# Prove $\lfloor x + n \rfloor= \lfloor x\rfloor + n : n \in \mathbb{Z}$

Prove $$\lfloor x + n \rfloor= \lfloor x\rfloor + n : n \in \mathbb{Z}$$

So far I have used that $$\lfloor x + n \rfloor - n \leq x < \lfloor x + n \rfloor - n + 1$$, but I don't know how to continue.

## 2 Answers

Just use the definition of $$\lfloor\, \cdot\,\rfloor$$: $$m=\lfloor x\rfloor\iff m\le x hence $$\;\lfloor x+n \rfloor=m+n$$.

Write $$x=k+r$$ where $$k=[x]$$ and $$0\leq r<1$$ Then we have $$[x+n] = [k+n+r] \underbrace{=}_{ \rm definition\; of \;[...]} k+n = [x]+n$$