# The power of two irrational numbers being rational

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There's a celebrated elementary result which claims that:

There are irrational numbers $$x$$ and $$y$$ for which $$x^y$$ is rational.

The standard proof goes like this. Now, we know that $$\sqrt{2}$$ is irrational, so let's consider $$r=\sqrt{2}^\sqrt{2}$$. Either $$r$$ is rational, or it is not. If it is rational, then we set $$x=\sqrt{2}$$, $$y=\sqrt{2}$$ and we are done. If $$r$$ is irrational, then set $$x=r$$ and $$y=\sqrt{2}$$. This means that $x^y=\left(\sqrt{2}^\sqrt{2}\right)^{\sqrt{2}}=\sqrt{2}^2=2$ which is rational.

This is a perfectly acceptable proof, but highly non-constructive, And for some people, the fact that the proof gives no information about the irrationality of $$\sqrt{2}^\sqrt{2}$$ is a fault.

So here's a lovely constructive proof I found on reddit . Set $$x=\sqrt{2}$$ and $$y=2\log_2{3}$$. The fact that $$y$$ is irrational follows from the fact that if $$y=p/q$$ with $$p$$ and $$q$$ integers, then $$2\log_2{3}=p/q$$ so that $$2^{p/2q}=3$$, or $$2^p=3^{2q}$$ which contradicts the fundamental theorem of arithmetic. Then:

\begin{eqnarray*} x^y&=&\sqrt{2}^{2\log_2{3}}\\\
&=&2^{\log_2{3}}\\\
&=&3. \end{eqnarray*}