Rational Numbers — Exercise 1(A)

Express $\dfrac{-64}{112}$ as a rational number with denominator 7.

Answer

To get the denominator 7, divide both the numerator and the denominator of the given rational number by the same number.

Note that 112 ÷ 7 = 16;

$\dfrac{-64}{112} = \dfrac{-64 \div 16}{112 \div 16} = \dfrac{-4}{7}$

Hence, $\mathbf{\dfrac{-64}{112} = \dfrac{-4}{7}}.$

Express $\dfrac{-48}{60}$ as a rational number with denominator 25.

Answer

First reduce the given rational number to its lowest terms, then multiply both the numerator and the denominator by the same number to get the denominator 25.

HCF of 48 and 60 = 12.

$\Rightarrow \dfrac{-48}{60} = \dfrac{-48 \div 12}{60 \div 12} = \dfrac{-4}{5} \\[1em] [5 \times 5 = 25] \\[1em] \Rightarrow \dfrac{-4}{5} = \dfrac{-4 \times 5}{5 \times 5} \\[1em] \Rightarrow \dfrac{-20}{25}$

Hence, $\mathbf{\dfrac{-48}{60} = \dfrac{-20}{25}}$.

Express the following rational number in standard form:

$\dfrac{-12}{30}$

Answer

A rational number $\dfrac{p}{q}$ is in standard form if $q \gt 0$ and $p, q$ have no common divisor other than 1.

Here the denominator 30 is already positive.

The greatest common divisor of 12 and 30 is 6.

Now,

$\dfrac{-12}{30} =\dfrac{-12 \div 6}{30 \div 6} = \dfrac{-2}{5}$

Hence, the standard form of $\mathbf{\dfrac{-12}{30} \text{ is } \dfrac{-2}{5}}$.

Express the following rational number in standard form:

$\dfrac{-14}{49}$

Answer

A rational number $\dfrac{p}{q}$ is in standard form if $q \gt 0$ and $p, q$ have no common divisor other than 1.

Here the denominator 49 is already positive.

The greatest common divisor of 14 and 49 is 7.

Now,

$\dfrac{-14}{49} = \dfrac{-14 \div 7}{49 \div 7} = \dfrac{-2}{7}$

Hence, the standard form of $\mathbf{\dfrac{-14}{49} \text{ is } \dfrac{-2}{7}}$.

Express the following rational number in standard form:

$\dfrac{24}{-64}$

Answer

A rational number $\dfrac{p}{q}$ is in standard form if $q \gt 0$ and $p, q$ have no common divisor other than 1.

Here the denominator -64 is negative, so first express the rational number with a positive denominator.

$\dfrac{24}{-64} = \dfrac{24 \times (-1)}{-64 \times (-1)} = \dfrac{-24}{64}$

The greatest common divisor of 24 and 64 is 8.

Now,

$\dfrac{-24}{64} = \dfrac{-24 \div 8}{64 \div 8} = \dfrac{-3}{8}$

Hence, the standard form of $\mathbf{\dfrac{24}{-64} \text{ is } \dfrac{-3}{8}}$.

Express the following rational number in standard form:

$\dfrac{-36}{-63}$

Answer

A rational number $\dfrac{p}{q}$ is in standard form if $q \gt 0$ and $p, q$ have no common divisor other than 1.

Here the denominator -63 is negative, so first express the rational number with a positive denominator.

$\dfrac{-36}{-63} = \dfrac{-36 \times (-1)}{-63 \times (-1)} = \dfrac{36}{63}$

The greatest common divisor of 36 and 63 is 9.

Now,

$\dfrac{36}{63} = \dfrac{36 \div 9}{63 \div 9} = \dfrac{4}{7}$

Hence, the standard form of $\mathbf{\dfrac{-36}{-63} \text{ is } \dfrac{4}{7}}$.

Compare the following pair of rational numbers:

$\dfrac{15}{32} \text{ and } \dfrac{17}{24}$

Answer

Write both the rational numbers with the LCM of their denominators as the common denominator and compare the numerators.

