Rational Numbers — Exercise 1.5

Represent the following rational number on the number line:

$\dfrac{11}{4}$

Answer

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

$\Rightarrow \dfrac{11}{4} \\[1em] \Rightarrow 2\dfrac{3}{4} \\[1em] \Rightarrow 2 + \dfrac{3}{4}$

So, $\dfrac{11}{4}$ lies between 2 and 3.

Divide the line segment between 2 and 3 into four equal parts. The third point of division from 2 represents $\dfrac{11}{4}$.

Hence, the point marked on the number line represents $\mathbf{\dfrac{11}{4}}$

Represent the following rational number on the number line:

$4\dfrac{3}{5}$

Answer

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

$\Rightarrow 4\dfrac{3}{5} = \dfrac{23}{5} \\[1em] \Rightarrow 4 + \dfrac{3}{5}$

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

Divide the line segment between 4 and 5 into five equal parts. The third point of division from 4 represents $4\dfrac{3}{5}$.

Hence, the point marked on the number line represents $\mathbf{4\dfrac{3}{5}}$

Represent the following rational number on the number line:

$\dfrac{-9}{7}$

Answer

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

$\Rightarrow \dfrac{-9}{7} \\[1em] \Rightarrow -1\dfrac{2}{7} \\[1em] \Rightarrow -1 + \left(\dfrac{-2}{7}\right)$

So, $\dfrac{-9}{7}$ lies between -1 and -2.

Divide the line segment between -1 and -2 into seven equal parts. The second point of division to the left of -1 represents $\dfrac{-9}{7}$.

Hence, the point marked on the number line represents $\mathbf{\dfrac{-9}{7}}$

Represent the following rational number on the number line:

$\dfrac{-2}{-5}$

Answer

First express the rational number with a positive denominator.

$\Rightarrow \dfrac{-2}{-5} \\[1em] \Rightarrow \dfrac{-2 \times (-1)}{-5 \times (-1)} \\[1em] \Rightarrow \dfrac{2}{5}$

As $\dfrac{2}{5}$ is a positive proper fraction, it lies between 0 and 1.

Divide the line segment between 0 and 1 into five equal parts. The second point of division from 0 represents $\dfrac{2}{5}$.

Hence, the point marked on the number line represents $\mathbf{\dfrac{2}{5}}$

Write the rational numbers for each point labelled with a letter:

Answer

The line segment between each pair of consecutive integers has been divided into seven equal parts, so each division represents $\dfrac{1}{7}$.

Counting the divisions from 0, the marked points are the 3rd, 7th, 8th, 12th and 13th points of division.

$\Rightarrow A = \dfrac{3}{7} \\[1em] \Rightarrow B = \dfrac{7}{7} = 1 \\[1em] \Rightarrow C = \dfrac{8}{7} \\[1em] \Rightarrow D = \dfrac{12}{7} \\[1em] \Rightarrow E = \dfrac{13}{7}$

Hence, $\mathbf{A = \dfrac{3}{7}, B = 1, C = \dfrac{8}{7}, D = \dfrac{12}{7} \text{ and } E = \dfrac{13}{7}}$

Write the rational numbers for each point labelled with a letter:

Answer

The line segment between each pair of consecutive integers has been divided into eight equal parts, so each division represents $\dfrac{1}{8}$.

Writing the marked values with denominator 8, we have

$\Rightarrow \dfrac{-5}{4} = \dfrac{-10}{8} \\[1em] \Rightarrow -1 = \dfrac{-8}{8} \\[1em] \Rightarrow \dfrac{-3}{4} = \dfrac{-6}{8}$

Counting the divisions to the left of 0, the marked points are the 12th, 11th, 7th, 4th and 3rd points of division.

$\Rightarrow T = \dfrac{-12}{8} = \dfrac{-3}{2} \\[1em] \Rightarrow S = \dfrac{-11}{8} \\[1em] \Rightarrow R = \dfrac{-7}{8} \\[1em] \Rightarrow Q = \dfrac{-4}{8} = \dfrac{-1}{2} \\[1em] \Rightarrow P = \dfrac{-3}{8}$

Hence, $\mathbf{T = \dfrac{-3}{2}, S = \dfrac{-11}{8}, R = \dfrac{-7}{8}, Q = \dfrac{-1}{2} \text{ and } P = \dfrac{-3}{8}}$

Which of the two rational numbers is greater in each of the following pairs?

