Question

Express $\sqrt{19}$ as a continued fraction, and find a series of fractions approximating to its value.

Answer 1

$\sqrt{19}=4+(\sqrt{19}-4)=4+\frac{3}{\sqrt{19}+4};$
$\frac{\sqrt{19}+4}{3}=2+\frac{\sqrt{19}-2}{3}=2+\frac{5}{\sqrt{19}+2};$
$\frac{\sqrt{19}+2}{5}=1+\frac{\sqrt{19}-3}{5}=1+\frac{2}{\sqrt{19}-3};$
$\frac{\sqrt{19}+3}{2}=3+\frac{\sqrt{19}-3}{5}=3+\frac{5}{\sqrt{19}+3};$
$\frac{\sqrt{19}+3}{5}=1+\frac{\sqrt{19}-2}{5}=1+\frac{3}{\sqrt{19}+2};$
$\frac{\sqrt{19}+2}{3}=2+\frac{\sqrt{19}-4}{3}=2+\frac{1}{\sqrt{19}+4};$
$\sqrt{19}+4=8+(\sqrt{19}-4)=8+......$
after this the quotients $2,1,3,1,2,8$ recur$;$ hence
$\sqrt{19}=4+\frac{1}{2+\frac{1}{1+\frac{1}{3+\frac{1}{1+..}}}}$