# A Neat Mathematical Proof

Here’s a quick mathematical proof that I came across on Quora. Rather than simply giving you the link, I’m going to repeat it with some of my own commentary because I like it so much:

Prove that there are two irrational numbers $a$ and  $b$ such that  $a^b$ is rational.

Proof: Consider two cases for the instance in which $a = b = \sqrt2$: either $a^b = \sqrt2^{\sqrt2}$ is rational, or it is irrational.

The first case is straight forward: if $a = b = \sqrt2$, then both  $a$ and $b$ are irrational. This case goes ahead and states that  $a^b = \sqrt2^{\sqrt2}$ is rational, so our proof suffices for this specific case.

The real fun, however, comes in the second case, in which $a^b = \sqrt2^{\sqrt2}$ is irrational. Where do we go from here?

[Aside: About eight months ago, I likely would have been bewildered at this point in the proof and would have promptly given up. But to quote Shakespeare’s _Cymbeline_, I will tell my past self to “Fear no more” (Act IV, Scene 2), for the next part of this proof is damn nifty.]

Think about it this way: in the second case, we suppose that  $\sqrt2^{\sqrt2}$ is irrational. If we set $\sqrt2^{\sqrt2} = a$ and set $\sqrt2 = b$, we can say that $a^b = \sqrt2^{\sqrt2^{\sqrt2}}$. This, in turn, means that $a^b =\sqrt2^{\sqrt2 \cdot \sqrt2} = \sqrt2^2 = 2$

As a result, we can say that if $a = b = \sqrt2$, then either  $a^b$ is rational, or $a^{b^b}$ is rational. And because in either case two irrational numbers  $a$ and  $b$ yield a rational number $a^b$, our proof is complete! Go on, celebrate with your friends and relatives. A Fields Medal is in your future. You could be the next Will Hunting for all I know.

It took me a long time to realize the beauty of mathematical proofs, largely because of how frustrating I first thought them to be. Enrich yourselves with simple proofs like these, and you’ll learn to appreciate the simplicity and logic that dictates higher mathematics. You’ll become better with proof writing as a consequence.