Without performing long division, determine which of the following rational numbers will have terminating decimals and which will be repeating: $\dfrac{7}{20}$, $\dfrac{4}{15}$ and $\dfrac{13}{250}$. Then check your answers by explicitly performing the long divisions and expressing these rational numbers as decimals.

To determine if a rational number $\dfrac{p}{q}$ in lowest form has a terminating decimal expansion, the prime factorisation of q must contain only 2's and/or 5's.

(i) $\dfrac{7}{20}$

Prime factorisation of 20 = 2² × 5.

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

Performing long division :

$\begin{array}{l} \phantom{20\overline{)}\,}0.35 \ 20\overline{\smash{\big)}\phantom{)}7.00} \ \phantom{20\overline{}}\underline{-60}\phantom{0} \ \phantom{20\overline{)})}100 \ \phantom{20\overline{}}\underline{-100} \ \phantom{20\overline{)}\,00}0 \end{array}$

⇒ $\dfrac{7}{20}$ = 0.35

(ii) $\dfrac{4}{15}$

Prime factorisation of 15 = 3 × 5.

Since the denominator has 3 (a prime other than 2 or 5), it is a repeating decimal.

Performing long division :

$\begin{array}{l} \phantom{15\overline{)}\,}0.2666\ldots \ 15\overline{\smash{\big)}\phantom{)}4.0000\ldots} \ \phantom{15\overline{}}\underline{-30} \ \phantom{15\overline{)})}100\phantom{0\ldots} \ \phantom{15\overline{)}}\underline{-90}\phantom{0\ldots} \ \phantom{15\overline{)}1)}100\phantom{\ldots} \ \phantom{15\overline{)}1}\underline{-90}\phantom{\ldots} \ \phantom{15\overline{)}11)}100 \ \phantom{15\overline{)}00}\underline{-90} \ \phantom{15\overline{)})111}10\ldots \end{array}$

⇒ $\dfrac{4}{15} = 0.2666\ldots = 0.2\overline{6}$

(iii) $\dfrac{13}{250}$

Prime factorisation of 250 = 2 × 5³.

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

Performing long division :

$\begin{array}{l} \phantom{250\overline{)}\,1}0.052 \ 250\overline{\smash{\big)}\phantom{)}13.000} \ \phantom{250\overline{)}}\underline{-0} \ \phantom{250\overline{)})}130\phantom{00} \ \phantom{250\overline{1},}\underline{-0} \ \phantom{250\overline{)})}1300\phantom{0} \ \phantom{250\overline{}}\underline{-1250} \ \phantom{250\overline{)}11)}500 \ \phantom{250\overline{)}))}\underline{-500} \ \phantom{250\overline{)}\,0000}0 \end{array}$

⇒ $\dfrac{13}{250}$ = 0.052

Hence, $\dfrac{7}{20} = 0.35$ (terminating), $\dfrac{4}{15} = 0.2\overline{6}$ (repeating) and $\dfrac{13}{250} = 0.052$ (terminating).