What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$

Question

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

Answer 1

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

$$\lfloor x \rfloor + \lfloor y \rfloor \le \lfloor x + y \rfloor \le \lfloor x \rfloor + \lfloor y \rfloor + 1$$

Answer 2

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 $$\lfloor x\rfloor+\lfloor y\rfloor\le \lfloor x+y\rfloor\le\lfloor x\rfloor+\lfloor y\rfloor+1.$$