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.
Show fractional part of integer multiplied by irrational is not equal to different integer multiplied by same irrational.
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$\begingroup$
real-analysis
analysis
floor-function
1 Answers
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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 $k\in\mathbb{R}\backslash\mathbb{Q}$.