Periodic Continued Fractions

Definition 5.4 (Periodic Continued Fraction)   A periodic continued fraction is a continued fraction $[a_0, a_1, \ldots, a_n, \ldots]$ such that

$$a_n = a_{n+h}$$

for some fixed positive integer $h$ and all sufficiently large $n$ . We call the minimal such $h$ the period of the continued fraction.

Example 5.4   Consider the periodic continued fraction $[1,2,1,2,\ldots] = [\overline{1,2}]$ . What does it converge to? We have

$$[\overline{1,2}] = 1+\frac{1}{\displaystyle 2+\frac{1}{\displaystyle 1+\frac{1}{\displaystyle 2+ \frac{1}{\displaystyle 1+\cdots}}}},$$

so if $\alpha=[\overline{1,2}]$ then

$$\alpha = 1 + \frac{1}{2+\displaystyle \frac{1}{\alpha}} = 1 + \fr... ...lpha+1}{\alpha}} = 1 + \frac{\alpha}{2\alpha+1} = \frac{3\alpha+1}{2\alpha+1}.$$

Thus $2\alpha^2 -2\alpha-1 = 0$ , so

$$\alpha = \frac{1+\sqrt{3}}{2}.$$

Theorem 5.4 (Periodic Characterization)   An infinite simple continued fraction is periodic if and only if it represents a quadratic irrational.

Proof. ( $\Longrightarrow$ ) First suppose that

$$[a_0, a_1, \ldots, a_n, \overline{a_{n+1},\ldots, a_{n+h}}]$$

is a periodic continued fraction. Set $\alpha=[a_{n+1},a_{n+2}, \ldots]$ . Then

$$\alpha = [a_{n+1},\ldots, a_{n+h}, \alpha],$$

so by Proposition 5.1.5

$$\alpha = \frac{\alpha p_{n+h} + p_{n+h-1}}{\alpha q_{n+h} + q_{n+h-1}}.$$

Here we use that $\alpha$ is the last partial quotient. Thus, $\alpha$ satisfies a quadratic equation with coefficients in  $\mathbb {Q}$ . Computing as in Example 5.4.4 and rationalizing the denominators, and using that the $a_i$ are all integers, shows that

$$[a_0, a_1, \ldots ] = [a_0, a_1, \ldots, a_n, \alpha]$$

$$= a_0 + \frac{1}{\displaystyle a_1 + \frac{1}{\displaystyle a_2 + \cdots + \frac{1}{\alpha}}}$$

is of the form $c+d\alpha$ , with $c,d\in\mathbb{Q}$ , so $[a_0, a_1, \ldots]$ also satisfies a quadratic polynomial over  $\mathbb {Q}$ .

The continued fraction procedure applied to the value of an infinite simple continued fraction yields that continued fraction back, so by Proposition 5.2.12, $\alpha\not\in\mathbb{Q}$ because it is the value of an infinite continued fraction.

( $\Longleftarrow$ ) Suppose $\alpha\in\mathbb{R}$ is an irrational number that satisfies a quadratic equation

$$a \alpha^2 + b\alpha + c = 0$$ (5.4.1)

with $a, b, c\in\mathbb{Z}$ and $a\neq 0$ . Let $[a_0, a_1, \ldots]$ be the continued fraction expansion of $\alpha$ . For each $n$ , let

$$r_n = [a_n, a_{n+1}, \ldots],$$

so

$$\alpha = [a_0, a_1, \ldots, a_{n-1}, r_n].$$

We will prove periodicity by showing that the set of $r_n$ 's is finite. If we have shown finiteness, then there exists $n, h>0$ such that $r_n = r_{n+h}$ , so

$$[a_0, \ldots, a_{n-1}, r_n] = [a_0, \ldots, a_{n-1}, a_n, \ldots, a_{n+h-1}, r_{n+h}]$$

$$= [a_0, \ldots, a_{n-1}, a_n, \ldots, a_{n+h-1}, r_{n}]$$

$$= [a_0, \ldots, a_{n-1}, a_n, \ldots, a_{n+h-1}, a_n, \ldots, a_{n+h-1}, r_{n+h}]$$

$$= [a_0, \ldots, a_{n-1}, \overline{a_n, \ldots, a_{n+h-1}}].$$

It remains to show there are only finitely many distinct $r_n$ . We have

$$\alpha = \frac{p_n}{q_n} = \frac{r_n p_{n-1} + p_{n-2}}{r_n q_{n-1} + q_{n-2}}.$$

Substituting this expression for $\alpha$ into the quadratic equation (5.4.1), we see that

$$A_n r_n^2 + B_n r_n + C_n = 0,$$

where

$$A_n = a p_{n-1}^2 + b p_{n-1} q_{n-1} + c q_{n-1}^2,$$

$$B_n = 2a p_{n-1} p_{n-2} + b(p_{n-1} q_{n-2} + p_{n-2} q_{n-1}) + 2c q_{n-1} q_{n-2},$$

$$C_n = a p_{n-2}^2 + b p_{n-2} q_{n-2} + c p_{n-2}^2.$$

Note that $A_n, B_n, C_n\in\mathbb{Z}$ , that $C_n = A_{n-1}$ , and that

$$B_n^2 - 4A_n C_n = (b^2- 4ac)(p_{n-1}q_{n-2} - q_{n-1}p_{n-2})^2 = b^2 - 4ac.$$

Recall from the proof of Theorem 5.2.10 that

$$\left\vert \alpha - \frac{p_{n-1}}{q_{n-1}}\right\vert < \frac{1}{q_n q_{n-1}}.$$

Thus

$$\left\vert \alpha q_{n-1} - p_{n-1}\right\vert < \frac{1}{q_n} < \frac{1}{q_{n-1}},$$

so

$$p_{n-1} = \alpha q_{n-1} + \frac{\delta}{q_{n-1}}$$   with $$\vert\delta\vert < 1.$$

Hence

$$A_n = a\left(\alpha q_{n-1} + \frac{\delta}{q_{n-1}}\right)^2 +b\left(\alpha q_{n-1} + \frac{\delta}{q_{n-1}}\right)q_{n-1} +c q_{n-1}^2$$

$$= (a\alpha^2 + b\alpha + c)q_{n-1}^2 + 2a\alpha\delta + a\frac{\delta^2}{q_{n-1}^2} + b\delta$$

$$= 2a\alpha\delta + a\frac{\delta^2}{q_{n-1}^2} + b\delta.$$

Thus

$$\vert A_n\vert = \left\vert 2a\alpha\delta + a\frac{\delta^2}{q_{n-1}^2} + b\delta\right\vert < 2\vert a\alpha\vert + \vert a\vert + \vert b\vert.$$

Thus there are only finitely many possibilities for the integer $A_n$ . Also,

$$\vert C_n\vert = \vert A_{n-1}\vert$$    and $$\quad \vert B_n\vert = \sqrt{b^2 - 4(ac-A_n C_n)},$$

so there are only finitely many triples $(A_n, B_n, C_n)$ , and hence only finitely many possibilities for $r_n$ as $n$ varies, which completes the proof. (The proof above closely follows [#!hardywright!#, Thm. 177, pg.144-145].) $\qedsymbol$