If $r\in(0,1)$ is rational and $m$ is a positive integer, then $[rm]+[(1-r)m]=m-1$ where $[\cdot]$ denotes the integral part

Question

Let $[\cdot]$ denote the integral part of a real number, that is the only integer $a$ satisfying

$a \leq x < a+1$.

Let $0<r<1$ be a rational number and $m \geq 0$ a positive integer.

How do you prove that the following equality?

$[rm]+[(1-r)m]=m-1$

This should be completely elementary but I don't manage!!

Answer 1

The claimed formula is incorrect. If $r=1/2$ and $m=2$ then the two integer parts sum to $m$ rather than $m-1$.

Answer 2

This will not be true always. Let $r=\frac 12, m=4$, then $[rm]+[(1-r)m]=4\neq m-1$. But this gives a hint: if $rm$ is integral, we have $[rm]+[(1-r)m]=rm+m-rm=m$, while if $rm$ is not integral we have $rm-[rm]+\{rm\}$ and $[rm]+[(1-r)m]=[rm]+[m-[rm]-\{rm\}]=[rm]+m-[rm]-1$ So the formula is correct for $rm$ non-integral, and the $-1$ should be removed for $rm$ integral.