Proof using pigeonhole and greatest integer (floor) function.

Question

The question is to prove that if m is a positive integer then, $$[mx] = [x] + \left[x+\frac{1}{m}\right] +\left[x+\frac{2}{m}\right] + \cdots + \left[x+\frac{(m-1)}{m}\right] $$ for $x \in \mathbb{R}$. Where $[x] =n$ such that $ n \leq x <n+1$

I'm given that the solution should use the pigeonhole principle. So I need to look for $n$ boxes where I'm trying to stuff $n+1$ things (my understanding of the pigeonhole principle). If I look at the fractional part of x I see $\{x\} \in [0,1)$ where $\{x\}$ is the fraction part of x. So $m\{x\} \in [0,m)$. I can break this interval into $[0,1) [1,2) \ldots [m-1,m)$ or I can divide everything by m and get $[0,\frac{1}{m}) [\frac{1}{m},\frac{2}{m}) \ldots [\frac{m-1}{m},1)$ and get m intervals of length $\frac{1}{m}$ which is my "n" boxes.

I am however stuck with regards to the "$n+1$" objects to put in the box and how to relate this observation to the question. Can anybody provide a hint or point out a flaw in what I have so far?

Answer 1

I wouldn’t use the pigeonhole principle: it’s not needed. You know that there is a unique integer $n$ such that $0\le n<m$ and $\frac{n}m\le\{x\}<\frac{n+1}m$. Now consider $x+\frac{k}m$, where $0\le k<m$:

$$x+\frac{k}m=\lfloor x\rfloor +\{x\}+\frac{k}m\;,$$ so

$$\lfloor x\rfloor+\frac{n+k}m\le x+\frac{k}m<\lfloor x\rfloor +\frac{n+1+k}m\;.\tag{1}$$

What does $(1)$ tell you about $\left\lfloor x+\frac{k}m\right\rfloor$? Specifically, for how many values of $k$ will it be $\lfloor x\rfloor$, and for how many will it be $\lfloor x\rfloor+1$? If you can answer that, you’ll have a good handle on the righthand side of your desired equation.

As for the lefthand side, start from the fact that $\lfloor mx\rfloor=\big\lfloor m(\lfloor x\rfloor+\{x\})\big\rfloor$.

Most of the Missing Detail: If $0\le k\le m-n-1$, $(1)$ tells us that

$$\lfloor x\rfloor\le\lfloor x\rfloor+\frac{n}m\le \lfloor x\rfloor+\frac{n+k}m\le x+\frac{k}m<\lfloor x\rfloor+\frac{n+k+1}m=\lfloor x\rfloor+1\;;$$

after getting rid of the excess baggage, we have $$\lfloor x\rfloor\le x+\frac{k}m<\lfloor x\rfloor+1\;,$$ and therefore $$\left\lfloor x+\frac{k}m\right\rfloor=\lfloor x\rfloor\;.\tag{2}$$

On the other hand, if $m-n\le k\le m-1$, $(1)$ implies that

$$\lfloor x\rfloor+1\le\lfloor x\rfloor+\frac{n+k}m\le x+\frac{k}m<\lfloor x\rfloor+\frac{n+1+k}m\le\lfloor x\rfloor+\frac{n+m}m<\lfloor x\rfloor+2$$

and hence that $$\left\lfloor x+\frac{k}m\right\rfloor=\lfloor x\rfloor +1\;.\tag{3}$$

Now $(2)$ holds for $m-n$ values of $k$, and $(3)$ holds for the other $n$ values of $k$, so

$$\begin{align*}\sum_{k=0}^{m-1}\left\lfloor x+\frac{k}m\right\rfloor&=(m-n)\lfloor x\rfloor+n(\lfloor x\rfloor+1)\\ &=m\lfloor x\rfloor+n\;. \end{align*}$$

But $$\begin{align*} \lfloor mx\rfloor&=\big\lfloor(m\lfloor x\rfloor+\{x\})\big\rfloor\\ &=\big\lfloor m\lfloor x\rfloor+m\{x\}\big\rfloor\\ &=m\lfloor x\rfloor+\lfloor m\{x\}\rfloor\;, \end{align*}$$

since $m\lfloor x\rfloor$ is an integer.

Now you need only show that $\lfloor m\{x\}\rfloor=n$ to finish the argument. Go back to the beginning to recall how $n$ was defined.