(i) $\dfrac{-1}{2}, \dfrac{4}{-5}$

(ii) $\dfrac{-9}{13}, \dfrac{-11}{-17}$

(iii) $\dfrac{-5}{11}, \dfrac{13}{-17}$

Answer

(i) First write each rational number with a positive denominator, then write them with a common denominator and compare the numerators.

$\dfrac{4}{-5} = \dfrac{4 \times (-1)}{-5 \times (-1)} = \dfrac{-4}{5}$

L.C.M. of 2 and 5 = 10

$\Rightarrow \dfrac{-1}{2} = \dfrac{(-1) \times 5}{2 \times 5} = \dfrac{-5}{10} \\[1em] \Rightarrow \dfrac{-4}{5} = \dfrac{(-4) \times 2}{5 \times 2} = \dfrac{-8}{10}$

As -8 < -5, so $\dfrac{-8}{10} \lt \dfrac{-5}{10}$

Hence, $\mathbf{\dfrac{-1}{2} \text{ is greater.}}$

(ii) First write each rational number with a positive denominator.

$\dfrac{-11}{-17} = \dfrac{-11 \times (-1)}{-17 \times (-1)} = \dfrac{11}{17}$

Here $\dfrac{11}{17}$ is a positive rational number and $\dfrac{-9}{13}$ is a negative rational number.

Every positive rational number is greater than every negative rational number.

Hence, $\mathbf{\dfrac{-11}{-17}}$ is greater.

(iii) First write each rational number with a positive denominator, then write them with a common denominator and compare the numerators.

$\dfrac{13}{-17} = \dfrac{13 \times (-1)}{-17 \times (-1)} = \dfrac{-13}{17}$

L.C.M. of 11 and 17 = 187

$\Rightarrow \dfrac{-5}{11} = \dfrac{(-5) \times 17}{11 \times 17} = \dfrac{-85}{187} \\[1em] \Rightarrow \dfrac{-13}{17} = \dfrac{(-13) \times 11}{17 \times 11} = \dfrac{-143}{187}$

As -143 < -85, so $\dfrac{-143}{187}$ < $\dfrac{-85}{187}$

Hence, $\mathbf{\dfrac{-5}{11}}$ is greater.

Arrange the following rational numbers in ascending order:

$\dfrac{-2}{3}, \dfrac{1}{-4}, \dfrac{-5}{7}, \dfrac{2}{11}$

Answer

First write each rational number with a positive denominator.

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

So, the given rational numbers are $\dfrac{-2}{3}, \dfrac{-1}{4}, \dfrac{-5}{7}, \dfrac{2}{11}$

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

$[\text{ LCM of } 3, 4, 7 \text{ and } 11 = 924] \\[1em] \Rightarrow \dfrac{-2}{3} = \dfrac{(-2) \times 308}{3 \times 308} = \dfrac{-616}{924} \\[1em] \Rightarrow \dfrac{-1}{4} = \dfrac{(-1) \times 231}{4 \times 231} = \dfrac{-231}{924} \\[1em] \Rightarrow \dfrac{-5}{7} = \dfrac{(-5) \times 132}{7 \times 132} = \dfrac{-660}{924} \\[1em] \Rightarrow \dfrac{2}{11} = \dfrac{2 \times 84}{11 \times 84} = \dfrac{168}{924}$

As -660 < -616 < -231 < 168

so $\dfrac{-660}{924}$ < $\dfrac{-616}{924}$ < $\dfrac{-231}{924}$ < $\dfrac{168}{924}$

$\Rightarrow \dfrac{-5}{7}$ < $\dfrac{-2}{3}$ < $\dfrac{-1}{4}$ < $\dfrac{2}{11}$

Hence, the given rational numbers in ascending order are $\mathbf{\dfrac{-5}{7}, \dfrac{-2}{3}, \dfrac{1}{-4}, \dfrac{2}{11}}$

Arrange the following rational numbers in descending order:

$\dfrac{-3}{10}, \dfrac{-2}{3}, \dfrac{4}{-5}, 0, \dfrac{7}{8}$

Answer

First write each rational number with a positive denominator.

