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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?

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  • $\begingroup$ I think that the condition (sufficient and necessary) is $ab\in\mathbb{Q}$. $\endgroup$ Dec 18, 2016 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, 2016 at 15:32
  • $\begingroup$ @Levent: How about either $ab\in\mathbb{Q}$ or $\frac{a}{b}\in\mathbb{Q}$ then? $\endgroup$ Dec 18, 2016 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, 2016 at 15:34
  • $\begingroup$ @Levent: $b$ cannot be $0$ since it's irrational. $\endgroup$ Dec 18, 2016 at 15:34

1 Answer 1

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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 $q\in\mathbb{Q}$.

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  • $\begingroup$ @barakmanos: to $a/b \in \mathbb Q$, unless $b = 0$. $\endgroup$
    – Santiago
    Dec 18, 2016 at 15:33
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    $\begingroup$ @Santiago: $b$ cannot be $0$ since it's irrational. $\endgroup$ Dec 18, 2016 at 15:34

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