Question

Express the following surds as continued fractions, and find the sixth convergent to $\sqrt{\frac{6}{5}}$.

Answer 1

We can write $\sqrt{\frac{6}{5}}=\frac{\sqrt{30}}{5}=1+\frac{\sqrt{30}-5}{5}$$=1+\frac{1}{\sqrt{30}+5}$
Lets take $\sqrt{30}+5=10+\sqrt{30}-5=10+\frac{5}{\sqrt{30}+5}$
$\Rightarrow \frac{\sqrt{30}+5}{5}=2+\frac{\sqrt{30}-5}{5}$
Therefore, the continued fraction is:-
$=1+\frac{1}{10+\frac{1}{2+...}}$
Convergents are:-
$1,\frac{11}{10},\frac{23}{21},\frac{241}{220},...$