Question

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

$\lfloor mx\rfloor =\lfloor x\rfloor +⌊x+\frac{1}{m}⌋+⌊x+\frac{2}{m}⌋+\dots +⌊x+\frac{m-1}{m}⌋$

Short answer

$\lfloor mx\rfloor =\lfloor x\rfloor +⌊x+\frac{1}{m}⌋+⌊x+\frac{2}{m}⌋+\dots +⌊x+\frac{m-1}{m}⌋$

Step-by-step solution

Step: 1

Let$x=n+(r/m)+\epsilon$ is a real number with $0\le \epsilon \le 1/m$.

The left-hand side is $\lfloor nm+r+m\epsilon \rfloor =nm+r$

Step: 2

On the right hand side, the terms$\lfloor x\rfloor$ through$\lfloor x+(m+r-1)/m\rfloor$ are all just n and the terms from$\lfloor x+(m-r)/m\rfloor$ on are all n+1. Therefore, the right-hand side is$(m-r)n+r(n+1)=nm+r,$ as well.

Step: 3

Hence$\lfloor mx\rfloor =\lfloor x\rfloor +⌊x+\frac{1}{m}⌋+⌊x+\frac{2}{m}⌋+\dots +⌊x+\frac{m-1}{m}⌋$