtake the first?
6. Let n equal the number of days elapsed since the departure
of the first boat, a the number of miles it goes per day, and b the
number of miles the second goes per day.
7. What will be the change in the question if n - 10, a = 6,
and b   4.
8. A courier, who goes 31- miles every 5 hours, is sent from a
certain place; when he had gone 8 hours another is sent after
him, and this one in order to overtake the first must go 221 miles,
every 3 hours. When will the second overtake the first?




Formulas for Equations of the First Degree.


107


9. When all the conditions of the preceding problem remain the
same, excepting that the first courier, besides the advantage of
starting earlier, has this also, that he travelled from a place 36 miles
farther on the road: in how many hours will they come together?
10. Make the problem general. Let the place from which the
first starts be situated a miles in advance on the road; further, let
the number of hours by which he had the start of the other be
equal to b; let the speed of the first be such that he goes c miles
in d hours, and the speed of the second such that he goes e miles
in f hours. In how many hours after the departure of the second
courier will they be together?
ns. (ab+bc)f hours.
~ns. e ---f C/ hours.
de-cf
11. In how many hours will they come together, when the first
courier, instead of starting from a place a miles in advance, starts
from one as many miles backwards? What must be done in order to adapt the solution of the preceding problem to this case?
12. Two bodies move in opposite directions; one runs c feet in
each second, the other C. The two places from which they start
at the same time are distant d feet from one another. When will
they meet?
13. In what time will the two bodies come together when that
which goes C feet each second runs after the other?
Is the problem as it is here stated always possible? What is required for it to be possible? What does the expression C  signify when C = c? What does it denote when C < c?
Equations of the Second Degree, having only one Unknown
Quantity.
90. HITHERTO I have been employed upon equations of the
first degree, or such as involve only the first power of the unknown quantities; but were the question proposed, To find a
number, which, multiplied by five times itself, will give a product
equal to 125; if we designate this number by x, five times the
same will be 5 x, and we shall have
5 2 - 125.
This is an equation of the second degree, because it contains x~2,
or the second power of the unknown quantity. If we free this
second power from its coefficient 5, we obtain




108


Elements of Algebra.


2   125                   O
=x  -,      or    x =25.
We cannot here obtain the value of the unknown  Meantity x,
as in art. 11., and the question amounts simply to this, to find a
number which, multiplied by itself, will give 25. It is obvious that
this number is 5; but it seldom  happens that the solution is
so easy; hence arises this new numerical question; to find a
number, which, multiplied by itself, will give a product equal to a
proposed number; or, which is the same thing, from the second
power of a number, to retrace our steps to the number from which
it is derived, and which is called the square root. I shall proceed,
in the first place, to resolve this question, as it is involved in the
determination of the unknown quantities, in all equations of the
second degree.
91. The method employed in finding or extracting the roots of
numbers, supposes the second power of such as are expressed by
only one figure to be known. See the nine primitive numbers with
their second powers written under them respectively.
1 2 3 4 5 6 7 8 9
1 4 9 16 25 36 49 64 81.
It is evident from this table, that the second power of a number
expressed by one figure, contains only two figures; 10, which is
the least number expressed by two figures, has for its square a
number composed of three, 100. In order to resolve the second
power of a number consisting of two figures, we must attend to the
method by which it is formed; for this purpose we must inquire,
how each part of the number 47, for example, is employed in the
production of the square of this number.
We may resolve 47 into 40 + 7, or into 4 tens and 7 units; if
we represent the tens of the proposed number by a, and the units
by b, the second power will be expressed by
(a + b) (a + b) = a2   2 a b + b2;
that is, it is made up of three parts, namely, the square of the tens,
twice the product of the tens multiplied by the units, and the square
of the units. In the example we have taken, a = 4 tens or 40
units, and b = 7; we have then
a2- 1600
2ab=     560
b2-    49
Total,    a2 + 2 a b + b2 = 2209 = 47 X 47.




4w


Equations of the Second Degree with one Unknown Quantity. 109
Now in order to return, by a reverse process, from the number
2209 to i    oft, we may observe, that the square of the tens,
1600, Iiino  figure, which denotes a rank inferior to hundreds,
and that Ais the greatest square, which the 22 hundreds, comprehended H'2209, contain; for 22 lies between 16 and 25, that is,
between the square of 4 and that of 5, as 47 falls between 4 tens
or 40, and 5 tens or 50.
We find, therefore, upon examination, that the greatest square
contained in 22 is 16, the root of which 4 expresses the number
of tens in the root of 2209; subtracting 16 hundreds, or 1600,
fromn 2209, the remainder 609 contains double the product of the
tens by the units, 560, and the square of the units 49. But as
double the product of the tens by the units has no figure inferior
to tens, it must be found in the two first figures 60 of the remainder 609, which contain also the tens arising from the square of
the units. Now, if we divide 60 by double of the tens 8, and
neglect the remainder, we have a quotient 7 equal to the units
sought. If we multiply 8 by 7, we have double the product of
the tens by the units,- 560; subtracting this from the whole remainder 609, we obtain a difference 49, which must be, and in
fact is, the square of the units.
This process may be exhibited thus;
22,09  47
16    [87
60,9
60,9
000
We write the proposed number in the manner of a dividend,
and assign for the root the usual place of the divisor. We then
separate the units and tens by a comma, and employ only the
two first figures on the left, which contain the square of the tens
found in the root. We seek the greatest square 16, contained in
these two figures, put the root 4 in its assigned place, and subtract
16 from 22. To the remainder we bring down the two other
figures, 09, of the proposed number, separating the last, which
does not enter into double the product of the tens by the units,
and divide the remainder on the left by 8, double the tens in the
root, which gives for the quotient the units 7. In order to col



110


Elements of Jlgebra.


lect into one expression the two last parts of the square contained
in 609, we write 7 by the side of 8, which gives  f, equal to
double the tens plus the units, or 2 a + b; this, multiplied by 7
or 6, reproduces 609 = 2 a b + 62, or double the product of the
tens by the units, plus the square of the units. This being subtracted leaves no remainder, and the operation shows, that 47 is
the square root of 2209.
If it were required to extract the square root of 324; the operation would be as follows;
3,24  18
1
22,4  28
22,4
000
Proceeding as in the last example, we obtain 1 for the place
of tens of the root; this doubled gives the number 2, by which
the two first figures 22 of the remainder are to be divided. Now
22 contains 2 eleven times, but the root can neither be more than
10, nor 10; even 9 is in fact too large, for if we write 9 by the
side of 2, and multiply 29 by 9, as the rule requires, the result is
261, which cannot be subtracted from 224. We are, therefore,
to consider the division of 22 by 2 only as a means of approximating the units, and it becomes necessary to diminish the quotient
obtained, until we arrive at a product, which does not exceed the
remainder 224. The number 8 answers to this condition, since
8 X 28 = 224; therefore, the root sought is 18.
By resolving the square of 18 into its three parts, we find
a2= 100
2 a b =   160
b2-   64
Total,          324 = 18 X 18,
and it may be seen, that the 6 tens, contained in the square of the
units, being united to 160, double the product of the tens by the
units, alters this product in such a manner, that a division of it by
double the tens will not give exactly the units.
92. It will not be difficult, after what has been said, to extract
the square root of a number, consisting of three or four figures;
but some further observations, founded upon the principles above




Equations of the Second Degree with one Unknown Quantity. 111
laid down, may be necessary to enable the reader to extract the
root of any number whatever.
No number less than 100 can have a square consisting of more
than four figures, since that of 100 is 10000, or the least number
expressed by five figures. In order, therefore, to analyze the
square of any number exceeding 100, of 473, for example, we
may resolve it into 470 + 3, or 47 tens plus 3 units. To obtain
its square from the formula,
a2 + 2 a b + b2,
we make a - 47 tens = 470 units, b = 3 units, then
a2 = 220900
2    a b   2820
ba=       9
Total,          223729 = 473 X 473.
In this example, it is evident that the square of the tens has no
figure inferior to hundreds, and this is a general principle, since
tens multiplied by tens, always give hundreds, (drith. 32).
It is therefore in the part 2237, which remains on the left of the
proposed number, after we have separated the tens and units, that
it is necessary to seek the square of the tens; and as 473 lies between 47 tens, or 470, and 48 tens, or 480, 2237 must fall between the square of 47 and that of 4; hence the greatest square
contained in 2237, will be the square of 47, or that of the tens of
the root. In order to find these tens, we must evidently proceed,
as if we had to extract the square root of 2237 only; but instead
of arriving at an exact result, we have a remainder, which contains
the hundreds arising from double the product of the 47 tens multiplied by the units.
The operation is as follows;
22,37,29  473
16        87
63,7     943
60,9
282,9
282,9
0
We first separate the two last figures 29, and in order to extract




112


Elements of Algebra.


the root of the number 2237, which remaiin, on the left, we further separate the two last figures 37 of this number; the proposed number is then divided into portions of two figures, beginning on the right and advancing to the left. Proceeding with the
two first portions as in the preceding article, we find the two first
figures 47 of the root; but we have a remainder 28, which, joined
to the two figures 29 of the last portion, contains double the product of the 47 tens by the units, and the square of the units. We
separate the figure 9, which forms no part of double the product of
the tens by the units, and divide 282 by 94, double the 47 tens;
writing the quotient 3 by the side of 94, and multiplying 943 by 3,
we obtain 2829, a number exactly equal to the last remainder, and
the operation is completed.
93. In order to show, by what method we are to proceed with
any number of figures, however great, I shall extract the root of
22391824. Whatever this root may be, we may suppose it
capable of being resolved into tens and units, as in the preceding
examples. As the square of the tens has no figure inferior to
hundreds, the two last figures 24 cannot make a part of it; we
may therefore separate them, and the question will be reduced
to this, to find the greatest square contained in the part 223918,
which remains on the left. This part consisting of more than
two figures, we may conclude, that the number, which expresses
the tens in the root sought, will have more than one figure; it
may therefore be resolved, like the others, into tens and units.
As the square of the tens does not enter into the two last figures
18 of the number 223918, it must be sought in the figures 2239,
which remain on the left; and since 2239 still consists of more
than two figures, the square, which is contained in it, must have
a root which consists of at least two; the number which expresses the tens sought will therefore have more than one figure;
it is then, lastly, in 22 that we must seek the square of that,
which represents the units of the highest place in the root required.  By this process, which may be extended to any length
we please, the proposed number may be divided into portions
of two figures from right to left; it must be understood, however, that the last figure on the left may consist of only one
figure.




Equations of the Second Degree with one Unknown Quantity. 113
Having divided the proposed number into portions as below,
we proceed with the three first portions, as 22,39,18,24 1 4732
in the preceding article; and when we   16         1 87
have found the three first figures 473 of the  63,99
root, to the remainder 189, we bring down  60,9     9462
the fourth portion 24, and consider the   301,8
number 18924, as containing double the    282 9
product of the 473 tens already found by   1892,4
the units sought, plus the square of these  1892 4
units.  We separate the last figure 4; di-  0000 0
vide those, which remain on the left, by 946, double of 473, and
then make trial of the quotient 2, as in the preceding examples.
Here the operation, in the present case, terminates; but it is
very obvious, that if we had one portion more, the four figures
already found 4732, would express the tens of a root, the units
of which would remain to be sought; we should proceed, therefore, to divide the remainder now found, together with the first figure of the following portion, by double of these tens, and so on
for each of the portions to be successively brought down.
94. If, after having brought down a portion, the remainder,
joined to the first figure of this portion, does not contain double
of the figures already found, a cipher must be placed in the root;
for the root, in this case, will have no units of this rank; the
following portion is then to be brought down, and the operation
to be continued as before. The example subjoined will illustrate
this case. The quantities to be subtracted are  49,42,09 1 703
not,put down, but the subtractions are suppos-  04,20, 9 1 1403
ed to be performed mentally, as in division.  0,00, 0
95. Every number, it will be perceived, is not a perfect square.
If we look at the table given, page 108, we shall see that between
the squares of each of the nine primitive numbers, there are intervals comprehending many numbers, which have no assignable
root; 45, for instance, is not a square, since it falls between 36
and 49. It very often happens, therefore, that the number, the
root of which is sought, does not admit of one; but if we attempt
to find it, we obtain for the result the root of the greatest square,
which the number contains. If we seek, for example, the root of
2276, we obtain 47, and have a remainder 67, which shows, that
the greatest square contained in 2276, is that of 47 or 2209.
llg.                 15




114


Elements of Algebra.


As a doubt may sometimes arise, after having obtained the
root of a number which is not a perfect square, whether the
root found be that of the greatest square contained in the number, I shall give a rule, by which this lmay be readily determined.
As the square of a + 6 is
a2 + 2 a b +-  2
if we make b = 1, the square of a + 1 will be
a2   2 a +,
a quantity which differs from a2, the square of a, by double of a
plus unity. Therefore, if the root found can be augmented by
unity, or more than unity, its square, subtracted from the proposed
number, will leave a eremainder at least equal to twice this root plus
unity.  Whenever this is not the case, tile root obtained will be,
in fact,\that of the greatest square contained in the number proposed.
96. Since a fraction is multiplied by another fraction, when
their numerators are multiplied together, and their denominators
together, it is evident that the product of a fraction multiplied by
itself, or the square of a fraction is equal to the square of its numerator, divided by the square of its denominator. Hence it follows, that to extract the square root of a fraction, we extract the
square root of its numerator and that of its denominator. Thus
the root of %f is 5, because 5 is the square root of 25, and 8 that
of 64.
It is very important to remark, that not only are the squares of
fractions, properly so called, always fractions, but every fractional
number which is irreducible (Arith. 59), will, when multiplied by
itself, give a fractional result, which is also irreducible.
97. This proposition depends upon the following; Every prime
number P, which will divide theproduct AB of two numbers A and
B, will necessarily divide one of these numbers.
Let us suppose, that it will not divide B, and that B is the greater; if we designate the entire part of the quotient by q, and the
remainder by B', we'have
B = qPP + B',
multiplying by a., we obtain
dB = qA P + aB',
and dividing the two members of this equation by P, we have
AB'          AB'
P     q- + +-p- 




Equations of the Second Degree with one Unknown Quantity. 115
from which it appears, that if IB be divisible by P, the product
B/' will be divisible by the same number. Now B', being the
remainder after tile division of B by P, must be less than P;
therefore B/ cannot be divided by P; if we divide P by B' we
have a quotient q' and a remainder B"; if further we divide P
by B", we have a quotient q" and a remainder B"', and so on,
since P is a prime number.
We have, therefore, the following series of equations;
P    q' B' + B", P= q" B" + B'", &c.
multiplying each of these by., we obtain
sP = q' IdB' + stB", sP = q" sIB" -+ sB"', &c.
dividing by P, we have
A B' AB    "    AB"              AR"'AB" 
=     -p- +                        +p,  =q"-p-+ -p-, 
From these results it is evident, that if.IB' be divisible by P, the
products sIB", JIB"/, &c. will also be divisible by it. But the
remainders B', B", B"', &c. are becoming less and less, continually, till they finally terminate in unity, for the operation exhibited
above may be continued in the same manner, while the remainder is
greater than 1, since P is a prime number. Now when the remainder becomes unity, we have the product A X 1, which must
be divisible by P; therefore s also must be divisible by P.
Hence, if the prime number P, which we have supposed not to
divide B, will not divide.1, it will not divide the product of these
numbers.
(This demonstration is taken principally from the Theorie des
Nombres of M. Legendre.)
98. Now when the fiaction - is irreducible, there is no prime
number which will divide, at the same time, 6 and a; but, from
the preceding demonstration, it is evident, that every prime number, which will not divide a, will not divide a X a, or a2; every
prime number; which will not divide b, will not divide b X b, or
b2; the numbers a2 and b6 are, therefore, in this case, prime to
6b2
each other; and consequently the square - of the fraction -, being
irreducible, as well as the fiaction itself, cannot become an entire
number.
99. From this last proposition it follows, that entire numbers,
except only such as are perfect squares, admit of no assignable root,




116


Elements of algebra.


either among whole numbers or fractions. Yet it is evident, that
there must be a quantity, which, multiplied by itself, will produce
any number whatever, 2276, for instance, and that in the present
case, this quantity lies between 47 and 48; for 47 X 47 gives a
product less than this number, and 48 X 48 gives one greater.
Dividing then the difference between 47 and 48 by means of
fractions, we may obtain numbers that, multiplied by themselves,
will give products greater than the square of 47, but less than
that of 48, and which will approach nearer and nearer to the
number 2276.
The extraction of the square root, therefore, applied to numbers which are not perfect squares, makes us acquainted with a
new species of numbers, in the same manner, as division gives
rise to fractions; but there is this difference between fractions
and the roots of numbers which are not perfect squares; that
the former, which are always composed of a certain number of
parts of unity, have with unity a common measure, or a relation
which may be expressed by whole numbers, which the latter
have not.
If we conceive unity to be divided into five parts, for example,
we express the quotient arising from the division of 9 by 5, or
-, by nine of these parts; - then, being contained five times in
unity, and nine times in 9, is the common measure of unity and
the fraction 9, and the relation these quantities have to each other
is that of the entire numbers 5 and 9.
Since whole numbers, as well as fiactions, have a common measure with unity, we say that these quantities are commensurable with
unity, or simply that they are commensurable; and since their relations or ratios, with respect to unity, are expressed by entire
numbers, we designate both whole numbers and fractions by the
common name of rational numbers.
On the contrary, the square root of a number, which is not a
perfect square, is incommensurable or irrational, because, as it
cannot be represented by any fraction, into whatever number of
parts we suppose unity to be divided, no one of these parts will be
sufficiently small to measure exactly, at the same time, both this
root and unity.
In order to denote, in general, that a root is to be extracted,
whether it can be exactly obtained or not, we employ the character V-, which is called a radical sign;




Equations of the Second Degree with one Unknown Quantity. 117


/16 is equivalent to 4,
v2 is incommensurable or irrational.
100. Although we cannot obtain, either among whole numbers
or fractions, the exact expression for  '2, yet we may approximate it, to any degree we please, by converting this number into
a fraction, the denominator of which is a perfect square. The
root of the greatest square contained in the numerator will then
be that of the proposed number expressed in parts, the value of
which will be denoted by the root of the denominator.
If we convert, for example, the number 2 into twenty-fifths,
we have  -. As the root of 50 is 7, so far as it can be expressed
in whole numbers, and the root of 25 exactly 5, we obtain 7, or
1 for the root of 2, to within one fifth.
101. This process, founded upon what was laid down in article
96., that the square of a fraction is expressed by the square of the
numerator divided by the square of the denominator, may evidently be applied to any kind of fraction whatever, and more
readily to decimals than to others. It is manifest, indeed, from
the nature of multiplication, that the square of a number expressed by tenths will be hundredths, and that the square of a number
expressed by hundredths will be ten thousandths, and so on; and
consequently, that the number of decimal figures in the square is
always double that-of the decimal figures in the root. The truth
of this remark is further evident fiom the rule observed in the
multiplication of decimal numbers, which requires that a product
should contain as many decimal figures, as there are in both the
factors. In any assumed case, therefore, the proposed number,
considered as the product of its root multiplied by itself, must
have twice as many decimal figures as its root.
From what has been said, it is clear, that in order to obtain
the square root of 227, for example, to within one hundredth, it
is necessary to reduce this number to ten thousandths, that is, to
annex to it four ciphers, which gives 2270000 ten thousandths.
The root of this may be extracted in the same mannner, as that
of an equal number of units; but to show that the result is hundredths, we separate the two last figures on the right by a comma.
We thus find that the root of 227 is 15,06, accurate to hundredths.
The operation may be seen below;




118


Elements of llgebra.


2,27,00,00 1 1506
12,7       25
2 00 00    3006
1964
If there are decimals already in the proposed number, they
should be made even. To extract, for example, the root of 51,7,
we place one cipher after this number, which makes it hundredths; we then extract the root of-51,70. If we proposed to
have one decimal more, we should place two additional ciphers
after this number, which would give 51,7000; we should then
obtain 7,19 for the root.
If it were required to find the square root of the numbers 2
and 3 to seven places of decimals, we should annex fourteen ciphers to these numbers; the result would be
V  = 1,4142136,     v/3    1,7320508.
102. When we have found more than half the number of
figures, of which we wish the root to consist, we may obtain the
rest simply by division. Let us take, for example, 32976; the
square root of this number is 181, and the remainder, 215. If
we divide this remainder 215, by 362, double of 181, and extend
the quotient to two decimal places, we obtain 0,59, which must
be added to 1S1; the result will be 181,59 for the root of 32976,
which is accurate to within one hundredth.
In order to prove that this method is correct, let us designate
the proposed number by JV, the root of the greatest square contained in this number by a, and that which it is necessary to add
to this root to nake it the exact root of the proposed number by
b; we have then
V= a2 + 2 a b + 62,
from which we obtain
N    a2 =2ab     b2;
dividing this by 2 a, we find
N- a2          b2
2a     -+ 2a'
From this result it is evident, that the first member may be
taken for the value of 6, so long as the quantity  - is less than a
unit of the lowest place found in b. But as the square of a number cannot contain more than twice as many figures as the number itself, it follows, that if the number of figures in a exceeds
double those in b, the quantity b will then be a fraction.
2- l the    bea  katin




Equations of the Second Degree with one Unknown Quantity. 119


In the preceding example, a = 181 units, or 18100 hundredths,
and consequently contains one figure more than the square of
b2
59 hundredths;.the fraction then 2  becomes in this case,
2 a
(59)2  s34S1
2 X 18100 )  36200' and is less than a unit of the second part
59, or than a hunidredth of a unit of the first.
103. This leads to a method of approximating the square root
of a number by means of vulgar fractions. It is founded on the
circumstance, that a, being the root of the greatest square contained in JN, b is necessarily a fraction, and 2- being much smaller than b, may be neglected.
If it were required, for example, to extract the square root of
2; as the greatest square contained in this number is 1, if we
subtract this, we have a remainder, 1. Dividing this remainder
by double of the root, we obtain; taking this quotient for the
value of the quantity b, we have, for the first approximation to
the root, 1 +-, or -. Raising tills root to its square, we find 9,
which, subtracted from 2 or 3, gives for a remainder -. In
this case the formula
N- a2          b2
2a'
becomes
I         bs
-   2 -b+     a-.
Substituting - y- for 6, we have for the second approximation
2     1 - - 7; taking the square of -17 we find 21  a quantity,
which still exceeds 2 or 2 8  Substituting 11 for a, we obtain
1           62
--  b + ---
12 X 34        2a
which gives
1     1
b                    T ---
12 X 34       408'
the third approximation will then be
17       1      17 X 34-1     577
12- 12 X 34         408       408'
This operation may be easily continued to any extent we
please. I shall give, in the Supplement to this treatise, other formulas more convenient for extracting roots in general.




120


1lements of.lgebra.


104. In order to approximate the square root of a fraction, the
method, which first presents itself, is, to extract, by approximation, the square root of the numerator and that of the denominator; but with a little attention it'will be seen, that we may avoid
one of these operations by making the denominator a perfect
square. This is done by multiplying the two terms of the proposed fraction by the denominator. If it were required, for example, to extract the square root of 3, we might change this
fraction into
'3X7     21
7 X 7- 49'
by multiplying its two terms by the denominator, 7. Taking the
root of the greatest square contained in the numerator of this
fraction, we have 4 for the root of, accurate to within 4.
If a greater degree of exactness were required, the fraction 4
must be changed by approximation or otherwise into another, the
denminator of which is tie square of a greater number than 7.
We shall have, for example, the root sought within 1, if we
convert 4 into 225ths, since 225 is the square of 15; thus the
fraction becomes 645 of one 225th, or —g-,, within -,  the root of
g95 falls between T and '-, but approaches nearer to the second
fraction than to the first, because 96 approaches nearer to a hundred than to 81; we have then I or 2 for the root of 4 within 1-.
By employing decimals in approximating the root of the numerator of the fraction 4-, we obtain 4,583 for the approximate
root of the numerator 21, which is to be divided by the root-of
the new denominator. The quotient thence arising, carried to
three places of decimals, becomes 0,655.
105. We are now prepared to resolve all equations involving
only the second power of the unknown quantity connected with
known quantities.
We have only to collect into one member all the terms containing
this power, to free it from the quantities, by which it is multiplied
(11); we then obtain the value of the unknown quantity by extracting the square root of each member.
Let there be, for example, the equation
x2- 8     4 -    2.
Making the divisors to disappear, we find first
15 2- 168- 84- 14 x2.
Transposing to the first member the term 14 x2, and to the second the term 168, we have




Equations of the Second Degree with one Unknown Quantity. 121


15x2+ 14wx2     84 + 168,
or                     29 x2 - 252,
and                       X2 — 2,2
X   - - 25
It should be carefully observed, that to denote the root of the
fraction?2-2, the sign v' is made to descend below the line,
which separates the numerator from the denominator. If it were
written thus,  25, the expression would designate the quotient
arising from the square root of the number 252 divided by 29; a
result different from  2s, which denotes, that the division is to
be performed before the root is extracted.
Let there be the literal equation
ax2 + b3 - c x2 + d3;
proceeding as with the above, we obtain successively
ax-x2     - d3-   b3,
x_ d3 -b
a-c
I would remark here, that in order to designate the square rootI would remark here, that in order to designate the square root
of a compound quantity, the upper line must be extended over the
whole radical quantity.
The root of the quantity 4 a2 b - 2 b3 + c3 is written thus,
/4 a b -2 b3 + c3,
or rather
^, (4 a2 b - 2 b3 + C3),
by substituting, for the line extended over the radical quantity, a
parenthesis including all the parts of the quantity, the root of which
is required. This last expression may often appear preferable to
the other (35).
In general, every equation of the second degree of the kind we
are here considering, may, by a transposition of its terms, be reduced to the form
p a:2
P- designating the coefficient, whatever it may be, of H.  We
then obtain
2.   aq
P


Alg.


16




122


Elements of Algebra.


106. With respect to numbers taken independently, this solution
is complete, since it is reduced to an operation upon the number either entire or fractional, which the quantity a-q represents, an
arithmetical operation leading always to an exact result, or to one,
which approaches the truth very nearly. But in regard to the
signs, with which the quantities may be affected, there remains,
after the square root is extracted, an ambiguity, in consequence of
which every equation of the second degree admits of two solutions,
while those of the first degree admit of only one.
Thus in the general equation x2 - 25, the value of x, being the
quantity, which, raised to its square, will produce 25, may, if we
consider the quantities algebraically, be affected either with the
sign + or —; for whether we take + 5, or -  5, for this value,
we have for the square
+5 X +5=+25, or -5x- 5 =+25;
we may therefore take
x- + 5,
or                      x   - 5.
For the same reason, from the general equation
2    a q
P'
we have
-=_+     a
UP
or
Both these expressions are comprehended in the following;
x =   - 1q:P
in which the double sign ~: shows, that the numerical value of
faqI
UP
may be affected with the sign + or
From what has been said, we deduce the general rule, that the
double sign ~= is to be considered as affecting the square root of
every quantity whatever.
It may be here asked, why x, as it is the square root of x2, is
not also affected with the double sign ~:? We may answer, first,




Equations of the Second Degree with one Unknown Quantity. 123
that the letter x, having been taken without a sign, that is, with
the sign +, as the representative of the unknown quantity, it is its
value when in this state, which is the subject of inquiry; and that,
when we seek a number x, the square of which is 6, for example,
there can be only two possible solutions; x = -+ {V, x = -
/b.   Again, if in resolving the equation x2 = b, we write d= x 
~ff 1/b, and arrange these expressions in all the different ways, of
which they are capable, namely,
+x —+/, -x V-,
+ x -vb,          x =+ v,
we come to no new result, since by transposing all the terms of
the equations - x = -  /-, - x   +- v    or which is the same
thing, by changing all the signs (57), these equations become identical with the first.
107. It follows from the nature of the signs, that if the second
member of the general equation
2     aq
P
were a negative number, the equation would be absurd, since the
square of a quantity affected either with the sign +  or -, having
always the sign -+ no quantity, the square of which is negative,
can be found either among positive or negative quantities.
This is what is to be understood, when we say, that the root of
a negative quantity is imaginary.
If we were to meet with the equation
x2 + 25 - 9,
we might deduce from it
x2 = 9- 25,
or                    x2     -16;
but, there is no number, which, multiplied by itself, will produce
- 16. It is true, that -  4 multiplied by + 4, gives - 16; but
as these two quantities have different signs, they cannot be considered as equal, and consequently their product is not a square.
This species of contradiction, which will be more fully considered
hereafter, must be carefully distinguished from that mentioned in
art. 58., which disappears by simply changing the sign of the unknown quantity; here it is the sign of the square x2, which is to
be changed.
108. To be complete, an equation of the second degree, with
only one unknown quantity, must have three kinds of terms, nare



124


Elements of Algebra.


ly, those involving the square of the unknown quantity, others containing the unknown quantity of the first degree, and lastly, such as
comprehend only known quantities. Tilhe following equations are
of this kind;
x2 -4x= 12,         4 x -  3 x2 = 4-2 x.
The first is, in some respects, more simple than the second,
because it contains only three terms, and the square of x is positive, and has only unity for a coefficient. It is to this last form,
that we are always to reduce equations of the second degree,
before resolving them; they may then be represented by this
formula,                x2 +-  x = q,
in whichp and q denote known quantities, either positive or negative.
It is evident, that we may reduce all equations of the second degree to this state, 1. by collecting into one member all the terms
involving x (10), 2. by changing the sign of each term of the
equation, in order to render that of x2 positive, if it was before
negative (57), 3. by dividing all the terms of the equation by the
multiplier of x2, if this square have a multipler (11), or by multiplying by its divisor, if it be divided by any number (12).
If we apply what has just been said to the equation
4x- x    c32 = 4- 2x,
we have, by collecting into the first member all the terms involving x,
- -x2 + 6 x - 4,
by changing the signs,
3 2 - 6 x -- - 4,
multiplying by the divisor 5,
3 x2- 30 x --    20,
dividing by the multiplier 3,
x2 - 10   =     2 0
If we now compare this equation with the general formula
X2 +  x    q,
we shall have
p      1o, O  q      2_ _.
109. In order to arrive at the solution of equations thus prepared, we should keep in mind what has been already observed
(34), namely, that the square of a.quantity, composed of two
terms, always contains the square of the first term, double the product of the first term multiplied by the second, and the square of




Equations of the Second Degree with one Unknown Quantity. 125
the second; consequently the first member of the equation
X2 + 2 a x + a2 = b,
in which a and b are known quantities, is a perfect square, arising
from x + a, and may be expressed thus,
(x + a) (x + a) = b.
If we take the square root of the first member and indicate that of
the second, we have
x +- a =   Vib,
an equation, which, considered with respect to x, is only of the
first degree; and from which we obtain, by transposition,
x = - a ~   /b.
An equation of the second degree may therefore be easily resolved, whenever it can be reduced to the form
x2 + 2 a x + a2 = b,
that is, whenever its first member is a perfect square.
But the first member of the general equation
X2 p-x = q
contains already two terms, which may be considered as forming
part of the square of a binomial; namely, x2, which is the square
of the first term x, and px, or double the first multiplied by the
second, which second is consequently only half of p, or pp. To
complete the square of the binomial x +- Ip, there must be also
the square of the second term, p; but this square may be formed, since p and -p are known quantities, and it may be added to
the first member, if, to preserve the equality of the two members,
it be added at the same time to the second; and this last member
will still be a known quantity.
As the square of p is ~p2, if we add it to the two members
of the proposed equation,
x2 -   x = q,
we shall have
x2 +pX +I P      =q +     P2.
The first member of this result is the square of x -+,p; taking
then the root of the two members, we have
x+ 1 p= -~I    q+    p2,  (106);
by transposition this becomes
X   -IP~ /q+ p2 
or which is the same thing
X   -    p+Vq + I p2,
2 P +    -t 4 




126


Elements of dlgebra.


and              x  — p-        q +-P2.
I have prefixed the sign + to the second term Ip, of the root
of the first member of the above equation, because the second
term of this member is positive; the sign - is to be prefixed in
the contrary case, because the square x2 -2 a x + a2 answers to
the binomial x - a.
Any equation whatever of the second degree may be resolved,
by referring it to the general formula,
x2 +px=q;
or more expeditiously, by performing immediately upon the equation the operations represented under this formula, which, expressed in general terms, are as follows;
To make the first member of the proposed equation a perfect
square, by adding to it, and also to the second, the square of half
the given quantity, by which the first power of the unknown quantity is multiplied; then to extract the square root of each member,
observing, that the root of the first member is composed of the unknown quantity, and half of the given number, by which the unknown quantity in the second term is multiplied, taken with the
sign of this quantity, and that the root of the second member must
have the double sign ~, and be indicated by the sign )/, if it cannot be obtained directly.
See this illustrated by examples.
110. To find a number such, that if it be multiplied by 7, and
this product be added to its square, the sum will be 44.
The number sought being represented by x, the equation will
evidently be
x2 + 7x = 44.
In order to resolve this equation, we take ', half of the coefficient 7, by which x is multiplied; raising it to its square we obtain
449; this added to each member gives
x2+ [7x+    4-9 = 44 +  -9;
reducing the second member to a single term, we have
x2 + 7 x +   9    2 2 5
The root of the first member, according to the rule given above, is
x + I, and we find for that of the second 15; whence arises the
equation
x +   —. I415
x + 2 --     2 X
from which we obtain
x — _    ~:,




Equations of the Second Degree with one Unknown Quantity. 127


or                  = -     + V_ =-  = 4,
XL,         1 7  15 - - 22 =  1.
The first value of x solves the question in the sense, in which it
was enunciated, since we have by this value
x2    16
7 x -28
sum                            44
As to the second value of x, since it is affected with the sign
-, the term 7 x, which becomes
7 X -11    — 77,
must be subtracted fiom x2, so that the enunciation of the question resolved by the number 1 1 is this,
To find a number such, that 7 times this number being subtracted
from its square, the remainder will be 44.
Thle negative value then here modifies the question in a manner,
analogous to what takes place, as we have already seen, in equations of the first degree.
[f we put the question, as enunciated above, into an equation, we
obtain
X - 7 x = 44,
this becomes, when resolved,
x2    7 x + -49 = 44 + 4,
x2 -7x+ 49- =      22 5
X    27 -  i 1 5
2       2
X      ~ 1 5
X    22 
= -- 7      -L- 4,
- 2     --   2 
The negative value of x becomes positive, as it satisfies precisely the new enunciation, and the positive value, which does not thus
satisfy it, becomes negative.
Hence we see, that in equations of the second degree, algebra
unites under the same formula two questions, which have a certain
analogy to each other.
111. Sometimes enunciations, which produce equations of the
second degree, admit of two solutions. The following is an example;
To find a number such, that if 15 be added to its square, the
sum will be equal to 8 times this number.
Let x be the number sought; the equation arising from the problem is then          x2 + 15 = 8 x.




126


Elements of./lgebra.


This equation reduced to the form prescribed in art. 108., becomes
x2-8 x = - 15,
x2 -Sx+ 16 ==-15+16,
x2-8 x + 16     -1,
x-    4~ ~ 1,
x- 4:= 1,
or                      x     5,
x=    3.
There are therefore two different numbers 5 and 3, which fulfil
the conditions of the question.
112. Questions sometimes occur, which cannot be resolved precisely in the sense of the enunciation, and which require to be
modified. This is the case, when the two roots of the equation
are negative, as in the following example,
x2 + 5 x + 6   2.
This equation, which denotes, that the square of the number
sought, augmented by 5 times this number, and also by 6, will give
a sum equal to 2, evidently cannot be verified by addition, as is
implied, since 6 already exceeds 2. Indeed, if we resolve it, we
find successively
x2 + 5x    -  4,
x2 +5x + 2 Y     =5 2- 4 
X +   2- f=~    
T     2 —
X  -   5 ----2 23   4.
From the sign -, with which the numbers 1 and 4 are affected, it
may be seen that the term 5 x must be subtracted from the others,
and that the true enunciation for both values is,
To find a number such, that if 5 times this number be subtracted
from its square, and 6 be added to the remainder, the result will be 2.
This enunciation furnishes the equation,
x2 - 5 x   6 =2,
which gives for x the two positive values 1 and 4.
113. Again, let the following problem be proposed;
To divide acnumber p into two parts, the product of which shall
be equal to q.
If we designate one of these parts by x, the other will be expressed by p - -x, and their product will be p x- x2; we have
then the equation         px -     = q,




Equations of the Second Degree with one Unknown Quantity. 129
or, changing the signs,
x2-px=-q;
resolving this last, we find
X = Ip ~ V/p2 —q.
If now we suppose
= 10,     q   21,
we have
x = 5 =: i/25-21,
or               x = 5   - 2,
x   7,
x   3,
that is, one of the parts will be 7, and the other consequently
10 -7, or 3.
If, on the contrary, we take 3 for x, the other part will be
10 -  3 or 7; so that the enunciation, as it stands, admits, strictly
speaking, of only one solution, since the second amounts simply
to a change in the order of the parts.
If we examine carefully the value of x in the question we have
been considering, we shall see that we cannot take any numbers
indifferently for p and q, for if q exceed t or the square of -p,
ap2
the quantity _ — q becomes negative, and we are presented with
that species of absurdity mentioned in art. 107.
If we take, for example,
p-= 10 and q =30,
we have
x = 5 =~: /25 -30 = 5 ~  /-5;
the problem then with these assumptions is impossible.
114. The absurdity of questions, which lead to imaginary roots,
is discovered only by the result, and we may wish to determine by
characters, which are found nearer to the enunciation, in what consists the absurdity of the problem, which gives rise to that of the
solution; this we shall be enabled to do by the following consideration.
Let d be the difference of the two parts of the proposed number; the greater part will be  + d, the less  -  (3); but it
has been proved (29, 30, & 34) that
Jilg.                 17                    [




130


Elements of Algebra.


