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We're all familair with this beautiful proof whether or not an irrational number to an irrational power can be rational. It goes something like this:

Take $(\sqrt{2})^{\sqrt{2}}$

If it's rational, then you proved it, if it's irrational, take $((\sqrt{2})^{\sqrt{2}} ){^\sqrt{2}} = 2$ and you've proved it.

I'm wondering if you can raise $\pi$ or $e$ to a certain non-trivial real power to make it rational? And if not, where is the proof that it can't be done?

p.s. - I almost left out the real part, but then I realized that $e^{i\pi} = -1$.

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I assume the word "non-trivial" in your question is meant to exclude $\pi^0 = 1$. –  Ilmari Karonen May 22 '13 at 21:37
    
What am I missing? wolframalpha.com/input/?i=sqrt+2%5Esqrt+2%5Esqrt+2 –  Double AA May 22 '13 at 23:47
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@DoubleAA You're missing Brackets... wolframalpha.com/input/?i=%28sqrt+2%5Esqrt+2%29%5Esqrt+2 –  Arjun Bajaj May 23 '13 at 0:01
    
@IlmariKaronen Indeed, I was trying to sound fancy, did I use it incorrectly? –  OmnipresentAbsence May 23 '13 at 14:36
    
Well... not incorrectly, just imprecisely. "Trivial" is one of the vaguest, most subjective and hand-wavey terms in mathematics; even in specific areas where it does have a reasonably well agreed-upon meaning, that meaning depends entirely on context. Anyway, if you were trying to sound like a working mathematician, I'd say you succeeded. However, I'd also say that's not necessarily something to be proud of. :) –  Ilmari Karonen May 23 '13 at 16:39

1 Answer 1

up vote 37 down vote accepted

Of course. Pick any positive rational $p$ and let $x=\log_\pi p$, then $\pi^x=p$.

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Derp, I am dumb. +1 –  OmnipresentAbsence May 22 '13 at 18:05
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Don't feel bad. It's easy to forget these things from time to time. –  Asaf Karagila May 22 '13 at 18:06
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@Patrick: How so? You can define the logarithm without exponentiation. Define $\ln x=\int_1^x\frac1tdt$, then define $\log_p x=\frac{\ln x}{\ln p}$. You can now prove that $p^{\log_p x}=x$. –  Asaf Karagila May 22 '13 at 18:22
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A geometric way of "seeing" this: The map $f: x\mapsto \pi^x$ is continuous and strictly increasing, so its graph $y=\pi^x$ passes through a lot of points where just one of the two coordinates (we want $y$) is rational. –  Jeppe Stig Nielsen May 22 '13 at 21:58
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@Jeppe: Except from rare occasions, the geometric way is never my way of seeing things! Thank you making that point, it wouldn't have crossed my mind. But the argument would be clearer if you state it as $f$ being surjective onto $(0,\infty)$, and therefore it hits every positive rational number. –  Asaf Karagila May 22 '13 at 22:00

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