An Arithmetic Progression Question

Question

Let $(a_{n})_{n\geq 1}$ be an arithmetic progression with $r>0$ such as $\vee n\geq 1,\vee x\in (a_{n},a_{n+1})$ we have $\left [ x \right ]=\left [ a_{n} \right ]$. Prove that ∃m$\in N,m\neq 0$ such as $r=\frac{1}{m}$. I figured out that $x\in (a_{n},a_{n}+r)$, but I don't know what to do from here

Answer 1

Consider what would happen if there were an $n\geq 1$ such that $a_n < k < a_{n+1}$ where $k$ is an integer. Also consider what the terms of the sequence near $k+1$ might be.

Remember the property $\forall n\geq 1,\forall x\in (a_{n},a_{n+1}), \left [ x \right ]=\left [ a_{n} \right ]$. It will matter somehow.