Does the floor of a number preserve order? For example, say you got x < y for some x and y. Then 
$\lfloor x \rfloor \geq \lfloor y \rfloor $ ?  Is it always the case? 
The reason why I am confused on this point is I was reading a solution posted on chegg which doesn't seem convincing. 
The question states to prove $\lfloor x + y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor$ for some real $x,y$. And their proof states that $x \geq \lfloor x \rfloor$ and $y \geq \lfloor y \rfloor$ then $x + y \geq \lfloor x \rfloor + \lfloor y \rfloor$. I understand this part, but then they said since the greatest integer function (floor) preserve order, then 
$\lfloor x + y \rfloor  \leq \lfloor  \lfloor x \rfloor \rfloor + \lfloor  \lfloor y \rfloor \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$  
which is not true. Assume it is true then since floor preserve order then the negation of $\geq$ is $<$ not $\leq$
if you guys can explain this it would be great! thank you

If $x<y$, then $\lfloor x\rfloor\leq\lfloor y\rfloor$. (Your inequality is flipped.)
