If is an infinite continued fraction, I want to define its value to be the limit of the convergents:
For this to make sense, I need to show that this limit exists.
In what follows, take as given an infinite continued fraction . Let
Lemma.
(a) .
(b) .
Proof. For (a),
For (b),
Lemma.
That is, the odd convergents get smaller, the even convergents get bigger, and any odd convergent is bigger than any even convergent.
Proof. If k is even, then
This shows that the even terms get bigger.
If k is odd, then
This shows that the odd terms get smaller.
Finally, I have to show any odd term is bigger than any even term. Note that it's true if the terms are adjacent:
Next, do the general case. Let be an odd term and let be an even term. Then
The first inequality is true because odd terms decrease. The second inequality is true because I just observed that an odd term is bigger than an adjacent even. Finally, the last inequality is true because even terms increase.
I've shown that any odd term is bigger than any even term, and that completes the proof.
Example. The first ten convergents for the golden ratio are:
The odd convergents get smaller, the even convergents get bigger, and any odd convergent is bigger than any even convergent.
Lemma. for all .
Proof. I'll induct on k. , so the result holds for . Take , and assume it holds for numbers . I'll prove that it holds for .
(The last inequality used .) This completes the induction step.
Theorem. Let be an infinite continued fraction with for , and let be the k-th convergent. Then
Proof. Consider the sequence of odd convergents
This is a decreasing sequence of numbers, and it's bounded below --- by any even convergent, for example. A standard result from analysis (see, for example, Theorem 3.14 of [1]) asserts that such a sequence must have a limit, so
Likewise, consider the sequence of even convergents:
This is an increasing sequence of numbers that's bounded above --- by any odd convergent, for example. The result from analysis mentioned above says that the sequence has a limit:
I have to show that the two limits agree.
The previous lemma implies that and , so
Now let . , so by the Squeezing Theorem of calculus,
Since the odd and even terms approach the same limit, exists.
Knowing this, I'm justified in defining
What can I say about its value?
Theorem. Let be an infinite continued fraction with for . Then is irrational.
Proof. Write for short. I want to show that x is irrational. Suppose on the contrary that , where p and q are integers. I will show this leads to a contradiction.
Since the odd convergents are bigger than x and the even convergents are smaller than x,
Then
Notice that this inequality is true for all k, and that the junk in the middle is an integer. But q is fixed, and , so if I make k sufficiently large eventually will become bigger than q. Then will be a fraction less than 1, and I have an integer caught between 0 and a fraction less than 1. Since this is impossible, x can't be rational.
Now I know that every infinite continued fraction made of positive integers represents an irrational number. The converse is also true true, and the next result gives an algorithm for computing the continued fraction expansion.
Theorem. Let be irrational. Let , and
Then
Proof.
Step 1. is irrational for .
Since x is irrational and , the result is true for .
Assume that and that the result is true for . I want to show that is irrational.
Suppose on the contrary that , where . Then
Now all the 's are clearly integers (since means they're outputs of the greatest integer function), so is the sum of an integer and a rational number. Therefore, it's rational, so is rational, contrary to the induction hypothesis.
It follows that is irrational. By induction, is irrational for all .
Step 2. The 's are positive integers for .
I already observed that the 's are integers.
Let . Since , the definition of the greatest integer function gives
But is irrational, so . Hence,
Since , this proves that the 's are positive integers for .
Step 3.
First, I'll get a formula for x in terms of the p's, q's, and a's.
Then I'll find and show that it's less than something which goes to 0.
To get the formula for x, start with
Do some algebra to get
Write out this equation for a few values of k:
Substituting the second equation of the set into the first gives
Substituting into this equation gives
Substituting into this equation gives
You get the idea. In general,
In other words, the 's are the "infinite tails" of the continued fraction.
Recall the recursion formulas for convergents:
The right sides only involve terms up to (and p's and q's of smaller indices). Therefore, the fractions
have the same p's and q's through index k.
Using the recursion formula for convergents, I get
Therefore,
Take absolute values:
Now
Therefore,
By an earlier lemma, and , so
Now , so by the Squeezing Theorem
This implies that
Example. I'll compute the continued fraction expansion of . Here are the first two steps:
Continuing in this way, I obtain:
Example. I'll compute the continued fraction expansion of .
Here are the first 5 terms:
In fact, the continued fraction expansion for is .
Theorem. The continued fraction expansion of an irrational number is unique.
Proof. Suppose
are two continued fractions for the irrational number x, where and for . I want to show that for all k.
Recall that:
Therefore,
Now
Thus, is an integer less than x, and the next larger integer is greater than x. This means that .
The same reasoning applies to the b's. Therefore, , so .
Hence,
I can continue in the same way to show that for all k.
Here's a summary of some of the important results on infinite continued fractions:
1. An irrational number has a unique infinite continued fraction expansion.
2. The algorithm for computing the continued fraction expansion of an irrational number x is:
Then
3. If is the continued fraction expansion of an irrational number, then is a positive integer for .
4. If is the continued fraction expansion of an irrational number and and are defined by the recursion formulas for convergents, then
Example. Here is the continued fraction expansion for .
Now
Thus, in this case,
Copyright 2010 by Bruce Ikenaga