How to teach students when they can and can't cancel factors in a fraction?

I mainly tutor adults in college algebra classes or lower. Sometimes an expression like $\dfrac {x+5}{5}$ will come up, and the students will say:

"We can cancel out the $5$'s and get $x+1$, right?"

I tell them that this is incorrect, and usually respond with a very ambiguous sentence:

"In order to be able to cancel out factors, everything in the numerator and the denominator must be multiplied by something"

I know that the students don't really get what I mean, and I just rely on examples from here. I know that a more correct response would be: "In order to cancel out factors, the numerator and denominator must consist exclusively of a single factor or a product of factors" (Though I'm not even sure of what a simple definition of factor would be here). I am just having a lot of trouble getting this concept into more understandable terms. The next time the students see $\dfrac {5+xy}{zy}$, they will want to cancel the $y$'s, because "they are being multiplied by something".

• definitely just a comment but maybe work in the use of parentheses in the numerator to emphasize that it is a quantity up there and not two separate expressions? Oct 24 '16 at 18:50
• I'd be very interested in an answer to this as well. My current method is just to start them off with pretty basic fractions like 4/8 (which we convert to 4*1/4*2 and then reduce 4/4 to 1/1 to make it 1/2) and then work up to fractions like (1+3)/3 to see if they want to reduce the 3's because they are together (since we can evaluate and see right/wrong) and then move over to expressions of the type (1+x)/x to see if they can make the jump (this is the tricky part because we've gone from concrete to abstract). I'd love a better way of doing this, because I feel this isn't a great way. Oct 24 '16 at 21:13
• A co-worker of mine teaches students to look out for "forms of one" which are any fraction with equal numerator and denominator. She never teaches them that anything cancels, but rather that when you have a "form of one" it can be rewritten as 1 for any values the expression is defined. It also helps to rewrite things like (1+x)/x as 1/x + x/x and show them that there is something that can happen with that x, but doing it that way doesn't remove the fraction just rearrange it. Oct 24 '16 at 23:15
• My belief is that many students see neither fractions not algebra as actually making sense and are just following rules. What to do about it is a whole other issue, of course! Oct 25 '16 at 19:39
• You ought to tell them that $$\frac{19}{9}=\require{cancel}\frac{1\cancel9}{\cancel9}=1$$by their logic. In this sort of manner, they assert that $$\frac{10+9}9=10?$$ Nov 12 '16 at 15:22

Instead of presenting "cancelling" as an arbitrary rule (which is often how students have seen it — or at least how they learned it — before), explain it in a way that shows what's actually going on. So, if we have $\frac{5x + 5}{5}$, instead of just "cancelling the 5's" (which sounds kind of arbitrary and mysterious), write out several steps, like: $$\frac{5x + 5}{5} = \frac{5 \cdot (x + 1)}{5 \cdot 1} = \frac{5}{5} \cdot \frac{x + 1}{1} = 1 \cdot \frac{x + 1}{1} = \frac{x + 1}{1} = x + 1.$$ Of course, you wouldn't want to give this level of detail every time, but I think it'd help to start out with this level of detail, and more gradually start omitting steps.

If you start with something like $\frac{x + 5}{5}$ instead, the same argument won't work, and I think a student is less likely to think $5(x + 1)$ is equal to $x + 5$ than to "cancel" incorrectly: "cancelling" is something many students do by reflex, without thinking about it, and writing out the steps in more detail leaves no reasonable choice but to think more carefully about what's going on. (You could still get errors like incorrectly factoring $x + 5$ as $(x + 1) \cdot 5$, but this at least reduces the problem to understanding that multiplication distributes over addition.) And if you go through the same process anyway, you get a true statement: $$\frac{x + 5}{5} = \frac{5 \cdot (\frac{1}{5} x + 1)}{5 \cdot 1} = \frac{5}{5} \cdot \frac{\frac{1}{5} x + 1}{1} = 1 \cdot \frac{\frac{1}{5} x + 1}{1} = \frac{\frac{1}{5} x + 1}{1} = \tfrac{1}{5} x + 1.$$

To reinforce this, avoid using mysterious words like "cancel" (and "move", as in "move to the other side") in your own explanations, and don't accept "cancelling" as an explanation from a student who's recently made mistakes when "cancelling" — even if they have a correct result, ask them what exactly they're doing when they're "cancelling". If they can't say, how do they know they can "cancel" like that at all?

I am not a teacher but it seems to me that "cancelling" is sort of obtuse. It should be divide both numerator and denominator by the same thing. Hopefully it is evident that 5 ~= (s+5). I had a friend who taught at the University of Illinois and had a college student answer on a test that cos(x)/x=cos( ).

