ALGEBRA FUNDAMENTAL OPERATIONS The primary operations in algebra are the same as in common arithmetic - namely, addition, subtraction, multiplication, and division ; and from the various combinations of these four, all the others are derived.

I. Addition.

Rule. Add together the coefficients of the quantities, prefix the common sign to the sum, and annex the letter or letters common to each term.

Rule. Add the positive coefficients into one sum, and the negative ones into another; then subtract the less of these sums from the greater, prefix the sign of the greater to the remainder and annex the common letter or letters as before Case 3. To add unlike quantities.

The reason of the rule for subtraction may be explainc,1 thus. Let it be required to subtract 2p - 32 from m+n. If we subtract 2p from m+ n, there will remain m+ n - 2p, but if we are to subtract 2p - 3q, which is less than 2p, it is evident that the remainder will be greater by a quantity equal to 3q; that is, the remainder will be IR n - 2.o + 3q, hence the reason of the rule is evident.

III. .3fultipl-leation, This rule, which is given by Diophantusl as the definition of + and -, may be said to constitute the basis of algebra as distinct from arithmetic.

If we admit the definitions given above, the rule may be demonstrated in the following way :-- + a x + b = + ab is assumed.

+a x - b will have the same value, whatever - b may be connected with, as it has when - b is connected with + b (Law 1).

Now +a x ( +b-b)= +ax +0=0 (Def.) But +ax( +b-b1= +ax +b, and +ax -b (Law 2).

+a x +b and +a x - b make up 0 • i.e., + ab and + a x -b make up O.

Now +ab-ab= 0, :. +a x - b = -ab.

Similarly -ax -b= +ab.

The examples of multiplication may be referred to two cases ; the first is when both the quantities are simple, and the second when one or both of them are compound.

Case 1. To multiply simple quantities.

Rule. Find the sign of the product by the general rule, and annex to it the product of the numeral coefficients; then set down all the letters, one after another, as in one word.

Case 2. To multiply compound quantities.

Rule. Multiply every term of the multiplicand by all the terms of the multiplier, one after another, according to the preceding rule, and collect their products into one sum, which will be the product required.

When several quantities are multiplica together so as to constitute a product, each of them is called a factor of that product : thus a, b, and c are factors of the product abc ; also, a + x and b - x are factors of the product (a + x) . (b - x).

The products arising from the continual multiplication of the same quantity are called powers of that quantity, which is called the root. Thus aa, acia, aaaa, &c., are powers of the root a. These powers are commonly expressed by placing above the root, towards the right hand, a figure, denoting how often the root is repeated. This figure serves to denominate the power, and is called its index or exponent. Thus, the quantity a being considered as the root, or as the first power of a, we have aa or a2 for its second power, aaa or a3 for its third power, aaaa or a4 for its fourth power, and so on.

The second and third powers of a quantity are generally called its square and cube.

By considering the notation of powers, and the rules for multiplication, it appears that powers of the same root are multiplied by adding their exponents. Thus a x a3 = a4, also x3 x x4= x7; and in general a"' x = a"'+".

When the quantities to be multiplied appear under a symmetrical form, the operation of multiplying them may sometimes be shortened by detached coefficients, by symmetry, and by general considerations suggested by the particular examples under consideration.

Ex. 1. Multiply x4 - 3x3 + 2x2 - Tx + 3 by x2 - 5x + 4. Here the powers of x occur in regular order, so that we need only write down the coefficients of the several terms during the operation, having it in our power to supply the x's whenever we require them ; we write, therefore, - The last line (for which the result might have been written down in full at once) is equivalent to x6 - 8x5 + 21x4 - 29x3+ 46x2 - 43x + 12 .

When any terms are wanting, they may be supplied by zeros ; thus, Ex. 2. Multiply x4 - 7x3+ x - 1 by X3 - X + 2.

We may take advantage of symmetry by two considerations either separately or combined.

(1.) Symmetry of a Symbol.

Ex. Find the sum of (a + b - 2c)2 + (a + c - 2b)2 + (b + c - 2a)2.

