Is there a number $x\neq0$ whose products with $\pi$ and with $e$ are both rational?

Question

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?

Answer 1

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 whether $\frac{\pi}{e}\in \mathbb Q$. As far as I know, this is an open problem.