Rational Numbers — Exercise 2(A)

Write down a rational number whose numerator is the largest number of two digits and denominator is the smallest number of four digits.

Answer

Largest number of two digits = 99

Smallest number of four digits = 1000

Hence, the required rational number = $\dfrac{99}{1000}$.

Write the numerator of each of the following rational numbers:

(i) $\dfrac{-125}{127}$

(ii) $\dfrac{37}{-137}$

(iii) $\dfrac{-85}{93}$

(iv) 2

(v) 0

Answer

(i) Numerator of $\dfrac{-125}{127}$ is -125.

(ii) Numerator of $\dfrac{37}{-137}$ is 37.

(iii) Numerator of $\dfrac{-85}{93}$ is -85.

(iv) 2 can be written as $\dfrac{2}{1}$, so its numerator is 2.

(v) 0 can be written as $\dfrac{0}{1}$, so its numerator is 0.

Write the denominator of each of the following rational numbers:

(i) $\dfrac{7}{-15}$

(ii) $\dfrac{-18}{29}$

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

(iv) -7

(v) 0

Answer

(i) Denominator of $\dfrac{7}{-15}$ is -15.

(ii) Denominator of $\dfrac{-18}{29}$ is 29.

(iii) Denominator of $\dfrac{-3}{4}$ is 4.

(iv) -7 can be written as $\dfrac{-7}{1}$, so its denominator is 1.

(v) 0 can be written as $\dfrac{0}{1}, \dfrac{0}{2}, \dfrac{0}{3}, \dfrac{0}{4}, \ldots$

Zero does not have any unique denominator, so its denominator can be any non-zero number.

Write down a rational number with numerator (-5) × (-4) and with denominator (28 - 27) × (8 - 5).

Answer

Numerator = (-5) × (-4) = 20

Denominator = (28 - 27) × (8 - 5) = 1 × 3 = 3

Hence, the required rational number = $\dfrac{20}{3}$.

(i) $\dfrac{-15}{1}$ in integer form is ............ .

(ii) $\dfrac{23}{-1}$ in integer form is ............ .

(iii) If $18 = \dfrac{18}{a}$ then a = ............ .

(iv) If $-57 = \dfrac{57}{a}$ then a = ............ .

Answer

(i) $\dfrac{-15}{1}$ in integer form is -15.

(ii) $\dfrac{23}{-1}$ in integer form is -23.

(iii) Given,

$\Rightarrow 18 = \dfrac{18}{a} \\[1em] \Rightarrow 18 \times a = 18 \times 1 \\[1em] \Rightarrow a = \dfrac{18}{18} \\[1em] \Rightarrow a = 1$

Hence, a = 1

(iv) Given,

$\Rightarrow -57 = \dfrac{57}{a}\\[1em] \Rightarrow -57 \times a = 57 \times 1 \\[1em] \Rightarrow a = \dfrac{57}{-57} \\[1em] \Rightarrow a = -1$

Hence, a = -1

Separate positive and negative rational numbers from the following:

$\dfrac{-3}{5}, \dfrac{3}{-5}, \dfrac{-3}{-5}, \dfrac{3}{5}, 0, \dfrac{-13}{-3}, \dfrac{15}{-8}, \dfrac{-15}{8}$

Answer

A rational number is positive if its numerator and denominator are both of the same sign, and negative if they are of opposite signs.

Positive rational numbers: $\dfrac{-3}{-5}, \dfrac{3}{5} \text{ and } \dfrac{-13}{-3}$

Negative rational numbers: $\dfrac{-3}{5}, \dfrac{3}{-5}, \dfrac{15}{-8} \text{ and } \dfrac{-15}{8}$

(0 is neither positive nor negative).

Find three rational numbers equivalent to

(i) $\dfrac{3}{5}$

(ii) $\dfrac{4}{-7}$

(iii) $\dfrac{-5}{9}$

(iv) $\dfrac{8}{-15}$

Answer

(i) Three rational numbers equivalent to $\dfrac{3}{5}$ are :

$\dfrac{3 \times 2}{5 \times 2} = \dfrac{6}{10}, \\[1em] \dfrac{3 \times 3}{5 \times 3} = \dfrac{9}{15}, \\[1em] \dfrac{3 \times 4}{5 \times 4} = \dfrac{12}{20}$

Hence, $\dfrac{6}{10}, \dfrac{9}{15} \text{ and } \dfrac{12}{20}$ are three rational numbers equivalent to $\dfrac{3}{5}$.

(ii) Three rational numbers equivalent to $\dfrac{4}{-7}$ are :

$\dfrac{4 \times 2}{-7 \times 2} = \dfrac{8}{-14}, \\[1em] \dfrac{4 \times 3}{-7 \times 3} = \dfrac{12}{-21}, \\[1em] \dfrac{4 \times 4}{-7 \times 4} = \dfrac{16}{-28}$

Hence, $\dfrac{8}{-14}, \dfrac{12}{-21} \text{ and } \dfrac{16}{-28}$ are three rational numbers equivalent to $\dfrac{4}{-7}$.

