Question

• Show that it’s possible \(a^b=c\) where \(a\) and \(b\) are irrational, and \(c\) is rational.

Solution

We know that \(\sqrt{2}\) is irrational, assuming we all know the proof.

We can try

$$\sqrt{2}^{\sqrt{2}}=c$$

Is this rational? We don’t know, but it doesn’t matter.

Outcome 1: If \(\sqrt{2}^{\sqrt{2}}\) is rational

Then great, we have found values for a pair of \(a\) and \(b\) such that \(c\) is rational.

Outcome 2: If \(\sqrt{2}^{\sqrt{2}}\) is irrational

If it is irrational, then we can raise it to the power of \(\sqrt{2}\) again.

$$ \begin{align*} \left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}&=\sqrt{2}^{\sqrt{2}\sqrt{2}}\\ &=2 \end{align*} $$

Either way, we are still able to find an example of one irrational number to the power of another, equating to a rational number.