A Peculiar, Non-Constructive Proof

2014-11-08 math

A friend of mine loves to trot out this oddity, and I always manage to forget it afterwards. Now that I have it in mind, I will take a minute to write it here for posterity.

Theorem: There exist $a, b \in R$ , irrational, such that $a^{b} \in Q$ .

In other words, the theorem asserts that an irrational raised to an irrational may be rational. This isn’t especially interesting and, while surprising, I don’t claim to have any sort of intuition violated here. However, the proof is slippery and clever.

Proof: I will take for granted that $\sqrt{2}$ is irrational, which is a standard first theorem in an analysis course.

Consider the quantity $x = {\sqrt{2}}^{\sqrt{2}} .$

There are two cases to be considered here:

$x$ is rational, so the theorem is true (with $a = b = \sqrt{2}$ ), or
$x$ is irrational, and we can compute $x^{\sqrt{2}} = {\sqrt{2}}^{2} = 2$

which also would show that the theorem holds (with $a = x$ , $b = \sqrt{2}$ ).

Therefore the theorem is proven either way, and we remain ignorant of whether $x = {\sqrt{2}}^{\sqrt{2}}$ is rational or not.