Rational Numbers — Exercise 2(B)

Mark the following pairs of rational numbers on the separate number lines:

(i) $\dfrac{3}{4} \text{ and }-\dfrac{1}{4}$

(ii) $\dfrac{2}{5} \text{ and } \dfrac{-3}{5}$

(iii) $\dfrac{5}{6} \text{ and } -\dfrac{2}{3}$

(iv) $\dfrac{2}{5} \text{ and } -\dfrac{4}{5}$

(v) $\dfrac{1}{4} \text{ and } -\dfrac{5}{4}$

Answer

(i) Since the denominator of each rational number is 4, divide each unit length between 0 and 1, and between 0 and -1, into four equal parts.

To represent $\dfrac{3}{4}$, move 3 parts to the right of 0; to represent $-\dfrac{1}{4}$, move 1 part to the left of 0.

(ii) Since the denominator of each rational number is 5, divide each unit length into five equal parts.

To represent $\dfrac{2}{5}$, move 2 parts to the right of 0; to represent $\dfrac{-3}{5}$, move 3 parts to the left of 0.

(iii) $-\dfrac{2}{3} = -\dfrac{4}{6}$, so both numbers have denominator 6. Divide each unit length into six equal parts.

To represent $\dfrac{5}{6}$, move 5 parts to the right of 0; to represent $-\dfrac{2}{3} = -\dfrac{4}{6}$, move 4 parts to the left of 0.

(iv) Since the denominator of each rational number is 5, divide each unit length into five equal parts.

To represent $\dfrac{2}{5}$, move 2 parts to the right of 0; to represent $-\dfrac{4}{5}$, move 4 parts to the left of 0.

(v) Since the denominator of each rational number is 4, divide each unit length into four equal parts.

To represent $\dfrac{1}{4}$, move 1 part to the right of 0; to represent $-\dfrac{5}{4}$, move 5 parts to the left of 0.

Compare:

(i) $\dfrac{3}{5} \text{ and } \dfrac{5}{7}$

(ii) $\dfrac{-7}{2} \text{ and } \dfrac{5}{2}$

(iii) -3 and $2\dfrac{3}{4}$

(iv) $-1\dfrac{1}{2}$ and 0

(v) 0 and $\dfrac{3}{4}$

(vi) 3 and -1

Answer

(i) On cross-multiplying $\dfrac{3}{5}$ and $\dfrac{5}{7}$:

3 × 7 = 21 and 5 × 5 = 25

Since 21 < 25,

Hence, $\dfrac{3}{5} \lt \dfrac{5}{7}$

(ii) $\dfrac{-7}{2}$ is a negative rational number and $\dfrac{5}{2}$ is a positive rational number.

Since every negative rational number is less than every positive rational number,

Hence, $\dfrac{-7}{2} \lt \dfrac{5}{2}$

(iii) -3 is a negative number and $2\dfrac{3}{4}$ is a positive number.

Since every negative number is less than every positive number,

Hence, $-3 \lt 2\dfrac{3}{4}$

(iv) $-1\dfrac{1}{2}$ is a negative number and 0 is neither positive nor negative.

Since every negative number is less than 0,

Hence, $-1\dfrac{1}{2} \lt 0$

(v) 0 is neither positive nor negative and $\dfrac{3}{4}$ is a positive number.

Since 0 is less than every positive number,

Hence, $0 \lt \dfrac{3}{4}$

(vi) 3 is a positive number and -1 is a negative number.

Since every positive number is greater than every negative number,

Hence, 3 > -1

Compare:

(i) $-\dfrac{1}{4}$ and 0

(ii) $\dfrac{1}{4}$ and 0

(iii) $-\dfrac{3}{8} \text{ and } \dfrac{2}{5}$

(iv) $\dfrac{-5}{8} \text{ and } \dfrac{7}{-12}$

(v) $\dfrac{5}{-9} \text{ and } \dfrac{-5}{-9}$

(vi) $\dfrac{-7}{8} \text{ and } \dfrac{5}{-6}$

(vii) $\dfrac{2}{7} \text{ and } \dfrac{-3}{-8}$

Answer

(i) $-\dfrac{1}{4}$ is a negative number and 0 is neither positive nor negative.

Since every negative number is less than 0,

Hence, $-\dfrac{1}{4} \lt 0$

(ii) $\dfrac{1}{4}$ is a positive number and 0 is neither positive nor negative.

Since 0 is less than every positive number,

Hence, $\dfrac{1}{4} \gt 0$

(iii) $-\dfrac{3}{8}$ is a negative number and $\dfrac{2}{5}$ is a positive number.

