# Find six rational numbers between 3 and 4 .

Question:

Find six rational numbers between 3 and 4 .

Solution:

There can be infinitely many rationals between 3 and 4 , one way is

to take them $3=\frac{21}{7}$ and $4=\frac{28}{7} .$      $(\because 6+1=7)$

First rational number between 3 and 4

$q_{1}=\left(\right.$ rational number between $\frac{21}{7}$ and $\left.\frac{28}{7}\right)=\frac{\frac{21}{7}+\frac{28}{7}}{2}=\frac{\frac{49}{7}}{2}=\frac{7}{2}$

$\therefore \quad \frac{21}{7}<\frac{7}{2}<\frac{28}{7}$

Second rational number between 3 and 4

$q_{2}=\left(\right.$ rational number between $\frac{21}{7}$ and $\left.\frac{7}{2}\right)=\frac{\frac{21}{7}+\frac{7}{2}}{2}=\frac{91}{28}$

$\therefore \quad \frac{21}{7}<\frac{91}{28}<\frac{7}{2}<\frac{28}{7}$

Third rational number between 3 and 4

$q_{3}=\left(\right.$ rational number between $\frac{7}{2}$ and $\left.\frac{28}{7}\right)=\frac{\frac{7}{2}+\frac{28}{7}}{2}=\frac{105}{28}$

$\therefore \quad \frac{21}{7}<\frac{91}{28}<\frac{7}{2}<\frac{105}{28}<\frac{28}{7}$

Similarly, $\frac{21}{7}<\frac{175}{56}<\frac{91}{28}<\frac{7}{2}<\frac{203}{56}<\frac{105}{28}<\frac{217}{56}<\frac{28}{7}$

Hence, the six rational numbers $\frac{175}{56}, \frac{91}{28}, \frac{7}{2}, \frac{203}{56}, \frac{105}{28}, \frac{217}{56}$ are all lying

between 3 and 4 .