Question

Find three rational numbers between root 3 and root 5

Answer 1

Dear student,
$$\begin{gathered} Thegivennumbersare\sqrt{3}and\sqrt{5} \\ Wecanfinditputbymeanmethodi.e\frac{r+s}{2} \\ Herer=\sqrt{3}ands=\sqrt{5} \\ \therefore \frac{\sqrt{3}+\sqrt{5}}{2} \\ thus\sqrt{3}<\frac{\sqrt{3}+\sqrt{5}}{2}<\sqrt{5} \\ Againapplyingformulabetween\sqrt{3}and\frac{\sqrt{3}+\sqrt{5}}{2} \\ weget,\frac{\sqrt{3}+\frac{\sqrt{3}+\sqrt{5}}{2}}{2} \\ =\frac{2\sqrt{3}+\sqrt{3}+\sqrt{5}}{4} \\ =\frac{3\sqrt{3}+\sqrt{5}}{4} \\ thus,\sqrt{3}<\frac{3\sqrt{3}+\sqrt{5}}{4}<\frac{\sqrt{3}+\sqrt{5}}{2}<\sqrt{5} \\ Againapplyingformulabetween\frac{3\sqrt{3}+\sqrt{5}}{4}and\frac{\sqrt{3}+\sqrt{5}}{2} \\ weget,\frac{\frac{3\sqrt{3}+\sqrt{5}}{4}+\frac{\sqrt{3}+\sqrt{5}}{2}}{2} \\ =\frac{3\sqrt{3}+\sqrt{5}+2\left(3\sqrt{3}+\sqrt{5}\right)}{8} \\ =\frac{3\sqrt{3}+\sqrt{5}+6\sqrt{3}+2\sqrt{5}}{8} \\ =\frac{9\sqrt{3}+3\sqrt{5}}{8} \\ thus,\sqrt{3}<\frac{9\sqrt{3}+3\sqrt{5}}{8}<\frac{3\sqrt{3}+\sqrt{5}}{4}<\frac{\sqrt{3}+\sqrt{5}}{2}<\sqrt{5} \\ Hencethethreenumberbetween\sqrt{3}and\sqrt{5}are \\ \frac{9\sqrt{3}+3\sqrt{5}}{8},\frac{3\sqrt{3}+\sqrt{5}}{4},\frac{\sqrt{3}+\sqrt{5}}{2} \end{gathered}$$
Regards