Since every negative number is less than every positive number,

Hence, $-\dfrac{3}{8} \lt \dfrac{2}{5}$

(iv) $\dfrac{7}{-12} = \dfrac{-7}{12}$

By Division Method,

$\begin{array}{l|r} 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$

LCM of 8 and 12 = 2 × 2 × 2 × 3 = 24

$\dfrac{-5 \times 3}{8 \times 3} = \dfrac{-15}{24} \text{ and } \dfrac{-7 \times 2}{12 \times 2} = \dfrac{-14}{24}$

Since -15 < -14,

Hence, $\dfrac{-5}{8} \lt \dfrac{7}{-12}$

(v) $\dfrac{5}{-9} = \dfrac{-5}{9} \text{ and } \dfrac{-5}{-9} = \dfrac{5}{9}$

$\dfrac{-5}{9}$ is negative and $\dfrac{5}{9}$ is positive.

Since every negative number is less than every positive number,

Hence, $\dfrac{5}{-9} \lt \dfrac{-5}{-9}$

(vi) $\dfrac{5}{-6} = \dfrac{-5}{6}$

By prime factorization,

$\begin{array}{l|r} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

LCM of 8 and 6 is 2 × 2 × 2 × 3 = 24

$\dfrac{-7 \times 3}{8 \times 3} = \dfrac{-21}{24} \text{ and } \dfrac{-5 \times 4}{6 \times 4} = \dfrac{-20}{24}$

Since -21 < -20,

Hence, $\dfrac{-7}{8} \lt \dfrac{5}{-6}$

(vii) $\dfrac{-3}{-8} = \dfrac{3}{8}$

On cross-multiplying $\dfrac{2}{7} \text{ and } \dfrac{3}{8}$:

$2 \times 8 = 16 \text{ and } 7 \times 3 = 21$

Since 16 < 21,

Hence, $\dfrac{2}{7} \lt \dfrac{-3}{-8}$

Arrange the given rational numbers in ascending order:

(i) $\dfrac{7}{10}, \dfrac{-11}{-30} \text{ and } \dfrac{5}{-15}$

(ii) $\dfrac{4}{-9}, \dfrac{-5}{12} \text{ and } \dfrac{2}{-3}$

Answer

(i) Writing each number with a positive denominator:

$\dfrac{-11}{-30} = \dfrac{11}{30} \text{ and } \dfrac{5}{-15} = \dfrac{-1}{3}$

So, the numbers are $\dfrac{7}{10}, \dfrac{11}{30} \text{ and } \dfrac{-1}{3}$.

By Division Method,

$\begin{array}{l|r} 2 & 10, 30, 3 \\ \hline 3 & 5, 15, 3 \\ \hline 5 & 5, 5, 1 \\ \hline & 1, 1, 1 \end{array}$

LCM of 10, 30 and 3 is 2 × 3 × 5 = 30

$\dfrac{7 \times 3}{10 \times 3} = \dfrac{21}{30}, \\[1em] \dfrac{11 \times 1}{30 \times 1} = \dfrac{11}{30}, \\[1em] \dfrac{-1 \times 10}{3 \times 10} = \dfrac{-10}{30}$

Since -10 < 11 < 21,

$\dfrac{-10}{30} \lt \dfrac{11}{30} \lt \dfrac{21}{30}$

Hence, $\dfrac{5}{-15} \lt \dfrac{-11}{-30} \lt \dfrac{7}{10}$.

(ii) Writing each number with a positive denominator:

$\dfrac{4}{-9} = \dfrac{-4}{9} \text{ and } \dfrac{2}{-3} = \dfrac{-2}{3}$

So, the numbers are $\dfrac{-4}{9}, \dfrac{-5}{12} \text{ and } \dfrac{-2}{3}$.

By Division Method,

$\begin{array}{l|r} 2 & 9, 12, 3 \\ \hline 2 & 9, 6, 3 \\ \hline 3 & 9, 3, 3 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}$

LCM of 9, 12 and 3 is 2 × 2 × 3 × 3 = 36

$\dfrac{-4 \times 4}{9 \times 4} = \dfrac{-16}{36}, \\[1em] \dfrac{-5 \times 3}{12 \times 3} = \dfrac{-15}{36}, \\[1em] \dfrac{-2 \times 12}{3 \times 12} = \dfrac{-24}{36}$

Since -24 < -16 < -15,

$\dfrac{-24}{36} \lt \dfrac{-16}{36} \lt \dfrac{-15}{36}$

Hence, $\dfrac{2}{-3} \lt \dfrac{4}{-9} \lt \dfrac{-5}{12}$.

