Question

solve this guys ..............​

Answer 1

question

insert three rational no between

$\frac{1}{4} and \frac{1}{2}$

solution

First rational no

$\frac{1}{2} ( \frac{1}{4} \times \frac{1}{2} )$

$\frac{1}{2} ( \frac{1}{8} )$

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

second rational no

$\frac{1}{2} \times ( \frac{1}{16} \times \frac{1}{4} )$

$\frac{1}{2} \times \frac{1}{64}$

$= \frac{1}{128}$

third rational no

$\frac{1}{2} \times ( \frac{1}{16} \times \frac{1}{2} )$

$\frac{1}{2} \times \frac{1}{32}$

$\frac{1}{64}$

therefore the three rational numbers are:

$\frac{1}{16} \: \: \frac{1}{128} \: \: \frac{1}{64} \: answer$