We've seen that quadratic irrationals correspond to periodic continued fractions. A periodic continued fraction may repeat eventually (like ) or repeat from the start (like ). In this section, I'll consider the second case.

* Definition.* A continued fraction of the form
is * purely periodic*.

I'll derive a criterion for a quadratic irrational to have a purely periodic continued fraction. It is a result of Galois from 1829 ([1]). Recall that if is a quadratic irrational, its conjugate is

* Theorem.* Let be a quadratic
irrational. The continued fraction for x is purely periodic if and
only if and .

* Proof.* ( ) Suppose and . Using the
general continued fraction algorithm and properties of conjugates, I
have

Note that for . But , so . Thus, for .

* Claim:* For all ,

I'll prove the claim using induction. First, , so by assumption .

Assume that . Then gives , so adding gives

Since the middle inequality shows , I have

This proves the claim by induction.

Now

Using the claim, I have

This inequality says that , and also that is an integer which differs from by less than 1. It follows that

Since x is a quadratic irrational, its continued fraction is periodic. Thus, there are indices i and j such that

Hence,

Thus, implies . Continuing to reduce indices in this way, I eventually obtain

Therefore,

Hence, x is purely periodic.

( ) Suppose x is a quadratic irrational that is purely periodic, so , where . Note that . I have

Hence,

So

The quadratic function has x and as its roots. I already know ; I need to show . It's enough to show that f has a root between -1 and 0: Since that root can't be x, it must be .

First, . Next,

Then and implies that there's a root between -1 and 0 by the Intermediate Value Theorem. As noted above, that root must be .

Thus, and .

For example, satisfies and . Its continued fraction is

On the other hand, , but does not lie between -1 and 0. Its continued fraction is

To motivate the next result, consider the following example.

* Example.* (a) Compute the numerators and
denominators of the convergents for .

(b) Compute the numerators and denominators of the convergents for .

(a)

(b)

Look at the numbers in the last two rows of the tables in the last example. They suggest the following result.

* Theorem.* Consider the continued fractions

Let and denote the numerator and denominator of the convergent for x.

Let and denote the numerator and denominator of the convergent for y.

Then:

(a)

(b)

* Remark.* By reversing the roles of x and y, it
also follows that

* Proof.* (a) We'll induct on n. For , consider the convergents tables
for and :

Then

The result holds for .

Assume that the result holds for n (that is, that it holds for a fraction with terms and its reverse). I need to prove the result for --- that is, for the fractions and , I have

Note that the primed p's and q's are for , not for .

I have

I'll apply the induction hypothesis to and . Note that and are the same for and . However, the p's and q's for and for are different, so I'll use double-primed p's and q's for . With this understanding, the induction hypothesis gives

Then using the last two equations, I have

Similarly,

By induction,

So

This proves the result for , so the result holds for all by induction.

(b) Recall that

It follows that and are in lowest terms. But and are convergents of a continued fraction, so they're in lowest terms as well.

is an equality between fractions in lowest terms, so and . Likewise, is an equality between fractions in lowest terms, so and .

This result relates a finite continued fraction and its "reverse" . The next result (also due to Galois) considers the relationship between the purely periodic continued fractions and .

* Theorem.* Let be a purely periodic quadratic irrational. Then
is purely periodic, and

* Proof.* The idea is to show that and are roots of the same quadratic
equation. This implies that they are conjugates.

Let be the convergent of x. Then

The convergents algorithm gives

Let

Let denote the convergent of y. The convergents algorithm gives

By the preceding theorem, , , , and . So

Thus, x and are roots of the quadratic , so they must be conjugates:

[1] É. Galois, Démonstration d'un théoréme sur les fractions continues périodiques, Annalles de mathématiques 19 (1828), 294--301.

Copyright 2019 by Bruce Ikenaga