Insert seven rational numbers between 2 and 3.

2 = $\dfrac{2}{1}$

3 = $\dfrac{3}{1}$

Since seven rational numbers are required between $\dfrac{2}{1}$ and $\dfrac{3}{1}$; multiply the numerator and the denominator of each rational number by 7 + 1 = 8.

$\dfrac{2}{1} = \dfrac{2 \times 8}{1 \times 8} = \dfrac{16}{8}$

and

$\dfrac{3}{1} = \dfrac{3 \times 8}{1 \times 8} = \dfrac{24}{8}$

Required rational numbers between $\dfrac{2}{1}$ and $\dfrac{3}{1}$ are : $\dfrac{16}{8} , \dfrac{17}{8} , \dfrac{18}{8} , \dfrac{19}{8} , \dfrac{20}{8} , \dfrac{21}{8} , \dfrac{22}{8} , \dfrac{23}{8} , \dfrac{24}{8}$

= $\dfrac{2}{1} , \dfrac{17}{8} , \dfrac{9}{4} , \dfrac{19}{8} , \dfrac{5}{2} , \dfrac{21}{8} , \dfrac{11}{4} , \dfrac{23}{8} , \dfrac{3}{1}$

= $2 , 2\dfrac{1}{8} , 2\dfrac{1}{4} , 2\dfrac{3}{8} , 2\dfrac{1}{2} , 2\dfrac{5}{8} , 2\dfrac{3}{4} , 2\dfrac{7}{8} , 3$

Hence, $2\dfrac{1}{8} , 2\dfrac{1}{4} , 2\dfrac{3}{8} , 2\dfrac{1}{2} , 2\dfrac{5}{8} , 2\dfrac{3}{4}$ and $2\dfrac{7}{8}$ lie between 2 and 3.