I am told that there is absolutely a counter-example for this floor function $$\lfloor x+y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor $$
With the domain of the discourse being all real numbers. I have tried many different examples but I simply cannot figure anything out. If anyone could provide some help I would be much obliged.
There cannot be counterexample (if you wrote the inequality exactly), since $$ \eqalign{ & \left\lfloor {x + y} \right\rfloor = \left\lfloor {\left\lfloor x \right\rfloor + \left\{ x \right\} + \left\lfloor y \right\rfloor + \left\{ y \right\}} \right\rfloor = \cr & = \left\lfloor {\left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor } \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor = \cr & = \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor \ge \left\lfloor x \right\rfloor + \left\lfloor y \right\rfloor \cr} $$
and by definition of the fractional part $$ 0 \le \left\{ x \right\} < 1 $$