an identity involving the floor function.

Question

Assume that $x$ is a real number and that $m$ and $n$ are positive integers. $[x]$ denotes the greatest integer that is $\leq x$. Can the following identity be true? $$[mnx] = m([nx]+1).$$ I tried to use $$[mnx]=[x]+[x+{\frac{1}{mn}}]+\cdots +[x+{\frac{mn-1}{mn}}],$$ and $$m([nx]+1)=m[x]+m[x+{\frac{1}{n}}]+\cdots +m[x+{\frac{n-1}{n}}]+m.$$ I cannot seem to compare these two expressions! I have no reason to believe that this identity is true. It is an expression that came up in the course of my thinking about another problem. Any insights would be useful!

Answer 1

A very useful tip when dealing with floor functions: remember the definition at all times!

Suppose $mnx = [mnx] + \alpha$. By definition $\alpha \in [0, 1)$. So $nx = \frac{[mnx]}m + \frac\alpha m$. Further suppose that $[mnx] = mk + r$, where $0\le r< m$. Then $nx = k + \frac{r + \alpha}{m}$. We would like to verify that $mk+r = m(k + \left[\frac{r + \alpha}{m}\right] + 1)$, which, upon taking $r \ne 0$, becomes false.

Answer 2

Let $\ell=\lfloor mnx\rfloor$, so that $\ell\le mnx<\ell+1$, and let $k=\lfloor nx\rfloor$, so that $k\le nx<k+1$, and hence $mk\le mnx<m(k+1)$. Thus,

$$\lfloor mnx\rfloor=\ell\le mnx<m(k+1)=m(\lfloor nx\rfloor+1)\;,$$

so in fact $\lfloor mnx\rfloor<m(\lfloor nx\rfloor+1)$.