Write the decimal expansions of the following numbers which have terminating decimal expansions:

$\begin{matrix} \text{(i)} & \dfrac{17}{8} \\[1.5em] \text{(ii)} & \dfrac{13}{3125} \\[1.5em] \text{(iii)} & \dfrac{7}{80} \\[1.5em] \text{(iv)} & \dfrac{6}{15} \\[1.5em] \text{(v)} & \dfrac{2^2 \times 7}{5^4} \\[1.5em] \text{(vi)} & \dfrac{237}{1500} \\[1.5em] \end{matrix}$

$\text{(i) } \dfrac{17}{8}$

The given number $\dfrac{17}{8}$ is in its lowest form.

Prime factorization of denominator 8:

$\begin{array}{l|l} 2 & 8 \ \hline 2 & 4 \ \hline 2 & 2 \ \hline & 1 \end{array}$

8 = 2 x 2 x 2 x 1
= 2³ x 5⁰    [∵ 5⁰ = 1]

$\dfrac{17}{8} = \dfrac{17}{2^3} \\[0.5em] = \dfrac{17 \times 5^3}{2^3 \times 5^3} \\[0.5em] = \dfrac{17 \times 125}{(2 \times 5)^3} \\[0.5em] = \dfrac{2125}{10^3} \\[0.5em] = 2.125 \\[0.5em] \bold{\therefore \dfrac{17}{8} = 2.125}$

$\text{(ii) } \dfrac{13}{3125}$

The given number $\dfrac{13}{3125}$ is in its lowest form.

Prime factorization of denominator 3125:

$\begin{array}{l|l} 5 & 3125 \ \hline 5 & 625 \ \hline 5 & 125 \ \hline 5 & 25 \ \hline 5 & 5 \ \hline & 1 \end{array}$

3125 = 5 x 5 x 5 x 5 x 5 x 1
= 5⁵ x 1
= 1 x 5⁵
= 2⁰ x 5⁵    [∵ 2⁰ = 1]

$\dfrac{13}{3125} = \dfrac{13}{5^5} \\[0.5em] = \dfrac{13 \times 2^5}{2^5 \times 5^5} \\[0.5em] = \dfrac{13 \times 32}{(2 \times 5)^5} \\[0.5em] = \dfrac{416}{10^5} \\[0.5em] = 0.00416 \\[0.5em] \bold{\therefore \dfrac{13}{3125} = 0.00416}$

$\text{(iii) } \dfrac{7}{80}$

The given number $\dfrac{7}{80}$ is in its lowest form.

Prime factorization of denominator 80:

$\begin{array}{l|l} 2 & 80 \ \hline 2 & 40 \ \hline 2 & 20 \ \hline 2 & 10 \ \hline 5 & 5 \ \hline & 1 \end{array}$

80 = 2 x 2 x 2 x 2 x 5
= 2⁴ x 5
= 2⁴ x 5¹

$\dfrac{7}{80} = \dfrac{7}{2^4 \times 5^1} \\[0.5em] = \dfrac{7 \times 5^3}{2^4 \times 5^1 \times 5^3} \\[0.5em] = \dfrac{7 \times 125}{2^4 \times 5^4} \\[0.5em] = \dfrac{7 \times 125}{(2 \times 5)^4} \\[0.5em] = \dfrac{875}{10^4} \\[0.5em] = 0.0875 \\[0.5em] \bold{\therefore \dfrac{7}{80} = 0.0875}$

$\text{(iv) } \dfrac{6}{15}$

GCD of numerator and denominator is 3. Reducing the number to its lowest form:

$\dfrac{6}{15} = \dfrac{\cancel{3} \times 2}{\cancel{3} \times 5} \\[0.5em] = \dfrac{2}{5}$

$\dfrac{6}{15} = \dfrac{2}{5} \\[0.5em] = \dfrac{2 \times 2}{2 \times 5} \\[0.5em] = \dfrac{4}{10} \\[0.5em] = 0.4 \\[0.5em] \bold{\therefore \dfrac{6}{15} = 0.4}$

$\text{(v) } \dfrac{2^2 \times 7}{5^4}$

$\dfrac{2^2 \times 7}{5^4} = \dfrac{2^2 \times 7 \times 2^4}{5^4 \times 2^4} \\[0.5em] = \dfrac{4 \times 7 \times 16}{(2 \times 5)^4} \\[0.5em] = \dfrac{448}{10^4} \\[0.5em] = 0.0448 \\[0.5em] \bold{\therefore \dfrac{2^2 \times 7}{5^4} = 0.0448}$

$\text{(vi) } \dfrac{237}{1500}$

GCD of numerator and denominator is 3. Reducing the number to its lowest form:

$\dfrac{237}{1500} = \dfrac{\cancel{3} \times 79}{\cancel{3} \times 500} \\[0.5em] = \dfrac{79}{500}$

$\dfrac{237}{1500} = \dfrac{79}{500} \\[0.5em] = \dfrac{79 \times 2}{500 \times 2} \\[0.5em] = \dfrac{158}{10^3} \\[0.5em] = 0.158 \\[0.5em] \bold{\therefore \dfrac{237}{1500} = 0.158}$