Here a2 occurs with 1 as a multiplier in the first square, with 1 as a multiplier in the second square, and with 4 as a multiplier in the third square, Gat is part of the result ; ab occurs with 2 as a multiplier in the first square, with - 4 in the second, and with - 4 in the third - Gab is part of the result.

But a2, b2, c2, are similarly circumstanced, as also ab, ac, be ; hence the whole result must be G(a2 + L2 + c2 - ab - . ac - be).

(2.) Symmetry of an Expression.

Ex. Find the sum of (a + b + c) (x + y + 2) + (a + b - c) (x- fFirst, the product of (a + b + c) by x + y + z is to be found by multiplying out term by term.' It is ax + ay + az + bx + by +bz + cx + cy + cz.

The product of (a + b - c) (x + y - z) is now simply written, down from the above, by changing the sign of every term which contains one only of the two quantities affected with a - sign, i.e., in this case c and z.

Lastly, the four products may be arranged below each other, the signs alone being written down ; thus, and the sum required is therefore 4ax + 4by + 4cz.

General Considerations.

Now a, b, c are similarly involved in (a + b + c)3 ; .'. Ls and c3 must appear along with a3, 3,2c, 3b2a, &c., along with 3a2b, and hence we can at once write down all the terms except that which contains abc. To obtain the coefficient of abc, we observe that if a, b, and c, are each equal to 1, (a + b + c)3 is reduced to 33 or 27. In other words, there are 27 terms, if we consider 3a2b and every similar expression as three terms; and as the terms preceding abc are in this way found to be 21 in number, we require Gabe to make up the full number 27; It is desirable to introduce here some examples of the application of the process of the substitution of a letter for any number or fraction to the properties of numbers, inequalities, &c.

Properties of Numbers.

Ex. 1. If unity is divided into any two parts, the difference of their squares is equal to the difference of the parts themselves.

Let x stand for one part ; 1 - x for the other.

i.e., the difference of the squares of the parts is equal to the difference of the parts.

Ex. 2. The product of three consecutive even numbers is divisible by 48.

Let 2n, 23t+2, 2n+4, be the three numbers their product is 8n(n + 1)(n + 2). Now, of three consecutive numbers, n, v.+ 1, n + 2, one must be divisible by 2, and one by 3, n(n + 1)(n + 2) is divisible by 6, whence the proposition.

Ex. 3. The sum of the squares of three consecutive odd numbers, when increased by 1, is divisible by 12, but never by 24.

Let 221, - 1, 2n +1, 2n + 3, be the three odd numbers.

The sum- of their squares when increased by 1 is 12n2+ 12n+ 12 =12(n2+n +1)=12(n. gt + 1 +1).

Now, either n or 77, + I is even, 2/(21, + 1) + 1 is odd ; hence the sum under consideration is 12 times an odd number, whence the proposition.

Additional Examples in Symmetry, (C.c.

Ex. 1. (a+S+c)2+(a+1) - c)2+(a+c-5)2+(b+c - a)9 = 4(a2+ 52+ c2).

This is written down at once, from observin,n that a2 occurs in each of the four expressions, and that tab occurs with a + sign in two, and with a - sign in the other two. There is no other form.

Ex. 2. (a+b+c)3+(a+b - c)3+(a+c - b)3+(b+c - a)3 2(a3+ 53+ c3)+6(a2b + a2c+ Ida+ b2c+ Oa+ e25)-12abc. 1st, a3 occurs + in three, and - in one term.

2d, 3a2b occurs + in three, and - in one term.

3d, When a, 5, c are all units, the number resulting is 30; there are 30 terms, and as (1st) and (2d) make up 42, there fall to be subtracted 12, i.e., the coefficient of abc is - 12.

Ex. 3. (ax + by + cz)2 + (ax+ cy +bz)2 +(bx + ay + ez)2 + (bx+ cy + az)2 + (cx + ay + bz)2 + (ex +by + az)2 = 2(a2+ 52 + ,2) (x2+ y2 + 12) + 4(a b ac+bc)(xy +xz+ yz).