/p +d\     p   d\    p2   dP
\;2    } +\2" 2J-~4      ~4 2
therefore, the product of the two parts of the proposed number,
p2
whatever they may be, will always be less than 4-, or than the
square of half their sum, so long as d is any thing but zero; when
d is nothing, each of the two parts being equal to, their product
will be only. It is then absurd to require it to be greater; and
it is just, that algebra should answer in a manner contradictory to
established principles, and thereby show, that what is sought does
not exist.
What has been proved concerning the equation
x-2 -p x   — q,
furnished by the preceding question, is true of all those of the
second degree, where q is negative in the second member, the only equations, which produce imaginary roots, since the term'-,
placed under the radical sign, preserves always the sign +-, whatever may be that of p. Indeed, it is evident that the equation
x- +px    -   q, or x2+ px+ q       0,
will admit of no positive solution, since the first member contains
only affirmative terms; and, to ascertain whether the unknown
quantity x can be negative, we have only to change x into -  y.
The unknown quantity y would then have positive values, which
would be furnished by the equation
y2_ py + q=0,      or y2-py       — q,
which is precisely the same as that in the preceding article; but
as the values of x can be real only when those of y would be so,
they become therfore imaginary in the case under consideration,
p2
when q exceeds ~.
It will be perceived then from what has been said, how, and for
what reason, when the known term of an equation of the second
degree is negative in the second member, and greater than the
square of half the coefficient of the first power of the unknown
quantity, this equation can have only imaginary roots.
115. The expressions
V/-b, a +     -b,
and, in general, those, which involve the square root of a negative




Equations of the Second Degree with one Unknown Quantity. 131


quantity, are called imaginary quantities*. They are mere symbols of absurdity, that take the place of the value, which we
should have obtained, if the question had been possible.
They are not, however, to be neglected in the calculation, because it sometimes happens, that when they are combined according to certain laws, the absurdity disappears, and the result becomes real. Examples of this kind will be found in the Supplemnent to this treatise.
116. As it is important that learners should have just ideas respecting all those analyticalfacts, which appear to be derived from
familiar notions, I have thought it proper to add some observations
to what has been said (106), on the necessity of admitting two solutions in equations of the second degree.
I shall show that, if there exists a quantity a, which, substituted
in the place of x, verifies the equation of the second degree,
x2 + p x = q, and is consequently the value of x, this unknown
quantity will still have another value. Now, if we substitute a for
x, the result will be a2 -p a = q; and since, by supposition, a
represents the value of x, q will be necessarily equal to the quantity a2 + p a; we may then write this quantity in the place of q,
in the proposed equation, which thus becomes
x2 — px = a2 +p a.
Transposing all the terms of the second member, we have
x2 q- p x -a2-pa =0,
which may be written,
xa -a2 + p (x - a) =0;
and because
x2 - a2    (x + a) (x- a), (34),
it is obvious, at once, that the first member is divisible by x- a,
and will give an exact quotient, namely, x +  a +p; we have
then,
x2 + px -q =      3 —a2 +p (x -a) = (x- a) (x      a + p).
Now it is evident, that a product is equal to zero, when any one
of its factors whatever becomes nothing; we shall have then
(x - a) (x   + +p) = 0,
not only when x - a = 0, which gives
x = a,


* It would be more correct to say, imaginary expressions, or symbols, as they are not quantities.




132


Elements of Algebra.


but also when x + a + p = 0, from which is deduced
x   - a -p.
Therefore, if a is one of the values of x, - a -p will necessarily be the other.
This result agrees with the two values comprehended in the
formula
x     2-~p~'   +'i/ p2;
for if we take for a the first value, -- p   -- /  + iP2 we obtain for the other
a —  = +    p — /q + ~ p2 -p          lp- v- q + 1P2
which is in fact the second value.
These remarks contain the germ of the general theory of equations of whatever degree, as will appear hereafter, when the subject will be resumed.
117. The difficulty of putting a problem into an equation, is
the same in questions involving the second and higher powers, as
in those involving only the first, and consists always in disentangling and expressing distinctly in algebraic characters all the conditions comprehended in the enunciation. The preceding questions present no difficulty of this sort; and, although the learner is
supposed to be well exercised in those of the first degree, I shall
proceed to resolve a few questions, which will furnish occasion for
some instructive remarks.
J. person employed two laborers, allowing them different wages;
thefirst received, at the end of a certain number of days, 96francs,
and the second, having worked six days less, received only 54
francs; if this last had worked the whole number of days, and the
other had lost six days, they would both have received the same
sum; it is required to find how many days each worked, and what
sum each received for a day's work.
This problem, which at first view appears to contain several unknown quantities, may be easily solved by means of one, because
the others may be readily expressed by this.
If x represent the number of days' work of the first laborer,
x - 6 will be the number of days' work of the second,
96fr.
96  will be the daily wages of the first,
54
z-6 the daily wages of the second;
if this last had worked x days, he would have earned




Equations of the Second Degree with one Unknown Quantity. 133
54     54 z
x X -- - or r
and the first working x - 6 days, would have received only
96     96 (x -6)
(x- 6), or       -6
The equation of the problem then will be
54x     96(a —6)
x   6       x
The first step is to make the denominators disappear; the equation then becomes
54 x2 = 96 (x - 6) (x - 6).
As the numbers 54 and 96 are both divisible by 6, the result may
be simplified by division; we shall then have
9x2 = 16 (x - 6) (x - 6).
This last equation may be prepared for solution according to the
rule given art. 108., but as the object of this rule is to enable us
with more facility to extract the root of each member of the equation proposed, it is here unnecessary, because the two members
are already presented under the form of squares; for it is evident,
that 9 x2 is the square of 3 x, and 16 (x - 6) (x - 6) the square
of 4 (x - 6). We have then
3x-=:~4 (x - 6);
from which may be deduced
3 x   4 x — 24, x= 24,
3x = — 4x +24, x =    2-.
By the first solution, the first laborer worked 24 days, and consequently earned  6 or 4 francs per day, while the second worked
only 18 days, and received k4 or 3 francs per day.
The second solution answers to another numerical question, connected with the equation under consideration, in a manner analogous to what was noticed in art. 111.
118..J banker receives two notes against the same person; the
first of 550 francs, payable in seven months, the second of 720
francs, payable in four months, and gives for both the sum of 1200
francs; it is required to find what is the annual rate of interest,
according to which these notes are discounted.
In order to avoid fractions in expressing the interest for seven
months and four months, we shall represent by 12 x the interest of
100 francs for one year; the interest for one month will then be x.
The present value of the first note will accordingly be found by
the proportion,




134


Elements of Algebra.


55000
100 + 7 I: 100:: 550: 10 —       (Idrith. 120.);
and the present value of the second note by the proportion,
72000
100+ 4 x: 100:7: 720: -
100 - 4 x
By uniting these values, we obtain for the equation of the problem,
55000       72000_
100 + 7 + 100+ 4 - - 1200.
Dividing each of the members by 200, we have
275         360
100+7x + 100     4x 4  6;
making the denominators disappear, we find successively,
275 (100 +4x)+ 360 (100 + 7 x)-6 (100 + 7 x) (100 +4 x),
27500 + 1 100x + 36000 + 2520x = 60000 + 6600x + 168x2,
which may be reduced to
168 x2 + 2980 x = 3500;
dividing by 2, we obtain
84 x2 + 1490 x = 1750,
which gives
1490     1750
X + -84 x       84
Comparing this equation with the formula,
x2 +px= q,
we have
1490         1750
P          84'  - 84;
and the expression
x=-2-p^ i4 +,
becomes
745         745. 745    1750
8-4x ---- 8 —*
84       4    84  84  8
Reducing the fractions, we have
745. 745 + 1750. 84  702025
84.84           84.84'
then, since the denominator of this fraction is a perfect square, we
have only to extract the square root of its numerator. If we stop
at thousandths, we find 837,869, for the root of 702025; this,
taken with the denominator 84, gives for the values of x
745   837,869   92,869
- ~84   84        84-'




Equations of the Second Degree with one Unknown Quantity. 135
745   837,869      1582,869
-   4 -8 -   84            4
The first of these values is the only one, which solves the question in the sense, in which it was enunciated. Dividing the denominator of this fiaction by 12, we have (.Arith. 54.)
92,869
12 x  -'-    = 13,267;
that is, the annual interest is at the rate of 13,27 nearly.
119. The following question deserves attention on account of
the character, which the expression for the unknown quantity presents.
To divide a number into two parts, the squares of which shall be
in a given ratio.
Let a be the given number,
m the ratio of the squares of its two parts,
x one of these parts;
the other will be a- x.
We shall then have, according to the enunciation,
x2
(a-x) (a-x)
This may be resolved in two ways; we may either reduce it to
the form x2 -p x-=, and then resolve it by the common
method; or since the fraction
(a -;) (a -x)
is a square, the numerator and denominator being each a square,
we thence conclude at once,
a -- 
x  - i (a - x) /m.
By resolving separately the two equations of the first degree comprehended in this formula, namely,
x    + (a - x) Vm
x   -(a- x) v/m,
we have
a m
1 +v/- a /n
1 -- /m
By the first solution, the second part of the number proposed is
a /m' a + a V/m - a m             a
a -- 1+m             1 + /m        - 1 +;_




136


Elements of Algebra.


and the two parts,
a /m          a
-  and,
1 + ~/m     1 ~+  /m
are both, as the enunciation requires, less than the number proposed.
By the second solution we have
a V/      a- a /m' + a Vm _       a
a +                                 1_ --.m  j
and the two parts are
a V/          a
---- and           -
Their signs being opposite, the number a is strictly no longer
their sum, but their difference.
If we make m = 1, that is, if we suppose that the squares of
the two parts sought are equal, we have
V/n = 1;
and the first solution will give two equal parts,
a      a
'2     2'
a conclusion, that is self-evident, while the second solution gives
for the results two infinite quantities (68), namely,
- -a   -a         a      a
1       -1 or -—, and  or.
This is necessary, for it is only by considering two quantities infinitely great, with respect to their difference a, that we can suppose the ratio of their squares equal to unity.
Now, let there be the two quantities x, and x - a, the ratio of
their squares will be
2;2
x2 - 2 a x - a2 
dividing the two terms of this fraction by x2, we obtain
1
2a    aa2
1        tbut it is evident, that the greater the number x, the less will be the
2 a  a2
fractions  a  a-, and the more nearly will the above ratio approach to -, or 1.
120. Now in order to compare the general method with that,
which we have just employed, we develope the equation




Equations of the Second Degree with one Unknown Quantity. 137


x2
(a- X) (a- x) =m
and we have, successively,
x2 = m (a -   ) (a -  ),
x2 == a2m -2 am x +{ m    2,
x2 -mx2       2amx = a m,
(1- mn) x2 + 2 a m x = a2 m,
x  2a m x   a2 m
y3| — m -- 1 m 1   m'
2 am              a2m
making         P                   =1 -m'
the general formula gives
^ a rn ~      a2 m2          a2 m
x   - m i                  +  am
-   1 —m       (I-m) (l-m)        1-me'
These values of x appear very different from those, which were
found above; yet they may be reduced to the same; and in this
consists the utility of the example, on which 1 am employed. It
will serve to show the importance of those transformations, which
different algebraic operations produce in the expression of quantities.
We must first reduce the two fractions comprehended under the
radical sign to a common denominator. This may be done by
multiplying the two terms of the second by 1 - m; we have then
a2M            a2       a m2 _  a   2   m ( - m)
(l —n) (1-   m)    1-m n      (1 --  ) (1-   )
a2 m2 +     - a 2 m a2 2     am n
(i —m) (l-m)      = (1-m) (I -m)'
The denominator being a square, it is only necessary to extract the
root of the numerator; we then have
a2 m3         a2 m     _/a2m m
(1rn-)(l-(   m — m -        1- m;
but the expression v/a2m may be further simplified.
It is evident that the square of a product is composed of the
product of the squares of each of its factors, for example,
bcd X bcd=b2, c2, d
and consequently the root of b2 c2 d2 is simply the product of the
roots b, c, and d, of the factors b2, c2, and d2.  Applying this
principle to the product a2 m, we see that its root is the product of
a, the root of a2, by v/m, which denotes the root of m, or that
4 a2 m = aim.


lsg.


18




138


Elements of Algebra.


It follows from these different transformations, that
am      a V/~
1-m m —
am -- a /m
or              x         1 
a m+a +  m
X - -    --       -
These expressions, however simple, are still not the same as
those given in the preceding article; if, moreover, we seek to verify them for the case, in which mr = 1, they become
- a -- a    0
X  -1  I     0'
- a       a    2a
x = 1-_   1 =     O 
We find, in the second, the symbol of infinity, as in the preceding
article, but the first presents this indeterminate form, 0, of which
we have already seen examples in articles 69. and 70.; and before
we pronounce upon its value, it is proper to examine, whether it
does not belong to the case stated in art. 70.; whether there is not
some factor common to the numerator and denominator, which the
supposition of m = I renders equal to zero.
--  m - a -  /m
The expression    -a-    -a
1- m
may be resolved into
a (-m +-   /-) _ a(/, -m
1     m         1- m
It is here evident, that the numerator does not become 0, except
by means of the factor   'm —m; we must therefore examine,
whether this last has not some factor in common with the denominator I -m. In order to avoid the inconvenience arising from
the use of the radical sign, let us make vm =- n, then taking the
squares, we have m = n2; the quantities, therefore,
/m ---m    and 1 —m
become                 n - n2 and 1 -n2,
but n-na = n (1-n), and I - n        = (1 -n) (      n)(34);
restoring to the place of n its value vm, we have
/   -    -fl = (1 -V 'm) V/m,
I -m=- (1 —/V     ) (1 + V  ),
and consequently,
a (s/mn-m) _     a(l - fon ) vm         a /m
1-m       -- (    /m)( +^/7n)        + /-m'
a result the same, as that found in art. 119.




Equations of the Second Degree with one Unknown Quantity. 139


In the same manner we may reduce the second value of x, observing that
-    a /- am     -a (1 -- +/mn) / -  _ - a /mn
1 —m-        (-1 -/ ) (+1 J V") - ' 1-  /  '
as in art. 119.'
It will be seen without difficulty, that we might have avoided
radical expressions in the preceding calculations, by taking m2
to represent the ratio, which the squares of the two parts of the
proposed number have to each other; m would then have been
the square root, which may always be considered as known,
when the square is known; but we could not have    perceived
fiom the beginning the object of such a change in a given term,
of which algebraists often avail themselves, in order to render
calculations more simple.  It is recommended to the learner,
therefore, to go over the solution again, putting m2 in the place
of m.
Examples in the Extraction of the Square Roots of Numbers.
1. V4096     = 64.
2. /6j(009   = 247.
3. V582i69   = 763.
4. /956484   - 978.
5. -57198969 = 7563.
6.   s       - 2.23606...
7. /13     =_ 3.60555...
8. V/22     = 4.69041...
9. V153     = 12.36931...
10. /7'.65   = 2.76586...
11. V7       = 1.32287...
* The example, which I have given at some length, corresponds
with a problem resolved by Clairaut, in his Algebra, the enunciation
of which is as follows; To find on the line, which joins any two luminous bodies, the point where these two bodies shine with equal light. I
have divested this problem of the physical circumstances, which are
foreign to the object of this work, and which only divert the attention from the character of the algebraic expressions. These expressions are very remarkable in themselves, and for this reason I have
developed them more fully, than they were done in the work referred to.




140


Elements of Algebra.


12. %/ 4    = 1.24721
13. yj1{~   =3.41869...
14. ~~      = 1.29099...
1 5. t,/7      0.93541.
Examples in Equations of the Second Degree involving only the
Second Power of the Unknown Quantity, and those which may
be solved without completing the Square.
1. 3x- 4    28    Ax2.  x   -.= 4,
2.x + y:Y:3i          =6          =3
cr~y  18.~- sx 4= 61  y ='t3.
3.                    x=~ - 5 Y  4 y  x  -4- 9,  y = 3.
4. 25 (I8- x)! - 16 x%4   x =10.
5 a + 5 X = 5 x +   4225i  X =~65.
4     x2
3   z-J-2
75 (x - 7)  48 (x-4)
7                            x --— 7  i  x= 19, or'g
x- 4        x - 
28 (X - 18)    63             2
8,             4            x1  72, or Iq
z       4 (x- 18)X
Examples in Equations of the Second Degree solved by completing
the Square.
1. x2+6x=27.        x = 3,   x=-9.
2. x2- 7 x + 3f = 0.    a    = 6   x -
3. 3xg-2x=65.        x = 5,    x =-4.
4. 622x- = 15 x 2 + 6384.  x = 22 4   x=18-.
5. 1i1x   3   2=-x 41      X= -2,        x=52.
6. x2 - 8 x = 14.  x = 4 +   Vjo- =9.4772...   x   4
v- v 5-0 =- 1.41T.72...
7. 118xc  2Lx2    20.   x= 118 +     2      47.0298.
2                    5
X    " ~~4' = C.-=47.0298...
8. 6x-30=3X2.         X1+       V-  9  x=    -.V-9.
9. z+6        7            14, 5         10.
z +60 -3x,   5 




Square Root of Algebraic Quantities.


141


8x        _20
10.       -6                 -10,        x —    2
+ 2           3         x= 1 
48       165
11.     3-    -  -    5.     x-+ 3         x-5.
31        16
126                 + 71.     x - =671,     x= 4.
2z -f-3        z ---
13. 2z63                            x = 13 2 2     x- 8.
70-x       25-3z        2            1 3-       1 =8.
14. 25  + 180       40 x,3
10x-         5l — -8       5*      =14          -   72 -18+xz   _20x+9            65
15.  ((3-   ) -  19 -7 x    4(3 -z)'     x=71T3           2
a2x2    2ax   f2                f2
16. --:     — 0,              Xa-.
J62 -    g     g2=              ag17.   b x +  3 a2      6a+ a b     2b2    b x        2 a -b
C,            C2             C          ac
3a+2b
be
ac      ( + b)            c +  /(c2+4ac)
18. c x +       a     +    X.    x=-        (+    )
l8.CX+~ 6~ d~b3        S        2(a+b)       '
=  c-,/(c  +4ac)
2 (a+(  b
19. axa2 + b2 +    - =a2 + 2 b c + 2 (b-c) xVa.
b- c+a               b-c- a
x =- v, -- _,  x    _
V/a                  %/a
Of the Extraction of the Square Root of Algebraic Quantities.
121. WE have sufficiently illustrated, by  the preceding articles, the manner of conducting the solution of literal questions.
We have given also an instance of a transformation, namely,
that of V/am  into a V/m, which is wortly of particular attention;
since, by means of it, we have been able to reduce the factors,
contained under a radical sign, to the smallest number possible,
and thus to simplify very much the extraction of the remaining
part of the root.
This transformation consists in taking the roots of all the factors
which are squares, and writing them  without the radical sign, as
multipliers of the radical quantity, and retaining under the radical
sign all those factors, which are not squares.
This rule supposes, that the student is already able to determine, whether an algebraic quantity is a square, and is acquainted




142


1Elements of.Algebra.


with the method of extracting the root of such a quantity. In order to this, it is necessary to distinguish simple quantities from polynomials.
122. It is evident, from the rule given for the exponents in multiplication, that the second power of any quantity has an exponent
double that of this quantity.
We have, for example,
al X a =2      a a2 X      a  a4,  a3 X a3 = 0a6, &c.
It follows then, that every factor, which is a square, must have an
exponent which is an even quantity, and that the root of this factor
is found by writing its letter with an exponent equal to half the
original exponent.
Thus we have
/a2 = (a1 or a,  4/a4 = a2,,/a6 = a3, &c.
With respect to numerical factors, their roots are extracted,
when they admit of any, by the rules already given.
Whence the factors a6, b4, c2, in the expression
4/64 a b64 c2,
are squares, and the number 64 is the square of 8; therefore, as
the expression proposed is the product offactors, which are squares,
it will have for a root the product of the roots of these several
factors (121); and, consequently,
/64 a6b4 c2    8 a3 b2 c.
123. In other cases, different from the above, zwe must endeavour to resolve the proposed quantity, considered as a product, into
two other products, one of which shall contain only such factors as
are squares, and the other those flctors which are not squares.
To effect this, we must consider each of the quantities separately.
Let there be, for example,
'"T72 a4 63 c5.
We see that among the divisors of 72, the following are perfect
squares, namely, 4, 9, and 36; if we take the greatest, we have
72    36 X 2.
As the factor a4 is a square, we se)arate it from the others; passing then to the factor b3, whiCh) is not a square, since 3 is an odd
number, we observe that this factor may be resolved into two others, b2 and b, the first of which is a square; we have then
b2 =b2. b;


it is obvious also that


C5 =  C4. c.




Square Root of.Igebraic Quantities.


143


By proceeding in the same manner with every letter, whose exponent is an odd number, the quantity is resolved thus,
72 a4 b3 c5 = 36. 2 a4 b. bc4. c;
by collecting the factors, which are squares, it becomes
36 a4b2 c4 X 2bc.
Lastly, taking the root of the first product and indicating that of
the second, we have
V72 a4 b3 c5  6 a2 b ca 2  b c.
See some examples of this kind of reduction, with the steps by
which they are performed;
\a3    L          aa       ab   a a
6-A b        2 b = ~  a ^    =    vab
6J75  b2  625 32       -- 6   25b2. 3 a
ab356 I3a 3Ob4. 3a 
6  98 ab  = 6.=*9-2   6, —49. =
6.5 b         33 a  30 b  13a
7       2      7     2 
aj2 nO'  a2 n     1a2 m2 - a2 m n
n2  a+?             2
m2               - 
(m2  + m n) -  ^/ m2 + m n.
It will be seen by the first of these examples, that the denominator of an algebraic fiaction may be taken from under the radical sign by being made a complete square, in the same manner as
we reduce the root of a numerical fraction (104).
124. We now proceed to tlhe extraction ot the square root of
polynomials. It must here be recollected, that no binomial is a
perfect square, because every simple quantity raised to a square
produces only a simple quantity, and the square of a binomial always contains three parts (34).
It would be a great mistake to suppose the binomial a + b to be
the square root of a2 + 62, although, taken separately, a is the root
of a2, and b that of b2; for the square of a + b, or a2 + 2 a b+ b,
contains the term + 2 a b, which is not found in the expression
a2 +b.
Let there be the trinomial
24 a2 3 c + 16 a4 c2 + 9 b6.
In order to obtain from this expression the three parts, which compose the square of a binomial, we must arrange it with reference to




144


Elements of Algebra.


one of its letters, the letter a, for example; it then becomes
16 a4 c2 +24a2b3c+9 6.
Now, whatever be the square root sought, if we suppose it arranged with reference to the same letter a, the square of its first term
must necessarily form the first term, 16 a4 c2, of the proposed
quantity; double the product of the first term of the root by the
second must give the second term, 24a2 b3c, of the proposed
quantity; and the square of the last term of the root must give
exactly the last term, 9 b6, of the proposed quantity. The operation may be exhibited, as follows;
16 a4 C2 + 24 a2 b3 c + 9 b6 4 a c + 3 b3 root
-16 a4 c2                     S 82 C+3 b3
+ 24 a2 b3c + 9 b6
-24 a b3 c     9 b6
0            0
We begin by finding the square root of the first term, 16 a4 C2,
and the result 4 a2 c (122) is the first term of the root, which is
to be written on the right, upon the same line with the quantity,
whose root is to be extracted.
We subtract from the proposed quantity, the square, 16 a4 ca,
of the first term, 4 a2 c, of the root; there remain then only the
two terms 24 a2 b3 c + 9 b6.
As the term 24 a2 b3 c is double the product of the first term
of the root, 4 a2 c, by the second, we obtain this last, by dividing
24 a2 63 c by 8 a2 c, double of 4 a2 c, which is written below the
root; the quotient 3 b3 is the second term of the root.
The root is now determined; and, if it be exact, the square of
the second term will be 9 66, or rather, double of the first term
of the root 8 a2 c together with the second 3 63, multiplied by the
second, will reproduce the two last terms of the square (91);
therefore we write + 3 63 by the side of 8 a2 c, and multiply
8 a2 c + 3 b3 by 3 b3; after the product is subtracted fiom tile two
last terms of the quantity proposed, nothing remains; and we
conclude, that this quantity is the square of 4 a2 c - 3 b3.
It is evident that the same reasoning and the sane process may
be applied to all quantities composed of three terms.
125. When the quantity, whose root is to be extracted, has
more than three terms, it is no longer the square of a binomial;
but if we suppose it the square of a trinomial, m + n + p, and




Square Root of llgebraic Quantities.


145


represent by I the sum m + n, this trinomial becoming now I + p,
its square will be
12+ 2 p + p2,
in which the square 12 of the binomial m + n, produces, when
developed, the terms m2- 2 mn-  + -n2.  Now, after we have
arranged the proposed quantity, the first term will evidently be
the square of the first term of the root, and the second will contain double the product of the first term of the root by the second
of this root; we shall then obtain this last by dividing the second
term of the proposed quantity by double the root of the first.
Knowing then the two first terms of the root sought, we complete
the square of these two terms, represented here by 12; subtracting this square from the proposed quantity, we have for a remainder
2 p+p2,
a quantity, which contains double the product of 1, or of the first
binomial m + n, by the remainder of the root, plus the square
of this remainder. It is evident, therefore, that we must proceed
with this binomial as we have done with the first term m of the
root.
Let there be, for example, the quantity
64 a2 b c+ 25 a2 b2 -40 a3 b + 16 a4  64 b2 c2 -80 ab2 c;
we arrange it with reference to the letter a, and make the same
disposition of the several parts of the operation as in the above
example.
16a4-40a3b+25a2b2-80ab2c+64b2c2       (4a2 -  5ab+8bc
+64a2bc                     8Sa2 -  5ab
-16a4                __                    82 - lOab+Sbc
1st rem. —40a3b+25a22 8 —Oab2c+64b2c2
+64a2bc
+40a3 —25a2 b2
2d rem...... -+64a bc —Oab2c+64b2c2
-64a2 bc+80ab2 c-64b2 c2
0         0        0
We extract the square root of the first term 16 a4, and obtain
4 a2 for the first term of the root sought, the square of which is
to be subtracted from the proposed quantity.
We double the first term of the root, and write the result, 8 a2,
under the root; dividing by this the term - 40a3 b, which begins the first remainder, we have - 5 a b for the second term of
llg.             19




146


Elements of Algebra.


the root; this is to be placed by the side of 8 a'; we then multiply the whole by this second term, and subtract the result from the
remainder, upon which we are employed.
Thus we have subtracted from the proposed quantity the square
of the binomial 4 a2 -  5a b; the second remainder can contain
only double the product of this binomial, by the third term of the
root, together with the square of this term; we take then double
the quantity 4 a2 - 5 a b, or
8 a2- 10 ab,
which is written under 8 a2 - 5 a b, and constitutes the divisor to
be used with the second remainder; the first term of the quotient,
which is 8 b c, is the third of the root.
This term we write by the side of 8 a2 -  10 a b, and multiply
the whole expression by it; the product being subtracted from the
remainder under consideration, nothing is left; the quantity proposed, therefore, is the square of
4 a2 -5 ab     8 bc.
The above operation, which is perfectly analogous to that, which
has been already applied to numbers, may be extended to any
length we please.
Examples in the Abbreviation and Transformation of Radical
Quantities.
1.     q-4 + 54 - -/6 - 4 V6.
2. 2 V8 - 7 s/18 + 5 Vf72 -,/50 = 8 V2.
3. v/12 + 2 v/7 + 3 V/75- 9 /48 =    - 13 i3.
i 4. 2 v3 + v-60 -v15 + v4    - 2 8 
3        3      3       3        3
5. 7 V/54 + 3 V16 + V/2 - 5 V128 = 8 y/2.
'q         3        3            _                 3
6. -/  - 2          %V-24- - 28 2 V 3 =8 V  - 23.
4        3        4      3
7.V +/32 +   2 y/40 = 2 y/2 + 4 /5.
8. 3 s/5 - 2 v2 + 3 /6 =    /45 -   8 + V54.
3               4     3              4
9. 5 V7 + 3 V/ + 2 V /     V875 + V/1 + V/8.
5       3             5       3
10. 4 vt + 3   2    5 3-   5T2 +    i54-   3 -.
11. V45 -3 v'8     + O /5a2c = (a- c) ^'/5c.
12. /18a5      /50a33 = (3 a2b+ 5 a b) /2ab.
3                           3                  3
13. /16 a3b +  /4a2 b -  /a b -  /54 a3b = a  b - a V 2b.
14. 1     -      3 +    ^c a2cd2  a (a2c     ad  |ic
\b d62         -     64e   -        d - -C 




Square Root of llgebraic Quantities.  147
3                       3 
27 a12r5 x  a2 X          a2 X"3
15.          2 -  fb = (3 at - 1)  -b
2        4 ac 
16. 36 2 %a3+ V/a5 c3-3 C -3a=     + 2a -JVa c.
C           bb
5  O   b 2 C3 _  5 a   b   3~ 
17.                     b + L   Va2 V.
Examples in the Extraction of the Square Root of Algebraic
Expressions.
2'
1.    a2 -ab + b2  - a c   b
2.  (a     6+6a3 x4 + 9 X8) = a3+ 3 X4.
19 8 a4 n6\          3a4  2 n3
3.l      +- 2 O" n   9      23
4. /(25 a62 bab_  + ab)CaI c C4) -.5a b.C2.
5. V(a2 + 2 a b +   2 a c  6 b2 +2 b c +c2) = a  + b + c.
6.lj(9x2-30ax-3a'.x+25a +5a3+4a/)
a2
3 9X2- _3ax-a,5a 
3x -  5a _ 
7.V(4 x4 + 8ax3+4a2 X2 +16 b2 x2 + 16 a b2 2 X +16 b4)
= 2 x2 + 2 a x + 4b 2.
E3.  ( 9 +6x-17 X2 - 28 X3 +t 49 X14): 3 + 2 x~ — 7 X2.
Ct2
9. J(9x -3ax3+6bx3+a       x  ~ab x2 + bi x2)
= 3 x2   ax + b x.
10. 4/ (4 a2x4  4 a b X3Z~+ (t a26b X2 z2+ b+2 X2z2
-4 a,2x  +4a b       _ )=ax26x z + 2a b z
11. v'(9 a2 - Ga b + 30 a c + 6 a d + 62 - 10 b c - 2 b d
+ 25 c2 + Il0 c d + d2) = 3 a -b + 5 c + d.
Examples in the Extraction of the Square Roots of Incomplete
Squares.
X2   X4     X6     5x8
1.  ~ (a       a -   -   -   -   -  3   -  5 12  
2. ~(a2 + X2) =.+a +   8 a    2       X  +~
2 a - SO + R,' - -a5 




148                 Elements of Algebra.
X     Z2     x3      5 X4
3.V   1-  x)  =1 —      --     -     - 
4  1y2       x3      5 z4
4.  (1 +x)  -1   2     8-  8 +  16    1283 +
Of the Formation of Powers and the Extraction of their Roots.
126. THE arithmetical operation, upon which the resolution of
equations of the second degree depends, and by which we ascend
from the square of a quantity to the quantity, from which it is
derived, or to the square root, is only a particular case of a
more general problem, namely, to find a number, any power of
which is known. The investigation of this problem leads to a
result, that is still termed a root, the different kinds being called
degrees; but the process is to be understood only by a careful examination of the steps by which a power is obtained, one operation being the reverse of the other, as we observe with respect to
division and multiplication, with which it will soon be perceived
that this subject has other relations.
It is by multiplication, that we arrive at the powers of entire
numbers (24), and it is evident, that those of fractions also are
formed by raising the numerator and denominator to the power
proposed (96).
So also the root of a fraction, of whatever degree, is obtained
by taking the corresponding root of the numerator and that of
the denominator.
As algebraic symbols are of great use in expressing every
thing, which relates to the composition and decomposition of quantities, I shall first consider how the powers of algebraic expressions
are formed, those of numbers being easily found by the methods
that have already been given (24).
Table of the first Seven Powers of JVumbersifrom I to 9.
IstI   2    3     4     5      6      7        8       9
2d      4    9    16   25      36    49       64      81
3  1    8  27    64   125    216    343     512      729
4thil  16  8    256   625 _296     2401    4096 1   6561
5thl 1  32 243 1024 3125    7770  168i07S - 276   59049
6tl] J  64 729 '34096 15625 466506 11764 9 2621441 531441
7th 1 1282187 1638478125279936 1235432097152 4782969




Formation of Powers and the Extraction of their Roots. 149
This table is intended particularly to show with what rapidity
the higher powers of numbers increase, a circumstance that will
be found to be of great importance hereafter; we see, for instance, that the seventh power of 2 is 128, and that of 9 amounts
to 4782969.
It will hence be readily perceived that the powers of fractions,
properly so called, decrease very rapidly, since the powers of the
denominator become greater and greater in comparison with
those of the numerator. The seventh power of a, for example,
is -8j, and that of - is only
1
4782969'
127. It is evident from what has been said, that in a product
each letter has for an exponent the sum of the exponents of its
several factors (26), that the power of a simple quantity is obtained by multiplying the exponent of each factor by the exponent of
this power.
The third power of a2 b3 c, for example, is found by multiplying
the exponents 2, 3, and 1, of the letters a, b, and c, by 3, the exponent of the power required; we have then a6 b9 c3; the operation may be thus represented,
a2 b3 c X a2 b3 c X a2 b3 c = a2.3 b3. 3 c1. 3.
If the proposed quantity have a numerical coefficient, this coefficient must also be raised to the same power; thus the fourth
power of 3 a b2 C5, is
81 a4 8 c20
128. With respect to the signs, with which simple quantities
may be affected, it must be observed, that every power, the exponent of which is an even number, has the sign +, and every power,
the exponent of which is an odd number, has the same sign as the
quantity from which it is formed.
In fact powers of an even degree arise from the multiplication
of an even number of factors; and the signs -, combined two and
two in the multiplication, always give the sign + in the product
(31). On the contrary, if the number of factors is uneven, the
product will have the sign -, when the factors have this sign,
since this product will arise fiom that of an even number of factors, multiplied by a negative factor.
129. In order to ascend from the power of a quantity, to the
root from which it is derived, we have only to reverse the rules




150


Elements of.dlgebra.


given above, that is, to divide the exponent of each letter by that,
which marks the degree of the root required.
Thus we find the cube root, or the root of the third degree, of
the expression a6 b9 c3, by dividing the exponents 6, 9, and 3, by
3, which gives
a2 b3 C.
When the proposed expression has a numerical coefficient, its
root must be taken for the coefficient of the literal quantity, obtained by the preceding rule.
If it were required, for example, to find the fourth root of
81 a4 b8 c20, we see, by referring to the table, art. 126., that 81 is the
fourth power of 3; then, dividing the exponent of each of the
letters by 4, we obtain for the result
3 a b c5.
When the root of the numerical coefficient cannot be found by
the table inserted above, it must be extracted by the methods to be
given hereafter.
130. It is evident, that the roots of the literal part of simple
quantities can be extracted, only when each of the exponents is
divisible by that of the root; in the contrary case, we can only
indicate the arithmetical operation, which is to be performed
whenever numbers are substituted in the place of the letters.
We use for this purpose the sign v/-; but to designate the degree of the root, we place the exponent as in the following expressions,
3     5
/a,   Va2,
the first of which represents the cube root, or the root of the
third degree of a, and the second the fifth root of a2.
We may often simplify radical expressions of any degree
whatever, by observing, according to art. 127., that any power
of a product is made up of the product of the same power of each of
the factors, and that, consequently, any root of a product is made
up of the product of the roots of the same degree of the several
factors.  It follows from  this last principle, tllht, if the quantity
placed under the radical sign have factors, which are exact powers
of the degree denoted by this sign, the roots of these factors may
be taken separately, and their product multiplied by the root of the
other factors indicated by the sign.
Let there be, for example,
5 _
V96 a5 b7 c11.




Formation of Powers and the Extraction of their Roots. 151


It may be seen tha
that
that
that
we have then


tj
96 = 32 X 3 = 25. 3,
a5    is the fifth power of a,
b7 -= b. b2
cll   C10. C;
-  o *  " 


96 a5 b7 c11 - 25 a5 65 c0 X 3 b2 c.
As the first factor, 25 a5 b5 c10, has for its fifth root the quantity
2 a b c2, the expression becomes
5__                5
V9(3 a5b7ci = 2 a b c2 V3bc.
131. As every even power has the sign + (128), a quantity,
affected with the sign -, cannot be a power of a degree denoted
by an even number, and it can have no root of this degree. It
follows from this, that every radical expression of a degree which
is denoted by an even number, and which involves a negative quantity, is imaginary, thus
4     6           8
V/- a, a,,a     b +- v/ - a b7,
are imaginary expressions.
We cannot, therefore, either exactly or by approximation, assign for a degree, the exponent of which is an even number, any
roots but those of positive quantities, and these roots may be affected indifferently with the sign + or -, because, in either case,
they will equally reproduce the proposed quantity with the sign +,
and we do not know to which class they belong.
The same cannot be said of degrees expressed by an odd number, for here the powers have the same sign as their roots (128);
and we must give to the roots of these degrees the sign, with which
the power is affected; and no imaginary expressions occur.
132. It is proper to observe, that the application of the rule
given in art. 129., for the extraction of the roots of simple quantities, by means of the exponent of their factors, leads to a more
convenient method of indicating roots, which cannot be obtained
algebraically, than by the sign /.
If it were required, for example, to find the third root of a5, it
is necessary, according to the rule given above, to divide the exponent 5 by 3; but as we cannot perform the division, we have
for the quotient the fractional number 5; and this form of the exponent indicates, that the extraction of the root is not possible in
the actual state of the quantity proposed. We may, therefore,
consider the two expressions




152


Elements of Algebra.


3          5_ 
v/a5  and a3
as equivalent.
The second, however, has this advantage over the first, that it
3
leads directly to a more simple form, which the quantity V~f-5 is capable of assuming; for if we take the whole number contained in
the fraction 5-, we have 1 + 2 as an equivalent exponent; consequently,
5                  2
a3 - a'+  a-1 X aX (25);
5
from which it is evident, that the quantity a3 is composed of two
3
factors, the first of which is rational, and the other becomes v/2.
The same result, indeed, may be obtained from the quantity
3
under the form V/, by the rule given in art. 130., but the fractional exponent suggests it immediately. We shall have occasion
to notice in other operations the advantages of fractional exponents.
We will merely observe for the present, that as the division of
exponents, when it can be performed, answers to the extraction of
roots, the indication of this division under the form of a fraction
is to be regarded as the symbol of the same operation; whence,
n             m
/ am~  and   an
are equivalent expressions.
We have rules then, which result from the assumed manner of
expressing powers, which lead to particular symbols, as, in art. 37.,
we arrived at the expression a~ - 1.
133. It may be observed here, that as we divide one power by
another, by subtracting the exponent of the latter from that of the
former (36), fractions of a particular description may readily be
reduced to new forms.
By applying the rule above referred to, we have
am
- = am-n o
an         I
but if the exponent n of the denominator exceed the exponent m
of the numerator, the exponent of the letter a in the second member will be negative.
If, for example, m = 2, n =- 3, we have
2 = a23 = a-';
a3-       -




Powers of Compound Quantities.


