If $\displaystyle x \in \mathbb{Q}$ show that there is a uniquely determined $\displaystyle n \in \mathbb{Z}$, such that $\displaystyle n \leq x \leq n+1$

I was thinking of simply showing this by contradiction, that if $\displaystyle m \neq n$ then the integer values would be different, but I got nowhere. I have a feeling that I should assume that $\displaystyle m \neq n$, and by manipulating it getting $\displaystyle n-m=0$, but I'm not too sure on how to go about it.