Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one

Question

Prove that $\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m-n}{p^k}\bigg\rfloor$ equal to zero or one for all $k,m,n\in \mathbb N$. where $m\geqslant n$ and $p$ is a prime.

What I did:

$\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor-\bigg\lfloor\frac{m}{p^k}-\frac{n}{p^k}\bigg\rfloor$

Let $x:=\frac{m}{p^k}$ and let $y:=\frac{n}{p^k}$

we get $\lfloor x\rfloor-\lfloor y\rfloor-\lfloor x-y\rfloor$

What should I do now?

Answer 1

Here's how you can prove it. This problem is easier if you generalize it to $$\lfloor x \rfloor +\lfloor y \rfloor +1 \ge \lfloor x+y \rfloor \ge \lfloor x \rfloor + \lfloor y \rfloor $$

This inequality can be proven if you set $x=n+\alpha, y=m+\beta$ where $0 \le \alpha, \beta \le 1$.

Now, since $[x+y]$ is an integer, this implies $\lfloor x+y \rfloor $ is $\lfloor x \rfloor + \lfloor y \rfloor$ or $\lfloor x \rfloor +\lfloor y \rfloor +1 $.

Answer 2

Note that $$ \bigg\lfloor\frac{m}{p^k}\bigg\rfloor\le \frac{m}{p^k}<\bigg\lfloor\frac{m}{p^k}\bigg\rfloor+1\quad\text{and}\quad \bigg\lfloor\frac{n}{p^k}\bigg\rfloor\le \frac{n}{p^k}<\bigg\lfloor\frac{n}{p^k}\bigg\rfloor+1. $$ Hence, $$ \frac{m}{p^k}-1-\frac{n}{p^k}<\bigg\lfloor\frac{m}{p^k}\bigg\rfloor-\bigg\lfloor\frac{n}{p^k}\bigg\rfloor<\frac{m}{p^k}-\frac{n}{p^k}+1. $$ Now you can do the same for the third term in your expression.

Answer 3

Since

$x = \left\lfloor x \right\rfloor + \left\{ x \right\}$

then we have $$ \eqalign{ & \left\lfloor {x - y} \right\rfloor = - \left\lceil {y - x} \right\rceil = \cr & = \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor + \left\lfloor {\left\{ x \right\} - \left\{ y \right\}} \right\rfloor = \cr & = \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor - \left[ {\left\{ x \right\} < \left\{ y \right\}} \right] \cr} $$ where $$ \left[ P \right] = {\rm Iverson bracket} = \left\{ \matrix{ 1\;{\rm if}\;P = TRUE \hfill \cr 0\;{\rm if}\;P = FALSE \hfill \cr} \right. $$ Thus $$ \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor - \left\lfloor {x - y} \right\rfloor = \left[ {\left\{ x \right\} < \left\{ y \right\}} \right] = 0\;OR\;1\quad \left| {\;\forall x,y\;{\rm real}} \right. $$