Find the rational numbers having the following decimal expansions:

Solution:

(i) $0 . \overline{3}$

Let $S=0 . \overline{3}$

$\Rightarrow \mathrm{S}=0.3+0.03+0.003+0.0003+0.00003+\ldots \infty$

$\Rightarrow \mathrm{S}=0.3\left(1+10^{-1}+10^{-2}+10^{-3}+10^{-4}+\ldots \infty\right)$

$\mathrm{S}$ is a geometric series with the first term, $a$, being 1 and the common ratio, $r$, being $10^{-1} .$

$\therefore \mathrm{S}=\frac{1}{1-r}$

$\Rightarrow \mathrm{S}=0.3\left(\frac{1}{1-10^{-1}}\right)$

$\Rightarrow \mathrm{S}=\frac{3}{9}=\frac{1}{3}$

(ii) $0 . \overline{231}$

Let $S=0 . \overline{231}$

$\Rightarrow \mathrm{S}=0.231+0.000231+0.000000231+\ldots \infty$

$\Rightarrow \mathrm{S}=0.231\left(1+10^{-3}+10^{-6}+\ldots \infty\right)$

It is a G.P.

$\therefore \mathrm{S}=0.231\left(\frac{1}{1-10^{-3}}\right)$

$\Rightarrow \mathrm{S}=\frac{231}{999}$

(iii) $3.5 \overline{2}$

Let $\mathrm{S}=3.5 \overline{\overline{2}}$

$\Rightarrow \mathrm{S}=3.5+0.02+0.002+0002+0.00002+\ldots \infty$

$\Rightarrow \mathrm{S}=3.5+0.02\left(1+10^{-1}+10^{-2}+10^{-3}+10^{-4}+\ldots \infty\right)$

It is a G.P.

$\therefore \mathrm{S}=3.5+0.02\left(\frac{1}{1-10^{-1}}\right)$

$\Rightarrow \mathrm{S}=3.5+\frac{0.2}{9}$

$\Rightarrow \mathrm{S}=\frac{317}{90}$

(iv) $0.6 \overline{8}$

Let $S=0.6 \overline{8}$

$\Rightarrow \mathrm{S}=0.6+0.08+0.008+0.0008+0.00008+\ldots \infty$

$\Rightarrow \mathrm{S}=0.6+0.08\left(1+10^{-1}+10^{-2}+10^{-3}+\ldots \infty\right)$

It is a G.P.

$\therefore \mathrm{S}=0.6+0.08\left(\frac{1}{1-10^{-1}}\right)$

$\Rightarrow \mathrm{S}=0.6+\frac{0.8}{9}$

$\Rightarrow \mathrm{S}=\frac{6.2}{9}$

$\Rightarrow \mathrm{S}=\frac{62}{90}=\frac{31}{45}$