Rational Numbers

What are rational numbers? Give four examples of each of positive rationals and negative rationals. Give an example of a rational number which is neither positive nor negative.

Answer

A rational number is a number that can be expressed in the form $\dfrac{p}{q}$, where p and q are integers and q ≠ 0. In $\dfrac{p}{q}$, we have numerator = p and denominator = q.

Examples of Positive Rational Numbers:

A rational number is positive if its numerator and denominator are both of the same sign (both positive or both negative).

(1) $\dfrac{3}{4}$

(2) $\dfrac{11}{20}$

(3) 5 (which is $\dfrac{5}{1}$)

(4) $\dfrac{-7}{-9}$

Examples of Negative Rational Numbers:

A rational number is negative if its numerator and denominator are of opposite signs.

(1) $-\dfrac{1}{2}$

(2) $\dfrac{-5}{8}$

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

(4)-10 (which is $\dfrac{-10}{1}$)

Example of a rational number which is neither positive nor negative is 0.

Which of the following are rational numbers?

(i) $\dfrac{5}{15}$

(ii) $\dfrac{-6}{23}$

(iii) 17

(iv) -25

(v) 0

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

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

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

(ix) $\dfrac{-23}{-37}$

(x) $\dfrac{-1}{7}$

Answer

(i) $\dfrac{5}{15}$

⇒ It is in $\dfrac{p}{q}$ form where both 5 and 15 are integers and q ≠ 0.

Hence, it is a rational number.

(ii) $\dfrac{-6}{23}$

⇒ Both -6 and 23 are integers and the denominator is not zero.

Hence, it is a rational number.

(iii) 17

⇒ An integer that can be written as $\dfrac{17}{1}$.

Hence, it is a rational number.

(iv) -25

⇒ A negative integer that can be written as $\dfrac{-25}{1}$.

Hence, it is a rational number.

(v) 0

⇒ Zero is a rational number because it can be written as $\dfrac{0}{1}$.

Hence, it is a rational number.

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

⇒ Division by zero is undefined; the denominator q cannot be 0.

Hence, it is not a rational number.

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

⇒ The denominator is zero, which is not allowed in the definition of a rational number.

Hence, it is not a rational number.

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

⇒ The numerator can be 0 as long as the denominator is a non-zero integer.

Hence, it is a rational number.

(ix) $\dfrac{-23}{-37}$

⇒ Both are integers and the denominator is not zero (this simplifies to $\dfrac{23}{37}$).

Hence, it is a rational number.

(x) $\dfrac{-1}{7}$

⇒ Both are integers and the denominator is not zero.

Hence, it is a rational number.

Write down the numerator and the denominator of each of the following rational numbers :

(i) $\dfrac{12}{17}$

(ii) $\dfrac{6}{-23}$

(iii) $\dfrac{-21}{5}$

(iv) 7

(v) -8

Answer

(i) We have:

$\dfrac{12}{17}$

Numerator = 12, Denominator = 17

(ii) We have:

$\dfrac{6}{-23}$

Numerator = 6, Denominator = -23

(iii) We have:

$\dfrac{-21}{5}$

Numerator = -21, Denominator = 5

(iv) We have:

7

It can be written as $\dfrac{7}{1}$

Numerator = 7, Denominator = 1

(v) We have:

-8

It can be written as $\dfrac{-8}{1}$

Numerator = -8, Denominator = 1

Which of the following are positive rational numbers?

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

(ii) $\dfrac{-13}{17}$

(iii) $\dfrac{-8}{-11}$

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

(v) $\dfrac{0}{-7}$

Answer

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

⇒ The numerator is negative and the denominator is positive (opposite signs), making it a negative rational number.

Hence, it is not a positive rational number.

(ii) $\dfrac{-13}{17}$

⇒ The signs are opposite, so it is a negative rational number.

Hence, it is not a positive rational number.

(iii) $\dfrac{-8}{-11}$

⇒ Both the numerator and denominator are negative, the signs cancel out ($\dfrac{-8}{-11} = \dfrac{8}{11}$), making the value positive.

Hence, it is a positive rational number.

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

⇒ The value is 0, and zero is neither positive nor negative.

Hence, it is not a positive rational number.

(v) $\dfrac{0}{-7}$

⇒ The value is 0, and zero is neither positive nor negative.

Hence, it is not a positive rational number.

