Another Proof — Try it, you’ll like it!

In the previous post, we proved that the product of two odd numbers is odd, and the product of two even numbers is a multiple of 4. This time we’ll prove that the square root of 2 is an irrational number. This is easy to follow, but it’s something that not everybody knows. Learn it & amaze your friends with your mathematical prowess!

√2 is an irrational number.

Proof (by contradiction): Suppose √2 is a rational number. Then there are two integers, a & b, such that a/b = √2 and a/b is in simplest form. (i.e. a and b have no common factors.) Then a²/b² = 2, so a² = 2b². That means that a² is even.

Since an odd number times an odd number is always odd, then if a² is even, then a itself must be even. (If a were odd, then (a)(a) would be odd.) Since a is even, then a² is a multiple of 4, because the product of two even numbers is a multiple of 4.

Since a² = 2b² and a² is a multiple of 4, b² must be even, and therefore b itself is even. But if b is even and a is even, then a/b is not in simplest form because a and b share a factor of 2. This contradicts our original assumption that a/b was in simplest form, so there are no two integers a and b such that a/b = √2. Therefore the √2 is irrational.

Not bad, huh? If you haven’t seen this proof before, then read through it a couple of times. If you read it two or three times, it begins to stick with you. If you write it down a couple of times, you will begin to think that it is so simple that it must be obvious to anyone. That’s one reason math teachers have trouble explaining things sometimes — we forget that what is obvious to us is not obvious to everyone, and that’s another reason people think math is hard. But that’s another post. . . .