Ex. 4. The difference of the squares of two consecutive numbers is equal to the sum of the numbers.

Ex. 5. The sum of the cubes of three consecutive numbers is divisible by the sum of the numbers.

Ex. 6. If x is an odd number, x5 - x is divisible by 24, and (x2+3) (x2+ 7) by 32.

Ex. 7. If (pq_ r)2 4(p2 g)(pr q)2 = then will 4(p2 - q)3= (2p3 - 3p1+2.)2, and 4(72 - P1)3= (273 - 3Pqr +r2)2.

Let the left hand side equal the right + u ; then multiplying out,

Now each of these three terms is a positive quantity, if it he not zero, and as the sum of three positive quantities cannot be equal to zero, it follows that each term must be separately equal to zero, The demonstrations of inequalities are of so simple and instructive a character, that a somewhat lengthened exhibition of them forms a valuable introduction to the higher processes of the science. In all that follows under this head, the symbols x, y, z stand for positive numbers or fractions, usually unequal.

Because (x - y)2 is +, whether x be greater or less than y, it follows that x2 - 2xy+ y2 is +, i.e., is some positive number or fraction, It will be remarked that wlp.::a x and y are equal, the inequality rises into an equality, and this is common to all inequalities of the character under discussion.

Ex. 6. The arithmetic mean of any number of quantities (all positive) is greater than the geometric.

(The arithmetic mean is the sum of the quantities divided by their number ; the geometric is that mot of their product which is represented by their number.) Let the quantities be denoted by xj, x2, x3.....-„, the numbers 1, 2, 3, placed under the x, indicating order only, so that x1 may be read the first x, x2 the second x, Sc. Exx14-x„ ample 1 gives - ,/x1x2, if we suppose the x and y of that example to be „,/x1, ,,/x2 of the present.

In the same way we prove the proposition for 8, 16, or any number of quantities which is a power of 2.

For any other number, such, for instance, as 5, the following process is employed : - The number is made up to 8 by the insertion of three quantities, each equal to the arithmetic mean of the other five, viz., Call this quantity y; then y3>x,x2 . . .

xi+x3+ ' • • x.s y or > 2' 3` 4'4 a• 5 Col% As a particular case, x3+ y3+23>3xyz.

Ex. 7. Given xix, ... x,, = y", to prove that (1 + x1) (1+x„)... (1 -1-x„)>(1+y)'.

The demonstration will be perfectly general in fact, though limited in form, if we suppose the number of quantities to be 5; in which case, x1x2 .. . x5= y5.

(1+ x3) (1 + x4)>(1 + (1 + x2)(1 + y) >(1 + Vx,y)2 (1+0(1+Y) = (1+ 's/Y!/)2 Multiplying these products together, and combining the right hand factors two and two, (1 +x,)(1 + x,) (1+;)(1 > -((1+ Nix1x2)(1 + NIX74)(1+,rir7y) (1 +y))2 > ((1 + ,YX1X,X.,X4)(1 + > (1 + 41x,x,x,,x4x.,y3)2 >(1 + (1 + xl) (1 + x2) . . . . (1 + x)> (1 + y)5 .

Ex. 8. If the sum of n fractions makes up 1, the sum of their reciprocals is greater than the square of their number.

Let xl+x2+ x„=1, then, - +- + . . . 1 + >7?, 1 (example 6).

xi x, x„ x1x2 . . . x„ But tiix,x3 . . . x„.< x3+ • x" (example 6)< 7?, >1/, Vxix2 ... x,, whence 1 - +-1 + ±-1 >n2.

x, x2 s„ x2"n+1(„ n+1( x+x3+ . . . x2.--i < Vw, + :7" 21/, nn Let the numerator and denominator of this fraction be designated by N and D. N may be divided into pairs of terms, at the same distance from either end, viz., 1+ x2", x2 + &c., with or without a middle term, each of which (after .1+:0") is, by example 4, less than that quantity; the middle term, if there be one, being less than (1 +x2"), in either case N<91"---4)--.1(1 + x2") . . . (1.) Again (example 6), D>n ",,,/xx3 . . . x2"-i >nx" . (2.) N n+1 ")• greater than - x + - x)' it is only necessary to multi- 2n ply up and reduce the result ; thus, n+ 1( , x + x--) 2n ks +x3+ . . . x2"-11+1 = 2n (2N - 1 -0") n+1 N < ra x (by 1) < N.

