Floor Function. Let x and y be rational numbers let x and y be rational numbers. 
A. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor 
    = \left \lfloor x+y \right \rfloor$
B. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor 
    \le \left \lfloor x+y \right \rfloor$
C. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor 
    \ge \left \lfloor x+y \right \rfloor$
D. None of the above.
Can anyone please explain why the answer is B? Thank You!

Let $\left \lfloor x \right \rfloor = x_{0} + x_{1}, x_{0} \in \mathbb{Z}, 0\leq x_{1}<1$
Similarily we define $\left \lfloor  y\right \rfloor$
Then $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor = x_{0} + y_{0}$
Now there are two cases to consider:
(i) $x_{1} + y_{1} < 1$
and
(ii) $x_{1} + y_{1} \geq 1$
Case (i):
$\left \lfloor x + y \right \rfloor = \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor = x_{0} + y_{0}$
Case (ii):
$\left \lfloor x + y \right \rfloor = x_{0} + y_{0} + 1 >  \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor $
Therefore $\forall x,y \in \mathbb{Q}$, $\left \lfloor x + y \right \rfloor \geq \left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor$ as required 
