rationality testing - Is $sqrt{2}^{sqrt{2}}$ rational?

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

6. 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?

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.