Whence the proposition.

Addltional Examples.

Ex. 1. + y + z)2 < 3(x2 + y2 + z2), and generally, (x+y+z)"<3"--1(e+y"+ z"). (See Induction.) Ex. 2. (x+y) (y+ z) (z+x)>8xyz < 3-(x3 + + z3).

Er. 3. r+y4+z4)>xyz(x+y+z).

Ex. 4. (0+62+ c2)( + y2 + z2)>(ax + by+ cz)2 Ex. 5. The arithmetic mean of the pth powers of n positive quantities is greater than the pth power of their mean, and also greater than the mean of their Combinations p together.

Ex. G. (ax +by + cz)2 +(ax+ cy+bz)2 +(bx + ay +12 + (bx + cy + azr + ex + ay >Tb + ac + bc)(xy + .xz+ yz, <6 a2 b2 z2) (x2 + .y2 4. z2).

It will be noted that the numerical multiplier of the second term of the powers of a +x already obtained is the same as the index. It is easy to see that this law is general. To demonstrate the fact formally we employ the method of induction.

The argument may be divided into four distinct steps1. Inference; 2. Hypothesis ; 3. Comparison; 4. Conclusion.

The first step, inference, is the discovery of the probable existence of a law.

The second step, hypothesis, is the assumption that that law holds to a certain point, up to which the opponent to the argument may be presumed to admit it.

The third step consists in basing on this assumption the demonstration of the law to a stage beyond what the opponent was prepared to admit.

The fourth step argues that as the law starts fair, and advances beyond a point at which any opponent is prepared to admit its existence, it is necessarily true.

Ex. 1. To prove that (a +x)"=a"+na"-zx +, &c.

By multiplication we get (a + x)4 = al+ 4a3x+, ne.

Let it be granted that (a +x)-=e+nia"--lx+, where m is the extreme limit to which the opponent will admit of its truth.

By multiplying the equals by a + x, we get (a +4"-1-1= a"'+'+ mex + , + a""x + , a"+' + (m+1)amx + , &,c., i.e., if the law be admitted true for ni it is proved true foi + 1 ; in other words, at whatever point the opponent compels us to limit our assumption, we can advance one step higher by argument.

Now, the law is true for 4, it is proved true for 5 and being true for 5, it is proved trac for 6, and so on, ad Ex. 2. The sum of the cubes of the natmul numbers iF the square of the sum of the numbers, /2. 3)2 Let us assume that ( 13+23+, &c., +x3= k 2 )• If this be so, then by adding (x+ 1)3 we get 13+23+ + (x+1)3= (x(x2 +1))2 + (X + 1)3 1, ((x+ 1) (xx+2)\2 Hence, if the law be true for any one number x, it is also true for x + 1.

IV. But it is true for 2, for 3, for 4,146c.

Ex. 3. To prove the inequality, (x 2)2 < 3 (x2 + y9. + 29).

Let us assume that (x + y+z)"‹ 3"--'(x"` + y"+e), then by multiplication we get (x y + z)"-"< 3"-1(x"+1+ y'''+' + z'+' + +y"'x + cez+ +y"z+z"`y).

Now, inequality, example 3, gives x"'y + rx< x"" + y'+1, + rx + x"'z + ex +rz+z,"y< 2(2'4' + y"'+' +e"), and (x + y+ 2)"'+1‹ 3,"(e+1 y"`+' +.s."+9, i.e., the law is true for m+ 1, if true for m; but it is tine for 2, it is always true.

