Does there exist a number $x\neq0$, such that $[x\cdot\pi\in\mathbb{Q}]\wedge[x\cdot{e}\in\mathbb{Q}]$?

I thought this question would be easy to answer, but it turns out otherwise.

Obviously $x\not\in\mathbb{Q}$, but that's just about the only obvious fact here.

The way I see it, there are three possible answers to this question:

  1. Yes (need to show an example)
  2. No (need to prove)
  3. Unknown

I suspect that the answer is either 2 or 3, but I'm not really sure how to continue.

I think that the following might be useful:

  • It is unknown if $[\pi+e \in\mathbb{Q}]\vee[\pi\cdot{e} \in\mathbb{Q}]$
  • It is known that $[\pi+e\not\in\mathbb{Q}]\vee[\pi\cdot{e}\not\in\mathbb{Q}]$

Does anybody have any idea how to proceed?

  • $\begingroup$ @GitGud: Thanks, I should obviously add except for $x=0$. $\endgroup$ – barak manos Feb 12 '15 at 11:10

If such $x$ exists, then $$ \frac{\pi}{e} = \frac{x \cdot\pi}{ x \cdot e}\in \mathbb Q$$ . Conversely, if $\frac{\pi}{e}\in \mathbb Q$ we can take $x = \frac{1}{e}$ and we have $$x\cdot \pi = \frac{\pi}{e}\in \mathbb Q $$ $$x\cdot e = 1\in \mathbb Q $$ So your question is equivalent to ask if $\frac{\pi}{e}\in \mathbb Q$.As far as I know, this is an open problem.

  • $\begingroup$ Nice! Just to conclude this using the facts that I have mentioned in my question, I believe that $[\pi\cdot{e}\in\mathbb{Q}\text{ unknown}]\implies[\frac{\pi}{e}\in\mathbb{Q}\text{ unknown}]$. $\endgroup$ – barak manos Feb 12 '15 at 11:39
  • $\begingroup$ @barakmanos Yes, because $\frac{\pi}{e}\in \mathbb Q \Rightarrow \pi e \not \in \mathbb Q$ $\endgroup$ – themaker Feb 12 '15 at 11:47
  • $\begingroup$ Because what? I don't understand, sorry. $\endgroup$ – barak manos Feb 12 '15 at 11:54
  • $\begingroup$ $\pi e$ and $\pi/e$ can't both be rational, because then their product, $\pi^2$, and their quotient, $e^2$, would be rational, and that's nonsense. But in principle either one of them could be rational (and the other then necessarily irrational). $\endgroup$ – Gerry Myerson Feb 12 '15 at 12:17

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