Does $\pi = re$ for some rational $r$? I assume the answer is no but cannot prove so.
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1Is this a completely random question, or do you have a compelling need to know? – KCd Dec 31 '14 at 02:19
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I'm trying to relate some group theory to transcendentals and this question arose. – Tom Gatward Dec 31 '14 at 02:20
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http://math.stackexchange.com/questions/456097/are-pi-and-e-algebraically-independent – jimjim Dec 31 '14 at 02:28
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Nobody knows. We don't even know if $\pi+e$ is irrational, letter alone what you are asking.
Tim
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1The irrationality of $\pi+e$ is not equivalent to the irrationality of $\frac{\pi}e$ - indeed, I believe it is a stronger statement, which means that us knowing nothing about it does not imply we know nothing of $\frac{\pi}e$. – Milo Brandt Dec 31 '14 at 02:23
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2If $\pi=re$ for some rational $r$, then $\pi+e=(r+1)e% and so $\pi+e$ is irrational. I cannot see an obvious way to go the other way. – Tim Dec 31 '14 at 02:27
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1I see that, and I suppose it does imply that not knowing whether $\pi+e$ is irrational means we don't know for sure that $\pi$ is a rational multiple of $e$ - but we could easily know "$\frac{\pi}e$ is irrational" without knowing "$\pi+e$ is irrational" (or with that being necessarily true). I believe this is an open problem and I know it is connected to the irrationality of $\pi+e$, but I think it is misleading to imply that the rationality of $\pi+e$ is a strictly weaker result. – Milo Brandt Dec 31 '14 at 02:31