# Division of Fractions

159. **TO DIVIDE ONE FRACTION BT ANOTHER, INVERT THE DIVISOR, AND THEN PROCEED AS IN MULTIPLICATION. (Art. 152.)**

Ex. 1. Divide $\frac{a}{b}$ by $\frac{c}{d}$. Ans. $\frac{a}{b}\cdot\frac{d}{c}=\frac{ad}{bc}$.

To understand the reason of the rule, let it be premised, that the product of any fraction into the same fraction inverted, is always a unit.

Thus $\frac{a}{b}\cdot\frac{b}{a}=\frac{ab}{ab}=1$. And $\left(\frac{d}{h+y}\right)\cdot\left(\frac{h+y}{d}\right)=1$.

But a quantity is not altered by multiplying it by a unit. Therefore, if a dividend be multiplied, first into the divisor inverted, and tfoen into the divisor itself, the last product wilt be equal to the dividend. Now, by the definition, (art. 112,) "division is finding a quotient, which.multiplied into the divisor will produce the dividend." And as the dividend multiplied into the divisor inverted is such a quantity, the quotient Is truly found by the rule.

This explanation will probably be best understood, by attending to the examples. In several which follow, the *proof* of the division will be given, by nmltiplying the quotient into the divisor. This will present, at one view, the dividend multiplied into the inverted divisor, and into the divisor itself.

2. Divide $\frac{m}{2d}$ by $\frac{3h}{y}$. Ans. $\frac{m}{2d}\cdot\frac{y}{3h}=\frac{my}{6dh}$.

Proof. $\frac{my}{6dh}\cdot\frac{3h}{y}=\frac{m}{2d}$ the dividend.

3. Divide $\frac{x+d}{r}$ by $\frac{5d}{y}$. Ans. $\frac{x+d}{r}\cdot\frac{y}{5d}=\frac{xy+dy}{5dr}$.

Proof. $\left[\frac{xy+dy}{5dr}\right]\cdot\frac{5d}{y}=\frac{x+d}{r}$ the dividend.

4. Divide $\frac{36d}{5}$ by $\frac{18h}{10y}$. Ans. $\frac{36d}{5}\cdot\frac{10y}{18h}=\frac{4dy}{h}$.

160. When a fraction is divided by an *integer*, the *denominator* of the fraction is multiplied into the integer.

Thus the quotient of $\frac{a}{b}$ divided by $m$, is $\frac{a}{bm}$.

For $m=\frac{1}{m}$; and by the last article, $\frac{a}{b}:\frac{m}{1}=\frac{a}{b}\cdot\frac{1}{m}=\frac{a}{bm}$.

So $\left[\frac{1}{a-b}\right]:h=\left[\frac{1}{a-b}\right]\cdot\frac{1}{h}=\frac{1}{ah-bh}$.

In fractions, multiplication is made to perform the office of division ; because division in the usual form often leaves a troublesome remainder: but there is no remainder in multiplication. In many cases, there are methods of shortening the operation. But these will be suggested by practice, without the aid of particular rules.

161. By the definition, "the *reciprocal* of a quantity, is the quotient arising from dividing a unit by that quantity."

Therefore the reciprocal of $\frac{a}{b}$ is $1:\frac{a}{b}=1\cdot\frac{b}{a}=\frac{b}{a}$. That is, *The reciprocal of a fraction U the fraction inverted.*

Thus the reciprocal of $\frac{b}{m+y}$ is $\frac{m+y}{b}$; the reciprocal of $\frac{1}{3y}$ is $\frac{3y}{1}$ or $3y$; the reciprocal of $\frac{1}{4}$ is $4$. Hence the reciprocal of a fraction whose numerator is $1$, is the denominator of the fraction.

Thus the reciprocal of $\frac{1}{a}$ is $a$; of $\frac{1}{a+b}$, is $a+b, \& c$.

162. A fraction sometimes occurs in the numerator or denominator of another fraction, as $\frac{\left[\frac{2}{3}a\right]}{b}$ It is often convenient,

o in the course of a calculation, to transfer such a fraction, from the numerator to the denominator of the principal fraction, or the contrary. That this may be done without altering the value, if the fraction transferred be*inverted*, is evident from the following principles:

First, *Dividing* by a fraction, is the same as *multiplying* by the fraction *inverted*. (Art. 159.)

Secondly, *Dividing the numerator* of a fraction has the same effect on the value, as *multiplying the denominator*; and multiplying tlie numerator has tnc same effect, as dividing the denominator. (Art 136.)

Thus in the expression [(3/5)a]/x the numerator of a/x is multiplied into $\frac{3}{5}$. But the value will be the same, if, instead of multiplying the numerator, we divide the denominator by $\frac{3}{5}$, that is, multiply the denominator by $\frac{5}{3}$.

Therefore $\frac{\left[\frac{3}{5}a\right]}{x}=\frac{a}{\frac{5}{3}x}$.

So $\frac{h}{\frac{7}{9}m}=\frac{\left[\frac{9}{7}h\right]}{m}$.

163. Multiplying the *numerator*, is in effect multiplying the *value* of the fraction. (Art. 134.) On this principle, a fraction may be cleared of a fractional coefficient which occurs in its numerator.

Thus $\frac{\left[\frac{3}{5}a\right]}{b}=\frac{3}{5}\cdot\frac{a}{b}=\frac{3a}{5b}$. And $\frac{\left[\frac{1}{5}a\right]}{y}=\frac{1}{5}\cdot\frac{a}{y}=\frac{a}{5y}$.

And $\frac{\frac{1}{3}h+\frac{1}{3}x}{m}=\frac{1}{3}\cdot\frac{h+x}{m}=\frac{h+x}{3m}$.

On the other hand, $\frac{3a}{7x}=\frac{3}{7}\cdot\frac{a}{x}=\frac{\left[\frac{3}{7}a\right]}{x}$.

164. But multiplying the *denominator*, by another fraction, it in effect *dividing* the value ; (Art. 135.) that is, it is *multiplying* the value by the fraction *inverted*. The principal fraction may therefore be cleared of a fractional coefficient, which occurs in its denominator.

Thus $\frac{a}{\left[\frac{3}{5}b\right]}=\frac{a}{b}:\frac{3}{5}=\frac{a}{b}\cdot\frac{5}{3}=\frac{5a}{3b}$.

On the other hand, $\frac{7a}{3x}=\frac{a}{\left[\frac{3}{7}x\right]}$.

164. b. The numerator or the denominator of a fraction, may be itself a fraction. The expression may be reduced to a more simple form, on the principles which have been applied in the preceding cases.