Finding supremum and infimum

Question

Let $A = \{\frac{x}{2} - \lfloor\frac{x+1}{2}\rfloor : x \in \mathbb{R} \}$

Does supremum and infimum of $A$ exist ? If the answer is yes then find them .

My try : I rewrite the expression $\frac{x}{2} - (\lfloor 2x \rfloor - \lfloor x \rfloor)$ but it doesn't help really . Also I substituted $x$ with $n + p $ which $n$ is an integer and $0\le p \lt 1$ and wasn't helpful again .

Answer 1

Hint: If $y = \frac{x+1}{2}$ then $\frac{x}{2}-\lfloor\frac{x+1}{2}\rfloor = y-\lfloor y\rfloor-1/2$. Can you the supremum and infimum?

Answer 2

Hint 1: Show that on $[-1,1)$, $\left\lfloor\frac{x+1}2\right\rfloor=0$.

Hint 2: Show that $\frac{x}2 - \left\lfloor\frac{x+1}2\right\rfloor$ has a period of $2$.