I've been working on Nested Fractions at the Khan Academy. Recently I was doing a routine problem and came to the correct conclusion but I realized I didn't understand why I wouldn't keep dividing. Here's what I begin with. I'm supposed to reduce it to an equivalent simplified expression. I wish I could link to the problem but I had moved on after posting here. Here's the start:

$\displaystyle \frac{(1+\frac xy)}{\frac xy}$

Part way through my problem now looks like:

$\displaystyle \frac{y\cdot(y+x)}{y(x)}$

I decide to "cancel out" the y's out side the parentheses. I end up with...

$\displaystyle \frac{y+x}{x}$

I understood this the be the right answer. But for a moment I was very tempted to cancel the remaining x's and be left with a y. Even though I do this canceling operation all the time I realized then that there is something about it I don't understand. So here I am. Can you explain why it is that I was able to cancel the y earlier but not the x near the end?

Much appreciated.

  • $\begingroup$ Can you post a link to the original question? Right now, what you say is the "question" is not a question at all. It is an expression. Questions end with a question mark. $\endgroup$
    – 5xum
    Commented Mar 11, 2015 at 13:28
  • 2
    $\begingroup$ You should definately write parentheses here. The way the first assignment is written down, it's just $1 + \frac xy \div \frac xy = 1 + 1 = 2$. You probably meant $(1+\frac xy)\div \frac xy$. For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – AlexR
    Commented Mar 11, 2015 at 13:28
  • $\begingroup$ Oh thanks! I looked around. You are absolutely correct about the problem format by the way. I'll see if I can make an edit. $\endgroup$ Commented Mar 11, 2015 at 13:42
  • $\begingroup$ 5xum, I tried to find the original question on the site. Normally really easy but no luck this time. Sorry! $\endgroup$ Commented Mar 11, 2015 at 13:58

3 Answers 3


In order to "cancel" something on the top with something on the bottom, the thing must be multiplied to the top, and multiplied to the bottom. Here, this x is added to something on top, so it cannot be cancelled immediately. Sometimes, you can factor(this is the reverse of distributing, where you kind of turn addition into multiplication) something in order to get it to "cancel". One way to think about this, is rather than "cancelling", try reducing to 1, so in this case $$\frac{y(y+x)}{y(x)}=\frac yy*\frac{y+x}x$$ and $\frac yy=1 $ if $y\neq 0$, so $$\frac yy*\frac{y+x}x=1*\frac{y+x}x=\frac{y+x}x$$ This is what is actually going on when you "cancel".

Also, in this case, you cannot factor the top in any productive way, so you can't reduce anything else to 1. Thus, this is as simple of an answer as you're going to get.

  • 1
    $\begingroup$ Oh I hadn't heard of the strategy of reducing to 1 and your example makes it very clear to me. In fact I'm a touch god smacked because some of this stuff I understood but hadn't made the connection. Thanks for your help. I'd upvote but I don't have the reputation to do so as I'm new to the site. $\endgroup$ Commented Mar 11, 2015 at 13:50
  • $\begingroup$ That isn't your fault. Unfortunately "cancelling" has become such a common term that even though it is misleading and doesn't get at the essence of what is going on, it is what is taught. There has been a push(at least in the small part of education that I see in my day-to-day) towards removing the word "cancelling" from teaching mathematics, and instead only talking of reducing to 1, but it will take time for the concept to become universal. $\endgroup$
    – SE318
    Commented Mar 11, 2015 at 13:53
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    $\begingroup$ Just a sidenote: In Slovenian language, we have a term (pokrajšati) that is specifficaly only used when you perform "reducing to $1$". For example, the term is not used if you simplify $a+b-a$ to $b$ (in English, I often heard it said in that case that the two $a$-s "cancel out"). Good luck with replacing the word in English, I think it is a very good idea. $\endgroup$
    – 5xum
    Commented Mar 11, 2015 at 14:02
  • $\begingroup$ @5xum Wow; you learn something new everyday I guess! $\endgroup$ Commented Mar 11, 2015 at 14:28

Short answer: you cannot cancel out things that are added, only things that are multiplied.

