Question

Convert in $\frac{p}{q}$ form :
(i) $0.\overline{47}$
(ii) $0.\overline{001}$
(iii) $0.\overline{9}$
(iv) $2.3\overline{5}$

Answer 1

i) Let $x=0.\overline{47}=\mathrm{0.4747.....}$

$\Rightarrow 100x=\mathrm{47.4747.......}$

$\Rightarrow 100x=47+\mathrm{0.4747......}$

$\Rightarrow 100x=47+0.\overline{47}$

$\Rightarrow 100x=47+x$

$\Rightarrow 99x=47$

$\Rightarrow x=\frac{47}{99}$

ii) Let $x=0.\overline{001}=\mathrm{0.001001.....}$

$\Rightarrow 1000x=\mathrm{001.001001.......}$

$\Rightarrow 1000x=1+\mathrm{0.001001......}$

$\Rightarrow 1000x=1+0.\overline{001}$

$\Rightarrow 1000x=1+x$

$\Rightarrow 999x=1$

$\Rightarrow x=\frac{1}{999}$

iii)Let $x=0.\overline{9}=\mathrm{0.9999.....}$

$\Rightarrow 10x=\mathrm{9.999.......}$

$\Rightarrow 10x=9+\mathrm{0.9999......}$

$\Rightarrow 10x=9+0.\overline{9}$

$\Rightarrow 10x=9+x$

$\Rightarrow 9x=9$

$\Rightarrow x=\frac{9}{9}=1$.

iv) $2.3\overline{5}=2.30+0.0\overline{5}$

Let $x=0.0\overline{5}=\mathrm{0.0555.....}\Rightarrow 10x=0.\overline{5}$

$\Rightarrow 100x=\mathrm{5.5555.......}$

$\Rightarrow 100x=5+\mathrm{0.5555......}$

$\Rightarrow 100x=5+0.\overline{5}$

$\Rightarrow 100x=5+10x$

$\Rightarrow 90x=5$

$\Rightarrow x=\frac{5}{90}$

$\therefore 2.3\overline{5}=2.30+\frac{5}{90}$