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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.

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 $$a = \sqrt 2^{\sqrt 2}, b = \sqrt 2$$.