Rational Numbers - Exercise 4(B)

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 $\dfrac{5}{9}$

Answer

(i) We have:

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

One number = $\dfrac{3}{-7} = \dfrac{3 \times (-1)}{-7 \times (-1)} = \dfrac{-3}{7}$

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

Since -3 < 1, therefore $\dfrac{-3}{7} \lt \dfrac{1}{7}$

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

(ii) We have:

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

One number = $\dfrac{11}{-18} = \dfrac{-11 \times (-1)}{18 \times (-1)} = \dfrac{-11}{18}$

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

$\dfrac{18}{-19} = \dfrac{-18 \times (-1)}{19 \times (-1)} = \dfrac{-18}{19}$

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.

Fill in the blanks with the correct symbol out of >, = or < :

(i) $\dfrac{-17}{4}$ ............... $\dfrac{-15}{4}$

(ii) 0 ............... $\dfrac{-1}{-2}$

(iii) $\dfrac{4}{-3}$ ............... $\dfrac{-8}{7}$

(iv) $\dfrac{-5}{12}$ ............... $\dfrac{7}{-16}$

(v) $\dfrac{-7}{8}$ ............... $\dfrac{-8}{9}$

(vi) $\dfrac{1}{-10}$ ............... $\dfrac{-4}{-5}$

Answer

(i) $\dfrac{-17}{4}$ < $\dfrac{-15}{4}$

(ii) 0 < $\dfrac{-1}{-2}$

(iii) $\dfrac{4}{-3}$ < $\dfrac{-8}{7}$

(iv) $\dfrac{-5}{12}$ > $\dfrac{7}{-16}$

(v) $\dfrac{-7}{8}$ > $\dfrac{-8}{9}$

(vi) $\dfrac{1}{-10}$ < $\dfrac{-4}{-5}$

Explanation

(i) Since the denominators are the same, we compare the numerators where -17 is less than -15.

(ii) $\dfrac{-1}{-2}$ simplifies to the positive rational number $\dfrac{1}{2}$, and zero is always less than any positive number.

(iii) L.C.M. of 3 and 7 is 21. After converting to a common denominator of 21, we compare $\dfrac{-28}{21}$ and $\dfrac{-24}{21}$, where -28 < -24.

(iv) L.C.M. of 12 and 16 is 48. After converting to a common denominator of 48, the fractions are $\dfrac{-20}{48}$ and $\dfrac{-21}{48}$, and since -20 > -21, the first is greater.

(v) L.C.M. of 8 and 9 is 72. After converting to a common denominator of 72, the fractions are $\dfrac{-63}{72}$ and $\dfrac{-64}{72}$, and -63 > -64 so $\dfrac{-63}{72}$ > $\dfrac{-64}{72}$.

(vi) $\dfrac{1}{-10}$ is a negative rational number while $\dfrac{-4}{-5}$ is positive, and any negative number is less than a positive one.

Arrange the following rational numbers in ascending order :

(i) $\dfrac{3}{4}, \dfrac{5}{8}, \dfrac{11}{16}, \dfrac{21}{32}$

(ii) $\dfrac{-2}{5}, \dfrac{7}{-10}, \dfrac{-8}{15}, \dfrac{17}{-30}$

(iii) $\dfrac{5}{-12}, \dfrac{-2}{3}, \dfrac{-7}{9}, \dfrac{11}{-18}$

(iv) $\dfrac{-4}{7}, \dfrac{13}{-28}, \dfrac{9}{14}, \dfrac{23}{42}$

Answer

(i) We have:

$\dfrac{3}{4}, \dfrac{5}{8}, \dfrac{11}{16}, \dfrac{21}{32}$

First we find the L.C.M.

L.C.M. of denominators 4, 8, 16, and 32 is 32.

Now, expressing each fraction with denominator 32:

$\dfrac{3}{4} = \dfrac{3 \times 8}{4 \times 8} = \dfrac{24}{32} \\[1em] \dfrac{5}{8} = \dfrac{5 \times 4}{8 \times 4} = \dfrac{20}{32} \\[1em] \dfrac{11}{16} = \dfrac{11 \times 2}{16 \times 2} = \dfrac{22}{32} \\[1em] \dfrac{21}{32} = \dfrac{21 \times 1}{32 \times 1} = \dfrac{21}{32} \\[1em]$

Clearly, $\dfrac{20}{32} \lt \dfrac{21}{32} \lt \dfrac{22}{32} \lt \dfrac{24}{32}$.

Therefore $\dfrac{5}{8} \lt \dfrac{21}{32} \lt \dfrac{11}{16} \lt \dfrac{3}{4}$

Hence, the ascending order is: $\dfrac{5}{8}, \dfrac{21}{32}, \dfrac{11}{16}, \dfrac{3}{4}$.

(ii) We have:

$\dfrac{-2}{5}, \dfrac{7}{-10}, \dfrac{-8}{15}, \dfrac{17}{-30}$

First, express each rational number with a positive denominator: $\dfrac{-2}{5}, \dfrac{-7}{10}, \dfrac{-8}{15}, \dfrac{-17}{30}$.

