Rational & Irrational Numbers — Exercise 1(B)

Write the additive inverse of :

(i) 5

(ii) -7

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

(iv) $\dfrac{-3}{17}$

(v) 0

(vi) $11\dfrac{5}{17}$

(vii) $-5\dfrac{3}{8}$

(viii) -37

(ix) 1

Answer

The additive inverse of a number is the number which, when added to the original number, results in zero.

(i) Let x be the additive inverse of 5, then :

⇒ 5 + x = 0

⇒ x = -5.

Hence, additive inverse of 5 = -5.

(ii) Let x be the additive inverse of -7, then :

⇒ -7 + x = 0

⇒ x = 7.

Hence, additive inverse of -7 = 7.

(iii) Let x be the additive inverse of $\dfrac{5}{9}$, then :

⇒ $\dfrac{5}{9}$ + x = 0

⇒ x = $-\dfrac{5}{9}$.

Hence,additive inverse of $\dfrac{5}{9} = -\dfrac{5}{9}$.

(iv) Let x be the additive inverse of $-\dfrac{3}{17}$, then :

⇒ $-\dfrac{3}{17}$ + x = 0

⇒ x = $\dfrac{3}{17}$.

Hence,additive inverse of $-\dfrac{3}{17} = \dfrac{3}{17}$.

(v) Let x be the additive inverse of 0, then :

⇒ 0 + x = 0

⇒ x = 0.

Hence, additive inverse of 0 is 0.

(vi) Let x be the additive inverse of $11\dfrac{5}{17}$, then :

⇒ $11\dfrac{5}{17}$ + x = 0

⇒ $\dfrac{192}{17}$ + x = 0

⇒ x = $-\dfrac{192}{17}$.

Hence, additive inverse of $11\dfrac{5}{17} = -\dfrac{192}{17}$.

(vii) Let x be the additive inverse of $-5\dfrac{3}{8}$, then :

⇒ $-5\dfrac{3}{8}$ + x = 0

⇒ $-\dfrac{43}{8}$ + x = 0

⇒ x = $\dfrac{43}{8}$.

Hence, additive inverse of $-5\dfrac{3}{8} = \dfrac{43}{8}$.

(viii) Let x be the additive inverse of -37, then :

⇒ -37 + x = 0

⇒ x = 37.

Hence, additive inverse of -37 = 37.

(ix) Let x be the additive inverse of 1, then :

⇒ 1 + x = 0

⇒ x = -1.

Hence, additive inverse of 1 = -1.

Write the multiplicative inverse of :

(i) 9

(ii) -1

(iii) $\dfrac{11}{16}$

(iv) $5\dfrac{1}{4}$

(v) $\dfrac{-2}{3}$

(vi) $17\dfrac{3}{20}$

(vii) $–18\dfrac{1}{2}$

(viii) –5

(ix) $\dfrac{-20}{41}$

Answer

The multiplicative inverse of a number is defined as a number that when multiplied by the original number gives the product as 1.

(i) Let the multiplicative inverse of 9, be x.

⇒ 9 × x = 1

⇒ x = $\dfrac{1}{9}$.

Hence, multiplicative inverse of 9 = $\dfrac{1}{9}$.

(ii) Let the multiplicative inverse of -1, be x.

⇒ -1 × x = 1

⇒ x = $-\dfrac{1}{1}$ = -1.

Hence, multiplicative inverse of -1 = -1.

(iii) Let the multiplicative inverse of $\dfrac{11}{16}$, be x.

⇒ $\dfrac{11}{16}$ × x = 1

⇒ x = $\dfrac{16}{11}$.

Hence, multiplicative inverse of $\dfrac{11}{16} = \dfrac{16}{11}$.

(iv) Let the multiplicative inverse of $5\dfrac{1}{4}$, be x.

⇒ $5\dfrac{1}{4}$ × x = 1

⇒ $\dfrac{21}{4}$ × x = 1

⇒ x = $\dfrac{4}{21}$.

Hence, multiplicative inverse of $5\dfrac{1}{4} = \dfrac{4}{21}$.

(v) Let the multiplicative inverse of $-\dfrac{2}{3}$, be x.

⇒ $-\dfrac{2}{3}$ × x = 1

⇒ x = $-\dfrac{3}{2}$.

Hence, multiplicative inverse of $-\dfrac{2}{3} = -\dfrac{3}{2}$.

(vi) Let the multiplicative inverse of $17\dfrac{3}{20}$, be x.

⇒ $17\dfrac{3}{20}$ × x = 1

⇒ $\dfrac{343}{20}$ × x = 1

⇒ x = $\dfrac{20}{343}$.

Hence, multiplicative inverse of $17\dfrac{3}{20} = \dfrac{20}{343}$.

(vii) Let the multiplicative inverse of $–18\dfrac{1}{2}$, be x.

⇒ $–18\dfrac{1}{2}$ × x = 1

⇒ $-\dfrac{37}{2}$ × x = 1

⇒ x = $–\dfrac{2}{37}$.

Hence, multiplicative inverse of $-18\dfrac{1}{2} = -\dfrac{2}{37}$.

(viii) Let the multiplicative inverse of –5, be x.

⇒ –5 × x = 1

⇒ x = $-\dfrac{1}{5}$.

Hence, multiplicative inverse of $-5 = -\dfrac{1}{5}$.

(ix) Let the multiplicative inverse of $\dfrac{-20}{41}$, be x.

⇒ $\dfrac{-20}{41}$ × x = 1

⇒ x = $-\dfrac{41}{20}$.

Hence, multiplicative inverse of $-\dfrac{20}{41} = -\dfrac{41}{20}$.

Represent each of the following on the number line :

(i) $\dfrac{3}{7}$

(ii) $\dfrac{16}{5}$

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

(iv) $-\dfrac{18}{11}$

(v) $-3\dfrac{1}{6}$

Answer

(i) On dividing,

$\dfrac{3}{7}$ = 0.428

(ii) On dividing,

$\dfrac{16}{5}$ = 3.2

(iii) On dividing,

$-\dfrac{4}{9}$ = -0.4444..

(iv) On dividing,

$-\dfrac{18}{11}$ = -1.6363..

(v) On dividing,

$-3\dfrac{1}{6} = -\dfrac{19}{6}$ = -3.166..

Find a rational number between $\dfrac{3}{5}$ and $\dfrac{7}{9}$.

Answer

Let x be a rational number between $\dfrac{3}{5}$ and $\dfrac{7}{9}$.

$\Rightarrow x = \dfrac{1}{2}\Big(\dfrac{3}{5} + \dfrac{7}{9}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{27 + 35}{45} \Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{62}{45} \Big) \\[1em] \Rightarrow x = \dfrac{31}{45} \\[1em]$

Hence, a rational number between $\dfrac{3}{5} \text{ and } \dfrac{7}{9} \text{ is } \dfrac{31}{45}$.

Find two rational numbers between :

(i) 2 and 3

(ii) $\dfrac{1}{3} \text{ and } \dfrac{2}{5}$

(iii) $\dfrac{3}{4} \text{ and } 1\dfrac{1}{5}$

(iv) –2 and 1

Answer

(i) Let the first rational number between 2 and 3 be x.

$\Rightarrow x = \dfrac{1}{2}\Big(2 + 3\Big) \\[1em] \Rightarrow x = \dfrac{1}{2} \times 5 \\[1em] \Rightarrow x = \dfrac{5}{2} \\[1em]$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2} \Big(\dfrac{5}{2} + 3\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{5+6}{2}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{11}{2}\Big) \\[1em] \Rightarrow y = \dfrac{11}{4} \\[1em]$

Hence, two rational numbers between 2 and 3 are $\dfrac{5}{2} \text{ and } \dfrac{11}{4}$.

(ii) Let the first rational number between $\dfrac{1}{3}$ and $\dfrac{2}{5}$ be x.

$\Rightarrow x = \dfrac{1}{2} \Big(\dfrac{1}{3} + \dfrac{2}{5}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2} \Big(\dfrac{5 + 6}{15}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2} \Big(\dfrac{11}{15} \Big) \\[1em] \Rightarrow x = \dfrac{11}{30}$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2} \Big(\dfrac{11}{30} + \dfrac{2}{5}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{11 + 12}{30}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{23}{30}\Big) \\[1em] \Rightarrow y = \dfrac{23}{60} \\[1em]$

Hence, two rational numbers between $\dfrac{1}{3}$ and $\dfrac{2}{5}$ are $\dfrac{11}{30} \text{ and } \dfrac{23}{60}$ .

(iii) Let the first rational number between $\dfrac{3}{4}$ and $1\dfrac{1}{5}$ be x.

$\Rightarrow x = \dfrac{1}{2}\Big(\dfrac{3}{4} + 1\dfrac{1}{5}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{3}{4} + \dfrac{6}{5}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{15 + 24}{20}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{39}{20} \Big) \\[1em] \Rightarrow x = \dfrac{39}{40}$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2}\Big(\dfrac{39}{40} + \dfrac{6}{5}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{39 + 48}{40}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{87}{40}\Big) \\[1em] \Rightarrow y = \dfrac{87}{80} \\[1em]$

Hence, two rational numbers between $\dfrac{3}{4}$ and $1\dfrac{1}{5}$ are $\dfrac{39}{40} \text{ and } \dfrac{87}{80}$ .

(iv) Let the first rational number between -2 and 1 be x.

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

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2}\Big[-\dfrac{1}{2} + (-2)\Big] \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-1-4}{2}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \times -\dfrac{5}{2} \\[1em] \Rightarrow y = -\dfrac{5}{4}.$

Hence, two rational numbers between -2 and 1 are $-\dfrac{1}{2} \text{ and } -\dfrac{5}{4}$.

Find three rational numbers between :

(i) 4 and 5

(ii) $\dfrac{1}{2} \text{ and }\dfrac{3}{5}$

(iii) –1 and 1

(iv) $2\dfrac{1}{3} \text{ and }3\dfrac{2}{3}$

(v) $-\dfrac{1}{2} \text{ and }\dfrac{1}{3}$

(vi) $-\dfrac{1}{3} \text{ and }\dfrac{1}{4}$

Answer

(i) Let the first rational number between 4 and 5 be x.

$\Rightarrow x = \dfrac{1}{2}(4 + 5) \\[1em] \Rightarrow x = \dfrac{1}{2} \times 9 \\[1em] \Rightarrow x = \dfrac{9}{2}.$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2}\Big(\dfrac{9}{2} + 5\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{9 + 10}{2}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{19}{2}\Big) \\[1em] \Rightarrow y = \dfrac{19}{4}$

Let the third rational number be z.

$\Rightarrow z = \dfrac{1}{2}\Big(\dfrac{9}{2} + 4\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{9 + 8}{2}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{17}{2}\Big) \\[1em] \Rightarrow z = \dfrac{17}{4}$

Hence, three rational numbers between 4 and 5 are $\dfrac{17}{4}, \dfrac{9}{2} \text{ and }\dfrac{19}{4}$.

(ii) Let the first rational number between $\dfrac{1}{2}$ and $\dfrac{3}{5}$ be x.

$\Rightarrow x = \dfrac{1}{2}\Big(\dfrac{1}{2} + \dfrac{3}{5}\Big)\\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{5 + 6}{10}\Big)\\[1em] \Rightarrow x = \dfrac{1}{2} \times \dfrac{11}{10} \\[1em] \Rightarrow x = \dfrac{11}{20}.$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2}\Big(\dfrac{11}{20} + \dfrac{3}{5}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{11 + 12}{20}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \Big(\dfrac{23}{20}\Big) \\[1em] \Rightarrow y = \dfrac{23}{40}.$

Let the third rational number be z.

$\Rightarrow z = \dfrac{1}{2}\Big(\dfrac{11}{20} + \dfrac{1}{2}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{11 + 10}{20}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2} \times \dfrac{21}{20} \\[1em] \Rightarrow z = \dfrac{21}{40} \\[1em]$

Hence, three rational numbers between $\dfrac{1}{2}$ and $\dfrac{3}{5}$ are $\dfrac{21}{40}, \dfrac{11}{20} \text{ and } \dfrac{23}{40}$.

(iii) Let the first rational number between -1 and 1 be x.

$\Rightarrow x = \dfrac{1}{2}(-1 + 1) \\[1em] \Rightarrow x = \dfrac{1}{2} \times 0 \\[1em] \Rightarrow x = 0.$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2}(0 + 1) \\[1em] \Rightarrow y = \dfrac{1}{2} \times 1 \\[1em] \Rightarrow y = \dfrac{1}{2}.$

Let the third rational number be z.

$\Rightarrow z = \dfrac{1}{2}[0 + (-1)] \\[1em] \Rightarrow z = \dfrac{1}{2} \times -1 \\[1em] \Rightarrow z = -\dfrac{1}{2}$

Hence, three rational numbers between -1 and 1 are $-\dfrac{1}{2}, 0 \text{ and } \dfrac{1}{2}$.

(iv) Let the first rational number between $2\dfrac{1}{3}$ and $3\dfrac{2}{3}$ be x.

