# Is $\sqrt{2}^{\sqrt{2}}$ rational? [duplicate]

In §2.2 of her essay on mathematical morality, Eugenia Cheng includes the following example:

1. Why is it possible for an irrational to the power of an irrational to be rational?

Here is a nice little proof that it is possible:

Consider $\sqrt{2}^{\sqrt{2}}$.

If it is rational, we are done.

If it is irrational, consider $$\left(\sqrt{2}^{\sqrt{2}}\right)^\sqrt{2} = \sqrt{2}^2=2.$$

However, as Cheng notes, this doesn't tell us whether $\sqrt{2}^{\sqrt{2}}$ itself is rational or not. So which is it?

## marked as duplicate by Martin R, Dietrich Burde, Asaf Karagila♦, Did, E.P.Jun 28 '16 at 11:50

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## 1 Answer

It is irrational (in fact it is transcendental). This can be shown using the Gelfond-Schneider theorem

By the theorem, where $a$ and $b$ are both algebraic with $a \neq 0,1$ and $b$ irrational, $a^b$ is transcendental. Transcendental numbers are necessarily irrational.