Thu, 26 Jan 2006
The square root of 2 is irrational
The usual proof goes like this. Suppose that √2 is rational; then there are integers a and b with a / b = √2, where a / b is in lowest terms. Then a2 / b2 = 2, and a2 = 2b2. Since the right-hand side is even, so too must the left-hand side be, and since a2 is even, a must also be even. Then a = 2k for some integer k, and we have 4k2 = 2b2, and so 2k2 = b2. But then since the left-hand side is even, so too must the right-hand side be, and since b2 is even, b must also be even. But since a and b are both even, a / b was not in lowest terms, a contradiction. So no such a and b can exist, and √2 is irrational.
There are some subtle points that are glossed over here, but that's OK; the proof is correct.
A number of years ago, a different proof occurred to me. It goes like this:
Suppose that √2 is rational; then there are integers a and b with a / b = √2, where a / b is in lowest terms. Since a and b have no common factors, nor do a2 and b2, and a2 / b2 = 2 is also in lowest terms. Since the representation of rational numbers by fractions in lowest terms is unique, and a2 / b2 = 2/1, we have a2 = 2. But there is no such integer a, a contradiction. So no such a and b can exist, and √2 is irrational.
This also glosses over some subtle points, but it also seems to be correct.
I've been pondering this off and on for several years now, and it seems to me that it seems simpler in some ways and more complex in others. These are all hidden in the subtle points I alluded to.
For example, consider fact that both proofs should go through just as well for 3 as for 2. They do. And both should fail for 4, since √4 is rational. Where do these failures occur? The first proof concludes that since a2 is even, a must be also. This is simple. And this is the step that fails if you replace 2 with 4: the corresponding deduction is that since a2 is a multiple of 4, a must be also. This is false. Fine.
You would also like the proof to go through successfully for 12, because √12 is irrational. But instead it fails, because the crucial step is that since a2 is divisible by 12, a must be also—and this step is false.
You can fix this, but you have to get tricky. To make it go through for 12, you have to say that a2 is divisible by 3, and so a must be also. To do it in general for √n requires some fussing.
The second proof, however, works whenever it should and fails whenever it shouldn't. The failure for √4 is in the final step, and it is totally transparent: "we have a2 = 4," it says, "but there is no such integer....oops, yes there is." And, unlike the first proof, it works just fine for 12, with no required fussery: "we have a2 = 12. But there is no such integer, a contradiction."
The second proof depends on the (unproved) fact that lowest-term fractions are unique. This is actually a very strong theorem. It is true in the integers, but not in general domains. (More about this in the future, probably.) Is this a defect? I'm not sure. On the one hand, one could be seen as pulling the wool over the readers' eyes, or using a heavy theorem to prove a light one. On the other hand, this is a very interesting connection, and raises the question of whether the corresponding theorems are true in general domains. The first proof also does some wool-pulling, and it's rather more complicated-looking than the second. And whereas the first one appears simple, and is actually more complex than it seems, the point of complexity in the second proof is right out in the open, inviting question.
The really interesting thing here is that you always see the first proof quoted, never the second. When I first discovered the second proof I pulled a few books off the shelf at random to see how the proof went; it was invariably the first one. For a while I wondered if perhaps the second proof had some subtle mistake I was missing, but I'm pretty sure it doesn't.
[ Addendum 20070220: a later article discusses an awesome geometric proof by Tom M. Apostol. Check it out. ]
[Other articles in category /math] permanent link