Question

Express the following surds as continued fractions, and find the fourth convergent to $\sqrt{a^2-a}$.

Answer 1

We can write $\sqrt{a^2-a}=a-1+(\sqrt{a^2-a}-\overline{a-1})$

$=a-1+\frac{a-1}{\sqrt{a^2-a}+a-1}$

Thus $\frac{\sqrt{a^2-a}+a-1}{a-1}=2+\frac{\sqrt{a^2-a}-(a-1)}{a-1}$

$=2+1\frac{1}{\sqrt{a^2-a}+a-1}$

$\Rightarrow \sqrt{a^2-a}+a-1=2(a-1)+(\sqrt{a^2-a}-\overline{a-1})$

$=2(a-1)+\frac{a-1}{\sqrt{a^2-a}+a-1}$

Therefore, $\sqrt{a^2-a}=(a-1)+\frac{1}{2}+\frac{1}{2(a-1)}+....$

The convergents are

$\frac{a-1}{1},\frac{2a-1}{2},\frac{4a^2-5a+1}{4a-3},\frac{8a^2-8a+1}{8a-4}$