The L.C.M. of denominators 5, 10, 15, and 30 is 30.

Now, expressing each fraction with denominator 30:

$\dfrac{-2}{5} = \dfrac{-2 \times 6}{5 \times 6} = \dfrac{-12}{30} \\[1em] \dfrac{-7}{10} = \dfrac{-7 \times 3}{10 \times 3} = \dfrac{-21}{30} \\[1em] \dfrac{-8}{15} = \dfrac{-8 \times 2}{15 \times 2} = \dfrac{-16}{30} \\[1em] \dfrac{-17}{30} = \dfrac{-17 \times 1}{30 \times 1} = \dfrac{-17}{30}$

Clearly, $\dfrac{-21}{30} \lt \dfrac{-17}{30} \lt \dfrac{-16}{30} \lt \dfrac{-12}{30}$.

Therefore $\dfrac{-7}{10} \lt \dfrac{-17}{30} \lt \dfrac{-8}{15} \lt \dfrac{-2}{5}$

Hence, the ascending order is: $\dfrac{7}{-10}, \dfrac{17}{-30}, \dfrac{-8}{15}, \dfrac{-2}{5}$.

(iii) We have:

$\dfrac{5}{-12}, \dfrac{-2}{3}, \dfrac{-7}{9}, \dfrac{11}{-18}$

Expressing with positive denominators: $\dfrac{-5}{12}, \dfrac{-2}{3}, \dfrac{-7}{9}, \dfrac{-11}{18}$.

The L.C.M. denominators of 12, 3, 9, and 18 is 36.

Now, expressing each fraction with denominator 36:

$\dfrac{-5}{12} = \dfrac{-5 \times 3}{12 \times 3} = \dfrac{-15}{36} \\[1em] \dfrac{-2}{3} = \dfrac{-2 \times 12}{3 \times 12} = \dfrac{-24}{36} \\[1em] \dfrac{-7}{9} = \dfrac{-7 \times 4}{9 \times 4} = \dfrac{-28}{36} \\[1em] \dfrac{-11}{18} = \dfrac{-11 \times 2}{18 \times 2} = \dfrac{-22}{36}$

Clearly, $\dfrac{-28}{36} \lt \dfrac{-24}{36} \lt \dfrac{-22}{36} \lt \dfrac{-15}{36}$.

Therefore $\dfrac{-7}{9} \lt \dfrac{-2}{3} \lt \dfrac{-11}{18} \lt \dfrac{-5}{12}$.

Hence, the ascending order is: $\dfrac{-7}{9}, \dfrac{-2}{3}, \dfrac{11}{-18}, \dfrac{5}{-12}$.

(iv) We have:

$\dfrac{-4}{7}, \dfrac{13}{-28}, \dfrac{9}{14}, \dfrac{23}{42}$

Expressing with positive denominators: $\dfrac{-4}{7}, \dfrac{-13}{28}, \dfrac{9}{14}, \dfrac{23}{42}$.

The L.C.M. of denominators 7, 28, 14, and 42 is 84.

Now, expressing each fraction with denominator 84:

$\dfrac{-4}{7} = \dfrac{-4 \times 12}{7 \times 12} = \dfrac{-48}{84} \\[1em] \dfrac{-13}{28} = \dfrac{-13 \times 3}{28 \times 3} = \dfrac{-39}{84} \\[1em] \dfrac{9}{14} = \dfrac{9 \times 6}{14 \times 6} = \dfrac{54}{84} \\[1em] \dfrac{23}{42} = \dfrac{23 \times 2}{42 \times 2} = \dfrac{46}{84}$

Clearly, $\dfrac{-48}{84} \lt \dfrac{-39}{84} \lt \dfrac{46}{84} \lt \dfrac{54}{84}$.

Therefore $\dfrac{-4}{7} \lt \dfrac{-13}{28} \lt \dfrac{23}{42} \lt \dfrac{9}{14}$.

Hence, the ascending order is: $\dfrac{-4}{7}, \dfrac{13}{-28}, \dfrac{23}{42}, \dfrac{9}{14}$.

Arrange the following rational numbers in descending order :

(i) $\dfrac{11}{12}, \dfrac{13}{18}, \dfrac{5}{6}, \dfrac{7}{9}$

(ii) $\dfrac{-11}{20}, \dfrac{3}{-10}, \dfrac{17}{-30}, \dfrac{-7}{15}$

(iii) $\dfrac{9}{-24}, -1, \dfrac{2}{-3}, \dfrac{-7}{-6}$

(iv) $\dfrac{7}{-10}, \dfrac{11}{15}, \dfrac{-17}{-30}, \dfrac{-2}{5}$

Answer

(i) We have:

$\dfrac{11}{12}, \dfrac{13}{18}, \dfrac{5}{6}, \dfrac{7}{9}$

The L.C.M. of denominators 12, 18, 6, and 9 is 36.

