Question

Find two rational numbers lying between $\frac{1}{3}\mathrm{and}\frac{1}{2}$.

Answer 1

$$\begin{gathered} \mathrm{Because}\frac{1}{3}\mathrm{is}\mathrm{less}\mathrm{than}\frac{1}{2},\mathrm{let}x=\frac{1}{3},y=\frac{1}{2}\mathrm{and}n=2. \\ \mathrm{Rational}\mathrm{numbers}\mathrm{between}x\mathrm{and}y\mathrm{are}x+d\mathrm{and}x+2d. \\ \mathrm{Here}, \\ d=\frac{y-x}{n+1}=\frac{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{3}}}{2+1}=\frac{1}{6\times 3}=\frac{1}{18} \\ \mathrm{Required}\mathrm{numbers}:x+d=\frac{1}{3}+\frac{1}{18}=\frac{6+1}{18}=\frac{7}{18}\mathrm{and}x+2d=\frac{1}{3}+\frac{2}{18}=\frac{4}{9} \end{gathered}$$