Proof $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $ I want to prove that $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $, I started proving but I got stuck.
Let $ x, y \in \mathbb{R} $. Therefore:
$ \lfloor{x}\rfloor \le x $
$ \lfloor{y}\rfloor \le y $
$ \Rightarrow $
$ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le x + y $

$ \lfloor{x}\rfloor $ is an integer $ \leq x;\ \lfloor{y}\rfloor $ is an integer $ \leq y.$
Therefore, $ \lfloor{x}\rfloor + \lfloor{y}\rfloor $ is an integer that is $ \leq x+y.$
Since $\lfloor{x+y}\rfloor$ is the greatest integer $ \leq x+y,$ and $ \lfloor{x}\rfloor + \lfloor{y}\rfloor $ is an integer that is $ \leq x+y,$ it follows that $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \leq \lfloor{x+y}\rfloor.$
