How can you prove that the square root of two is irrational? I have read a few proofs that $\sqrt{2}$ is irrational.
I have never, however, been able to really grasp what they were talking about.
Is there a simplified proof that $\sqrt{2}$ is irrational?

There is also a proof of this theorem that uses the well ordering property of the set of positive integers, that is in a non empty set of positive integers there is always a least element. The proof follows the approach of proof by contradiction but uses the well ordering principle to find the contradiction :) -

Let us assume $\sqrt{2}$ is rational, hence it can be written down in
  the form $\sqrt{2}=a/b$ assuming that both $a$ and $b$ are positive
  integers in that case if we look at the set $S = \{k\sqrt{2} \mid k,
k\sqrt{2}\text{ are integers}\}$ we find that it's a non empty set of
  positive integers, it's non empty because $a = b\sqrt{2}$ is in the
  above set. Now using the Well ordering principle we know that every
  set of positive integers which is non-empty has a minimum element, we
  assume that smallest element to be $s$ and let it equal to $s =t\sqrt{2}$. Now an interesting thing happens if we take the difference
  between the following quantities $s\sqrt{2} - s = (s-t)\sqrt{2} =
s(\sqrt{2} - 1)$ which is a smaller element than $s$ itself, hence
  contradicting the very existence of $s$ being the smallest element.
  Hence we find that $\sqrt{2}$ is irrational.

I know the proof but I am still amazed at how the author came up with the set assumption. Sometimes such assumptions make you feel kinda dumb :). If anyone has some insight regarding how to come up with such assumptions kindly post your answer in the comment, otherwise I would just assume that it was a workaround.