(iii) Three rational numbers equivalent to $\dfrac{-5}{9}$ are :

$\dfrac{-5 \times 2}{9 \times 2} = \dfrac{-10}{18}, \\[1em] \dfrac{-5 \times 3}{9 \times 3} = \dfrac{-15}{27}, \\[1em] \dfrac{-5 \times 4}{9 \times 4} = \dfrac{-20}{36}$

Hence, $\dfrac{-10}{18}, \dfrac{-15}{27} \text{ and } \dfrac{-20}{36}$ are three rational numbers equivalent to $\dfrac{-5}{9}$.

(iv) Three rational numbers equivalent to $\dfrac{8}{-15}$ are :

$\dfrac{8 \times 2}{-15 \times 2} = \dfrac{16}{-30}, \\[1em] \dfrac{8 \times 3}{-15 \times 3} = \dfrac{24}{-45}, \\[1em] \dfrac{8 \times 4}{-15 \times 4} = \dfrac{32}{-60}$

Hence, $\dfrac{16}{-30}, \dfrac{24}{-45} \text{ and } \dfrac{32}{-60}$ are three rational numbers equivalent to $\dfrac{8}{-15}$.

Which of the following are not rational numbers:

(i) -3

(ii) 0

(iii) $\dfrac{0}{4}$

(iv) $\dfrac{8}{0}$

(v) $\dfrac{0}{0}$

Answer

A number $\dfrac{p}{q}$ is a rational number only if q ≠ 0.

(i) -3 = $\dfrac{-3}{1}$, which is a rational number.

(ii) 0 = $\dfrac{0}{1}$, which is a rational number.

(iii) $\dfrac{0}{4} = 0$, which is a rational number.

(iv) $\dfrac{8}{0}$ has denominator 0, so it is not a rational number.

(v) $\dfrac{0}{0}$ has denominator 0, so it is not a rational number.

Hence, $\dfrac{8}{0}\text{ and }\dfrac{0}{0}$ are not rational numbers.

Express each of the following integers as a rational number with denominator 7:

(i) 5

(ii) -8

(iii) 0

(iv) -16

(v) 7

Answer

(i) $5 = \dfrac{5 \times 7}{1 \times 7} = \dfrac{35}{7}$

(ii) $-8 = \dfrac{-8 \times 7}{1 \times 7} = \dfrac{-56}{7}$

(iii) $0 = \dfrac{0 \times 7}{1 \times 7} = \dfrac{0}{7}$

(iv) $-16 = \dfrac{-16 \times 7}{1 \times 7} = \dfrac{-112}{7}$

(v) $7 = \dfrac{7 \times 7}{1 \times 7} = \dfrac{49}{7}$

Express $\dfrac{3}{5}$ as a rational number with denominator:

(i) 20

(ii) -20

(iii) 45

(iv) 25

(v) -35

Answer

(i) $\dfrac{3}{5} = \dfrac{3 \times 4}{5 \times 4} = \dfrac{12}{20}$

(ii) $\dfrac{3}{5} = \dfrac{3 \times (-4)}{5 \times (-4)} = \dfrac{-12}{-20}$

(iii) $\dfrac{3}{5} = \dfrac{3 \times 9}{5 \times 9} = \dfrac{27}{45}$

(iv) $\dfrac{3}{5} = \dfrac{3 \times 5}{5 \times 5} = \dfrac{15}{25}$

(v) $\dfrac{3}{5} = \dfrac{3 \times (-7)}{5 \times (-7)} = \dfrac{-21}{-35}$

Express $\dfrac{4}{7}$ as a rational number with numerator:

(i) 12

(ii) -12

(iii) -16

(iv) -20

(v) 20

Answer

(i) $\dfrac{4}{7} = \dfrac{4 \times 3}{7 \times 3} = \dfrac{12}{21}$

(ii) $\dfrac{4}{7} = \dfrac{4 \times (-3)}{7 \times (-3)} = \dfrac{-12}{-21}$

(iii) $\dfrac{4}{7} = \dfrac{4 \times (-4)}{7 \times (-4)} = \dfrac{-16}{-28}$

(iv) $\dfrac{4}{7} = \dfrac{4 \times (-5)}{7 \times (-5)} = \dfrac{-20}{-35}$

(v) $\dfrac{4}{7} = \dfrac{4 \times 5}{7 \times 5} = \dfrac{20}{35}$

Find x, such that:

(i) $-\dfrac{2}{3} = \dfrac{6}{x}$

(ii) $\dfrac{7}{-4} = \dfrac{x}{8}$

(iii) $\dfrac{3}{7} = \dfrac{x}{-35}$

(iv) $\dfrac{-48}{x} = 6$

(v) $\dfrac{36}{x} = 3$

(vi) $\dfrac{-27}{x} = 9$

Answer

(i) Given,

$\Rightarrow -\dfrac{2}{3} = \dfrac{6}{x} \\[1em] \Rightarrow -2 \times x = 3 \times 6 \\[1em] \Rightarrow -2x = 18 \\[1em] \Rightarrow x = \dfrac{18}{-2} \\[1em] \Rightarrow x = -9$

