What we will do is prove it is irrational. To do this, we will use a technique called proof by contradiction, by assuming the opposite and later getting into confusion. In this case, we will assume √2 is rational.
√2 = p/q
Assume p/q is in lowest terms, as we can write any rational number in lowest terms. In order to get rid of that square root sign and deal with easier numbers, let's square both sides.
(√2)^2 = (p/q)^2
2 = p^2/q^2
Now, let's multiply both sides by q^2 to get rid of the fraction.
q^2(2) = q^2(p^2/q^2)
2q^2 = p^2
p^2 = 2q^2
This means that p^2 must be even. In that case, p is even because an even squared is always even, an odd squared is always odd. So, we know p is even, or of the form 2a.
Let's plug 2a in for p.
(2a)^2 = 2q^2
4a^2 = 2q^2
2a^2 = q^2
q^2 = 2a^2
In this case, we are running into the same thing. Here, q^2 is even, meaning that q is even. So, we have p as even and q as even. However, we said that p/q is in lowest terms. This leads us to say that √2 is irrational.
I would never have thought that you could actually prove a number to be irrational. I'd assume you can with other numbers, such as π, e, the golden ratio, all of those guys. That is definitely one of my favorite proofs.