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.