# Why does the Denominator of the Denominator go to the Numerator?

Quite blindly I've learnt a basic rule about fractions:

The Denominator of the Denominator goes to the numerator.

I'm confused about it and I'll give an example as to why. Imagine the following:

1/2/2

Now, if the denominator of the denominator went to the numerator this would be 2/2 which is 1
and that's the what everyone agrees upon.

But what I feel is that since 1/2 is 0.5 ,

1/2/2 = 0.5/2 = 0.25

So, I hope you can see why I feel confused.
I apologize if this question seems stupid

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$1/(2/2)\neq 1/2/2$. Check the priority of operators. –  Clement C. May 14 '13 at 15:42

There are two confusions here. One is that the expression 1/2/3, when there are no parentheses, is defined to be (1/2)/3, which is different from the expression 1/(2/3), and that is why you got the wrong answer in your post. The second is why "the denominator of the denominator goes to the numerator", and it is not something you need to memorize, instead you can do the following

$$\dfrac1{\left(\frac23\right)} = \dfrac1{\left(\frac23\right)}\cdot 1 = \dfrac1{\left(\frac23\right)}\cdot \dfrac33 = \dfrac{1\cdot 3}{\left(\frac23\right)\cdot 3} = \dfrac{3}{\frac{2}3\cdot 3} = \dfrac32.$$

That is, if the denominator in the denominator is 3, just multiply the entire fraction by $\frac33$, and this will cancel the denominator in the denominator.

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Thank you for understanding. –  Nick May 17 '13 at 13:03
I have never seen it like this...I love algebra! –  zerosofthezeta Jul 30 '13 at 7:07

You need to be careful with the order of operations:

$$1/2/2 = (1/2)/2 = 1/4$$ while $$1/(2/2) = 2/2 = 1$$

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So, when the order of operations is not defined I just go from top to bottom? –  Nick May 14 '13 at 17:07
Programming languages (which tend to "linearize" expressions) generally compute left to right, except for powers (right to left). But parentesis are cheap, better overuse to avoid misunderstandings. –  vonbrand May 14 '13 at 20:32
@Nick Yes, perform the operation that comes first. –  al0 May 14 '13 at 20:50

Just convert such divisions into multiplication like this, to avoid confusion.

$$\dfrac{\dfrac{k}{m}}{\dfrac{p}{q}}=\dfrac{k}{m} \cdot \dfrac{q}{p}$$

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Sadly, I know I should do this but the problem is I don't understand how this happens. And to me, when some vital piece of information is missing, what I am being told is that wizards are the reason it happens. –  Nick May 14 '13 at 17:11