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

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


6 Answers 6


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

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    $\begingroup$ And one of the most beautifully frustrating ones, since even at the end we do not know what proved our theorem! $\endgroup$
    – RKD
    Jan 31, 2012 at 1:35
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    $\begingroup$ That's awesome. I've never seen a proof quite like that one. $\endgroup$ Jan 31, 2012 at 5:08
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    $\begingroup$ This is especially noteworthy because it's an extremely simple existence proof: we proved that such a number exists without proving what that number is! $\endgroup$ Jan 31, 2012 at 18:26
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    $\begingroup$ Out of curiosity, is $A$ rational or irrational? (Just in the case we'd like to find a constructive proof of this example.) $\endgroup$
    – Petr
    Oct 28, 2012 at 11:33
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    $\begingroup$ @PetrPudlák The Gelfond-Schneider theorem establishes that $A$ is transcendental (and in particular, irrational). $\endgroup$ Apr 5, 2013 at 7:24

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?

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    $\begingroup$ Ba-dum-chhhh. Short and sweet. $\endgroup$ Jan 31, 2012 at 1:28
  • $\begingroup$ It's well known that e is irrational, but what about $\log 2 = \ln 2$? $\endgroup$
    – Myself
    Jan 31, 2012 at 1:33
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    $\begingroup$ OTOH, proving that $e$ and $\log 2$ are irrational is not trivial. $\endgroup$
    – lhf
    Jan 31, 2012 at 1:34
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    $\begingroup$ @myself: if $\ln 2 = a/b$ then $e$ a solution of $x^a-2^b=0$. But $e$ is transcendental (not algebraic) so $a/b$ cannot be rational. $\endgroup$
    – Henry
    Jan 31, 2012 at 1:38
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    $\begingroup$ The phrase "such numbers have infinitely many values" is a contradiction in terms. A value is a number, so every number has (is) a unique value. You mean infinitely many values can be assigned to the expression in question. Although in this case (positive real base) one usually does consider the value uniquely defined. Would you say that $-\exp(-2\pi^2)$ is a possible value of $e^{i\pi}$ because $1+2i\pi$ is a possible value of $\ln e$? $\endgroup$ Jan 31, 2012 at 11:16

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 $x^y = r$.


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.

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    $\begingroup$ How do you know your $x$ is irrational? $\endgroup$
    – GEdgar
    Feb 3, 2012 at 22:34
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    $\begingroup$ Actually, I just realized that my example is a specific case of robert's answer. In any case, @GEdgar, I'm quite sure that $x$ is irrational but as long as I (or someone else) does not prove it I'll keep a softer phrasing. $\endgroup$
    – Itamar
    Feb 5, 2012 at 21:39

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

  • $\begingroup$ This looks quite a lot like the accepted answer to this post. $\endgroup$
    – Mr Pie
    May 27, 2018 at 7:56
  • $\begingroup$ it should be as common as the $\sqrt{2}^{\sqrt{2}}$ 'trick" $\endgroup$
    – orangeskid
    May 27, 2018 at 8:36

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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