Hence, x = -9

(ii) Given,

$\Rightarrow \dfrac{7}{-4} = \dfrac{x}{8}\\[1em] \Rightarrow -4 \times x = 7 \times 8\\[1em] \Rightarrow -4x = 56 \\[1em] \Rightarrow x = \dfrac{56}{-4} \\[1em] \Rightarrow x = -14$

Hence, x = -14

(iii) Given,

$\Rightarrow \dfrac{3}{7} = \dfrac{x}{-35}\\[1em] \Rightarrow 7 \times x = 3 \times (-35) \\[1em] \Rightarrow 7x = -105 \\[1em] \Rightarrow x = \dfrac{-105}{7} \\[1em] \Rightarrow x = -15$

Hence, x = -15

(iv) Given,

$\Rightarrow \dfrac{-48}{x} = 6\\[1em] \Rightarrow 6x = -48 \\[1em] \Rightarrow x = \dfrac{-48}{6} \\[1em] \Rightarrow x = -8$

Hence, x = -8

(v) Given,

$\Rightarrow \dfrac{36}{x} = 3\\[1em] \Rightarrow 3x = 36 \\[1em] \Rightarrow x = \dfrac{36}{3} \\[1em] \Rightarrow x = 12$

Hence, x = 12

(vi) Given,

$\Rightarrow \dfrac{-27}{x} = 9\\[1em] \Rightarrow 9x = -27 \\[1em] \Rightarrow x = \dfrac{-27}{9} \\[1em] \Rightarrow x = -3$

Hence, x = -3

Express each of the following rational numbers to the lowest terms:

(i) $\dfrac{12}{15}$

(ii) $\dfrac{-120}{144}$

(iii) $\dfrac{-48}{-72}$

(iv) $\dfrac{14}{-56}$

Answer

(i) By Prime Factorization,

$\begin{array}{l|r} 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

HCF of 12 and 15 = 3.

$\dfrac{12}{15} = \dfrac{12 ÷ 3}{15 ÷ 3} = \dfrac{4}{5}$

Hence, $\dfrac{12}{15} = \dfrac{4}{5}$

(ii) By Prime Factorization,

$\begin{array}{l|r} 2 & 120 \\ \hline 2 & 60 \\ \hline 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

HCF of 120 and 144 = 2 × 2 × 2 × 3 = 24.

$\dfrac{-120}{144} = \dfrac{-120 ÷ 24}{144 ÷ 24} = \dfrac{-5}{6}$

Hence, $\dfrac{-120}{144} = \dfrac{-5}{6}$

(iii) $\dfrac{-48}{-72} = \dfrac{48}{72}$

By Prime Factorization,

$\begin{array}{l|r} 2 & 48 \\ \hline 2 & 24 \\ \hline 2 & 12 \\ \hline 2 & 6 \\ \hline 3 & 3 \\ \hline & 1 \end{array} \quad \text{ and } \quad \begin{array}{l|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

HCF of 48 and 72 = 2 × 2 × 2 × 3 = 24.

$\dfrac{48}{72} = \dfrac{48 ÷ 24}{72 ÷ 24} = \dfrac{2}{3}$

Hence, $\dfrac{-48}{-72} = \dfrac{2}{3}$

(iv) By Prime Factorization,

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

HCF of 14 and 56 = 2 × 7 = 14.

$\dfrac{14}{-56} = \dfrac{14 ÷ 14}{-56 ÷ 14} = \dfrac{1}{-4} = -\dfrac{1}{4}$

Hence, $\dfrac{14}{-56} = -\dfrac{1}{4}$

Express each of the following rational numbers in the standard form.

(i) $\dfrac{-7}{-8}$

(ii) $\dfrac{5}{-12}$

(iii) $\dfrac{-7}{-20}$

(iv) $\dfrac{4}{-9}$

Answer

A rational number is in standard form if its denominator is positive and the rational number is in its lowest terms.

(i) $\dfrac{-7}{-8} = \dfrac{-7 \times (-1)}{-8 \times (-1)} = \dfrac{7}{8}$

Hence, standard form of $\dfrac{-7}{-8}\text{ is }\dfrac{7}{8}$.

(ii) $\dfrac{5}{-12} = \dfrac{5 \times (-1)}{-12 \times (-1)} = \dfrac{-5}{12}$

Hence, standard form of $\dfrac{5}{-12}\text{ is }\dfrac{-5}{12}$.

(iii) $\dfrac{-7}{-20} = \dfrac{-7 \times (-1)}{-20 \times (-1)} = \dfrac{7}{20}$

Hence, standard form of $\dfrac{-7}{-20}\text{ is }\dfrac{7}{20}$.

(iv) $\dfrac{4}{-9} = \dfrac{4 \times (-1)}{-9 \times (-1)} = \dfrac{-4}{9}$

Hence, standard form of $\dfrac{4}{-9}\text{ is }\dfrac{-4}{9}$.