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## A interesting proof on there exists irrational numbers a and b such that a^b is rational

I found a question on mathoverflow, it asks “is there  exists irrational numbers $a$ and $b$ such that $a^b$ is rational”.

Obviously, if you let $a=e$, $b=\log 3$, then $a^b=e^{\log 3}=3$ is rational.

But someone gave a very interesting proof.

You don’t have to know whether $\sqrt{2}^{\sqrt{2}}$ is rational or not, you can just get the answer. That is a clever trick.

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Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao