Find three rational numbers lying between

$x=\frac{3}{5}$ and $y=\frac{7}{8}$

n = 3

$d=\frac{(y-x)}{n+1}=\frac{\frac{7}{8}-\frac{3}{5}}{3+1}=\frac{11}{40} \times \frac{1}{4}=\frac{11}{160}$

Rational numbers between $x=\frac{3}{5}$ and $y=\frac{7}{8}$ will be

$(x+d),(x+2 d), \ldots,(x+n d)$

$\Rightarrow\left(\frac{3}{5}+\frac{11}{160}\right),\left(\frac{3}{5}+2 \times \frac{11}{160}\right),\left(\frac{3}{5}+3 \times \frac{11}{160}\right)$

$\Rightarrow\left(\frac{107}{160}\right),\left(\frac{118}{160}\right),\left(\frac{129}{160}\right)$

$\Rightarrow\left(\frac{107}{160}\right),\left(\frac{59}{80}\right),\left(\frac{129}{160}\right)$

There are infinitely many rational numbers between two given rational numbers.
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