Question

$\left[\frac{\left[x\right]}{n}\right]=\left[\frac{x}{n}\right]$. Where [.] represents greatest integer function and $n\in N$.

• A. False
• B. True

Answer 1

The correct option is B

True

Given,
$\left[\frac{\left[x\right]}{n}\right]=\left[\frac{x}{n}\right]$ $n\in N$.

First solving LHS:
Let $\left[x\right]=nq+r$ $.......\left(1\right)$
where $q$ and $r$ are quotient and remainder respectively.
Now, putting the value of $\left(1\right)$ in LHS we get,

$\Rightarrow \left[\frac{\left[x\right]}{n}\right]$ = $\left[\frac{nq+r}{n}\right]$

$\Rightarrow$ $\left[\frac{\left[x\right]}{n}\right]=[q+\frac{r}{n}]$

We know that,
$0\le r<n$

$\Rightarrow 0\le \frac{r}{n}<1$

$\therefore \frac{r}{n}$ is the fractional part of $\left[\frac{\left[x\right]}{n}\right]$

$\Rightarrow$ $\left[\frac{\left[x\right]}{n}\right]=q$

Now solving RHS:
we know that
$x=\left[x\right]+${$x$}
putting this in RHS we get,
$\left[\frac{x}{n}\right]=\left[\frac{\left[x\right]+\left\{x\right\}}{n}\right]$
$\Rightarrow \left[\frac{x}{n}\right]=\left[\frac{nq+r+\left\{x\right\}}{n}\right]$ $.....\left(2\right)$ $(\because [x]=nq+r)$

$\because 0\le r\le n-1$ $......\left(3\right)$
Also, $0\le \left\{x\right\}<1$ $.......\left(4\right)$
On adding (3) and (4) we get,
$\Rightarrow 0\le r+\left\{x\right\}<n$

$\Rightarrow 0\le \frac{r+\left\{x\right\}}{n}<1$
Now, from (2) we get,

$\Rightarrow \left[\frac{x}{n}\right]=[q+\frac{r+\left(x\right)}{n}]=q$
$\Rightarrow \left[\frac{\left[x\right]}{n}\right]=\left[\frac{x}{n}\right]$
Hence, the given statement is true.