​ Express the following in the form p/q, where p and q are integers and q \ne 0. ​

i. Let x = \( 0.\overline 6\) \( \Rightarrow\) x = 0.6666......  .....(1)

ii.

Multiply both sides by 10,

10x = 0.6666 \( \times\) 10

10x = 6.6666.....  .........(2)

Subtracting (1) from (2), we get

\( \cfrac{10\mathrm x = 6.6666...\\-\mathrm x = 0.6666...}{9\mathrm x = 6}\)

9x = 6

x = \( \frac{6}{9}=\frac{2}{3}\)

Therefore, on converting \( 0.\overline 6\) \( =\frac{2}{3}\) which is in the \( \frac{p}{q}\) form,

(i) Let x = \( 0.\overline{47}\) \( \Rightarrow\) x = 0.47777.....  ......(a)

Multiply both sides by 10, we get

10x = 4.7777.....  ......(b)

Subtract the equation (a) from (b),we get

\(\cfrac{\begin{matrix}10\mathrm x = 4.7777....\\-\mathrm x = 0.47777....\end{matrix}}{9\mathrm x = 4.3}\)

x = \( \frac{4.3}{9}\frac{\times 10}{\times 10}\) \( \frac{43}{90}\)

x = \( \frac{43}{90}\)

Therefore, on converting \( 0.\overline{47}=\frac{43}{90}\) in the \( \frac{p}{q}\) form.

(iii) Let x = \( 0.\overline{0001}\) \( \Rightarrow\)   ........(a)

multiply both sides by 1000 (because the number of recurring decimal number is 3)

1000 \( \times\) x = 1000 \( \times\) 0.001001.....

So, 1000x = 1.001001......  .......(b)

Subtract the equation (a) from (b),

\( \cfrac{\begin{matrix}1000\mathrm x =1.001001...\\-\mathrm x = 0.001001...\end{matrix}}{999\mathrm x =1}\)

\( \rightarrow\) x = \( \frac{1}{999}\)

Therefore, on converting \( 0.\overline{001}\) = \( \frac{1}{999}\) which is in the \( \frac{p}{q}\) form.