First, let's assume that sqrt(2) is rational. That is, it can be expressed in the form a/b, where a and b are integers and the fraction is simplified as far as possible.
So we have sqrt(2) = a/b --> 2 = a2/b2 --> a2=2b2. Since b2 must also be an integer, doubling it to a2 must be an even number. Only even numbers square to give even numbers, so a is also even. Let a = 2n, then a2 = 2b2 = 4n2 --> b2 = 2n2. From this we can see that b2 is even, so b must be too. We've now established that both a and b are even, but this means that the original fraction wasn't simplified as far as possible. This is a contradiction, so we can conclude that the assumption that sqrt(2) is rational is incorrect.