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It is not even know if $\pi+e$ is rational, and the same is true for other similar expressions involving $\pi$ and $e$, but does this have an impact?

If it were, for example, proven that $\pi=ae$ or $\pi=a+e$ for a rational number $a$ would there be any consequences beyond it being a bit weird?

Aditya Hase
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Tim
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  • Not much relationship to rationality, but $e^{i\pi} = -1$. – Yiyuan Lee Dec 09 '14 at 03:48
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    As with a lot of these kinds of questions where the result itself isn't so amazing, it's more about the math developed to answer the question. It would likely give us a better way of determining relations between transcendental numbers. – Cameron Williams Dec 09 '14 at 03:49
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    @CameronWilliams: No, you know that at least one of $\pi + e$ and $\pi - e$ is irrational. It is unlikely that either is rational. It's also true that at least one of $\pi + e$ and $\pi e$ is irrational. – Robert Israel Dec 09 '14 at 03:52
  • @RobertIsrael Oh doh. Thank you for that. Removed the stupid comment. – Cameron Williams Dec 09 '14 at 03:53
  • What about $$\frac\pi\pi=e\cdot\left(\frac{1}{e}\right)$$ – Aditya Hase Dec 09 '14 at 05:39

1 Answers1

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If $\pi e$ were rational, the limit points of $\exp(n! i)$ as $n \to \infty$ would be a finite set of roots of unity.

EDIT: Note that from $e = \sum_j 1/j!$ we get $n!e = (integer) + r$ where $$0 < r = \sum_{j=n+1}^\infty \dfrac{n!}{j!} < \sum_{k=1}^\infty \dfrac{1}{(n+1)^k} = \dfrac{1}{n}$$ so $\exp(2 i n!\pi e) \to 1 $. If $\pi e = a/b$ where $a$ and $b$ are positive integers, then $\exp(2in! a) = (\exp(2in! \pi e))^b \to 1$ so any limit point of $\exp(in!)$ is a $2a$'th root of unity.

Alexander Gruber
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Robert Israel
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