To divide one fractional quantity by another.

RULE.

158. Multiply the dividend by the reciprocal of the divisor, or which is the same, invert the divisor, and proceed, in every respect, as in multiplication of algebraic fractions; and the product thus found will be the quotient required.

When a fraction is to be divided by an integral quantity; the process is the reverse of that in multiplication; or, which is the same, multiply the denominator by the integral, (Art. 120), or divide the numerator by it. The latter mode is to be preferred, when the numerator is a multiple of the divisor.

159. But it is, however, frequently more simple in practice to divide mixed quantities by one another, without reducing them to improper fractions, as in division of integral quantities, especially when the division would terminate.

3

Ex. 5. Divide x1 — § x3 + x2 - 1x by x2-x. x2 -1x) x1 — x3 + 11 x 2 — — x (x2 — 3 x + 1

3

8

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§ VII. RESOLUTION OF ALGEBRAIC FRACTIONS OR

QUOTIENTS INTO INFINITE SERIES.

160. An infinite series is a continued rank, or progression of quantities, connected together by the signs or; and usually proceeds according to some regular, or determined law.

Thus, +++÷+}+/+, &c.

Or, -+-+-+, &c.

In the first of which, the several terms are the reciprocals of the odd numbers 1, 3, 5, 7, &c. ; and in the latter the reciprocals of the even numbers, 2, 4, 6, 8, &c., with alternate signs.

161. We have already observed (Art. 96), that if the first or leading term of the remainder, in the division of algebraic quantities, be not divisible by the divisor, the operation might be considered as terminated; or, which is the same, that the integral part of the quotient has been obtained. And, it has also been remarked, (Art. 89), that the division of the remainder by the divisor can be only indicated, or expressed, by a fraction: thus, for xample, if we have to divide a°by a+1, we write for the quotient : This, however, does not prevent us from attempting the division according to the rules that have been given, nor from continuing it as far as we please, and we shall thus not fail to find the true quotient, though under different forms.

1

a+1

162. To prove this, let us actually divide a or 1, by 1-a, thus ;

аз

1- -α

may be exhi

=1+a+a2+ ; =1+a+a2 +a3 +

1 α

1 a

Now, by considering the first of these formula,

which is 1+: and observing that 1

1 --a

=

we

α

a

1

-a

-

If we follow the same process with regard to the second expression, that is to say, if we reduce the integral part 1+a to the same denominator, 1-a.

In the third formula of the quotient, the integers 1+a+a2 reduced to the denominator 1

Therefore each of these formulæ is in fact the

value of the proposed fraction

163. This being the case, we may continue the series as far as we please, without being under the