Show that $\left[\alpha\right]+\left[\alpha+\frac{1}{m}\right]+\cdots + \left[\alpha+\frac{m-1}{m}\right] = \left[m\alpha\right]$
Floor function properties: $[2x] = [x] + [ x + \frac12 ]$ and $[nx] = \sum_{k = 0}^{n - 1} [ x + \frac{k}{n} ] $