Problem 1477

Solving according to formula (13) we get
$$\begin{aligned} \sqrt{8} & =2+(\sqrt{8}-2)=2+4 /(\sqrt{8}+2) \\ & =\langle 2,(\sqrt{8}+2) / 4\rangle \\ & =\langle 2,1+(\sqrt{8}-2) / 4\rangle=\langle 2,1, \sqrt{8}+2\rangle \\ & =\langle 2,1,4+(\sqrt{8}-2)\rangle \\ & =\langle 2,1,4,(\sqrt{8}+2) / 4\rangle \end{aligned}$$

This returns to the case of $\langle 2,(\sqrt{8}+2) / 4\rangle$, therefore, the numbers repeat, yielding
$$\sqrt{8}=\langle 2,1,4,1,4,1,4, \cdots\rangle .$$

We also obtain (why)
$$\begin{array}{l} (\sqrt{8}+2) / 4=\langle 1,4,1,4,1,4, \cdots\rangle \\ \sqrt{8}+2=\langle 4,1,4,1,4,1, \cdots\rangle \end{array}$$

Proof: Without loss of generality, assume $(a, b)=1$ (why). From formula (3) in §2, we know there exists a unique $n$ such that
$$k_{n} \leqslant b<k_{n+1}$$

First, we prove that $b=k_{n}$. If not, there must be
$$k_{n}<b<k_{n+1}$$

and $a / b$ is not a convergent fraction (why), thus
$$\left|h_{n} / k_{n}-a / b\right| \geqslant 1 /\left(b k_{n}\right).$$

From $b<k_{n+1}$, Theorem 6(i) implies
$$\left|\xi_{0} k_{n}-h_{n}\right| \leqslant\left|\xi_{0} b-a\right|,$$

Thus, from equations (30), (31), and condition (27), we get
$$\begin{aligned} 1 /\left(b k_{n}\right) & \leqslant\left|h_{n} / k_{n}-a / b\right| \leqslant\left|\xi_{0}-h_{n} / k_{n}\right|+\left|\xi_{0}-a / b\right| \\ & <1 /\left(2 b k_{n}\right)+1 /\left(2 b^{2}\right) \end{aligned}$$

From the above inequality, we deduce $b<k_{n}$, which is a contradiction. Therefore, $b=k_{n}$ must hold. Furthermore, from the right half of the above inequality (which is valid due to condition (27) and equation (31)), we get
$$\left|h_{n} / k_{n}-a / k_{n}\right|<1 / k_{n}^{2},$$

which implies $\left|h_{n}-a\right|<1 / k_{n}$. Thus, $a=h_{n}$. Therefore, $a / b=h_{n} / k_{n}$ is a convergent fraction of $\xi_{0}$. Proof completed.