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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?

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The classic answer involves $\sqrt{2}^{\sqrt{2}}$. See for instance – lhf Jan 31 '12 at 1:31
up vote 126 down vote accepted

There is a classic example here. Consider $A=\sqrt{2}^\sqrt{2}$. Then A is either rational or irrational. If it is irrational, we have $A^\sqrt{2}=\sqrt{2}^2=2$.

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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! – Ravi Jan 31 '12 at 1:35
That's awesome. I've never seen a proof quite like that one. – Ken Williams Jan 31 '12 at 5:08
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
Out of curiosity, is $A$ rational or irrational? (Just in the case we'd like to find a constructive proof of this example.) – Petr Pudlák Oct 28 '12 at 11:33
@PetrPudlák The Gelfond-Schneider theorem establishes that $A$ is transcendental (and in particular, irrational). – Ragib Zaman Apr 5 '13 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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Ba-dum-chhhh. Short and sweet. – Louis Wasserman Jan 31 '12 at 1:28
It's well known that e is irrational, but what about $\log 2 = \ln 2$? – Myself Jan 31 '12 at 1:33
OTOH, proving that $e$ and $\log 2$ are irrational is not trivial. – lhf Jan 31 '12 at 1:34
@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 1:38
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

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 $x^y = r$.

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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.

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How do you know your $x$ is irrational? – GEdgar Feb 3 '12 at 22:34
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 5 '12 at 21:39

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