153


a2
but by another method of simplifying the fraction -3, we find it is
1
equal to -, the expressions
1
and   a-,
a
are therefore equivalent.
In general, we obtain by the rule for the exponents,
am,
a   -- am —n - a-n
amqand by another method
am     1
an+n -   an;
it follows from this, that the expressions
1
-   and  a-n,
an
are equivalent.
In fact, the sign -, which precedes the exponent n, being
taken in the sense defined in art. 62., shows that the exponent in
question arises from a fraction, the denominator of which contains
the factor a, n times more than the numerator, which fraction is
indeed a-; we may, therefore, in any case which occurs, substitute one of these expressions for the other.
as bs
The quantity 2 -d3 for example, being considered as equivalent to
1 1
a2b5 X C- X,
may be reduced to the following form,
a2 b c-2 d-3;
that is, we may transfer to the numerator all the factors of the denominator, by giving to their exponents the sign -.
Reciprocally, when a quantity contains factors, which have negative exponents, we may convert them into a denominator, observing
merely to give to their exponents the sign +; thus the quantity
a2 b5 C-2 d-3,
becomes
a2 b5
c2 d3'
Of the Formation of the Powers of Compound Quantities.
134. WE shall begin this section by observing, that the powers
of compound quantities are denoted by including these quantities.1g.                  20




154


Elements of Algebra.


in a parenthesis, to which is annexed the exponent of the power.
The expression
(4 a — 2 a b + 5 b)3
for example, denotes the third power of the quantity,
4 a2 -2 a b + 5 b2.
This power may also be expressed thus,
4 a2- 2 a b +5 5   3.
135. Binomials next to simple quantities are the least complicated, yet if we undertake to form powers of these by successive
multiplications, we in this way arrive only at particular results, as,
in art. 34., we obtained the second and third power; thus
(x+ a)2 = x2    2 ax + a2,
(x + a)3 =  3 + 3 a x2 + 3 (t2x + a3,
(x + a)4 = x4 + 4 a x3 + 6 a 2 x2 + 4 a3 x + a4,
&c.
It is not easy from this table to fix upon the law, which determines the value of the numerical coefficients. But by considering how the terms are multiplied into each other, we perceive,
that the coefficients have their origin in reductions depending on
the equality of the factors, which form a power. This is rendered very evident by an arrangement, which prevents these reductions from taking place.
It is sufficient for this purpose to give to the several binomials to
be multiplied different second terms. If we take, for example,
x + a, x +-b, x + c, x - d, &c.
by performing the multiplications indicated below, and placing in
the same column the terms, which involve the same power of x,
we shall immediately find, that
(x + a) (x+b) =    2 + ax + ab
+ bx
(x + a) (x + b ( + c) =x 3 + ax + abx + abc
+ bx2 + acx
+ cx2 + bcx
(x+a) (x + b) (x + c (x + d) = x4 + ax3 + abx2 + abcx+abcd
+ bx3 + acx2 + abdx
- CX3 + adx2 + acdx
- dx2 + bcx2 + bcdx
+ bdx2
+ cdx2




Powers of Compound Quantities.


155


Without carrying these products any further, we may discover
the law according to which they are formed.
By supposing all the terms involving the same power of x,
and placed in the same column, to form only one, as, for example,
X3 + b x3 + c x3 + d x3 = (a + b + c + d) x3,
&c.
(1.) [e find in each product one term more than there are units
in the number offactors.
(2.) The exponent of x in the first term is the same as the number offactors, and goes on decreasing by unity in each of the following terms.
(3.) The greatest power of x has unity for its coefficient; the
following, or that, whose exponent is one less, is multiplied by the
sum of the second terms of the binomials; that, whose exponent is
two less, is multiplied by the sum of the different products of the
second terms of the binomials taken two and two; that whose exponent is three less, is multiplied by the sum of the differentproducts of the second term of the binomials, taken three and three,
and so on; in the last term, the exponent of x, being considered
as zero (37), is equal to that of the first diminished by as many
units as there are factors employed, and this term contains the product
of all the second terms of the binomials.
It is manifest, that the form of these products must be subject
to the same laws, whatever be the number of factors; as may be
shown by other evidence beside that from analogy.
136. It will be seen immediately, that the products, of which
we are speaking, must contain the successive powers of x, from
that, whose exponent is equal to the number of factors employed,
to that, whose exponent is zero.  To present this proposition under
a general form, we shall express the number of factors by the letter m; the successive powers of x will then be denoted by
X   m, X-l  I     c-2  &C.
We shall employ the letters.X, B,      C,.......... Y,
to express the quantities, by which these powers, beginning with
x"-l, are to be multiplied; but as the number of terms, which depends on the particular value given to the exponent, will remain indeterminate, so long as this exponent has no particular value, we
can write only the first and last terms of the expression, designating the intermediate terms by a series of points.




156


Elements of Algebra.


The formula then
xm +    xm-1                          + Bxm-+ Cx m-.....  Y,
represents the product of any number m of factors,
x + a, x  - b, x + c, x + d, &c.
If we multiply this by a new factor x + 1, it becomes
m —l +. m +   B xm-1+      Cxa-2.....
+   lxm +lM.-l+lBxm      -2.... + I Y 
It is evident, 1. that if./ is the sum of the m second terms
a, b, c, d, &c.. + I will be that of the m + I second terms
a, b, c, d, &c. 1, and that consequently the expression employed
to denote the coefficient will be true for the product of the degree
m + 1, if it is true for that of the degree m.
2. If B is the sum of the products of the m quantities a, b, c, d,
&c. taken two and two, B + 1 A will express that of the products
of the m - 1 quantities a, b, c, d, &c. l, taken also two and two;
for A. being the sum of the first, 1 d will be that of their products
by the new quantity introduced l; therefore the expression employed will be true for the degree m + 1, if it is for the degree m.
If C is the sum of the products of the m quantities a, b, c, d,
&c. taken three and three, C + IB will be that of the products
of the m + 1 quantities a, b, c, d, &c. 1, taken also three and three,
since I B, from what has been said, will express the sum of the
products of the first taken two and two, multiplied by the new
quantity introduced l; therefore, the expression employed will be
true for the degree mn + 1, if it is true for the degree m.
It will be seen, that this mode of reasoning may be extended to
all the terms, and that the last, I Y, will be the product of m + 1
second terms.
The propositions laid down in art. 135., being true for expressions of the fourth degree, for example, will be so, according to
what has just been proved, for those of the fifth, for those of the
sixth, and, being extended thus from one degree to another, they
may be shown to be true generally.
It follows from this, that the product of any number whatever
m, of binomial factors x + a, x + b, x + c, x + d, &c. being
represented by
Xm + i xm-l + B xm-2 +   C xm     + &c..A will always be the sum of the im letters a, b, c, &c., B that of
the products of these quantities, taken two and two, C that of the
produces of the quantities, taken three and three, and so on.




Powers of Compound Quantities.


157


To comprehend the law of this expression in a single term, I take
one, whose place is indeterminate, and which may be represented
by.NJ xn-1.
This term will be the second, if we make n = 1, the third,
if we make n = 2, the eleventh, if we make n = 10, &c. In the
first case, the letter JN will be the sum of the m letters a, b, c,
&c. in the second, that of their products, when taken two and two;
in the third, that of their products, when taken ten and ten; and in
general, that of their products, taken n and n.
137. To change the products
(x + a) (x + b), (x + a) (x + b) (x + c),
( + a) ( + b) (x +c) (x+d), Sc.
into powers of x +- a, namely, into
(x + a)2      (x + a)3,
(x + a)4,     &c.
it is only necessary to make, in the developement of these products,
a   b,             a - b = c,
a= b = c = d,      &c.
All the quantities, by which the same power of x is multiplied,
become in this case equal; thus the coefficient of the second term,
which in the product
(x + a) (x + b) (x + c) (x + d) is a + b + c + d,
is changed into 4 a; that of the third term in the same product,
which is,
ab + a c + a d + b c ~ b d + c d,
becomes 6 a2.  Hence it is easy to see, that the coefficients of
the different powers of x will be changed into a single power of a,
repeated as many times as there are terms, and distinguished by
the number of factors contained in each of these terms.  Thus,
the coefficient V', by which the power xm-n is multiplied, will, in
the general developement, be that power of a denoted by n, or a",
repeated as many times, as we can form different products by
taking in every possible way a number n of letters from among a
number m; to find the coefficient of the term containing xa-n then
is reduced to finding the number of these products.
138. In order to perform the problem just mentioned, it is
necessary to distinguish arrangements or permutations from products or combinations. Two letters, a and b, give only one product, but admit of two arrangements, a b and b a; three letters,
a, b, c, which give only one product, admit of six arrangements
(88), and so on.




158


Elements of Allgebra.


To take a particular case, I will suppose the whole number of
letters to be nine, namely,
a, b, c, d, e, f, g, h, i,
and that it is required to arrange them in sets of seven. It is
evident, that if we take any arrangement we please, of six of
these letters, ab c d ef, for example, we may join successively to
it each of the three remaining letters, g, h, and i; we shall then
have three arrangements of seven letters, namely,
abcdefg,        abcdefh,       abcdefi.
What has been said of a particular arrangement of six letters,
is equally true of all; we conclude, therefore, that each arrangement of six letters will give three of seven, that is, as many as
there remain letters, which are not employed. If, therefore, the
number of arrangements of six letters be represented by P, we
shall obtain the number consisting of seven letters by multiplying
P by 3 or 9 - 6. Representing the numbers 9 and 7 by n and
n, and regarding P as expressing the number of arrangements,
which can be furnished by mn letters, taken n- 1 at a time, the
same reasoning may be employed; we shall thus have for the
number of arrangements of n letters,
P (m -(n      1)),   or    P (m -n+      ).
This formula comprehends all the particular cases, that can
occur in any question. To find, for example, the number of
arrangements, that can be formed out of m letters, taken two and
two, or two at a time, we make n = 2, which gives
n — = 1;
we have then
P=m;
for P will in this case be equal to the number of letters taken one
at a time; there results then from this
m (m- 2      1),    or       (m-1),
for the number of arrangements taken two and two.
Again, taking
P = m (m- 1)       and    n = 3,
we find for the number of arrangements, which m letters admit of,
taken three and three,
m (m- 1) (m- 3 + 1)=m (m- 1) (m-2).
Making
P= m (m —) (m-2)         and  n= 4,




Powers of Compound Quantities.


159


we obtain
m (m-   1 (m-2) (m-3)
for the number of arrangements taken four and four. We may
thus determine the number of arrangements, which may be formed
from any number whatever of letters.*
139. Passing now from the number of arrangements of n letters, to that of their different products, we must find the number
of arrangements, which the same product admits of. In order to
this, it may be observed, that if in any of these arrangements, we
put one of the letters in the first place, we may form of all the
others as many permutations, as the product of n - 1 letters
admits of. Let us take, for example, the product a b c d cfg,
composed of seven letters; we may, by putting a in the first place,
write this product in as many ways, as there are arrangements in
the product of six letters b c d efg; but each letter of the proposed product may be placed first.  Designating then the number
of arrangements, of which a product of six letters is susceptible,
by Q, we shall have 9 X 7 for that of tie arrangements of a
product of seven letters. It follows from this, that if Q designate
the number of arrangements, which may be formed frorfi a product
of n -  1 letters, Q n will express the number of arrangements of
a product of n letters.
* In these arrangements it is supposed by the nature of the inquiry,
that there are no repetitions of the same letter; but the theory of
permutations and combinations, which is the foundation of the doctrine of chances, embraces questions in which they occur. The effect
may be seen in the example we have selected, by observing, that we
may write indifferently each of the 9 letters a, b, c, d,'e,f, g, h, i, after
the product of 6 letters a b c d e f. Designating, therefore, the number of arrangements, taken six at a time, by P, we shall have P X 9
for the number of arrangements, taken 7 at a time. For the same
reason, if P denote the number of arrangements of m letters, taken
n - 1 at a time, that of their arrangements, when taken n at a time,
will be P m.
This being admitted, as the number of arrangements of m letters,
taken one at a time, is evidently m, the number of arrangements,
when taken 2 and 2, will be m X m, or m2, when taken 3 and 3, the
number will be m X m X m, or m3; and lastly, m" will express the
number of arrangements, when they are taken n and n.




160


Elements of Algebra.


Any particular case is readily reduced to this formula; for making
n = 2, and observing, that when there is only'one letter, Q = 1,
we have 1 X 2 = 2 for the number of arrangements of a product of
two letters. Again, taking Q = 1 X 2 and n = 3, we have
I X 2 X 3 = 6 for the number of arrangements of a product of
three letters; further, making Q = 1 X 2 X 3 and n = 4, there
result 1 X 2 X 3X 4, or 24 possible arrangements in a product
of four letters, and so on.
140. What we have now said being well understood, it will
be perceived, that by dividing the whole number of arrangements obtained from m letters, taken n at a time, by the number
of arrangements of which the same product is susceptible, we
have for a quotient the number of the different products, which
are formed by taking in all possible ways n factors among
these m letters. This number will, therefore, be expressed by
P(m-n+1)          1i
- (m — n + 1);* which being considered in connexion with
Qn
P (nm-n+ 1)
what was laid down in art. 1.37., will give P    n  +  )an m- n
for the term containing xm"-n in the developement of (x + a)".
It is evident, that the term which precedes this will be expressed by - a"n- xm-'+   for in going back towards the first
term, the exponent of x is increased by unity, and that of a
diminished by unity; moreover, P and Q are the quantities,
which belong to the number n- 1.
P
141. If we make   - JM, the two successive terms indicated
above, become
M a-'1 xm-n+l and M     (m n  1) an xm-n.
n
* It may be observed, that if we make successively
n - 2,    n = 3,   n = 4, &c.
the formula P (m — n  1) becomes
Qn
m(m-1) m(m-l) (m —2)          m(m —l) (m-2) (m-3)
1.2   '       1.2.3                1.2.3.4           &.
numbers, which express respectively, how many combinations may
be made of any number m of things, taken two and two, three and
three, four and four, &c.




Powers of Compound Quantities.


161


These results show how each term in the developement of (x + a)m,
is formed from the preceding.
Setting out fiom the first term, which is,X", we arrive at the
second, by making n     1; we have Jl 1= 1, since x"~ has only
unity for its coefficient; the result then is   X  a xn-1, or
a.Xm-l. In order to pass to the third term, we make.M -l,
and n = 2, and we obtain m (m1l   ) a2 xnt-    The fourth is
found by supposing.M  =  n (m2   ) and n = 3, which gives
mn (    -   - 1) (  2) a3 x-3, and so on; whence we have the formnula
(x + a)m-= - x', +- -        a + m ( l  2 1 r -2
+ m (m-1) (- 2) a3 m-3 + &.
+. tl.3
which may be converted into this rule.
To pass from one term to the following, we multiply the numerical
coefficient by the exponent of x in the first, divide by the number
which marks the place of this term, increase by unity the exponent of
a, and diminish by unity the exponent of x.
Although we cannot determine the number of terms of this
formula without assigning a particular value to m; yet, if we
observe the dependence of the terms upon each other, we can
have no doubt respecting the law of their formation, to whatever
extent the series may be carried.  It will be seen, that
m (m-I ) (          -  2).... (m --  + l) a  man xm-n
I. 2. 3....      n
expresses the term, Mhich has n terms before it.
This last formula is called the general term of the series
+    a x-1 + m (n -     1)a2 x7-2+ &c.
because if we make successively
n = 1,   n = 2,    n = 3, &c.
it gives all the terms of this series.
142. Now, if (x + a)5 be developed, according to the rule
given in the preceding article; the first term being
x5 or a~x5 (37),
lag.                  21




162


Elements of Algebra.


the second will be
a1 X4 or   5 a x4,
the third
5X 4a.3x or 10a2x3,
the fourth
10X3 a3a     or  10 a3 x,
3
the fifth
10X2
10- -  a4x  or   5a4 x
4
the sixth
5 Xla5      or  a5
Here the process terminates, because in passing to the following
term it would be necessary to multiply by the exponent of x in the
sixth, which is zero.
This may be shown by the formula; for the seventh term,
having for a numerical coefficient
m (m — ) (m -2) (m -3) (m -4) (m-5)
1. 2.3.4. 5. 6
contains the factor m - 5, which becomes 5 - 5 = 0; and this
same factor entering into each of the subsequent terms, reduces it
to nothing.
Uniting the terms obtained above, we have
(x -- a)5 = x5 + 5 a x4 +  10 a2 x3 +  10 a3 x2- + 5a4 x + a5.
143. Any power whatever of any binomial may be developed
by the formula given in art. 141. If it were required for example
to form the sixth power of 2 3 - 5 a3, we have only to substitute
in the formula the powers of 2 x3 and - 5 a3 respectively for those
of x and a; since, if we make
2 3 - x' and    5a3 - a',
we have
(2 3 - 5 a3)6 = (x + a')6 =
ax6 + 6 a'  5 +  15a'/2x4 + 20 a'3x3
+ 15 a'4 X2+ 6a'5S' x+ a'6 (141),
and it is only necessary to substitute for x/ and a' the quantities,
which these letters designate. We have then




Powers of Compound Quantities.


163


(2 x3)6 +  6 (- 5 a3) (2 x3)5 + 15- 5 a3)2 (2 x3)4
+ 20 (- 5 a3)3 (2 x3)3 + 15- 5 a3)4 (2 x3)2
+   6 (- 5 a3)5 (2 X3) + (- 5 a3)6
or
64 x8 - 960 a3 x15  + 6000 a6 x12
- 20000 a~ x9 + 37500 a12 x6
- 37500 a15 x3 + 15625 a18.
The terms produced by this developement are alternately positive
and negative; and it is manifest, that they will always be so, when
the second term of the proposed binomial has the sign -.
144. The formula given in art. 141, may be so expressed as to
facilitate the application of it in cases analogous to the preceding.
Since:m          Zm         2;m
m —1 --       X n —2 -   '   i —3 -  C.
- 3;     '       - 2   -  3'"  c
the formula may be written
A + ml a X   nM m (m- 1) aa   + 
+T - ax +         1.72        + m&c
which may be reduced to
1    m a    m (m - -l) a   n (m -l(m-      2) a3     )
x,   T   q-Z X2^              1. 2     3    3 - +&' 
by insulating the common factor xw. In applying this formula,
the several steps are, toform the series of numbers,
m m-1 m-2 m-3
1'   2   '   3       4, &
to multiply the first by the fraction a, then thisproduct by the second
a
and also by the fraction -, then again this last result by the
third and by the fraction -, and so on; to unite all these terms,
and add unity to the sum; and lastly, to multiply the whole by the
factor Xm.
In the example (2 3- 5 a3)6, we must write (2 x3)6 in the
5 a3         a
place of "m, and - -  in that of -. I shall leave the application
of the formula as an exercise for the learner.f


t The formula for the developement of (x + a)m answers for all
values of the exponent m, and is equally applicable to cases in which




164


Elements of.lgebra.


145. We may easily reduce the developement of the power of
any polynomial whatever, to that of the powers of a binomial, as
may be shown with respect to the trinomial a + b - c, the third
power for instance being required.
First, we make b -- c = m, we then obtain
(a + b +    c   ( +  c)3 =- a3 +13   3 =  n +   33  a n m2 + i;
substituting for nz the binomial b +- c, which it represents, we have
(a+   + c)3 = a3 + 3a2 ( + c) + 3 a (6 +c)2 + (b      c)3.
It only remains for us to develope the powers of the binomial
b + c, and to perform the multiplications, which are indicated;
we have then
a3 + 3a2 +- 3ab2 - b3
+ 3a2 c + 6abc + 3b2 c
+ 3 a C2   3 b c2
3
+ C3
Of the Extraction of the Roots of Compound Quantities.
146. HAVING explained the formation of the powers of compound quantities, I now pass to the extraction of their roots, beginning with the cube root of numbers.
In order to extract the cube root of numbers, we must first become acquainted with the cubes of numbers, consisting of only
one figure; these are given in the second line of the following
table;
1    2     3     4    5     6     7    8     9
1    8    27    64  125   216  343   512  729
and the cube of 10 being 1000, no number consisting of three figures can contain the cube of a number consisting of more than one.
The cube of a number consisting of two figures is formed in a
manner analogous to that, by which we arrive at the square; for
if we resolve this number into tens and units, designating the first
by a, and the second by 6, we have
(a+-b)3 = a3 + 3a2 b+-3ab2 a     b3.
the exponent is fractional or negative.  This property, which is
very important, is demonstrated in a note to the last part of the
Cambridge course of Mathematics on the Differential and Integral
Calculus.




Extraction of the Roots of Compound Quantities.


165


Hence it is evident, that the cube, or third power, of a number composed of tens and units, contains four parts, namely, the cube of the
tens, three times the square of the tens multiplied by the units, three
times the tens multiplied by the square of the units, and the cube of
the units.
If it were required to find the third power of 47, by making
a = 4 tens or 40, b = 7 units, we have
a2 - 64000
3 a2 b = 33600
3 a b2 - 5880
b3 __      343
Total,              103823 = 47 X 47 X 47.
Now to go back from the cube 103823 to its root 47, we begin
by observing that 64000, the cube of the 4 tens, contains no significant figure inferior to thousands; in seeking the cube of the tens
therefore, we may neglect the hundreds, the tens, and the units of
the number 103823. Pursuing, therefore, a method similar to that
employed in extracting the square root, we separate, by a comma,
the first three figures on the right; the greatest cube contained
in 103 will be the cube of the tens. It is evi-  103,823 47
dent from the table, that this cube is 64, the  64    48
root of which is 4; we therefore put 4 in the  398,23
place assigned for the root. We then subtract 64 firom 103;
and by the side of the remainder, 39, bring down the last three
figures. The whole remainder, 3985'3, contains still three parts
of the cube, namely, three times the square of the tens multiplied
by the units, or 3aa2 b, three times the tens multiplied by the
square of the units, or 3 a b2, and the cube of the units, or b3. If
the value of the product 3 a2 b were known, we might obtain the
units b, by dividing this product by 3 a2, which is a known quantity, the tens being now found; but, on the supposition that the
product, 3 a2b, is unknown, we readily perceive, that it can have
no figure inferior to hundreds, since it contains the factor a2, which
represents the square of the tens; it must, therefore, be found in
the part 398, which remains on the left of the number 39823,
after the tens and units have been separated, and which contains,
besides this product, the hundreds arising from the product, 3 a b2
of the tens by the square of the units, and from the cube b3, of
the units.




166


Elements of Jlgebra.


If we divide 398 by 48, which is triple the square of the tens,
3 a2 or 3 X 16, we obtain 8 for the quotient; but from what precedes, it appears that we ought not to adopt this figure for the
units of the root sought, until we have made trial of it, by employing it in forming the last three parts of the cube, which must
be contained in the remainder 39823. Making b = 8, we find
3 a2 b == 38400
3 ab2 = 7680
b3 =-   512
Total,      46592.
As this result exceeds 39823, it is evident that the number 8 is too
great for the units of the root. If we make a similar trial with 7,
we find that it answers to the above conditions; 47 therefore is the
root sought.
Instead of verifying the last figure of the root in the manner
above described, we may raise the whole number expressed by
the two figures, immediately to a cube; and this last method is
generally preferred to the other.  Taking the number 48 and
proceeding thus, we find
48 X 48 X 48 = 110592.
As the result is greater than the proposed number, it is evident,
that the figure 8 is too large.
147. What we have laid down in the above example may be
applied to all cases, where the proposed number consists of more
than three figures and less than seven.  Having separated the
first three figures on the right, we seek the greatest cube in the
part which remains on the left, and write its root in the usual
place; we subtract this cube from the number to which it relates,
and to the remainder bring down the last three figures; separating now the tens and the units, we proceed to divide what
remains on the left, by three times the square of the tens found;
but before writing down the quotient as a part of the root, we
verify it by raising to the cube thle number consisting of the tens
known, together with this figure under trial.  If the result of
this operation is too great, the figure for the units is to be diminished; we then proceed in the same manner with a less figure,
and so on, until a root is found, the cube of which is equal to the
proposed number, or is the greatest contained in this number, if
it does not admit of an exact root. As we have often remainders
that are very considerable, I will here add to what has been said, a




Extraction of the Roots of Compound Quantities.


167


method, by which it may be soon discovered, whether or not the
unit figure of the root be too small.
The cube of a + b, when b = 1, becomes that of a + 1,
or                 a3+ -3a2 + 3a+ 1,
a quantity, which exceeds a3, the cube of a, by
3a2 +3a + 1.
Hence it follows, that whenever the remainder, after the cube root
has been extracted, is less than three times the square of the root,
plus three times the root, plus unity, this root is not too small.
148. In order to extract the root of 105823817, it may be observed, that whatever be the number of figures in this root, if we
resolve it into units and tens, the cube of the tens cannot enter into
the last three figures on the right, and must consequently be found
in 105823. But the greatest cube contained in 105823 must have
more than one figure for its root; this root then may be resolved
into units and tens, and, as the cube of the tens has no figure inferior to thousands, it cannot enter into the three last figures 823.
If, after these are separated, there remain more than three figures
on the left, we may repeat the reasoning just employed, and thus,
dividing the number proposed into portions of three figures each,
proceeding from right to left, and observing that the last portion
may contain less than three figures, we come at length to the place
occupied by the cube of the units of the highest order in the root
sought.
Having thus taken the preparatory steps, we seek, by the rule
given in the preceding article, the cube root of the two first portions on the left, and find for the result 47; 105,823,817 473
we subtract the cube of this number from the  61     48
two first portions, and to the remainder 2000  41 8,23  6627
bring down the following portion 817. The 103 823
number 2000817 will then contain the last  2 0008,17
three parts of the cube of a number, the tens of 105 823 817
which are 47, and the units remain to be found. 000 000 000
These units are therefore obtained as in the example given in
the preceding article, by separating the last two figures on the
right of the remainder, and dividing the part on the left by 6627,
triple the square of 47. Then making trial with the quotient 3,
arising from this division, by raising 473 to a cube, we obtain for
the result the proposed number, since this number is a perfect
cube.




168


Elements of.,qlebra.


The explanation, we have given, of the above example, may
take the place of a general rule. If the number proposed had
contained another portion, we should have continued tie operation,
as we have done for the third; and it is to be recollected always,
that a cipher must be placed in the root, if the number to be divided
on the left of the remainder happen not to contain the number used
as a divisor; we should then bring down the following portion, and
proceed with it as with the preceding.
149. Since the cube of a fraction is found by multiplying this
fraction by its square, or which amounts to the same thing, by taking
the cube of the numerator and that of the denominator; reversing
this process, we arrive at the root, by extracting the root of the new
numerator and that of the new denominator. The cube of 5, for
example, is -25; taking the cube root of 125 and of 216, we
find A.
We always proceed in this way, when the numerator and denominator are perfect cubes; but when this is not the case, we may
avoid the necessity of extracting the root of the denominator, by
multiplying the two terms of the proposed fraction by the square of
this denominator. The denominator thence arising, will be the
cube of the original denominator; and it will be only necessary
then to find the root of the numerator. If we have, for example,
3, by multiplying the two terms of this fraction by 25, the square of
the denominator, we obtain
75
5 X 5 X 5'
The root of the denominator is 5; while that of 75 lies between
4 and 5. Adopting 4, we have 4 for the cube root of " to within
one fifth. If a greater degree of accuracy be required, we must
take the approximate root of 75, by the method 1 shall soon proceed to explain.
If the denominator be already a perfect square, it will only be
necessary to multiply the two terms of the fraction by the square
root of this denominator. Thus in order to find the cube root of 4,
we multiply the two terms by 3, the square root of 9; we thus
obtain
12
3   3    3'
Taking the root of the greatest cube 8, contained in 12, we
have i for the root sought, within one third.




Extraction of the Roots of Compound Quantities.


169


150. It follows from wlat has been demonstrated in art. 97.,
that the cube root of a number, which is not a perfect cube, cannot
be expressed exactly by any fraction however great may be the
denominator; it is therefore an irrational quantity, though not of the
same kind with the square root; for it is very seldom that one of
them can be expressed by means of the other.
151. We may obtain the approximate cube root by mea s of
vulgar fractions. The mode of proceeding is analogous to that
given for finding the square root (103); but, as it may be readily
conceived, and is besides not the most eligible, I shall not stop to
explain it.
A better method of employing vulgar fractions for this purpose
consists in extracting the root in fractions of a given kind. Thus,
if it were required to find, for example, the cube root of 22, within
a fifth part of unity, observing that the cube of I is T1, we reduce
22 to 2, 7555; then taking the root of 2750, so far as it can be
expressed in whole numbers, we have y, or 24, for the approximate
root of 22.
152. It is the practice of most persons, however, in extracting
the cube root of a number, by approximation, to convert this
number into a decimal fraction, but it is to be observed, that this
fraction must be either thousandths or millionths, or of some higher
denomination; because when raised to the third power, tenths
become thousandths, and thousandths nillionths, and in general,
the number of decimal figures found in the cube, is triple the number
contained in the root. From this it is evident, that we must place
after the proposed number three times as many ciphers, as there
are decimal places required in the root.  The root is then to be
extracted according to the rules already given, and the requisite
number of decimal figures to be distinguished in the result.
If we would find, for example, the cube root of 327, within a
hundredth part of unity, we must write six ciphers after this number,
and extract the root of 327000000 according to the usual method.
This is done in the following manner;
327,000,000   688
216           108
1110,00       13872
3144 32
125 680,00
325 660 672
339 328w
I 339 328.Alg.


22




170


Elements of Algebra.


Separating two figures on the right of the result for decimals,
we have 6,88; but 6,89 would be more exact, because the cube of
this last number, although greater than 327, approaches  itmore
nearly than that of 6,88.
If the proposed number contain decimals already, before we
proceed to extract the root, we must place on the right as many
ciphers, as will be necessary to render the number of decimal
figures a multiple of 3. Let there be, for example, 0,07, we
must write 0,070, or 70 thousandths, which gives for a root 0,4.
In order to arrive at a root exact to hundredths, we must annex
three additional ciphers, which gives 0,070000. The root of the
greatest cube contained in 70000 being 41, that of 0,07 becomes
0,41, to within a hundredth.
153. Hitherto I have employed the formula for binomial quantities only in the extraction of the square and cube roots of numbers; this formula leads to an analogous process for obtaining
the root of any degree whatever. I shall proceed to explain this
process, after offering some remarks upon the extraction of roots,
the exponent of which is a divisible number.
We may find the fourth root by extracting the square root twice
successively; for by taking first the square root of a fourth power,
a4, for example, we obtain the square, or a2, the square root of
which is a, or the quantity sought.
It is obvious also, that the eighth root may be obtained by
extracting the square root three times successively, since the
square root of a8 is a4, and that of a4 is a2, and lastly, that of
a2 is a.
In the same manner it may be shown, that all roots of a degree,
designated by any of the numbers 2, 4, 8, 16, 32, &c. that is, by
any power of 2, are obtained by successively extracting the square
root.
Roots, the exponents of which are not prime numbers, may be
reduced to others of a degree less elevated; the sixth root, for
examnple, may be found by extracting the square and afterwards
the cube, root. Thus, if we take a6 and go through this process
with it, we find by the first step a3, and by the second a; we may
also take first the cube root, which gives a2, and afterwards tilhe
square root, whence we have a, as before.
154. 1 now proceed to treat of the general method, which I
shall apply to roots of the fifth degree. The illustration will be




Extraction of the Roots of Compound Quantities.


171


rendered more easy, if we take a particular example; and by
comparing the different steps with the methods given, for the
extraction of the square and the cube root, we shall readily perceive, in what manner we are to proceed in finding roots of any
degree whatever.
Let it be required then to extract the fifth root of 231554007.
Now the least number, it may be observed, consisting of 2 figures,
that is 10, has in its fifth power, which is 100000, six figures; we
therefore conclude, that the fifth root of the number proposed
contains at least two figures; this root may then be represented
by a + b, a denoting the tens and b the units. The expression for
the proposed number will then be
(a + b)5 -:a5 + 5 a4 b+ 10a3 62 + &c.
I have not developed all the terms of this power, because it is
sufficient, as will be seen immediately, that the composition of the
first two be known.
Now it is evident, that as a5, or the fifth power of the tens of
this root, can have no figure, that falls below hundreds of thousands, it does not enter into the last five figures on the right of the
proposed number; we, therefore, separate these five figures.
If there remained more than five figures on the left, we should
repeat the same reasoning, and thus separate the proposed number
into portions of five figures each, proceeding from the right to the
left. The last of these portions on the left, will contain the fifth
power of the units of the highest order found in the root.
We find, by forming the fifth powers of  2315,540071 47
numbers consisting of only one figure, that  1024
2315 lies between the fifth power of 4, or  1291 5,4007  1280
1024, and that of five, or 3125. We take, therefore, 4 for the
tens of the root sought; then subtracting the fifth power of this
number, or 1024, from the first portion of the proposed number,
we have for a remainder 1291. This remainder, together with
the following portion, which is to be brought down, must contain
5 a4 b + 10 a3 b2 + &c. which is left, after a5 has been subtracted from (a + b)5; but among these terms, that of the highest
degree is 5 a4b, or five times the fourth power of the tens multiplied by the units, because it has no figure, which falls below tens
of thousands. In order to consider this term by itself, we separate the last four figures on the right, which make no part of it,
and the number 12915, remaining on the left, will contain this




172


Elements of lgebra.


term, together with the tens of thousands arising fiom the succeeding terms. It is obvious, therefore, that by dividing 12915
by 5 a4, or five times the fourth power of the four tens already
found, we shall only approximate the units. The fourth power
of 4 is 256; five times this gives 1280; if we divide 12915 by
1280, we find 10 for the quotient, but we cannot put more than
9 in the place of the root, and it is even necessary, before we
adopt this, to try whether the whole root 49, which we thus
obtain, will not give a fifth power greater than the proposed
number. We find indeed by pursuing this course, that the number 49 must be diminished by two units, and that the actual root
is 47, with a remainder 2209000; for the fifth power of 47 is
229345007; that is, the exact root of the proposed number falls
between 47 and 48.
If there were another portion still, we should bring it down
and annex it to the remainder, resulting from the subtraction of
the fifth power found as above, from the first two portions, and
proceed with this whole remainder, as we did with the preceding,
and so on.
After what has been said, it will be easy to apply the rules,
which have been given, as well in extricating the square and cube
root of fractions, as in approximating the roots of imperfect powers
of these degrees.
155. We may by processes, founded on the same principles,
extract the roots of literal quantities.  Tie following example
will be sufficient to illustrate the method, which is to be employed,
whatever be the degree of the root required.
We found in art. 143., the sixth power of 2 x3 - 5 a3; we
shall now extract the root of this power. The process is as follows;
64 x18 - 960 a3 x15 + 6000 a6 x12 - 20000 a9 x9 2 3 - 5 a3
+ 37500 a2 x6- 37500 a15 x23     15 -
+ 15625 a18
-64X'8
rem.   -960 a3 x15 + &c.
The quantity proposed being arranged with reference to the
letter x, its first term must be the sixth power of the first term
of the root arranged with reference to the same letter; taking then
the sixth root of 64 x18, according to the rule given in art. 129., we
have 2 x3 for the first term of the root required.




Equations with Two Terms.


173


If we raise this result to the sixth power, and subtract it from
the proposed quantity, the remainder must necessarily commence
with the second term, produced by the developement of the sixth
power of the first two terms of the root. But, in the expression
(a + b)6 = a6 + 6 a5b + &kc.
this second term is the product of six times the fifth power of the
first term of the root by the second; and if we divide it by 6 a5,
the quotient will be tle second term b.
We must, therefore, take six times the fifth power of the first
term 2 x3 of the root, which gives
6 X 32x15 or 192x'5,
and divide, by this quantity, the term - 960 a3 x15, which is the
first term of the remainder, after the preceding operation; the
quotient - 5 a3 is the second term of the root. In order to verify
it, we raise the binomial 2 x3 - 5 a3 to the sixth power, which we
find is the proposed quantity itself.
If the quantity were such as to require anotler term in the root,
we should proceed to find, after the manner above given, a second
remainder, which would begin with six times the product of the
fifth power of the first two terms of the root by the third, and
which consequently being divided by 6 (2 x3 - 5 a3)5, the quotient
would be this third term of the root; we should then verify it by
taking the sixth power of the three terms. The same course might
be pursued, whatever number of terms might remain to be found.
Of Equations with Two Terms.
156. EVERY equation, involving only one power of the unknown
quantity, combined with known quantities, may always be reduced
to two terms, one of which is made up of all those which contain
the unknown quantity, united in one expression, and the other
comprehends all the known quantities collected together. This
has been already shown with respect to equations of the second
degree, art. 105., and nmay be easily proved concerning those of
any degree whatever.
If we have, for example, the equation
3 x5 - a5 b2 = b4 c3 + a x5
by bringing all the terms involving x into one member, we obtain
a2 x5 — a cx5 = b4 c3 + a5 b2,
or              (a2 - ac) x5 = b4 C3 + a5 b.




174


Elements of.Algebra.


Now if we represent the quantities
a2- a c by p, b4 c3 + a5 b2 by q,
the preceding equation becomes
px5 — q;
freeing x5 from the quantity, by which it is multiplied, we have
X5 =qE
P
whence we conclude
5
In general, every equation with two terms being reduced to the
form
p Xm   q
gives
Xp
taking the root then of the degree m of each member, we have
7= s  i.
UP
157. It must be observed, that if the exponent m is an odd
number, the radical expression will have only one sign, which
will be that of the original quantity (131).
When the exponent m is even, the radical expression will have
the double sign 4-; it will in this case be imaginary, if the quantity q is negative, and the question will be absurd, like those of
which we have seen examples in equations of the second degree (131).
See some examples.
The equation           x5   -  1024,
gives
5 __1024__           4,
the exponent 5 being an odd number.
The equation
X4 = 625,
4
gives               x = ~ V/625 =-     5,
as the exponent 4 is even.




Equations with Two Terms.


175


Lastly, the equation
X4   -    16,
which gives
4
x- = 4- v/ - 16,
leads only to imaginary values, because while the exponent 4 is
even, the quantity under the radical sign is negative.
158. I shall here notice an analytical fact, which deserves attention on account of its utility, as well in the remaining part of the
present treatise, as in the Supplement, and which is sufficiently
remarkable in itself; it is this, that all the expressions x - a,
x2 -  a2, x3 - a3, and in general xm -  am (m being any positive
whole number), are exactly divisible by x - a. This is obvious
with respect to the first. We know that the second
x2-   a2 =(x + a) (x -   a) (34).
and the others may be easily decomposed by division. If we
divide xm - a"m by x -  a, we obtain for a quotient
xm-1 + a m    + a2 xm-3 + &c.
the exponent of x, in each term, being less by unity than in the
preceding, and that of a increasing in the same ratio. But instead
of pursuing the operation through its several steps, I shall present
immediately to view the equation
z- am =n xm-1 + a xm-2 + a2 xm3.. + am-2 x + am —'
which may be verified by multiplying the second member by x -a.
It then becomes
Xm + a xim-1 + a2 xm-2.......... + a- 2 X2 + am-' x
a         - a2 xm-2 - a3 m-3..........   am- x -am;
all the terms in the upper line, after the first, being the same, with
the exception of the signs, as those preceding the last in the lower
line, there only remains after reduction xm -  am, that is, the dividend proposed.
It must be observed, that the term a2 xm-2, in the upper line, is
necessarily followed by the term a3 Xm-3, which is destroyed by
the corresponding term in the lower line; and that, in the same
manner we find, in the lower line, before the term a'-i x, a term
- am-2 x2, which destroys the corresponding one in the upper line.
These terms are not expressed, but are supposed to be comprehended in the interval denoted by the points.




176


Elements of Algebra.


