Question

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

Answer 1

$x^2-6y^2=1$

$\sqrt{6}=1+\sqrt{6}-1$

$=1+\frac{1}{\frac{1}{\sqrt{6}-1}}=1+\frac{1}{\frac{1+\sqrt{6}}{5}}$

$=1+\frac{1}{\frac{1+1+\sqrt{6}-1}{5}}$

$=1+\frac{1}{1+\frac{1}{5}+\frac{1}{\frac{5}{\sqrt{6}}-5}}$

$=1+\frac{1}{\frac{6}{5}+\frac{1}{\frac{5}{\sqrt{6}}-\mathrm{5....}}}$