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

- 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. - 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. - 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. - 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:- Assume that there exist integers a, b with \(b \neq 0\) such

that \((a/b)^2=2\). - Show that we may assume \(a,b > 0\).
- Observe that if such an expression exists, then there must be

one in which \(b\) is as small as possible. - Show that \((\frac{2b-a}{a-b})^2 = 2\).
- Show that \(2b-a>0,a-b>0\).
- Show that \(a-b<b\), a contradiction.

- Assume that there exist integers a, b with \(b \neq 0\) such
- The fifth proof uses the rational root theorem on the

polynomial \(x^2-2=0\).