# 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.

• At the beginning, you write $n$ where I think that you want $p$. (Else $n$ is irrelevant, and the claim is false if $0<p<1$.) – Toby Bartels Sep 17 '18 at 20:53

## 3 Answers

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

• Thanks a lot.{}{}{}{}{}{}{}{} – hamam_Abdallah Sep 17 '18 at 21:18
• @robjohn: Thanks! Now it's fixed. – mwt Sep 17 '18 at 22:15
• (+1) Now it looks correct! – robjohn Sep 17 '18 at 22:15

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

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

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 $$\left\lfloor\frac{\lfloor px\rfloor}p\right\rfloor=\lfloor x\rfloor\tag7$$

• Thanks all for the time you took. Merci. – hamam_Abdallah Sep 17 '18 at 22:44