$\dfrac{4}{-5} = \dfrac{4 \times (-1)}{-5 \times (-1)} = \dfrac{-4}{5}$

So, the given rational numbers are $\dfrac{-3}{10}, \dfrac{-2}{3}, \dfrac{-4}{5}, 0, \dfrac{7}{8}$

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

$[\text{ LCM of } 10, 3, 5 \text{ and } 8 = 120] \\[1em] \Rightarrow \dfrac{-3}{10} = \dfrac{(-3) \times 12}{10 \times 12} = \dfrac{-36}{120} \\[1em] \Rightarrow \dfrac{-2}{3} = \dfrac{(-2) \times 40}{3 \times 40} = \dfrac{-80}{120} \\[1em] \Rightarrow \dfrac{-4}{5} = \dfrac{(-4) \times 24}{5 \times 24} = \dfrac{-96}{120} \\[1em] \Rightarrow 0 = \dfrac{0}{120} \\[1em] \Rightarrow \dfrac{7}{8} = \dfrac{7 \times 15}{8 \times 15} = \dfrac{105}{120}$

As 105 > 0 > -36 > -80 > -96

so $\dfrac{105}{120} \gt \dfrac{0}{120} \gt \dfrac{-36}{120} \gt \dfrac{-80}{120} \gt \dfrac{-96}{120}$

$\Rightarrow \dfrac{7}{8} \gt 0 \gt \dfrac{-3}{10} \gt \dfrac{-2}{3} \gt \dfrac{-4}{5}$

Hence, the given rational numbers in descending order are $\mathbf{\dfrac{7}{8}, 0, \dfrac{-3}{10}, \dfrac{-2}{3}, \dfrac{4}{-5}}$

Find three rational numbers between -2 and -1.

Answer

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

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

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

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

Hence, three rational numbers between -2 and -1 are $\mathbf{\dfrac{-7}{4}, \dfrac{-3}{2} \text{ and } \dfrac{-5}{4}}$

Write five rational numbers which are smaller than -4.

Answer

Every rational number lying to the left of -4 on the number line is smaller than -4.

As -9 < -8 < -7 < -6 < -5 < -4, each of these numbers is smaller than -4.

There are infinitely many such rational numbers; for example $\dfrac{-9}{2}, \dfrac{-13}{3}, \dfrac{-17}{4}$ are also smaller than -4.

Hence, five rational numbers smaller than -4 are -5, -6, -7, -8 and -9.

Find twenty rational numbers between $\dfrac{-3}{7} \text{ and } \dfrac{2}{3}$

Answer

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

$[\text{ LCM of } 7 \text{ and } 3 = 21] \\[1em] \Rightarrow \dfrac{-3}{7} = \dfrac{-3 \times 3}{7 \times 3} = \dfrac{-9}{21} \\[1em] \Rightarrow \dfrac{2}{3} = \dfrac{2 \times 7}{3 \times 7} = \dfrac{14}{21}$

We note that -8, -7, -6, …, 12, 13 are the integers between -9 and 14

i.e. -9 < -8 < -7 < -6 < … < 12 < 13 < 14

Hence, twenty rational numbers between $\mathbf{\dfrac{-3}{7} \text{ and } \dfrac{2}{3}}$ are

$\mathbf{\dfrac{-8}{21}, \dfrac{-7}{21}, \dfrac{-6}{21}, \dfrac{-5}{21}, \dfrac{-4}{21}, \dfrac{-3}{21}, \dfrac{-2}{21}, \dfrac{-1}{21}, \dfrac{0}{21}, \dfrac{1}{21}, \dfrac{2}{21}, \dfrac{3}{21}, \dfrac{4}{21}, \dfrac{5}{21}, \dfrac{6}{21}, \dfrac{7}{21}, \dfrac{8}{21}, \dfrac{9}{21}, \dfrac{10}{21}, \dfrac{11}{21}}$.

Find six rational numbers between $\dfrac{-1}{2} \text{ and } \dfrac{5}{4}$

Answer

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

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

We note that -1, 0, 1, 2, 3, 4 are the integers between -2 and 5

i.e. -2 < -1 < 0 < 1 < 2 < 3 < 4 < 5

Hence, six rational numbers between $\mathbf{\dfrac{-1}{2} \text{ and } \dfrac{5}{4}}$ are $\mathbf{\dfrac{-1}{4}, 0, \dfrac{1}{4}, \dfrac{1}{2}, \dfrac{3}{4} \text{ and } 1}$

Identify the rational number which is different from the other three. Explain your reasoning

$\dfrac{-5}{11}, \dfrac{-1}{2}, \dfrac{-4}{9}, \dfrac{-7}{3}$

Answer

Compare each rational number with -1 to see where it lies on the number line.

$\dfrac{-5}{11}, \dfrac{-1}{2} \text{ and } \dfrac{-4}{9}$ are all proper fractions with a negative sign, so each of them lies between -1 and 0 on the number line.

But $\dfrac{-7}{3} = -2\dfrac{1}{3}$ is an improper fraction with a negative sign, and it lies between -3 and -2 on the number line.

Hence, $\mathbf{\dfrac{-7}{3}}$ is different from the other three.