# Why does $\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$?

As much as it embarasses me to say it, but I always had a hard time understanding the following equality:

$$\frac{a}{\frac{b}{x}} = x \times \frac{a}{b}$$

I always thought that the left-hand side of the above equation was equivalent to

$$\frac{a}{\frac{b}{x}} = \frac{a}{b} \div \frac{x}{1} = \frac{a}{b} \times \frac{1}{x}$$

What am I doing wrong, here?

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Do a few calculations where the fractions collapse, like $a=48$, $b=20$, $x=5$. – André Nicolas Dec 5 '12 at 5:33
You have to distinguish between $\dfrac{a}{\left(\dfrac{b}{x}\right)}$ and $\dfrac{\left(\dfrac{a}{b}\right)}{x}$. E.g., $8/(4/2)\neq (8/4)/2$. – Jonas Meyer Dec 5 '12 at 5:34

$\frac{a}{b}\times\frac{1}{x}=\frac{a}{b}\div x=\frac{\frac{a}{b}}{x}\neq\frac{a}{\frac{b}{x}}=a\div\frac{b}{x}=a\times \frac{x}{b}$

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Delectably symmetric answer. – Thomas Dec 5 '12 at 9:28

You’re treating the fraction $$\frac{a}{b/x}$$ as if it were $$\frac{a/b}{1/x}\;.$$ If you apply the rule invert the denominator and multiply to $$\frac{a}{\frac{b}x}=\frac{a}{b/x}\;,$$ you get $$a\cdot\frac{x}b=\frac{ax}b\;.$$

To see that this really is correct, remember that the statement that $\dfrac{p}q=r$ means that $p=qr$. Thus, if $\dfrac{a}{b/x}$ really is $\dfrac{ax}b$, we should find that

$$a=\frac{b}x\cdot\frac{ax}b\;,$$

which you can check is indeed the case.

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The problem is that $\frac{a}{\frac{b}{x}}=\frac{\frac{a}{b}}{\frac{1}{x}}=x\frac{a}{b}$ because $\frac{1}{\frac{1}{x}}=x$.

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$\frac{a}{\frac{b}{x}}$ means a divided by $\frac{b}{x}$.

Note that

$$\frac{b}{x}\frac{x}{b} =1$$

This means that

$$a \frac{b}{x}\frac{x}{b} = a$$

Now divide both sides by $\frac{b}{x}$ and you get

$$a\frac{x}{b} = \frac{a}{ \frac{b}{x}}$$

The mistake you make is confusing $\frac{a}{\frac{b}{x}}$ with $\frac{\frac{a}{b}}{x}$. In general

$$\frac{a}{\frac{b}{x}}\neq \frac{\frac{a}{b}}{x}$$

as they have different meanings.

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It might help to use the definition of equality of fractions, which says that two fractions $a/b$ and $c/d$ are equal if and only if $ad = bc$.

EDIT: I think that it is always a bad idea to use the notation $\frac{a}{\frac{b}{x}}$, even though the relative length of the bars supposedly makes it unambiguous. In my experience there is a strong correlation between people who use this notation and people who are confused. So it may help to avoid writing such things.

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$$\frac{a}{\large\frac{b}{x}} = \large\frac {x}{\not x} \frac{a}{\frac{b}{\not x}} = x\cdot \frac{a}{b}$$

Multiplying by $\dfrac xx = 1$ does not change the expression; but by multiplying numerator and denominator by $x$, the numerator becomes $ax$ and the denominator becomes $x\cdot \dfrac{b}{x} = b$

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It seems that your problem is mostly notational. For example,

$\dfrac{\big(\frac{a}{b}\big)}{c} = \dfrac{a}{bc}$, but $\dfrac{a}{\big( \frac{b}{c} \big)} = \dfrac{ac}{b}$, and these mean different things.

As for the intuition, I have always found that in doubt, you should try with 'natural' feeling numbers. For example, try $a =1, b = c = 2$, and you should be convinced of the differences.

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It's really a question of which fraction is "inside" another fraction. Another way of looking at your top equation is:

$$\frac{a}{\frac{b}{x}}=a \div \frac{b}{x} = a \times \frac{x}{b} = \frac{ax}{b}$$

Hope that helps.

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