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!
 A: The fact that $\left\lfloor x\right \rfloor\leq x$ should be clear (let me know if it isn't). Now add this to the same inequality with $y$ to get:
$$\left\lfloor x\right \rfloor+\left\lfloor y\right \rfloor\leq x+y$$
Now, taking the floor of both sides does not change the left-hand side, and gives your answer.
A: 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 
A: If $x$ and $y$ are rational (or real) numbers, then there must exist numbers $h$ and $k$ such that $0 \le h,k < 1$,  $x = \left \lfloor x \right \rfloor + h$, and  $y = \left \lfloor y \right \rfloor + k$.
So:
\begin{align}
   \left \lfloor x + y \right \rfloor
   &= \left \lfloor
         \left \lfloor x \right \rfloor + h +
         \left \lfloor y \right \rfloor + k
     \right \rfloor \\
   &= \left \lfloor x \right \rfloor +
      \left \lfloor y \right \rfloor 
      + \left \lfloor h+k \right \rfloor\\
   &\ge \left \lfloor x \right \rfloor +
      \left \lfloor y \right \rfloor \\
\end{align}
A: For any $x$ then there is an integer $n$ so that $n \le x < n + 1$.  We call $n = [x]$.
so $[x] \le x < [x]+ 1; [y] \le y < [y] + 1$ so $[x]+ [y] \le x + y < [x] + [y] + 2$.
so there are two possibilities:
$[x] + [y] \le x + y < [x] + [y] + 1$ and so $[x+y] = [x]+ [y]$.  This will happen if $(x -[x]) + (y - [y]) < 1$.  e.g. $x = 2.3$ $y = 7.4$.
of $[x]+ [y] + 1 \le x+y < [x]+ [y]+ 1 + 1$ and $ [x+y] = [x]+[y] + 1$.  This will happen if $(x -[x]) + (y - [y]) \ge 1$. e.g $x = 2.5$ and $y=7.6$. 
So $[x+y] = \{[x]+[y], [x]+[y]+1\}$.
In either case... $[x]+ [y] \le [x+y]$.