Long answer: When you are canceling out factors in division, you are really doing two things: first you are factoring them out, then you are using the fact that $\frac aa$ is equal to $1$ if $a\neq 0$. The result is that, if $A$, $B$ and $C$ are some (nonzero) expressions, then

$$\frac{A\cdot B}{A\cdot C} = \frac{B}{C}$$

which is a direct consequence of the fact that $$\frac{ab}{cd} = \frac{a}{c} \cdot \frac{b}{d}$$ for arbitrary values of $a,b,c,d$, since it means that $$\frac{A\cdot B}{A\cdot C} = \frac AA \frac BC = 1\cdot \frac BC = \frac BC.$$

In your case, you have the expression $$\frac{y(y+x)}{y\cdot x}$$

You can use the rule I cited by seting $A=y$, $B=(y+x)$ and $C=x$

You cannot use the rule for the $x$-es, because the top (numerator) of the fraction is not of the form $x\cdot A$, where $A$ is some expression.

  • $\begingroup$ Ah that's interesting. I definitely don't think of a taking everything in the parentheses and establishing that as one separate and distinct variable. But I suppose that's true too! Thanks, this has been enlightening. $\endgroup$ Commented Mar 11, 2015 at 14:02
  • $\begingroup$ @AdamGarner I also do not think of it as a "separate variable". I think of $A$ as an "expression". It is "something", I don't care what it is. I just know that if the same expression appears above and below, and it is multiplied by two other expressions and there is only the multiplication, no addition/subtraction, then I can remove the expression. (note: only multiplication between expressions. I do not care what is inside the expression (there may be a lot of addition and subtraction). $\endgroup$
    – 5xum
    Commented Mar 11, 2015 at 14:15

The key fact about "canceling" which is easy to miss (unless it is taught very, very well) is that it is performed strictly by standard arithmetic operations. It may sometimes look like a matter of finding matching symbols and crossing them out, but when that occurs it is only a byproduct of the operations that are applied.

Canceling terms in the numerator and denominator of a fraction or ratio is done by means of multiplication or division. First of all, can we take the two facts $\frac ab \cdot \frac cd = \frac{ac}{bd}$ and $\frac aa = 1,$ and combine them like this:

$$ \frac mn = 1 \cdot \frac mn = \frac aa \cdot \frac mn = \frac {am}{an}.$$

Now, you can use this in two ways. One way is, you can run it backwards to show that it is OK to cancel the two $a$s in the expression $\frac {am}{an}.$ But another way that I find sometimes easier to understand is, if I want (for example) to cancel an $x$ in the numerator and denominator of $\frac {xp}{xq}$, I can take $a = \frac 1x$ in the example above, so that

$$ \frac {xp}{xq} = \frac {1/x}{1/x} \cdot \frac {xp}{xq} = \frac {\frac 1x \cdot xp}{\frac 1x \cdot xq} = \frac pq.$$

Some people would say to divide both the numerator and denominator by $x$, but I prefer to think of it as multiplying by $\frac 1x$; it's the same thing, of course, but I find the distributive law of multiplication over addition is better ingrained in my head than the corresponding rule for division, so I'll do a multiplication when I can.

In practice, once you have mastered the technique you can usually just write the first and last expressions without the ones in between, but sometimes it is still helpful to remember the in-between steps, especially the one with $\frac 1x \cdot xp$ and $\frac 1x \cdot xq$.

So if you have an expression $$\frac{y + x}{x},$$ in fact you can try to "cancel" an $x$ if you do it correctly:

$$\frac{y + x}{x} = \frac{\frac 1x(y + x)}{\frac 1x \cdot x}.$$

So the question is what you get from this. It depends on how you handle $\frac 1x(y + x)$; do you apply the distributive law or not? If you don't use the distributive law, you get

$$\frac{\frac 1x(y + x)}{\frac 1x \cdot x} = \frac{\frac 1x(y + x)}{1} = \frac 1x(y + x) = \frac{y + x}{x},$$

that is, you end up where you started out. If you use the distributive law, you get

$$\frac{\frac 1x(y + x)}{\frac 1x \cdot x} = \frac{\frac yx + 1}{1} = \frac yx + 1.$$

Of course you could have gotten the same result using the fact that $\frac{a+b}{c} = \frac ac + \frac bc$, like this:

$$\frac{y + x}{x} = \frac yx + \frac xx = \frac yx + 1.$$

So we see that in this case, canceling $x$ did not do anything particularly useful. But it's not always easy to tell whether you'll find a cancellation useful until you've tried it.


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