Show fractional part of integer multiplied by irrational is not equal to different integer multiplied by same irrational.

Question

I'm not sure where to start with this problem and would like some help. If we let $ m,n \in \mathbb{Z}, m \neq n. k \in \mathbb{R}\backslash\mathbb{Q} $ and let $ \{x\} = x - \lfloor x\rfloor $. Now prove that $ \{mk\} \neq \{nk\} $. I think I should prove it by contraption. So assume $\{mk\} = \{nk\} \implies mk - \lfloor mk\rfloor = nk - \lfloor nk\rfloor \implies k(m-n) = \lfloor mk\rfloor - \lfloor nk\rfloor$. But I'm not sure where to go from here.

Answer 1

Notice that $m-n$ and $\lfloor mk\rfloor-\lfloor nk\rfloor$ are both integers, and that $m-n\neq0$ since $m\neq n$. Then, dividing both sides by $m-n$ in $k\left(m-n\right)=\lfloor mk\rfloor - \lfloor nk\rfloor$ gives $k=\frac{\lfloor mk\rfloor - \lfloor nk\rfloor}{m-n}\in\mathbb{Q}$.

Therefore, by contraposition, $\left\{mk\right\}\neq\left\{nk\right\}$ whenever $m,n\in\mathbb{Z}$, $m\neq n$, and $k\in\mathbb{R}\backslash\mathbb{Q}$.