This is a proof by contradiction. Firstly, assume sqrt(2) is rational, i.e can be represented as the irreducible fraction m/n where m and n are integers. We have sqrt(2)=m/n. Squaring, and multiplying both sides by n2, we get m2 = 2*n2.
This tells us that m2 is even. Now the only way to get an even square is to have its root also even, because even*even=even and odd*odd=odd. So m must also be even. This means that we can write m = 2*k where k is another integer.
So now we can rewrite m2 = 2*n2 as (2*k)2 = 2*n2 = 4 * k2. Halving both sides of this, we get n2 = 2 * k2.
This tells us that n2 is even. So n must also be even by the same reasoning as given above. So we can write n = 2 * j.
So if m is even and n is even, then m/n is not an irreducible fraction. And this argumentation can go on for ever. So the assumption that sqrt(2) is rational must be wrong, thus sqrt(2) is irrational. Q.E.D.
Now go visit my blog please, or look at other interesting maths stuff :-)