Prove an equality with floor function.

Question

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.

Answer 1

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 $\left\lfloor \frac {\lfloor py \rfloor}{p} \right\rfloor = 0$.

Answer 2

$\lfloor px\rfloor \le px<\lfloor px\rfloor +1$

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 $p\lfloor x \rfloor \le \lfloor px \rfloor < px < \lfloor px \rfloor +1 < p\lfloor x \rfloor + p$ ....

Answer 3

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, $$ \left\lfloor\frac{\lfloor px\rfloor}p\right\rfloor\ge\lfloor x\rfloor\tag4 $$

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

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