Question

Find six rational numbers between 3 and 4.

Answer 1

One rational number between 3 and 4

= $\frac{1}{2}(3+4)\frac{7}{2}$

$(3+\frac{7}{2})=\frac{1}{2}\left(\frac{6+7}{2}\right)=\frac{1}{2}\times \frac{13}{2}=\frac{13}{4}Onerationalnumberbetween\frac{7}{2}and4\frac{1}{2}(\frac{7}{2}+\frac{4}{1})=\frac{1}{2}\left(\frac{7+8}{2}\right)=\frac{1}{2}\left(\frac{15}{2}\right)=\frac{15}{4}Onerationalnumberbetween3and\frac{13}{4}=\frac{1}{2}(3+\frac{13}{4})=\frac{1}{2}\left(\frac{12+13}{4}\right)=\frac{1}{2}\times \frac{25}{4}=\frac{25}{8}Onerationalnumberbetween\frac{15}{4}and4=\frac{1}{2}(\frac{15}{4}+\frac{4}{1})=\frac{1}{2}\left(\frac{15+16}{4}\right)=\frac{1}{2}\times \frac{31}{4}=\frac{31}{8}Onerationalnumberbetween3and\frac{25}{4}=\frac{1}{2}(3+\frac{25}{8})$

$=\frac{1}{2}\left(\frac{24+25}{8}\right)=\frac{1}{2}\times \frac{49}{8}=\frac{49}{16}\therefore Sixrationalnumbersbetween3and4are\frac{49}{16},\frac{25}{8},\frac{13}{4},\frac{7}{2},\frac{15}{4},\frac{31}{8}$