IV. Division.

This rule is derived from the general rule for the signs in multiplication, by considering that the quotient must be such a quantity as, when multiplied by the divisor, shall produce the dividend, with its proper sign.

This definition of division is the same as that of a fraction; hence the quotient arising from the division of one quantity by another may be expressed by placing the dividend above a line, and the divisor below it ; but it may also be often real-cal to a more simple form by the following rules.

Rule. Divide the coefficient of each term of the dividend by the coefficient of the divisor, and expunge out of each term the letter or letters in the divisor : the result is the quotient.

Ex. Divide 16a3xy - 28a2xz2+ 4a2x3 by 4a"-x.

The process requires no explanation. It is founded on Laws II. and III., together with the rule of signs.

The quotient is 4ay - 7z2 + x2.

If the divisor and dividend be powers of the same quantity, the division will evidently be performed by subtracting the exponent of the divisor from that of the dividend. Thus a5, divided by a', has for a quotient a5-3= a'.

Case 2. When the divisor is simple, but not a factor of the dividend.

Rule. The quotient is expressed by a fraction, of which the numerator is the dividend, and the denominator the divisor.

Thus the quotient of 3ab3, divided by 2mbc, is the fraction • It will sometimes happen that the quotient found thus may be reduced to a more simple form, as shall be explained when we come to treat of fractions.

Case 3. When the divisor is compound.

Rule. The terms of the dividend are to be arranged in the order of the powers of some one of its letters, and those of the divisor according to the powers of the same letter. The operation is then carried on precisely as for division of numbers.

To illustrate this rule, let it be required to divide Sa2+ 2ab - 15b2 by 2a + 3b, the operation will stand thus :

By proceeding in all respects as before, we find - 5b for the second term of the quotient, and no remainder: the operation is therefore finished, and the whole quotient is 4a - 5b.

The following examples will also serve to illustrate the manner of applying the rule.

3a - b)3a3- 12a2 - a2b + 10ab - 2b2(a2 - 4a + 2b 3a3 - a2b - 12a2 + 1 Oab - 12a2 + 4ab Gab - 2b2 Gab - 9b2 +x + x - x+ x2 + 52 - X3 + X3 Sometimes, as in this last example, the quotient will never terminate ; in such a case it may either be considered as an infinite series, the law according to which the terms are formed being in general sufficiently obvious; or the quotient may be completed as in arithmetical division, by annexing to it a fraction (with its proper sign), the numerator of which is the remainder, and denominator the divisor Thus the completed quotient, in last example, is If x be small compared with unity, the remainders, as we advance, continually become smaller and smaller. If, on the other hand, x be large compared with unity, the remainders continually become larger and larger. In this case the quotient is worthless. To obtain a quotient which shall be of any practical value, we must reverse the order of arrangement, putting - x +1 in place of 1 - x. The division then becomes - x +1)11 1 - x x +7, As it is generally the largest of the quantities that we desire to divide out, we observe that, in order to effect this, we have had to begin with that quantity. Hence the Rule - The terms of the divisor and dividend are to be arranged according to the powers of that letter which it is wished (if possible) to divide out.

We have spoken as if magnitude alone was the circumstance which should determine the precedence of the letters in a division. In the more advanced processes of algebra there are other circumstances which give precedence to certain letters, such, for example, as the fact that x may and often does stand for the phrase" quantity," whilst a stands for sonic determinate numerical quantity. This leads us to exhibit a proposition in division of the greatest value and most extensive application. It is as follows :- To prove this proposition we shall employ the following Amom : - If two expressions in x are identical in form and value, but one multiplied out farther than the other, we may write any numerical quantity we please in place of x in both, and the results will be equal.

For example, (x - 1)2 + (x - 1) - 3 is identical with x2- 2(x +1)+ x -1 ; and it is evident that if we write any number (say 1) for x, the results are the same in both.

We now proceed to prove the proposition.

Let the dividend be x" + pf-4 + qx"- 2, &c., where n is a whole number, and p, q, &c., positive or negative numerical quantities.

