Question

Express the following surds as continued fractions, and find the sixth convergent to $\sqrt{13}$.

Answer 1

$\sqrt{13}=1+\sqrt{13}-1$

$=1+\frac{12}{\sqrt{13}+1}$

$=1+\frac{12}{1+(1+\frac{12}{\sqrt{13}+1})}$

$=1+\frac{12}{2+\frac{12}{\sqrt{13}+1}}$

$=1+\frac{12}{2+\frac{12}{1+1+\frac{12}{\sqrt{13}+1}}}$

$=1+\frac{12}{2+\frac{12}{2+\frac{12}{\sqrt{13}+\mathrm{1...}}}}$

$⟹\sqrt{13}=1,\frac{13}{7},\frac{26}{5},...$