Show that $\left[\alpha\right]+\left[\alpha+\frac{1}{m}\right]+\cdots + \left[\alpha+\frac{m-1}{m}\right] = \left[m\alpha\right]$

Question

Let $m$ be a positive integer and let $\alpha$ be a real number. Show that $$\left[\alpha\right]+\left[\alpha+\frac{1}{m}\right]+\cdots + \left[\alpha+\frac{m-1}{m}\right] = \left[m\alpha\right]$$ where $\left[\ \ \right]$ is the integral part function.

Remark: Attempt to prove this result by induction but I failed in the inductive step, I have been unable use the induction hypothesis.

Answer 1

You don't have to use induction.

Suppose $\frac{l}{m}\leqslant \alpha-\lfloor \alpha \rfloor<\frac{l+1}{m}, \:0\leqslant l<m,\:l\in \Bbb{N}$. Then $$ \:l\leqslant m(\alpha-\lfloor \alpha \rfloor)<l+1 \quad\text{and so}\quad \lfloor m\alpha \rfloor=l+m\lfloor \alpha \rfloor $$ Moreover $$ \left\lfloor \alpha +\frac{k}{m}\right\rfloor-\lfloor \alpha \rfloor=\begin{cases}0,\quad0\leqslant k<m-l,\:k\in \Bbb{N} \\1,\quad m-l\leqslant k<m \end{cases} $$ Hence $$ \sum_{k=0}^{m-1}\left\lfloor \alpha +\frac{k}{m}\right\rfloor=(m-l)\lfloor \alpha \rfloor+ l+l\lfloor \alpha \rfloor=l+m\lfloor \alpha \rfloor=\lfloor m\alpha \rfloor $$

Answer 2

This is known as Hermite's identity. A nice proof is due to Matsuoka "On a Proof of Hermite's Identity", AMM 71:10 (1964), pp. 1115, DOI: 10.2307/2311413.

Let:

$\begin{align} f(x) = \lfloor m x \rfloor - \lfloor x \rfloor - \left\lfloor x + \frac{1}{m} \right\rfloor - \left\lfloor x + \frac{2}{m} \right\rfloor - \dotsb - \left\lfloor x + \frac{m - 1}{m} \right\rfloor \end{align}$

For real $\alpha, \beta$ it is $\lfloor \alpha + 1 \rfloor - \lfloor \beta + 1 \rfloor = \lfloor \alpha \rfloor - \lfloor \beta \rfloor$, so that:

$\begin{align} f\left( x + \frac{1}{m} \right) &= \lfloor m x + 1 \rfloor - \left\lfloor x + \frac{1}{m}\right\rfloor - \left\lfloor x + \frac{2}{m} \right\rfloor - \dotsb - \left\lfloor x + \frac{m - 1}{m} \right\rfloor - \lfloor x + 1 \rfloor \\ &= \lfloor m x \rfloor - \lfloor x \rfloor - \left\lfloor x + \frac{1}{m} \right\rfloor - \left\lfloor x + \frac{2}{m} \right\rfloor - \dotsb - \left\lfloor x + \frac{m - 1}{m} \right\rfloor \\ &= f(x) \end{align}$

On the other hand, for $0 \le x < 1 /m$ it is $f(x) = 0$, and so $f(x) = 0$ for all $x$, as was to be proved.