<s>[INST] How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \lfloor\frac{\lfloor x\rfloor}{n}\rfloor=\lfloor\frac{x}{n}\rfloor$. How to prove or disprove $\forall x\in\Bbb{R}, \forall n\in\Bbb{N},n\gt 0\implies \left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{n}\right\rfloor$.
So we want to prove $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$ and $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
Since $\lfloor x\rfloor\le x$, we can just start from here and prove 
$\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\le\left\lfloor\frac{x}{n}\right\rfloor$
But for $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor\ge\left\lfloor\frac{x}{n}\right\rfloor$, I have no idea how to start.
 [/INST] Since $n$ is a natural number we can divide $x$ (with remainder) by $n$ in order to express $x = nb +r_1$ for some $b \in \mathbb{N}$ and $r_1 < n$. Now $\lfloor \frac{x}{n} \rfloor = \lfloor \frac{nb+r}{n} \rfloor = \lfloor b + \frac{r}{n} \rfloor = b$ since $r<n$. On the other hand we have that $\lfloor x \rfloor = bn + r_2$ for the $\textbf{same}$ b and with $r_2<n$. This is because n cannot divide the fractional part of x since it is a natural number. Then $\lfloor \frac{\lfloor x \rfloor}{n} \rfloor =  \lfloor \frac{ bn + r_2}{n} \rfloor = b$. Therefore, $\lfloor \frac{\lfloor x \rfloor}{n} \rfloor = \lfloor \frac{x}{n} \rfloor$.
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