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.
 A: Let $x = k + y$, where $k\in\mathbb{Z}$ is the integer part of $x$ ($k = \lfloor x \rfloor$) and $y\in[0,1)$ is its fractional part. 
Then, for LHS:
$$
\left\lfloor \frac {\lfloor px \rfloor}{p} \right\rfloor = 
\left\lfloor \frac {\lfloor pk + py \rfloor}{p} \right\rfloor = 
\left\lfloor \frac {pk + \lfloor py \rfloor}{p} \right\rfloor = 
\left\lfloor k + \frac {\lfloor py \rfloor}{p} \right\rfloor = 
k + \left\lfloor \frac {\lfloor py \rfloor}{p} \right\rfloor.
$$
However for every $y\in[0,1)$ we have $\lfloor py \rfloor \leq py < p$. 
Thus $\left\lfloor \frac {\lfloor py \rfloor}{p} \right\rfloor = 0$.
A: $\lfloor px\rfloor \le px<\lfloor px\rfloor +1$
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 $p\lfloor x \rfloor \le \lfloor px \rfloor < px < \lfloor px \rfloor +1 < p\lfloor x \rfloor + p$ ....
A: Since
$$
x=\lfloor x\rfloor+\{x\}\tag1
$$
and $\lfloor x\rfloor\in\mathbb{Z}$, we have
$$
\lfloor px\rfloor=p\lfloor x\rfloor+\lfloor p\{x\}\rfloor\tag2
$$
so
$$
\frac{\lfloor px\rfloor}p=\lfloor x\rfloor+\frac{\lfloor p\{x\}\rfloor}p\tag3
$$
Therefore,
$$
\left\lfloor\frac{\lfloor px\rfloor}p\right\rfloor\ge\lfloor x\rfloor\tag4
$$

Since
$$
\lfloor px\rfloor\le px\tag5
$$
we have
$$
\left\lfloor\frac{\lfloor px\rfloor}p\right\rfloor\le\lfloor x\rfloor\tag6
$$

Inequalities $(4)$ and $(6)$ give
$$
\left\lfloor\frac{\lfloor px\rfloor}p\right\rfloor=\lfloor x\rfloor\tag7
$$
