Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division:

(i) $\dfrac{3}{50}$

(ii) $\dfrac{2}{9}$

(i) Given, $\dfrac{3}{50}$

Prime factorisation of 50 = 2 × 5².

Since the denominator has only 2's and 5's, it has a terminating decimal expansion.

By long division :

$\begin{array}{l} \phantom{50\overline{)}\,}0.06 \ 50\overline{\smash{\big)}\phantom{)}3.00} \ \phantom{50\overline{}}\underline{-0}\phantom{.00} \ \phantom{50\overline{)}1}30\phantom{.0} \ \phantom{50\overline{)})}\underline{-0}\phantom{.0} \ \phantom{50\overline{)}1}300 \ \phantom{50\overline{,}}\underline{-300} \ \phantom{50\overline{)}\,00}0 \end{array}$

⇒ $\dfrac{3}{50} = \dfrac{3 \times 2}{50 \times 2} = \dfrac{6}{100} = 0.06.$

Hence, $\dfrac{3}{50} = 0.06$ (terminating).

(ii) Given, $\dfrac{2}{9}$

Prime factorisation of 9 = 3².

Since the denominator has 3 (a prime other than 2 or 5), it has a non-terminating repeating decimal expansion.

By long division :

$\begin{array}{l} \phantom{9\overline{)}\,}0.222\ldots \ 9\overline{\smash{\big)}\phantom{)}2.000\ldots} \ \phantom{9\overline{}}\underline{-0}\phantom{.000\ldots} \ \phantom{9\overline{)}1}20\phantom{00\ldots} \ \phantom{9\overline{)}}\underline{-18}\phantom{00\ldots} \ \phantom{9\overline{)}\,00}20\phantom{0\ldots} \ \phantom{9\overline{)}1}\underline{-18}\phantom{0\ldots} \ \phantom{9\overline{)}\,000}20\phantom{\ldots} \ \phantom{9\overline{)}11}\underline{-18}\phantom{\ldots} \ \phantom{9\overline{)})1111}2\ldots \end{array}$

⇒ $\dfrac{2}{9} = 0.2222\ldots = 0.\overline{2}$.

Hence, $\dfrac{2}{9} = 0.\overline{2}$ (non-terminating, repeating).