What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$ What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$?    
My effort: 
We have $x+y-1\lt \lfloor x+y\rfloor \leq x+y$. What can I do after this? Please help
 A: For some real number $0  \le \alpha < 1,\quad x = \lfloor x \rfloor + \alpha$
That is really all you need to show that $$x-1 < \lfloor x \rfloor \le x < \lfloor x \rfloor + 1$$
and
$$\lfloor \lfloor x \rfloor + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$$
One thing you can use it for is, for some $0 \le \alpha, \beta < 1$
\begin{align}
   \lfloor x + y \rfloor
   &= \lfloor
      (\lfloor x \rfloor + \alpha) +
      (\lfloor y \rfloor + \beta)
      \rfloor \\
   &= \lfloor x \rfloor + \lfloor y \rfloor +
      \lfloor \alpha + \beta \rfloor \\
\end{align}
Since $0 \le \alpha + \beta < 2$, it follows that $0 \le \lfloor \alpha + \beta \rfloor \le 1$. So
$$\lfloor x \rfloor + \lfloor y \rfloor 
  \le \lfloor x + y \rfloor
   \le \lfloor x \rfloor + \lfloor y \rfloor + 1$$
A: As by definition, $\,\lfloor x\rfloor\le x <\lfloor x\rfloor+1$, we deduce
$$\lfloor x\rfloor+\lfloor y\rfloor\le x+y<\lfloor x\rfloor+\lfloor y\rfloor+2, $$
whence
$$\lfloor x\rfloor+\lfloor y\rfloor\le \lfloor x+y\rfloor\le\lfloor x\rfloor+\lfloor y\rfloor+1.$$
