Which of the two rational numbers is greater in each of the following pairs?

(i) $\dfrac{3}{-7}$ or $\dfrac{1}{7}$

(ii) $\dfrac{11}{-18}$ or $\dfrac{-5}{18}$

(iii) $\dfrac{7}{10}$ or $\dfrac{-9}{10}$

(iv) 0 or $\dfrac{-3}{4}$

(v) $\dfrac{1}{12}$ or 0

(vi) $\dfrac{18}{-19}$ or 0

(vii) $\dfrac{7}{8}$ or $\dfrac{11}{16}$

(viii) $\dfrac{11}{-12}$ or $\dfrac{-10}{11}$

(ix) $\dfrac{-13}{5}$ or -4

(x) $\dfrac{17}{-6}$ or $\dfrac{-13}{4}$

(xi) $\dfrac{7}{-9}$ or $\dfrac{-5}{8}$

(xii) $\dfrac{-3}{-8}$ or $\dfrac{5}{9}$

(i) We have:

$\dfrac{3}{-7}$ and $\dfrac{1}{7}$

One number = $\dfrac{3}{-7} = \dfrac{3 \times (-1)}{-7 \times (-1)} = \dfrac{-3}{7} \quad \text{[Making the denominator positive]}$

The other number = $\dfrac{1}{7}$.

Since -3 < 1, therefore $\dfrac{-3}{7} \lt \dfrac{1}{7} \quad \text{[Both rational numbers have same denominator]}$

Hence, $\dfrac{1}{7}$ is greater.

(ii) We have:

$\dfrac{11}{-18}$ and $\dfrac{-5}{18}$

One number = $\dfrac{11}{-18} = \dfrac{-11 \times (-1)}{18 \times (-1)} = \dfrac{-11}{18} \quad \text{[Making the denominator positive]}$

The other number = $\dfrac{-5}{18}$.

Since -11 < -5, therefore $\dfrac{-11}{18} \lt \dfrac{-5}{18}$.

Hence, $\dfrac{-5}{18}$ is greater.

(iii) We have:

$\dfrac{7}{10}$ and $\dfrac{-9}{10}$

Since 7 > -9, therefore $\dfrac{7}{10} \gt \dfrac{-9}{10}$.

Hence, $\dfrac{7}{10}$ is greater.

(iv) We have:

0 and $\dfrac{-3}{4}$

Since $\dfrac{-3}{4}$ is a negative rational number, we have $\dfrac{-3}{4} \lt 0$.

Hence, 0 is greater.

(v) We have:

$\dfrac{1}{12}$ and 0

Since $\dfrac{1}{12}$ is a positive rational number, we have $\dfrac{1}{12} \gt 0$.

Hence, $\dfrac{1}{12}$ is greater.

(vi) We have:

$\dfrac{18}{-19}$ and 0

$\dfrac{18}{-19} = \dfrac{-18 \times (-1)}{19 \times (-1)} = \dfrac{-18}{19} \hspace{2cm}\text{[Making the denominator positive]}$

Since, $\dfrac{-18}{19}$ is a negative rational number, we have $\dfrac{-18}{19} \lt 0$.

Hence, 0 is greater.

(vii) We have:

$\dfrac{7}{8}$ and $\dfrac{11}{16}$

L.C.M. of denominators 8 and 16 is 16.

$\dfrac{7}{8} = \dfrac{7 \times 2}{8 \times 2} = \dfrac{14}{16}$.

Now, we have:

$\dfrac{14}{16}$ and $\dfrac{11}{16}$

Clearly 14 > 11, and so $\dfrac{14}{16} \gt \dfrac{11}{16}$ i.e., $\dfrac{7}{8} \gt \dfrac{11}{16}$

Hence, $\dfrac{7}{8}$ is greater.

(viii) We have:

$\dfrac{11}{-12}$ and $\dfrac{-10}{11}$

$\dfrac{11}{-12} = \dfrac{11 \times (-1)}{-12 \times (-1)} = \dfrac{-11}{12}$

L.C.M. of denominators 12 and 11 is 132.

