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 $x^2-2=0$.