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 $$x^2-2=0$$.