Question

Express the following surds as continued fractions, and find the sixth convergent to $\sqrt{11}$.

Answer 1

$\sqrt{11}=1+\sqrt{11}-1$

$=1+\frac{10}{1+\sqrt{11}}$

$=1+\frac{10}{1+1+\frac{10}{1+\sqrt{11}}}$

$=1+\frac{10}{2+\frac{10}{1+1+\frac{10}{1+\sqrt{11}}}}$

$=1+\frac{10}{2+\frac{10}{2+\frac{10}{1+1+\frac{10}{1+\sqrt{11}}}}}$

$⟹\sqrt{11}=[1,2,2,2,...]$

Convergents of $\sqrt{11}=1,\frac{10}{11},...$