Question

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

• A. True
• B. False

Answer 1

The correct option is A

True

In $\left[\frac{\left[x\right]}{n}\right]$ We want to devide [x] by m first.

Let the quotient be q and the remainder be r .

$\Rightarrow$ [x] = nq + r ............(1)

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

= q + $\frac{r}{n}$

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

We know , 0 $\le$ r < n

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

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

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

x = [x] + {x} ({x} - is fractional part of x)

= nq + r + {x} (from (1) we have [x] = nq + r)

Consider r + {x}

We know 0 $\le$ < n => 0 $\le$ r $\le$ n - 1

Also, 0 $\le$ {x} < 1

$\Rightarrow$ = 0 $\le$ r + {x} < n

$\Rightarrow$ 0 $\le$ $\frac{r+\left(x\right)}{n}$ $\le$ 1

Consider $\left[\frac{x}{n}\right]=\left[\frac{\left[x\right]+\left(x\right)}{n}\right]$

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

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

0 $\le$ $\frac{r+\left(x\right)}{n}$ < 1 $\to$ [from (1) ]

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

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