The continued fraction [1; overline 2] for sqrt(2)

Since $1<\sqrt{2}<2$, [F3] gives $a_0=1$. Then [F3, F4, given, algebra] ${\alpha }_1=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1,$ so $2<{\alpha }_1<3$ and $a_1=2$. Moreover $\frac{1}{{\alpha }_1-2}=\frac{1}{\sqrt{2}-1}={\alpha }_1,$ so every later complete quotient is again ${\alpha }_1$. Thus the algorithm never terminates and produces the digits $1,2,2,\dots$; by [F4] its continued fraction converges to the original number. Hence $\sqrt{2}=[1;\bar{2}]\mathrm{.}$

Applying [F5] to the digits from step 1.1 gives [F5, step 1.1, algebra] $(p_0,q_0)=(1,1),(p_1,q_1)=(3,2),(p_2,q_2)=(7,5),(p_3,q_3)=(17,12),$ so the convergents begin $1,\frac{3}{2},\frac{7}{5},\frac{17}{12}\mathrm{.}$ The same recurrence gives $q_4=29$. For the displayed pairs one checks $3\cdot 1-1\cdot 2=1,7\cdot 2-3\cdot 5=-1,17\cdot 5-7\cdot 12=1,$ exactly as [F1] predicts.

The error formula [F2] now gives [F2, step 2.1, algebra] $\mid \sqrt{2}-\frac{3}{2}\mid =\frac{3-2\sqrt{2}}{2}<\frac{1}{10},\mid \sqrt{2}-\frac{7}{5}\mid =\frac{5\sqrt{2}-7}{5}<\frac{1}{60},$ and similarly $\mid \sqrt{2}-\frac{17}{12}\mid =\frac{17-12\sqrt{2}}{12}<\frac{1}{348}\mathrm{.}$ So the concrete convergents alternate around $\sqrt{2}$ with the expected quality of approximation.