159. This leads to very important consequences, relative to the
equation with two terms x" =.
If we designate by a the number, which is obtained by directly
extracting the root according to the rules given in art. 154., we
have
q - am   or xm      am;
transposing the second member we obtain
xm    am    o.
The quantity xm -  am is divisible by x - a, and we have by the
preceding article
xm - am = (x- a) (xm-1 +,a xm-2..... + am-2 x + am-1).
This last result, which vanishes when x  a, is also reduced to
nothing, if we have
Xm-1 + a xm-2. + am-2 X + am-i    O. (116);
and, consequently, if there exists a value of x, which satisfies this
last equation, it will satisfy also the equation proposed.
These values have with unity very simple relations, which may
be discovered by making x = a y; then the equation xm - a' 0
becomes
am ym_ am    O, or ym      1 -- 0,
and we obtain the values of x, by multiplying those of y by the
number a.
The equation ym - 1 = 0, gives in the first place
m_
y!=/  1, y     =  = 1;
then by dividing ym - 1 by y - 1, we have
ym-1+ ym-2 + ym-3.....+ y2 + y + 1
Taking this quotient for one of the members, and zero for the
other, we form the equation on which the other values of y depend; and these values will, in the same manner, satisfy the
equation
ym-   1 = 0, or ym = 1,
that is, their power of the degree m will be unity.
Hence we infer the fact, singular at first view, that unity may
have many roots beside itself. These roots, though imaginary, are
still of frequent use in analysis. I can, however, exhibit here only




I


Equations with Two Terms.               177
those of the four first degrees, as it is only for these degrees, that
we can resolve, by preceding observations, the equation
ym-l + ym-2.....       - 1 - 0,
from which they are derived.
(1.) Let m = 2, we have
y2 -1 =,
whence we obtain
y=+     1, y     — 1.
(2.) By making m = 3, we have
-    1 = 0,
whence we deduce
y — 1,
then                y2 + y +  1    O.
This last equation being resolved, gives
-1 + -    -_ 3.-1 -      /   3
Y          2       ' Y          2;
thus we have for this degree the three roots
1 + V-3              -1 -         3
y==1,  y       2       ' Y2
The last two are imaginary; but if we take the cube, forming that
of the numerator, by the rule given in art. 34., and observing that
the square of  ~  3 being -  3, its cube is -  3  - -  3, we still
find y3 -- 1 in the same manner as when we employ the root
y=1.
(3.) Taking m = 4, we have
y4    1 = 0,
from which we deduce
y= 1,
then               y3 +    + y +   1 - 0.
We are not, at present, furnished with the means of resolving this
equation; but observing that
y4_   1        (y +  1) (y2 —  1),
we have successively
y2- 1 = p,      2 +-  1=0
whence
y=-+l, y=-l, y —+v1 —, y = --- —l.
Two of these values only are real; and the other two imaginary.
Alg.                     23




178


Elements of Algebra.


This multiplicity of roots of unity is agreeable to a general law
of equations, according to which any unknown quantity admits of
as many values, as there are units in the exponent denoting the
degree of the equation, by which this unknown quantity is determined; and when the question does not admit of so many real
solutions, the number is completed by purely algebraic symbols,
which being subjected to the operations, that are indicated, verify
the equation.
Hence it follows, that there are two kinds of expressions or
values for the roots of numbers; the first, which we shall term the
arithmetical determination, is the number which is found by the
methods explained in art. 154., and which answers to each particular case; the second comprehends negative values and imaginary
expressions, which we shall designate by the term algebraic determinations, because they consist merely in the combination of algebraic signs.
Of Equations which may be resolved in the same manner as those
of the Second Degree.
160. THESE are equations, which contain only two different
powers of the unknown quantity, the exponent of one of which is
double that of the other. Their general formula is
a2m +    p  = q,
p and q being known quantities.
Now if we take xm for the unknown quantity, and make xm  u,
we have
x2m =. u2,
whence
u2 + p u    q,
u-    -    ip ~ /q+4p2   (109);
restoring xm1 in the place of u, we have
xm =- -     d=/2 p  n   ^+ 4 p2,
an equation consisting of two terms, since the expression
I p ~E VTq + I p2
as it implies only known operations, to be performed on given
quantities, must be regarded as representing known quantities.
Designating the two values of this expression by a and a', we
have
xm = a and xm = a',




Equations with Two Terms.


179


from which we obtain
m           m
x = ha and x =,/a7.
If the exponent m be even, instead of the two values given
above, we shall have four, since each radical expression may take
the sign ~=; then
x =  + va, x == +     a/',
m m
x = -/, x = -^/
and these four values will be real, if the quantities a and a' are
positive.
All the values of x may be comprehended under one formula,
by indicating directly the root of the two members of the equation
Xm = --   p ~+ / + p
which gives
m
X =            ~ -  p i 1/q+ i p2 -The following question produces an equation of this kind.
161. To resolve the number 6 into two such factors, that the sum
of their cubes shall be 35.
Let x be one of these factors, the other will be 6; then taking
216
the sum of their cubes x3 and  6, we have the equation
3 216
x3 + -- = 35,
which may be reduced to
x6 +216 = 35X 3,
or                 x6 — 35 x3    -216.
If we consider x3 as the unknown quantity, we obtain, by the
rule given for equations of the second degree,
a?3 =  3 5     3:52 -L 16.
c3    _ ~.() 2 )       216.
By going through the numerical calculations, which are indicated, we find
___W2    1 225
V'()2 -216=v - 1      = 
and consequently,
X3 =   5 +     - 9  = 54 -27,
X 3   3 -,    =-  1 =    8,




180


Elements of.lgebra.


whence
3
x = v27 = 3,
3
x _         2
The first value gives for the second factor A or 2, while the second
value presents 6 or 3; we have, therefore, in the one case 3 and
and 2 for the factors sought, and in the other 2 and 3. These
two solutions differ only in the order of the factors of the given
number 6.
162. The equations, we have been considering, are also comprehended under the general law given in art. 159.; for the values
m   m
of Va, v/a  are to be multiplied by the roots of unity belonging to
the degree denoted by the exponent m.
Applying what has been said to the equation,
6 - 353 = — 216,
we find the six following roots;
c = 1 X 3,                 x = 1 X 2,
x -=     + -       X 3, x -                  X 2,
2                        2
-        1 —  3-3
x-         2          3    X 3, x2 —         X 2,:2
of which the first two only are real.
Calculus of Radical Expressions.
163. THE great number of cases, in which no exact root can
be found, and the length of the operation necessary for obtaining
it by approximation, have led algebraists to endeavour to perform
immediately upon the quantities subjected to the radical sign, the
fundamental operations, intended to be performed upon their roots.
In this way we simplify the expression as much as possible, and
leave the extracting of the root, which is a more complicated process, to be performed last, when the quantities are reduced to the
most simple state, which the nature of the question will allow.
The addition and subtraction of dissimilar radical quantities can
take place only by means of the signs + and -.  For example,
the sums
3     5       3      3,/i  -+ /a,    /a + /vb




Calculus of Radical Expressions.


181


and the differences
3      5      3      3
V/a -V /a,     V/a -V /b,
can be expressed only under their present form.
The same cannot be said of the expression
3       3          5 c 3
4a /2    +   /- 16a3 b - d    2tb,
because the radical quantities, of which it is composed, become
similar, when they are reduced to their more simple forms, according to the method explained in art. 130.  First, we have
3          3                  3
/ 16 a3 b =-/ 8 a3. 2 b  or  2 a v/2 b
3          3                  3
/2a6Ob  = /a6. 2b      or   a2 V2b;
the quantity, therefore, becomes
3           3 _    5a2 c 3
4 a  'I2b + 2 a /        d -b  Vb,
which gives, when reduced,
3       Sac 3                      a 3
6 a /2 b     l-  '2b  or   (6 d- 5 c)    /b.
164. With respect to other operations the calculus of radical
quantities depends upon the principle already referred to, namely;
that a product, consisting of several factors, is raised to any power
by raising each of the factors to this power.  So also, by suppressing the radical sign, prefixed to a quantity, we raise this quantity
to the power denoted by the exponent of this sign.  For example,
7
/a raised to the seventh power, is a simply, since this operation,
7
being the reverse of that which is indicated by the sign V, merely
restores the quantity a to its original state.
According to the principles here laid down, if, for example, in
the expression
7      7
/a X vb,
we suppress the radical signs, the result a b will be the seventh
power of the above product; and taking the seventh root, we find
7 _    7     7
V/a X  /b    %/ab.
This reasoning, which may be applied to all similar cases, shows,
that in order to multiply two radical expressions of the same degree
together, we must take the product of the quantities under the radical sign, observing to place it under a sign of the same degree.




182


Elements of tlgebra.


We have by this rule
3 V2a 3 X 7    5 a3 b c = 21 /lOa4 b4 c
21 a2 b2 /10c;
4 /a2 _ b2 X     /a2 -  b2 = 4 V(a2 - b2) (a2     b2) 
4 /a4    b4;
5                5
5a4 4__      X          d2
5
2 a9 -a3 66    a2 b3 c2 + 5 C2
-   a4- b4             d2
5
a3 (2 a6 - b6)   b3 dC+
a4  b4     X   d2 (a - 2+- 2)
_  a3b3 c2  2 a6 - b6
d2         a2 - b2'
since
a4-   b4  - (a2 + b2) (a2 -  b2)
7
V'a
165. As the seventh power of the expression -a, for example,
V/b
is b it will be seen, by taking the seventh root of this last result, that
7      7
V7 '
'/b
Hence to divide a radical quantity by another of the same degree,
we must take the quotient arising from the division of the quantities
under the radical sign, recollecting to place it under a sign of the
same degree.
We find by this rule, that
/6 ab _     6 a b
4/a / -  '-  O-3a
^/~3a            = %a  - b;
V/a-2-b"       2 _  2-       - 62
/ a-+ b      --   +b -- /a-b;
5         5        5 _
Va4 b       a_ 4 b  _  a4
5- -b3 '   c2 —    b2c2 ',%63 C2




Calculus of Radical Expressions.


183


166. It follows from the rule, given in art. 164., for the multiplication of radical quantities of the same degree, that to raise a
radical quantity to any power whatever, we have only to raise to
thispower the quantity under the radical sign, observing that the
5
result must take the same sign; thus to raise v/ab, for example to
the third power is to take the product
5       5       5
Vaab X   /a b X  /a b,
and as the radical signs are all of the same degree, the quantities
to which they belong are to be multiplied together, and the radical sign to be prefixed to the product, which gives
5
/53 b3.
7
In the same manner V/a2 b3 raised to the fourth power, gives
7
/a8 b2, which may be reduced to
7
a b,/a 5,
by resolving a8 b12 into-a7 b7 X a b5, and taking the root of the factor a7b7 (130).
It may be observed, that when the exponent belonging to the
radical sign is divisible by that of the power to which the proposed
quantity is to be raised, the operation is performed by dividing the
first exponent by the second. For example,
(6)_2    3 _
\/a =- Va,
because ~ = 3.
6
Indeed v/a denotes a quantity, which is six times a factor in a,
3
and the quantity V/a, which is obtained by dividing 6 by 2, being
only three times a factor in a, is consequently equivalent to the
product of two of the first factors, and is therefore the second
6
power of one of these factors, or of x/a.
The same reasoning may be applied to all similar cases, as in
the following'example;
12 _\3     4
\*/a2 b)3 = Va2 b.
167. 'If we reverse the methods given in the preceding article,
we shall be furnished with rules for extracting the roots of radical
quantities.




184


Elements ojt Agebra.


We perceive, by attending to the rule first stated, that if the
exponents of the quantities under the radical sign are divisible by
that of the root required, the operation may be performed as if there
were no radical sign, only it is to be observed, that the result must
be placed under the original sign.
We find, for example, that
3                 5
4 ___     3 ___5, /a6 ~- =V N,/a = Vca,2.
4           3
3           4        3
%/a4 b8 --  /a4 b8 = %/a b2.
From the second rule given in the preceding article, it is evident,
that the general method for finding the root of radical quantities,
is to multiply the exponent belonging to the radical sign by that of
the root, which is to be extracted.
By this last rule, we find, that
3
f5-    15
/ Va4 = */a4.
5
In fact /a4 is a quantity, which is five times a factor in a4
(24, 129); but the cube root of /4, being also three times a factor in this last quantity, is found 5 X 3 times or 15 times a factor
3
5      1.5
in the first a4; therefore 4/V4  /a4 4. In the same manner it
5
3^     15
might be shown, that v^ / - v/4.
168. Since by multiplying the exponent of a quantity under a
radical sign, by any number (166), we raise the root which is indicated, to the power denoted by this number, and by multiplying
also the exponent belonging to the radical sign, by the same number (167), we obtain for the result a root of a degree equal to that
of the power which was before formed, it is evident, that this
second operation reduces the proposed quantity back to its original
state.
5                                    35
The expression,,/a3 for example, may be changed into va2l,
by multiplying the exponents 5 and 3 by 7; for multiplying the
exponent of a3 by 7, we have, making use of the radical sign,
5
/a-1, the seventh power of the proposed radical quantity, and
multiplying by 7 the exponent 5 belonging to the radical sign in




Calculus of Radical Expressions.


185


5
the expression Va2i, we obtain the seventh root of the former
result; this last process, therefore, restores the expression to its
original value.
169. By this double operation, we reduce to the same degree any
number of radical quantities of different degrees, by multiplying, at
the same time, the exponent belonging to each radical sign, and those
of the quantities under this sign, by the product of the exponents
belonging to all the other radical signs. That the new exponents,
which are thus found for the radical signs, are the same, is obvious
at once, since they arise from the product of all the exponents
belonging to the original radical signs; and after what has been
said above, it is evident that the value of each radical quantity is
the same as before.
By this rule we transform
5            7
V/a3 b2  and  Vc4 d3,
35            35
into               v/a2l b14  and  Vc20 d15.
In the same manner, the three quantities,
3      5        7
v/a 2, Va2 c3, V/b4 C3
become respectively
105      105       105
V/a35 b70, V/a42 c63, Vb&60 C45.
If we meet with numbers, under the radical signs, we shall be
led, in applying this rule, to raise them to the power denoted by
the product of the exponents belonging to the other radical signs.
170. In the samne way, we may place under a radical sign a
factor which is without one, by raising it to the power denoted by
the exponent which accompanies this sign.
We may change, for example,
5            3       3
a2 into 4/a10, and 2 a V/b into V8 a3 b.
171. After having, by the transformation explained above,
reduced any radical quantities whatever, to the same degree, we
may apply to them the rules, given in articles 164. and 165. for
the multiplication and division of radical quantities of the same
degree.
Let there be the general expressions
m   b     n,
^/aP be X 4gbr c9;


Aig.


24




186


Elements of Slgebra.


we change (169)
nm        n
/aP b,     /b c',
into
mn         mun
anPbnQ, 4/bmr c/Ls
then by the rule given in art. 164. we have
mn _       mn           m
/ anP bnq X ^/bmr Cms =-  /anP bnq-+m"r cms
for the product of the proposed radical quantities.
We have also by the rule, art. 165.
m        mn          mn          mn
/aP bq _   /anp bnq     anp nq a     nP  b n  -mr
n        mn ___         mr Cms        C nms
r/bT Cs   /bmr Cms
Remarks on some peculiar cases, which occur in the Calculus of
Radical Quantities.
172. THE rules to which we have reduced the calculus of
radical quantities, may be applied without difficulty, when the
quantities employed are real. But they might lead the learner
into error with regard to imaginary quantities, if they are not
accompanied with some remarks upon the properties of equations
with two terms.
For example, the rule laid down in art. 164. gives directly
/ -a X V - a           -= -ax -a X-  aat2;
and if we take + a for v/2, we evidently come to an erroneous
result, for the product V' —a X v — a, being the square of,v'a, must be obtained by suppressing the radical sign, and is
therefore equal to -  a.
Bezout has obviated this difficulty, by observing, that when we
do not know by what method the square a2 has been formed, we
must assign for its root both + a and - a; but when, by means
of steps already taken, we know which of these two quantities
multiplied by itself produced a2, we are not allowed, in going back
to the root, to take the other quantity. This is evidently the case
with respect to the expression / — a X v - a; here we know,
that the quantity a2, contained under the radical sign in the expression /v'a  arises from -  a multiplied by -  a; the ambiguity,




Calculus of Radical Expressions.


187


therefore, is prevented, and it will be readily seen, that in taking
the root, we are limited to - a.
The difficulty above mentioned would present itself in regard to
the product /'a X Ala, if we were not led, by the circumstance of
there being no negative sign in the expression, to take immediately
the positive value of v'a. In this case, since a2 arises fiom + a
multiplied by + a, its root must necessarily be + a.
There can be no doubt with respect to examples of the kind we
have been considering; but there are cases, which can be clearly
explained only by attending to the properties of equations with two
terms.
173. If, for example, it were required to find the product
4
va   /;-1; reducing the second of these radical expressions to the
same degree with the first (169), we have
4     4            4     4         4
Va X v/( -1)2.= V/a X -V/ +     = Va,
a result which is real, although it appears evident, that the quantity
4
V/a multiplied by the imaginary quantity v' - 1, ought to give an
imaginary product. It must not be supposed, however, that the
4
expression V/a is in all respects false, but only that it is to be taken
in a very peculiar sense.
4
In fact, Va, considered algebraically, being the expression for
the unknown quantity x, in the equation with two terms,
x4   a = 0,
admits of four different values (159); for if we make a -= a4 by
4
taking a to represent the numerical value of V/a, considered independently of its sign, or the arithmetical determination of this quantity, we have the four values
a X +1,      X -   1, a X + V' 1, aX —'V- 1,
the third of which is precisely the product proposed.
By a little attention, it will be readily perceived, whence the
ambiguity, of which we have been speaking, arises. The second
power +  1 of the quantity -  1 under the radical sign, as it may
arise as well from + 1 X + 1, as from -  1 X -  1, causes the
4
quantity /l1 to have two values, which are not found in v - 1.
m      n
In general, the process by which the product /a X v/' is




188                Elements of Algebra.
formed, is reduced to that of raising this product to the power
m n; for if we represent it by z, that is, if we make
m      n
/a X Vb = z,
by raising the two members of this equation, first to the power m,
we have
a Vbm = zm
again, raising it to the power n, we obtain
an bm = zmn
This product, therefore, being determined only by means of its
power of the degree m n, or by an equation of this degree with
two terms, must have mn values (159). This will be perceived
m      n
at once, if we reflect that the expressions Va and v/', being nothing but the values of the unknown quantities x and y, in the equations with two terms,
m -a = O, y - b = O,
and, consequently, admitting of m and of n determinations, we
have, by uniting the several m determinations of x, with the several n determinations of y, m n determinations of the product
required.
When we are employed upon real quantities, there is no difficulty in finding the values, because the number of those, that are
real, is never more than two (157), which differ only in the sign.
174. If we use the transformation explained in art. 159., the
difficulty will be confined to the roots of +  1 and -  1; for if we
make x = a t and y =   u, a and f denoting the numerical values
n  n
of Va, b' considered without regard to the sign, the equations
xI:= a = O,     y:F b = 0,
become
tm::F 1 -O,   u1n:F 1 = 0,
whence
m          n                    m         n
in which a p represents the product of the numbers /-a, Vb, or the
arithmetical determination of the root of the degree m n of the
number a" bm.




Calculus of Fractional Exponents.


189


If we would give a determinate value to the product of the
m _      n
radical quantities / ~ a, / i- b, by fixing the degree of the radical signs, we must obtain from the equations
t     l:=: 1 = 0, us =F 1. 0,
m       n
the several expressions for /: 1,  /  1, and combine them in a
suitable manner.
To conclude, these operations are not often required, except in
some very simple cases, of which the following are the principal;
(1.) v-a X v-=b         V/a X   /b (/ ~1 X V-);
I suppress the radical sign in the expression V - 1, and obtain
1/ — a X  / — b     -  /ab X -  1 = -   ab.
4         4        4     4
(2.)        / - a X V - b     V a b (,/ _  )2;
I do not here multiply -  1 by - 1, because this would lead to
the ambiguity mentioned in art. 173.; but observing, that the
square of the fourth root is simply the square root, we have
4         4        4 _      4
V-a X V -b = Vab X '-1.
6        6         6       6__        6       3
(3.)    -a X   /-  b =b    ab X, (V-1)2 =V/ab X V6                   6
/ab X     — 1 —1       ab.
The results will be thus found to be alternately real and imaginary.
Calculus of Fractional Exponents.
175. IF we substitute in the place of the radical signs, their
corresponding fractional exponents (132), and apply immediately
the rules for the exponents, we shall obtain the same results, as
those furnished by the methods employed in the calculus of radical
quantities.
If we transform, for example,
5 _    5
V/a3 2, / a3 c2,
into
3 2        3
a by,      as ca,
we ave
Va3 Xb x Vac      = a b X ab c =




190                 Elements of Algebra.
3     I  3 2 2  6  2  2
as -  5 b C5 = a= b5 C5;
then, since X =  1 +  5, and, consequently,
a5= a    +- }    a X a5   (25),
2I  2               5
and a5 b5 c5 is equivalent to va b c2, we have
5        5          5
V/a3b X   / a3 c2 =  ab a b2 c2
a result which is not only exact, but is reduced to its most simple
form.
m         n___  n
Let there be the general example,/aP bq x 4/br c; the radical expressions here employed may be transformed into
p q    r.
am bm, b" cn,
we then have, according to the rules for exponents, (25),
p q     r s      p q    r s
am m x b bm cn = am br + n c.
Now in order to add the fractions q -, we must reduce them to
m n
the same denominator; and to give uniformity to the results, we
must do the same with respect to the fractions  n;, we obtain
io n
by this means,
np nq+mr ms
amnb mn cMn 
and placing this result under the radical sign, we have
m                      mn nm
/aP bM X )  / br c8 =-  /anp b-,mr c?.
176. The manner of performing division is equally simple, we
have for example
5    3 2
-/'5   aX b5  b/      =P  ab        _ a     (38))
5 --       r= 14. a3            )
/a4 c     a~ c5   a5    5 C
which may be reduced to
2
b5e
as c
this placed under the radical sign becomes




Calculus of Radical Expressions.        191
5        5
/ a3 b2  -b
5  --- 
/a4 c
We have in general,
m          p q     p     r
4/aP bq   ambm    am7bm   n
n         r s        s;
A/br cs   bn cn      cn
reducing the fractional exponents to the same denominator, in
order to perform the subtraction, which is required, we find
m          np nq —mr  mn
/ aP bq   amn b mn    _      bnq-mr
n            ms      -    Cnms
/6r CS       can
It is obvious, that the reduction of fractional exponents to the
same denominator, answers here to the reduction of radical expressions to the same degree, and leads to precisely the same
results (171).
177. It is also very evident, by the rule given in art. 127., that
m _n          p n    np   m _
/aP     -  ka) = am =     /a-P,
and by the rule laid down in art. 129., that
n   ___   n
Im         p      p    mn
j4a/a  =       = am? = a/aP.
The calculus of fractional exponents affords one of the most
remarkable examples of the utility of signs, when well chosen.
The analogy which prevails among exponents, both fractional and
entire, renders the rules, that are to be followed with respect to the
latter, applicable also to the former; but a particular investigation
is necessary in each case, when we use the sign y-, because it
has no connexion with the operation that is indicated. The further we advance in algebra, the more fully shall we be convinced
of the numerous advantages, which arise from the notation by
exponents, introduced by Descartes.
Examples in the Formation of Powers of Compound Algebraic
Expressions.
_-.
1. (a - b)3-a3 --- 3a2b + 3ab-       b3.
2. (4 -3 b)3 = 64 -    144 b +  108 b - 27 b3.




192


192            ~~~Elements of./flgebra.


3. (5 -4 X)4 = 625 - 2000x + 2400 x2 - 1.280 X3- 256X4.
4. (a3 +3 ab)4 =a 12 +12a"' b +54 a8b +108a'6b3
+ 81 a4b 4.
5. (5 a2 c9-d'- 4a bd2)4 = 625 a8 c'd4 — 2000 a7b6c d5
+ 2400 a~ b c4d6-_ 1280 a5bV c d7 + 256 a4 6b d8.
6. (3ac-2bd)5z=243a~c5-SlOa4c4bd+ 1080 a3 c3 62 d2
-720 a 2 C2 b3 d3 + 240 a c 6'1d4 - 32 bV d5.
7. (v'+V)4= a2l+ 6 ab +6 + (4 a+ 4 b) Vab.
8. (a+ b +c)3 =a3+ 3a'b + 3ac+ 3ab2 +6abc
+ 3 ac2 + b3 + 3 b c + 3b6c2 + C3.
9. (a+2b+ C)3-=a3+ 6a2b+3a2c+ 12ab2+l12abc
+ 3 a c2 + 8 b3 + 1.2 b2 c + 6 6 c2 + C3.
10. (a + b + c + d)2 = a,2 + 2 ab + 2 ac + 2a(id + 6 2
+ 2. b c + 26b d + c2 + 2 c d + d2.
n (n-i) (n-2) an- b3 +........
IYX 2 X 3
12. Make n=-       -1 n{ =.=2
Examples in the Extraction of the Cube Roots of Numbers.
1. VI12)67   23.           /.2. V843   = 96.:3. V119101= 106.
3
4. V2607 = 1 35
3
5. S         22842..
6. V5.8   = 1.79670..
3
3  _
9. Vj        0.87358....
10.           0.94103..
3 _
1 1. V3g   = 1. 56049..




Calculus of Radical Expressions.


193


Examples in the Extraction of the Cube Roots of dlgebraic
Expressions.
1. V'(6x2+x3+S + 12x) =x+ 2.
3
2. y9(294 a x2-84 a2' x+8a3-343x3) =2 a —7x.
3
3. V (X6 -6CX5~ 12c2 x4 - 8 c3 X3) = x2 _ 2 -CX.
3
4. V (a3'f - 6 a2m I  + 12 am+ 2x2- 8 a3X 3n)
=am- 2 a Xn.
3. 4a3 C3      3a2 c   3ab    _  X     ac     b
b3    ~~~b ~5+         3      b     C11 ~)
-xs ---  C- x+    cc4 ~4f_  Z
t~ 4j~b~+  3 a2 bo-  3 O4 b  a6.)b3 +2 +                 2 —4 x6
a2           a2
b + ~a2 C2    b + 2 C2 X2'
7. V(a3+3a2b+3a2c +3 a2 + 6 abc+ 3 a c + b3
+ 3 bV c + 3 b c2 + c3) = a + b + c.
8. *,/(27 z6 - 54 a z5 + 63 a2 z4 - 44 a3z3 + 21 a4 z2
- 6 a-5 z + a6) = 3 z2 - 2 a z + a2
9.  (8x6+48 c x5 +60 C2 X4 _S0oc3 X3 -_90 C4 X2
+ 108 c5 x-27 c6)= 2 x+ 4 c x- 3 c2.
10.; (,3 _,3) =.,  X3  A6   5z9    10 x12
y (a3- x3) = a32      9 a   81 a-   243 aI11.;(a3 f Ic3) = a  3  16  5 X9  10) X,'I 1.  (a~l + X3)  a +  8! a2  +  N3 all 
3          X   X~~2  5 X3  10 X4
12.V     IX)=a3X             a81a8243
x   22  5t3   10 X4
13. V (1 +x)    1+ 3      +81      24 3 +.
Examples in the Multiplication of Radical Expressions.
1.  x 4/X   = ab c/ Xyz.
3      3      3      3.2. V4 X 7 V/  X I  =-5  V10
AIg.             25




194


194            ~~~Elements of.hlgebtad


6    6
3. 4 x 2y.3-X V7/2 Sv.
4. 5 v5 >K 7 Xv >Kv/ = 140.
5. cy,/a x d N/a = a cd.
3    4    12 __ _ _
6. V2>(,/ vX V5 =V648000.
3    6    8    24
7. v2 X~ VX-. V3 =v2 56 
U2    8     24
a 4     amJ   a3m+2
3       6          6
a c  b cd  bIb'de I_ 1  ac3
d  eX     a2 c5 - bd bd3 e
1O. (V5+ 2 V~+ 3v,~o) X 2 V5=L0+4 v35+6 V5iJ
11. (3 + vS5) X( (2 - )= 1 - -
12. (7 + 2 -v/) X (9 -  = v~  3 - 17 N/~
14. (9 + 2 viiio) X (9 -2 VTh) -- 4 1.
15. (2V%8 + 3V%5- -7 V~)X (Vfi2- 5 VFO- 2V
-- 174 + 42 VIO
3     3      3   3           3     3
16. (V,/2V6-)X (3 V, — V3) =12 + 3 20-6~~
3
3      ~3 _3            3
17. (2 V5+ V,2) X (2 +j~-/) =4 + 2V2/~+ N/18+ 6V
1.(Va- - Vb ) X (-,-a+ )=a - b.
19.(cva- + d v'b) X (cyv-a- d y/i) a= -c2-6d2.
21. (    3e+V + ~)=    a +    3i? +  ac + 2  b~
4  4    ~~~4     4
+ 2 V-a + 2 V'b c.
$Pxamptes in the Dirvision of Radical Expressj?fl  m~~~~~~?
b?a-db        a4




Calculus of Radical Expressions,15


105


3
3_.2. 2 a623 4yVOb C5d = 14
15
3     5a7
3         6_
4. 4V%/122 V5= 2v*
5. cyi,(a2-x2)-~-y(a+x)  ~~~a  x).
6. y/(ab2-b2c) — <aC b
7.  /2,32 4   '  
8. (v+4 vi78- 3-8 -  ~-v %/ = v~ + 4v-v
3
1jO. l-&(v~ + 2) zz2 -,,3.
12. (  +   V)  (22 +12  42 6
37. 7 -   7%,5 ( IY  - V,3)  63  -  +173.+63v
14, (6- 3+14  T    4
42  4..  29  *~4        4 /6
bav, a a  ac


9




196


196            ~~~Elements of Algebra.


