Question

$\left[x\right]\left\{x\right\}=x$, where $[\cdot ]$ and $\{\cdot \}$ denote the greatest integer function and fractional part, respectively.
Which of the following is true

• A. $x=\frac{{\left[x\right]}^2}{\left[x\right]-1}$, where $\left[x\right]$ is any non-positive integer.
• B. $x=\frac{{\left[x\right]}^3}{\left[x\right]-1}$, where $\left[x\right]$ is any non-positive integer.
• C. $x=\frac{{\left[x\right]}^4}{\left[x\right]-1}$, where $\left[x\right]$ is any non-positive integer.
• D. $x=\frac{\left[x\right]}{\left[x\right]-1}$, where $\left[x\right]$ is any non-positive integer.

Answer 1

Answer: (A)

$x=\frac{{\left[x\right]}^2}{\left[x\right]-1}$, where $\left[x\right]$ is any non-positive integer.

We know that,
$x=\left[x\right]+\left\{x\right\}.........\left(i\right)$

Thus, we have

$\left[x\right]\left\{x\right\}=\left[x\right]+\left\{x\right\}$

$\Rightarrow$ $\left\{x\right\}=\frac{\left[x\right]}{\left[x\right]-1}$ (ii)

Here, in Eq. (ii), $\left[x\right]\ne 1$

But, if $\left[x\right]=1,$ then given equation

$\Rightarrow$ $\left\{x\right\}=x$

which is true only when $xϵ[0,1)$ (iii)

and $\left[x\right]=1\Rightarrow xϵ[0,1)$ (iv)

$\therefore$ From Eqs. (iii) and (iv) no value of $x$,

when $\left[x\right]=1$ (v)

Now, from Eq. (ii),

$\left\{x\right\}=\frac{\left[x\right]}{\left[x\right]-1}$

when $\left[x\right]\ne 1$

Again, as we know $\left\{x\right\}ϵ[0,1)$

$\therefore$ $0\le \frac{\left[x\right]}{\left[x\right]-1}<1$

ie,
$\frac{\left[x\right]}{\left[x\right]-1}<1$
and $\frac{\left[x\right]}{\left[x\right]-1}\ge 0$

ie, $\frac{1}{\left[x\right]-1}<0$ and
$\{\left[x\right]\le 0or\left[x\right]>1\}$

ie, $\left[x\right]<1$ and $\{\left[x\right]\le 0or\left[x\right]>1\}$ (using number line rule)

ie, $\left[x\right]\le 0$

$\therefore x=\left[x\right]+\left\{x\right\}$

$\Rightarrow$ $x=\left[x\right]+\frac{\left[x\right]}{\left[x\right]-1}$

$x=\frac{{\left[x\right]}^2}{\left[x\right]-1}$, where $x$ takes values less than $1$.

$\therefore$ $x=\frac{{\left[x\right]}^2}{\left[x\right]-1}$, where $x<1$

or $x=\frac{{\left[x\right]}^2}{\left[x\right]-1}$, where $\left[x\right]$ is any non-positive integer