Arrange the given rational numbers in descending order:

(i) $\dfrac{5}{8}, \dfrac{13}{-16} \text{ and } \dfrac{-7}{12}$

(ii) $\dfrac{3}{-10}, \dfrac{-13}{30} \text{ and } \dfrac{8}{-20}$

Answer

(i) Writing each number with a positive denominator:

$\dfrac{13}{-16} = \dfrac{-13}{16}$

So, the numbers are $\dfrac{5}{8}, \dfrac{-13}{16} \text{ and } \dfrac{-7}{12}$.

By Division Method,

$\begin{array}{l|r} 2 & 8, 16, 12 \\ \hline 2 & 4, 8, 6 \\ \hline 2 & 2, 4, 3 \\ \hline 2 & 1, 2, 3 \\ \hline 3 & 1, 1, 3 \\ \hline & 1, 1, 1 \end{array}$

LCM of 8, 16 and 12 is 2 × 2 × 2 × 2 × 3 = 48

$\dfrac{5 \times 6}{8 \times 6} = \dfrac{30}{48}, \\[1em] \dfrac{-13 \times 3}{16 \times 3} = \dfrac{-39}{48}, \\[1em] \dfrac{-7 \times 4}{12 \times 4} = \dfrac{-28}{48}$

Since 30 > -28 > -39,

$\dfrac{30}{48} \gt \dfrac{-28}{48} \gt \dfrac{-39}{48}$

Hence, $\dfrac{5}{8} \gt \dfrac{-7}{12} \gt \dfrac{13}{-16}$.

(ii) Writing each number with a positive denominator:

$\dfrac{3}{-10} = \dfrac{-3}{10} \text{ and } \dfrac{8}{-20} = \dfrac{-2}{5}$

So, the numbers are $\dfrac{-3}{10}, \dfrac{-13}{30} \text{ and } \dfrac{-2}{5}$.

By Division Method,

$\begin{array}{l|r} 2 & 10, 30, 5 \\ \hline 3 & 5, 15, 5 \\ \hline 5 & 5, 5, 5 \\ \hline & 1, 1, 1 \end{array}$

LCM of 10, 30 and 5 is 2 × 3 × 5 = 30

$\dfrac{-3 \times 3}{10 \times 3} = \dfrac{-9}{30}, \\[1em] \dfrac{-13 \times 1}{30 \times 1} = \dfrac{-13}{30}, \\[1em] \dfrac{-2 \times 6}{5 \times 6} = \dfrac{-12}{30}$

Since -9 > -12 > -13,

$\dfrac{-9}{30} \gt \dfrac{-12}{30} \gt \dfrac{-13}{30}$

Hence, $\dfrac{3}{-10} \gt \dfrac{8}{-20} \gt \dfrac{-13}{30}$.

Fill in the blanks:

(i) $\dfrac{5}{8} \text{ and } \dfrac{3}{10}$ are on the .......... side of zero.

(ii) $-\dfrac{5}{8} \text{ and } \dfrac{3}{10}$ are on the .......... sides of zero.

(iii) $-\dfrac{5}{8} \text{ and } -\dfrac{3}{10}$ are on the .......... side of zero.

(iv) $\dfrac{5}{8} \text{ and } -\dfrac{3}{10}$ are on the .......... sides of zero.

Answer

(i) $\dfrac{5}{8} \text{ and } \dfrac{3}{10}$ are both positive, so they are on the same side of zero.

(ii) $-\dfrac{5}{8}$ is negative and $\dfrac{3}{10}$ is positive, so they are on the opposite sides of zero.

(iii) $-\dfrac{5}{8} \text{ and } -\dfrac{3}{10}$ are both negative, so they are on the same side of zero.

(iv) $\dfrac{5}{8}$ is positive and $-\dfrac{3}{10}$ is negative, so they are on the opposite sides of zero.

Insert three rational numbers between:

(i) $-\dfrac{2}{3} \text{ and } \dfrac{3}{4}$

(ii) $\dfrac{5}{7} \text{ and } \dfrac{7}{9}$

(iii) $-\dfrac{5}{8} \text{ and } -\dfrac{1}{6}$

Answer

(i) By Division Method,

$\begin{array}{l|r} 2 & 3, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$

LCM of 3 and 4 = 2 × 2 × 3 = 12

$-\dfrac{2}{3} = \dfrac{-2 \times 4}{3 \times 4} = \dfrac{-8}{12} \text{ and } \dfrac{3}{4} = \dfrac{3 \times 3}{4 \times 3} = \dfrac{9}{12}$

Now, $\dfrac{-8}{12} \lt \dfrac{-7}{12} \lt \dfrac{-6}{12} \lt \dfrac{-5}{12} \lt \dfrac{9}{12}$

Hence, $\dfrac{-7}{12}, \dfrac{-6}{12} \text{ and } \dfrac{-5}{12}$ are three rational numbers between $-\dfrac{2}{3} \text{ and } \dfrac{3}{4}$.

