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

(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 $-\dfrac{3}{16}, -\dfrac{1}{24} \text{ and } \dfrac{5}{48}$.