3. a  ~Xa     8  a   8F
12
4. a-! b     X aw bc =a'f-5b   l.7 __  8  1~2  3  2  38. 7   0o  _
5. Va12 X Va3 )KX 4  a5a-f-C a3= a' 05-aS y 52.
8     ~~~~3
lc y2)3  (C2 -  2) -f  2    I         5 3
6 (4Ja + x)8 X ~  ~      c-y2 a + x)
(t+ X)24(c2-y2
4
12
a-6 a2 cl'2=c4br
8. (VW3 +F2) X (VanW- VbP), (a4 + b 5) X (a 4  bV)
3    4     _   5_
= a2b-  - a V-a -  _Vb.
(b) Division.
m  p     in  p     mq+np
a  n - a q     q =a    na.
3.5  c a-'     c
2. c a"-+d a 0=  d       1
d 12
3. a b -+a,b             =- Vb3.
9  2     2....9  Li  60.a  2b3        d~    a3 b2c  b1b6d'20
cod     ~     d3  w     4al~l
12 
a3b2c g, d4 
4     6         124 
~~~~~~3~~~ ~  2
(Va    a + a           VaF


w




Calculus of Radical Expressions.


197


(c) Powers of Powers.
/. a\   -       1          _bmp     1
1 2a   p         1          MP   nq
2. a  n) - =            p   an= Va.
3. (a- b3)- =       Va.
40
4. (a.2 b-2 C 5)-  4      a20 C4 
20
5. ((a )         -a 
6      5             6
6. v (a3 b /a 3bc)5 = a3 b Vc.
6
7 (   c2 d  )        _  f(a+  )3
3,~ (                   - C4 d2 
(a+b b)2d
4.     v-/  3   8
8.    a V-     =Va4 b. 
General Theory of Equations.
178. EQUATIONS of the first and second degree are, properly
speaking, the only ones, which admit of a complete solution; but
there are general properties of equations of whatever degree, by
which we are able to solve them, when they are numerical, and
which lead to many conclusions, of use in the higher parts of
algebra. These properties relate to the particular form, which
every equation is capable of assuming.
An equation in its most general form must contain all the powers
of the unknown quantity, from that of the degree of the equation to
the first degree, multiplied each by some known quantity, together
with one term wholly known.
*    A general equation of the fifth degree, for example, contains all
the powers of the unknown quantity, from the first to the fifth;
and if there are several terms involving the same power of the




198


Elements of 3lgebra.


unknown quantity, we must suppose them to be united in one;
according to the method given for equations of the second degree,
art. 108. All the terms of the equation are then to be brought
into one member, as in the article above referred to; the other
member will necessarily be zero; and when the first term is negative, it is rendered positive by changing the signs of all the terms
of the equation,
In this way we obtain an expression similar to the following;
nx5 + px4 + qx-2 + rx- 2 + -sx + t = 0,
in which it is to be observed, that the letters n, p, q, r, s, t, may
represent negative as well as positive numbers; then dividing the
whole by n, in order that the first term may have only unity for its
coefficient, and making
p - P     q        r-        s        t
_~p,                         -- ~'.   S  T
P-        -= -_ -, --     T,
n         n        n         n        u
we have
X 5 + Px4 +  Qx r3 + R x2 + Sx +   T = 0.
In future, I shall suppose, that equations have always been prepared as above, and shall represent the general equation of any
degree whatever by
x+Px- +         P  Qxn-+2        Tx + U = 0.
The interval denoted by the pdmts may be filled up, when the
exponent n takes a determinate value.
Every quantity or expression, whether real or imaginary, which,
put in the place of tie unknown quantity x in an equation prepared
as above, renders the first member equal to zero, and which consequently satisfies the question, is called the root of the proposed
equation; but as the inquiry does not at present relate to powers,
this acceptation of the term root is more general, than that, in
which it has hitherto been used (90, 129).
179. Take a proposition analogous to those given in articles
116 and 159, and one which may be regarded as fundamental,
If the root of any equation whatever,
x +- P x'1- + Q x"- 2.     +.... + Tx + U = 0,
be represented by a, the first member of this equation may be exactly
divided by x - a.
Indeed, since a is one value of x, we have necessarily,  4
an + Pa-1 -- +  an-2....+     Ta +   U= 0,




General Theory of Equations.


and, consequently,
U= -    an _-  P an- -  Q a"-2.....-    Ta,
so that the equation proposed is precisely the same as
n + p an-1 +    Q x-2..... +   Tx 
-a-         P a7'-1 Q_ a-2 - a.. - -Ta ~    0,
which may be reduced to
n -   an +  P (n-1 - an-1) +     q (n-2 _   an-2) a........................ +                - -  T (x —a)  -
As the quantities
Xn -  a    n-1 -,       xn —2, an- -2 - 2 a2...... X - -a,
are each divisible by x -  a (158), it is evident, that the first
member of the proposed equation is made up of terms, all of which
are divisible by this quantity, and may consequently be divided by
x -  a, as the enunciation of the proposition requires.*
180. To form the quotient we have only to substitute for the
quantities
nn n- a",  1 —1  - a    n-1,  n-2_- an-2,.....  -  a,
the quotients, which are obtained by dividing these quantities by
x — a, and which are respectively..................,...........
* D'Alembert has proved the same proposition in the following
manner.
If we conceive the first member of the proposed equation to be
divided by x - a, and the operation continued until all the terms
involving x are exhausted, the remainder, if there be any, cannot contain x. If we represent this remainder by R, and the quotient to
which we arrive by Q, we have necessarily
+n + P   '         +c-1..... +&C. =  Q ( - a) + R.
Now if we substitute a in the place of x, the first member is reduced
to nothing, since a is the value of x; the term Q (x - a) is also
nothing, because the factor x - a becomes zero; we Inust, therefore,
have R - 0, and it is so, independently of the substitution of a; for,
as this remainder does not contain x, the substitution cannot take
place, and it still preserves the value it had before.
Hence it follows, that in every case, R - 0, and that, consequently,
n + P xy-1 i + Q xn-2 &c.
is exactly divisible by z - a.




200


Elements of Algebra.


+t1 + ax"-2 + a2     "3 +...       an-1
an-2 + a    -3.....- an-2
a x~-3.....  -+- an-s,
n —3        + a3n —3
+ 1.
Arranging the result with reference to the powers of x, we have
xlt-1 + a x7n-2 +  a2 x-3... +    a-I,
+ P xn-2 + P a x1-3....    + P at-2,
+   Qxn3          + Q an-3,
+ T.
181. It is evident from the rules of division simply, that if the
first member of the equation,
n + Pxn'-l +    Q Xn 2 + &c. =0,
be divided by x - a, the quotient obtained will be exhibited under
the following form,
+n-1 +, P' n- +  Q  - x34 + &c.
P', Q,, &c. representing known quantities different from P, Q, &c.
we have then
X +- P n"-l1 + &c. = (x -   a) (xa-l 1  P/xn1-2 + &c.);
and according to what was observed in art. 116., the proposed
equation may be verified in two ways, namely, by making
x - a == 0, or Xn-1 + P1 Xun — + &c. - O.
Now if the equation
n-l + P/ Xn-2 +- &c. = 0
has a root b, its first member will be divisible by x-  b; we have
then
xn-l + P' x^-2 + &c. =(x -b) (x"-2 + P", xn-3 + &c.),
and, consequently,
x" 4 P      +x1n-l + &c. = ( -a) (x - b) (x7-2 + P/" lX-3 + &c.);
the equation proposed may, therefore, be verified in three ways,
namely, by making
— a = 0, or x-b = 0, or x"-2 + P/ n-3 +- &c. = 0.
If the last of these equations has a root, c, its first member may
still be decomposed into two factors,
X -  C, Xn-3 + P/" aXn-4 + &c. = 0;
we then have
"n + P X'n- + &c.
= (        ( - a-        ) -  ( 3 + P/"' +C-4 + &c.);




General Theory of Equations.


201


from which it is obvious, that the proposed equation may be verified in four ways, namely, by making
x —  a = 0, x -b = Ox X-  c=0, x -3 + P"/ xn-4 + &C.      0.
Pursuing the same reasoning, we obtain successively factors of
the degrees
n -  4, n -   5, n-   6, &c.;
and if each of these factors, being put equal to zero, is susceptible
of a root, the first member of the proposed equation is reduced to
the form
(x -  a) (x-  b) (x -  c) (x -  d)..... (x-  1);
that is, it is decomposed into as many facto s of the first degree,
as there are units in the exponent n, which denotes the degree of
the equation.
The equation
" + P xn-i + &c. - 0,
may be verified in n ways; namely, by making
x -  a =0, orx-0b =, or x  -   b  0, or c  -  d = 0,
or lastly,              x - I = 0.
It is necessary to observe, that these equations are to be regarded as true only when taken one after the other, and there arise
manifest contradictions from the supposition, that they are true at
the same time. In fact, from the equation x - a = 0, we obtain
cx = a, while x -  b = 0 gives x = b, results, which are inconsistent, when a and b are unequal quantities.
182. If the first member of the proposed equation,
n + P xn-1 + &c. = 0,
be decomposed into n factors of the first degree,
x --- a, x -  b, x  c, x -  d.....     -i. x 1,
it cannot be divided by any other expression of this degree.
Indeed, if it were possible to divide it by a binomial x -  a, different from the former ones, we should have
n + P Xn-1 +- &c. = (x -a) (Xn-1 + p n —2 + &C.)
and, consequently,
(x -  a) (x -  ) (x -  c) (x -  d).... (x -- l)
= (x -- ) (Xl-1 + p Xn-2 + &c.);
now by changing x into a, this becomes
alg.              26




202


Elements of Jlgebra.


(-   na) (a-  b) (o-  c) (a-d)..... (-      I)
=(-    ( a) ( -1 + p cn-2 + &C.);
The second member vanishes by means of the factor a -, which
is nothing; this is not the case with respect to the first, which is
the product of factors, all of which are different from zero, so long
as a differs from the several roots a, b, c, d... 1. The supposition we have made then is not true; therefore, an equation of
any degree whatever does not admit of more binomial divisors of
the first degree, than there are units in the exponent denoting its
degree, and consequently cannot have a greater number of roots.*
183. An equation regarded as the product of a number of
factors,
x -  a, x -  b, x - c, x - d, &c.,
equal to the exponent of its degree, may take the form of the product exhibited in art. 135., with this modification, that the terms
will be alternately positive and negative.
If we take four factors, for example, we have
X4- ax3 +- abx -     a b cx + a b c d  0
-  b x3 + a c x2 - abdx
-  c 3 + a d x2 -  a c dx
-- dx3    b c x2-  b cdx
+ bdx2
+ cd x2
The second terms of the binomials x -  a, x -  b, x - c, &c.
being the roots of the equation, taken with the contrary sign, the
properties enumerated in art. 135., and proved generally in art.
136., will, in the present case, be as follows,
The coefficient of the second term, taken with the contrary sign,
will be the sum of the roots;
The coefficient of the third term will be the sum of the products of
the roots, taken two and two;
The coefficient of the fourth term, taken with the contrary sign,
will be the sum of the products of the roots, multiplied three and
three, and so on, the signs of the coefficients of the even terms
being changed;


* This demonstration is taken from the Annales de Mathematiqices
published by M. Gergonne. See vol. iv, pp. 209, 210, note.




General Theory of Equations.


203


The last term, subject also to this law, will be the product of all
the roots.
Making, for example, the product of the three factors
x -  5, x + 4, x + 3,
equal to zero, we form the equation
x3 + 2 2 - 22 x -   60 = 0,
the roots of which are
+ 5,- 4, - 3;
we have for their sum
5-4-3 =-2;
for the sum of their products, taken two and two,
+ 5 X-4 + 5 X -3-4 X - 3 = —20- 15 + 12 =-23,
and for the product of the three roots,
+ 5 X -4 X-3 = 60.
In this way we form the coefficients, 2 -  23, -  60, changing the
signs of those for the second and fourth terms.
If we make the product of the factors
x - 2, x -  3, and x + 5,
equal to zero, the equation thence arising
x3 - 19 x + 30 = p,
as it has no term involving x2, the power immediately inferior to
that of the first term, wants the second term; and the reason is,
that the sum of the roots, which, taken with the contrary sign,
forms the coefficient of this term, is here
2 + 3 - 5,
or zero, or in other words, the sum of the positive roots is equal to
that of the negative.*
184. We have proved (182), that an equation, considered as
arising from the product of several simple factors, or factors of the
first degree, can contain only as many of these factors, as there
are units in the exponent n denoting the degree of this equation;
but if we combine these factors two and two, we form quantities of
the second degree, which will also be factors of the proposed
equation, the number of which will be expressed by
n (n-)     (140).
1.2


* See note at the end of this treatise.




204


Elements of Algebra.


For example, the first member of the equation
X4 -  ax3 + ab x2 -   a b c x- abcd = 0
-  b x3 + a c x -  a b d x
-  c x3 +   d ad c  -  a c d x
-  d x3 +     2 b c2 -bc dx
+- bdX2
+ cdx2
being the product of
(x -  a) X (x -  b) X ( -   c) X (x - d),
may be decomposed into factors of the second degree, in the six
following ways;
(x    a) (x -  b) X (x -  c) (x -  d)
(x - a) ( -   c) X ( -   b) (x -  d)
(x -  a) (x - d) X (x -b) (x -c)
(x -b) (x -c) X (x - a) (x - d)
(x -b) (x -   d) X (x -   a) (x -  c)
(x -  c) (x -  d) X (x -  a) (x - b);
whence it appears, that an equation of the fourth degree may have
six divisors of the second.
By combining the simple factors three and three, we form quantities of the third degree for divisors of tie proposed equation;
for an equation of the degree n the number will be
n (n   1) (n -2)
1.2.3
and so on.
Of Elimination among Equations exceeding the First Degree.
185. THE rule given in art. 78., or the method pointed out in
art. 84., is sufficient in all cases, for eliminating in two equations
an unknown quantity, which does not exceed the first degree,
whatever may be the degree of the others; and the rule of art. 78.
is applicable, even when the unknown quantity is of the first degree
in only one of the proposed equations.
If we have, for example, the equations
ax2 + bxy + cy2 = m2,
x2 + xy =- ',
taking, in the second, the value of y, which will be
n2  2;2
y    -    2
-'h2.'




Equations exceeding the First Degree.


205


and substituting this value and its square, in the place of y and y2
in the first equation, we obtain a result involving only x.
186. If both of the proposed equations involved the 'second
power of each of the two unknown quantities, the above method
could be applied in resolving only one of the equations, either with
respect to x or y.
Let there be, for example, the equations
a x2 + b x y -- c y2  m2,
x2 +  y2 =2n2;
the second gives
y - = - /12 - x.
Substituting this value of y, and its square in the first, we obtain
a x2 ~- b x V/n2  x'2 + c (?I2 - X2) - m2.
Our purpose appears to be answered, since we' have arrived at a
result, which does not involve the unknown quantity y, but we are
unable to resolve the equation containing x, without reducing it to a
rational form, by making the radical sign, under which the unknown
quantity is found, to disappear.
It will be readily seen, that if tllis radical expression stood
alone in one member, we mighl t make the radical sign to disappear
by raising this member to a square. Collecting together all the
rational terms then in one member, by transposing the terms: b x /i^2 - x2 and m=, we have
ax 2 +   c (al2 -   X )  '- m =   -=-  b X./2 — ~ j2
taking the square of each member we form the equation
a2x4 +c2 (n 2 —x2 )2 +m4              )     -2 
+2acx2 (n2_x2)     2ac2x2     m2(n2-x2)               x),
which contains no radical expression.
The method, we have just employed for making the radical sign
to disappear, deserves attention, on account of the frequent occasion
we have to apply it; it consists in insulating the quantity found
under the radical sign, and then raising the two members of the
proposed equation to the power (denoted by the degree of this sign.
187. The complicated nature of this process, which increases
in proportion to the number of radical expressions, added to the
difficulty of resolving one of the proposed equations with reference to one of the unknown quantities, a difficulty, which is often
insurmountable in the present state of algebra, has led those, who
have cultivated this science, to seek a method of effecting the




206


Elements of.lpebra.


elimination without this; so that the resolution of the equations
shall be the last of the operations required for the solution of the
problem.
In order to render the operation more simple, we reduce equations with two unknown quantities to the form of equations with
only one, by presenting only that, which we wish to eliminate.
If we have, for example,
x2 +axy+bx =cy2 + dy +e,
we bring all the terms into one member, and arrange them with
reference to x; the equation then becomes
X2 - (ay - b) x-   cy- dy - e - 0;
abridging this, by making
ay + b = P,      cy -dy-e,
we have
s + Px +       -=O.
The general equation of the degree m with two unknown
quantities must contain all the powers of x and y, which do not
exceed this degree, as well as those products, in which the sum of
the exponents of x and y does not exceed m; this equation then
may be represented thus;
x'm+ (a+ by)xm-l+(c+dy+ey2 )xn-2+(f+gy+ hy2 + ky 3)x;-m3
+(p+qy+ry2... +ulym-l)x+p+qy+r'y2.. +v'ym=0.
No coefficient is assigned to xm in this equation, because we may
always, by division, free any term of an equation we please, from
the number, by which it is multiplied. Now if we make
a+by=P, c+dy +ey2             Q, f+gy+hy2+ky3=R,
p + q y.. + U ym-1      T, p + q' y....+ v' y = U,
the above equation takes the following form,
xm + p      xm-1 + Q xm-2+ R xm-3...... +.  Tx + U = 0.
188. It should be observed, that we may immediately eliminate
x in the two equations of the second degree,
x2 +    x+ Q= 0, x2 + P        + Q = 0,
by subtracting the second from the first. This operation gives
(P-P')x+ Q-          ' =0,
Q-QI
whence                x    — _ p
substituting this value in one of the two proposed equations, the
first for example, we find




Equations exceeding the First Degree.


207


(Q — Q)2 P (Q — Q')
(P-P'/)2      p — p     +-   -o;
making the denominators to disappear, we have
(Q-     ')- P (P -P') (Q- q') + Q(P -P')2 -,
then developing the two last terms, and making the reduction
(Q- Q,)2 + (-p,) (        Q,-  Q p,) = o.
We have then only to substitute for P, Q, P', and Q', the particular values which answer to the case under consideration.
189. Before proceeding further, I shall show, how we may
determine, whether the value of any one of the unknown quantities satisfies at the same time the two equations proposed. In
order to make this more clear, I shall take a particular example;
the reasoning employed will, however, be of a general nature.
Let there be the equations
3 + 3 x2 y + 3     x  - 98 = 0..... (1),
x2+ 4zy -2y2       -  10 = 0.....(2),
which we shall suppose furnished by a question, that gives y = 3.
In order to verify this supposition, we must substitute 3 in the
place of y, in the proposed equation; we have then
3 +- 9 x2 + 27 x- 98 = 0..... (a),
x2+ 12x- 28 =..... (b),
equations, which must present the same value of x, if that, which
has been assigned to y, be correct. If the value of x be represented by a, the equation (a) and the equation (b) will, according
to what has been proved in art. 179., both of them be divisible
by x-a; they must, therefore, have a common divisor, of
which x - a forms a part; and in fact, we find for this common
divisor x -2 (48); we have therefore a =2. Thus the value
y= 3 fulfils the conditions of the question, and corresponds to
x -2.
If there remained any doubt, whether or not the common divisor of the equations (a) and (b) must give the value of x, we
might remove it, by observing, that these equations reduce themselves to
(x2+ 11 x + 49) (x- 2)     0,
(x+ 14) (x- 2) =0,
from which it is evident, that they are verified by putting 2 in the
place of x.
190. The method I have just explained, for finding the value of
x, when that of y is known, may be employed immediately in the
elimination of x.




208


Elements oJ Algebra.


Indeed, if we take the equations (1) and (2), and go through
the process necessary for determining whether they have a common divisor involving x, instead of finding one, we arrive at a
remainder, which contains only the unknown quantity y and
numbers, that are given; and it is evident, that if we put in the
place ofy its value 3, this remainder will vanish, since by the same
substitution, the equations'(1) and (2) become the equations (a)
and (b), which have a common divisor. Forming an equation,
therefore, by taking this remainder and zero for the two members,
we express the condition, which the values of y must fulfil, in order
that the two given equations may admit, at the same time, of the
same value for x.
The adjoining table presents the several steps of the operation
relative to the equations,
x3 + 3   3X    x y  y  98 -  0,
x2+4xy -2y2       -10= 0,
on which we have been employed in the preceding article. We
find for the last divisor,
(9y2 + 10) x- 2y3- 10y-98;
and the remainder, being taken equal to zero, gives
43 y6 + 345 y4- 1960 y3 + 750 y2 - 2940 y - 4302 = 0,
an equation, which admits, besides the value y = 3 given above,
of all the other values of y, of which the question proposed is susceptible.
The remainder above mentioned being destroyed, that preceding
the last becomes the common divisor of the equations proposed;
and being put into an equation, gives the value of x when that of y
is introduced. Knowing, for example, that y =3, we substitute
this value in the quantity
(9y2 + 10) -2yy3 -    10y-98;
then taking the result for one member, and zero for the other, we
have the equation of the first degree
91 x-182 = 0, or x = 2.
191. The operation to which the above equations have been
subjected, furnishes occasion for several important remarks.
First, it may happen that the value of y reduces the remainder
preceding the last to nothing; in this case, the next higher remainder, or that which involves the second power of x, becomes
the common divisor of the two proposed equations. Introducing
then into this the value of y, and putting it equal to zero,




x3+3x2y+3y2x-98 x2+4xy-2y2-1O
-x3-4x y+2yx +I lOx x    y
- 12 y+5y'x+lOx-98
+ X2y+4.y5x-2 2y-lOy
1 st rein...+ (9Y2 + 1 O)x-2y3-lOy-98
x2-f- 4xy -2y~-  l 0l(9y2+IO) x.-2yP31IOy-98S
or rather (9y52+10)x-+36x&A_8I 1 o      01O  (9y2+ IO)x-2yP —l Oy ---9-S 
-(9y2+ 1O)x2+ 2xy3+98x                 x+387,3+50y+98
+ l Oxy________
+3Sxy3-18y4 _I1Oy2_10O
+ 5Oxn
+98X
or rather  (38y3+50y+98)(oy2+10)x-162y6-1 170y4 -2000y2-10
rem  (S3+5y+9) (~2+10O)x+ 76y%+ 480y4~3920y3+ 500y2+588Oy+9604......8-6y- 69O114+39.2O93~-150y2+5880y+8604 -inPutting this remainder equal to zero, then dividirng all its trsb,adcagn   h   in
inorder to mnake. the first term positive, we have b ersy2,adcnggthsis
43y6+345y4-l960y3+750y2_294Oy-4302=0,




210


Elements of Algebra.


we have an equation of the second degree, involving only x, the
two values of which will correspond to the known value of y. If
this value still reduce to nothing the remainder of the second degree, we must go back to the preceding, or that into which the
third power of x enters, because this, in the case under consideration, becomes the common divisor of the two proposed equations;
and the value of y will correspond to the three values of x. In
general, we must go back until we arrive at a remainder, which is
not destroyed by substituting the value of y.
It may sometimes happen, that there is no remainder, or that
the remainder contains only known quantities.
In the first case, the two equations have a common divisor
independently of any determination of y; they assume then the
following form,
P X D=O, Q X D-=0,
D being the common divisor. It is evident, that we satisfy both
the equations at the same time, by making in the first place
D = 0; and this equation will enable us to determine one of the
unknown quantities by means of the other, when the factor D
contains both; but if it contains only given quantities and x, this
unknown quantity will be determinate, and the other will remain
wholly indeterminate. With respect to the factors, which do not
contain x, they are found by what is laid down in art. 50.
Next, if we make at the same time
P=0,     Q-=0,
we have still two equations, which will furnish solutions of the question proposed.
Let there be, for example,
(a x + b y - c) (mx + ny- d) = 0,
(a'x + by - c') (mx + n y- d)= 0;
by supposing, first, the second factor, common to the two equations, to be nothing, we have with respect to the unknown quantities x and y only the equation
mx + ny-d = 0,
and in this view the question will be indeterminate; but if we
suppress this factor, we are furnished with the equations
ax + by - c      0, a'x + b'y-c'= 0,
or           ax + by     c, a' x + b' y = c';
and in this case the question will be determinate, since we have as
many equations as unknown quantities.




Equations exceeding the First Degree.


211


When the remainder contains only given quantities, the two
proposed equations are contradictory; for the common divisor, by
which it is shown that they may both be true at the same time,
cannot exist, except by a condition which can never be fulfilled.(D)
This case corresponds to that mentioned in art. 68., relative to
equations of the first degree.*
192. If then we have any two equations,
ax' + P xm      Q      + R...+      T +  T U = 0,
x + P/ X-1 + - Q' x-2 + R' x-3.... + Y x + Z' = 0,
where the second unknown quantity, y, is involved in the coefficients, P, Q, &c. P', Q', &c. in seeking the greatest common
divisor of their first members, we resolve them into other more
simple expressions, or come to a remainder independent of c,
which must be made equal to zero.
This remainder will form the final equation of the question
proposed, if it does not contain factors foreign to this question;
but it very often begins with polynomials involving y, by which the
highest power of xc, in the several quantities, that have been successively employed as divisors, is multiplied, and we arrive at a
result more complicated than that which is sought, should be. In
order to avoid being led into error with respect to the values of y
arising from these factors, the idea, which first presents itself, is,
to substitute immediately in the equations proposed each of the
values furnished by the equation involving y only; for all the
values, which give a common divisor to these equations, necessarily belong to the question, and the others must be excluded. It
will be perceived also, that the final equation will become incom* It will be readily perceived, by what precedes, that the problem
for obtaining the final equation from two equations with two unknown
quantities, is, in general, determinate; but the same final equation
answers to an infinite variety of systems of equations with two unknown quantities. Reversing the process, by which the greatest
common divisor of two quantities is obtained, we may form these
systems at pleasure; but as this inquiry relates to what would be of
little use in the elementary parts of mathematics, and would lead me
into tedious details, I shall not pursue it here. Researches of this
nature must be left to the sagacity of the intelligent reader, who will
not fail, as occasion offers, of arriving at a satisfactory result.




212


Elements of.lgebra.


plete, if we suppress in the operation any factor involving y; but
all these circumstances together occasion some inconvenience in
the application of the above method,' and lead me to prefer the
method given by Euler, which I shall explain in the following
article.
193. Let there be the equations
x3 + P x2 + Qx + R      - 0,
O 4 + P, X3 + Qx3 + R' x + S' = 0;
representing by x —  a the factor, which must be common to both,
when y is determinate in a proper sense, we may consider the
first as the product of x - a by the factor of the second degree,
x2 - p x + q, and the second as the product of x -  a by the
factor of the third degree x3 + p' x2 + q' x + r'/, p and q, p', q'
and r' being indeterminate coefficients. We have then
x3+Px2+ qx+R=(x- a)(x 2+ px+q),
4 +- P'x3 + QX2 + R'x + S'- (x-a) (x3 +p'x2 + q'x + r').
Exterminating the binomial (x - a), in the same manner as an
unknown quantity of the first degree (84), we find
(x3 + p x +   Q x + R) (X3 + p/ 2 + q Ix + r)
== (4 + P/ 3     q   2 + X2  R'x + S') (x2 + p x + q) 
a result, which must verify itself without any particular value being
assigned to x; this cannot take place, however, unless the first
member be composed of the same terms as the second; we must,
therefore, after performing the multiplications, which are indicated,
put the coefficients belonging to each power of x in one member,
respectively, equal to those belonging to the same power in the
other. In this way we obtain the following equations;
P+p/=P'+p        Rp'+   q'+Pr'=S' +Rlp+       'q
Q+Pp'+q'= Q+P'p+q           Rq'+ Qr'= Sp+R'q
R+ Q~p'+Pq'+r'=R'+ Q'p+P'q             Rr'=S'q.
As we have here six equations, and only five indeterminate quantities, namely, p, q, p', q', and r', all of which are of the first
degree, these quantities may be exterminated; we shall thus arrive
at an equation, which, involving only the quantities P, Q, R, P',


* On this subject see a memoir of M. Bret, in the 15th number of
Journal de PEcole Polytechnique, also one of M. Lefebure, 3d number, vol, ii. of the Correspondance of the same school.




Equations Exceeding the First Degree.


213


', R', and S', will express a condition necessarily implied in the
conditions of the question, and which, consequently, will be the
final equation in y.*
Should this equation be identical, it follows, that the proposed
equations have at least one factor of the form x - a, whatever y
may be; on the contrary, if the final equation contain only known
quantities, the proposed equations are contradictory.
When the final equation takes place, we obtain the factor x - a
by dividing the first of the proposed equations by the polynomial
x2 + p x + q; we find for the quotient
x + P-p,
and neglect the remainder, because it must necessarily be reduced
to nothing, when we substitute in the place of y a value obtained
from the final equation. Putting the above quotient equal to zero,
we find
x =    - P,
and this value of x will be known, or at least will be expressed by
* The method of Euler, explained here, amounts to multiplying
each of the proposed equations by a factor, the coefficients of which
are indeterminate, putting the products equal, and disposing the
coefficients in such a manner, that the terms containing the unknown
quantity destroy each other. In this form it is presented in his
Introduction to the Analysis of Ifinites. The exponent, which denotes the degree of the products, being designated by k, that of the
factors is k - m for the equation of the degree m, and k - n for that
of the degree n. The first term of each of these factors, having unity
for a coefficient, the one contains k - m indeterminate coefficients,
and the other k - n. The sum of the products contains a number k
of terms involving x; but it is necessary to destroy k - I terms only,
because that, which contains the highest power of x, vanishes of
itself. It follows from this, that the whole number 2 k - m - n of
indeterminate coefficients must be equal to k - 1, and consequently
k = m + n - 1; we must, therefore, multiply the equation of the
degree m by a factor of the degree n - 1, that of the degree n by a
factor of the degree m - 1, and put the products equal, term to term,
a method similar to that given in the text. It may be observed, that
this former method of Euler contains the germ of that developed by
Bezout in his Theorie des Equations Algebriques.




214


Elements of Algebra.


means of y, if we substitute for p its value deduced from the equations of the first degree, formed above.
This expression assumes, in general, a fractional form, so that
we have x — l O, or Jx -   M  == 0; and it may be seen in this
case, that the values of y, which would cause JM and nJ to vanish
at the same time, would verify the preceding equation independently of x; this takes place in consequence of the fact, that by
means of these values, the proposed equations would acquire a
common factor of a degree above the first. It would not be difficult to go back to the immediate conditions in which this circumstance is implied; but the limits I have prescribed to myself in the
present treatise do not permit me to enter into details of this kind.
194. Now let there be the equations
2 -+ Px + Q       O, x2 - PX + - Q'=O;
the factors, by which x —  r is multiplied, will be here of the first
degree, or x + p and x +- p' simply; in this case,
R = 0, R' = 0, S'= 0, q = o, q' = 0,         - = 0,
and we have
P+P= P I        +p     )    ( P-pI = P             P/
Q             + + p        orp  <  Pp-P   = Q      Q
QP' =Q'P                 Q/ - p-Qp =0.
From the first two equations we obtain
P (P- P')P     (Q-   Q')
P - --— P - P/,   (P- P) P/ -(Q        Q')
P                -   P/P
Substituting these values in the third, we have
(P- P) 'P -(Q- Q') Q= (P -P')P'Q-(Q-                      ) Q
or    (P -   P') (P    -    P/) + (Q-      Q)2 = 0.
Now if in the equation
x = p-P;
we put in the place of p, its value found above, we have
Q- Q
P     P"~
x = — p    -_p-.
195. In order to aid the learner, I shall indicate the operations
necessary for eliminating x in the two equations
x3+Px2+Qx+R=O, =            x3+P'x2     -Q   'x+ R' = 0.




Equations exceeding the First Degree.


215


In this case, we have
S'=O,     r'=O     (193),
and are furnished with these five equations;
P+p'      =P'     + P,
Q+Pp'+q'          = Q' +P' +q,
R + Q        + P q'-R'    +  Q p + P/ q,
Rp'+-Qq=R' pp+          Q'q,
Rq'    R'q,
which may take the following form,
P — 2~ P       -- P',
P'p-Pp+           q-q' = Q —Q,
Qp- Qp'+ P'q-Pq' R- R',
Rip-.Rp' +      Q'q -Qq'       0,
R'q -Rq' = 0.
We may, by the rules given in art. 88., obtain immediately from
any four of these equations, the values of the unknown quantities
p, p', q and q'; but the simple form, under which the first and the
last of the equations are presented, enables us to arrive at the
result, by a more expeditious method.   In order to abridge the
expressions, we make
P-   P     e,  Q       ' _Q e',  R-R' = e";
and proceed to deduce from the first and last of the proposed
equations,
p=p-_e, q'= R 
then substituting these values in the three others, and making the
denominator R to disappear, we have
(P —P) Rp+ (R          -    R')q =R(e' -P e)... (a),
(Q'-   Q) Rp + (RP'-PR') q = R (e" -         Qe)... (b),
(R' - R) Rp + (RQ'-       QR') q = -   R     e...... (c).
If now we obtain, from the equations (a) and (b), the values of p
and q (88), and suppress the factor R, which will be common to
the numerators and the denominator, we have
(e - P e) (RP'- PR') -(R- R') (e" - Q e)
P - (P - P) (RPI - PR) - (R -R) (Q/ -- Q)
(P' - P) (e" - Qe) R - R(e'- Pe) (Q'-       Q)
q     (P' - P) (RP'- PR')-(R-R') (Q - Q) 
putting these values in the equation (c), we obtain a final equation,
divisible by R, and which may be reduced to




216


Elements of Algebra.


(R'- R) [(e' - P e) (RP- P R') - (R - R') (e" - Qe)]
+ (RQ'-    QR') [(PI'- P) (e — Q e) - (e - P e) (Q' -   ()]
=-Re [(P'- P) (RP' -PR') - (R -          R) (Q/-    Q)];
it only remains then to substitute for the letters e, e', e", the quantities they represent.
196. If we have the three unknown quantities x, y, and z, and
are furnished with an equal number of equations, distinguished
by (1), (2), and (3); in order to determine these unknown quantities, we may combine, for example, the equation (1) with (2) and
with (3), to eliminate x, and then exterminate y from the two results, which are obtained. But it must be observed, that by this
successive elimination, the three proposed equations do not concur,
in the same manner, to form the final equation; the equation (1)
is employed twice, while (2) and (3) are employed only once;
hence the result, to which we arrive, contains a factor foreign to
the question (84). Bezout, in his Theorie des Equations, has
made use of a method, which is not subject to this inconvenience,
and by which he proves, that the degree of the final equation, resulting from the elimination among any number whatever of complete equations, containing an equal number of unknown quantities,
and quantities of any degrees whatever, is equal to the product of the
exponents, which denote the degree of these equations. M. Poisson,
has given a demonstration of the same proposition more direct
and shorter than that of Bezout; but the preliminary information,
which it requires, will not permit me to explain it here; it will be
found in the Supplement.  At present, I shall observe simply,
that it is easy to verify this proposition in the case of the final
equations presented in articles 194. and 195. If we suppose the
proposed equations given in those articles to be complete, the
unknown quantity y enters of the first degree into P and P', of
the second degree into Q and Q', of the third into R and R';
hence it follows, that e will be of the first degree, e' of the second,
and e" of the third, and that the terms of the highest degree found
in the products indicated in the final equation given in art. 194.,
will have 4, or 2. 2, for an exponent, and those of the final equation art. 195., will have 9 or 3. 3.




Commensurable Roots and equal Roots of Numerical Equations. 217


Of Commensurable Roots and the equal Roots of Numerical
Equations.
197. HAVING made known the most important properties of
algebraic equations, and explained the method of eliminating the
unknown quantities, when several occur, I shall proceed to the
numerical resolution of equations with only one unknown quantity,
that is, to the finding of their roots, when their coefficients are
expressed by numbers.*
I shall begin by showing, that when the proposed equation has
only whole numbers for its coefficients, and that of its first term is
unity, its real roots cannot be expressed by fractions, and consequently can be only whole numbers, or numbers, that are incommensurable.
In order to prove this, let there be the equation
Xn + Pxn1 +    Q X2....... +  Tx+ U=O,
in which we substitute for x an irreducible fraction; the equation then becomes, +,Pa-l +, an-2          a*    T      7    0
reducing all the terms to the same denominator, we have
an + P an-1 b +  Qan-2 b2......+ Ta ba-1 + Ubn       0,
which is equivalent to
an + b ( P an-1 + Q a-2 b.....+ T a bn-2 + U b- )         0.
The first member of this last equation consists of two entire parts,
one of which is divisible by b, and the other is not (98), since it is
a 
supposed, that the fraction b is reduced to its most simple form,
or that a and b have no common divisor; one of these parts cannot
therefore destroy the other.
198. After what has been said, we shall perceive the utility of
making the fractions of an equation to disappear, or of rendering
* There is no general solution for degrees higher than the fourth;
properly speaking, it is only that for the second degree, which can
be regarded as complete. The expressions for the roots of equa.
tions of the third and fourth degree are very complicated, subject to
exceptions, and less convenient in practice than those, which I am
about to give; I shall resume the subject in the Supplement.
ilg.              28




218


Elements of Algebra.


its coefficients entire numbers, in such a manner, however, that
the first term  may have only unity for its coefficient.  This is
done by making the unknown quantity proposed, equal to a new
unknown quantity divided by the product of all the denominators of
the equation, then reducing all the terms to the same denominator,
by the method given in art. 52.
Let there be, for example, the equation
ax2   bx   c
x+       + +;
m      n 'p
we take x = —Y, and introducing this expression for x into the
proposed equation, we obtain'
y3        a y       by     c
m3n3 p3   m3 n2 p2   m n'2 p  p O
as the divisor of the first term contains all the factors found in
the other divisors, we may multiply by this divisor and thus reduce each term to its most simple expression; we find then
y3 + a npy2     b m2 np2y + cm3 n3p     = 0.
If the denominators, m, n, p, have common divisors, it is only
necessary to divide y by the least number, which can be divided at
the same time by all the denominators.  As these methods of
simplifying expressions will be readily perceived, I shall not stop to
explain them; I shall observe only, that if all the denominators
were equal to m, it would be sufficient to make x = Y.
m
The proposed equation, which would be in this case,
ax2    bx    c
-~ +y+ -=0,
m     m    m
then becomes
y3    a y4 b0 c,
m3 -+  3 + m2 m 
and we have
y3 + a y2 + b m y + m2 c = 0.
It is evident, that the above operation amounts to multiplying
all the roots of the proposed equations by the number m, since
x =     M gives y = mx.
199. Now since, if a be the root of the equation
n + Pxn —+ Qxn-2.... + Tx + U = 0,
we have
U=-          P  n-' - an-s..... -     Ta (179),




Commensurable Roots and equal Roots of Numerical Equations. 219
it follows, that a is necessarily one of the divisors of the entire
number U, and consequently, when this number has but few divisors, we have only to substitute them successively in the place of
x, in the proposed equation, in order to determine, whether or not
this equation has any root among whole numbers.
If we have, for example, the equation
X3-   6 xs - 27x. -  38 = 0,
as the numbers
1, 2,   19, 3S,
are the only divisors of the number 39, we make trial of these,
both in their positive and negative state; and we find, that the
whole number + 2 only satisfies the proposed equation, or that
x = 2. We then divide the proposed equation by x -  2; putting the quotient equal to zero, we form the equation
x -   4x + 19 = 0,
the roots of which are imaginary; and resolving this, we find that
the proposed equation has three roots,
x- 2, x =2 + v-5, x             2 -        5.
200. The method just explained, for finding the entire number,
which satisfies an equation, becomes impracticable, when the last
term of this equation has a great number of divisors; but the
equation
U = -   (t1 -  P (a'1 -   a"1-....  -  Ta,
furnishes new conditions, by means of which the operation may be
very much abridged. In order to make the process more plain, I
shall take, as an example, the equation
4 + p x3 + Qx + Rx + S=0.
The root being constantly represented by a, we have
a4 + P a3 +    a2 - +  a + S =0,
S =- R a -       a2    P a3 - a4,
from which we obtain
R-     - Qa - Pa2 - a3.
a
It is evident from this last equation, that - must be a whole number.
a
Bringing R into the first member, we have
+R        -      a - a-Pa -a3;




220                 Elements of Algebra.
abridging the expression by making a + R = R', and dividing the
two members of the equation
R — Q-     a - P - a3
by a, we have
=         - P a -    a2,
whence we conclude, that    must also be a whole number.
a
RI
Transposing Q, and making -  + Q- Q=, then dividing the
two members by a, we obtain
Q,
P    a,
a
whence we infer, that -  must be a whole number.
a
Lastly, bringing P into the first member, making  + P = PF',
and dividing by a, we have
^   =-1.
a
Putting together the above mentioned conditions, we shall perceive that the number a will be the root of the proposed equation,
if it satisfy the equations
S +R = R'
a
RI +
7 + 2& Q
_ + P      P
a
p/
-- +  1 = 0,
in such a manner, as to make R', Q', and P' whole numbers.
Hence it follows, that in order to determine, whether one of the
divisors a of the last term S can be a root of the proposed equation, we must,
1st. Divide the last term by the divisor a, and add to the quotient
the coefficient of the term involving x;
2d. Divide this sum by the divisor a, and add to the quotient the
coefficient of the term involving x2;




Commensurable Roots and equal Roots of NVumerical Equations. 221


3d. Divide this sum by the divisor a, and add to the quotient the
coefficient of the term involving x3;
4th. Divide this sum by the divisor a, and add to the quotient
unity, or the coefficient of the term involving x4; the result will
become equal to zero, if a is, in fact, the root.
The rules given above are applicable, whatever be the degree
of the equation; it must be observed, however, that the result will
not become equal to zero, until we arrive at the first term of the
proposed equation.*
201. In applying these rules to a numerical example, we may
conduct the operation in such a manner as to introduce the several
trials with all the divisors of the last term, at the same time.
