<s>[INST] Prove an equality with floor function. Let $p\in \Bbb N \ne 0$ and $x\in \Bbb R$.
prove that
$$\left\lfloor \frac {\lfloor px \rfloor}{p} \right\rfloor=\lfloor x\rfloor$$
I tried using the double inequality
$$\lfloor px\rfloor \le px<\lfloor px\rfloor +1$$
and divided by $p$ but a small problem remains.
 [/INST] $\lfloor px\rfloor \le px<\lfloor px\rfloor +1$
Right. So $\frac {\lfloor px\rfloor}p \le \frac{px}p<\frac {\lfloor px\rfloor +1}p$
$\frac {\lfloor px\rfloor}p \le x<\frac {\lfloor px\rfloor}p + \frac 1p$
...
But perhaps more to the point.
$\lfloor x\rfloor \le x < \lfloor x\rfloor + 1$ 
$p\lfloor x\rfloor \le px < p\lfloor x\rfloor + p$.  
$p\lfloor x\rfloor \le \lfloor px \rfloor \le px <  \lfloor px \rfloor + 1 \le p\lfloor x\rfloor + p$
$\lfloor x\rfloor \le \frac {\lfloor px \rfloor}p \le x < \lfloor x\rfloor + 1$
.... or to go back to your original idea:
Not $\lfloor x \rfloor \le x$ so $p\lfloor x \rfloor \le px$ but $\lfloor px \rfloor$ is the largest possible integer equal or less than $px$ so
$p\lfloor x \rfloor \le \lfloor px \rfloor$
So you have $p\lfloor x \rfloor \le \lfloor px \rfloor < px < \lfloor px \rfloor +1 < p\lfloor x \rfloor + p$ ....
</s>