Which of the following are negative rational numbers?

(i) $\dfrac{-16}{5}$

(ii) $\dfrac{-10}{-11}$

(iii) -21

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

(v) 17

Answer

(i) $\dfrac{-16}{5}$

⇒ The numerator is negative and the denominator is positive (opposite signs).

Hence, it is a negative rational number.

(ii) $\dfrac{-10}{-11}$

⇒ Both are negative, so the signs cancel out ($\dfrac{10}{11}$), making it positive.

Hence, it is not a negative rational number.

(iii) -21

⇒ It can be written as $\dfrac{-21}{1}$. The signs are opposite.

Hence, it is a negative rational number.

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

⇒ The value is 0, which is neutral (neither positive nor negative).

Hence, it is not a negative rational number.

(v) 17

⇒ It is a positive integer (can be written as $\dfrac{17}{1}$).

Hence, it is not a negative rational number.

Find four rational numbers equivalent to each of the following :

(i) $\dfrac{3}{10}$

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

(iii) $\dfrac{6}{-13}$

(iv) 9

(v) -1

Answer

(i) $\dfrac{3}{10}$

We have:

$\dfrac{3}{10} = \dfrac{3 \times 2}{10 \times 2} = \dfrac{3 \times 3}{10 \times 3} = \dfrac{3 \times 4}{10 \times 4} = \dfrac{3 \times 5}{10 \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow \dfrac{3}{10} = \dfrac{6}{20} = \dfrac{9}{30} = \dfrac{12}{40} = \dfrac{15}{50}$

∴ Four rational numbers equivalent to $\dfrac{3}{10}$ are:

$\dfrac{6}{20}, \dfrac{9}{30}, \dfrac{12}{40} , \dfrac{15}{50}$

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

We have:

$\dfrac{-5}{9} = \dfrac{(-5) \times 2}{9 \times 2} = \dfrac{(-5) \times 3}{9 \times 3} = \dfrac{(-5) \times 4}{9 \times 4} = \dfrac{(-5) \times 5}{9 \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow \dfrac{-5}{9} = \dfrac{-10}{18} = \dfrac{-15}{27} = \dfrac{-20}{36} = \dfrac{-25}{45}$

∴ Four rational numbers equivalent to $\dfrac{-5}{9}$ are:

$\dfrac{-10}{18}, \dfrac{-15}{27}, \dfrac{-20}{36} , \dfrac{-25}{45}$

(iii) $\dfrac{6}{-13}$

We have:

$\dfrac{6}{-13} = \dfrac{6 \times 2}{(-13) \times 2} = \dfrac{6 \times 3}{(-13) \times 3} = \dfrac{6 \times 4}{(-13) \times 4} = \dfrac{6 \times 5}{(-13) \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow \dfrac{6}{-13} = \dfrac{12}{-26} = \dfrac{18}{-39} = \dfrac{24}{-52} = \dfrac{30}{-65}$

∴ Four rational numbers equivalent to $\dfrac{6}{-13}$ are:

$\dfrac{12}{-26}, \dfrac{18}{-39}, \dfrac{24}{-52} , \dfrac{30}{-65}$

(iv) 9

Since $9 = \dfrac{9}{1}$, we have:

$\dfrac{9}{1} = \dfrac{9 \times 2}{1 \times 2} = \dfrac{9 \times 3}{1 \times 3} = \dfrac{9 \times 4}{1 \times 4} = \dfrac{9 \times 5}{1 \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow 9 = \dfrac{18}{2} = \dfrac{27}{3} = \dfrac{36}{4} = \dfrac{45}{5}$

∴ Four rational numbers equivalent to 9 are:

$\dfrac{18}{2}, \dfrac{27}{3}, \dfrac{36}{4} , \dfrac{45}{5}$

(v) -1

Since $-1 = \dfrac{-1}{1}$, we have

$\dfrac{-1}{1} = \dfrac{(-1) \times 2}{1 \times 2} = \dfrac{(-1) \times 3}{1 \times 3} = \dfrac{(-1) \times 4}{1 \times 4} = \dfrac{(-1) \times 5}{1 \times 5}$ [Multiply the numerator and denominator by 2, 3, 4, and 5]

$\Rightarrow -1 = \dfrac{-2}{2} = \dfrac{-3}{3} = \dfrac{-4}{4} = \dfrac{-5}{5}$

∴ Four rational numbers equivalent to -1 are:

