Five proofs that the squrate root of 2 is irrational

This weekend I went through a few proofs that the square root of 2 is irrational.

1. The elementary proof is the one that everyone sees. You assume
   that \(\sqrt{2}\) can be represented as \(\frac{p}{q}\) where \(p\)
   and \(q\) share no common factors, that is \(\gcd(p,q)=1\). By
   multiplying out both sides you're lead to the result that both
   \(p\) and \(q\) must have 2 as a factor. See
   Wikipedia for the proof.
2. The second proof is a stronger result, and relies on the
   fundamental theorem of arithmetic to prove that \(\sqrt{p}\)
   where \(p\) is a prime number must be irrational.
3. The third proof again relies on the fundamental theorem of
   arithmetic and states that \(\sqrt[k]{n}\), the k-th root of a
   number \(n\) that is not a perfect k-th power is irrational.
4. The fourth proof is one that I hadn't seen before. It is
   elementary and of course Wikipedia has it. I like the one I found
   in Galois Theory by Ian Stewart:
   1. Assume that there exist integers a, b with \(b \neq 0\) such
      that \((a/b)^2=2\).
   2. Show that we may assume \(a,b > 0\).
   3. Observe that if such an expression exists, then there must be
      one in which \(b\) is as small as possible.
   4. Show that \((\frac{2b-a}{a-b})^2 = 2\).
   5. Show that \(2b-a>0,a-b>0\).
   6. Show that \(a-b<b\), a contradiction.
5. The fifth proof uses the rational root theorem on the
   polynomial \(x^2-2=0\).