Answer 2

Hint: it's clear upon writing $x$ in radix $m$, e.g. for $\:m = 10\:$ and $\:x = 12.34\:$ we have

$\qquad \lfloor 12.{\color{red}3}4\rfloor + \lfloor 12.44\rfloor + \lfloor 12.54\rfloor +\:\cdots\:+ \lfloor {\color{red}{13.04}}\rfloor + \lfloor {\color{red}{13.14}}\rfloor + \lfloor {\color{red}{13.24}} \rfloor $

$\quad =\ 10\cdot 12 + {\color{red}3}\ =\ \lfloor 10\cdot 12.34 \rfloor$

So the "boxes" for the box principle are the $m$ possible digits following the decimal point.

Notice how the choice of radix representation greatly clarifies the innate structure.

Answer 3

For $m\in\mathbb{Z}$, define $$ f_m(x)=\lfloor mx\rfloor-m\lfloor x\rfloor\tag{1} $$ For $k\in\mathbb{Z}$, $\lfloor x\rfloor-k=\lfloor x-k\rfloor$; thus, because $m\lfloor x\rfloor\in\mathbb{Z}$, we have $$ \begin{align} f_m(x) &=\lfloor mx\rfloor-m\lfloor x\rfloor\\ &=\lfloor mx-m\lfloor x\rfloor\rfloor\\ &=\lfloor m(x-\lfloor x\rfloor)\rfloor\tag{2} \end{align} $$ Since $0\le x-\lfloor x\rfloor<1$, we get that $0\le\lfloor m(x-\lfloor x\rfloor)\rfloor\le m-1$; that is, $$ 0\le f_m(x)\le m-1\tag{3} $$ Furthermore, for $k\in\mathbb{Z}$, $$ \begin{align} f_m\left(x+\frac{k}{m}\right) &=\left\lfloor m\left(x+\frac{k}{m}\right)\right\rfloor-m\left\lfloor x+\frac{k}{m}\right\rfloor\\ &=\left\lfloor mx\vphantom{\frac{k}{m}}\right\rfloor+k-m\left\lfloor x+\frac{k}{m}\right\rfloor\\ &\equiv\lfloor mx\rfloor+k-m\lfloor x\rfloor\pmod{m}\\ &=f_m(x)+k\tag{4} \end{align} $$

Therefore, $(3)$ and $(4)$ show that for $k=0,1,\dots,m-1$, $f_m\left(x+\frac{k}{m}\right)$ takes on each value $0,1,\dots,m-1$. Thus, $$ \sum_{k=0}^{m-1}f_m\left(x+\frac{k}{m}\right)=\frac{m(m-1)}{2}\tag{5} $$ Relating $(5)$ and $(1)$ yields $$ \begin{align} \frac{m(m-1)}{2} &=\sum_{k=0}^{m-1}\left\lfloor m\left(x+\frac{k}{m}\right)\right\rfloor-m\left\lfloor x+\frac{k}{m}\right\rfloor\\ &=\sum_{k=0}^{m-1}\left\lfloor mx\vphantom{\frac{k}{m}}\right\rfloor+k-m\left\lfloor x+\frac{k}{m}\right\rfloor\\ &=m\lfloor mx\rfloor+\frac{m(m-1)}{2}-\sum_{k=0}^{m-1}m\left\lfloor x+\frac{k}{m}\right\rfloor\tag{6} \end{align} $$ Simplifying equation $(6)$ yields $$ \lfloor mx\rfloor=\sum_{k=0}^{m-1}\left\lfloor x+\frac{k}{m}\right\rfloor\tag{7} $$ As far as I see, step $(5)$, where we fill up all the equivalence classes $\!\!\!\pmod{m}$, is as close to a pigeonhole as we get.

Answer 4

The following is a classical proof which I love, I'm surprised noboby mentioned it yet.

Let $$ f(x)= [x] + \left[x+\frac{1}{m}\right] +\left[x+\frac{2}{m}\right] + \cdots + \left[x+\frac{(m-1)}{m}\right] -[mx] $$

Then $f(x)=0 \forall x \in [0, \frac{1}{m}]$, since all terms are $0$.

Moreover, $$f(x+\frac{1}{m})=[x+\frac{1}{m}] + \left[x+\frac{2}{m}\right] +\left[x+\frac{3}{m}\right] + \cdots + \left[x+1\right] -[mx+1]=[x+\frac{1}{m}] + \left[x+\frac{2}{m}\right] +\left[x+\frac{3}{m}\right] + \cdots + \left[x\right]+1 -[mx]-1 = f(x)$$

Therefore $f$ is periodic with period $\frac{1}{m}$ and zero on an half open interval of length $\frac{1}{m}$. It is trivial to deduce that $f \equiv 0$.