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*}

 
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