# Condition on two irrational numbers

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?

• I think that the condition (sufficient and necessary) is $ab\in\mathbb{Q}$. – barak manos Dec 18 '16 at 15:31
• It is not correct. Consider $a=b=\pi$. Clearly $ab\not\in\mathbb{Q}$ but if you choose $c=1/\pi$ it works. – Levent Dec 18 '16 at 15:32
• @Levent: How about either $ab\in\mathbb{Q}$ or $\frac{a}{b}\in\mathbb{Q}$ then? – barak manos Dec 18 '16 at 15:32
• @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. – Levent Dec 18 '16 at 15:34
• @Levent: $b$ cannot be $0$ since it's irrational. – barak manos Dec 18 '16 at 15:34

## 1 Answer

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 $q\in\mathbb{Q}$.

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