(Dis)prove: $\left\lceil\frac{\left\lceil\frac{x}2\right\rceil}2\right\rceil=\left\lceil\frac{x}4\right\rceil\ \forall x\in\Bbb R$

Question

I don't how to prove if $\frac{x}2\in\Bbb Z$.

Let $\frac{x}2=n\in\Bbb Z$.

$\begin{aligned}\left\lceil\frac{\left\lceil\frac{x}2\right\rceil}2\right\rceil&=\left\lceil\frac{x}4\right\rceil\\\left\lceil\frac{n}2\right\rceil&=\left\lceil\frac{x}2\cdot\frac12\right\rceil\\\left\lceil\frac{n}2\right\rceil&=\left\lceil\frac{n}2\right\rceil\end{aligned}$

Let $\frac{x}2, b\in\Bbb R,\ 0<b<1,\ a\in\Bbb Z$.

$m=a+b$

case 1: $0<b<0.5$

$0<b/2<0.25$

$\left\lceil\frac{\lceil a+b\rceil}2\right\rceil=\left\lceil\frac{a+b}2\right\rceil$ What should $\frac{a}2$ be?

Answer 1

Write $x=4m-a$ with $m\in\Bbb Z$ and $a\in[0,4)$. Then $$\left\lceil\frac x4\right\rceil =\left\lceil m-\frac a4\right\rceil =m,$$ $$\left\lceil\frac x2\right\rceil =\left\lceil 2m-\frac a2\right\rceil =\begin{cases}2m&0\le a<2\\2m-1&2\le a<4\,\end{cases}$$ and so $$ \left\lceil\frac{ \left\lceil\frac x2\right\rceil}2\right\rceil=\left\lceil m\quad\textit{or}\quad m-\frac12\right\rceil=m$$