GENERAL RELATIONS IN DIVISION. 83. The algebraic value of a quotient depends upon the comparative values and relative signs of the dividend and divisor. Now if either the dividend or the divisor be changed with respect to its value or sign, the quotient will undergo a change, according to a certain law. As these mutual relations are frequently concerned in algebraic investigations, we present them in this place, considering first the law of change with respect to absolute value; and second, the law of change with respect to algebraic signs. 1st. Change of value, 84. In any case of exact division, the quotient is composed of those factors of the dividend which are not included among the factors of the divisor. It is evident, therefore, that if we introduce a new factor into the dividend, the divisor remaining the same, we shall introduce the same factor into the quotient; and if we exclude a factor from the dividend, the divisor remaining the same, we shall exclude this factor from the quotient. Again, if we introduce a factor into the divisor, we shall exclude it from the quotient; and if we exclude a factor from the divisor, we shall introduce it into the quotient,—the dividend remaining the same in both cases. Hence we have the following general principles : I. Multiplying the dividend multiplies the quotient, and dividing the diviilend divides the quotient. II. Multiplying the divisor divides the quotient, and dividing the divisor multiplies the quotient. III. Multiplying or dividing both dividend and divisor by the same quantity does not change the quotient. 2d. Change of signs. 85. To show in what manner the sign of the quotient is affected by changing the sign of dividend or divisor, we observe that two signs can have only three relations, as follows: + + Now if one of the signs, only, in any of these couplets, be changed, the relation of the signs in that couplet will be changed, either from like to unlike, or from unlike to like; but if both of the signs in any couplet be changed, their relation will not be altered. Hence, I. Changing the sign of either dividend or divisor, changes the sign of the quotient. II. Changing the signs of both dividend and divisor, does not alter the sign of the quotient. RECIPROCALS, ZERO POWERS, AND NEGATIVE EXPONENTS 0-C 86. The Reciprocal of a quantity is the quotient obtained by di 1 viding unity by that quantity. Thus, is the reciprocal of x; 1 is the reciprocal of a-c. 87. In dividing any power of a quantity by any other power of the same quantity, we subtract the exponent of the divisor from the exponent of the dividend, to obtain the exponent of the quotient. Thus, a' --a’=a6--=a'; a';a=ar-s=a* And in general, we have a" m saman an -9 If in this expression n=m, the exponent of the quotient will be 0; and if n>m, the exponent of the quotient will be negative. Thus, a ao a =a*-=a-', etc. a 88. It has been found useful for certain purposes in Algebra, to employ the notation, a', a-, a-?, a-', etc. We will therefore proceed to interpret the meaning of zero and negative exponents, in general. Let a represent any quantity, and m the exponent of any power whatever. Then by the rule of division, am But the quotient obtained by dividing any quantity by itself must be equal to 1. That is a” 1 Therefore, by Ax. 7, we have a=1 Hence, 1. Any quantity having a cipher for its exponent is equal to unity. Again, by the rule of division we have ao But we have already shown that a'=1. Substituting this value for the dividend, we obtain the quotient in another form; thus, a'. 1 a” Therefore, by Ax, 7, we have 1 a Hence, 2. Any quantity having a negative exponent is equal to the reciprocal of that quantity with an equal positive exponent. DIVISIBILITY OF QUANTITIES IN THE FORM OF am +bm. 89. There are certain cases of exact division of quantities in the form of a" + or am-m, which have important application: These may be exhibited in four general problems, as follows: 1.- Divide a” +8m by a+b. am + im latb -Ba-) +1+ im =+2am-+48-) Now if this operation be continued, it is evident from the form which the first and second remainders assume, that when the expo a" lami -am 1st rem. a m-a 2d rem, a 1st rem. nent, m, is an odd number, the mth remainder will be —6"(a*-*_*)=-6a-6°)=-1"(1-1)=0 and the division will therefore be exact. But if m be even, the mth remainder will be +2"(am-m+*-*)=+"a+)=+(1+1)=+26" and the division will not be exact. Hence, T'he sum of the same powers of two quantities is divisible by the sum of the quantities, if the exponent is odd, but not otherwise. 2.-Divide a" + by a—b. a" -3 m-? tamb =+bam' +67-) =+b(am-+6m-s) tomam-m tem-m)=+6mCao +80=+**(1+1)=+26" and the division can not, therefore, be exact. Hence, , The sum of the same powers of two quantities is never divisible by the difference of the quantities. 3.-Divide am_lm by a+b. Commencing the division, we have latt + laman? im -am-tom-) 2d rem. m ат 1st rem. -am-7 2d rem. tam-72— =+b(am-2_6–) If this operation be continued, then it is evident that when m is odd; the mth remainder will be -mam-m+2mm)=-"(ao+8)=-2"(1+1)=-27 and the division can not be exact. But if m be even, the mth re. mainder will be +bmcrmm_/mm)=+Bm(a°—2°)=+3m(1-1)=0 and the division in this case will be exact. Hence, ат lab a 1st rem. 2d rem. I'he difference of the same powers of two quantities is divisible by the sum of the quantities, if the exponent is even, but not otherwise. 4.—Divide am_1, by ab. Commencing the division, we have 7 a” -18 am-tam-6 im =+b(am-1—3~—-) im =tv'(am-"-/M-) If this operation be continued, then it is evident that whether m be odd or even, the mth remainder will be +Bm(am-m—7m-m)=+bm(a°—6°)=+(1-1)=0 and the division will therefore be exact. Hence, The difference of the same powers of two quantities is always divisible by the difference of the quantities. 90. If we continue ibe division in the 1st, 3d, and 4th of the preceding problems, then in the cases of exact division, the form of the quotients will be as follows: am +6 -amatamoy-am-+28+......-alm-·+7m-' (1) amām -am-b+am-sb--am-*13+ a+b talm-?_m~() ambm =am-tam-tam-88% +um+2+...... talm-*+m-(3) -B 91. By giving particular values to m in (1), (2) and (3), we obtain the following results, which may be useful for reference: a+2 =a'-ab+3 (1) a+b =a'-a'l+ab-6. =a'-abta'bi--a'l+ab8 ma' atb m-1 a a =a-6 |