$\dfrac{-2}{2}, \dfrac{-3}{3}, \dfrac{-4}{4}, \dfrac{-5}{5}$

Write each of the following rational numbers with positive denominator :

(i) $\dfrac{16}{-21}$

(ii) $\dfrac{1}{-5}$

(iii) $\dfrac{-7}{-12}$

(iv) $\dfrac{5}{-1}$

(v) $\dfrac{-6}{-1}$

Answer

(i) We have:

$\dfrac{16}{-21} = \dfrac{16 \times (-1)}{-21 \times (-1)} = \dfrac{-16}{21}$

∴ The rational number with a positive denominator is $\dfrac{-16}{21}$.

(ii) We have:

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

∴ The rational number with a positive denominator is $\dfrac{-1}{5}$.

(iii) We have:

$\dfrac{-7}{-12} = \dfrac{(-7) \times (-1)}{(-12) \times (-1)} = \dfrac{7}{12}$

∴ The rational number with a positive denominator is $\dfrac{7}{12}$.

(iv) We have:

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

∴ The rational number with a positive denominator is $\dfrac{-5}{1}$.

(v) We have:

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

∴ The rational number with a positive denominator is $\dfrac{6}{1}$.

Express $\dfrac{4}{9}$ as a rational number with numerator

(i) 24

(ii) -20

Answer

(i) Numerator 24

We know that 4 x 6 = 24. So, we multiply the numerator and the denominator of the given number by 6:

$\dfrac{4}{9} = \dfrac{4 \times 6}{9 \times 6} = \dfrac{24}{54}$

∴ The rational number with numerator $24$ is $\dfrac{24}{54}$.

(ii) Numerator -20

We know that 4 x (-5) = -20. So, we multiply the numerator and the denominator of the given number by -5:

$\dfrac{4}{9} = \dfrac{4 \times -5}{9 \times -5} = \dfrac{-20}{-45}$

∴ The rational number with numerator -20 is $\dfrac{-20}{-45}$.

Express $\dfrac{3}{8}$ as a rational number with denominator

(i) 48

(ii) -32

Answer

(i) Denominator 48

Clearly, 8 x 6 = 48.

So, we multiply the numerator as well as the denominator of the given rational number by 6.

$\dfrac{3}{8} = \dfrac{3 \times 6}{8 \times 6} = \dfrac{18}{48}$

∴ The rational number with denominator $48$ is $\dfrac{18}{48}$.

(ii) Denominator -32

Clearly, 8 x (-4) = -32.

So, we multiply the numerator as well as the denominator of the given rational number by -4.

$\dfrac{3}{8} = \dfrac{3 \times (-4)}{8 \times (-4)} = \dfrac{-12}{-32}$

∴ The rational number with denominator $-32$ is $\dfrac{-12}{-32}$.

Express $\dfrac{-6}{11}$ as a rational number with numerator

(i) -36

(ii) 42

Answer

(i) Numerator -36

Clearly, (-6) x 6 = -36.

So, we multiply the numerator as well as the denominator of the given rational number by 6.

$\dfrac{-6}{11} = \dfrac{(-6) \times 6}{11 \times 6} = \dfrac{-36}{66}$

∴ The rational number with numerator $-36$ is $\dfrac{-36}{66}$.

(ii) Numerator 42

Clearly, (-6) x (-7) = 42.

So, we multiply the numerator as well as the denominator of the given rational number by -7.

$\dfrac{-6}{11} = \dfrac{(-6) \times (-7)}{11 \times (-7)} = \dfrac{42}{-77}$

∴ The rational number with numerator $42$ is $\dfrac{42}{-77}$.

Express $\dfrac{2}{-7}$ as a rational number with denominator

(i) 42

(ii) -28

Answer

(i) Denominator 42

Clearly, (-7) x (-6) = 42.

So, we multiply the numerator as well as the denominator of the given rational number by -6.

$\dfrac{2}{-7} = \dfrac{2 \times (-6)}{-7 \times (-6)} = \dfrac{-12}{42}$

∴ The rational number with denominator 42 is $\dfrac{-12}{42}$.

(ii) -28

Clearly, (-7) x 4 = -28.

So, we multiply the numerator as well as the denominator of the given rational number by 4.

