Express the following in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \ne 0$.
$(i)$ $0.\overline{6}$
$(ii)$ $0.4\overline{7}$
$(iii)$ $0.\overline{001}$

(N/A) $(i)$ Let $x = 0.\overline{6} = 0.6666\ldots$
Since there is one repeating digit,we multiply both sides by $10$:
$10x = 6.6666\ldots$
Subtracting $x$ from $10x$:
$10x - x = 6.6666\ldots - 0.6666\ldots$
$9x = 6$
$x = \frac{6}{9} = \frac{2}{3}$
Thus,$0.\overline{6} = \frac{2}{3}$.
$(ii)$ Let $x = 0.4\overline{7} = 0.4777\ldots$
Multiply by $10$:
$10x = 4.777\ldots$ $(1)$
Multiply by $100$:
$100x = 47.777\ldots$ $(2)$
Subtracting $(1)$ from $(2)$:
$100x - 10x = 47.777\ldots - 4.777\ldots$
$90x = 43$
$x = \frac{43}{90}$
Thus,$0.4\overline{7} = \frac{43}{90}$.
$(iii)$ Let $x = 0.\overline{001} = 0.001001\ldots$ $(1)$
Since there are three repeating digits,multiply by $1000$:
$1000x = 1.001001\ldots$ $(2)$
Subtracting $(1)$ from $(2)$:
$1000x - x = 1.001001\ldots - 0.001001\ldots$
$999x = 1$
$x = \frac{1}{999}$
Thus,$0.\overline{001} = \frac{1}{999}$.