8.3: Continued fractions

Definition: Continued fractions

A simple continued fraction is of the form, denoted by $[a_0,a_1,\ldots]$,

$$a_0 + \frac{1}{a_1+\frac{\displaystyle 1}{\displaystyle a_2+ \ldots}},$$ where $a_0$, $\ a_1$, $\ a_2$, $\ldots$ $\in \mathbb{Z}$. Continued fraction has been studied extensively, but we will only explore some of them in this class.

Example $\PageIndex{1}$:

A simple finite continued fraction

$$\frac{1}{2}=[1,1]=0+\frac{1}{1+\frac{1}{1}}$$

A simple infinite continued fraction: Golden Ratio

$$\phi =\frac{1+\sqrt{5}}{2}=[1,1,\ldots]=1+\frac{1}{1+\frac{1}{1+\ldots}},$$ which can be found using $x=1+\dfrac{1}{1+x}$.

 

 

Note

Rational numbers have a simple finite continued fraction, and irrational numbers have an infinite continued fraction.

 Writing Continued Fractions

Let’s explore the process of writing continued fractions using two examples:

Example $\PageIndex{2}$

Starting with the fraction

$$\frac{47}{17} = 2 + \frac{13}{17}$$

This expression indicates that

$$\frac{47}{17} \text{ is equivalent to the whole number } 2 \text{ plus a fraction } \frac{13}{17}.$$

Next, we can express it as

$$\frac{47}{17} = 2 + \frac{1}{\frac{17}{13}}$$

Here, we’ve written

$$\frac{47}{17} \text{ as the whole number } 2 \text{ plus the reciprocal of the fraction } \frac{17}{13}.$$

Continuing, we have

$$\frac{47}{17} = 2 + \frac{1}{1 + \frac{4}{13}}$$

Now, the continued fraction form includes the whole number $2$ and a fraction where the denominator is the sum of $1$ and $\frac{4}{13}$.

$$\frac{47}{17} = 2 + \frac{1}{1 + \frac{1}{13}}$$ Here, we express $$\frac{47}{17} \text{ as } 2 \text{ plus a fraction where the denominator itself is a continued fraction.}$$ We can further express it as $$\frac{47}{17} = 2 + \frac{1}{1 + \frac{1}{3 + \frac{1}{4}}} \tag{9}$$ The final form illustrates the continued fraction expansion of $\frac{47}{17}$ with multiple levels of nesting.

Example $\PageIndex{3}$

Now, let’s consider the square root of 2:

$$\sqrt{2} = 1 + (\sqrt{2} - 1) \tag{10}$$ Here, we expressed that the square root of 2 equals $1 + (\sqrt{2} - 1)$. Next, we write: $$\sqrt{2} = 1 + \frac{1}{1 + \sqrt{2}} \tag{11}$$ In this form, the square root of 2 is written as 1 plus the reciprocal of the quantity $1 + \sqrt{2}$. Now replace $\sqrt{2}$ with what we proved it to equal in equation (11): $$\sqrt{2} = 1 + \frac{1}{1 + 1 + \frac{1}{1 + \sqrt{2}}} \tag{12}$$ Simplifying gives us: $$\sqrt{2} = 1 + \frac{1}{2 + \frac{1}{1 + \sqrt{2}}} \tag{13}$$ Continue with replacing $\sqrt{2}$ with what we proved it to equal in equation (11): $$\sqrt{2} = 1 + \frac{1}{2 + \frac{1}{1 + 1 + \frac{1}{1 + \sqrt{2}}}} \tag{14}$$ This leads to: $$\sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{1 + \sqrt{2}}}} \tag{15}$$ Finally, we have: $$\sqrt{2} = 1 + \frac{1}{2 + \frac{1}{2 + \frac{1}{\ldots}}} \tag{16}$$ This final form illustrates the continued fraction expansion of the square root of 2 with repeated nesting. The ellipsis ($\ldots$) represents an infinite continuation of the pattern.

Example $\PageIndex{4}$

Using the Euclidean algorithm to find a simple finite continued fraction

Let's explore the following example:

Consider $\dfrac{2520}{154}$.

By Euclidean algorithm we have,

$2520=({16})(154)+56$

$154=(2)(56)+42$

$56=(1)(42)+14$

$42=(3)(14)+0$.

The quotients give us the simple finite continued fraction $[16, 2, 1, 3]$. That is

$$\frac{2520}{154}= 16 + \frac{1}{2+ \frac{1}{1+ \frac{1}{3}}}.$$

Continued fractions are more structured and easier to manipulate for certain mathematical operations, especially in number theory problems. While familiar, decimals can be challenging for analysis and computation.