Prove that if $m$ is a positive integer and $x$ is a real number...

Question

Q: Prove that if $m$ is a positive integer and $x$ is a real number, then

$$ \lfloor mx \rfloor=\lfloor x \rfloor+\lfloor x+\frac{1}{m} \rfloor+\lfloor x+\frac{2}{m} \rfloor+...+\lfloor x+\frac{m-1}{m} \rfloor. $$

Let $x=n+\frac{r}{m}+c$, where n is an integer, r is a nonnegative integer less than m, and c is a real number with $0\le c<\frac{1}{m}$.

If we substitute this x to $\lfloor mx \rfloor$, we get

$\lfloor mx \rfloor=\lfloor m(n+\frac{r}{m}+c) \rfloor=\lfloor mn+r+mc \rfloor$.

Because $c<\frac{1}{m},$ $mc$ is smaller than 1. Hence, $\lfloor mn+r+mc \rfloor=mn+r.$

Moving on to the next function, we get

$\lfloor x+\frac{1}{m} \rfloor = \lfloor n+\frac{r}{m}+c+\frac{1}{m} \rfloor=\lfloor n+\frac{r+1}{m}+c \rfloor$.

$\lfloor x+\frac{2}{m} \rfloor = \lfloor n+\frac{r}{m}+c+\frac{2}{m} \rfloor=\lfloor n+\frac{r+2}{m}+c \rfloor$.

...

$\lfloor x+\frac{m-1}{m} \rfloor = \lfloor n+\frac{r}{m}+c+\frac{m-1}{m} \rfloor=\lfloor n+\frac{m+r-1}{m}+c \rfloor$.

In order to solve this countably infinite functions, we have to split them into two groups. The first group owns functions that $ \frac{r}{m}+c+\frac{y}{m}<1,$ so $\lfloor x+\frac{y}{m} \rfloor=n.$

Second group contains functions that $ \frac{r}{m}+c+\frac{y}{m} \ge 1,$ so $\lfloor x+\frac{y}{m} \rfloor=n+1.$

Then, we have to know where the number starts to become 1. We can notice this by:

$ \frac{r+y}{m}+c = 1$

$r+y=m(1-c)$

$y=m(1-c)-r$

Hence, we now know that

$\lfloor x+\frac{m(1-c)-r}{m} \rfloor=\lfloor n+1 \rfloor=n+1.$

Therefore, from $\lfloor x \rfloor$ to $\lfloor x+\frac{m(1-c)-r-1}{m} \rfloor$, all the values are n. From $\lfloor x+\frac{m(1-c)-r}{m} \rfloor$ to $\lfloor x+\frac{m-1}{m} \rfloor$, all the values are n+1.

But then I am stuck at this part. I have no idea how to move on to the next step to solve this question.

Please Help Me!!

Answer 1

Since there are $m$ terms in any series from $0$ to $m-1$, you have that $$\begin{align} \lfloor mx\rfloor&=(m-k)\times n+k\times (n+1) \\ &=mn+k \end{align}$$ You have only to prove that $k=r$. You will see that $k$ can be determined as follows. $$\begin{align} k&=\lfloor (m-1)-(m(1-c)-r-1)\rfloor \\ &=\lfloor m-1-m+mc+r+1\rfloor \\ &=\lfloor mc+r\rfloor \end{align}$$ And since $c<\frac{1}{m}$, $mc<1$. So, $k=\lfloor mc+r \rfloor = r$.

Answer 2

I totally understood this question.

From $\lfloor x+\frac{(1-c)m-r}{m} \rfloor$, the value becomes $n+1$.

Hence, the number of $n$ is

$m(1-c)-r=m-mc-r$. Because the number of n has to be an integer and $mc<1$, it becomes: $m-r$.

The number of $n+1$ is

$(m-1)-(m(1-c)-r-1)=(n+1)(mc+r)$.

Following the exact same logic that is explained above, we get: $mc+r=r$.

Thus, if we add them together,

$n(m-r)+(n+1)r=mn-nr+nr+r=mn+r$

Hence, proved.