Floor function lemma Let $x\in\mathbb{R}$ and $n\in\mathbb{Z}$ with $n\gt0$. Show that:$$\lfloor\lfloor{x}\rfloor/n\rfloor=\lfloor{x/n}\rfloor$$ I did referenced this post Floor Function Proof but the answers, even the marked answer are not quite right. The marked answer has a proof that leads to $b\in\{0,1/n,...,(n-1)/n\}$ and $b+\frac{c}{n}\in\{\frac{c}{n},\frac{c+1}{n},...,\frac{c+(n-1)}{n}\}\subseteq[0,1)$. I don't understand why we can get these results, especially the second. I appreciate any help. 
 A: $k = \lfloor x/n\rfloor$ is characterized by $k\in \mathbb{Z}$ and
$$k \le x/n \lt k+1$$
As $n> 0$, it follows that
$$n k\le x\lt nk + n$$
As $n k\in\mathbb{Z}$, it follows that
$$n k\le \lfloor x\rfloor\le x\lt n k+n$$
hence
$$k\le \lfloor x\rfloor/n\lt k+1$$
hence
$$\lfloor\lfloor x\rfloor/n\rfloor = k = \lfloor x/n\rfloor$$
A: $\lfloor\lfloor x\rfloor/n\rfloor$ is the largest integer $a$ with
$a\le \lfloor x\rfloor/n$, that is with $an\le\lfloor x\rfloor$.
As $an$ is an integer, $an\le\lfloor x\rfloor$ if and only if $an\le x$, that is $a\le x/n$. The largest integer with $a\le x/n$ is
$\lfloor x/n\rfloor$. Therefore $\lfloor\lfloor x\rfloor/n\rfloor=\lfloor x/n\rfloor$.
A: Here is yet another type of proof,$%
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\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
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%$ this time using the following definition of the floor function: $\;\floor x\;$ is the integer such that $$
k \le \floor x \;\equiv\; k \le x
$$ for all integers $\;k\;$.  This allows us investigate the integer lower bounds of the left hand side, and work towards the right hand side: for any integer $\;k\;$,
$$\calc
    k \le \floor{\floor x / n }
\op\equiv\hint{by the above definition}
    k \le \floor x / n
\op\equiv\hint{arithmetic, using $\;n > 0\;$}
    k \times n \le \floor x
\op\equiv\hint{by the above definition}
    k \times n \le x
\op\equiv\hint{arithmetic, using $\;n > 0\;$}
    k \le x / n
\op\equiv\hint{by the above definition}
    k \le \floor{x / n}
\endcalc$$
So both sides have the same integer lower bounds, and since they are both integers, they are therefore equal.
$%
\endgroup
%$
A: Here is another simple proof for this problem:
Since $\frac{x}{n}=\frac{-x}{-n}$, we may assume without loss of generality that $n>0$.
$\lfloor x\rfloor =x-\{x\}$, where $0\leq \{x\}<1$ is the fractional part of $x$,  is an integer and by Euclid's theorem, there are integers $q$ and $r$ such that
$\lfloor x\rfloor =qn+r$, where $0\leq r<n$. Then
$$
\frac{x}{n}=\frac{[x]+\{x\}}{n}=q+\frac{r+\{x\}}{n}
$$
Since $0\leq r + \{x\}\leq n-1+\{x\}<n$, we have that $0\leq \frac{r}{n}\leq \frac{r+\{x\}}{n}<1$. Consequently
$$
q=\Big[\frac{x}{n}\Big]=\Big[\frac{[x]}{n}\Big]$$
