Show $\left \lfloor{x}\right \rfloor + \left \lfloor{y}\right \rfloor \leq \left \lfloor{x+y}\right \rfloor$ for every $x, y \in \mathbb{R}$

$\left \lfloor{x}\right \rfloor + \left \lfloor{y}\right \rfloor \leq \left \lfloor{x+y}\right \rfloor$ for every $x, y \in \mathbb{R}$

Got a question to show this inequality but I have no clue how to formalize a proof for this. It seems pretty obvious (summing two numbers before we get rid of the non integer parts will always be greater or equal than the sum of the integer parts), how would one write this formally (ideally using elementary number theory techniques).

• You have $\lfloor x \rfloor + \lfloor y \rfloor \leq x+y$ and the left hand side is an integer. – copper.hat Feb 7 '18 at 3:27
• $\lfloor{x+y}\rfloor$ is going to take the greatest number $\le x+y$, while the right left hand side does this twice. – Yash Jain Feb 7 '18 at 3:32

We always have $\lfloor x \rfloor \le x$, which implies that $$\lfloor x \rfloor + \lfloor y \rfloor \le x + y.$$ By definition, $\lfloor x+y \rfloor$ is the greatest integer less than or equal to $x+y$. But $\lfloor x \rfloor + \lfloor y \rfloor$ is an integer less than or equal to $x+y$, which implies that $$\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x+y\rfloor.$$
$x = \left \lfloor{x}\right \rfloor + \alpha\\ y = \left \lfloor{y}\right \rfloor + \beta$
where $\alpha, \beta \geq 0$.
$x + y = \left \lfloor{x}\right \rfloor + \left \lfloor{y}\right \rfloor + \alpha + \beta\\ \left \lfloor{x+y}\right \rfloor = \left \lfloor{x}\right \rfloor + \left \lfloor{y}\right \rfloor + \left \lfloor{\alpha + \beta}\right \rfloor$
$\implies\left \lfloor{x+y}\right \rfloor \geq \left \lfloor{x}\right \rfloor + \left \lfloor{y}\right \rfloor$