Rational Numbers — Exercise 1(D)

Find a rational number between the following pair of rational numbers:

$\dfrac{7}{10} \text{ and } \dfrac{10}{17}$

Answer

If $x$ and $y$ are two rational numbers, then $\dfrac{1}{2}(x + y)$ is a rational number lying between them.

LCM of 10 and 17 = 170

Finding the required rational number,

$\dfrac{1}{2} \times \left(\dfrac{7}{10} + \dfrac{10}{17}\right) = \dfrac{1}{2} \times \left(\dfrac{7 \times 17 + 10 \times 10}{170}\right) \\[1em] = \dfrac{1}{2} \times \dfrac{119 + 100}{170} \\[1em] = \dfrac{1}{2} \times \dfrac{219}{170} \\[1em] = \dfrac{219}{340}$

Hence, $\mathbf{\dfrac{219}{340}}$ is a rational number lying between $\mathbf{\dfrac{7}{10} \text{ and } \dfrac{10}{17}}$.

Find a rational number between the following pair of rational numbers:

$1\dfrac{3}{8}$ and 2

Answer

If $x$ and $y$ are two rational numbers, then $\dfrac{1}{2}(x + y)$ is a rational number lying between them.

First convert the mixed fraction into an improper fraction.

$1\dfrac{3}{8} = \dfrac{11}{8}$

Finding the required rational number,

$\dfrac{1}{2} \times \left(\dfrac{11}{8} + 2\right) = \dfrac{1}{2} \times \dfrac{11 + 2 \times 8}{8} \\[1em] = \dfrac{1}{2} \times \dfrac{27}{8} \\[1em] = \dfrac{27}{16} = 1\dfrac{11}{16}$

Hence, $\mathbf{\dfrac{27}{16}}$ is a rational number lying between $\mathbf{1\dfrac{3}{8} \text{ and } 2}$.

Find a rational number between the following pair of rational numbers:

$\dfrac{-3}{5} \text{ and } \dfrac{-4}{7}$

Answer

If $x$ and $y$ are two rational numbers, then $\dfrac{1}{2}(x + y)$ is a rational number lying between them.

Finding the required rational number,

LCM of 5 and 7 = 35

$\dfrac{1}{2} \times \left(\dfrac{-3}{5} + \dfrac{-4}{7}\right) = \dfrac{1}{2} \times \left(\dfrac{(-3) \times 7 + (-4) \times 5}{35}\right) \\[1em] = \dfrac{1}{2} \times \dfrac{-21 - 20}{35} \\[1em] = \dfrac{1}{2} \times \dfrac{-41}{35} \\[1em] = \dfrac{-41}{70}$

Hence, $\mathbf{\dfrac{-41}{70}}$ is a rational number lying between $\mathbf{\dfrac{-3}{5} \text{ and } \dfrac{-4}{7}}$.

Find a rational number between the following pair of rational numbers:

-2 and $\dfrac{-17}{21}$

Answer

If $x$ and $y$ are two rational numbers, then $\dfrac{1}{2}(x + y)$ is a rational number lying between them.

Finding the required rational number,

$\dfrac{1}{2} \times \left(-2 + \dfrac{-17}{21}\right) = \dfrac{1}{2} \times \dfrac{(-2) \times 21 - 17}{21} \\[1em] = \dfrac{1}{2} \times \dfrac{-42 - 17}{21} \\[1em] = \dfrac{1}{2} \times \dfrac{-59}{21} \\[1em] = \dfrac{-59}{42} = -1\dfrac{17}{42}$

Hence, $\mathbf{\dfrac{-59}{42}}$ is a rational number lying between $\mathbf{-2 \text{ and } \dfrac{-17}{21}}$.

Find three rational numbers between:

4 and $4\dfrac{2}{3}$

Answer

Write both the numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

$\Rightarrow 4 = \dfrac{12}{3} \\[1em] \Rightarrow 4\dfrac{2}{3} = \dfrac{14}{3}$

To insert 3 rational numbers, multiply both the numerator and the denominator of each rational number by 3 + 1 i.e. 4.

$\Rightarrow \dfrac{12}{3} = \dfrac{12 \times 4}{3 \times 4} = \dfrac{48}{12} \\[1em] \Rightarrow \dfrac{14}{3} = \dfrac{14 \times 4}{3 \times 4} = \dfrac{56}{12}$

As 48 < 49 < 50 < 51 < 56,

$\therefore \dfrac{48}{12} \lt \dfrac{49}{12} \lt \dfrac{50}{12} \lt \dfrac{51}{12} \lt \dfrac{56}{12}$

Hence, three rational numbers between 4 and $\mathbf{4\dfrac{2}{3}}$ are $\mathbf{\dfrac{49}{12}, \dfrac{50}{12} \text{ and } \dfrac{51}{12}}$.

