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 $a,b\in{\Bbb R}$, irrational, such that $a^b\in{\Bbb Q}$. 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: 1. $x$ is rational, so the theorem is true (with $a=b=\sqrt{2}$), or 2. $x$ 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.