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

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?

• @GitGud: Thanks, I should obviously add except for $x=0$. – 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.
• Nice! Just to conclude this using the facts that I have mentioned in my question, I believe that $[\pi\cdot{e}\in\mathbb{Q}\text{ unknown}]\implies[\frac{\pi}{e}\in\mathbb{Q}\text{ unknown}]$. – barak manos Feb 12 '15 at 11:39
• @barakmanos Yes, because $\frac{\pi}{e}\in \mathbb Q \Rightarrow \pi e \not \in \mathbb Q$ – themaker Feb 12 '15 at 11:47
• $\pi e$ 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). – Gerry Myerson Feb 12 '15 at 12:17