# 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.

$k = \lfloor x/n\rfloor$ 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 $$\lfloor\lfloor x\rfloor/n\rfloor = k = \lfloor x/n\rfloor$$
$\lfloor\lfloor x\rfloor/n\rfloor$ 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 $\lfloor\lfloor x\rfloor/n\rfloor=\lfloor x/n\rfloor$.