• The last one is super funny
– Ovi
Oct 28 '16 at 18:10
• It is "well known" that $\displaystyle \frac{\sin x}{n} = 6$. Nov 3 '16 at 19:29
• @MichaelJoyce $$\require{cancel}\frac{\text{si}\cancel{\text{n}}x}{\cancel n}=\text{six}=6$$ Nov 12 '16 at 15:20

@DanielHast's answer has it, but here is another tack to take, especially for students who aren't yet in a position to be all that 'trustful' of algebraic manipulations, or to comfortably follow the teasing out of common factors in top and bottom.

After the student performs such a cancellation, move to some scratch paper and present them with something (adjust the numbers to suit their mental-math capacities) like:

• Q: Add four and nine for me.
• A: Uh, thirteen.
• Q: OK, now divide by four.
• A: Um, three... and a quarter. 3.25.
• Q: Great - now, write out what 'four plus nine' looks like for me.
• A: (writes) '$4 + 9$'
• Q: And divide it all by four.
• A: (writes) '$\frac{4 + 9}{4}$'

Now refer back to their cancellation, and carry it out on the expression that they have written. Take pains to show that what you're doing here is the same as what they had done there:

$$\frac{4 + 9}{4} = 1+9 = 10 \neq 3.25$$

We now have a wonderful and approachable contradiction to resolve, which the student is likely to find accessible with some guidance.

• I've done something like this. It seems to help them understand in the short term, though I don't know how much it helps them stop making cancellation errors in the long term. Oct 28 '16 at 1:48
• This is what I was going to write. However, bringing fractions (or decimals) into it makes it unnecessarily complex. Try just writing out (7+35)/7 and asking them to solve it. Possible wrong answers before they get it include 35, 36, 5, 12, and finally 6. Oct 29 '16 at 3:53

For what it's worth, here's the slide on that topic from my college algebra class: Note that I start by checking a counterexample of that common reduction error. In accordance with my theme that the essence of algebraic thought is to think about inverse operations, I try to connect with the intuition that it makes sense that division only cancels with multiplying (not addition or any other operation). I agree with zipirovich that it's important to be previously drilling students on the names of things, so that we can concisely state the important difference at this point (factors vs. terms). Then I run a few exercises (with answers hidden until another mouse-click by me).

This is no silver bullet in my experience; students still commonly have trouble on later days -- perhaps they find fractions so unpleasant that they just shut off that part of the brain whenever rational expressions show up. But connecting with the inverse operations gets a few heads nodding, and perhaps as important, continues the broad theme of the course that hopefully pays off on later days.

• (Hopefully the "check with value" doesn't only use $x=1.5$...) Oct 29 '16 at 18:32
• @BenjaminDickman: As is probably the case for most of us, when I ask students for a random number, they always respond with a positive integer. :-) Oct 29 '16 at 19:13

Frankly, I don't know how to teach this topic from scratch, but then we're not discussing that here, are we? It sounds like your situation is that of remediation — your students already have some concept of reducing fractions, but it happens to be wrong. I constantly find myself in a similar situation when my college students, say in a Calculus class, do exactly the same thing.

My approach is to ask them the meaning of the two words: terms and factors. Funny thing: most students will say that they know the words, but they can't describe their meaning. (In my book, that means that they don't really know the words, but I digress.) So I give a few examples, leading my students to the following working definition: terms are things that are added or subtracted and factors are things that are multiplied. Not perfectly accurate, but in my experience it helps. Then I reiterate that the rule for reducing fractions states that fractions are reduced by common factors, and emphasize that to use it correctly we need to know what factors are, which we now do. There's still more: I show examples that to be a factor of the numerator, this quantity must be factored out, i.e. the numerator must be this quantity times everything else; same for the denominator, of course. That helps to explain why we can't reduce $y$'s from $\frac{5+xy}{zy}$ — because in the numerator $y$ is only in one of the terms, not a factor of the numerator. Note how at this point knowing the word "term" comes into play!

I agree with some other answers that "cancelling" is a misleading word. As a non-native speaker I resort to Wikipedia and find that reducing or simplifying would be the preferrable word here.

That said, I'd try to make clear to the students in question that reducing means dividing both numerator and denominator by the same value ($\neq 0$). If one attempts to reduce a fraction like $\frac{x+5}{5}$ and the divisor $5$ comes to mind, they should check what $5:5$ is and what $(x+5):5$ is and if the result is helpful at all.

Another way would be to reverse the "reduction" and multiply the new/wrong numerator and denominator by the value that was used for the "reduction". They would then (hopefully) get $(x+1) \cdot 5 = 5x+5 \neq x+5$ for the numerator and see their error.

It all boils down to proper application of the distributive law.