Question

Express the following surds as continued fractions, and find the sixth convergent to $\sqrt{\frac{7}{11}}$.

Answer 1

We can write $\sqrt{\frac{7}{1}}=\frac{7}{\sqrt{77}};\frac{\sqrt{77}}{7}=1+\frac{\sqrt{77}-7}{7}$

Therefore, $\frac{\sqrt{77}}{7}=1+\frac{4}{\sqrt{77}+7}$

$\Rightarrow \frac{\sqrt{77}+4}{4}=3+\frac{\sqrt{77}-5}{4}=3+\frac{13}{\sqrt{77}+5}$

$\Rightarrow \frac{\sqrt{77}+5}{13}=1+\frac{\sqrt{77}-8}{13}=1+\frac{1}{\sqrt{77}+8}$

$\Rightarrow \sqrt{77}+8=16+\sqrt{77}-8=16+\frac{13}{\sqrt{77}+8}$

$\Rightarrow \frac{\sqrt{77}+8}{13}=1+\frac{\sqrt{77}-5}{13}=1+\frac{4}{\sqrt{77}+5}$

$\Rightarrow \frac{\sqrt{77}+5}{4}=3+\frac{\sqrt{77}-7}{4}=3+\frac{7}{\sqrt{77}+7}$

$\Rightarrow \frac{\sqrt{77}+7}{7}=2+\frac{\sqrt{77}-7}{7}=2+\frac{4}{\sqrt{77}+7}$

Therefore, the continued fraction is:-

$=\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{16+\frac{1}{1+\frac{1}{3+\frac{1}{2+...}}}}}}}$

and the convergents are :-

$1,\frac{3}{4},\frac{4}{5},\frac{67}{84},\frac{71}{89},\frac{280}{351},...$