For the equation
X4 -  9 3 + 23 x2 -  20 x + 15    0,
the operation is, as follows;
+ 15,+    5, +   3, +   1,-   1,-    3,-   5,-15,
+   1, +  3, +   5,+15,      15,- -  5,-   3,-    1,
- 19, - 17,    1-    5,- 5,- 35, - 25, - 23, - 21,
-   5-     5 5, + 35,
+ 18, +   18, + 58,
+   6, +  18,- 58,
-   3, +   9,- 67,
-   1, +   9, + 67,
0.
All the divisors of the last term 15 are arranged, in the order of
magnitude, both with the sign +  and -, and placed in the same
line; this is the line occupied by the divisors a.
The second line contains the quotients arising from the number
15, divided successively by all its divisors; this is the line for the..  S
quantities -.
a
* It would not be difficult to prove by means of the formula for the
quotients given in art. 180., that the quantities -, —, taken with
a  a   a
the contrary sign, and with the order inverted, are the coefficients of
the quotient arising from the polynomial
4 + P3 + Qa 2+Rx +S
divided by z - a, and which is, consequently,
a       a R'  S
a;3 -_    x2 -;   -- 
a      a      a




222


Elements of Algebra.


The third line is formed by adding to the numbers found in the
preceding the coefficient - 20, by which x is multiplied; this is
the line for the quantities R' = - + R.
The fourth line contains the quotients of the several numbers in
the preceding, divided by the corresponding divisors; this is the
line for the quantities -. In forming this line, we neglect all the
numbers, which are not entire.
The fifth line results from the numbers, written in the preceding,
added to the number 23, by which x2 is multiplied; this line contains the quantities Q'.
The sixth line contains the quotients arising from the numbers in
the preceding, divided by the corresponding divisors; it comprehends the quantities -.
a
The seventh line comprehends the several sums of the numbers
in the preceding, added to the coefficient - 9, by which X3 is
Q,
multiplied; in this line are found the quantities - + P.
a
Lastly, the eighth line is formed, by dividing the several numbers
in the preceding by the corresponding divisors; it is the line for. As we find -  1 only in the column, at the head of which +3
a
stands, we conclude, that the proposed equation has only one commensurable root, namely, +3; it is, therefore, divisible by x - 3.*
The divisors + 1 and -  I may be omitted in the table, as it is
easier to make trial of them, by substituting them immediately in
-the proposed equation.
202. Again, let there be, for example,
x3 - 7 x2 + 36 = 0.
Having ascertained, that the numbers + 1 and -  1 do not
satisfy this equation, we form the table subjoined, according to the
preceding rules, observing that, as the term involving x is wanting
in this equation, x must be regarded as having 0 for a coefficient;
we must, therefore, suppress the third line, and deduce the fourth
immediately from the second.
* Forming the quotient according to the preceding note, we find
Z3 _ 6 x-2  5 z - 5




Commensurable Roots and EqualRoots of JNumerical Equations. 223
+36,+18,+12,+9,+6,+4,+ 3,+ 2,- 2,- 3,-4,-G,-9, —12, -18,-36
+ 1,+ 2,-+- 3,+-4,-;,9,+1o,+18, — 18,-12,-9, —6,-4, — 3,- 2,- 1
+ L, + 4, + 9, + 9, + 4, + 1,
-6,     3, + 2, +2, — 3,-6,
-1,-1,         -1, + 1,+ 1,
0,   0,        0.
We find in this example three numbers, which fulfil all the
conditions, namely, + 6, + 3, and -  2. Thus we obtain, at the
same time, the three roots, which the proposed equation admits
of; we conclude then, that it is the product of three simple factors,
x - 6, x - 3, and x + 2.
203. It may be observed, that there are literal equations, which
may be transformed, at once, into numerical ones.
If we have, for example,
y3 + 2py2      33 p2y + 14p3= 0,
making y = p x, we obtain
p 3 + 2p3 x2 - 33p3x + 14p3 = 0,
a result, which is divisible by p3, and may be reduced to
3+ 2x2 -33x+ 14 = 0.
As the commensurable divisor of this last equation is x + 7,
which gives x = - 7, we have
y   -  7p.
The equation involving y is among those which are called
homogeneous equations, because taken independently of the numerical coefficients, the several terms contain the same number of
factors.*
204. When we have determined one of the roots of an equation,
we may take for an unknown quantity the difference between this
root and any one of the others; by this means we arrive at an
equation of a degree inferior to that of the equation proposed, and
which presents several remarkable properties.
Let there be the general equation
Xm +Px m-1+ Q xm-2 + R     m-3... + Tx+ U           =o,
and let a, b, c, d, &c. be its roots; substituting a + y in the place
of x, and developing the powers, we have
* For a more full account of the commensurable divisors of equations, the reader is referred to the third part of the Elemens d'Algebre of Clairaut. This geometer has treated of literal as well as
numerical equations.




224


Elemnents of.lgebra.


a"   +mar-ly         -+   Mn-   1)    ar-y+... +y
2
+P am-l+(m-1)P am-2y + (m-)(m-)pam-3y2+.
+ Q am-2+(m-2) Q am-y+((m2)(          3) Qamy2+...        =0.
I~f`  f2  &am y2~t=0.
+R am-3+(m-3)R am-4y+(m-)(m-4)Ram —y2+...
+Ta + Ty
+U
The first column of this result, being similar to the proposed
equation, vanishes of itself, since a is one of the roots of this
equation; we may, therefore, suppress this column, and divide all
the remaining terms by y; the equation then becomes
m a     +    m(am-1)      am-y... + ym-'
-mare-]          '-2-   a
+ (m -   1) P am-2 + (m-l)(m-2) p     m-y
+ (r -2) Q a-3 + (m-2) (m-3) Q ar_4        + 
2       Qam)  y +... '0.
+ (m - 3) R a"-4 + (m-3) (m-4) R am-5y +
2
+T T
This equation has evidently for its m- 1 roots
y =b-a, y -c- a, y = d-a.....&c.
I shall represent it by
B y +             C y2....  +l y-.......(d),
abridging the expressions, by making
mam-l     -     (m - 1)Pam-2+  (m- 2) Q am3.....      T  2,
m (m -  1) a-2 + (m -    1) (m-  2) P am-3.....        =,
&c.,
and I shall designate by Vthle expression
am- + Pam-1 + q a-2....... + Ta + U.
205. If the proposed equation has two equal roots; if we have,
for example, a = b, one of the values of y, namely, b - a, becomes nothing; the equation (d) will therefore be verified, by
supposing y = 0; but upon this supposition all the terms vanish,
except the known term.l; this last must, therefore, be nothing of




Commensurable Roots and equal Roots of Numerical Equations. 225


itself; the value of a must, therefore, satisfy, at the same time, the
two equations
V —     and.J = 0.
When the proposed equation has three roots equal to a, namely,
a = b = c, two of the roots of the equation (d) become nothing,
at the same time, namely, b - a and c - a. In this case the
equation (d) will be divisible twice successively by y-0 (179)
or y; but this can' happen, only when the coefficients 3. and B
are nothing; the value of a must then satisfy, at the same time, the
three equations
V==0,. =0, B =0.
Pursuing the same reasoning, we shall perceive, that when the
proposed equation has four equal roots, the equation (d) will have
three roots equal to zero, or will be divisible three times successively
by y; the coefficients,.1, B, and C, must then be nothing, at the
same time, and consequently the value of a must satisfy at once the
four equations,
V-=, 1=0, BO         =0,   C=0.
By means of what has been said, we shall not only be able to
ascertain, whether a given root is found several times among the
roots of the proposed equation, but may deduce a method of determining, whether this equation has roots repeated, of which we
are ignorant.
For this purpose, it may be observed, that when we have.- = 0, or
7n am-i + (m - 1) P am-2 + (m -2)    a-3...+ T = 0,
we may consider a as the root of the equation
mx-l + (n      1) Pxm-2+ (m-_2) Q        m-.-3.  + T  O,
x representing, in this case, any unknown quantity whatever; and
since a is also the root of the equation V  0, or
xm + P xm-l + &c. = 0,
it follows (189), that x - a is a factor common to the two above
mentioned equations.
Changing in the same manner a into x in the quantities, B, C,
&c. the binomial x - a becomes likewise a factor of the two
new equations, B = 0, C = 0, &c. if the root a reduces to nothing
the original quantities, B, C, &c.
What has been said with respect to the root a, may be applied to
every other root, which is several times repeated; thus, by seeking,.lg.               29




226


Elements of Jllgebra.


according to the method given for finding the greatest common
divisor, the factors common to the equations,
F=0, a =0, B =0,          C= 0,&c.,
we shall be furnished with the equal roots of the proposed equation,
in the following order;
The factors common to the first two equations only, are twice
factors in the equation proposed; that is, if we find for a common
divisor of V= 0 and J = 0, an expression of the form (x = a)
(x - 6), for example, the unknown quantity x will have two values
equal to c, and two equal to 6, or the proposed equation will have
these four factors,
(x- a), (x -    ), (x - 6), (x-6).
The factors common, at the same time, to the first three of the
above mentioned equations form triple factors in the proposed
equation; that is, if the former are presented under the form
(x - a) (x - 6), the latter will take the form, (x -  )3 (a- 6)3,
This reasoning may easily be extended to any length we please.
206. It may be remarked, that the equation.J = 0, which, by
changing a into x, becomes
m xa-1 + (m - 1) P xn1-2 + (m - 2) Q   -3.. + T = 0
is deduced immediately fiom the equation V= 0, or from the
proposed equation,
xm+p      lX —1 + Q2-....+ Tx + U= O,
by multiplying each term of this last by the exponent of the power
of x, which it contains, and then (diminishing this exponent by
unity. We may remark here, that the term U, which is equivalent to UX x~, is reduced to nothing in this operation, where it
is multiplied by 0. The equation B = 0 is obtained from. = 0,
in the same manner as / = 0 is deduced fiom V = 0; C = 0
is obtained from B = 0, in the same manner as this from. = 0,
and so on.*
* It is shown, though very imperfectly, in most elementary treatises, that the divisor common to the two equations V-= 0 and A= 0
contains equal factors raised to a power less by unity than that of
the equation proposed; this may be readily inferred from what precedes; but for a demonstration of this proposition we refer the reader
to the Supplement, where it is proved in a manner, which appears to
me to be simple and new.




Commensurable Roots and equal Roots of Numerical Equations. 227
207. To illustrate what has been said, by an example, I shall
take the equation
X5-   13 4 +- 67 x3 -  171,' + 216 x -  108 = 0;
the equation. = 0 becomes, in this case,
5 X4-  52x3 + 201 2 -    342 x + 216 = 0;
the divisor common to this and the proposed equation is
X3 -  8x2 + 21 x-    18.
As this divisor is of the third degree, it must itself contain several
factors; we must therefore seek, whether it does not contain some
that are common to the equation B  = 0, which is here
20 x3 -  156.2 + 402 x -  342 - 0.
We find, in fact, for a result x - 3; the proposed equation then
has three roots equal to 3, or admits of (x -  3)3 among the number of its factors. Dividing the first common divisor by x - 3, as
many times as possible, that is, in this case twice we obtain x -2.
As this divisor is common only to the proposed equation, and to the
equation A = 0, it can enter only twice into the proposed equation. It is evident then, that this equation is equivalent to
( -   3)3 (x -  2)2 = 0.
208. As the equation (d) gives the difference between b and the
several other roots, when b is substituted for a, the difference
between c and the others, when c is substituted for a, &c. and
undergoes no change in its form by these several substitutions,
retaining the coefficients belonging to the equation proposed, it may
be converted into a general equation, which shall give all the
differences between the several roots combined two and two. For
this purpose, it is only necessary to eliminate a by means of the
equation
an + p am-1 +  Q a,-2.      +.... +  Ta +  U= 0;
for the result being expressed simply by the coefficients, and exhibiting the root under consideration in no form whatever, answers
alike to all the roots.
It is evident, that the final equation must be raised to the degree
m (m -  1); for its roots
a -b, a -c, a-d, &c.
b-a, b -c, b-d, &c.
c-a,     c   b, c -d, &c.




228


Elements of Algebra.


are equal in number to the number of arrangements, which the m
letters, a, b, c, &c. admit of when taken two and two. Moreover,
since the quantities
a- b and b - a, a- c and c - a, b - c and c - b, &c.
differ only in the sign, the roots of the equation are equal, when
taken two and two, independently of the signs; so that if we have
y = a, we shall have, at the same time, y = -. Hence it follows, that this equation must be made up of terms involving only
even powers of the unknown quantity; for its first member must
be the product of a certain number of factors of the second degree
of the form
-- 2 = ( —a) (y +       )   (184);
it will, therefore, itself be exhibited under the form
y2n + p y 2n-2 - _  y2n4.... + t y2 +-  = 0.
If we put y2 = z, this becomes
z"- pzn     +- q z.....+ tz + U =         O;
and as the unknown quantity z is the square of y, its values will be
the squares of the differences between the roots of the proposed
equation.
It may be observed, that as the differences between the real roots
of the proposed equation are necessarily real, their squares will be
positive, and consequently the equation in z will have only positive
roots, if the proposed equation admits of those only which are real.
Let there be, for example, the equation
X3 -  7 x + 7 — Q;
putting x = a + y, we have
a3 + 3 a.  y + 3 ay2 + y3)
-7    -- 7                        0.
+7
Suppressing the terms a3 -  7 a + 7, which, from their identity
with the proposed equation, become nothing when united, and
dividing the remainder by y, we have
3 a2+ 3a y+y2 -      7   0 O;
eliminating a by means of this equation and the equation
a3 -  7a + 7 =0,
we have
y6__ 42 y4 + 441 y2 -        49;




Commensurable Roots and equal Roots of Numerical Equations. 229
putting z  y2, this becomes
z3- 42z2 + 441 z — 49 = 0.
209. The substitution of a + y in the place of x in the equation
x + p xm-1 + Q      -2..... +  U= 0 (204),
is sometimes resorted to also in order to make one of the terms of
this equation to disappear. We then arrange the result with reference to the powers of y, which takes the place of the unknown
quantity x, and consider a as a second unknown quantity, which is
determined by putting equal to zero the coefficient of the term we
wish to cancel; in this way we obtain
ym + ma ym~l     m(m — 1 ) a2.....+ am
+-  Pym-   + (m-    1)Paym-2..       + Pam"-1.
+            Qym-2.       + Qam_-2
+Ut u
If the term we would suppress be the second, or that which
involves y"-l, we make m a + P = 0, from which we deduce
P
a = --. Substituting this value in the result, there remain only
the terms involving
ym,  y-,    ym~3,  &c.
Hence it follows, that we make the second term of an equation to
disappear, by substitutingfor the unknown quantity in this equation
a new unknown quantity, united with the coefficient of the second
term taken with the sign contrary to that originally belonging to it,
and divided by the exponent of the first term.
Let there be, for example, the equation
x3 + 6 x -  3x + 4 = 0;
we have by the rule
x =y-     - = y-   2;
substituting this value, the equation becomes
y3 -   y2 + 12y_-    8
+ 6y2-    24y + 24      0
-   3y+ 6 
+ 4J
which may be reduced to
y3 -15y + 26 = 0,
in which the term involving y2 does not appear. We may cause




230


Elements of Algebra.


the third term, or that involving ym-2, to disappear by putting equal
to zero the sum of the quantities, by which it is multiplied, that is,
by forming the equation
m( —1) a       (m     ) Pa +     =0.
Pursuing this method, we shall readily perceive, that the fourth
term will be made to vanish by means of an equation of the third
degree, and so on to the last, which can be made to disappear
only by means of the equation
am + P am-l +   Q am-2..... +   U= 0,
perfectly similar to the equation proposed.
It is not difficult to discover the reason of this similarity. By
making the last term of the equation in y equal to zero, we suppose, that one of the values of this unknown quantity is zero; and
if we admit this supposition with respect to the equation x = y + a,
it follows that x = a; that is, the quantity a, in this case, is necessarily one of the values of x.
210. We have sometimes occasion to resolve equations into
factors of the second and higher degrees. I cannot here explain
in detail the several processes, which may be employed for this
purpose; one example only will be given.
Let there be the equation
x3 - 243      1       1 x + 12  x  7 = 0,
in which it is required to determine the factors of the third degree; I shall represent one of these factors by
x3 + p 2 + q x + r,
the coefficients p, q, and r, being indeterminate. They must be
such, that the first member of the proposed equation will be exactly divisible by the factor
a3 + p x2 + q x +- r,
independently of any particular value of x; but in making an actual
division, we meet with a remainder
_  (p3 -_  p q _ 24 p + r -  12)x2
-  (p2 q -p r -  q -   24 q + 11) x
-  (p2r r-     - 24 r -  7),
an expression, which must be reduced to nothing, independently of
x, when we substitute for the letters, p, q, and r, the values that
answer to the conditions of the question. We have then




Resolution of JNumerical Equations by lpproximation. 231


p3 - 2pq-24         - + r — 12 -  0,
P2 q -     r-  q2-  24q + 11 = 0,
p2r -  qr - 24 r -   7 =0.
These three equations furnish us with the means of determining
the unknown quantities, p, q, and r; and it is to a resolution of
these, that the proposed question is reduced.
Of the Resolution of Numerical Equations by 3lpproximation.
211. HAVING completed the investigation of commensurable
divisors, we must have recourse to the methods of finding roots by
approximation, which depend on the following principle;
When we arrive at two quantittics, which, substituted in the place
of the unknown quantity in an elquation, lead to two results with
contrary signs, we may infer, that one of the roots of the proposed
equation lies between these two quantities, and is consequently real.
Let there be, for example, the equation
x-3 13x2+ 7x -       =0;
if we substitute, successively, 2 and 20 in the place of x, in the
first member, instead of being reduced to zero, this member
becomes, in tile former case, equal to -  31, and in the latter, to
- 2939; we may therefore conclude, that this equation has a
real root between 2 and 20, that is, greater than 2 and less
than 20.
As there will be fiequent occasion to express this relation, I
shall employ the signs > and <, wlich algebraists have adopted
to denote the inequality of two nagnlittudes, placing the greater of
two quantities opposite the opening of the lines, and the less against
the point of meeting. Thus I shall write
x > 2, to denote, that x is greater than 2,
x < 20, to denote, that x is less than 20.
Now in order to prove what has been laid down above, we may
reason in the following manner. Bringing together the positive
terms of the proposed equation, and also those which are negative,
we have
X3 + 7 x -  (13 x2 + 1),
a quantity, which will be negative, if we suppose x = 2, because,
upon this supposition,
X3 + 7x < 13 x2 + 1,




232


Elements of algebra.


and which becomes positive, when we make x = 20, because,
in this case,
x3 + 7x>    13x2 + 1.
Moreover, it is evident, that the quantities
x3+7x     and   13x2+1,
each increase, as greater and greater values are assigned to x, and
that, by taking values, which approach each other very nearly, we
may make the increments of the proposed quantities as small as
we please. But since the first of the above quantities, which was
originally less than the second, becomes greater, it is evident, that
it increases more rapidly than the other, in consequence of which
its deficiency is made up, and it comes at length to exceed the
other; there must, therefore, be a point at which the two magnitudes are equal.
The value of x, whatever it be, which renders
x3+ 7 x = 13 x2 + 1,
and such a value has been proved to exist, gives
3 + 7x - (13X2 + 1) = 0,
or              X3 - 13 x2 + 7 x - 1 = 0,
and must necessarily, therefore, be the root of the equation proposed.
What has been shown with respect to the particular equation
X3-_ 13 x2 + 7 x    l 1= 0,
may be affirmed of any equation whatever, the positive terms of
which I shall designate by P, and the negative by N. Let a be
the value of x, which leads to a negative result, and b that which
leads to a positive one; these consequences can take place only
upon the supposition, that by substituting the first value, we have
P < JV, and by substituting the second, P > N; P, therefore,
from being less, having become greater than AN, we conclude as
above, that there exists a value of x, between a and b, which gives
P  _ N.*
* The above reasoning, though it may be regarded as sufficiently
evident, when considered in a general view, has been developed by
M. Encontre in a manner, that will be found to be useful to those,
who may wish to see the proofs given more in detail.
1. It is evident, that the increments of the polynomials P and N




Resolution of Numerical Equations by 1tpproximation.  233
The statement here given seems to require, that the values assigned to x should be both positive or both negative, for if they
may be made as small as we please. Let
P =c ax   +        Gxn......+  x,
m being the highest exponent of x; if we put a +- y in the place ofx,
this polynomial takes the form
A+ By+ Cy2.....+          T ym,
the coefficients, A, B, C,.... T, being finite in number and having
a finite value; the first term A will be the value the polynomial P
assumes, when x - a; the remainder,
By + Cy.... +  Ty   - y (B + Cy.... Tym-1),
will be the quantity, by which the same polynomial is increased when
we augment by y the value x - a. This being admitted, if S designate the greatest of the coefficients, B, C,.... T, we have
B + Cy.... +    Ty"n-1 <  S(1 + y.... + y"-1);
now
+1  y *....  ym-l=1 — -        (158);
y —
therefore,
(1 — ym)
y (B + Cy....      Tym-)< Sy      ( 
- y
and, consequently, the quantity by which the polynomial P is increassy(1 - y")
ed, will be less than any given quantity m, if we make -y (-  y)
S -y
less than this last quantity; this is effected by making 1 —Y   m
because, in this case, = yS-  being < 1, the quantity  ( y  )
equal to               ill necessarily be ess than the quantity m
equal to      - - -, will necessarily be less than the quantity m,
1-y       - y
which is indefinitely small.
2. If we designate by h the increment of the polynomial P, and by
k that of the polynomial N, the change, which will be produced in
the value of their diflerence, will be h - k, and may be rendered
smaller than a given quantity, by making smaller than this same
quantity the increment, which is the greater of the two; we may,
therefore, in the interval between  - a and x  b, take values,
which shall make the difference of the polynomials P and N change
by quantities as small as we please, and since this difference passes in
this interval from positive to negative, it may be made to approach as
near to zero as we choose. See Annales dc Mathemnatiques pures et
appliquees, published by M. Gergonne, vol. iv. p. 210,
g       /.        30




234


Elements of Algebra.


have different signs, that which is negative produces a change in
the signs of those terms of the proposed equation, which contain
odd powers of the unknown quantity, and, consequently, the expressions P and JV are not formed in the same manner, when we
substitute one value, as when we substitute the other. This difficulty vanishes if we make x = O; in this case, the proposed equation reduces itself to its last term, which has necessarily a sign
contrary to that of the result arising from the substitution of one or
the other of the above mentioned values. Let there be, for example, the equation
4 -2x 3 -      3C2 -- 15  -  3 = 0,
the first member of which, when we put
x =- 1     and  x = 2,
becomes +  12 and -  45. If we suppose x = 0, this member is
reduced to - 3; substituting, therefore,
x-= 0   and  x =-    1,
we arrive at two results with contrary signs; but putting - y in
the place of x, the proposed equation is changed to
y4 +. 2y3   3 y2 + 15 y    3 = 0,
and we have
P    y4 + 2 y3+ 15 y, JV='    3 S + 3,
whence
P < NV, when y = 0,
P > JN, when y =      1.
Reasoning as before, we may conclude, that the equation in y has
a real root, found between 0 and +- 1; whence it follows, that the
root of the equation in x lies between 0 and - 1, and, consequently, between + 2 and -  1.
As every case the proposition enunciated can present, may be
reduced to one or the other of those which have been examined,
the truth of this proposition is sufficiently established.
212. Before proceeding further, I shall observe, that whatever
be the degree of an equation, and whatever its coefficients, we may
always assign a number, which, substituted for the unknown quantity, will render the first term greater than the sum of all the others.
The truth of this proposition will be immediately apparent from
what has been intimated of the rapidity, with which the several
powers of a number greater than unity increase (126); since the
highest of these powers exceeds those below it more and more in




Resolution of Numerical Equations by.pproximation. 235
proportion to the increased magnitude of the number employed, so
that there is no limit to the excess of the first above each of the
others. Observe, moreover, the method by which we may find a
number that fulfils the condition required by the enunciation.
It is evident, that the case most unfavorable to the supposition
is that, in which we make all the coefficients of the equation negative, and each equal to the greatest, that is, when, instead of
xm + pX l + Q aX-2.... + Tx + U =O,
we take
xM -    s   - S -I.... SX-  x -  S  0,
S representing the greatest of the coefficients, P, Q,..... T, U.
Giving to the first member of this equation the form
x' -  S (Xm-1 +  M-2.....+   1),
we may observe, that
xam-1 +  m-2.. + _        -1   (158);
the preceding expression then may be changed into
S(x     l)              Sxm       S
-- m   -—., or into x' —       + ~
X - 1           xz-1     x-1
If we substitute.M for x, this becomes
IM SM      S
a quantity, which evidently becomes positive, if we make
Now if we divide each member of this equation by im", we have
1=       1 or       S +  1.
By substituting therefore for x the greatest of the coefficients
found in the equation, augmented by unity, we render the first term
greater than the sum of all the others.
A smaller number may be taken for.M, if we wish simply to
render the positive part of the equation greater than the negative;  *
for to do this, it is only necessary to render the first term greater
than the sum arising from all the others, when their coefficients are
each equal, not to the greatest among all the coefficients but to the




236


Elements of Algebra.


greatest of those which are negative; we have, therefore, merely to
take for Jl this coefficient augmented by unity.*
Hence it follows, that the positive roots of the proposed equation are necessarily comprehended within 0 and S +  1.
In the same way we may discover a linit to the negative roots;
for this purpose we tnust substitute - y for x, in the proposed
equation, and render the first term positive, if it becomes negative
(178). It is evident, that by a transformation of this kind, the
positive values of y answer to the negative values of x, and the
reverse. If R be the greatest negative coefficient after this change,
R +   1 will form a limit to the positive values of y; consequently
-       R-  1 will form that of the negative values of x.
Lastly, if we would find for the smallest of the roots a limit
approaching as near to zero as possible, we may arrive at it by
substituting - for x in the proposed equation, and preparing the
equation in y, which is thus obtained, according to the directions
given in art. 178. As the values of y are the reverse of those of
IA, the greatest of the first will correspond to the least of the second,
and, reciprocally, the greatest of the second to the least of the first.
If, therefore, S' +  1 represent the highest limit to the values of y,
that is, if
y< S+      1,
which gives
< S'+      l,
we shall have, successively,
1     <   (S' +  ) x,   < x.
Indeed, it is very evident, that we may, without altering the
relative magnitude of two quantities separated by the sign < or
>, multiply or divide them by the same quantity, and that we
may also add the same quantity to or subtract it from each side of
the signs < and >, which possess, in this respect, the same properties as the sign of equality.
* In the Resolution des equations numeriques, by Lagrange, there
are formulas, which reduce this number to narrower limits, but what
has been said above is sufficient to render the fundamental propositions for the resolution of numerical equations independent of the
consideration of infinity.




Resolution of NJumerical Equations by Approximation. 237


213. It follows from what precedes, that every equation of a
degree denoted by an odd number has necessarily a real root affected
with a sign contrary to that of its last term; for if we take the
number JI such, that tie sign of the quantity
Jlm +- PJIM'I-1 +  QJ7Mt-2.-..... +  TM'~ U,
depends solely on that of its first term Jl"1, the exponent m being
an odd number, the term JMA" will have the same sign as the number MJ (128). This being admitted, if the last term U has the
sign  -, and we make x=     -.1, we shall arrive at a result
affected with a sign contrary to that, which the supposition of
x =0 would give; from which it is evident, that tlie proposed
equation has a root between 0 and -  11, that is, a negative root.
If the last term  U has the sign -, we make x = — - JM; the
result will then have a sign contrary to that given by the supposition of x = 0, and in this case, the root will be found between 0
and +- J7, that is, it will be positive.
214. When the proposed equation is of a degree denoted by
an even number, as the first term Ailm remains positive, whatever
sign we give to.J, we are not, by the preceding observations,
furnished with the means of proving the existence of a real root,
if the last term has the sign +, since, whether we make x = 0,
or x =   -~ -M, we have always a positive result. But when this
term is negative, we find, by making
x = + M, x=O, x= —M,
three results, affected respectively with the signs  -, -, and —,
and, consequently, the proposed equation has, at least, two real
roots in this case, the one positive, found between JM and 0, the
other negative, between 0 and - Ml; therefore, every equation of
an even degree, the last term of which is negative, has at least two
real roots, the one positive and the other negative.
215. I now proceed to the resolution of equations by approximation; and in order to render what is to be offered on this
subject more clear, 1 shall begin with an example.
Let there be the equation
X4- 4 x3- 3 x+ 27      0;
the greatest negative coefficient found in this equation being - 4,
it follows (212), that the greatest positive root will be less than
5. Substituting - y for x, we have
y4 + 4y3 + 3 y + 27 = 0;




238.Elements of dlgebra.


and as all the terms of this result are positive, it appears, that
y must be negative; whence it follows, that x is necessarily positive, and that the proposed equation can have no negative roots;
its real roots are, therefore, found between 0 and + 5.
The first method, which presents itself for reducing the limits,
between which the roots are to be sought, is to suppose successively
x = 1, x- = 2, x- = 3, x- 4
and if two of these numbers, substituted in the proposed equation,
lead to results with contrary signs, they will form new limits to
the roots. Now if we make
x -- 1, the first member of the equation becomes + 21,
x    2..+....................        5,
x —.......................-               9,
x =  4......................      15;
it is evident, therefore, that this equation has two real roots, the
one found between 2 and 3, and the other between 3 and 4. To
approximate the first still nearer, we take the number 2,5, which
occupies the middle place between 2 and 3 (3rith. 129), the
present limits of this root; making then x = 2,5, we arrive at the
result
+ 39,0625 - 62,5 - 7,5 + 27 =- 3,9375;
as this result is negative, it is evident, that the root sought is between 2 and 2,5. The mean of these two numbers is 2,25; taking
x = 2,3, we have the root sought within about one tenth of its value,
and shall approximate the true root very fast by the following
process, given by Newton.
We make x = 2,3 + y; it is evident, that the unknown quantity y amounts only to a very small fraction, the square and higher
powers of which may be neglected; we have then
=X4       (2,3)4 +  4 (2,3)3y
- 4 3 - - 4 (2,3)3 - 12 (2,3)2 y
-3x    =- 3(2,3) -     3y;
substituting these values, the proposed equation becomes
-0,5839-     17,812 y = 0,
which gives
0,5839
Y   - ~-17,812




Resolution of NV'umerical Equations by.pproximation. 239
Stopping at hundredths, we obtain for the result of the first operation
y = - 0,03 and x = 2,3 - 0,03 = 2,27.
To obtain a new value of x, more exact than the preceding, we
suppose x = 2,27 + y'; substituting this value in tle proposed
equation and neglecting all the powers of y' exceeding the first, we
find
-0,04595359 -    18,046468 y' = 0,
whence
0,04595359         0
18,046468 =-0,0025,
and, consequently, x = 2,2675. We may, by pursuing this process, approximate, as nearly as we please, the true value of x.
If we seek the second root, contained between 3 and 4, by the
same method, we find, stopping at the fourth decimal place,
x   3,6797.
216. We may ascertain the exactness of the method above
explained, by seeking the limit to the values of the terms, which
are neglected.
If the proposed equation were
m +P +      + Q   m-2...... + Tx + U= 0,
substituting a - y for x, we should have for the result the first
of the equations found in art. 204., because a being not the root of
the equation, but only an approximate value of x, cannot reduce to
nothing the quantity
am+ Pa-l+      Qam-2....... +      Ta+ U.
Representing this last by V, we have, instead of the equation (d)
above referred to, the following
A      B   y2     C    y3         +y=;
v+y+          y2+-I.3Y3 +.....        ym=
from which we obtain
B          C...
y=-     V —.2 y. 2.3 y......   y
V     B y2      C y3                  ym
Y-     -A    1.2A     1.2.3A    *A t.'
Neglecting the powers of y exceeding the first, we have
V
Y=- A,




240


Elements of Algebra.


and this value differs from the real value of y by
By2         Cy3               ym
1. 2A.2.3A.*            A
If a differs from the true value of x only by a quantity less
than - a, the above mentioned error becomes less than that,
which would arise from putting - a in the place of y, which would
P
give
B                Ca\2   \3 a           /am
Ap            1.2.3 A \p). 2. A p 
Finding the value of this quantity, we shall be able to determine,
whether it may be neglected when considered with reference to, and if it be found too large, we must obtain for a a number,
which approaches nearer to the true value of x.
To conclude, when we have gone through the calculation with
several numbers, y, y', y", &c. if the results thus obtained form a
decreasing series, an approximation is certain.
217. The method we have employed above, is called the.Method by successive Substitutions.  Lagrange has considerably
improved it.*  He has remarked, that by substituting only entire
numbers, we may pass over several roots without perceiving them.
In fact, if we have, for example, the equation
(x -  ) (x-   ) (x-3) (x - 4) =0,
by substituting for x the numbers, 0, 1, 2, 3, &c. we shall pass
over the roots |- and I, without discovering that they exist; for we
shall have
(0 -) (0-     ) (0-3) (0 -4)       +r  X    X 3 X 4,
(1- - ) (1-   ) ( 1-3) ( -4)        - +  X 1 X 2 X 3,
results affected by the same sign. It will be readily perceived,
that this circumstance takes place in consequence of the fact,
that the substitution of I for x changes at the same time the
signs of both the factors, x -  2, and x -,, which pass fiom the
negative state, in which they are when 0 is put in the place of
x, to the positive; but if we substitute for x a number between
~ and ~, the sign of the factor x -  alone will be changed, and
we shall obtain a negative result.
" See Resolution des EyLuations Nunteriques.




Resolution of Numerical Equations by ipproximation. 241
We shall necessarily meet with such a number, if we substitute,
in the place of x, numbers, which differ from each other by a
quantity less than the difference between the roots 1 and -. If, for
example, we substitute 1, 4,4, 4 -, &c. there will be two changes
of the sign.
It may be objected to the above example, that when the fiactional coefficients of an equation have been made to disappear, the
equation can have for roots only either entire or irrational numbers,
and not fractions; but it will be readily seen, that the irrational
numbers, for which we have, in the example, substituted fractions
for the purpose of simplifying the expressions, may differ from each
other by a quantity less than unity.
In general, the results will have the same sign, whenever the
substitutions produce a change in the sign of an even number of
factors.*  To obviate this inconvenience we must take the numbers to be substituted, such, that the difference between the smallest limit and the greatest, will be less than the least of the differences, which can exist between the roots of the proposed equation; by this means the numbers to be substitued will necessarily fall between the successive roots, and will cause a change in
the sign of one factor only. This process does not presuppose the
smallest difference between the roots to be known, but requires
only, that the limit, below which it cannot fall, be determined.
In order to obtain this limit, we form the equation involving the
squares of the differences of the roots (208).
Let there be the equation
zn   p z-1   - qz n-2.... +    + u    - =0.... (D),
to obtain the smallest limit to the roots, we make (212) z —; we
have then the equation
In + 1 I_ + q a-       **+ t   + u o,
or, reducing all the terms to the same denominator,
1 -+       q v2..... + t Vn-1 + u v'   0,.then disengaging v",
* Equal roots cannot be discovered by this process, when their
number is even; to find these we must employ the method given in
art. 205..//lg.           31




242


Elements of Algebra.


v 4     n-1 ++ q V2 +- v +- = O;
v*u V...                    0;
U            U      U     U
and if - represent the greatest negative coefficient found in this
U
equation, we shall have
r
-+1
u
It is only necessary to consider here the positive limit, as this alone
relates to the real roots of the proposed equation.
Knowing the limit
1   _    u
u          u
-         + U'
less than the square of the smallest difference between the roots of
the proposed equation, we may find its square root, or at least, take
the rational number next below this root; this number, which I
shall designate by k, will represent the difference which must exist
between the several numbers to be substituted. We thus form the
two series,
0, + k, + 2 k, +3 k, &c.
— k, -2k, - 3 k, &c.
from which we are to take only the terms, comprehended between
the limits to the smallest and the greatest positive roots, and those
to the smallest and the greatest negative roots of the proposed
equation.  Substituting these different numbers, we shall arrive at
a series of results, which will show by the changes of the sign that
take place, the several real roots, whether positive or negative.
218. Let there be, for example, the equation
x3-7 x + 7 = 0,
from which, in art. 208., was derived the equation
Z3 — 42 z2 + 441 z -   49 = 0;
1
making z    -=, and, after substituting this value, arranging the result with reference to v, we have
v3   9v2 +   4  v-     = 0,
from which we obtain
< 10, z > -




Resolution of Numerical Equations by Approximation. 243


1
we must, therefore, take k = or <  -=. This condition will be
fulfilled,if we make k =-; but it is only necessary to suppose
k =.; for by putting 9 in the place of v in the preceding equation, we obtain a positive result, which must become greater, when
a greater value is assigned to v, since the terms v3 and 9 v2 already
destroy each other, and 42 v exceeds -.
The highest limit to the positive roots of the proposed equation
X3 -7 x + 7 = 0,
is 8, and that to the negative roots -  8; we must, therefore, substitute for x the numbers
-_1._2 *    3  4              24
2        334. _ 2 4
We may avoid fractions by making x  3; for in this case the
differences between the several values of x' will be triple of those
between the values of x, and, consequently, will exceed unity; we
shall then have only to substitute, successively,
0,   1,  2,    3,.......    24,
-  1,- 2, - 3,.......-      24,
in the equation
X'3 - 63 x' + 189 = 0.
The signs of the results will be changed between + 4 and + 5,
between + 5 and + 6, and between - 9 and -   10, so that we
shall have for the positive values,
x/ > 4 and < 5 I            > A and <  5
x' > 5 and < 6 } whence   x >  5 and < ~
and the negative value of x' will be found between - 9 and - 10
that of x between - 9- and -  o~.
Knowing now the several roots of the proposed equation within
-, we may approach nearer to the true value by the method explained in art. 215.
219. The methods employed in the example given in art. 21 5
and in the preceding article, may be applied to an equation of any
degree whatever, and will lead to values approaching the several
real roots of this equation. It must be admitted, however, tha
the operation becomes very tedious, when the degree of the proposed equation is very elevated; but in most cases it will be unnecessary to resort to the equation (D), or rather its place may be




244


Elements of JIlgebra.


supplied by methods, with which the study of the higher branches
of analysis will make us acquainted.*
I shall observe, however, that by substituting successively the
numbers 0, 1, 2, 3, &c. in the place of x, we shall often be lead
to suspect the existence of roots, that differ from each other by a
quantity less than unity. In the example, upon which we have
been employed, the results are
+ 7, + 1, + 1, + 13,
which begin to increase after having decreased from + 7 to + 1.
From this order being reversed it may be supposed, that between
the numbers + 1 and + 2 there are two roots either equal or
nearly equal. To verify this supposition, the unknown quantity
should be multiplied. Making x- =, we find
y3 _  700 y + 7000 - 0,
an equation, which has two positive roots, one between 13 and 14,
and the other between 16 and 17.
The number of trials necessary for discovering these roots is not
great; for it is only between 10 and 20, that we are to search for
y; and the values of this unknown quantity being determined in
whole numbers, we may find those of x within one tenth of unity.
220. When the coefficients in the equation proposed for resolution are very large, it will be found convenient to transform this
equation into another, in which the coefficients shall be reduced to
smaller numbers. If we have, for example
x4 _  80x3 +  1998 2- 14937 x + 5000 = 0,
we may make x == 10 z; the equation then becomes
z4- 8 z3 + 19,98 z2 -   14,937 z + 0,5 = 0.
If we take the entire numbers, which approach nearest to the
coefficients in this result, we shall have
-    S z3 + 20 z2 -  15 z + 0,5 = 0.
It may be readily discovered, that z has two real values, one between 0 and 1, the other between 1 and 2, whence it follows, that
those of the proposed equation are between 0 and 10, and 10 and
20.


* A very elegant method, given by Lagrange for avoiding the use
of the equation (D) may be found in the Trait6 de la Resolution des
Equations numeriques.




Resolution of JNumerical Equations by approximation. 245
I shall not here enter into the investigation of imaginary roots, as
it depends on principles we cannot at present stop to illustrate;
I shall pursue the subject in the Supplement.
221. Lagrange has given to the successive substitutions a form
