Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

$\begin{matrix} \text{(i)} & \dfrac{13}{3125} \\[1.5em] \text{(ii)} & \dfrac{17}{8} \\[1.5em] \text{(iii)} & \dfrac{23}{75} \\[1.5em] \text{(iv)} & \dfrac{6}{15} \\[1.5em] \text{(v)} & \dfrac{1258}{625} \\[1.5em] \text{(vi)} & \dfrac{77}{210} \\[1.5em] \end{matrix}$

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

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]

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{13}{3125}$ has a terminating decimal expansion.

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

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 1
= 2³ x 5⁰    [∵ 5⁰ = 1]

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{17}{8}$ has a terminating decimal expansion.

$\text{(iii) } \dfrac{23}{75}$

The given number $\dfrac{23}{75}$ is in its lowest form.

Prime factorization of denominator 75:

$\begin{array}{l|l} 3 & 75 \ \hline 5 & 25 \ \hline 5 & 5 \ \hline & 1 \end{array}$

75 = 3 x 5 x 5 x 1
= 3 x 5² x 1
= 3 x 5² x 2⁰    [∵ 2⁰ = 1]

Denominator has a prime factor 3 other than 2 or 5.

∴ The given number $\dfrac{23}{75}$ has a non-terminating repeating decimal expansion.

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

Both numerator and denominator contain common factor 3. Reducing the number to its lowest form:

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

The Denominator 5 = 2⁰ x 5¹

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{6}{15}$ has a terminating decimal expansion.

$\text{(v) } \dfrac{1258}{625}$

The given number $\dfrac{1258}{625}$ is in its lowest form.

Prime factorization of denominator 625:

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

625 = 5 x 5 x 5 x 5 x 1
= 5⁴ x 1
= 1 x 5⁴
= 2⁰ x 5⁴    [∵ 2⁰ = 1]

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{1258}{625}$ has a terminating decimal expansion.

$\text{(vi) } \dfrac{77}{210}$

Both numerator and denominator contain common factor 7. Reducing the number to its lowest form:

$\dfrac{77}{210} = \dfrac{\cancel{7} \times 11}{\cancel{7} \times 30} \\[0.5em] = \dfrac{11}{30}$

Prime factorization of denominator 30:

$\begin{array}{l|l} 2 & 30 \ \hline 3 & 15 \ \hline 5 & 5 \ \hline & 1 \end{array}$

30 = 2 x 3 x 5

Denominator has a prime factor 3 other than 2 or 5.

∴ The given number $\dfrac{77}{210}$ has a non-terminating repeating decimal expansion.