If $x\in\mathbb{R}-\mathbb{Q}$, I can prove that $$x-\lfloor x\rfloor,\ 2x-\lfloor 2x\rfloor,\ \dots,\ nx - \lfloor nx\rfloor $$ are all distinct. If we assume otherwise then $$ (i-j)x = \lfloor ix\rfloor -\lfloor jx\rfloor$$ for $i\neq j$, and therefore $x\in\mathbb{Q}$.
Is this true if we consider $x\in \mathbb{Q}$?
If $x=\frac{p}{q}$, then, for $k,n\in\mathbb Z$:
$$(k+nq)x-\lfloor (k+nq)x\rfloor=kx+np-\lfloor kx+np\rfloor=kx+np-\lfloor kx\rfloor - np = kx-\lfloor kx\rfloor$$
thus the values $kx-\lfloor kx\rfloor$ are never all distinct if $x$ is rational, and they repeat with the period at most $q$.