Exercise 4.1.10 (Find $x$ such that $\left \lfloor \frac{\lfloor x \rfloor}{m} \right \rfloor \ne \left \lfloor \frac{x}{m} \right \rfloor$)

Proof. Let $m\in ℝ\setminus \mathbb{N}$. We search $x\in ℝ$ such that

• We suppose first that $m>0$.

  Put $x=m$. Then $\lfloor \frac{x}{m}\rfloor =1$. But $0\le \lfloor x\rfloor =\lfloor m\rfloor <m$, since $m$ is not a positive integer, thus $0\le \frac{\lfloor x\rfloor }{m}<1$, so $\lfloor \frac{\lfloor x\rfloor }{m}\rfloor =0\ne 1=\lfloor \frac{x}{m}\rfloor$.

• Suppose now that $m<0$ (integer or not). Take $x=\lfloor m\rfloor +\mu$, where $\mu$ is chosen such that $\{m\}<\mu <1$ (here $\{m\}=m-\lfloor m\rfloor$ is the fractional part of $m$). Such a $\mu$ exists, since $\{m\}<1$, for instance $\mu =(1+\{m\})∕2$, so that

  $$x=\lfloor m\rfloor +\mu =\lfloor m\rfloor +(1+m-\lfloor m\rfloor )∕2.$$

  Then

  $$\lfloor m\rfloor \le m<m+\mu =x<\lfloor m\rfloor +1,$$

  thus $\lfloor x\rfloor =\lfloor m\rfloor .$

  Since $m<0$, the division by $m$ gives

  $$\frac{\lfloor m\rfloor }{m}\ge 1>\frac{x}{m}.$$

  Since $\frac{x}{m}<1$,

  $$\begin{array}{lll}\hfill \lfloor \frac{x}{m}\rfloor \le 0.& & \hfill \text{(1)}\end{array}$$

  Moreover $\frac{\lfloor x\rfloor }{m}=\frac{\lfloor m\rfloor }{m}\ge 1$, thus

  $$\begin{array}{lll}\hfill \lfloor \frac{\lfloor x\rfloor }{m}\rfloor \ge 1.& & \hfill \text{(2)}\end{array}$$

  Then the inequations (1) and (2) show that

  $$\lfloor \frac{\lfloor x\rfloor }{m}\rfloor \ne \lfloor \frac{x}{m}\rfloor .$$