Let the quotient, when this is divided by x - a, be Q, the remainder, which does not contain x, It; then x"-l-px"--4+7.2?-2+ = Q(x - +It by the definition of Division.

Now this equality is in reality an identity in terms of the axiom. If then we write a in place of x, the results will be equal ; this gives a" + pa"-I + qa"-2 + &c. = Q.0 + R R,

Examples.

Ex. 1. If a be any whole number, x"- a" is divisible by x- a without remainder.

For the remainder, by the proposition, is - = 0 . Ex. 2. If a be an even number, x" - a" is divisible by x +a without remainder.

For the remainder is ( - a)" - a" = 0, since n is even.

Observe that the divisor here has to be changed to x - ( - a), so that - a stands in place of the a of the proposition.

Ex. 3. If n be an odd number, x" + a" is divisible by x +a without remainder.

For the remainder is ( - a)" + a" = 0, because n is odd.

Ex. 4. To prove that 4b2c2 - (52 + c2 - a -2 )is divisible by -a+b+c; and hence to resolve it into simple factors. Here the x - a of the proposition is replaced by a-(b+c) (the negative sign of the whole divisor being of no consequence).

To determine the remainder, therefore, we'write b+cin place of a in the dividend, or thing to be divided ; the result is, 47)20 _ + + = 0 , hence 452e- - (b2 +c2a2)2 is divisible by - a + b+ c.

Now, since the dividend contains only squares of a, and 1), and c, any change in the sign of a, or 5, or c, produces no change in the dividend. What we have just proved then becomes (putting - a for a) the following : - 4b2c2- (52 + c2 - a2)2 is divisible by a+ b +c .

This last becomes (putting - b for b, and then - c for 0: - 452,2_ (52+ e2 _ -2, a ) is divisible by a - b+ c, and by a + 1)- c. Hence finally, (b2 + c2 a2\2= ) (a + b + c) (- a + b + c) (a - b + (a + b - c).

The above example is a good exercise for the student. The result may be more simply arrived at by employing a proposition of very great value and frequent use - that the difference of the squares of two quantities is the product of the suns and difference of the quantities.

Ex. 5. To prove that (1 - a2) (1 - 52) (1 - c2) - (c + ab) (1) + ac) (a + bc) is divisible by 1 4- abc.

It is simpler here to write a single letter x for abc, whereby the given quantity becomes which is obviously under the form p - p, when - 1 is written for x, and is divisible by 1 +x.

Ex. 6. Prove that (x2- x +1) (x4 - x2 +1) (x2 - x4 +1) (x16 - xS + (x2' - x" +1) is the quotient of x"+x2"+ I by x2+ x +1 ; 25 being any power of 2.

The divisor (x2 + x + 1) being multiplied by x2 - x + 1 gives x4 + x2 + 1 ; which, being again multiplied by x4- x2+ 1, gives xs + x4+1 ; and so on to the end.

Additional Examples in Division.

Ex. 1. Divide 1 - 10x3+15x4- 6x5 by (1 - X)3.

We must first multiply out (1 - x)3, and then divide the given expression by the product, 1 - 3x + 3x2 - x'. The quotient is 1 + 3x + 6x2.

Ex. 2. Divide 65x2y2 - (x4 + 64y4) by x2 - 7xy - 8y2.

We must arrange dividend and divisor in terms of powers of one of the letters, say x; the division will then assume the form by the product of x2 - 1, x2 - 2. Here we observe that x2- 1 is the product of x +1, x- 1.

Now (Art. 20), x2+ 3x+ 2 is divisible by x+ 1, and x2- 5x + 4 by x - 1. Hence, if the product is divisible by x2- 1, x2 - 2, without remainder, the third factor, x4+ 5x2 - 14 must be divisible by x2 - 2, which is found to be the case. The quotient required is therefore the product of (x + 2) (x - 4) (x2 + 7) = x4 - - x2 - 14x - 56.

In the example before us we write, -