Reminds me of a cool proof I saw recently that there are two numbers a and b such that a and b are both irrational, but a^b is rational:
Take sqrt(2)^sqrt(2), which is either rational or not. If it's rational, we're done. If not, consider sqrt(2) ^ (sqrt(2) ^ sqrt(2)). Since (a^b)^c = a^bc, we get sqrt(2) ^ (sqrt(2))^2 = sqrt(2)^2 = 2, which is rational!
It feels like a bit of a sleight of hand, since we don't actually have to know whether sqrt(2)^sqrt(2) is rational for the proof to work.
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