Question

Find limits of the error when $\frac{268}{65}$ is taken for $\sqrt{17}$.

Answer 1

We can write $\sqrt{17}=4+\sqrt{17}-4$
$=4+\frac{1}{\sqrt{17}+4}$
$\Rightarrow \sqrt{17}+4=8+\sqrt{17}-4=8+\frac{1}{\sqrt{17}+4}$
The continued fraction is:-
$=4+\frac{1}{8+\frac{1}{8+....}}$
The convergents are:-
$4,\frac{33}{8},\frac{268}{65},\frac{2177}{528},...$
Therefore, the error is taking $\frac{268}{65}$ is less than $\frac{1}{{\left(65\right)}^2}$ and greater than $\frac{1}{{2\left(528\right)}^2}$.