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

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

José Carlos Santos
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John Hoffman
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    The classic answer involves $\sqrt{2}^{\sqrt{2}}$. See for instance http://en.wikipedia.org/wiki/Law_of_the_excluded_middle#Examples. – lhf Jan 31 '12 at 01:31
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    http://www.math.purdue.edu/pow/fall2011/pdf/problem10.pdf; solution at http://www.math.purdue.edu/pow/fall2011/pdf/solution10.pdf . – whuber Jan 31 '12 at 16:47
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    https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem – Simply Beautiful Art Apr 28 '17 at 15:23

6 Answers6

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

Sage Stark
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Potato
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    And one of the most beautifully frustrating ones, since even at the end we do not know what proved our theorem! – RKD Jan 31 '12 at 01:35
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    That's awesome. I've never seen a proof quite like that one. – Ken Williams Jan 31 '12 at 05:08
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    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! – BlueRaja - Danny Pflughoeft Jan 31 '12 at 18:26
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    Out of curiosity, is $A$ rational or irrational? (Just in the case we'd like to find a constructive proof of this example.) – Petr Oct 28 '12 at 11:33
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    @PetrPudlák The [Gelfond-Schneider](http://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem) theorem establishes that $A$ is transcendental (and in particular, irrational). – Ragib Zaman Apr 05 '13 at 07:24
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    sorry if this is really obvious, but I think don't get it. So raising to irrationals always gives rationals? – Charlie Parker Sep 12 '14 at 22:28
  • @Pinocchio: No - "usually", in a handwaving sense, an irrational number raised to an irrational power is irrational, but sometimes it is rational. – Henry Apr 01 '15 at 06:55
  • I stared at this excellent example blankly for 5 minutes before realising you mean "If A is irrational then we have $A^{\sqrt2}=2$ is rational instead." – samerivertwice Nov 20 '18 at 04:11
  • What if I told you I don't accept the law of excluded middle? – Adrian May 16 '19 at 09:56
  • [More detailed explanation for those having trouble following this very terse proof.](http://mathforum.org/library/drmath/view/51618.html) – 6005 Jan 06 '20 at 19:23
  • Sir what if $A$ is rational? Then it gives required number. – Akash Patalwanshi May 14 '20 at 06:59
  • @6005 even after reading that, im still struggling. what is the signifiance behind the last sentence? – Trajan Oct 20 '20 at 17:28
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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?

Will Jagy
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    Ba-dum-chhhh. Short and sweet. – Louis Wasserman Jan 31 '12 at 01:28
  • It's well known that e is irrational, but what about $\log 2 = \ln 2$? – Myself Jan 31 '12 at 01:33
  • @Myself: See http://math.stackexchange.com/questions/15285/is-the-natural-log-of-n-rational – Jonas Meyer Jan 31 '12 at 01:34
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    OTOH, proving that $e$ and $\log 2$ are irrational is not trivial. – lhf Jan 31 '12 at 01:34
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    @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. – Henry Jan 31 '12 at 01:38
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    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 $\ln e$? – Marc van Leeuwen Jan 31 '12 at 11:16
  • The link to Niven's article does not work anymore :/ – BarbaraKwarc Feb 22 '17 at 23:09
  • @BarbaraKwarc here is their page http://www.maa.org/press/ebooks/irrational-numbers available ebook or paperback. Also some preview at that site – Will Jagy Feb 22 '17 at 23:14
  • *THUD* <-- That was the sound of my head bashing against the paywall :q But thanks anyway... – BarbaraKwarc Feb 23 '17 at 06:42
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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 $x^y = r$.

Robert Israel
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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.

Itamar
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    How do you know your $x$ is irrational? – GEdgar Feb 03 '12 at 22:34
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    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. – Itamar Feb 05 '12 at 21:39
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For example: $$\sqrt{2}^{2\log_2 3} = 3$$

orangeskid
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  • This looks quite a lot like the accepted answer to [this post](https://math.stackexchange.com/questions/449431/rational-number-to-the-power-of-irrational-number-irrational-number-true?rq=1). – Mr Pie May 27 '18 at 07:56
  • it should be as common as the $\sqrt{2}^{\sqrt{2}}$ 'trick" – orangeskid May 27 '18 at 08:36
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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.

b_jonas
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