5
$\begingroup$

Consider two irrational numbers $a$ and $b$. Are there well known sufficent conditions on the relation between $a$ and $b$ that would allow me to conclude that there is $c \in \mathbb{R}$, $c \neq 0$, such that both $ca$ and $cb$ are rational?

$\endgroup$
  • $\begingroup$ I think that the condition (sufficient and necessary) is $ab\in\mathbb{Q}$. $\endgroup$ – barak manos Dec 18 '16 at 15:31
  • $\begingroup$ It is not correct. Consider $a=b=\pi$. Clearly $ab\not\in\mathbb{Q}$ but if you choose $c=1/\pi$ it works. $\endgroup$ – Levent Dec 18 '16 at 15:32
  • $\begingroup$ @Levent: How about either $ab\in\mathbb{Q}$ or $\frac{a}{b}\in\mathbb{Q}$ then? $\endgroup$ – barak manos Dec 18 '16 at 15:32
  • $\begingroup$ @barakmanos Yes it works and it is equivalent to say that $a=qb$ for some $q\in\mathbb{Q}$. I prefer the second one since it does not require $b$ to be nonzero. $\endgroup$ – Levent Dec 18 '16 at 15:34
  • $\begingroup$ @Levent: $b$ cannot be $0$ since it's irrational. $\endgroup$ – barak manos Dec 18 '16 at 15:34
8
$\begingroup$

If $a=qb$ for some $q\in\mathbb{Q}$ then the condition clearly holds. For the other implication, say there exists $0\neq c\in\mathbb{R}$ with $ca$ and $cb$ are rational. Then let $q=ca/cb$. Then $a=qb$.

Therefore there exists a $0\neq c\in\mathbb{R}$ with $ca$ and $cb$ rational iff $a=qb$ for some $q\in\mathbb{Q}$.

$\endgroup$
  • $\begingroup$ @barakmanos: to $a/b \in \mathbb Q$, unless $b = 0$. $\endgroup$ – Santiago Dec 18 '16 at 15:33
  • 2
    $\begingroup$ @Santiago: $b$ cannot be $0$ since it's irrational. $\endgroup$ – barak manos Dec 18 '16 at 15:34

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.