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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

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As far as I can tell, this is one of the problems by Hilbert, and is quite difficult to answer. However, Gelfond-Schneider tells us that, if the base is algebraic, then it is always transcendental. – awllower Feb 6 '13 at 10:51
@awllower, do you have a reference to which of Hilbert's problem this is related to? – Ittay Weiss Feb 6 '13 at 10:54
@IttayWeiss In the link. – awllower Feb 6 '13 at 10:55
Hilbert's seventh is not quite what is being asked here. – Ittay Weiss Feb 6 '13 at 10:57
Yes, I know, so I did not post it as an answer. And I shall reformulate the comment as : related to one of the problems of Hilbert. – awllower Feb 6 '13 at 11:02

$2^2$ 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.

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By the theorem of GFelfond and Schneider, then, one can easily conclude that $\sqrt2 ^{\sqrt2}$ is even transcendental. – awllower Feb 6 '13 at 10:54
It was Gelfond. Apology here. – awllower Feb 6 '13 at 11:02
@IttayWeiss- How can 1 prove that √2^√2 is an irrational number???? – KAVISH Feb 7 '13 at 16:53
it's not that easy @KAVISH. The Gelfond-Schneider theorem implies that it is in fact transcendental. – Ittay Weiss Feb 7 '13 at 19:20

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.

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it also has the advantage of being a really nice example of proving a statement of the form $P\vee Q$ without actually proving either $P$ or $Q$. – Ittay Weiss Feb 6 '13 at 11:10

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