Let's prove that there are two irrational numbers, call them \(a\) and \(b\), such that \(a^b\) is a rational number! And let's do it in a tweet.

I am excited to tell you that I *just* released the alpha version of my “Pydont's” book, a book that compiles all my “Pydon't” articles. You can get the book at leanpub: leanpub.com/pydonts.

Let \(a = b = \sqrt 2\). If \(a^b\) is rational, then we are done; if not, take \(a = \sqrt 2^{\sqrt 2}\) and \(b = \sqrt 2\). Notice how \(a^b = (\sqrt2 ^ \sqrt2)^{\sqrt 2} = (\sqrt 2)^2 = 2\) which is obviously rational. So we have shown such \(a\) and \(b\) exist, even though we don't know if we should take \(a = b = \sqrt 2\) or \(a = \sqrt 2^{\sqrt 2}, b = \sqrt 2\).

And the actual tweet with the proof:

Twitter proof:

— Mathspp (@mathsppblog) July 23, 2020

Let a=b=√2. If a^b is rational then there is nothing to be done. Assume a^b is irrational. Redefine a=√2^√2. Notice how a^b = (√2^√2)^√2 = (√2)^2 = 2 which is clearly rational. QED.https://t.co/3sFinyzJwu

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