3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @GitGud: Thanks, I should obviously add except for $x=0$. $\endgroup$ – barak manos Feb 12 '15 at 11:10
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.