Can an irrational number raised to an irrational power be rational?
If it can be rational, how can one prove it?
$\endgroup$ 36 Answers
$\begingroup$There is a classic example here. Consider $A=\sqrt{2}^\sqrt{2}$. Then $A$ is either rational or irrational. If it is irrational, then we have $A^\sqrt{2}=\sqrt{2}^2=2$.
$\endgroup$ 12 $\begingroup$Yes, it can, $$ e^{\log 2} = 2 $$
Summary of edits: If $\alpha$ and $\beta$ are algebraic and irrational, then $\alpha^\beta$ is not only irrational but transcendental.
Looking at your other question, it seems worth discussing what happens with square roots, cube roots, algebraic numbers in general. I heartily recommend Irrational Numbers by Ivan Niven, NIVEN.
So, the more precise question is about numbers such as $$ {\sqrt 2}^{\sqrt 2}.$$ For quite a long time the nature of such a number was not known. Also, it is worth pointing out that such expressions have infinitely many values, given by all the possible values of the expression $$ \alpha^\beta = \exp \, ( \beta \log \alpha ) $$ in $\mathbb C.$ The point is that any specific value of $\log \alpha$ can be altered by $2 \pi i,$ thus altering $\beta \log \alpha$ by $2 \beta \pi i,$ finally altering the chosen interpretation of $\alpha^\beta.$ Of course, if $\alpha$ is real and positive, people use the principal branch of the logarithm, where $\log \alpha$ is also real, so just the one meaning of $\alpha^\beta$ is intended.
Finally, we get to the Gelfond-Schneider theorem, from Niven page 134: If $\alpha$ and $\beta$ are algebraic numbers with $\alpha \neq 0, \; \alpha \neq 1$ and $\beta$ is not a real rational number, then any value of $\alpha^\beta$ is transcendental.
In particular, any value of $$ {\sqrt 2}^{\sqrt 2}$$ is transcendental, including the "principal" and positive real value that a calculator will give you for $\alpha^\beta$ when both $\alpha, \; \beta$ are positive real numbers, defined as usual by $ e^{\beta \log \alpha}$.
There is a detail here that is not often seen. One logarithm of $-1$ is $i \pi,$ this is Euler's famous formula $$ e^{i \pi} + 1 = 0.$$ And $\alpha = -1$ is permitted in Gelfond-Schneider. Suppose we have a positive real, and algebraic but irrational $x,$ so we may take $\beta = x.$ Then G-S says that $$ \alpha^\beta = (-1)^x = \exp \,(x \log (-1)) = \exp (i \pi x) = e^{i \pi x} = \cos \pi x + i \sin \pi x $$ is transcendental. Who knew?
$\endgroup$ 9 $\begingroup$If $r$ is any positive rational other than $1$, then for all but countably many positive reals $x$ both $x$ and $y = \log_x r = \ln(r)/\ln(x)$ are irrational (in fact transcendental), and $x^y = r$.
$\endgroup$ $\begingroup$Consider, for example, $2^{1/\pi}=x$, where $x$ should probably be irrational but $x^\pi=2$. More generally, 2 and $\pi$ can be replaced by other rational and irrational numbers, respectively.
$\endgroup$ 2 $\begingroup$For example: $$\sqrt{2}^{2\log_2 3} = 3$$
$\endgroup$ 2 $\begingroup$Let me expand orangeskid's answer, both because I think it teaches us something useful, and it might be the easiest elementary proof for this question.
The proof that $ \sqrt{2} $ is irrational is well-known, so I will not repeat it here.
But there's a proof just as simple showing that $ \log 3 / \log 2 $ is irrational. Suppose on contrary that $ \log 3 / \log 2 = p / q $ where p and q are integers. Since $ 0 < \log 3 / \log 2 $, we can choose $ p $ and $ q $ both as positive integers. The equality then rearranges to $ 3^q = 2^p $. But here, the left hand side is odd and the right hand side is even, so we get a contradiction.
That gives a positive answer to the original question:$$ \big(\sqrt{2}\big)^{2 \log 3 / \log 2} = 3 $$
Thanks to Rand al'Thor, who mentioned this problem in SE chat and thus inspired this answer.
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