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

(i) $$0.\overline{6}$$

(ii) $$0.\overline{47}$$

(iii) $$0.\overline{001}$$

Asked by Vishal kumar | 2 years ago |  347

##### Solution :-

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.

Answered by Vishal kumar | 2 years ago

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