Floor function lemma

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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 $x$, and $\lfloor x\rfloor$ is the largest integer smaller that $x$. 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 $$ q=\Big[\frac{x}{n}\Big]=\Big[\frac{[x]}{n}\Big]$$

Answer 4

Here is yet another type of proof,$% \require{begingroup} \begingroup \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} %$ 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.

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