periodic continued fractions represent quadratic irrationals

This article shows that infinite simple continued fractions that are eventually periodic correspond precisely to quadratic irrationals.

Throughout, we will freely use results on convergents to a continued fraction; see that article for details.

Definition 1.

A periodic simple continued fraction is a simple continued fraction

$$[a_0,a_1,a_2,\mathrm{\dots }]$$

such that for some $k\ge 0$ there is $m>0$ such that whenever $r\ge k$, we have $a_r=a_{r+m}$. Informally, a periodic continued fraction is one that eventually repeats. A purely periodic simple continued fraction is one for which $k=0$; that is, one whose repeating period starts with the initial element.

If

$$[a_0,a_1,\mathrm{\dots },a_{k-1},a_k,\mathrm{\dots },a_{k+j-1},a_k,\mathrm{\dots },a_{k+j-1},a_k,\mathrm{\dots }]$$

is a periodic continued fraction, we write it as

$$[a_0,a_1,\mathrm{\dots },a_{k-1},\overline{a_k,\mathrm{\dots },a_{k+j-1}}].$$

Theorem 1.

If

$$\alpha =[a_0,a_1,\mathrm{\dots },a_r,\overline{b_1,\mathrm{\dots },b_t}]$$

is a periodic simple continued fraction, then $\alpha$ is a quadratic irrational $p\mathrm{+}q\mathit{}\sqrt{d}$ for $p\mathrm{,}q$ rational and $d$ squarefree. Conversely, every such quadratic irrational is represented by such a continued fraction.

Proof.

The forward direction is pretty straightforward. Given such a continued fraction, let $\beta$ be the ${(r+1)}^{\mathrm{st}}$ complete convergent, i.e.

$$\beta =[\overline{b_1,\mathrm{\dots },b_t}]$$

Note first that $\beta$ must be irrational since the continued fraction for any rational number terminates. Then the article on convergents to a continued fraction shows that

$$\beta =\frac{\beta p_t+p_{t-1}}{\beta q_t+q_{t-1}}$$

where the $p_i,q_i$ are the convergents to the continued fraction for $\beta$. Thus

$$q_t{\beta }^2+(q_{t-1}-p_t)\beta -p_{t-1}=0$$

and thus $\beta$ is irrational and satisfies a quadratic equation so is a quadratic irrational. A simple computation then shows that $\alpha$ is as well.

In the other direction, suppose that $\alpha$ is a quadratic irrational satisfying

$$a{\alpha }^2+b\alpha +c=0$$

and with continued fraction representation

$$\alpha =[a_0,a_1,\mathrm{\dots }]$$

Then for any $n>0$, we have

$$\alpha =\frac{t_n{p}_{n-1}+p_{n-2}}{t_n{q}_{n-1}+q_{n-2}}$$

where the $t_i$ are the complete convergents of the continued fraction, so that from the quadratic equation we have

$$A_n{t}_n^2+B_n{t}_n+C_n=0$$

where

$$A_n=a{p}_{n-1}^2+b{p}_{n-1}{q}_{n-1}+c{q}_{n-1}^2$$

$$B_n=2a{p}_{n-1}{p}_{n-2}+b(p_{n-1}{q}_{n-2}+p_{n-2}{q}_{n-1})+2c{q}_{n-1}{q}_{n-2}$$

$$C_n=a{p}_{n-2}^2+b{p}_{n-2}{q}_{n-2}+c{q}_{n-2}^2=A_{n-1}$$

Note that $A_n\ne 0$ for each $n>0$ since otherwise

$$a{p}_{n-1}^2+b{p}_{n-1}{q}_{n-1}+c{q}_{n-1}^2=0$$

so that

$$a{\left(\frac{p_{n-1}}{q_{n-1}}\right)}^2+b\frac{p_{n-1}}{q_{n-1}}+c=0$$

and the quadratic equation would have a rational root, contradicting the fact that $\alpha$ is irrational.

The remainder of the proof is an elaborate computation that shows we can bound each of $A_n,B_n,C_n$ independent of $n$. Assuming that, it follows that there are only a finite number of possibilities for the triples $(A_n,B_n,C_n)$, so we can choose $n_1,n_2,n_3$ such that

$$(A_{n_1},B_{n_1},C_{n_1})=(A_{n_2},B_{n_2},C_{n_2})=(A_{n_3},B_{n_3},C_{n_3})$$

Then each of $t_{n_1},t_{n_2},t_{n_3}$ is a root of (say)

$$A_{n_1}{t}^2+B_{n_1}t+C_{n_1}$$

so that two of them must be equal. But then $t_{n_1}=t_{n_2}$ (say), and

$$t_{n_1}=[a_{n_1},a_{n_1+1},\mathrm{\dots }]$$

$$t_{n_2}=[a_{n_2},a_{n_2+1},\mathrm{\dots }]$$

and the continued fraction is periodic.

We proceed to find the bounds. We know that

so that

$$\alpha -\frac{p_{n-1}}{q_{n-1}}=\frac{ϵ}{q_{n-1}^2}$$

for some $ϵ$ (depending on $n$) with . Thus

$A_n$ $=a{p}_{n-1}^2+b{p}_{n-1}{q}_{n-1}+c{q}_{n-1}^2$

$=a{\left(\alpha q_{n-1}+{\displaystyle \frac{ϵ}{q_{n-1}}}\right)}^2+b{q}_{n-1}\left(\alpha q_{n-1}+{\displaystyle \frac{ϵ}{q_{n-1}}}\right)+c{q}_{n-1}^2$

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

$=2a\alpha ϵ+a{\displaystyle \frac{{ϵ}^2}{q_{n-1}^2}}+bϵ$

so that

and thus also

It remains to bound $B_n$. But

$$B_n^2-4A_n{C}_n={(2A_n{t}_n+B_n)}^2$$

Substituting the values of $A_n,B_n$ on the right, and using the fact that

$$t_n=-\frac{p_{n-2}-x{q}_{n-2}}{p_{n-1}-x{q}_{n-1}}$$

we get after a computation

$\sqrt{B_n^2-4A_n{C}_n}$ $=(p_{n-1}{q}_{n-2}-p_{n-2}{q}_{n-1}){\displaystyle \frac{2ax{p}_{n-1}+b{p}_{n-1}+bx{q}_{n-1}+2c{q}_{n-1}}{p_{n-1}-x{q}_{n-1}}}$

$=\pm {\displaystyle \frac{(2ax+b)p_{n-1}+(2a{x}^2+2bx+2c)q_{n-1}-(2a{x}^2+bx)q_{n-1}}{p_{n-1}-x{q}_{n-1}}}$

$=\pm {\displaystyle \frac{(p_{n-1}-x{q}_{n-1})(2ax+b)}{p_{n-1}-x{q}_{n-1}}}=\pm (2ax+b)$

so that

$$B_n^2-4A_n{C}_n={(2ax+b)}^2=b^2-4ac$$

Thus

and we have thus bounded all of $A_n,B_n,C_n$ independent of $n$. ∎

Title	periodic continued fractions represent quadratic irrationals
Canonical name	PeriodicContinuedFractionsRepresentQuadraticIrrationals
Date of creation	2013-03-22 18:04:40
Last modified on	2013-03-22 18:04:40
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	9
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 11Y65
Classification	msc 11A55
Defines	periodic continued fraction
Defines	purely periodic