Monday, October 3, 2011

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