# Day 49: Root Two Is Irrational

We’ve been thinking about different classes of numbers for a while in seventh grade. The time was ripe to discuss why, exactly, $\sqrt{2}$ is irrational. For homework, I asked my students to do some research online to find a proof that they could noodle through, at least pretty well. Most of them came to class today with somewhere between iffy and so-so grasp of the usual algebraic argument. This is what I expected. One student had wrapped his head around it enough that he came to the board and worked it through from memory with just a little help—though he said there were parts where he wasn’t quite sure what was going on.

There are a lot of different proofs of $\sqrt{2}$‘s irrationality. A fabulous list can be found here; 7 and 8”’ are faves. But something has been buzzing around in my head that I wanted to work out with my students. I’ve probably seen this or something very much like it before, but my class liked it very much and found it way more satisfying than the algebraic proof. One student said that I should add it to the Wikipedia article!

The argument ran like this:

There are only four kinds of rational numbers. odd/odd, odd/even, even/odd, and even/even. Of course, even/even can be reduced to one of the other three kinds, so we only need to consider these. We’re going to show that none of these kinds of fractions could be $\sqrt{2}$—that is, that none of them squared is 2. Note that another way of saying that a fraction is equal to 2 is that its top is double its bottom.

Well, when you square odd/odd, you get another fraction that’s odd/odd, and clearly the top is not the double of the bottom—after all, an odd number can’t be the double of a whole number.

The same goes for odd/even.

So the last category standing is even/odd. When this is squared, we get even/odd—so it looks like it might be possible for the top to be the double of the bottom. But consider this: when an even number is squared, the result is a multiple of 4. (Yeah?) And a multiple of 4 is never the double of an odd number. So $\sqrt{2}$ can’t be a fraction that’s even/odd.

But then there’s no option left! So $\sqrt{2}$ is irrational.

This feels more concrete and less “magic” than the usual proof. What do you think?