Find three rational numbers between:

$\dfrac{-1}{2} \text{ and } \dfrac{-1}{4}$

Answer

Write both the rational numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

$[\text{ LCM of } 2 \text{ and } 4 = 4] \\[1em] \Rightarrow \dfrac{-1}{2} = \dfrac{-1 \times 2}{2 \times 2} = \dfrac{-2}{4} \\[1em] \Rightarrow \dfrac{-1}{4} = \dfrac{-1}{4}$

To insert 3 rational numbers, multiply both the numerator and the denominator of each rational number by 3 + 1 i.e. 4.

$\Rightarrow \dfrac{-2}{4} = \dfrac{-2 \times 4}{4 \times 4} = \dfrac{-8}{16} \\[1em] \Rightarrow \dfrac{-1}{4} = \dfrac{-1 \times 4}{4 \times 4} = \dfrac{-4}{16}$

As -8 < -7 < -6 < -5 < -4,

$\therefore \dfrac{-8}{16} \lt \dfrac{-7}{16} \lt \dfrac{-6}{16} \lt \dfrac{-5}{16} \lt \dfrac{-4}{16}$

Hence, three rational numbers between $\mathbf{\dfrac{-1}{2} \text{ and } \dfrac{-1}{4}}$ are $\mathbf{\dfrac{-7}{16}, \dfrac{-6}{16} \text{ and } \dfrac{-5}{16}}$.

Find three rational numbers between:

2 and 3

Answer

Write both the numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

To insert 3 rational numbers between 2 and 3, multiply both the numerator and the denominator of each number by 3 + 1 i.e. 4.

$\Rightarrow 2 = \dfrac{2 \times 4}{1 \times 4} = \dfrac{8}{4} \\[1em] \Rightarrow 3 = \dfrac{3 \times 4}{1 \times 4} = \dfrac{12}{4}$

As 8 < 9 < 10 < 11 < 12,

$\therefore \dfrac{8}{4} \lt \dfrac{9}{4} \lt \dfrac{10}{4} \lt \dfrac{11}{4} \lt \dfrac{12}{4}$

Hence, three rational numbers between 2 and 3 are $\mathbf{\dfrac{9}{4}, \dfrac{10}{4} \text{ and } \dfrac{11}{4}}$.

Find five rational numbers between:

$\dfrac{3}{5} \text{ and } \dfrac{2}{3}$

Answer

Write both the rational numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

$[\text{ LCM of } 5 \text{ and } 3 = 15] \\[1em] \Rightarrow \dfrac{3}{5} = \dfrac{3 \times 3}{5 \times 3} = \dfrac{9}{15} \\[1em] \Rightarrow \dfrac{2}{3} = \dfrac{2 \times 5}{3 \times 5} = \dfrac{10}{15}$

To insert 5 rational numbers, multiply both the numerator and the denominator of each rational number by 5 + 1 i.e. 6.

$\Rightarrow \dfrac{9}{15} = \dfrac{9 \times 6}{15 \times 6} = \dfrac{54}{90} \\[1em] \Rightarrow \dfrac{10}{15} = \dfrac{10 \times 6}{15 \times 6} = \dfrac{60}{90}$

As 54 < 55 < 56 < 57 < 58 < 59 < 60

$\therefore \dfrac{54}{90} \lt \dfrac{55}{90} \lt \dfrac{56}{90} \lt \dfrac{57}{90} \lt \dfrac{58}{90} \lt \dfrac{59}{90} \lt \dfrac{60}{90}$

Hence, five rational numbers between $\mathbf{\dfrac{3}{5} \text{ and } \dfrac{2}{3}}$ are $\mathbf{\dfrac{55}{90}, \dfrac{56}{90}, \dfrac{57}{90}, \dfrac{58}{90} \text{ and } \dfrac{59}{90}}$.

Find five rational numbers between:

-2 and $-1\dfrac{1}{2}$

Answer

Write both the numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

First convert the mixed fraction into an improper fraction and write both the numbers with a common denominator.

$\Rightarrow -1\dfrac{1}{2} = \dfrac{-3}{2} \\[1em] \Rightarrow -2 = \dfrac{-4}{2}$

To insert 5 rational numbers, multiply both the numerator and the denominator of each rational number by 5 + 1 i.e. 6.

