If you like playing around with numbers, here is a question for you to ponder over: is it possible to obtain a rational number by raising an irrational number to the power of another irrational number?

The answer, rather surprisingly, is YES.

And here is a classic proof to establish the fact beyond all doubt -

Let us consider the irrational numbers $\sqrt{2}$ and $\sqrt{3}$.

Now, let us look at the number $\sqrt{3}^{\sqrt{2}}$, which is obtained by raising an irrational number to the power of another irrational number.

If the number $\sqrt{3}^{\sqrt{2}}$ is rational, then we are already successful in proving our claim.

Otherwise, if $\sqrt{3}^{\sqrt{2}}$ is irrational, then let us look at the number $(\sqrt{3}^{\sqrt{2}})^{\sqrt{2}}$, which is again obtained by raising an irrational number to the power of another irrational number.

However, $(\sqrt{3}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{3}^{(\sqrt{2}.\sqrt{2})} = (\sqrt{3})^{2} = 3$, which is a rational number. Thus, in this case also, our claim is proven.

Hence, in either case, we have managed to establish that it is indeed possible to obtain a rational number by raising an irrational number to the power of another irrational number.

Notice that the above proof, attributed to Dov Jarden, is *non-constructive* in nature, which means that it does not provide us explicitly with a rational number obtained by raising an irrational to the power of another. However, in case you are someone who doesn't feel satisfied without seeing an explicit example, consider the number $(\sqrt{3})^{\log_{3}{4}}$.

We know that $\sqrt{3}$ is irrational. Let us convince ourselves that $\log_{3}{4}$ is also irrational.

Suppose it was rational. Then, we could have written it in the form $\frac{p}{q}$, where $p$ and $q$ are co-prime integers. In that case, we would have had:

$$ \log_{3}{4} = \frac{p}{q} \Leftrightarrow 3^{\frac{p}{q}} = 4 \Leftrightarrow 3^{p} = 4^{q} $$

which is clearly a contradiction, since $3^p$ is odd while $4^q$ is even. Hence, the number $\log_{3}{4}$ must be irrational.

Now, $(\sqrt{3})^{\log_{3}{4}} = 3^{\frac{1}{2}\log_{3}{4}} = 3^{\log_{3}{2}} = 2$, which is rational. Thus, it is further proof that our assertion is indeed correct.