Can an irrational number raised to an irrational power be rational?

Question

Can an irrational number raised to an irrational power be rational?

If it can be rational, how can one prove it?

Answer 1

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 $A^\sqrt{2}=\sqrt{2}^2=2$.

Answer 2

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?

Answer 3

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 $x^y = r$.

Answer 4

For example: $$\sqrt{2}^{2\log_2 3} = 3$$

$\bf{Added:}$ Let's use Gelfond-Schneider theorem and show that there exist $\alpha$, $\beta$ transcendental with $\alpha^\beta$ rational. Indeed, consider the $\alpha>1$ satisfying $\alpha^{\alpha} = 3$. Then $\alpha$ is not rational ( easy to see), and hence it must be transcendental.

Consider the bijection $[1, \infty) \ni x \overset{\phi}{\mapsto} x^x \in [1, \infty)$. It is not hard to check that if $r$ and $\phi(r)$ are rational, then $r$ is an integer. Hence we have lots of example as above.

Answer 5

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.

Answer 6

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.

Answer 7

$e^{\ln(2)}$, when $e$ is the base of the natural logarithm.

Hermite proved that $e$ is transcendental.

https://web.math.utk.edu/~freire/m400su06/transcendence%20of%20e.pdf

$e$ is a real number. If it were rational, it would be the solution of some $ax+b=0$ for integers $a$ and $b$, which is impossible per Hermite. So $e$ is irrational.

To prove $\ln(2)$ is irrational, we assume coprime integers $p$ and $q$ such that $\frac{p}{q}=\ln(2)$

As $e^{\ln(2)}=2$, we can substitute $\frac{p}{q}=\ln(2)$ to get

$$e^{\frac pq}=2 $$

Raise both sides to the $q$th power.

$$e^p=2^q$$

Now, let

$$x^p-2^q=0$$

This is a binomial equation in terms of $x$ with degree $p$, and $e$ would be a solution, contradicting Hermite above. So $\ln(2)$ is irrational.