[Math] Inequalities with floor function

Question

I need some help with this exercise, I'm pretty new solving this exercises.

$$ \lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$$

I know that I had to use the formal definition of the floor function, which is:

$$ \lfloor x \rfloor = \max {\{ m \in \Bbb Z \mid m \le x\}}$$

Answer 1

Use that $x=\lfloor x\rfloor +\alpha,\,\alpha\in[0,1)$ and $y=\lfloor y\rfloor +\beta,\,\beta\in[0,1)$

$$\lfloor x+y\rfloor =\lfloor \lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta\rfloor\ge \lfloor x\rfloor+\lfloor y\rfloor $$ because $\lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta\ge \lfloor x\rfloor+\lfloor y\rfloor$ and therefore $\lfloor x+y\rfloor$ is at least $\lfloor x\rfloor+\lfloor y\rfloor$

For the other inequality: $$\lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta < \lfloor x\rfloor+\lfloor y\rfloor+2\Rightarrow $$ From here you see that one integer $m\in\mathbb Z$ that satisfies $$(*)\quad\quad m\leq \lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta $$ is $m=\lfloor x\rfloor+\lfloor y\rfloor$ and the greatest possible is $\lfloor x\rfloor+\lfloor y\rfloor+1$, because already $\lfloor x\rfloor+\lfloor y\rfloor+2>\lfloor x\rfloor+\lfloor y\rfloor+\alpha+\beta$ -see $(*)$. Therefore $$\lfloor x+y\rfloor =\lfloor \lfloor x\rfloor+\alpha+\lfloor y\rfloor +\beta\rfloor\leq \lfloor x\rfloor+\lfloor y\rfloor+1$$