* Definition.* Let , ... be real numbers, with
, ..., positive. A * finite continued
fraction* is an expression of the form

To make the writing easier, I'll denote the continued fraction above by . In most cases, the 's will be integers.

* Remark.* In some cases people have considered
continued fractions where the numerators don't have to be 1. For
example,

In this case, they refer to continued fractions where the numerators
are all 1 as * simple continued fractions*.

I will only be considering continued fractions where the numerators are all 1. So to save writing, I won't use the adjective "simple", and use the phrase "continued fraction" to mean a continued fraction with numerators all eqaul to 1.

* Example.* Find a continued fraction expansion
for .

I can do this by repeated division:

In short form, .

Now repeated division is what you do in executing the Euclidean algorithm and so a little bit of thought should convince you that you can express any rational number as a finite continued fraction in this way. Specifically,

Notice that the successive quotients 2, 1, 3, and 4 are the numbers in the continued fraction expansion.

* Example.* Find a finite continued fraction
expansion for .

I have been asking for *a* finite continued fraction
expansion, not *the* finite continued fraction expansion. In
fact, the continued fraction expansion of a rational number is not
unique. For example,

And in general,

* Lemma.* Every finite continued fraction with
integer terms represents a rational number.

* Proof.* If , then is rational.

Inductively, suppose that a finite continued fraction with "levels" is a rational number. I want to show that

By induction,

So

This is the sum of two rational numbers, which is rational as well.

I'd like eventually to discuss * infinite continued
fractions* --- things that look like

In preparation for this, I'll look at the effect of
*truncating* a continued fraction.

* Definition.* The * convergent* of the
continued fraction is

Note that for , .

* Example.* Find the first four convergents of
.

And as well.

It is tedious to have to compute convergents by doing the algebra to simplify an enormous fraction. The next result gives an algorithm for computing the convergents of a continued fraction. It's important for theoretical reasons, too --- I'll need it for several of the proofs that follow. For the theorem, I won't assume that the 's are integers, since I will need the general result later on.

* Theorem.* Let , where and . Let

Then the k-th convergent of is .

* Proof.* First, note that

I got this by regarding the last two terms as a single term.

Note also that , ..., and , ..., are the same for these two fractions, since they only differ in the k-th term.

Now I'll start the proof --- it will go by induction on k. For ,

And for ,

Suppose , and assume that result holds through the k-th convergent. Then

Now this is the k-th convergent of a continued fraction, so by induction this is , where and refer to (as opposed to ). But what are the and for this fraction? They're given inductively by

Now , , , are the same for and , as I noted at the start. On the other hand, the k-th term of is . So

(The next to the last equality also follows by induction.) This shows that the result holds for , so the induction step is complete.

* Example.* Compute for .

There is a pattern to the computation of the p's and q's which makes things pretty easy. To get the next p, for instance, multiply the current a by the last p and add the next-to-the-last p.

Notice that in the last example, the convergents oscillate, and that the fractions which give the convergents are always in lowest terms. We'll see that this holds in general.

* Example.* Find for .

Again, notice that the convergents oscillate, and that the fractions for the convergents are always in lowest terms.

I'll prove that the convergent fractions are in lowest terms first.

* Theorem.* Let , where and . Let

Then

* Proof.* I'll induct on k. For ,

Take , and assume the result holds for k. Then

This proves the result for , so the general result is true by induction.

* Corollary.* Let , where and . Let

Then is in lowest terms for .

* Proof.*
implies that .

* Example.* The Golden Ratio has the
*infinite* continued fraction expansion .

(a) Find the first 10 convergents.

(b) Find an integer linear combination of 89 and 55 that is equal to 1.

(a)

(You might have noticed that the p's and q's are the Fibonacci numbers!)

In fact, . In this case, you can see formally that should be . Let

The roots are . Since the fraction is positive, take the positive root to obtain .

(b) Applying the Theorem to the last two rows of the table, I have

Copyright 2022 by Bruce Ikenaga