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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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  • $\begingroup$ $1/(2/2)\neq 1/2/2$. Check the priority of operators. $\endgroup$ – Clement C. May 14 '13 at 15:42
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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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  • $\begingroup$ Thank you for understanding. $\endgroup$ – Nick May 17 '13 at 13:03
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    $\begingroup$ I have never seen it like this...I love algebra! $\endgroup$ – zerosofthezeta Jul 30 '13 at 7:07
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Let me try to explain the formula $$ \frac {\frac ab}{\frac cd}=\frac ab \cdot\frac dc\:\:\:\:\:(*)$$ without too much abstraction. I intend to use only the definition of multiplication of fractions and the rules of solving equations that you study in the school. Let us denote the left side of the previous equality by $x.$ Then $$x=\frac {\frac ab}{\frac cd}.$$ Multiplying both sides of this equality by ${\frac cd}$ we obtain $${\frac cd}x= {\frac ab}.$$ Now multiplying both sides of the equation by ${\frac dc}$ we get $${\frac dc\frac cd}x= {\frac dc\frac ab}.$$ Which is $${\frac {dc}{cd}}x= {\frac ab\frac dc}.$$ Or, equivalently, $$ x= {\frac ab\frac dc}.$$ This proves that the formula $(*)$ must be true.

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  • $\begingroup$ Nice use of temp. variable. This resolves my uneasiness with @Inceptio's answer from 3 years ago :) $\endgroup$ – Nick May 7 '16 at 12:30
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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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  • $\begingroup$ So, when the order of operations is not defined I just go from top to bottom? $\endgroup$ – Nick May 14 '13 at 17:07
  • $\begingroup$ 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. $\endgroup$ – vonbrand May 14 '13 at 20:32
  • $\begingroup$ @Nick Yes, perform the operation that comes first. $\endgroup$ – al0 May 14 '13 at 20:50
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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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  • $\begingroup$ 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. $\endgroup$ – Nick May 14 '13 at 17:11
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One way to avoid the problem is to convert everything to multiplication using $\left(\backslash x \equiv \times \dfrac 1x \right)$:

$a/b/c = a \times \dfrac 1b \times \dfrac 1c = \dfrac{a}{bc}$

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  • $\begingroup$ You've redefined ambiguous notation :D I like it. $\endgroup$ – Nick May 7 '16 at 12:26
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This is how I think about it, it's just a more generalized version of Samuel's answer:

$$\dfrac {\dfrac ab}{\dfrac cd}$$

$$=\dfrac {\dfrac ab}{\dfrac cd} \cdot 1$$

$$=\dfrac {\dfrac ab}{\dfrac cd} \cdot \dfrac {\dfrac dc}{\dfrac dc}$$

$$=\dfrac{\dfrac ab \cdot \dfrac dc}{1}$$

$$= \dfrac ab \cdot \dfrac dc$$

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1/2/2

According to the order of operations 1/2/2 is 1/2 divided by 2 which is the same as 1/2 divided by 2/1 and in this case the denominator of the denominator is actually 1. Look 1/2 is the numerator and 2/1 is the denominator and the denominator of the denominator 2/1 is 1. So the denominator of the denominator goes to the numerator:

1/2 divided by 2/1 = 1*1/2 divided by 2 = 1/2 divided by 2 = 0.5/2 = 0.25

But what if there were parentheses:

1/(2/2)

the denominator of the denominator goes to the numerator:

2 -the denominator of the denominator 2/2- goes to the numerator 1

1/(2/2) = 2/2 = 1

See there is no confusion just the denominator of the denominator in the problem: 1/2/2 is not 2 it is 1 and in the problem 1/(2/2) it is 2.

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Hmmm. I don't have much to add. My precursors have covered it pretty good. But, "in a word", multiplication and division are inverse operations. So, division by x and multiplication by the reciprocal of x, $\frac {1}{x} $, are the same. Finally, the reciprocal of $\frac {a}{b} $ is $\frac {b}{a} $. It can be a little confusing. Just follow it through. ..

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