Without actual division, find whether the following rational numbers are terminating decimals or recurring decimals :

$\begin{matrix} \text{(i)} & \dfrac{13}{45} \\[1.5em] \text{(ii)} & -\dfrac{5}{56} \\[1.5em] \text{(iii)} & \dfrac{7}{125} \\[1.5em] \text{(iv)} & -\dfrac{23}{80} \\[1.5em] \text{(v)} & -\dfrac{15}{66} \\[1.5em] \end{matrix}$

In case of terminating decimals, write their decimal expansions.

$\text{(i) } \dfrac{13}{45}$

The given number $\dfrac{13}{45}$ is in its lowest form.

Prime factorization of denominator 45:

$\begin{array}{l|l} 3 & 45 \ \hline 3 & 15 \ \hline 5 & 5 \ \hline & 1 \end{array}$

45 = 3 x 3 x 5 x 1
= 3² x 5 x 1

Denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\dfrac{13}{45}$ is recurring decimal.

$\text{(ii) } -\dfrac{5}{56}$

The given number $-\dfrac{5}{56}$ is in its lowest form.

Prime factorization of denominator 56:

$\begin{array}{l|l} 2 & 56 \ \hline 2 & 28 \ \hline 2 & 14 \ \hline 7 & 7 \ \hline & 1 \end{array}$

56 = 2 x 2 x 2 x 7 x 1
= 2³ x 7 x 1

Denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $-\dfrac{5}{56}$ is recurring decimal.

$\text{(iii) } \dfrac{7}{125}$

The given number $\dfrac{7}{125}$ is in its lowest form.

Prime factorization of denominator 125:

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

125= 5 x 5 x 5 x 1
= 5³ x 1
= 5³ x 2⁰

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

$\dfrac{7}{125} = \dfrac{7}{2^0 × 5^3} \\[1.5em] = \dfrac{7 × 2^3}{2^3 × 5^3} \\[1.5em] = \dfrac{56}{(2 × 5)^3} \\[1.5em] = \dfrac{56}{(10)^3} \\[1.5em] = \dfrac{56}{1000} \\[1.5em] = 0.056$

∴ The given number $\dfrac{7}{125}$ is a terminating decimal and its decimal expansion is 0.056.

$\text{(iv) } \dfrac{-23}{80}$

The given number $-\dfrac{23}{80}$ 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 x 1
= 2⁴ x 5¹

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

$-\dfrac{23}{80} = -\dfrac{23}{2^4 × 5^1} \\[1.5em] = -\dfrac{23 × 5^3} {2^4 × 5^4} \\[1.5em] = -\dfrac{23 × 125}{(2 × 5)^4} \\[1.5em] = -\dfrac{2875}{(10)^4} \\[1.5em] = -\dfrac{2875}{10000} = -0.2875$

∴ The given number $-\dfrac{23}{80}$ is a terminating decimal and its decimal expansion is -0.2875.

$\text{(v) } -\dfrac{15}{66}$

The given number $-\dfrac{15}{66}$ is in its lowest form.

Prime factorization of denominator 66:

$\begin{array}{l|l} 2 & 66 \ \hline 3 & 33 \ \hline 11 & 11 \ \hline & 1 \end{array}$

66 = 2 x 3 x 11 x 1
= 2 x 3 x 11

Denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $-\dfrac{15}{66}$ is recurring decimal.