Question

Express the following surds as continued fractions, and find the sixth convergent to $\sqrt{22}$.

Answer 1

$\sqrt{22}=1+\sqrt{22}-1$

$=1+\frac{21}{1+\sqrt{22}}$

$=1+\frac{21}{1+1+\frac{21}{1+\sqrt{22}}}$

$=1+\frac{21}{2+\frac{21}{1+\sqrt{22}}}$

$=1+\frac{21}{2+\frac{21}{2+\frac{21}{1+\sqrt{22}}}}$

$=1+\frac{21}{2+\frac{21}{2+\frac{21}{2+..}}}$