L.C.M. of 32 and 24 = 96

$\Rightarrow \dfrac{15}{32} = \dfrac{15 \times 3}{32 \times 3} = \dfrac{45}{96} \\[1em] \Rightarrow \dfrac{17}{24} = \dfrac{17 \times 4}{24 \times 4} = \dfrac{68}{96}$

As 45 < 68, so $\dfrac{45}{96} \lt \dfrac{68}{96}$.

Hence, $\mathbf{\dfrac{15}{32} \lt \dfrac{17}{24}}$.

Compare the following pair of rational numbers:

$\dfrac{10}{11} \text{ and } \dfrac{17}{18}$

Answer

Write both the rational numbers with the LCM of their denominators as the common denominator and compare the numerators.

L.C.M. of 11 and 18 = 198

$\Rightarrow \dfrac{10}{11} = \dfrac{10 \times 18}{11 \times 18} = \dfrac{180}{198} \\[1em] \Rightarrow \dfrac{17}{18} = \dfrac{17 \times 11}{18 \times 11} = \dfrac{187}{198}$

As 180 < 187, so $\dfrac{180}{198} \lt \dfrac{187}{198}$

Hence, $\mathbf{\dfrac{10}{11} \lt \dfrac{17}{18}}$.

Compare the following pair of rational numbers:

$\dfrac{-5}{12} \text{ and } \dfrac{-3}{4}$

Answer

Write both the rational numbers with the LCM of their denominators as the common denominator and compare the numerators.

L.C.M. of 12 and 4 = 12

$\Rightarrow \dfrac{-5}{12} = \dfrac{-5}{12} \\[1em] \Rightarrow \dfrac{-3}{4} = \dfrac{(-3) \times 3}{4 \times 3} = \dfrac{-9}{12}$

As -9 < -5, so $\dfrac{-9}{12} \lt \dfrac{-5}{12}$.

Hence, $\mathbf{\dfrac{-5}{12} \gt \dfrac{-3}{4}}$.

Compare the following pair of rational numbers:

$\dfrac{-7}{24} \text{ and } \dfrac{9}{-20}$

Answer

Write both the rational numbers with the LCM of their denominators as the common denominator and compare the numerators.

First express $\dfrac{9}{-20}$ with a positive denominator.

$\dfrac{9}{-20} = \dfrac{9 \times (-1)}{-20 \times (-1)} = \dfrac{-9}{20}$

L.C.M. of 24 and 20 = 120

$\Rightarrow \dfrac{-7}{24} = \dfrac{(-7) \times 5}{24 \times 5} = \dfrac{-35}{120} \\[1em] \Rightarrow \dfrac{-9}{20} = \dfrac{(-9) \times 6}{20 \times 6} = \dfrac{-54}{120}$

As -54 < -35, so $\dfrac{-54}{120} \lt \dfrac{-35}{120}$.

Hence, $\mathbf{\dfrac{-7}{24} \gt \dfrac{9}{-20}}$.

Arrange in ascending order:

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

Answer

All the denominators are already positive, so write each rational number with the LCM of the denominators as the common denominator and arrange the numerators in an increasing order.

LCM of 6, 9, 12 and 18 = 36

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

As 26 < 28 < 30 < 33

So, $\dfrac{26}{36} \lt \dfrac{28}{36} \lt \dfrac{30}{36} \lt \dfrac{33}{36}$

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

Hence, the given rational numbers in ascending order are $\mathbf{\dfrac{13}{18}, \dfrac{7}{9}, \dfrac{5}{6} \text{ and } \dfrac{11}{12}}$.

Arrange in ascending order:

$\dfrac{5}{-7}, \dfrac{-9}{14}, \dfrac{-5}{6} \text{ and } \dfrac{7}{-12}$

Answer

First write each rational number with a positive denominator, then write each of them with the LCM of the denominators as the common denominator and arrange the numerators in increasing order.

