Prove the following identity holds for all real numbers $x$:
$$\lfloor4x\rfloor=\lfloor x \rfloor +\left\lfloor x+\frac14 \right\rfloor+\left\lfloor x+\frac24\right\rfloor+\left\lfloor x+\frac34\right\rfloor$$
I understand that $\lfloor4.3\rfloor$ would be $4$ and $\lfloor-2.4\rfloor$ would be $-3$
I am trying to prove this by cases. I think that I should prove each case first such as $\lfloor x\rfloor$ first and then $\left\lfloor x+\frac14\right\rfloor$ and on but I'm having trouble proving it in generality
Best Answer
Hint: Separate all real numbers into a few groups, depending on the relation of their fractional part to the intervals formed by $0$, $\frac14$ , $\frac12$ , $\frac34$ and $1$.