Question

Find limits of the error when $\frac{916}{191}$ is taken for $\sqrt{23}$.

Answer 1

We know $\sqrt{23}=4+\sqrt{23}-4$
$\Rightarrow 4+\frac{7}{\sqrt{23}+4}$
$\Rightarrow \frac{\sqrt{23}+4}{7}=1+\frac{\sqrt{23}-3}{7}=1+\frac{2}{\sqrt{23}+3}$
$\Rightarrow \frac{\sqrt{23}+3}{2}=3+\frac{\sqrt{23}+3}{2}=3+\frac{7}{\sqrt{23}+3}$
$\therefore$ the continued fraction is:-
$=4+\frac{1}{1+\frac{1}{3+\frac{1}{1+...}}}$
and the convergents are :-
$\Rightarrow \frac{4}{1},\frac{5}{1},\frac{19}{4},\frac{24}{5},\frac{211}{44},\frac{235}{49},\frac{916}{191},\frac{1151}{240},....$
Therefore, the error in taking $\frac{916}{191}$ is less than $\frac{1}{{\left(191\right)}^2}$ and greater than $\frac{1}{{2\left(240\right)}^2}$.