# Is $\text{rational}^{\text{irrational}}$ rational or irrational?

Is the number $\text{rational}^{\text{irrational}}$ rational or irrational?

For example $2^{\sqrt{2}}$: is it rational or irrational?

I tried using a logarithm but it didn't work. It seems by superficial studying that it will be irrational. But what is the proof?

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Consider $2^{\log_37}$. – Andrés E. Caicedo Apr 21 '14 at 16:25
As for the specific number you mention, it's irrational but the proof is very far from trivial - see en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem . – Steven Stadnicki Apr 21 '14 at 16:26
As for your example, $2^{\sqrt2}$, it is irrational (in fact, transcendental). The proof is far from easy. See here. – Andrés E. Caicedo Apr 21 '14 at 16:26

Given any two rational numbers, $a,b$, note that $a^{\log_a(b)} = b$ is rational but $\log_a(b)$ is generally not rational.

For example $2^{\log_2(3)} = 3$ but $\log_2(3)$ is not rational.

So a rational number taken to an irrational power can easily be rational.

However, if $a\neq 0,1$ is rational (in fact even algebraic) and $b$ is an irrational algebraic number, then it is known that $a^b$ is actually transcendental. This is called the Gelfond–Schneider theorem.

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This is essentially the Gelfond-Schneider Theorem which says:

$$\text{If } a,b \text{ are algebraic}, a \neq 0,1 \text{ and } b \in \mathbb{R} \setminus \mathbb{Q} \text{ then } a^b \text{ is transcendental.}$$

Now every transcendental number is also irrational, and every rational number is algebraic. $\sqrt{2}$ is also algebraic so in this case, yes $2^\sqrt{2}$ is irrational and transcendental.

But (I believe) a non-zero algebraic number to the power of a transcendental number can either be rational or irrational, so a rational to the power of an irrational may be rational, only if the irrational is also transcendental. For concrete examples note that $\log_2 (3)$ is irrational, but $2^{\log_2 (3)} = 3$, however I suspect $2^\pi$ is irrational.

Examples of algebraic numbers are say $\sqrt{n}$ or $\sqrt{3 + \sqrt{2}}$ or anything that is the root of a polynomial (that is in fact the definition), whereas transcendental numbers are more like $\pi$ or $e$.

Hopefully this was helpful for you, but in that particular example, yes $2^\sqrt{2}$ is irrational.

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"But (I believe) it is an open problem to show whether an algebraic number to the power of a transcendental number is irrational." It is not open, look at my first comment. – Andrés E. Caicedo Apr 21 '14 at 16:39
So \/2^\/2 will irrational and transcendental? – Satvik Mashkaria Apr 21 '14 at 16:48
I maybe misphrased that, I meant an algebraic to a transcendental is EVER irrational, obviously it can be rational but I couldn't find a proof that (say) $2^\pi$ is irrational even though I'd assume it was. But either way, I'll edit it to make what I meant more clear. – CameronJWhitehead Apr 21 '14 at 16:49
@SatvikMashkaria, $\sqrt{2} ^ {\sqrt{2}}$ is in fact transcendental yes, it is the square root of the Gelfond-Scneider Constant, which has its own wikipedia page en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_constant – CameronJWhitehead Apr 21 '14 at 16:53
Nope. Not yet. Probably you want the conclusion to be "can either be rational or transcendental." – Andrés E. Caicedo Apr 21 '14 at 17:05