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.
 A: Let $[\frac{x}{n}]=q$ then $q\leqslant \frac{x}{n}< q+1$. Then $nq\leqslant x < n(q+1)$. Hence $nq\leqslant [x] < n(q+1)$ and $q\leqslant [\frac{[x]}{n}]< q+1$. Thus $[\frac{[x]}{n}]=q$
A: 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, $\lfloor \frac{\lfloor x \rfloor}{n} \rfloor = \lfloor \frac{x}{n} \rfloor$.
A: 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
$$\left\lfloor{\lfloor x\rfloor\over n}\right\rfloor=\left\lfloor{kn+r\over n}\right\rfloor=\left\lfloor k+{r\over n}\right\rfloor=k=\left\lfloor k+{r+\epsilon\over n}\right\rfloor=\left\lfloor {kn+r+\epsilon\over n} \right\rfloor=\left\lfloor {x\over n} \right\rfloor$$
A: Write $x$ as a fractional number in the base $10$.
$$x=314.159\cdots$$ 
so that by dropping the decimals
$$\lfloor x\rfloor=314.$$
Then shifting the decimal dot
$$\frac x{10}=31.4159\cdots$$ and 
$$\frac{\lfloor x\rfloor}{10}=31.$$
Obviously, these two numbers have the same integer part (which is the integer part of $x$ but the units digit).
This is a general proof, as the base $10$ can be replaced by the base $n$ and $x$ by any sequence of digits.
A: We know that integers can be pulled in and out of the floor function. Then WLOG we may assume $0\le x<n$ (otherwise subtract a suitable multiple of $n$). Obviously $0\le\lfloor x\rfloor\le x$.
Then dividing by $n$ and taking the floor,
$$0\le\left\lfloor \frac{\lfloor x\rfloor}n\right\rfloor\le\left\lfloor  \frac xn\right\rfloor<1.$$
