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

Question

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.

Answer 1

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

Answer 2

Just use the definition of $\lfloor\, \cdot\,\rfloor$: $$m=\lfloor x\rfloor\iff m\le x <m+1,\quad\text{so }\;m+n\le x+n<m+n+1,$$ hence $\;\lfloor x+n \rfloor=m+n$.