Insert three rational numbers between:

$\dfrac{8}{11}$ and $\dfrac{4}{9}$

As we know that, for any two rational numbers $\dfrac{a}{b}$ and $\dfrac{c}{d}$, $\Big(\dfrac{a + c}{b + d}\Big)$ is also a rational number with its value lying between $\dfrac{a}{b}$ and $\dfrac{c}{d}$.

Given numbers = $\dfrac{8}{11}$ and $\dfrac{4}{9}$

$= \dfrac{8}{11}, \dfrac{8 + 4}{11 + 9} ,\dfrac{4}{9} \\[1em] = \dfrac{8}{11}, \dfrac{12}{20}, \dfrac{4}{9} \\[1em] = \dfrac{8}{11}, \dfrac{3}{5}, \dfrac{4}{9} \\[1em] = \dfrac{8}{11}, \dfrac{8 + 3}{11 + 5}, \dfrac{3}{5}, \dfrac{4}{9} \\[1em] = \dfrac{8}{11}, \dfrac{11}{16}, \dfrac{3}{5}, \dfrac{4}{9}\\[1em] = \dfrac{8}{11}, \dfrac{11}{16}, \dfrac{3}{5},\dfrac{3 + 4}{5 + 9}, \dfrac{4}{9}\\[1em] = \dfrac{8}{11}, \dfrac{11}{16}, \dfrac{3}{5},\dfrac{7}{14}, \dfrac{4}{9}\\[1em] = \dfrac{8}{11}, \dfrac{11}{16}, \dfrac{3}{5},\dfrac{1}{2}, \dfrac{4}{9}$

Hence, required rational numbers between $\dfrac{8}{11}$ and $\dfrac{4}{9}$ are : $\dfrac{11}{16},\dfrac{3}{5}$ and $\dfrac{1}{2}$