How can you prove that the square root of two is irrational?

You use a proof by contradiction. Basically, you suppose that $\sqrt{2}$ can be written as $p/q$. Then you know that $2q^2 = p^2$. As squares of integers, both $q^2$ and $p^2$ have an even number of factors of two. $2q^2$ has an odd number of factors of 2, which means it can't be equal to $p^2$.

Another method is to use continued fractions (which was used in one of the first proofs irrationality of $\displaystyle \pi$).

Instead of $\displaystyle \sqrt{2}$, we will consider $\displaystyle 1 + \sqrt{2}$.

Now $\displaystyle v = 1 + \sqrt{2}$ satisfies

$$v^2 - 2v - 1 = 0$$

i.e

$$v = 2 + \frac{1}{v}$$

This leads us to the following continued fraction representation

$$1 + \sqrt{2} = 2 + \cfrac{1}{2 + \cfrac{1}{2 + \dots}}$$

Any number with an infinite simple continued fraction is irrational and any number with a finite simple continued fraction is rational and has at most two such simple continued fraction representations.

Thus it follows that $\displaystyle 1 + \sqrt{2}$ is irrational, and so $\displaystyle \sqrt{2}$ is irrational.

Exercise: Show that the Golden Ratio is irrational.

More information here: http://en.wikipedia.org/wiki/Continued_fraction

If $\sqrt 2$ were rational, we could write it as a fraction $a/b$ in lowest terms. Then $$a^2 = 2 b^2.$$ Look at the last digit of $a^2$. It has to be $0$, $1$, $4$, $5$, $6$ or $9$. Now look at the last digit of $2b^2$. It has to be $0$, $2$ or $8$. As $a^2$ and $2b^2$ are the same number, its last digit must be $0$. But that's only possible if $a$ ends in $0$ and $b$ ends in $0$ or $5$. Either way both $a$ and $b$ are multiples of $5$ contradicting $a/b$ being in lowest terms.