Express the following in the form $\frac {p}{q}$, where $p$ and $q$ are integers and $q \ne 0$.

$(i)$ $0 . \overline{6}$

$(ii)$ $0 . 4\overline{7}$

$(iii)$ $0 . \overline{001}$

$(i)$ Let $x=0 . \overline{6}=0.6666 \ldots$

since, there is one repeating digit.

$\therefore$ We multiply both sides by $10$,

$10 x =(0.666 \ldots) \times 10$   or   $10 x =6.6666 \ldots$

$\therefore \quad 10 x - x =6.6666 \ldots-0.6666 \ldots $       or        $\quad 9 x =6$

or $\quad x=\frac{6}{9}=\frac{2}{3}$

Thus, $\quad 0 . \overline{6}=\frac{2}{3}$

$(ii)$ Let $\quad x=0.4 \overline{7}=0.4777 \ldots$

$\therefore $ $10 x=10 \times(0.4777 \ldots)$

or                      $10 x=4.777$                 ........... $(1)$

and                  $100 x=47.77$                 ........... $(2)$

Subtracting $(1)$ from $(2)$, we have

$100 x-10 x=(47.777 \ldots)-(4.777 \ldots)$

$90 x=43$   or     $x=\frac{43}{90}$

Thus, $\quad 0.4 \overline{7}=\frac{43}{90}$

$(iii)$  Let $x=0 . \overline{001}=0.001001$ .........  $(1)$

Here, we have three repeating digits after the decimal point, therefore we multiply by $1000.$

$\therefore $       $1000 x =1000 \times(0 . \overline{001})=1000 \times 0.001001 \ldots$

Or                         $1000 x =1.001001$

Subtracting $(1)$ from $(2)$, we have

              $1000 x - x =(1.001 \ldots)-(0.001 \ldots)$

or           $999 x=1$                             $\therefore $ $x=\frac{1}{999}$

Thus,            $0 . \overline{001}=\frac{1}{999}$