1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.