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 $a,b\in \mathbb{R}$, irrational, such that ${a}^{b}\in \mathbb{Q}$.

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 $x={\sqrt{2}}^{\sqrt{2}}\mathrm{.}$

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.