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Is it known whether $\log 2\pi$ is rational (where the base of the logarithm is $e$)? Or algebraic?

Martin Sleziak
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echinodermata
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1 Answers1

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You're trying to see if $(2\pi)^q=e^p$ for some $p,q\in\Bbb Z$, in particular this means $e,\pi$ are not algebraically independent (over $\Bbb Q$). This is not known to the date.

Pedro
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  • Although I'd be willing to place a rather large wager that they are independent. – user7530 Apr 02 '14 at 01:54
  • Thanks. So if the answer is that $\log 2\pi$ is rational, then it would imply a currently unknown result. However if $\log 2\pi$ is irrational, then this negative answer would not be strong enough to resolve the algebraic independence or non-independence of $\pi$ and $e$ (i.e., even though $(2\pi)^q = e^p$ would be impossible, there still might be a polynomial relationship between $\pi$ and $e$ of another form). Is the negative answer also unproven? – echinodermata Apr 02 '14 at 02:02
  • To my knowledge, it's unproven. So is the question of rationality of even more basic expressions such as $\pi + e$ or $\pi e$. – user7530 Apr 02 '14 at 02:40
  • @user7530 there is a proof that $\pi + e$ and $\pi e $ cannot both be rational. – mick Feb 10 '24 at 20:19