$\Rightarrow x = \dfrac{1}{2}\Big(2\dfrac{1}{3} + 3\dfrac{2}{3}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{7}{3} + \dfrac{11}{3}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2} \times \dfrac{18}{3} \\[1em] \Rightarrow x = \dfrac{1}{2} \times 6 \\[1em] \Rightarrow x = 3$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2}\Big(3 + \dfrac{11}{3}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{9 + 11}{3}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \times \dfrac{20}{3} \\[1em] \Rightarrow y = \dfrac{10}{3}.$

Let the third rational number be z.

$\Rightarrow z = \dfrac{1}{2}\Big(3 + \dfrac{7}{3}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{9 + 7}{3}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2} \times \dfrac{16}{3} \\[1em] \Rightarrow z = \dfrac{8}{3}.$

Hence, three rational numbers between $\dfrac{7}{3}$ and $\dfrac{11}{3}$ are $\dfrac{8}{3}, 3 \text{ and } \dfrac{10}{3}$.

(v) Let the first rational number between $-\dfrac{1}{2}$ and $\dfrac{1}{3}$ be x.

$\Rightarrow x = \dfrac{1}{2}\Big(-\dfrac{1}{2} + \dfrac{1}{3}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{-3 + 2}{6}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{-1}{6} \Big) \\[1em] \Rightarrow x = \dfrac{-1}{12}$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-1}{12} + \dfrac{-1}{2}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-1 - 6}{12}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-7}{12}\Big) \\[1em] \Rightarrow y = -\dfrac{7}{24} \\[1em]$

Let the third rational number be z.

$\Rightarrow z = \dfrac{1}{2}\Big(\dfrac{-1}{12} + \dfrac{1}{3}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{-1 + 4}{12}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{3}{12}\Big) \\[1em] \Rightarrow z = \dfrac{1}{8} \\[1em]$

Hence, three rational numbers between $\dfrac{-1}{2}$ and $\dfrac{1}{3}$ are $\dfrac{-7}{24}, \dfrac{-1}{12} \text{ and } \dfrac{1}{8}$.

(vi) Let the first rational number between $-\dfrac{1}{3}$ and $\dfrac{1}{4}$ be x.

$\Rightarrow x = \dfrac{1}{2}\Big(-\dfrac{1}{3} + \dfrac{1}{4}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{-4 + 3}{12}\Big) \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(\dfrac{-1}{12} \Big) \\[1em] \Rightarrow x = -\dfrac{1}{24}.$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2} \Big(\dfrac{-1}{24} + \dfrac{-1}{3}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-1 - 8}{24}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2} \times \dfrac{-9}{24} \\[1em] \Rightarrow y = -\dfrac{9}{48} = -\dfrac{3}{16}.$

Let the third rational number be z.

$\Rightarrow z = \dfrac{1}{2}\Big(\dfrac{-1}{24} + \dfrac{1}{4}\Big) \\[1em] \Rightarrow z = \dfrac{1}{2}\Big(\dfrac{-1+6}{24} \Big) \\[1em] \Rightarrow z = \dfrac{1}{2} \times \dfrac{5}{24} \\[1em] \Rightarrow z = \dfrac{5}{48}$

Hence, three rational numbers between $-\dfrac{1}{3}$ and $\dfrac{1}{4}$ are $-\dfrac{3}{16}, -\dfrac{1}{24} \text{ and } \dfrac{5}{48}$.

Find four rational numbers between 4 and 4.5.

Answer

Let a = 4, b = 4.5 and n = 4

Difference between consecutive rational numbers =

$\dfrac{b - a}{n + 1} = \dfrac{4.5 - 4}{4 + 1} = \dfrac{0.5}{5} = 0.1$

Rational numbers between 4 and 4.5 are :

⇒ a + d, a + 2d, a + 3d, a + 4d

⇒ 4 + 0.1, 4 + 0.2, 4 + 0.3, 4 + 0.4

⇒ 4.1, 4.2, 4.3, 4.4

Hence, four rational numbers between 4 and 4.5 are 4.1, 4.2, 4.3 and 4.4.

Find six rational numbers between 3 and 4.

Answer

Let a = 3, b = 4 and n = 9

Difference between consecutive rational numbers =

$\dfrac{b - a}{n + 1} = \dfrac{4 - 3}{9 + 1} = \dfrac{1}{10} = 0.1$

Rational numbers between 3 and 4 are :

⇒ a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d

⇒ 3 + 0.1. 3 + 0.2, 3 + 0.3, 3 + 0.4, 3 + 0.5, 3 + 0.6

⇒ 3.1, 3.2, 3.3, 3.4, 3.5, 3.6

Hence, six rational numbers between 3 and 4 are 3.1, 3.2, 3.3, 3.4, 3.5, 3.6.