Problem about $[x]$ 
$$[x]-2[x/2]\leq 1$$
  Equivalently, $[x]-2[x/2]$ assumes only the values 0 and 1.

It seems easy, but I don't know how to prove it...

Say $x=n+r$, with $0\leq r\lt 1$, and $n\in\mathbb{Z}$. 
If $n=2k$ is even, then $\frac{x}{2} = k+(r/2)$, $0\leq (r/2)\lt 1$, so $[x]=2k$, $[x/2]=k$, and $[x]-2[x/2] = 2k-2k = 0$.
If $n=2k+1$ is odd, then $x/2 = k+(0.5+(r/2))$. Since $0\leq r\lt 1$, then $0\leq \frac{r}{2}\lt 0.5$, so $[x/2] = k$. Hence $[x]-2[x/2] = 2k+1 - 2k = 1$.