$\dfrac{2}{-7} = \dfrac{2 \times 4}{-7 \times 4} = \dfrac{8}{-28}$

∴ The rational number with denominator $-28$ is $\dfrac{8}{-28}$.

Express $\dfrac{-48}{36}$ as a rational number with numerator

(i) -4

(ii) 8

Answer

(i) -4

We know that (-48) ÷ 12 = -4.

So, we divide the numerator and the denominator of the given number by 12:

$\dfrac{-48}{36} = \dfrac{-48 \div 12}{36 \div 12} = \dfrac{-4}{3}$

∴ The rational number with numerator $-4$ is $\dfrac{-4}{3}$.

(ii) 8

We know that (-48) ÷ (-6) = 8.

So, we divide the numerator and the denominator of the given number by -6:

$\dfrac{-48}{36} = \dfrac{-48 \div (-6)}{36 \div (-6)} = \dfrac{8}{-6}$

∴ The rational number with numerator $8$ is $\dfrac{8}{-6}$.

Express $\dfrac{78}{-117}$ as a rational number with numerator

(i) -6

(ii) 2

Answer

(i) -6

We know that 78 ÷ (-13) = -6.

So, we divide the numerator and the denominator of the given number by -13:

$\dfrac{78}{-117} = \dfrac{78 \div (-13)}{-117 \div (-13)} = \dfrac{-6}{9}$

∴ The rational number with numerator $-6$ is $\dfrac{-6}{9}$.

(ii) 2

We know that 78 ÷ 39 = 2.

So, we divide the numerator and the denominator of the given number by 39:

$\dfrac{78}{-117} = \dfrac{78 \div 39}{-117 \div 39} = \dfrac{2}{-3}$

∴ The rational number with numerator $2$ is $\dfrac{2}{-3}$.

Write each of the following rational numbers in standard form :

(i) $\dfrac{56}{32}$

(ii) $\dfrac{16}{-40}$

(iii) $\dfrac{-36}{54}$

(iv) $\dfrac{-22}{-77}$

(v) $\dfrac{78}{-65}$

(vi) $\dfrac{-95}{114}$

(vii) $\dfrac{-69}{115}$

(viii) $\dfrac{155}{-217}$

Answer

(i) $\dfrac{56}{32}$

The HCF of 56 and 32 is 8.

Divide the numerator and denominator by 8:

$\dfrac{56}{32} = \dfrac{56 \div 8}{32 \div 8} = \dfrac{7}{4}$

∴ The standard form is $\dfrac{7}{4}$.

(ii) $\dfrac{16}{-40}$

First, multiply the numerator and denominator by -1 to make the denominator positive:

$\dfrac{16 \times (-1)}{-40 \times (-1)} = \dfrac{-16}{40}$

The HCF of 16 and 40 is 8.

Divide the numerator and denominator by 8:

$\dfrac{-16 \div 8}{40 \div 8} = \dfrac{-2}{5}$

∴ The standard form is $\dfrac{-2}{5}$.

(iii) $\dfrac{-36}{54}$

The HCF of 36 and 54 is 18.

Divide the numerator and denominator by 18:

$\dfrac{-36 \div 18}{54 \div 18} = \dfrac{-2}{3}$

∴ The standard form is $\dfrac{-2}{3}$.

(iv) $\dfrac{-22}{-77}$

First, multiply the numerator and denominator by -1 to make the denominator positive:

$\dfrac{-22 \times (-1)}{-77 \times (-1)} = \dfrac{22}{77}$

The HCF of 22 and 77 is 11.

$\dfrac{22 \div 11}{77 \div 11} = \dfrac{2}{7}$

∴ The standard form is $\dfrac{2}{7}$.

(v) $\dfrac{78}{-65}$

First, multiply the numerator and denominator by -1 to make the denominator positive:

$\dfrac{78 \times (-1)}{-65 \times (-1)}$ = $\dfrac{-78}{65}$

The HCF of 78 and 65 is 13.

$\dfrac{-78 \div 13}{65 \div 13} = \dfrac{-6}{5}$

∴ The standard form is $\dfrac{-6}{5}$.

(vi) $\dfrac{-95}{114}$

The HCF of 95 and 114 is 19.

$\dfrac{-95 \div 19}{114 \div 19} = \dfrac{-5}{6}$

∴ The standard form is $\dfrac{-5}{6}$.

(vii) $\dfrac{-69}{115}$

The HCF of 69 and 115 is 23.