$\Rightarrow \dfrac{5}{-7} = \dfrac{5 \times (-1)}{-7 \times (-1)} = \dfrac{-5}{7} \\[1em] \Rightarrow \dfrac{7}{-12} = \dfrac{7 \times (-1)}{-12 \times (-1)} = \dfrac{-7}{12}$

So, the given rational numbers are $\dfrac{-5}{7}, \dfrac{-9}{14}, \dfrac{-5}{6} \text{ and } \dfrac{-7}{12}$.

Now write each rational number with the LCM of the denominators as the common denominator.

LCM of 7, 14, 6 and 12 = 84

$\Rightarrow \dfrac{-5}{7} = \dfrac{(-5) \times 12}{7 \times 12} = \dfrac{-60}{84} \\[1em] \Rightarrow \dfrac{-9}{14} = \dfrac{(-9) \times 6}{14 \times 6} = \dfrac{-54}{84} \\[1em] \Rightarrow \dfrac{-5}{6} = \dfrac{(-5) \times 14}{6 \times 14} = \dfrac{-70}{84} \\[1em] \Rightarrow \dfrac{-7}{12} = \dfrac{(-7) \times 7}{12 \times 7} = \dfrac{-49}{84}$

As -70 < -60 < -54 < -49,

So $\dfrac{-70}{84} \lt \dfrac{-60}{84} \lt \dfrac{-54}{84} \lt \dfrac{-49}{84}$

$\Rightarrow \dfrac{-5}{6} \lt \dfrac{-5}{7} \lt \dfrac{-9}{14} \lt \dfrac{-7}{12}$

Hence, the given rational numbers in ascending order are $\mathbf{\dfrac{-5}{6}, \dfrac{5}{-7}, \dfrac{-9}{14} \text{ and } \dfrac{7}{-12}}$.

Arrange in ascending order:

$-2, \dfrac{1}{3}, \dfrac{-13}{6} \text{ and } \dfrac{8}{-3}$

Answer

First write each rational number with a positive denominator, then write each of them with the LCM of the denominators as the common denominator and arrange the numerators in increasing order.

$\Rightarrow -2 = \dfrac{-2}{1} \\[1em] \Rightarrow \dfrac{8}{-3} = \dfrac{8 \times (-1)}{-3 \times (-1)} = \dfrac{-8}{3}$

So, the given rational numbers are $\dfrac{-2}{1}, \dfrac{1}{3}, \dfrac{-13}{6} \text{ and } \dfrac{-8}{3}$.

Now write each rational number with the LCM of the denominators as the common denominator.

LCM of 1, 3, 6 and 3 = 6

$\Rightarrow \dfrac{-2}{1} = \dfrac{(-2) \times 6}{1 \times 6} = \dfrac{-12}{6} \\[1em] \Rightarrow \dfrac{1}{3} = \dfrac{1 \times 2}{3 \times 2} = \dfrac{2}{6} \\[1em] \Rightarrow \dfrac{-13}{6} = \dfrac{-13}{6} \\[1em] \Rightarrow \dfrac{-8}{3} = \dfrac{(-8) \times 2}{3 \times 2} = \dfrac{-16}{6}$

As -16 < -13 < -12 < 2,

so $\dfrac{-16}{6} \lt \dfrac{-13}{6} \lt \dfrac{-12}{6} \lt \dfrac{2}{6}$

$\Rightarrow \dfrac{-8}{3} \lt \dfrac{-13}{6} \lt -2 \lt \dfrac{1}{3}$

Hence, the given rational numbers in ascending order are $\mathbf{\dfrac{8}{-3}, \dfrac{-13}{6}, -2 \text{ and } \dfrac{1}{3}}$.

Arrange in ascending order:

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

Answer

First write each rational number with a positive denominator, then write each of them with the LCM of the denominators as the common denominator and arrange the numerators in increasing order.

