This is mostly curiosity on my part. I assume experts would have some up-to-date info.
- Are $1$, $\pi$ and $e$ linearly independent over $\mathbb{Q}$?
- Does the set $\{ m\pi+ne;\;\;m,n\in\mathbb{Z}\}$ contain any algebraic numbers?
Thanks.
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asked Jul 16, 2020 at 12:24
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1$\begingroup$ see mathoverflow.net/q/33817/11260 $\endgroup$– Carlo BeenakkerCommented Jul 16, 2020 at 12:28
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1$\begingroup$ @CarloBeenakker I checked that question before I posed mine. I do not understand what Wilkie's work has to do with the specific question. I am quite familiar with 0-minimality. Then I asked about linear independence. $\endgroup$– Liviu NicolaescuCommented Jul 16, 2020 at 14:17
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1$\begingroup$ on a trivial note, the set $\{ m\pi+ne;\;\;m,n\in\mathbb{Z}\}, m^2+n^2 \ne 0$ can contain at most one algebraic number with $(m,n) =1$ of course and $mn \ne 0$ for any such $\endgroup$– ConradCommented Jul 16, 2020 at 15:30
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Almost certainly $1, \pi, e$ are linearly independent over $\mathbb Q$, and also over the algebraic numbers (so in particular the only algebraic number in your set $\{m\pi + n e: m,n \in \mathbb Z\}$ is $0$ which comes from $m=n=0$). This would follow from Schanuel's conjecture, I think. But it is only a conjecture: we don't even know that $e + \pi$ is irrational.
answered Jul 16, 2020 at 15:10
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3$\begingroup$ Yes, Schanuel's conjecture implies that $\{ 1, i \pi, e, e^{i \pi} \}$ has transcendence degree 2, so $e$ and $\pi$ are algebraically independent (which is stronger than linear independence over the algebraic numbers). $\endgroup$ Commented Jul 16, 2020 at 23:37