which has this advantage, that it shows immediately what approaches we make to the true root by each of the several operations, and which does not presuppose the value to be known within
one tenth.
Let a represent the entire number immediately below the root
sought; to obtain this root, it will be only necessary to augment a
by a fraction; we have, therefore, x = a +  -.  The equation
Y
involving y, with which we are furnished by substituting this value
in the proposed equation, will necessarily have one root greater
than unity; taking b to represent the entire number immediately
below this root, we have for the second approximation x = a +  b.
But b having the same relation to y, which a has to x, we may, in
the equation involving y, make y = b + -, and y' will necessarily
be greater than unity; representing by b' the entire number immediately below the root of the equation in y', we have
1    bb + l
y - b +-    6-    ' 
substituting this value in the expression for x, we have
b'
x - a + b b
for the third approximation to x. We may find a fourth by making
y' = b' +  1; for if b" designate the entire number immediately
below y", we shall have
1    b'b" - 1
y' = b+    -      b —l'
whence
b"       b b' b" +b" + b
y      b + b' b" + 1    b' b" + 1
b' b - + 1
x - a + 6 6b' 6t +6- +b'
and so on.




246                Elements of Algebra.
222. I shall apply this method to the equation
X3-   7 x + 7 = 0.
We have already seen (218), that the smallest of the positive roots
of this equation is found between 4 and ", that is, between 1 and
2; we make, therefore, x =  1 + -; we shall then have
y3-   4 y3 + 3y +  1 =0.
The limit to the positive roots of this last equation is 5, and by
substituting, successively, 0, 1, 2, 3, 4, in the place of y, we immediately discover, that this equation has two roots greater than unity,
one between 1 and 2, and the other between 2 and 3.  Hence
x =  1 + }, and    x=    1 +,
that is,
x    2, and    x-= -.
These two values correspond to those, which were found above
between -: and 5, and between E and  -, and which differ from
each other by a quantity less than unity.
In order to obtain the first, which answers to the supposition of
y = 1, to a greater degree of exactness, we make
yy  l
we then have
y3 -  2 y2    y +  1 = 0.
We find in this equation only one root greater than unity, and that
is between 2 and 3, which gives
=   1 +  - =  2
whence
x    1 +  2.
Again, if we suppose y' = 2 + -, we shall be furnished with the
equation
y/-3 /_   2 _  4 y"_ 1 = 0;
we find the value of y" to be between 4 and 5; taking the smallest
of these numbers, 4, we have
y' =          y-    l+!- 2-,    =  1 +T 13  -  22
It would be easy to pursue this process, by making y" = 4 +- y,
and so on.




Resolution of NJVumerical Equations by approximation. 247
I return now to the second value of x, which, by the first approximation, was found equal to ', and which answers to the supposition of y - 2.  MIaking y = 2 +     and substituting this
expression in the equation involving y, we have, after clianging the
signs in order to render the first term positive,
y3 + y      2 y'-   1 = 0.
This equation, like the corresponding one in the above operation,
has only one root greater than unity, which is found between 1 and
2; taking y' = 1, we have
y    3,   x =4.
Again assuming
=  + 1
we are furnished with the equation
y//3 _ 3 y//2 _  4 y// _  1 = 0,
in which y" is found to be between 4 and 5, whence
y  -- 5    y -  1 4    X = — 1 9
We may continue the process by making y" - 4 +- y,    and
so on.
The equation x3 - 7 x + 7 = 0 has also one negative root,
between - 3 and - 4. In order to approach it more nearly, we
make x    - 3 - -; which gives
y3- 20 y2    9 y - I = 0, y > 20 and < 21,
whence
x - - 3 -     -2   6 1
-- -2 
To proceed further, we may suppose y =- 20 +   1 &c., we
shall then obtain, successively, values more and more exact.
The several equations transformed into equations in y, y', ye/,
&c. will have only one root greater than unity, unless two or more
roots of the proposed equation are comprehended within the same
limits a and a + 1; when this is the case, as in the above example,
we shall find in some one of the equations in y, y', &c. several
values greater than unity. These values will introduce the different
series of equations, which show the several roots of the proposed
equation, that exist within the limits a and a + 1.




248


Elements of Algebra.


The learner may exercise himself upon the following equation;
X3 —2x-5 =0,
the real root of which is between 2 and 3; we find for the entire
values of y, y', &c.
10, 1, 1, 2, 1, 3, 1, 1, 12, &c.
and for the approximate values of x,
2  21  23  44  11   1 55  576  731  1307  16415
Of Proportion and Progression.
223. ARITHMETIC introduces us to a knowledge of the definition and fundamental properties of proportion and equidifference,
or of what is termed geometrical and arithmetical proportion. I
now proceed to treat of the application of algebra to the principles
there developed; this will lead to several results, of which frequent
use is made in geometry.
I shall begin by observing, that equidifference and proportion
may be expressed by equations.   Let., B, C, D, be the four
terms of the former, and a, b, c, d, the four terms of the latter;
we have then
b   d
B -.  =D -     C (.Erith. 127),a =  (rith. cIl),
equations, which are to be regarded as equivalent to the expressions.. B: C. D,    a:b:: c:d,
and which give.A+D = B+ C,          ad= bc.
Hence it follows, that, in equidifference, the sum of the extreme
terms is equal to that of the means, and in proportion, the product
of the extremes is equal to the product of the means, as has been
shown in Arithmetic (127, 11 3), by reasonings, of which the above
equations are only a translation into algebraic expressions.
The reciprocal of each of the preceding propositions may be
easily demonstrated; for from the equations
- +D= B+- C,          ad= b c,
we return at once to
J)-C= B-A, b -
a c




Proportion and Progression.


249


and, consequently, when four quantities are such, that two among
them give the same sum, or the same product, as the other two, the
first are the means and the second the extremes (or the converse)
of an indifference or proportion.
When B = C, the equidifference is said to be continued; the
same is said of proportion, when b = c. We have in this case
+D       2 B,     ad= b2;
that is, in continued equidifference, the sum of the extremes is equal
to double the mean; and in proportion, the product of the extremes
is equal to the square of the mean. From this we deduce
B=A +D          b=a d;
the quantity B is the middle or mean arithmetical proportional between A and D, and the quantity b the mean geometrical proportional
between a and d.
The fundamental equations,
b d
B-      -qD- C, =-,
a    c
lead also to the following;
d
C-      - D —B, a- -
from which it is evident, that we may change the relative places
of the means in the expressions.2. B: C. D, a: b:: c: d, and in
this way obtain A. C: B. D, a: c:: b: d. In general, we may
make any transposition of the terms which is consistent with the
equations
+ D = B +      C and a d = bc (Arith. 114.)
I have now done with equidifference, and shall proceed to
consider proportion simply.
224. It is evident, that to the two members of the equation
b   d
- -    we may add the same quantity m, or subtract it from them;
a    c
so that we have
b         d
-: mr= - ~ m;
a         c
reducing the terms of each member to the same denominator, we
obtain
b:tm a    d:m c
— Zg ~ 32~


J8g.


32




250


Elements of Algebra.


an equation, which may assume the form
c    dim c
a    b i= m a'
and may be reduced to the following proportion,
b ~ ma:d     m c:: a: c;
cd
and as -   -, we have likewise
a   b'
di:: mc    d
b ~m- n a  b'
or                b = m a: d imc::b: d.
These two proportions may be enunciated thus; The first consequent plus or minus its antecedent taken a given number of times, is
to the second consequent plus or minus its antecedent taken the same
number of times, as the first term is to the third, or as the second is
to the fourth.
Taking the sums separately and comparing them together, and
also the differences, we obtain
d+ m c _ c      d-m c      c
b + ma   a'      -ma a'
whence we conclude
d + mc    d- nz c
b + ma    b- ma'
that is,
b + ma: d -+mc:: b -m a:d - mc;
or rather, by changing the relative places of the means
b +ma:b —ma::d + mc: d —mc;
and if we make m = I, we have simply
b - a: b -a:: d    c: d- c,
which may be enunciated thus;
The sum of the first two terms is to their difference as the sum of
tha last two is to their difference.
225. The proportion a: b:: c: d may be written thus;
a:c::b:d;
we have then
c        d
-   m      - im,
whence
c =zE M a  d~ n b
_          F      I




Proportion and Progression.


and lastly,
c -ma:d -mb::a:b or::c:d,
from which it follows, that the second antecedent plus or minus the
first taken a given number of times, is to the second consequent plus
or minus the first taken the same number of times, as any one of the
antecedents whatever is to its consequent.
This proposition may also be deduced immediately from that
given in the preceding article; for by changing the order of the
means in the original proportion
a: b:: c: d,
and applying the proposition referred to, we obtain, successively,
a:c::b d,
c ima: d     mb::a:b or::c:d,
and denominating the letters, a, b, c, d, in this last proportion,
according to the place they occupy in the original proportion, we
may adopt the preceding enunciation.
Making m - 1, we obtain the proportions
c ~ a: d ~- b::a:b::c:d
c + a: c -a::d + b: d -;
whence it appears, that the sum or difference of the antecedents is
to the sum or difference of the consequents, as one antecedent is to
its consequent, and that the sum of the antecedents is to their difference as that of the consequents is to their difference.
In general, if we have
b =d     fh
' g, &c.,
and make -    q, we shall have
and make b    q, we shall have
d       f       h        C
c   q, -    Q, s   =q,&c.,
which gives
b =a q, d    cq, f= e q, h =g q, &c.
then, by adding these equations member to member, we obtain
b + d +f+ h = aq + cq + eq + gq,
or       b+d+f+ h               -q(a+  c +  e  -+g).




252


Elements of Algebra.


whence it follows, that
b+d+f+h               b
a +c +- e +g          a
This result is enunciated thus; in a series of equal ratios,
a: b:: c  d:: e: f:: g: h, &c.
the sum of any number whatever of antecedents is to the sum of a
like number of consequents, as one antecedent is to its consequent.
226. If we have the two equations
b    d          _ h
a-  —,  and   -—,
a    c         e   g
and multiply the first members together and the second together,
the result will be
bf    dh
ae    cg
an equation equivalent to the proportion
ae: bf:: cg  d h,
which may be obtained also by multiplying the several terms of the
proportion
a  b:cd,
by the corresponding ones in the proportion
e:f::g:h.
Two proportions multiplied thus term by term are said to be multiplied in order; and the products obtained in this way, are, as
will be seen, proportional; the new ratios are the ratios compounded
of the original ratios (drith. 123).
It will be readily perceived also, that if we divide two proportions term by term, or in order, the result will be a proportion.
227. If we have
b    d
a    c'
we may deduce from it
bm    dm
am  ~m 
which gives
am:bm    cm  din;
whence it follows, that the squares, the cubes, and, in general, the
similar powers offour proportional quantities are also proportional.
The same may be said of fractional powers, for, since




Proportion and Progression.


253


m     m
lb.  b
a a     a
and
m    m 
therefore,
m rm
m  = '
/a    1vc
or
m    m     m    m
Va    b: V:/: /d,
if a: b:: c: d; that is, the roots of the same degree of four proportional quantities, are also proportional.
Such are the leading principles in the theory of proportion.
This theory was invented for the purpose of discovering certain
quantities by comparing them with others. Latin names were for
a long time used to express the different changes or transformations, which a proportion admits of. We are beginning to relieve
the memory of the mathematical student from so unnecessary a
burden; and this parade of proportions might be entirely superseded by substituting the corresponding equations, which would
give greater uniformity to our methods, and more precision to our
ideas.
228. We pass from proportion to progression by an easy transiti-n. After we have acquired the notion of three quantities in
continued equidifference, the last of which exceeds the second, as
much as this exceeds the first, we shall be able, without difficulty,
to represent to ourselves an indefinite number of quantities, a, b, c,
d, &c., such, that each shall exceed the preceding one, by the
same quantity 6, so that
b = a + S, c = b +   S, d = c + S, e = d + 6, &c.
A series of these quantities is written thus;
-a. b. c. d. e.f, &c.
and is termed an arithmetical progression; I have thought it proper, however, to change this denomination to that of progression by
differences. (See Arith. art. 127, note.)




254


Elements of Algebra.


We may determine any term whatever of this progression, without employing the intermediate ones.  In fact, if we substitute for
b its value in the expression for c, we have
c = a + 2;
by means of this last, we find
d = a +   3 8, then e =a + 4 S,
and so on; whence it is evident, that representing by I the term,
the place of which is denoted by n, we have
I = a +   (n -  1)8.
Let there be, for example, the progression
-  3. 5. 7. 9. 11. 13. 15. 17, &c.
here the first term a = 3, the difference or ratio  == 2; we find
for the eighth term
3 +(8-     1)2 = 17,
the same result, to which we arrive by calculating the several
preceding terms.
The progression we have been considering is called increasing;
by reversing the order, in which the terms are written, thus,
-. 17. 15. 13. 119. 9.7. 5. 3.  1.- - 3, &c.
we form a decreasing progression. We may still find any term
whatever by means of the formula a + (n -  1) A, observing only,
that a is to be considered as negative, since, in this case, we must
subtract the difference from any particular term in order to obtain
the following.
229. We may also, by a very simple process, determine the sum
of any number whatever of terms in a progression by differences.
This progression being represented by
-      a.b............   i. k. 1,
and S denoting the sum of all the terms, we have
S = a + b + c.......... + i + k + 1.
Reversing the order, in which the terms of the second member of
this equation are written, we have still
S =l - +k + i.........- +c + b + a.
If we add together these equations, and unite the corresponding
terms, we obtain
2S=(a+-  )+(b+-k)+(c+i)....+(i+c) +(k-+b)+(l+(a);
but by the nature of the progression, we have, beginning with the
first term,




Proportion and Progression.


255


a +   = b, b +- a =   c,.......i. i +  -= k, k  +  = I,
and, consequently, beginning with the last
1      --  k, c -  = i....... c - -=b, b -    = a;
by adding the corresponding equations, we shall perceive at once,
that
a +   = b + k = c + i, &c.
and, consequently, that
2 S    n (a + I);
whence it follows, that
S    n (a +I)
2- 
Applying this formula to the progression
-3. 5. 7. 9 &c.
we find for the sum of the first eight terms
(3     17) 8  80.
230. The equation
= a + (n-     1)S,
together with
S    (a + 1)n
2
furnishes us with the means of finding any two of the five quantitities, a, S, n, 1, and S, when the other three are known; I shall
not stop to treat of the several cases, which may be presented.
231. From proportion is derived progression by quotients or
geometrical progression, which consists of a series of terms, such,
that the quotient arising from the division of one term by that which
precedes it, is the same, from whatever part of the series the two
terms are taken. The series
- 2: 6: 18:54: 162: &c.
45: 15: 5:: &c.
are progressions of this kind; the quotient or ratio is 3 in the first,
and - in the other; the first is increasing, and the second decreasing.  Each of these progressions forms a series of equal ratios, and
for this reason is written, as above.
Let
a, b, c, d,..........  k, 1,
be the terms of a progression by quotients; making  = q, we have
by the nature of the progression,




256


Elements of lAlgebra.


b   'c  d    e        I
-a = b       c     *        '
or     b   aq, c=b q, d =cq, e=d q,... I = kq.
Substituting, successively, the value of b in the expression for c,
and the value of c in the expression for d, &c., we have
b= a, c= aq2, d=aq3, e= a q4,... = aq-l,
taking n to represent the place of the term 1, or the number of
terms considered in the proposed progression.
By means of the formula I = a (qn-   we may determine any
term whatever, without making use of the several intermediate
ones. The tenth term of the progression
2:6:18:&c.,
for example, is equal to 2 X 39 = 39366,
232. We may also find the sum of any number of terms we
please of the progression
a:b:c:d, &c.
by adding together the equations
b = a q, c = b q, d = c q, e= dq,....     kq;
for the result will be
b+c+d+e... +l= (a+b+c+d... +k)q;
and representing by S the sum sought, we have
b+c+d+e....- +          =-  S-  a,
a + b + c +      d.... + k = S   -1,
whence                S-a - q (S- 1),
and, consequently,
S= q -    t.
q-1
t The truth of this result may be rendered very evident, independently of analysis. If it were required, for example, to find the sum
of the progression
2: 6: 18:54: 162,
multiplying by the ratio, we have
- 6: 18: 54: 162: 486.
The first series being subtracted from this gives 486 - 2, equal to so
many times the first series, as is denoted by the ratio minus one,
that is




Proportion and Progression.


257


In the above example, we find for the sum of the first ten terms
of the progression
2:6:18: &c.
2 X 31' -2:310    1 -  59048.
233. The two equations,
-=a qn-1,    S       -a
comprehend the mutual relations, which exist among the five
quantities, a, q, n, 1, and S, in a progression by quotients, and
enable us to find any two of these quantities, when the other three
are given.
234. If we substitute a qn-l in the place of 1, in the expression
for S, we have
s      - I
q —1
When q is a whole number, the quantity qf1 will become greater
and greater in proportion to the increased magnitude of the number
n; and S may be made to exceed any quantity whatever, by assigning a proper value to n, that is, by taking a sufficient number of
terms in the proposed progression. But if q is a fraction, represent1
ed by -, we have
a (      i)ta a m     - -        a- 
1                 m-1      m-         -
_  1
m
2 + 6 +18 +54 + 162 =          X62 —  2
If we multiply by the ratio q the general series
— a:b:c:d:e......1,
we have          a q:b:cq:dq:eq.....g.
Then, because b = a q, &c., the second series minus the first is lq -a,
equal to so many times the first series, as is denoted by the ratio minus
one.
Hence     a + b + c + d + e..... + I     I    a.
t Multiplying the numerator and denominator by - m.
Alg.             33




i58


Elements of Algebra.


and it is evident, that as the number n becomes greater, the term
a
m —n  will become smaller, and, consequently, the value of S will
approach nearer and nearer to the quantity  a  from which it will
differ only by
(m- 1) mn-1
therefore, the greater the number of terms we take in the proposed progression, the more nearly will their sum approach td
a mquantity less
a ml. It may even differ fronm       by a quantity less than
any assignable quantity, without ever becoming in a rigorous sense
equal to it.
am
The quantity  --, which I shall designate by L, forms, we
perceive, a limit, to which the particular sums represented by S,
approach nearer and nearer.
Applying what has been said to the progression
* @  1.:-:.:1.: I1o  &c.
we have
a =, q    -=  -,
m
whence
-2  L     am 
m- I-;
and the greater the number of terms we take in the above progression, the nearer their sum will approach to an equality with 2.
We have, in fact,
1                      _   1 -. 2- I,
1.
1L+ 1  += 1                  = 2     Ilq-     -~+,....        -, D
i+     +    +             = _   2 —,
- L  +  4t +  8+ -   -   = 2 -3 1.
B                     A2
&c.
The expression for Li may be considered as the sum of the
decreasing progression by quotients, continued to infinity, and it is
thus, that it is usually presented; but in order to form a clear idea
of it, we must represent it in a limited view.




Proportion and Progression.


259


235. We may obtain from the expression
S  a (q   1)
_     2q —1  '
all the terms of the progression, of which it denotes the sum  for,
if we divide qn -  1 by q-  1 (158), we find
- =        - 1   1      -- + q - 3 ++  4.- *  +  -
which gives
S==a a   aq    a q2..... +..  a q1-.
We may employ the value of L for the same purpose; in this
case, m is to be divided by m-  1, as follows;
m         m- 1,- m +    1  +-+      + 4- 1  - _&c.
1
X        2 -1     1
1     1
m2 [ m2
&c.
We begin, by dividing, according to the usual method, by the
first term, and find 1 for the quotient; we multiply this quotient
by the divisor and subtract the product from the dividend; then,
dividing the remainder by the first term of the divisor, we obtain
for the quotient, and have - for a remainder; we go through
the same process with this remainder as with the preceding. Pursuing this method, we soon discover the law, to which the several
particular quotients are subjected, and perceive that the expression
--- is equivalent to the series
1    1     1
1+   I +  1 +    3 + &o.
continued to infinity.  Substituting for m its value -, and multiplying by a, we find as before
a + a q - aq2. +aqa3 + &c.
for the progression of which L represents the limit.




260


Elements of Algebra.


236. The series
1     1    1
+m +       +     + &.
is considered as the value of the fraction,whenever it is conm- 1
verging, that is, when the terms, of which it is composed, become
smaller and smaller the further they are removed from the first.
Indeed, if we make the division cease successively at the first,
second, third...... remainder, we have
the quotients 1         and the remainders 1
1                         1
I+-                           1
m     In
1     1m m2
&c.                           &c.
the former of which approach the true value, exactly in proportion
as the latter are diminished; and this takes place, only when m
exceeds unity. In all other cases we must have regard to the
remainders, which, increasing without limit, make it evident, that
the quotients are departing further and further from the true value.
To render this clear, we have only to make, successively,
m = 2, m =   1, m =. Upon the first supposition, we have
= 2 = 2     1 +   +    +   +     + &c.
and it has been shown (234), that the series, which constitutes the
second member, approaches, in fact, nearer and nearer to 2.
The second supposition leads us to
m
n —1       - '  - + t -{ +  1 + 1 + 1    1 &c.
This result, 1 + 1 + 1 + 1 + 1 &c., continued to infinity, presents in reality an infinite quantity, as the nature of the expression
- implies; yet if we neglect the remainders in this example, we
are led into an absurdity; for since the divisor, multiplied by the
quotient, must produce the dividend, we have
1-(1      1 1  + 1 +.....)0;
but the second member is strictly reduced to nothing, we have
therefore 1 = 0.
The third supposition leads to consequences not less absurd, if
we neglect the remainders, and consider the series, which is obtain



Proportion and Progression.


261


ed, as expressing the value of the fraction, from which it is derived.
Making m = -, we find
h=i- -  -1     1 + 2+4+8+ 16 + &c.,
which is evidently false.
There will be no contradiction of this kind, if we observe, that,
in the second case, the remainders
1   I 1
1a, n — 2 n,3 &c.
are each equal to 1, and that, since they do not diminish, they can
never be neglected, to whatever extent the series is continued. If
we add, therefore, one of these remainders to the second member
of the equation
1 =(1 +    1 +  1 +  1 +  1 +...... )0,
the equation becomes true. In the third case, the remainders,
1   1    1
1 -, -, &c.
form the increasing progression, 1, 2, 4, 8, 16, &c. and, if we add
to the several quotients the fiactions, arising from the corresponding remainders, the exact expressions for --- will be
+1
1 + m    l
-+   I + m 1 
1         1
71 +   m   (in -2 1)
1 m  l m   "r 2 (rn - 1)
&c.,
each of which gives -  1, when m -.
m              n
If we take m = - n, the fi'action m-i becomes; the
series, which is produced by developing this fraction, assumes the
form
1      -  n +  -, + &c.,
and making n =     1, we have
1 -     1   - 1 -  1 + -  1 + &c.,
a series, which becomes alternately 1 and 0, and which, consequently, as often exceeds, as it falls below, the true value of




262


Elements of.lgebra.


n, equal in this case to i-; but as the above series is not converging, it cannot give this true value; and we must, therefore,
take into consideration the remainder, at whatever term we stop.
If we suppose, in the preceding series, n - 2, we have
1 -, +  } -  - +  r-  &c.,
a series, in which the particular sums, 1, 1, 3, 3, &c. are, alter,
nately, smaller and greater than the true value of  which is
-, but to which they approach continually, because the proposed
series is converging.
Although diverging series, that is, those, the terms of which go
on increasing, continue to depart further and further from the true
value of the expressions from which they are derived, yet considered as developements of these expressions, they may serve to show
such of their properties, as do not depend on their summation.
237. If we continue any process of division in algebra, according to the method pursued above (235), with respect to the quantities m and m -  1, the quotient will always be expressed by an
infinite series composed of simple terms. Infinite series are also
formed by extracting the roots of imperfect powers, and continuing
the operation upon the several successive remainders; but they
are obtained more easily by means of the formula for binomial
quantities, as will be shown in the Supplement, where I shall treat
of the more common series.
Examples in arithmetical Progressions.
Let a denote the first, I the last term, n the number of terms, 5
the common difference, and S the sum of an arithmetical progression.
= a+(n-      1),    S   (a+)        (2a + (n-l)d)
1. Given a =  1, 8 = 1, n =  14; then I = 14, S = 105.
2. Given a = 2, 5 - 3, n =   17; then I = 50, S = 442.
3. Given a =7,, n=   16; then I  103, S -142.
4, Given a = 2, d    1=, n = 100; then l = 35-, S - 1900,
5. Given a, 5     -, n =26; then I = 37, S = 6041.
6. Given a=,   = 1, n =n 13; then I= 20,     = 139.
7. Given a =- -      7,  -  3, n  8; then I - 14, S = 28.
8. Given a = - 6, S = a, n = 30; then I =- 152, S == 1461,




Theory of Exponential Quantities and of Logarithms. 263


9. Given a =,     - 1, n _ 20; then I     -   1 7
S    -  133.
10. Given a = 3,    =-    25, n =  15; then I =    361,
S =- 247'~.
11. Given a = 0, S, n = 11; then 1 = 5, S = 271.
12. Given a - -  10,    -       2, n =6; then  -20,
S   -     90.
13. Given a   -, 8 = -     n = 25; then I = - 213,
S —     2814.
Examples in Geometrical Progression,


Let a be the first term, q the common ratio, I the last term, it
the number of terms, and S the sum of all the terms; then
I = a qn-1   S - q        - a (q-  1 )
' ~     q —l        q1. Given a - 1, q    2, n = 7; then I = 64, S = 127,
2. Given a = 4, q = 3, n =   10; then I = 78732,
S =  118096.
3. Given a- 9, q= -, n        =  7; then I  =  258-~ 7,
S =- 591 74'.
4. Given a = 6{, q =-, n = 8; then 1 =   1064 4,
S = 3074 4
6. Given a    6,,      q  n = 6; then - 12=17 S= 19374. 
6. Given a    5, q = 4, n - 9; then 1 = 327680,
S = 436905.
7. Given a = 8, q     1, n = 15; then I =,
S-   152047
8. Given a = 3, q =, n = 8; then I    = — 575 32-~9Y
S =      6 2 9 4
9. Given a -=, q =, n =  11; then 1 = —   76,
S-=   T 45 3 a 7.
9J  -- 11t5e  1   --


Theory of Exponential Quantities and of Logarithms.
238. IN the several questions we have resolved thus far, the
unknown quantities have not been made subjects of consideration
as exponents; this will be requisite, however, if we would determine the number of terms in a progression by quotients, of which
the first term, the last term, and the ratio are given. In fact, we
are furnished by a question of this kind with the equation




264


Elements of algebra.


I = aqn-1 (231),
in which n will be the unknown quantity; abridging the expression, by making n -  I  --, we have I = a q. This equation
cannot be resolved by the direct methods hitherto explained; and
quantities like x cannot be represented by any of the signs already
employed. In order to present this subject in a more clear light,
I shall go back to state, according to Euler, the connexion which
exists between the several algebraic operations, and the manner, in
which they give rise to new species of quantities.
239. Let a and b be two quantities, which it is required to add
together; we have
a + b = -c;
and in seeking a or b from this equation, we find
a    c -  b, b = c-a;
hence the origin of subtraction; but when this last operation cannot be performed in the order in which it is indicated, the result
becomes negative.
The repeated addition of the same quantity gives rise to multiplication; a representing the multiplier, b the multiplicand, and c
the product, we have
ab = c,
whence we obtain
C      C
and hence arises division, and fractions, in which this division terminates, when it cannot be performed without a remainder.
The repeated multiplication of a quantity by itself produces the
powers of this quantity; if b represent the number of times a is a
factor in the power under consideration, we have
ab = c.
This equation differs essentially from the preceding, as the quantities a and b do not both enter into it of the same form, and hence
the equation cannot be resolved in the same way with respect to
both. If it be required to find a, it may be obtained by simply
extracting the root, and this operation gives rise to a new species
of quantities, denominated irrational; but b must be determined by
peculiar methods, which I shall proceed to illustrate, after having
explained the leading properties of the equation ab = c.




Theory of Exponential Quantities and of Logarithms. 265


240. It is evident, that if we assign a constant value greater than
unity to a, and suppose that of b to vary, as may be requisite, we
may obtain successively for c all possible numbers.  Making
b = 0, we have c = I; then since b increases, the corresponding
values of c will exceed unity rmore and moie, and may be rendered
as great as we please. The contrary will be the case, if we suppose
b negative; the equation ab = c being then changed into a-b" = c,
or -  = c, the values of c will go on decreasing, and may be rendered indefinitely small. We may, therefore, obtain from the same
equation all possible positive numbers, whether entire or fractional,
upon the supposition, that a exceeds unity. The same is true, if
we have a < 1; only the order, in which the values stand, will be
reversed; blt if we suppose a = 1, we shall always find c = 1,
whatever value be assigned to b; we must, therefore, consider the
observations which follow, as applying only to cases, in which a
differs essentially from unity.
In order to express more clearly, that a has a constant value,
and that the two other quantities b and c are indeterminate, I shall
represent them by the letters x and y; we then have the equation
ax = y, in which each value of y answers to one value of x, so that
either of these quantities may be determined by means of the other.
241. This fact, that all numbers may be produced by means of
the powers of one, is very interesting, not only when considered in
relation to algebra, but also on account of the facility with which it
enables us to abridge numerical calculations. Indeed, if we take
another number y', and designate by x' the corresponding value of
A, we shall have a' = y', and, consequently, if we multiply y by
y', we have
yy' = a; X a' = az+";
if we divide the same, the one by the other, we find
y    ax 
lastly, if we take the mth power of y, and the nthl root, we have
ye   (az)m-  = am
for the one, and
1       1     z
y" = (aCt =    l"
for the other..lg.              34




266


Elements of Algebra.


It follows from the first two results, that knowing the exponents
x and x' belonging to the numbers y and y', we may, by taking
their sum, find the exponent which answers to the product yy',
and by taking their difference, that which answers to the quotient
Y. From the last two equations it is evident, that the exponent
y
belonging to the mth power of y may be obtained by simple multiplication, and that which answers to the nth root, by simple division.
Hence it is obvious, that by means of a table, in which, against
the several numbers y, are placed the corresponding values of x, y
being given, we may find x, and the reverse; and the multiplication of any two numbers is reduced to simple addition, because,
instead of employing these numbers in the operation, we may add
the corresponding values of x, and then seeking in the table the
number, to which this sum answers, we obtain the product required. The quotient of the proposed numbers may be found, in the
same table, opposite the difference between the corresponding
values of x, and, therefore, division is performed by means of subtraction.
These two examples will be sufficient to enable us to form an
idea of the utility of tables of the kind here described, which have
been applied to many other purposes since the tine of Napier, by
whom they were invented. The values of x are termed logarithms,
and, consequently, logarithms are the exponents of the powers, to
which a constant number must be raised, in order that all possible
numbers may be successively deducedfrom it.
The constant number is called the base of the table or system of
logarithms.
I shall, in future, represent the logarithm of y by 1 y; we have
then x = 1 y, and since y = a", we are furnished with the equation y = ay.
242. As the properties of logarithms are independent of any
particular value of the number a, or of their base, we may form an
infinite variety of different tables by giving to this number all possible values, except unity. Taking, for example, a =  10, we
have y = (lIly, and we discover at once, litat the numbers
1,        100,  1000,  10000, 100000, &c.,
which are all powers of 10, have, for logarithms, the numbers
0,    1,    2,    3,    4,    6,    &C.




Theory of Exponential Quantities and of Logarithms. 267


The properties mentioned in the preceding article may be verified in this series; thus if we add together the logarithms of 10
and 1000, which are 1 and 3, we perceive, that their sum, 4, is
found directly under 10000, which is the product of the proposed
numbers.
243. The logarithms of the intermediate numbers, between 1
and 10, 10 and 100, 100 and 1000, &c. can be found only by
approximation. To obtain, for example, the logarithm of 2, we
must resolve the equation (10)2 = 2, by the mettlod given in art.
221., finding first the entire number approaching nearest to the
value of x. It is obvious at once, that x is between 0 and 1, since
(10)~ =  I, (10)1 = 10; we make therefore x =  -, the equation
then becomes (IO0) = 2, or 10 = 22; now z is found between 3
and 4; we suppose, therefore, z = 3 + -, and hence x
1          1         1
10- =2      '  23 X  - S X 2';
or                     2  = y = {,
or, lastly,
2 =( ):
As the value of z' is between 3 and 4, we make
1
z'=3 +     -;
we have then
1            1
2 = (4)    = (.)",
whence we obtain
(4)     - 2 (6    ) __,8 or (1, 8)-  5;
and after a few trials we discover that z" is between 9 and 10.
The operation may be continued further; but as I have exhibited
this process merely to show the possibility of finding the logarithms
of all numbers, I shall confine myself to the supposition of z" = 9;
we have then, going back through the several steps,
Z/ -- 2 8. _ 93    =  8
- - X    -  2 8 Z  - 8 ) 
This value of x, reduced to decimals, is exact to the fourth figure,
as it gives
x = 0,30107.




268


Elements of Algebra.


By calculations carried to a greater degree of exactness, it is found,
that
x = 0,5010300,
the decimal figures being extended to seven places.
Regarding this value of x as an exponent, we must conceive
the number 10 to be raised to the power denoted by the number
3010300, and the root of the result to be taken for the degree
denoted by 10.000000; we thus arrive at a number approaching
3010300
very nearly to 2; that is (10)~-wO-  -= 2, very nearly; the first
member is a little greater than 2; but (10) T  ~ ~ W  ~ is smaller.*
* The method explained in this article becomes impracticable,
when the numbers, the logarithms of which are required, are large;
another method, however, which may be very useful, is given by
Long, an English geometer, in the Philosophical Transactions for the
year 1724, No. 339.
As the process for determining x in the equation (10) = y is
very laborous, we may, reversing the order, furnish ourselves with the
several expressions for x, then forming a table of the values of y corresponding to those of x, we shall afterwards, as will be perceived, be
able, in any particular case, to determine z by means of y.
We take first for z the values co:nprehended between 0,1 and 0,9;
we have then only to determine the value of y, which answers to
x = 0,1, or (10) T, because the several other values of y, namely,
(10).T0, (10) ~, &~c.
are the 2d, 3d, &C. powers of the first.
By extracting the square root, we discover at once, that
(10)t or   (10)   =. 3,162277660;
then taking the fifth root of this result, we have
(10) I= 1,258925412.
By a similar process, we deduce from
(10)T' - 1,258925412,
the value of
4 (10)J =    (10)7 = (10)T ~ = 1,122018454;
then taking the fifth root, we have
(10)T0    1,023292992;




Theory of Exponential Quantities and of Logarithms. 269


244. By multiplying the logarithm of 2, successively by 2, 3, 4,
&c., we obtain logaritlls of the numbers, 4, 8, 16, &c., which are
the 2d, 3d, 4th, &c. powers of 2.
and raising the result to the 2(1, 3d..... 9th powers, we obtain the
values of y, corresponding to those of z comprehended between 0,01
and 0,09.
It will be readily seen, that by this method, we may also find the
values ofy for those of x between 0,001 and 0,009, between 0,0001
and 0,0009; thus we shall be furnished with the following table.


Log. I   Nat. Num.   I


0,9 7,943282347
8 6,309573445
7 5,011872336
63,981071706
53,162277660
42,511886432
31,995262315
2 1,54893193
1 1,258925412


Log.    Nat. Num.
0,00009 -1,00207254
81,000184224
71,000161194
6 1,000138165
5 1,00115136
41,000092106
31,000069080
21,000046053
1,000023026


0,09 1,230268771  0,000009 1,000020724
81,202264435         8 1,000018421
71,174897555         71,000016118
61,148153621         61,000013816
5 1,1220184545 1,000011513
41,096478196         41,000009210
31,071519305         31,000006908
21,047128548         21,000004605
1 1,023292992        1 1,000002302
0,009 1,020939484 0,0000009 1,000002072
8 1,018591388        8 1,000001842
71,016248694         71,000001611
61,013911386         61,000001381
5 1,011579454        5 1,000001151
41,009252886         4 1000000921
31,006931669         3 1,000000690
2 1,004615794        2 1,000000460
1 1,002305238        11,000000230
0,0009 1,002074475 0,00000009 1,000000207
8 1,001843766        8 1,000000184
7 1,001613109        7,000000161
6 1,001382506        6 1,000000138
51,001151956         51,000000115
4 1,000921459        4 1,000000092
3 1,00069 1015       3 1,000000069
2 1,000460623        2 1,000000046
] 1,000230285        1,000000023


By means of this table, we may find the logarithm of any number
whatever, by dividing it by 10 a sufficient number of times.  To




270


Elements of Algebra.


By adding to the logarithm of 2 the logarithms of 10, 100,
1000, &c. we obtain those of 20, 200, 2000, &c.; it is evident,
therefore that if we have the logarithms of the former numbers, we
obtain, for example, the logarithm of 2549, we first divide this number by (10)3 or 1000, which is the greatest power of 10 it contains;
we have then
2549 =(10)3 X 2,549;
we then seek in the table the power of 10 immediately below 2,549,
and find
(10)0,4  2,511886432;
dividing 2,549 by this last number, we have
2,549 = (10)~,4 X 1,014775177,
Again seeking in the table the power of 10 immediately below
1,014775177, we find
(10)0006, - 1,013911386;
then dividing the preceding quotient 1,014775177 by this number, we
obtain a third quotient 1,000851742. This process is to be continued,
until we arrive at a quotient, which differs from unity only in those
decimal places we propose to neglect.
If we consider, in the present case, the third quotient as equal to
unity, the proposed number will be resolved into factors, which will
be powers of 10, for we shall have
2549 =(10)3 X (10)0,4 X (10)~,~~6 =(10)3,406,
from which it is evident, that 3,406 is the logarithm of the number
2549. By extending the divisions to 7 in number, this logarithm will
be found to be 3,406369.
The same table enables us with still more ease to find a number by
means of its logarithm, as in the following example.
Let 2,547 be the given logarithm; the number sought will be
(10)2,547 = (10)2 X (10)0'5 X (10)00,4 X (10)0'007;
it will, therefore, be equal to the product of the numbers
(10)2  = 100,
(10)0,5   3,162277660,
(10)0,04 - 1,096478196,
(10)0,007 =  1,016248694,
taken from the table; and will, consequently, be
2,547 = 1. 352,357.
A table of the same kind with the above, but much more extended,
has been published in England, by Dodson, the object of which is to
furnish the means of finding the number answering to a given logarithm.




Theory of Exponential Quantities and of Logarithms. 271
may find the logarithms of all numbers composed of them, which
latter can be only powers or products of the former. The number
210, for example, being equal to
2 X 3 X 5 X 7,
its logarithm is equal to
12 +13 +15+17,
and since 5 =- L, we have
15 = 110 -   12.
245. Logarithms, which are always expressed by decimals,
are composed of two parts, namely, the units placed on the left
of the comma, and tile decimal figures found on the right. The
first of these is called the characteristic, because in the logarithms
under consideration, which are adapted to the supposition of
a =  10, and which are called common logarithms, this part shows
to what order of units the number corresponding to the logarithm belongs. The several logarithms of the numbers between
1 and 10, as they are between 0 and 1, have, necessarily, 0 for
their characteristic; those of the numbers between 10 and 100,
have 1 for their characteristic; those of the numbers between
100 and 1000 have 2; in general, the characteristic of a logarithm contains as many units, as the proposed number has figures,
minus one.
246. It is important also to remark, that the decimal part of the
logarithms of numbers, which are decuple the one of the other, is
the same; for example,
the logarithm of   54360   is   4,7352794,
5436        3,7352794,
543,6      2,7352794,
54,36     1,7352794,
5,436    0,7352794;
for, as each of these numbers is the quotient of that which precedes
it, divideli by 10, the logarithm of the one is found by taking an
unit from the characteristic of that of the other (241,242).
247. According to what has been said in art. 240., the logarithms
of fractional nu.tbers are, up)n our present hypothesis, negative;
and we may easily deduce them from those of entire numbers,
if we observe that a fraction represents the quotient arising from
the division of the numerator by the denominator.  When the
numerator is less than the denominator, its logarithm is also less




272


Elements of./lgebra.


than that of the denominator, and, consequently, if we subtract the
latter from the former, the result will be negative.
In order to obtain the logarithm of the fraction A, for example,
we subtract from 0, which denotes the logarithms of 1, the fraction
0,3010300, which represents that of 2; the result is
-0,3010300.
If we subtract from 0 the number 1,3010300, which is the logarithm
of 20, we have the logarithm of J-, equal to
-1,3010300.
The logarithm of 3 being 0,4771213, that of 2- will be
0,3010300 - 0,4771213 -- - 0,1760913.
248. It is evident from the manner in which the logarithms of