$\dfrac{13}{-28} = \dfrac{13 \times (-1)}{-28 \times (-1)} = \dfrac{-13}{28}$

So, the given rational numbers are $\dfrac{-13}{28}, \dfrac{-23}{42}, \dfrac{-4}{7} \text{ and } \dfrac{-9}{14}$.

Now write each rational number with the LCM of the denominators as the common denominator.

LCM of 28, 42, 7 and 14 = 84

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

As -54 < -48 < -46 < -39

so $\dfrac{-54}{84} \lt \dfrac{-48}{84} \lt \dfrac{-46}{84} \lt \dfrac{-39}{84}$

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

Hence, the given rational numbers in ascending order are $\mathbf{\dfrac{-9}{14}, \dfrac{-4}{7}, \dfrac{-23}{42} \text{ and } \dfrac{13}{-28}}$.

Represent the following number on the number line:

$\dfrac{5}{6}$

Answer

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

As $\dfrac{5}{6}$ is a positive proper fraction, it lies between 0 and 1.

Divide the unit length between 0 and 1 into 6 equal segments by making 5 equally spaced marks. Starting from 0, count forward 5 segments to the right. Label this specific position as point A, representing $\dfrac{5}{6}$ on the number line, as shown below :

Hence, the point A on the number line represents $\mathbf{\dfrac{5}{6}}$.

Represent the following number on the number line:

$\dfrac{14}{3}$

Answer

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

Convert the improper fraction into a mixed fraction to find the two integers between which it lies.

$\dfrac{14}{3} = 4\dfrac{2}{3}$

So, $\dfrac{14}{3}$ lies between 4 and 5.

Divide the unit length between 4 and 5 into 3 equal segments by making 2 equally spaced marks. Starting from the integer 4, count forward (to the right) 2 segments. Label this specific position as point B, representing $4\dfrac{2}{3}$ i.e. $\dfrac{14}{3}$, as shown below :

Hence, the point B on the number line represents $\mathbf{\dfrac{14}{3}}$.

Represent the following number on the number line:

$\dfrac{-3}{7}$

Answer

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

As $\dfrac{-3}{7}$ is a negative proper fraction, it lies between -1 and 0.

Divide the unit length between 0 and -1 into 7 equal segments by making 6 equally spaced marks. Starting from 0, count backward (to the left) 3 segments. Label this specific position as point C, representing $\dfrac{-3}{7}$, as shown below :

Hence, the point C on the number line represents $\mathbf{\dfrac{-3}{7}}$.

Represent the following number on the number line:

$\dfrac{-17}{5}$

Answer

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

Convert the improper fraction into a mixed fraction to find the two integers between which it lies.

$\dfrac{-17}{5} = -3\dfrac{2}{5}$

So, $\dfrac{-17}{5}$ lies between -3 and -4.

Divide the unit length between -3 and -4 into 5 equal segments by making 4 equally spaced marks. Starting from the integer -3, count backward (to the left) 2 segments. Label this specific position as point D, representing $-3\dfrac{2}{5}$ i.e. $\dfrac{-17}{5}$, as shown below :

Hence, the point D on the number line represents $\mathbf{\dfrac{-17}{5}}$.

Represent the following number on the number line:

$-2\dfrac{2}{7}$

Answer

Draw a number line and mark a point O on it, representing the integer 0. Each point of division represents an integer and each length between two successive integers is a unit length.

The given number is already a mixed fraction, so we can directly find the two integers between which it lies.

$-2\dfrac{2}{7} = \dfrac{-16}{7}$

So, $-2\dfrac{2}{7}$ lies between -2 and -3.

Divide the unit length between -2 and -3 into 7 equal segments by making 6 equally spaced marks. Starting from the integer -2, count backward (to the left) 2 segments. Label this specific position as point E, representing $-2\dfrac{2}{7}$ i.e. $\dfrac{-16}{7}$, as shown below :

Hence, the point E on the number line represents $\mathbf{-2\dfrac{2}{7}}$.