Now, expressing each fraction with denominator 36:

$\dfrac{11}{12} = \dfrac{11 \times 3}{12 \times 3} = \dfrac{33}{36} \\[1em] \dfrac{13}{18} = \dfrac{13 \times 2}{18 \times 2} = \dfrac{26}{36} \\[1em] \dfrac{5}{6} = \dfrac{5 \times 6}{6 \times 6} = \dfrac{30}{36} \\[1em] \dfrac{7}{9} = \dfrac{7 \times 4}{9 \times 4} = \dfrac{28}{36}$

Clearly, $\dfrac{33}{36} \gt \dfrac{30}{36} \gt \dfrac{28}{36} \gt \dfrac{26}{36}$. Therefore $\dfrac{11}{12} \gt \dfrac{5}{6} \gt \dfrac{7}{9} \gt \dfrac{13}{18}$.

Hence, the descending order is: $\dfrac{11}{12}, \dfrac{5}{6}, \dfrac{7}{9}, \dfrac{13}{18}$.

(ii) We have:

$\dfrac{-11}{20}, \dfrac{3}{-10}, \dfrac{17}{-30}, \dfrac{-7}{15}$

First, express each with a positive denominator: $\dfrac{-11}{20}, \dfrac{-3}{10}, \dfrac{-17}{30}, \dfrac{-7}{15}$.

The L.C.M. of denominators 20, 10, 30, and 15 is 60.

Now, expressing each fraction with denominator 60:

$\dfrac{-11}{20} = \dfrac{-11 \times 3}{20 \times 3} = \dfrac{-33}{60} \\[1em] \dfrac{-3}{10} = \dfrac{-3 \times 6}{10 \times 6} = \dfrac{-18}{60} \\[1em] \dfrac{-17}{30} = \dfrac{-17 \times 2}{30 \times 2} = \dfrac{-34}{60} \\[1em] \dfrac{-7}{15} = \dfrac{-7 \times 4}{15 \times 4} = \dfrac{-28}{60}$

Clearly, $\dfrac{-18}{60} \gt \dfrac{-28}{60} \gt \dfrac{-33}{60} \gt \dfrac{-34}{60}$. Therefore $\dfrac{-3}{10} \gt \dfrac{-7}{15} \gt \dfrac{-11}{20} \gt \dfrac{-17}{30}$.

Hence, the descending order is: $\dfrac{3}{-10}, \dfrac{-7}{15}, \dfrac{-11}{20}, \dfrac{17}{-30}$.

(iii) We have:

$\dfrac{9}{-24}, -1, \dfrac{2}{-3}, \dfrac{-7}{-6}$

Expressing with positive denominators: $\dfrac{-9}{24}, \dfrac{-1}{1}, \dfrac{-2}{3}, \dfrac{7}{6}$.

The L.C.M. of denominators 24, 1, 3, and 6 is 24.

Now, expressing each fraction with denominator 24:

$\dfrac{-9}{24} = \dfrac{-9 \times 1}{24 \times 1} = \dfrac{-9}{24} \\[1em] -1 = \dfrac{-1 \times 24}{1 \times 24} = \dfrac{-24}{24} \\[1em] \dfrac{-2}{3} = \dfrac{-2 \times 8}{3 \times 8} = \dfrac{-16}{24} \\[1em] \dfrac{7}{6} = \dfrac{7 \times 4}{6 \times 4} = \dfrac{28}{24}$

Clearly, $\dfrac{28}{24} \gt \dfrac{-9}{24} \gt \dfrac{-16}{24} \gt \dfrac{-24}{24}$. Therefore $\dfrac{7}{6} \gt \dfrac{-9}{24} \gt \dfrac{-2}{3} \gt -1$.

Hence, the descending order is: $\dfrac{-7}{-6}, \dfrac{9}{-24}, \dfrac{2}{-3}, -1$.

(iv) We have:

$\dfrac{7}{-10}, \dfrac{11}{15}, \dfrac{-17}{-30}, \dfrac{-2}{5}$

Expressing with positive denominators: $\dfrac{-7}{10}, \dfrac{11}{15}, \dfrac{17}{30}, \dfrac{-2}{5}$.

The L.C.M. of denominators 10, 15, 30, and 5 is 30.

Now, expressing each fraction with denominator 30:

$\dfrac{-7}{10} = \dfrac{-7 \times 3}{10 \times 3} = \dfrac{-21}{30} \\[1em] \dfrac{11}{15} = \dfrac{11 \times 2}{15 \times 2} = \dfrac{22}{30} \\[1em] \dfrac{17}{30} = \dfrac{17 \times 1}{30 \times 1} = \dfrac{17}{30} \\[1em] \dfrac{-2}{5} = \dfrac{-2 \times 6}{5 \times 6} = \dfrac{-12}{30}$

Clearly, $\dfrac{22}{30} \gt \dfrac{17}{30} \gt \dfrac{-12}{30} \gt \dfrac{-21}{30}$. Therefore $\dfrac{11}{15} \gt \dfrac{17}{30} \gt \dfrac{-2}{5} \gt \dfrac{-7}{10}$.

Hence, the descending order is: $\dfrac{11}{15}, \dfrac{-17}{-30}, \dfrac{-2}{5}, \dfrac{7}{-10}$.