Exponentiation when the exponent is irrational

Question

I am just curious about what inference we can draw when we calculate something like $$\text{base}^\text{exponent}$$ where base = rational or irrational number and exponent = irrational number

Answer 1

An example I have always liked of $$\textbf{irrational}^{\textbf{irrational}} = \textbf{rational}$$ is the following: \begin{equation} 2 = \sqrt{2}^2= (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}. \end{equation} So either $\alpha = \sqrt{2}^{\sqrt{2}}$ is rational, or $\alpha$ is irrational, and then $\alpha^{\sqrt{2}}$ is rational.

PS @IttayWeiss in his post has a much more precise statement. This has the advantage of being elementary.

Answer 2

$2^2$ is rational while $2^{1/2}$ is irrational. Similarly, $\sqrt 2^2$ is rational while $\sqrt 2^{\sqrt 2}$ is irrational (though it is not so easily proved), so that pretty much settles all cases. Much more can be said when the base is $e$. The Lindemann-Weierstrass Theorem asserts that $e^a$ where $a$ is a non-zero algebraic number is a transcendental number.