Values $nx - [nx]$ are distinct for an irrational number

Question

When reading the proof for Dirichlet's Approximation Theorem, I came across the following statement:

If $x$ is irrational, then $nx - [nx]$ are distinct for all $n \in \mathbb{Z}$.

I don't really understand why. Can someone explain to me, preferably with a proof. Thanks.

Answer 1

Suppose $nx-[nx]=mx-[mx]$. Then $$(n-m)x=[nx]-[mx]\in\mathbb{Z}$$ But an integer multiple of an irrational number can only be an integer if it is $0$, so we must have $$n=m$$ Therefore it is distinct for all $n$.

Answer 2

We have to prove that for distinct $i,j, \{ix \} \ne \{jx\}$ where $\{x\}$ denotes the fractional part and $x$ is irrational.

If $\{ix\} = \{ij\}$, then $ix-ij = \lfloor ix \rfloor - \lfloor ij \rfloor$ or $x = \frac{\lfloor ix \rfloor - \lfloor ij \rfloor}{i-j} \in \mathbb{Q}$ which is a contradiction.