Floor plus One

Theorem
Let $$x \in \R$$.

Then:
 * $$\left \lfloor {x + 1} \right \rfloor = \left \lfloor {x} \right \rfloor + 1$$

where $$\left \lfloor {x} \right \rfloor$$ is the floor function of $$x$$.

Proof
$$\lfloor x + 1 \rfloor = n \Rightarrow n \le x + 1 \le n + 1 \Rightarrow n-1 \le x \le n \Rightarrow \lfloor x \rfloor = n-1 \Rightarrow \lfloor x+1 \rfloor = \lfloor x \rfloor + 1$$. In general, $$ \lfloor x + n \rfloor = \lfloor x \rfloor + n, n \in \Z $$