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Is $\log \pi $ a rational number? That is, are there non-zero integers $a, b$ s.th. $\pi^a = e^b$ ?

ParaH2
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Hans Engler
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    I don't believe it is known. The algebraic independence of $\pi$ and $e$ is still an open problem. – Najib Idrissi Mar 28 '14 at 18:55
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    Of course you can always take a=b=0. – awllower Mar 28 '14 at 18:57
  • @nik: doesn't this count as a dependency: $e^{i \pi} + 1 = 0$? – barak manos Mar 28 '14 at 18:58
  • @barakmanos Doesn't algebraic independence require algebraic expressions, e.g. $\mathbb{Q}[\mathrm{e},\pi]$. I don't think that one thing to the power of the other is allowed. – Fly by Night Mar 28 '14 at 19:01
  • @awllower: Thank you, I just specified the problem a bit more. – Hans Engler Mar 28 '14 at 19:01
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    possible duplicate of Are $\pi$ and $e$ algebraically independent? –  Mar 28 '14 at 19:07
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    Well, if it was known that $\log \pi$ is rational, it would probably be known to virtually everyone. No way it would escape universal attention. So my guess is either "no" or "it's unknown". – Dan Shved Mar 28 '14 at 19:12
  • @FlybyNight I think $x^a-y^b$ is a polynomial in two variables? – awllower Mar 28 '14 at 19:12
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    I fail to see how not knowing whether $e$ and $\pi$ are algebraically independent entails not knowing whether $\log \pi$ is rational. If the logarithm is irrational (which is reasonable to assume), then it doesn't tell us much about algebraic independence. Or does it? – Dan Shved Mar 28 '14 at 19:20
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    Even if we knew that $\log \pi$ were irrational, it wouldn't prove that $\pi$ and $e$ were algebraically independent (though the converse does hold). This isn't a strict duplicate. That said, I doubt that $\log\pi$ has been proven to be irrational. – mjqxxxx Mar 28 '14 at 19:21
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    @T.Bongers: It's not a duplicate. If $\pi$ and $e$ are algebraically independent, then $\log(\pi)$ is irrational. The converse isn't true. – Najib Idrissi Mar 28 '14 at 19:32
  • @awllower You're absolutely right that $x^a-y^b$ is a polynomial in $x$ and $y$ when $a$ and $b$ are non-negative integers. However, the example of $\mathrm{e}^{\mathrm{i}\pi} + 1$ is as much of a polynomial in $\mathbb{Q}[\pi,\mathrm{e}]$ as $x^y$ is a polynomial in $\mathbb{Q}[x,y]$. In one dimension: $x^2+1$ is a polynomial but $2^x+1$ and $x^x+1$ are not. – Fly by Night Mar 28 '14 at 20:27
  • @FlybyNight Sorry, I thought you were referring to nik. Apology here. – awllower Mar 29 '14 at 07:15
  • Maybe using $\log_\pi(x)$ ... – Déjà vu Jan 30 '16 at 15:59

1 Answers1

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This is still an open problem. The algebraic independence of $\pi$ and $e$ is unknown, i.e., it is unknown whether there exist algebraic numbers $n_0, n_1, \ldots, n_k$ such that

$$ n_{0} \pi^{0} + \ldots + n_{k-1} \pi^{k-1} + n_k \pi^{k} = e. $$

Klangen
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  • This fails to answer the question; even if they are algebraically dependent, it will not imply that the answer to the question is "yes". Furthermore, for all we know it may be possible to prove that the answer is "no" without being able to establish algebraic independence. – user21820 Jun 10 '19 at 14:34