Connecting Greek Ladders and Continued Fractions - Continued Fractions

A continued fraction is a fraction of the form

$$a_1 + \frac{b_1}{a_2 +\frac{b_2}{a_3 + \frac{b_3}{a_4 + \frac{b_4}{\ddots}}}}$$

where $a_i$ and $b_i$ are real or complex numbers. There may be a finite number of terms or infinitely many. Much investigation has been devoted to continued fractions in the case where $b_i = 1$ for all $i$, the $a_i$ are all integers, and $a_i >0$ for $i\ge 2$. Such fractions are called simple continued fractions. We use the shorthand notation $[a_1,a_2,a_3,a_4,\dots]$ to represent the simple continued fraction

$$a_1 + \frac{1}{a_2 +\frac{1}{a_3 + \frac{1}{a_4 + \frac{1}{\ddots}}}}$$

The notation $[a_1,a_2,\dots,a_n]$ represents a finite simple continued fraction whose value can be computed easily. For example, $[1,4,1,2] = \frac{17}{14}.$ This terminology and notation is consistent with that in [5].

In the case of a continued fraction which is not simple we will use the shorthand notation $[a_1; b_1, a_2; b_2, a_3; \dots]$. We define the $n$th convergent of a continued fraction as

$$S_n = [a_1; b_1, a_2; \dots ; b_{n-1},a_n].$$

For example $[1;2,3;3,4]$ represents

$$1 + \frac{2}{3 +\frac{3}{4}} = \frac{23}{15}.$$

We say the continued fraction $[a_1; b_1, a_2; b_2, a_3; \dots]$ converges provided $\lim_{n\rightarrow\infty}{S_n}$ exists. As an example, consider the simple continued fraction $[1,2,2,2,\dots,]$ which corresponds to

$$1 + \frac{1}{2 +\frac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}}$$

In Section 3.2 of [5], the author proves that if this continued fraction converges, then it converges to $\sqrt{2}$. As a brief summary of that work, note that for $n \ge 2$ we have

$$S_n = [1,2,2,\dots,2] \ \ ({\rm with}\ n-1 \ 2{\rm{s}})$$

$$= 1 + \frac{1}{2 +\frac{1}{2 + \frac{1}{2 + \frac{1}{\ddots \frac{1}{2+\frac{1}{2}}}}}}$$

$$= 1 + \frac{1}{1 +\left( 1+ \frac{1}{2 + \frac{1}{2 + \frac{1}{\ddots \frac{1}{2+\frac{1}{2}}}}}\right)}$$

But by definition of the convergents, the expression in parentheses is $S_{n-1}.$ Thus,

$$S_n = 1+\frac{1}{S_{n-1}+1}\quad\quad{\rm{(Equation}}\ 1).$$

If we assume that $\lim_{n \rightarrow \infty} S_n$ exists and is equal to $L$, then letting $n \rightarrow \infty$ yields

$$L=1+\frac{1}{L+1}.$$

Solving for $L$ we see that $L= \sqrt{2}$. Thus, we know that if the simple continued fraction $[1,2,2,2,\dots]$ converges, it converges to $\sqrt{2}$. For a proof that this limit exists, see Section 3.6 of [5]. Thus, the convergents of this continued fraction provide us with a sequence of rational numbers which can be used as more and more accurate approximations to $\sqrt{2}$.

For an algorithm that produces a simple continued fraction converging to a given real number, see Section 3.2 of [5]. Briefly, the algorithm proceeds as follows: let $x$ be a real number. Let

$$a_1=\lfloor x \rfloor\ {\rm and}\ r_1 = x-a_1.$$

For $n \ge 2$,

$$a_n = \bigg\lfloor{\frac{1}{r_{n-1}}}\bigg\rfloor\ {\rm and}\ r_n = \frac{1}{r_{n-1}}-a_n.$$

Then $[a_1,a_2,a_3,\dots]$ is a simple continued fraction that converges to $x$. The reader can check that the simple continued fraction $[1,2,2,2,\dots]$ converging to $\sqrt{2}$ is the result of this algorithm.

Moreover, for a real number $x$, the representation of $x$ as a simple continued fraction is unique. That is, if $x$ is expressed as a simple continued fraction by $[a_1, a_2, a_3, \dots]$ and by $[b_1, b_2, b_3, \dots]$, then $a_k=b_k$ for all $k$. A proof by induction can be found in many number theory texts, such as Theorem 12.16 of [8] and Theorem 15.6 of [2].