# Proof of dividing fractional expressions

For dividing two fractional expressions, how does the division sign turns into multiplication? Is there a step by step proof which proves

$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}=\frac{ad}{bc}?$$

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There is something assumed about the order of operations in the notation $a/b \div c/d$. I see this as a problem in pedagogy when fractions are first introduced in schools. Somehow we have to assume that the slash bars are to be done before the division symbol, even though each of those symbols stands for "divide". If I were you, I'd always stay in the habit of writing fractions as $\frac{a}{b}$ rather than $a/b$.

end rant.

Traditionally, $\frac{a}{b} \div \frac{c}{d}$ means: Beginning with the quantity $\frac{a}{b}$, divide this by the quantity $\frac{c}{d}$. So let's think of a small example not dealing with fractions first:

$$8 \div 2.$$

Now, we all know the answer is $4$. But I want to draw your attention to the fact that this expression is the same as:

$$8 \cdot \frac{1}{2}.$$

What happened? Well division by $2$ is the same as multiplication by $\frac{1}{2}$, because $\frac{1}{2}$ is the multiplicative inverse, or reciprocal of $2$. By defintion, the reciprocal of a number $x$ is a number $y$ such that $x\cdot y = 1$. Every nonzero number has a reciprocal. If $x \neq 0$, then its reciprocal is $\frac{1}{x}$ (because $x \cdot \frac{1}{x} = 1$). So, for $x \neq 0$:

$$8 \div x = 8 \cdot \frac{1}{x}.$$

Now, if $x$ is a fraction, such as $\frac{c}{d}$, then the most natural way to write the reciprocal of $x$ is by "flipping" the fraction. The reciprocal of $\frac{c}{d}$ is $\frac{d}{c}$ (well... because $\frac{c}{d}\cdot \frac{d}{c} = \frac{cd}{cd} = 1$). Thus:

$$8 \div \frac{c}{d} = 8 \cdot \frac{d}{c}.$$

Finally, it does not matter so much what the first factor of the expression is. It could be another fraction:

$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc}.$$

Hope this helps!

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Division is normally the inverse operation to multiplication, so - as I have explained to one of my daughters this week - dividing by $n$ is the same as multiplying by $1/n$. In one sense this is a totally trivial observation, but in another it involves a much deeper understanding of what is going on.

It follows that division, where it makes sense, can be replaced by multiplication by the reciprocal (or inverse), which is the precise case my daughter wanted to understand. Getting hold of this opens up a world of algebraic possibilities, so it is well worth doing.

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The fraction $x = \frac{a}{b}$ is the solution to $b \cdot x = a$. Similarly $y = \frac{c}{d}$ solves $d \cdot y = c$. Now, multiply the left-hand-side with $b \cdot x$ and the right hand-size with $a$ which preserves the equality since $a = b \cdot x$. We get

$$\begin{eqnarray} (b \cdot x) \cdot ( d \cdot y) &=& c \cdot a \\ (b \cdot d) \cdot (x \cdot y) & = & (c \cdot a) \end{eqnarray}$$ Thus $( x \cdot y)$ is the fraction $\frac{a \cdot c}{b \cdot d}$.

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Write $\frac{a}{b} \div \frac{c}{d}$ as $$\frac{\ \frac{a}{b}\ }{\frac{c}{d}}.$$ Suppose you wanted to clear the denominator of this compound fraction. You could try multiplication by $\frac{d}{c}$, but you'll have to multiply the top and the bottom of the fraction to avoid changing it. So, you end up with $$\frac{\ \frac{a}{b}\ }{\frac{c}{d}} = \frac{\ \frac{a}{b}\ }{\frac{c}{d}} \cdot \frac{\ \frac{d}{c}\ }{\frac{d}{c}} = \frac{\ \frac{a}{b} \cdot \frac{d}{c}\ }{\frac{c}{d} \cdot \frac{d}{c}} = \frac{\ \frac{ad}{bc}\ }{1} = \frac{ad}{bc}.$$

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HINT $\rm\ \ x/z\ =\ y\ \iff\ x\ =\ y\ z\:.\$ Thus $\rm\ \dfrac{a}b\!\!\ \bigg/\!\! \dfrac{c}d\ =\ \dfrac{a}b\!\!\dfrac{d}c\ \iff\ \dfrac{a}{b}\ =\ \dfrac{a}b\!\!\dfrac{d}c\!\!\dfrac{c}{d}$

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Suppose $\frac{a}{b}$ and $\frac{c}{d}$ are fractions. That is, $a$, $b$, $c$, $d$ are whole numbers and $b\ne0$, $d\ne0$. In addition we require $c\ne0$.

Let $\frac{a}{b}\div\frac{c}{d}=A$. Then by definition of division of fractions , $A$ is a unique fraction so that $A\times\frac{c}{d}=\frac{a}{b}$.

However, $(\frac{a}{b}\times\frac{d}{c})\times\frac{c}{d}=\frac{a}{b}\times(\frac{d}{c}\times\frac{c}{d})=\frac{a}{b}\times(\frac{dc}{cd})=\frac{a}{b}\times(\frac{dc}{dc})=\frac{a}{b}$.

Then by uniqueness (of $A$), $A=\frac{a}{b}\times\frac{d}{c}$.

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