Now, expressing each fraction with denominator 132:

$\dfrac{-11}{12} = \dfrac{-11 \times 11}{12 \times 11} = \dfrac{-121}{132} \\[1em] \dfrac{-10}{11} = \dfrac{-10 \times 12}{11 \times 12} = \dfrac{-120}{132}$.

Now, we have:

$\dfrac{-121}{132}$ and $\dfrac{-120}{132}$

Since -121 < -120, and so $\dfrac{-121}{132} \lt \dfrac{-120}{132}$ i.e., $\dfrac{11}{-12} \lt \dfrac{-10}{11}$

Hence, $\dfrac{-10}{11}$ is greater.

(ix) We have:

$\dfrac{-13}{5}$ and $\dfrac{-4}{1}$

L.C.M. of denominators 5 and 1 is 5.

= $\dfrac{-4 \times 5}{1 \times 5} = \dfrac{-20}{5}$.

Now, we have:

$\dfrac{-13}{5}$ and $\dfrac{-20}{5}$

Since -13 > -20, and so $\dfrac{-13}{5} \gt \dfrac{-20}{5}$ i.e., $\dfrac{-13}{5} \gt \dfrac{-4}{1}$

Hence, $\dfrac{-13}{5}$ is greater.

(x) We have:

$\dfrac{17}{-6}$ and $\dfrac{-13}{4}$

$\dfrac{17}{-6} = \dfrac{17 \times (-1)}{-6 \times (-1)} = \dfrac{-17}{6}$

L.C.M. of denominators 6 and 4 is 12.

$\dfrac{-17}{6} = \dfrac{-17 \times 2}{6 \times 2} = \dfrac{-34}{12} \\[1em] \dfrac{-13}{4} = \dfrac{-13 \times 3}{4 \times 3} = \dfrac{-39}{12}$.

Now, we have:

$\dfrac{-34}{12}$ and $\dfrac{-39}{12}$

Since -34 > -39, and so $\dfrac{-34}{12} \gt \dfrac{-39}{12}$ i.e., $\dfrac{17}{-6} \gt \dfrac{-13}{4}$

Hence, $\dfrac{17}{-6}$ is greater.

(xi) We have:

$\dfrac{7}{-9}$ and $\dfrac{-5}{8}$

$\dfrac{7}{-9} = \dfrac{7 \times (-1)}{-9 \times (-1)} = \dfrac{-7}{9}$

L.C.M. of denominators 9 and 8 is 72.

Now, expressing each fraction with denominator 72:

$\dfrac{-7}{9} = \dfrac{-7 \times 8}{9 \times 8} = \dfrac{-56}{72} \\[1em] \dfrac{-5}{8} = \dfrac{-5 \times 9}{8 \times 9} = \dfrac{-45}{72}$

Now, we have:

$\dfrac{-56}{72}$ and $\dfrac{-45}{72}$

Since -56 < -45, and so $\dfrac{-56}{72} \lt \dfrac{-45}{72}$ i.e., $\dfrac{7}{-9} \lt \dfrac{-5}{8}$

Hence, $\dfrac{-5}{8}$ is greater.

(xii) We have:

$\dfrac{-3}{-8}$ and $\dfrac{5}{9}$

$\dfrac{-3 \times -1}{-8 \times -1} = \dfrac{3}{8}$.

L.C.M. of denominators 8 and 9 is 72.

$\dfrac{3}{8} = \dfrac{3 \times 9}{8 \times 9} = \dfrac{27}{72} \\[1em] \dfrac{5}{9} = \dfrac{5 \times 8}{9 \times 8} = \dfrac{40}{72}$.

Now, we have:

$\dfrac{27}{72}$ and $\dfrac{40}{72}$

Since 27 < 40, and so $\dfrac{27}{72} \lt \dfrac{40}{72}$ i.e.,$\dfrac{-3}{-8} \lt \dfrac{5}{9}$

Hence, $\dfrac{5}{9}$ is greater.