# Prob

### denominator fraction fractions denominators

PROB. Y. - To Reduce Fractions having different Denominators to others of the same value which shall haze a common Denominator.

Rule. Multiply each numerator separately into all the denominators except its own for the new numerators, and all the denominators together for the common denominator.

Ex. 1. Reduce - a - x and +x - to fractions of equal value, a havinn a common denominator_

Rule. Reduce the fractions to a common denominator, and add or subtract their numerators; and the sum or difference placed over the common denominator is the sum or remainder required..

In practice, however, it is generally better to separate the process into two or more parts analogous to the addition or subtraction of sums of money, where the pounds are added to the pounds, the shillings to the shillings, and the result afterwards combined.

The numerator will consist of the sum of two products, the one containing +x, exactly in the same way that the other contains – x. If, then, we write down one of these products, and double the even powers of x in it, omitting the odd powers, we shall obtain the required result. The product of the denominators again may be readily obtained by regarding it as that of the difference and sum of 1 +x2 and x+ x3. As such processes are of constant occurrence, we will indicate the work in full.

none of the denominators being zero, then 1= m = n. Multiply the first quantity by 1, and subtract, there results l + –1+m n, which, when substituted in the first 1n quantity, gives m = n, whence the proposition.

x"– athe denominator x2 – a2 is the product of x+ a and x – a.

is the sum of the fractions whose denominaHence' x2– a2 tors are x +a and x – a.

A Let x2–a22x x+a + x– - B a' where A and B are quantities which involve a only, not x, since x2 does not appear in the numerator of the sum.

2x A(x–a)+B(x+a) By addition, x2– a2– 2x =A(x – a) + B(x + a).

To obtain A and B from this equality, we remark that the equality is an identity, as in Art. 20. We may, therefore, deal with it in either of two ways : 1. Make the x's on the left hand side to coincide with the x's on the right, and the a's in like manner. 2. As in Art. 20, write anything we please in place of x on both sides. We will in this example take the first method, and illustrate the second method by the subsequent examples. We get 2= A + B, 0 =A – B ; A =B=1, and the result is

The reader will observe that we have treated a:6 as if it were not itself a fraction. In fact, in the application of the subject before us, the letters a and b stand for arithmetical quantities, and the fraction is simply an arith• metical fraction, as contradistinguished from an algebraical fraction like •

of. ritou. v o xtumply r racctons.

tlule. Multiply the numerators of the fractions for the numerator of the product, and the denominators for the denominator of the product.

The demonstration follows at once from the definition )f a fraction given in Art. 26; thus since -6a x b = a, -(-1° x d=c, flute. Multiply the dividend by the reciprocal of the divisor, the product will be the quotient required.

This rule requires no demonstration.

Examples in Multiplication and Division of Fractions.

, a b a2 a b a2-b2 Ex. 1. Multiply -b by by -a-2742. Since - - = - the product is a2-b2 a2 a2 a x = - • abaa-b2 ab b 1-3-3x+2 x2+2x+1 Ex. 2. Multiply x3+2x2+2x+1 by e - x 4.

Because the numerator of the first fraction, and the denominator of the second both become 0, when 1 is written for x, each is divisible by x -1 (Art. 20). In the same way the denominator of the first fraction, and the Restore symmetry by writing -c for c; the numerator of the sum is (x - 3b)(x - 3c)+ (x - 3a)(x - 3c)+ (x- 3a) (x - 3b) = 3(x2 - 2(a + b + c)x + 3(ab + ac + bc)). But = 2(a + b + c), whence the first and second terms make up 0 ; and 1 - + b - + - = 0, is the third term divided by abc, a, c the sum required is 0.

Ex. 3. Given that (a2 + bc) (b2 + ac) (c2 + ab) + (a2 - bc) (b2 - ac) (c2 - ab) = 0, when multiplied out and reduced, may be written a3 + b3 + 03+ abc = 0, prove that (a2 + bc) (b2+ ac) (c2 + ab)- (a2- bc)(b2 - ac) (c2 - ab)= 0, may be reduced to - a' + IP - + c - + abc - 0. The latter given ,equality, by dividing it by a2bc x b2ac x c2ab. becomes

- ' are written in place of a, c. Now the former result is an ± b3 + abc = 0, the