$\dfrac{-69 \div 23}{115 \div 23} = \dfrac{-3}{5}$

∴ The standard form is $\dfrac{-3}{5}$.

(viii) $\dfrac{155}{-217}$

First, multiply the numerator and denominator by -1 to make the denominator positive:

$\dfrac{155 \times (-1)}{-217 \times (-1)}$ = $\dfrac{-155}{217}$

The HCF of 155 and 217 is 31.

$\dfrac{-155 \div 31}{217 \div 31} = \dfrac{-5}{7}$

∴ The standard form is $\dfrac{-5}{7}$.

Find the value of x such that :

(i) $\dfrac{-2}{3} = \dfrac{14}{x}$

(ii) $\dfrac{8}{-3} = \dfrac{x}{6}$

(iii) $\dfrac{5}{9} = \dfrac{x}{-27}$

(iv) $\dfrac{11}{6} = \dfrac{-55}{x}$

(v) $\dfrac{15}{x} = -3$

(vi) $\dfrac{-36}{x} = 2$

Answer

(i) $\dfrac{-2}{3} = \dfrac{14}{x}$

We have,

$\dfrac{-2}{3} = \dfrac{14}{x} \\[1em] \Rightarrow (-2) \times x = 3 \times 14 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{3 \times 14}{-2} \\[1em] \Rightarrow x = 3 \times (-7) = -21$

Hence, x = -21.

(ii) $\dfrac{8}{-3} = \dfrac{x}{6}$

We have,

$\dfrac{8}{-3} = \dfrac{x}{6} \\[1em] \Rightarrow (-3) \times x = 8 \times 6 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{8 \times 6}{-3} \\[1em] \Rightarrow x = 8 \times (-2) = -16$

Hence, x = -16.

(iii) $\dfrac{5}{9} = \dfrac{x}{-27}$

We have,

$\dfrac{5}{9} = \dfrac{x}{-27} \\[1em] \Rightarrow 9 \times x = 5 \times (-27) \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{5 \times (-27)}{9} \\[1em] \Rightarrow x = 5 \times (-3) = -15$

Hence, x = -15.

(iv) $\dfrac{11}{6} = \dfrac{-55}{x}$

We have,

$\dfrac{11}{6} = \dfrac{-55}{x} \\[1em] \Rightarrow 11 \times x = 6 \times (-55) \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{6 \times (-55)}{11} \\[1em] \Rightarrow x = 6 \times (-5) = -30$

Hence, x = -30.

(v) $\dfrac{15}{x} = -3$

Since $-3 = \dfrac{-3}{1}$, we have

$\dfrac{15}{x} = \dfrac{-3}{1} \\[1em] \Rightarrow (-3) \times x = 15 \times 1 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{15}{-3} = -5$

Hence, x = -5.

(vi) $\dfrac{-36}{x} = 2$

Since $2 = \dfrac{2}{1}$, we have

$\dfrac{-36}{x} = \dfrac{2}{1} \\[1em] \Rightarrow 2 \times x = (-36) \times 1 \quad \text{[cross multiplication]} \\[1em] \Rightarrow x = \dfrac{-36}{2} = -18 \\[1em]$

Hence, x = -18.

State whether the given statement is true or false :

(i) The quotient of two integers is always a rational number.

(ii) Every rational number is a fraction.

(iii) Zero is the smallest rational number.

(iv) Every fraction is a rational number.

Answer

(i) False
Reason — While a rational number is defined as the quotient of two integers $\dfrac{p}{q}$, the denominator q cannot be zero. If the divisor (the second integer) is zero, the quotient is undefined and is not a rational number.

(ii) False
Reason — Fractions typically refer to a part of a whole and are usually represented as a ratio of two natural numbers. Rational numbers include negative values (like $-\dfrac{2}{3}$) and integers (like -5), which are not traditionally considered fractions in their standard form.

(iii) False
Reason — Rational numbers include negative values. Therefore, any negative rational number (such as -1 or $-\dfrac{1}{2}$) is smaller than zero. Because the set of rational numbers extends infinitely in the negative direction, there is no "smallest" rational number.

(iv) True
Reason — A fraction is represented as $\dfrac{a}{b}$ where a and b are whole numbers and b ≠ 0. Since whole numbers are also integers, every fraction fits the definition of a rational number.

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

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.

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