(ii) By Division Method,

$\begin{array}{l|r} 3 & 7, 9 \\ \hline 3 & 7, 3 \\ \hline 7 & 7, 1 \\ \hline & 1, 1 \end{array}$

LCM of 7 and 9 = 3 × 3 × 7 = 63

$\dfrac{5}{7} = \dfrac{5 \times 9}{7 \times 9} = \dfrac{45}{63} \\[1em] \dfrac{7}{9} = \dfrac{7 \times 7}{9 \times 7} = \dfrac{49}{63}$

Now, $\dfrac{45}{63} \lt \dfrac{46}{63} \lt \dfrac{47}{63} \lt \dfrac{48}{63} \lt \dfrac{49}{63}$

Hence, $\dfrac{46}{63}, \dfrac{47}{63} \text{ and } \dfrac{48}{63}$ are three rational numbers between $\dfrac{5}{7} \text{ and } \dfrac{7}{9}$.

(iii) By Division Method,

$\begin{array}{l|r} 2 & 8, 6 \\ \hline 2 & 4, 3 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$

LCM of 8 and 6 is 2 × 2 × 2 × 3 = 24

$-\dfrac{5}{8} = \dfrac{-5 \times 3}{8 \times 3} = \dfrac{-15}{24} \\[1em] -\dfrac{1}{6} = \dfrac{-1 \times 4}{6 \times 4} = \dfrac{-4}{24}$

Now, $\dfrac{-15}{24} \lt \dfrac{-14}{24} \lt \dfrac{-13}{24} \lt \dfrac{-12}{24} \lt \dfrac{-4}{24}$

Hence, $\dfrac{-14}{24}, \dfrac{-13}{24} \text{ and } \dfrac{-12}{24}$ are three rational numbers between $-\dfrac{5}{8} \text{ and } -\dfrac{1}{6}$.

Insert four rational numbers between:

(i) $-\dfrac{3}{8} \text{ and } -\dfrac{5}{6}$

(ii) $-\dfrac{4}{5} \text{ and } \dfrac{2}{3}$

(iii) -3 and 6

(iv) 0 and 6

Answer

(i) $\dfrac{5}{-6} = -\dfrac{5}{6}$

By Division Method,

$\begin{array}{l|r} 2 & 8, 6 \\ \hline 2 & 4, 3 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$

LCM of 8 and 6 = 2 × 2 × 2 × 3 = 24

$-\dfrac{5}{6} = \dfrac{-5 \times 4}{6 \times 4} = \dfrac{-20}{24} \\[1em] -\dfrac{3}{8} = \dfrac{-3 \times 3}{8 \times 3} = \dfrac{-9}{24}$

Now, $\dfrac{-20}{24} \lt \dfrac{-19}{24} \lt \dfrac{-18}{24} \lt \dfrac{-17}{24} \lt \dfrac{-16}{24} \lt \dfrac{-9}{24}$

Hence, $\dfrac{-19}{24}, \dfrac{-18}{24}, \dfrac{-17}{24} \text{ and } \dfrac{-16}{24}$ are four rational numbers between $-\dfrac{3}{8} \text{ and } -\dfrac{5}{6}$.

(ii) By Division Method,

$\begin{array}{l|r} 3 & 5, 3 \\ \hline 5 & 5, 1 \\ \hline & 1, 1 \end{array}$

LCM of 5 and 3 = 3 × 5 = 15

$-\dfrac{4}{5} = \dfrac{-4 \times 3}{5 \times 3} = \dfrac{-12}{15} \\[1em] \dfrac{2}{3} = \dfrac{2 \times 5}{3 \times 5} = \dfrac{10}{15}$

Now, $\dfrac{-12}{15} \lt \dfrac{-11}{15} \lt \dfrac{-10}{15} \lt \dfrac{-9}{15} \lt \dfrac{-8}{15} \lt \dfrac{10}{15}$

Hence, $\dfrac{-11}{15}, \dfrac{-10}{15}, \dfrac{-9}{15} \text{ and } \dfrac{-8}{15}$ are four rational numbers between $-\dfrac{4}{5} \text{ and } \dfrac{2}{3}$.

(iii) The integers between -3 and 6 are -2, -1, 0, 1, 2, 3, 4 and 5, each of which is a rational number.

Hence, -2, -1, 0 and 1 are four rational numbers between -3 and 6.

(iv) The integers between 0 and 6 are 1, 2, 3, 4 and 5, each of which is a rational number.

Hence, 1, 2, 3 and 4 are four rational numbers between 0 and 6.