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I saw a demonstration that there was an irrational number that raised to another irrational number resulted in a rational number. The demonstration was as follows:

We know that $\sqrt{2}$ is irrational. If $\sqrt{2}^\sqrt{2}$ is rational, we are done. Otherwise, we have $(\sqrt{2}^\sqrt{2})^\sqrt{2} = \sqrt{2}^2 = 2$, which is rational, and we are done.

It's a interesting demonstration, because it is not necessary to determine the numbers in order to show that they exist. However, I would like to know if, after all, it is possible to determine whether $\sqrt{2}^\sqrt{2}$ is rational or irrational. I don't know how could I aproache this problem.

Alma Arjuna
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1 Answers1

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By this theorem: https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem

it follows that $\sqrt{2}^\sqrt{2}$ is actually $transcendental$ (which means it is not even algebraic, which is much stronger than saying that it is not rational.)

For a proof of the result that $\sqrt{2}^\sqrt{2}$ is irrational, see https://mathoverflow.net/questions/138247/prove-that-sqrt2-sqrt2-is-an-irrational-number-without-using-a-theorem

JimN
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