$\Rightarrow \dfrac{-4}{2} = \dfrac{-4 \times 6}{2 \times 6} = \dfrac{-24}{12} \\[1em] \Rightarrow \dfrac{-3}{2} = \dfrac{-3 \times 6}{2 \times 6} = \dfrac{-18}{12}$

As -24 < -23 < -22 < -21 < -20 < -19 < -18,

$\therefore \dfrac{-24}{12} \lt \dfrac{-23}{12} \lt \dfrac{-22}{12} \lt \dfrac{-21}{12} \lt \dfrac{-20}{12} \lt \dfrac{-19}{12} \lt \dfrac{-18}{12}$

Hence, five rational numbers between -2 and $\mathbf{-1\dfrac{1}{2}}$ are

$\mathbf{\dfrac{-23}{12}, \dfrac{-22}{12}, \dfrac{-21}{12}, \dfrac{-20}{12} \text{ and } \dfrac{-19}{12}}$.

Find five rational numbers between:

-3 and -2

Answer

Write both the numbers with a common denominator, then multiply the numerator and the denominator of each by one more than the number of rational numbers required.

To insert 5 rational numbers between -3 and -2, multiply both the numerator and the denominator of each number by 5 + 1 i.e. 6.

$\Rightarrow -3 = \dfrac{-3 \times 6}{1 \times 6} = \dfrac{-18}{6} \\[1em] \Rightarrow -2 = \dfrac{-2 \times 6}{1 \times 6} = \dfrac{-12}{6}$

As -18 < -17 < -16 < -15 < -14 < -13 < -12

$\therefore \dfrac{-18}{6} \lt \dfrac{-17}{6} \lt \dfrac{-16}{6} \lt \dfrac{-15}{6} \lt \dfrac{-14}{6} \lt \dfrac{-13}{6} \lt \dfrac{-12}{6}$

Hence, five rational numbers between -3 and -2 are

$\mathbf{\dfrac{-17}{6}, \dfrac{-16}{6}, \dfrac{-15}{6}, \dfrac{-14}{6} \text{ and } \dfrac{-13}{6}}$.

Find 10 rational numbers between $\dfrac{-3}{4} \text{ and } \dfrac{5}{6}$.

Answer

The given rational numbers have different denominators, so first write them with a common denominator.

$[\text{ LCM of } 4 \text{ and } 6 = 12] \\[1em] \Rightarrow \dfrac{-3}{4} = \dfrac{-3 \times 3}{4 \times 3} = \dfrac{-9}{12} \\[1em] \Rightarrow \dfrac{5}{6} = \dfrac{5 \times 2}{6 \times 2} = \dfrac{10}{12}$

We note that -8, -7, -6, …, 8, 9 are the integers between -9 and 10

i.e. -9 < -8 < -7 < -6 < -5 < -4 < -3 < -2 < -1 < 0 < 1 < … < 9 < 10

Choosing any ten of these integers as numerators over the common denominator 12, we get the required rational numbers.

Hence, ten rational numbers between $\mathbf{\dfrac{-3}{4} \text{ and } \dfrac{5}{6}}$ are

$\mathbf{\dfrac{-8}{12}, \dfrac{-7}{12}, \dfrac{-6}{12}, \dfrac{-5}{12}, \dfrac{-4}{12}, \dfrac{-3}{12}, \dfrac{-2}{12}, \dfrac{-1}{12}, \dfrac{0}{12} \text{ and } \dfrac{1}{12}}$.

Find 12 rational numbers between -1 and 2.

Answer

To insert 12 rational numbers, write both the numbers with the denominator 12 + 1 i.e. 13.

$\Rightarrow -1 = \dfrac{-1 \times 13}{1 \times 13} = \dfrac{-13}{13} \\[1em] \Rightarrow 2 = \dfrac{2 \times 13}{1 \times 13} = \dfrac{26}{13}$

We note that -12, -11, -10, …, 24, 25 are the integers between -13 and 26

i.e. -13 < -12 < -11 < -10 < … < -2 < -1 < 0 < … < 25 < 26

Choosing any twelve of these integers as numerators over the common denominator 13, we get the required rational numbers.

Hence, twelve rational numbers between -1 and 2 are

$\mathbf{\dfrac{-12}{13}, \dfrac{-11}{13}, \dfrac{-10}{13}, \dfrac{-9}{13}, \dfrac{-8}{13}, \dfrac{-7}{13}, \dfrac{-6}{13}, \dfrac{-5}{13}, \dfrac{-4}{13}, \dfrac{-3}{13}, \dfrac{-2}{13} \text{ and } \dfrac{-1}{13}}$.