<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] One way to do it is to write $x=kn+r+\epsilon$, where $0\le r\le n-1$ is an integer and $0\le\epsilon\lt1$, so that $0\le r+\epsilon\lt n$.  Then
$$\left\lfloor{\lfloor x\rfloor\over n}\right\rfloor=\left\lfloor{kn+r\over n}\right\rfloor=\left\lfloor k+{r\over n}\right\rfloor=k=\left\lfloor k+{r+\epsilon\over n}\right\rfloor=\left\lfloor {kn+r+\epsilon\over n} \right\rfloor=\left\lfloor {x\over n} \right\rfloor$$
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