fractions are obtained, that, considered independently of their
signs, they belong (241) to the quotients, arising from the division
of the denominator by the numerator, and, consequently, answer to
the number, by which it is necessary to divide unity in order to
obtain the proposed fraction. Indeed, 2, for example, may be
exhibited under the form 3 and I = 1 3 - 1 2 = 0,1760913.
It would be inconvenient, in order to find the value of a fraction,
to which a given negative logarithm belongs, to employ the number
to which the same logarithm answers when positive, since it would
be necessary to divide unity by this number; but if we subtract
this logarithm from 1, 2, 3, &c. units, the remainder will be the
logarithm of a number, which expresses the fraction sought, when
reduced to decimals, since this subtraction answers to the division
of the numbers, 10, 100, 1000, &c. by the number to which the
proposed logarithm belongs.
Let there be, for example, - 0,3010300; if, without regarding
the sign, we take this logarithm from 1, or 1,0000000, the remainder, 0,6989700, being the logarithm of 5, shows, that the
fraction sought is equal to 0,5, since we supposed unity to be
composed of 10 parts.
If, in seeking the logarithm of a fraction, we conceive unity to
be made up of 10, or 100, or 1000, &c. parts, or which amounts
to the same thing, if we augment the characteristic of the logarithm of the numerator by a number of units sufficient to enable us
to subtract that of the denominator from it, we obtain in this way a
positive logarithm, which may be employed in the place of that
indicated above.




Theory of Exponential Quantities and of Logarithms. 273


In order to introduce uniformity into our calculations, we most
frequently augment the characteristic of the logarithm of the numerator by 10 units. If we do this with respect to the fraction -, for
example, we have
10,3010300 - 0,4771213   - 9,8239087.
It will be readily seen, that this logarithm exceeds the negative
logarithml  -  0,1760913 by 10 units, and that, consequently,
whenever we add it to others, we introduce 10 units too much
into the result; but the subtraction of these ten units is easily
performed, and by performing it we effect at the same time the
*subtraction of 0,1760913. Let NJ be the number, to which we
add the positive logarithm 9,8239087; the result of the operation
will be represented by
N + 10 -0,1760913.
and if we subtract 10, we have simply
JV-   0,1760913.
According to the preceding observations, we cause addition to
take the place of subtraction, by employing, instead of the number
to be subtracted, its arithmetical complement, that is, what remains,
when this number is subtracted from one of the numbers, 10, 100,
1000, &c., a result which is obtained by taking the units of the
proposed numbter from 10 and the several other figures from 9,
We add this complement to the number, from which the proposed
logarithm is to be subtracted, and fiom the sum subtract an unit of
the same order as the complement.
It is evident, that if the complement is repeated several times,
we must subtract, after the addition, as many units of the same
order with the complement, as there are in the number, by which
it is multiplied; and, for the same reason, if several complements
are employed, we must subtract for each an unit of the same order,
or as many units as there are complements, if they are all of the
same order.
Sometimes this subtraction cannot be effected; in this case, the
result is the arithmetical complement of the logarithm of a fraction,
and answers in the tables to the expression of tis fraction reduced
t, decimals. If 10 units remain to be taken from the characteristic, as is most frequently the case, the result is the same as if we had
multiplied by 10000000000, the numerator of the fraction sought,
in order to render it divisible by the denominator; the character-.atg.              35




274


Elements of Algebra.


istic of the logarithm of the quotient shows the highest order of the
units contained in this quotient, considered with reference to those
of the dividend. In 9,8239087, the characteristic 9 shows, that
the quotient must have one figure less than the number, by which
we have multiplied unity; and, consequently, if we separate 10
figures for decimals, the first significant figure on the left will be
tenths; and we shall find only hundredths, thousandths, &c., for
the numbers the arithmetical complements of which have 8, 7, &c.
for their characteristics.
249. What has been said respecting the system of logarithms, in
which a - 10, brings into view the general principles necessary
for understanding the nature of the tables; for more particular
information the learner is referred to the tables themselves, which
usually contain the requisite instruction relating to their arrangement and the method of using them.  I will merely mention the
tables of Callet, stereotype edition, and those of Borda, as very
complete and very convenient.
250. If we have the logarithm of a number y for a particular
value of a, or for a particular base, it is easy to obtain the logarithm
of the same number in any other system. If we have a =- y;
for another base., we have Ax = y, X being different from x;
hence we deduce.x =- ax. Taking the logarithms according to
the system, the base of which is a, we have
1./x = 1 a;
now 1 a" = x by hypothesis, and 1 /x =  X1. A(241); therefore,
X1.     = x, or X -= I-; but if we consider A as a base, X will
be the logarithm of y in the system founded on this base; if, therefore, we designate this last by Ly, in order to distinguish it from
the other, we have
ly
Ly     I-A'
and we find the logarithm of y in the second system, by dividing its
logarithm taken in the first by the logarithm  of the base of the
second system.
The preceding equation gives also -1  = 1 -; from which it is
by 
evident, that whatever be the number y, there is between the logarithms I y and L y, a ratio invariably represented by 1./




Theory of Exponential Quantities and of Logarithms.  275
251. In every system the logarithm of 1 is always 0, since
whatever be the value of a we have always a~ - 1. As logarithms
may go on increasing indefinitely, they are said to become infinite
at the same time with the corresponding numbers; and as, when y
1   -
is a fractional number, we have y = --  = a-", it is evident, that
in proportion as y becomes smaller, x in its negative state becomes
greater, but we can never assign for x a number, which shall render y strictly nothing.  In this sense it is said, that the logarithm
of zero is equal to an infinite negative quantity, as we find in many
tables.
252. I now proceed to give some examples of the use, which
may be made of logarithms in finding the numerical value of formulas. It follows from  what is said in art. 241., and from the
definition of logarithms, by which we are furnished with the equation al' - y, that
I (dB) = 1    +   B,     l(   ) \= l- 1B,
1    1
1     \m=m,.= n           dlJ.
n
Applying these principles to the formula
A2 /32~ -   2
5
C./D3 EF
which is very complicated, we find
1 (/2 v/B - c) = 1 [2 v(B + C) (B - C)]
21/ +      (B +   C) +    1(B-    C),
5
I(CvD3EF)-        C+   3 ID 1     I+ E +  IF,
and, consequently,
/12 /vB'2  C2
5          =
\ C   /D3EF 
213-+ Il(B +q C)+ I(B-C) -C- I1D —IE -                     F.
If we take the arithmetical complements of 1 C,  1 D,  1 E,  1.F,
designating them by C', D', E', F', instead of the preceding
result, we have
21. — +  \(B+       +   1 (B - C  ) (B -   C) '  D' + El'+F',
only we must observe to subtract from the sum as many units of
the same order with the complements, as there are complements




276


Elements of Algebra.


taken, that is 4. When we have found the logarithm of the proposedj formula, the tables will show the number, to which this
logarithm belongs, which will be the value sought.
253. Logarithms are of the most frequent use in finding the
fourth term of a proportion.  It is evident, that if a: b:: c: d we
have
be
d =, whence  Id = 16 -+ 1 c-1-a;
a
that is, the logarithm of thefourth term sought is equal to the sum
of the logarithms of the two means, diminished by the logarithm of
the known extreme, or rather, to the sum of the logarithms of the
means, plus the arithmetical complement of the logarithm of the
known extreme.
254. If we take the logarithms of each member of the equation
- = -, which presents the character of a proportion, we have
a    C
lb -la     ld — lc (252);
whence it follows, that the four logarithms
la. lb:lc. ld
form an equidifference (223.)
The series of equations,
b    c    d    e
^a - 6 =-      d&c. (231)
leads also to
lb —a -la                     c l -lb =  d — c =  e --   d, &c.,
and hence we infer, that the progression by quotients,
-- a: b: c: d: e, &c.
corresponds to the progression by differences,
-    la. lb. lIc. ld. le, &c.,
and, consequently, the logarithms of numbers in progression by
quotients, form a progression by differences.
255. If we have the equation bz = c, we may easily resolve it
by means of logarithms; for as  bz is equal to z lb, we have
1 c 
z b = 1 c, and, consequently, z = 1-,. The equation bc  d may
be resolved in the same manner; making c-. = u, we have
l d          I
bu = d, ulb =Id, u —,  or ld
or IlbI




Theory of Exponential Quantities and of Logarithms. 277
Again taking the logarithms, we find
zlc =I       l -  l d-l 11b  and   z =        -   IIc
In this last expression, l b represents the logarithm of the logarithm of b, and is found by considering this logarithm as a number.
z
The quantities, bZ, b6, and all which are derived from them, are
called exponential quantities.
Application to finding the Logarithms of Literal Expressions.
1. log    = logf + logg -   log c -  log d.
2. log am bx cP -- m log a + n log b + p log c.
am b-'n
3. log   d     =mloga-nlog b-p log c - q log d.
m    P
4. log ab    q c =     log a  log b + log c.
5. log   am* b- c --   log a   log b + -  log c.
nn 
6. log  a   -   log a + n   log c - log b - log d.
b -i -''  -     n
7. log (a + b)   = n log (a + b) + m log c —log (c + d)
(c + d) vd3
-  log d.
8. log ( +   )     -         m  log (a + b").
9. log        -      1log (a + b).
v(a+b)
10. loa       lg (a -         +  ) + m log (a -   x).
11. A log a = log al.
a1n bm
12. n log a + m log b-p log c = log.
13. n log (a + y) + logc — m log (a- y)   log  (a + y)
(a -yja
14. 1 log (2 a + 3 b) -  log c = log  (2 a-  3b)
iVc2




278


Elements of tlgebra 


Application to finding the Logarithms of Numerical Expressions.
1. log (93 X 3514) = 5.51428.
2. log (628 X 493) = 5.49080.
5. log 4 = 0.09691.
4. log 379 = 0.74596.
5. log 1-4 - 0.66900.
6. log 15-= 1.19728.
7. log  1   76 = 3.24757.
138
S. log 315 - 7.15681.
9. log 527 = 18.87219.
10. log (,)14 = 5.15167.
11. log (12)32 = 12.16675.
12. log /5 = 0.34948.
13. log /73567 = 2.43334.
3
14. log /135 = 0.71011.
100
15. log /13= 0.011139.
7
16. log v/l   = 0,08200.
17
17. log v(954)12 = 2.10321.
Calculation of Numerical Expressions by Logarithms.
7
1. 8 = 1.34590....
4
2. V35246 = 13.70179....
5
3. v/3 = 0.95932...
7
4. /171 -   - 1.19074....
13
5. v/338 = 1.14605....
6. ( )21 = 11.86322...
7. (2)9 = 11767.35....
8. (4)123   3.16810....




Formulas relating to Interest.


279


5   4
9. v(v 6) = 1.29569....
1716.5
10. 253  7165 - 2016.914...
8       6
11. V(21 + Vi9) =   1.4768....
Questions relating to the Interest of Mloney.
256. THE principles of progression by quotients and of logarithms will be found to occur in the calculations relating to interest.
To understand what I have to offer on this subject it must be recollected, that the income derived from a sum of money employed in
trade, or in executing some productive work, will be in proportion
to the frequency with which it is exchanged in either case. Hence
it follows, that he, who borrows a sum of money for any purpose,
ought, upon returning this money at the expiration of a given time,
to allow the lender a premium equivalent to the profits, which he
might have received, if he had employed it himself.  Such is the
view in which the subject of interest presents itself. In order to
determine the interest of any sum, we compare this sum with 100
dollars taken as unity, having fixed the premitlum, which ought to
be allowed for this last at the end of a particular term, one year for
example. I shall not here consider those things, which, in the
different kinds of speculation, occasion the rise and fall of interest;
this belongs to the elements of political and commercial arithmetic,
which should be preceded by some account of the doctrine of
chances. My object in what follows is simply to resolve certain
questions, which refer themselves to progression by quotients.
To present this subject in a general point of view, I shall suppose the annual premium, allowed for a sum 1, to be represented
by r, r being a fraction; it is evident, that the interest of a sum
100, for the same time, will be 100 r, that of any sum whatever a
will be denoted by a r; if we designate this last by a, we have
cc = a r.
By means of this formula, it is easy to find the interest of any sum
whatever, when that of 100 or of any other sum, for a known time,
is given; questions of this kind belong to what is called simple
interest.
257. But if the lender, instead of recieving annually the interest




280


Elements of.2lgebra.


of his money, leaves it in the hands of the borrower to accumulate,
together with the original sum, during the following year, the value
of the whole at the end of this year may be found in the following
manner. The original sum being a, if we add to it the interest
a r, it becomes at the end of the first year
a + a r    a (1 + r),
Now if we make
a (1   r)= a,
the interest of the sum a' for one year being a' r, that of the sum
a (I + r) will. be, for a second year, a r (I + r); and as, at the
end of the first year, the principal a augmented by the interest,
becomes a (1 + r), the principal a' amounts at the end of the
second year, to
a' (1 + r) = a (1 + r)2 - a",
If the lender does not now withdraw the sum a", but leaves it to
accumulate during a third year, at the end of this, it will become,
according to what precedes,
a/ ( -+ r) = a (1 + r)3    a"'.
It will be readily perceived, that a"' will become at the end of the
fourth year
a"'(1 + r)    a (1 + r)4,
and so on; and that, consequently, the sum first lent, and the
several sums due at the end of the first, second, third, fourth, &c.
years, form the following progression by quotients 
a: a (- +r):a (a  + r)2: a (I + r)3: a (1 + r)4: &c.
of which the quotient is I + r, and the general term
a (1 + r) =
the number n representing the number of years, during which the
interest is suffered to accumulate,
If the rate of interest be 5 per cent., for example, that is, if for
100 dollars during one year 105 dollars are paid back; we have
100r= 5, or r = T           2, - -,  and  1 + r =   -.
If we would know to what the sum a amounts, when left to accu,
mulate during 25 years, we have
n = 25, and a \2 -instead of the original sum. The 25th power of a may be easily




Formulas relating to Interest.


281


found by means of logarithms, since we have (252)
1, 212 _ 251 2 = 25 (121 -120)        0,5297322,
which gives
(51\25
( 21)  = 3,386  nearly,. = 3,386 a;
and hence it may be readily seen, that 1000 dollars will in this
way amount at compound interest to 3386 dollars, at the end of
25 years.
If the suni lent were for 100 years, we should have
21\100
s =a(W0)      =131 a
nearly; thus 1000 dollars would produce, at the end of this period, a sum of 131000 dollars nearly. These examples will be
sufficient to show with what rapidity sums accumulate by means of
conpc(und interest.
258. The equation
a (1 + r)',
gives rise to four questions; the first, which is to find., when
a, r, and n, are known, presents itself, whenever we seek the
amount of the principal at the end of a number n of years. I
have already given an example of this.
The second, which is to find r, when a,., and n, are known,
occurs whenever it is required to determine the rate of interest by
means of the original sum, the whoae amount that has become due,
and the time during which it has been accumulating; we have in
this case
1 + r =    At.
1 a
The third, which is to find a, when A, r, and n are known, the
formula for which is
A
a   (1 + r)n'
has for its object to determine the principal, which it is necessary
to emnploy in order to be entitled after a number n of years, to a
Sum I.
The fourth, which is to find i2, when A, a, and r are known,
can be resolved only by means of logarithms (238, 252). Taking.Ag.             36




282


Elements of Algebra.


the logarithm of each member of the proposed equation, we have
d=la+nl(l +-r),
whence
1A -la
1 (1 +  r)
By means of this last equation we determine how many years the
principal a must remain at interest in order to amount to a sum S.
To illustrate this by an example, I shall suppose that it is required to find the time in which the original sum will be doubled,
the rate of interest being 5 per cent.; we have.= 2 a, 1./ =     a+ -12,
and, consequently,
12       12       0,3010300 
1 -   - 1 21-120 - 0,0211893 - 
nearly.
259. The following question is one of the most complicated,
that we meet with relating to this subject. We suppose, that the
lender during a number n of years, adds each year a new sum,
to the amount of this year; it is required to find what will be the
value of these several sums, together with the compound interest
that may thence arise at the expiration of the term proposed. Let
a, b, c, d,.... k, be the sums added the first, second, third,
fourth, &c. years; the sum a remaining in the hands of the borrower during a number n of years, amounts to
a( + r)";
the sum b, which remains n - 1 years only, becomes
6(1 +r)n-l,
the sum c, which remains n -2 years only, becomes
c(1 + r)-2,
and so on; the last sum, k, which is employed only one year,
becomes simply
k(1 +r);
we have, therefore,
A = a(l + r)n + b (1 + r)n-1 + c (1 + r)'- 2.... +k(l + r).
By calculating the several terms of the second mumber separately,
we obtain the value of A.




Formulas relating to Annuities.        283
The operation is very much simplified when
a =b =c-d.... =         k,
for in this case we have.~ = a (1 + r)i + a(1 + r)- + a(l + r)-2...+ a(1 + r) 
the second member of this equation forms a progression by quotients, of which the first term is a (1 - r), the last term a (1 -- r)n,
the quotient 1 -- r, and the sum, consequently,
a (l+r)+'- a(1+ r) (
we have, therefore, in this case,
- a ( +r) [( + r)n-1]
r
This equation gives rise also to four questions corresponding to
those mentioned in connexion with the equation,   - a (I + r).
260. By reversing the case we have been considering, we may
represent those annual sums, or sums due at stated intervals,
called annuities; here the borrower discharges a debt with the
interest due upon it, by different payments made at regular periods. These payments, made by the borrower before the debt
in question is discharged, may be considered, as sums advanced
to the lender toward the discharge of the debt, the value of
which sums will depend upon the interval of time between the
payment and the expiration of the annuity. Thus, if we represent each sum by a, the first payment, which will take place
n- 1 years before the expiration of the term of the annuity,
referred to this time, is worth a (1 + r)'-l; the second referred
to the same epoch, is worth only a (1 +   r)'-2; the third,
a ( 4- r)-3, and so on to the last, which amounts only to the
value of a. But on the other hand, the sum lent being represented
by., will be worth in the hands of the borrower, after n years,.J (1 + r)", which must be equal to the amount of the several
payments advanced by him to the lender; we have, therefore,
A,(1 + r)n    a( +r    + n-+ a ((1 + r)n-2 +a( +)n3... +a,
or taking the sum of the progression, which constitutes the second
member
(1 + r)" = a[( +) —1]
r




284


Elements of dlgebra.


an equation, in which we may take for the unknown quantity, successively, the quantity., which I shall call the value of the annuity, because it is the sum, which it represents, the quantity a,
which is the quota of the annuity, the quantity r, which is the rate
of interest, and lastly, the quantity n, which denotes the term of
the annuity. In order to find this last we must have recourse to
logarithms. We first disengage (1 + r)n, which gives
(t + r)n =then taking the logarithms, we have
nl(l + r)    la -     (a -   - r),
whence
l a - 1 (a - A r)
l(1 +r)
261. To give an instance of the application of the above formulas, I shall take the following question;
To find what sum must be paid annually to cancel in 12 years a
debt of 100 dolls. with the interest during that time, the rate of
interest being 5 per cent.
In this example the quantities given are
A =     100, n =  12, r =20
and the annuity a is required to be found; resolving the equation
J (1 + r)n _ a    + r)- 1]
with reference to the letter a, we have
A r (I + r)Ar(l+ r)n
The values of the letters,., r, and n, are to be substituted in this
expression; and it will be found most convenient in the first place
to calculate, by the help of logarithms, the quantity (1 + r)",
which becomes (}i)12; and
(2 )12   1,795S6.
By means of this value we obtain *
100. '. 1,79586  5. 1,79586
a =   1,79586 - 1       0,79586;
and, determining the values of this last expression either directly
or by means of logarithms, we find
a = 11,2826;




Formulas relating to Annuities.


285


an annuity of 11,28 dolls., therefore, is necessary to cancel in 12
years a debt of 100 dolls., the rate of interest being 5 per cent.
262. I am prevented firom entering into further details on this
subject by the limits I have prescribed myself in this treatise; I
will merely add, therefore, that in order to compare the values of
different sums, as they concern the person, who pays or receives
them, they must be reduced to the same epoch, that is, we must
find what they would amount to when referred to the same date.
A banker, for instance, owes a sum a payable in n years; as an
equivalent he gives a note, the nominal value of which is represented by b, and which is payable in p years, the first sum at the
time the note is given, is worth only  a., because it must be
considered as the original value of a principal, which amounts to a
at the expiration of n years; the sum b, for the same reason, is
worth at the time the note is given 1; the difference
(1  1 —?)P
a           b
( + r)n    (1 + r)P
represents, therefore, according as it is positive or negative, what
the banker ought to give or receive by way of balance; if this balance is not to be paid until after a number of years denoted by q,
c representing its value at the time the exchange is made, it will
amount at the expiration of this term, to
c (t + r)q;
so that it will be equivalent to
((     -            (     +,- —,=   (I + )(r)-,- b (L + r)-P.
The several sums, a,b,..... k, in art. 259., were reduced to
the time of the payment of the sum./, and in art. 260., each of the
payments, as well as the sum A, was referred to the time, when
the annuity was to cease
Questions relating to Interest and Annuities.
1. A capital of 50001. stands at 4 per cent. compound interest.
What will it amount to in 40 years?    Ans. 24005.1031.
2. What will 32001. amount to at 3 per cent. in 80 years?
Ains. 34050.841.




286


Elements of.lgebra.


3. How long must a capital a remain at the interest p to become
as much as a capital a' at the interest p' for n years?
~ns. log a' + n log p' - log ayears,dns. -6-*.        -ogp - years.
log p
4. How long must 36001. remain at 5 per cent. compound
interest so that it may become as much as 50001. at 4 per cent. for
12 years?                          J/ins. 16 years, nearly.
5. What is the amount of capital which at the interest p for n
years is of equal value with a capital a' for n' years at the interest
p?./Ans. log a = log a' + n'. log p' - n log p.
6. What is the amount of a capital which stands at 4 per cent.
that 15 years hence it may be equal in value with 45001. at 6 per
cent. for 9 years?                   Ins. 42211., nearly.
7. What is the rate of interest that a capital a in n years may
be equal to a capital a' in n' years at the interest p'?
ns. lg   log a' + n' log p' - log a
Ans. log p — n
~ -      ~n
8. How long must a capital stand at 4 per cent. compound
interest that it may double itself; and how long that it may be
tripled.
Ans. It doubles itself in between 17 and 18 years, and triples
itself in between 28 and 29 years.
9. An usurer lent a person 6001. and drew up for the amount a
bond payable in 3 years bearing no interest. What did he take
per cent. reckoning compound interest?.ns. 10 per cent., nearly.
10. A capital of 8001. increased in the space of 6 years to
36001. What did it gain per cent.?
Ans. 28- per cent., nearly.
11. A person enjoys an annuity of 5001. for six years. How
much ready money can a person give him for this annuity, calculating 3- per cent.?.Ans. 26641. 5s. 10d.
12. What is the present value of an annuity of 3501. assigned
for 8 years at 4 per cent.?   Ans. 23561. 9s. 2d., nearly.
13. A debt due at this present time amounting to 12001. is to
be discharged in seven yearly and equal payments. What is the
amount of these payments if the interest be calculated at 4 per
cent.?                                Ans. 2001., nearly.
14. A person wishes to obtain an annuity of 20001. for 345801.
For how many years can this annuity be granted him, computing
the interest at 4 per cent.?,     Ins. About 30 years.




NOTES.
(Referred to Page 86.)
IN articles 66 and 75 I have interpreted the negative solutions by
the examination of the equation, which they immediately verify, as I
had done before, and this method appeared to me always exact, as
the object is merely to show, that these solutions have a rational
sense, since they resolve questions analogous to the one proposed;
but there are often several ways of forming these questions, and the
following, which was communicated to me by M. Francais, a distinguished geometer, Professor at the School of Artillery of AMayence,
seemed to me more simple, than tlat given in these Elements.
l" Ie thinks, that we ought to leave out of the enunciation of the
question of art. 65. the idea of the departure of the couriers, and to
suppose them to have been travelling from an indefinite time; the
question then would be stated thus.  Two couriers travel the same
route in the same direction C' A B C (page 77); after they have proceeded, each a certain time, one Jinds himself in A at the instant that
the other is in B; their distance and rate of going are known; it is
asked at what point of the route they will encounter each other?
This enunciation leads to the same equation, as that of art. 65.;
but "the continuity of the motion being once established, the negative solution admits of an explanation without the necessity of changing the direction of one of the couries. Indeed, since their motion
does not commence at the points A and B, but both, before arriving
at these points, are supposed to have been going in the same manner
for an indefinite time from C' toward B, it is easy to conceive, that
the courier, who at this point is in advance of the one at A, who
travels slower, must at a certain time have been behind him and overtaken him before his arrival at the point A. The sign - then indicates (as in the application of Algebra to Geometry) that the distance
AR' is to be taken in a direction opposite to AR, which is regarded
as positive. The change to be made in the enunciation, to render
the negative solution positive, is reduced to supposing, that the two




288


JNotes.


couriers must have come together before their arrival at the point A,
instead of its taking place afterward."
Indeed, when we place the point R' between A and C, instead of
putting it between A and B, we find AB = BR' - AR, whence
results the equation y - x = a, instead of x - y - a, which we
first obtained; and there is no need of changing the sign of c, the
second equation remaining z-  -.
c — -
M. Francais applies not less happily these considerations to the
case of art. 75., by substituting, for the couriers, moveable bodies,
subjected to a continued motion commencing from an indefinite time.
He enunciates the problem thus; " Two moveable bodies are carried
uniformly in a straight line CB (page 85) one in the direction BC,
and the other in the direction CB with given velocities; that, which is
carried in the first direction, is found in B, a known number of hours
before the other has arrived at A; it is asked, at what point of the
indefinite straight line BC their meeting takes place?
" The solution z =  -48mls. implies, that the two moveable bodies
met at the point R, before that, which is carried from C towards B,
had reached the point A, and that the second, which moves from B
towards C, was at the point C, where he is found when the other is at
the point A."
The position assigned to the point R, verifies itself by observing,
that there results from it AC = BC - AB = c d - a, instead of
a + c d, as first obtained (page 85,) and, consequently,
x    cd- a    x
b        c
an equation which gives x - 48.
In this manner there is no change to be made in the direction of
the motion; indeed there is a difference in the circumstances of the
problem, and as I said before, this proves, that there are several physical questions corresponding to the same mathematical relations.
But the enunciations, here given, have the advantage of not breaking
the law of continuity, and this is derived from the consideration of
lines, which represent in a manner the most simple and general, the
circumstances of a change of sign in magnitudes. (Sec the Elementary Treatise of Trigonometry and Application of Algebra to Geometry.)




(Note referred to Page 203.)


IT may be thoughlt, that in order to discover the roots of any equation of the fourth degree
X4 + 1P 3 + q( a2 + r X + S - 0,
it would be sufficient to compare it with the product of article 183.,
observi!g to put equal to each other tlhe quantities by which the same
power of x is multipllied  and it is in thiis manner that most elementary writers think to demonstrate, that an cyquation of any degree
whatever is the /prodlrt of' as ml,,y simple factors, as there are units
in thle c.ponent of its de.r)e. It will be seen by what follows, that
the reasoning by which tlis is attempted to be proved, is defective.
We stated the proplosition with qualification in article 182., because it
is necessary, in order to estabiish it unconditionally, to show that an
equation of whatever degree has a root, real or imaginary, which is
not easily done in an elementary work, and which happily is not
necessary. Some remarks relating to this subject may be found in
the Supllement.
By forming the equations,
a - b - c -       = p,
a b + a c + a d(   bc + b d +   cd_ q,
- a b c - a b d - a c d - b c (-,
a b c e1  s,
in order to deduce from them the value of tlle letters, a, b, c, d7, the
roots of the proposdl ecqlation, tle calculationll wolll be very complicated, if, in tlle determlillatiol of tlle unknown quantities, a, b, c, d,
we adopt the metlod of article 78; but if we multiply the first of the
above equations by ea3, the second by aC", tlhe third by a, and add these
three products to the fourtl, member to member, we shall have
-  (t4 =-  ) (3 + q a` -- Tr a --  s,
from which we derive, by simple transposition,
-T- 1p +  -,  q t -f- + a r - + s - 0.
This equation contains only a, but it is entirely similar to the one
proposed. The difficulty of obtalining, therefore, is the same as that
of obtaiining x.
(4 Thus," says Castillon (Metm. td  Berlin, annnee 17S9,) "c it is
shown in every work on algebra, that an equation of any degree we
please, is formed of several simple binonials, but it is not so evident
that an equation, formed by the multiplication of several simple binomials, can have such coefficients as we please."
Alg                37




X90.Notes.
If, instead of multiplying the first three equations in a, b, c, d, by
a3, a2, and a, respectively, we multiply them by b3, b", and b, or by
c3, C2, c, or d3, d2, d, and add the products to the fourth equation, we
shall have in the first
- b4 _    Y 3 + q Vb + i- b + s,
in the second
-    C4 =  p3 + q c2  r c + s,
in the third
-       d4 = p d3 +  d2 + r d + s;
from which it follows, that we are conducted to the same equation in
the case of a, in that of b, &c. Indeed the quantities, a, b, c, d, being
all disposed in the same manner in each equation, it is not to be supposed that one should be determined by a different operation from
that of the others; and, in general, if in the investigation of several
unknown quantities, we are obliged to employ for each the same reasonings, the same operations, and the same known quantities, all
these quantities will necessarily be roots of the same equation.




QUESTIONS FOR PRACTICE
IN
LACROIX'S ALGEBRA.
1..iddition,./lrt. 18.
1. Add the quantities x + y z + 42 - 29 x  z - 9../Ins. 33 -28 x.
2. Add 147a~23b-a-b+2a..ns. 148 a + 22 b.
3. Add 1 11a  I1 a b + 11 a b c- I1Ia 1  a b -1 1 a b c.
dns. i-2 a b.
4. Add 43 a - 27c - 20 a ~ 7 c - 61 b - 21 a + 57 b
+ 20 C..lns. 2 a -4 b.
5. Add a + 9 d + a - 7 c + S x - a + 6 d + 6 c - 7 v
14 d.
Ans. a - c + d + x.
(6. Add 7   a b c +Ga b + 5 c- a b c + 21 x + 9 c -27 x y
+ 8 a b c + 10x - 93z +   10 b+  3 t x —2a b- c + 5 xy
- a b c + 33 z..ns. 13 a b c + 4 a b 6+13 c + 10 b + 62 x - 22 x y- 60z.
HI. Subtraction, Art. 20.
7. From   6 x - S y + 3
subtract 2x + 9y -2.        dns. 4x- 17y + 5.
8. From    5 x y -S
subtract-3~y~.                Ans. -S3y.x.-.9.
9. From   4xy   x    x-fxy
subtract  2xy +  + xy.
Ans. 2 x y - x -2.
I10. Fromi  5 x+ x - 8 -4 b
subtract 6xC) x 1 +4 b - x..ns.2+-     b 8 b.




292                questionsfor Practice.
I1. Frorn    148 a + 47 a b -23 a b c +1-x
subtract  99 a - 47 a b -  a b c - 2 + -4 x.
Jiqm. 19 a + 94 a 6 - 15 a b c + 3 - 5 x.
12. From     7 b - S c + -32  y      - 436- III c + a
subtract  - 500 b -22 a - 87 x y - 7 c..dns. 464 b + 23 a + I 10 c + 413 x y.
III.  Iiut1ilwlication,.lrt. 32.
13. Multiply 12 a x by 3 c.                /iIns. 36 a2' x
14. Multiply 3 x y y- S + 2 xyz by x y..ilhs. 3 x.2y2  Sxy+ 2x2 y2 Z.
15. Multiply 12 x2 - 4 y2 by - 2 xA.
sins. -24 x4 + 8 cc2 y2.
16. Ml1ultiply x3 + X2 y + x y-) + y3 by x - y..s7ns. c4 -_y4.
17. Multiply  cc2 +  y + y2 by x2 - x y +9y2.
sins. X4 + cc2 y2 + y4,
IS. Multiply 3c2 - 2 x y + 5 by cx 2+2cxy - 3.
sins. 3 xc4 + 4x3 y - 4 xr2 - 4x0c  2 + 16 x y - 15.
19. 1,V1ultiply 3 il + 2x y2 -+- 3 y3 by 2 c3 -3c2y2 + 5 3.
sins. 6C " - 5 x5 Y2 + 21crtY3  O  - 6  4+ it25 + 15y6.
1V. Division,./rt. 46.
20. Divide 0Iy - 15 y2 - 5.yby 5 y..ln s. 2c2   y 3y  1.
21. Divide 3 a2 - 15 + 6 a + 3 b by 3 a.
5 +
sins, a -    + 2 b
a        a
22. Divide 6c44- 96 by 3 x - 6.
sins. 2cc3 + 4 XC2 + 8cc+ 16.
23. Divide 48 X3 -_76 ax2 - 64 a2,cc + 105 a3by 2 x - 3 a.
sins. 24 c2 -2 acc x -35 a2.
V. Reduction of Fractions, sirt. 50 and 52.
24. What is the greatest common measure of c x +  2
c a2 + a2 z
sins. c + x.




Questions for Practice.


293


-;2- 1
25. What is the greatest common measure of -—?
bx/ + Y
J1ns. x + 1.
X2 — y,2
26. What is the greatest common divisor of '4 —./ns. x2 -y2;4 - b4
27. Reduce6_         to its lowest terms.
lins. x2 -  b2 gr. c. d. and    - lowest terms.
5 5        + - -- 10 4 x 4- 5.3 X2
28. Reduce                     -      to its lowest terms.
a3 x + '2 O ax2 -+ 2 a -x3 -- x4
2lns. a + x gr. c. d. au.d, -+   a~ x  lowest terms.
b'  ~( a-Z -V- 8 x:2 +   13
29           3  Rz          2..Reduce, and a +     to equivalent fractions having
a common denominator.
9a   Sax       ' c, 2 + 2  a24 
~0Ans.                 R dc and..
1 a2      x2 a + O 50
30. Reduce^   a —, and    -to fractions having a common
denominator.
3x + 3 -a  2ox-+2a3         6   +" x-~6a2
bn  ac (Id 
6I - {- fl  6a'  6x+6a  '    i and 6-6a
b   c     d
31. Reduce 2               t- and   to fractions having a common de32. What is the prodduct of- and           a
VI. Multiplication and Division of F ractions, Art. 51.
32  What is the product of and
a     a +- c
As. x2- a
a2+. ac'




294


Questionsfor Practice.


2 x 3 a b    3 a c
33. What is the product of, and 26
a    C       2b.dns. 9 a x.
34. What is the product of  6 b2 and xo + b9
b~~~  (I   J    YA-   b 
2:4 -4
's'P c-ftbC2'
x  xb
35. What is the quotient of  dividcdl by
39
lins. II~
2x
36. What is the quotient of  +  divided by  x
6 
4x~
37. What is the quotient of  4-       divided by
q e   -2 b x+b 62
X2 + b x.ns. x + -
x-, b
VII..Iddition and Subtraction of Fractionzs, /lrt. 53.
38. Add x +   Li to 3 x +       
3
Jim 4x  l10x- 17
2 x 72;    2     1
39. Add         and        together.
169 x, + 12        49 x   1
J~ns.          or 2 x +     + F
60+ 60
40. Add together 4w,,and 2 +
158x o        23 x
An 5or 3~2+   45
41. From        subtract            Ans. dx+ad-c
b           d'                    b d
3x  ~~~~~22 ~~~13x
42. From 3   subtract 2x                        ns
7         -C- I                          7
X 2 -x- a
43. From 3w +     subtract wn.ls. 2x+ C+ bx- ab
b 7; -b




Questionsfor Practice.


295


VIII. Problems in Simple Equations, Art. 82,
44. In 5 x -  15 = 2 x + 6 to find the value of x.
J1ns. x   7.
45. In 3y- 2 + 24      31 to find y.
/ins.  = 3.
46. In the eqlations +  + Sy    31 and y +    + 10 x =
192 to find x and y.
zins. x = 19 and y = 3.
47. Out of a cask of wine, which had leaked away one third,
21 gallons were drawn, and then being gauged it was found to be
half full: how much did it hold?,1ns. 126 gallons.
48. What two numbers are those, whose difference is 7 and
sum 33?                                  Jns. 13 and 20.
49. What number is that fiom which if:5' be subtracted, two
thirds of the remainder will be 40?:        Is. 65.
50. At a certain election 375 persons voted, and the candidate
chosen had a majority of 91 votes: how many voted for each
candidate?         Jns. 233 for one, and 142 for the other.
51. A post is -- in the mud, - in the water, and 10 feet above
the water: what is its lwhole length?.dns. 24 feet.
52. A man arriving at Paris, spent the first day 1 of the money
he brought with him, the second day -, and the third day 5, after
which lie had only 26 crowns left: how much did lie have on
arriving at Paris?.Rns. 120 crowns.
53. A horse said to a mule, if I give you one of my sacks we
shall be equally loaded, if I take one of yours I shall have twice as
much as you: how many sacks had each?
/lns. The horse 7 and the mule 5.
54. A man being asked how many crowns le had, replied, if
you add together a half, a third, and a quarter of what I have, the
sum will exceed the number of crowns I have by one: what was
the number?                                   sins. 12.
55. A privateer running at the rate of 10 miles an hour discovers a ship 18 miles off making way at the rate of 8 miles an hours:
how many miles can the ship run before being overtaken?.dns. 72 miles, or 9 hours.




296


Questions for Practice.


56. A hare is 50 leaps before a grey-hound, and takes 4 leaps
to the grey-hound's 3; but two of the grey-hound's leaps are as
much as tllree of the hare's: how many leaps must the grey-hound
take to catch tlhe hare?                        dns. 300.
57. A person being asked his age, replied, that - of his age
multiplied by i-' of his age would give a product equal to his age:
what was his age?                                Ans. 16.
58. A person has a lease for 99 years; and being asked how
much of it was already expired, answcred, tliat two tlirds of the
time past w'acl equal to four fifilis of tile time to colme: whlt was
the tilne )ast?.i:ns. 54 years.
59. Tlere is a fisl whose tail weighs 9!l)s., his head weiglhs as
muc1h a  is s tail andl Ihilf his boti, alld his body weighs as much
as his head and tail: xwhlit is the whole weight of tle fisl?
lIns. 72 lbs.
60. There is a certain numnber, consisting of two digits, the sum
of which  ligits is 5; and if 9 be added to the number itself the
digits will be inverted: what is the number?      lns. 23.
61. A person found, upon beginning the study of his profession,
that ] of Iis life hitherto had passed before lie cotmmenced his
education, 3 under a p)rivate teacher, -' at a public sclool, and four
years at the university: what was his age?.lns. 21 years.
62. To find a number such, tllat vwetler it be (diviled into two
or three equal parts, tile continued product of its parts shall be
equal to the same quantity.                       Jns. 6{.
63. A person has two horses and a saddle worth 501.: now if
the saddle be put on tle back of tle first horse, it will njake his
value double tlat of the second; but if it be put on tlhe back of the
second, it will make his value triple that of tle first: what is the
value of each horse?        ins. one 301. and the other 401.
64. To divide the number 90 into four such parts that if the
first be increased by 2, the second ditminished by 2, thle third
multiplied by 2, and the fourtht dividedl by 2, the sumn, difference,
product, and quotient, shall e;ch equal thle same quantity.
Jins. The parts are 1S, 22, 10, and 40.
65. By his will a father disposed of his property as follows;
namely, to his oldest son he gave 100 dollars of the property, and




Questions for Practice.:


297


a tenth part of the residue; to the second, 200 dollars and a tenth
part of the residue; to a third, 300 dollars and a tenth part
of the residue; and so on to the last, always increasing the
sum first paid out by 100 dollars. It appeared that the portions of
all the children were alike. Required the value of the property,
the number of children, and the portion of each child..Ins. The estate was 8100 dollars, the children 9, and the portion of each 900 dollars.
66. A and B have the same income; A is extravagant and
contracts an annual debt amounting to I of his income; but B
lives upon 4 of his; at the end of 10 years, B lends A money
enough to pay off his debts, and has 1601. to spare: what is their
income?                                      dins. 2801.
67. A person passed 1 of his age in childhood, y- in youth, 4
and 5 years besides in matrimony, at the end of which time he
had a son, who died 4 years before his father, and reached only
half his father's age; at what age did the father die?
Ains. 84.
68. A shepherd, driving a flock of sheep in time of war, meets
with a company of soldiers, who plunder him of half his flock and
half a sheep over; and he receives the same treatment from a
second, third, and fourth company, each succeeding company
plundering him of half the flock the last had left and half a sheep
beside, insomuch that in the end he had only 8 sheep left: how
many sheep had he in the beginning?           Ins. 143,
69. A person, fifteen years after he was married, being asked
the age of himself and of his wife at the time of their marriage,
replied, that he was then thrice as old as his wife, but that now he
was only twice as old: what were their ages?.ins. He was 45 and she 15.
70. It is required to find two numbers such that the first added
to half the second shall make 20, and the second added to one
third the first shall also make 20..ns. 12 and 16.
71. Two travellers, distant 154 miles, set out at the same time
to meet each other, the one proceeding at the rate of 3 miles in 2
hours, and the other at the rate of 5 miles in 4 hours: how long
and how far did each travel before they met?
Ans. The time was 56 hours, the speces travelled 84 and 70
miles..ig.             38




298


Questions for Practice.


IX. Formation of Powers and Extraction of Roots.
72. What is the square root of 9 x2? (.rt. 122.)
Ans. 3 x.
73. What is the square root of 94 y?     ns. 3 z
4a3             2aV~
74. What is the square root of a4 +- 4 a3 X +- 6 a  2 + 4 a x3 -
x4? (Art. 124.)                  Ans. a2 + 2 a x + x.
3      xa   1
75. What is the square root of a4 - 2 x3 + - 2 - 2 -+  6.Ans. xa2 -  + i,
76. What is the third power of - 8 x2 y3? (Art. 127.)
Ans. - 512 x6y9.
77. What is the fifth root of- 32 x5 y10? (rt. 129.)
Ans. -2x y2.
78. What is the fourth power of xa - a? (rt. 141.)
AIns. a4 - 4x3 a +- 6x 2 a2 - 4 x a3+ a4.
79. What is the square of a2 + 2 a x + -2? (Art. 145.)
A/ns. a4 + 4 a3  + 6 a2 x; + 4 a x3 + X4.
80. What is the cube root of x6 - 6 a5 + 15 x4 - 20 x3 +
15xg2-  6 x-+ 1   (Art. 155.)       Ans. 2 - 2 x + 1.
81, What is the fifth root of 32 5- - 30 x4 +- 80 3 - 40 xa
l+ 10x-, 1?                             iAns. 2 x- 1,




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