# The power of two irrational numbers being rational

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*}