Question

Express the following surds as continued fractions, and find the sixth convergent to $\frac{1}{\sqrt{21}}$.

Answer 1

We can write $\sqrt{21}=4+\sqrt{21}-4$
$=4+\frac{5}{\sqrt{21}+4}$
$\Rightarrow \sqrt{21}-4=\frac{5}{\sqrt{21}+4}$
$\Rightarrow 1+\frac{\sqrt{21}-1}{5}=\frac{5}{\sqrt{21}+4}$
$\Rightarrow 1+\frac{4}{\sqrt{21}+1}=\frac{5}{\sqrt{21}+4}$
$\Rightarrow \frac{\sqrt{21}+1}{4}=1+\frac{\sqrt{21}-3}{4}=1+\frac{3}{\sqrt{21}+3}$
$\Rightarrow \frac{\sqrt{21}+3}{3}=2+\frac{\sqrt{21}-3}{3}=2+\frac{4}{\sqrt{21}+3}$
$\Rightarrow \frac{\sqrt{21}+3}{4}=1+\frac{\sqrt{21}-1}{4}=1+\frac{5}{\sqrt{21}+1}$
$\Rightarrow \frac{\sqrt{21}+1}{5}=1+\frac{\sqrt{21}-4}{5}=1+\frac{1}{\sqrt{21}+4}$
$\Rightarrow \sqrt{21}+4=8+\sqrt{21}-4$
Therefore, the continued fraction is $:-$
$\frac{1}{4+\frac{1}{1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{1+\frac{1}{8+...}}}}}}}$
and the convergents are :-
$\frac{1}{4},\frac{1}{5},\frac{2}{9},\frac{5}{23},\frac{7}{32},\frac{12}{55},...$