Division by Zero?

I am currently teaching a summer term algebra class. Over the course of five weeks, we manage to get through an entire semester’s worth of material. The pace is pretty grueling, and I wonder at my students’ capacity to keep up, but at the end of two weeks, we seem to be doing okay.

Yesterday, we started in on rational expressions and equations.[1] One of the first thing that students learn about fractions in elementary school is that \(\frac{x}{0}\) is undefined. A fraction of that form represents a division by 0, which is not allowed. When dealing with variable expressions and equations, however, such divisions are often easy to hide, and can cause serious problems.

To drive this point home to them, I spent about 15 minutes yesterday working through the following “proof” with them yesterday.[2]

  1. We start by declaring that we have two numbers that are, in fact, the same number. Call them x and y, and let

    \(x = y\).

  2. By the multiplicative property of equality, we can multiply both sides of the equation by any number we like, and it will not change the truth value of the statement. Because I happen to like the number x, let’s multiply both sides of the equation by it, giving us

    \(x^2 = xy\).

  3. By the additive property of equality, we can also add or subtract anything we like to both sides of the equation. For instance, I could subtract \(y^2\), if I liked:

    \(x^2-y^2 = xy-y^2\).

  4. Since we just finished a chapter on factoring polynomials in class, the next thing that I might want to do is factor both sides of the equation. On the left side, I have a difference of squares which can be pretty easily factored, while on the right I can factor a y from each term. This gives us

    \((x+y)(x-y) = y(x-y)\).

  5. Now look! We have a common factor of \(x-y\) on each side of the equation. Why don’t we cancel that out?

    \(x+y = y\).

  6. Now for a bit of trickery: in step 1, we defined x and y to be equal. Because they are equal, I can substitute one for the other if I like. I don’t really like y, so instead of writing y, I am going to write x. This gives us

    \(x + x = x\).

  7. After a minor simplification, we have

    \(2x = x\).

  8. Finally, by dividing each side by x, we have

    \(2 = 1\).


Can you find the error?

When I got to the end of this “proof,” many of my students looked very confused. I gave them a little time to talk it over, and eventually they hit upon the error.

The problem is in step 5. I stated that we were “canceling” a factor of \(x-y\) from each side of the equation. But what do I mean by canceling? In this context, I am dividing each side of the equation by \(x-y\). This is a perfectly valid step, just so long as \(x-y\) is not zero. However, look at the statement in step 1. I said that x and y are equal, which implies that \(x-y=0\). In step 5, we divided by zero!

I like to think that this example brings home the point that we have to be very careful when dealing with rational expressions. Variables allow us to generalize ideas, but they can also be used to obfuscate. When working with any kind of equation or expression, we need to be vigilant, and make sure that each of our steps makes sense. On the surface, it seems like step 5 is perfectly reasonable, but when we recheck our assumptions, it falls apart.


A rational expression is like a more generalized fraction—it is an fractional expression with polynomials in the numerator and denominator.
This proof is an old mathematical chestnut. If you haven’t seen it before, read over it, and try to spot the “trick” before reading on.
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