State whether each of the following statements is true or false:

(i) Every whole number is a rational number.

(ii) Every integer is a rational number.

(iii) $\dfrac{2}{3} = \dfrac{2+4}{3+4}$

(iv) $\dfrac{-5}{6} \lt \dfrac{-6}{5}$

(v) $\dfrac{-3}{-4}$ is a negative rational number.

(vi) 0 is a whole number but it is not a rational number.

Answer

(i) Every whole number $p$ can be written as $\dfrac{p}{1}$, which is of the form $\dfrac{p}{q}$ where $p, q$ are integers and $q ≠ 0$. For example, $0 = \dfrac{0}{1} \text{ and } 7 = \dfrac{7}{1}$

Hence, the given statement is true.

(ii) Every integer $p$ can be written as $\dfrac{p}{1}$, which is of the form $\dfrac{p}{q}$ where $p, q$ are integers and $q ≠ 0$. For example, $-5 = \dfrac{-5}{1} \text{ and } 5 = \dfrac{5}{1}$.

Hence, the given statement is true.

(iii) An equivalent rational number is obtained by multiplying (or dividing) both the numerator and the denominator by the same non-zero number, not by adding the same number to both.

$\Rightarrow \dfrac{2 + 4}{3 + 4} = \dfrac{6}{7} \\[1em] \Rightarrow \dfrac{2}{3} ≠ \dfrac{6}{7} \quad [\text{as } 2 \times 7 ≠ 3 \times 6]$

Hence, the given statement is false.

(iv) Write both the rational numbers with the LCM of their denominators as the common denominator.

L.C.M. of 6 and 5 = 30

$\Rightarrow \dfrac{-5}{6} = \dfrac{(-5) \times 5}{6 \times 5} = \dfrac{-25}{30} \\[1em] \Rightarrow \dfrac{-6}{5} = \dfrac{(-6) \times 6}{5 \times 6} = \dfrac{-36}{30}$

As -36 < -25, so $\dfrac{-6}{5} \lt \dfrac{-5}{6}$.

Hence, the given statement is false.

(v) A rational number is positive if its numerator and denominator are either both positive or both negative.

$\dfrac{-3}{-4} = \dfrac{-3 \times (-1)}{-4 \times (-1)} = \dfrac{3}{4}$, which is a positive rational number.

Hence, the given statement is false.

(vi) 0 is a whole number and it can also be written as $\dfrac{0}{1}$, which is of the form $\dfrac{p}{q}$ where $q ≠ 0$, so 0 is a rational number as well.

Hence, the given statement is false.

Represent $\dfrac{13}{5} \text{ and } \dfrac{-13}{5}$ on the number line.

Answer

Draw a number line and mark a point O on it, representing the integer 0. Mark the integers to the right (1, 2, 3) and to the left (-1, -2, -3).

Convert each of the given improper fractions into a mixed fraction to find the two integers between which it lies.

$\Rightarrow \dfrac{13}{5} = 2\dfrac{3}{5} \\[1em] \Rightarrow \dfrac{-13}{5} = -2\dfrac{3}{5}$

So, $\dfrac{13}{5}$ lies between 2 and 3, and $\dfrac{-13}{5}$ lies between -2 and -3.

Divide the unit length between 2 and 3, as well as the unit length between -2 and -3, into 5 equal segments by making 4 equally spaced marks in each interval.

For $\dfrac{13}{5}$ : Starting from the integer 2, count forward (to the right) 3 segments. Label this specific position as point A, representing $2\dfrac{3}{5}$ i.e. $\dfrac{13}{5}$.

For $\dfrac{-13}{5}$ : Starting from the integer -2, count backward (to the left) 3 segments. Label this specific position as point B, representing $-2\dfrac{3}{5}$ i.e. $\dfrac{-13}{5}$, as shown below :

Hence, the points A and B on the number line represent $\mathbf{\dfrac{13}{5} \text{ and } \dfrac{-13}{5}}$.