Infinite Continued Fractions and Irrational Numbers

Recall from the Convergence of Infinite Continued Fractions page that if $\langle a_0; a_1, a_2, ... \rangle$ is an infinite simple continued fraction with $a_0 \in \mathbb{Z}$ and $a_n \in \mathbb{N}$ for $n \geq 1$ then $\langle a_0; a_1, a_2, ... \rangle = \lim_{n \to \infty} r_n$ exists. We will now show that the limit must be an irrational number.

Theorem 1: Let $\langle a_0; a_1, a_2, ... \rangle$ be an infinite simple continued fraction with $a_0 \in \mathbb{Z}$ and $a_n \in \mathbb{N}$ for $n \geq 1$. Then $\langle a_0; a_1, a_2, ... \rangle$ represents an irrational number.

• Proof: Let $\displaystyle{x = \lim_{n \to \infty} r_n}$ and suppose instead that $x \in \mathbb{Q}$. Then there exists $a, b \in \mathbb{Z}$ with $b \neq 0$ for which $x = \frac{a}{b}$. Recall that for all $j$ we have that:

(1)

\begin{align} \quad r_{2j} < x < r_{2j+1} \end{align}

• So for all $j$ we have that:

(2)

\begin{align} \quad 0 < x - r_{2j} < r_{2j+1} - r_{2j} = \frac{(-1)^{2j}}{k_{2j+1}k_{2j}} = \frac{1}{k_{2j+1}k_{2j}} \quad (*) \end{align}

• But also observe that that since $x \neq r_{2j}$ for each $j$ (since $r_{2j} < x$ for all $j$) we have that $ak_{2j} - bh_{2j} \geq 1$. So:

(3)

\begin{align} x - r_{2j} = \frac{a}{b} - \frac{h_{2j}}{k_{2j}} = \frac{ak_{2j} - bh_{2j}}{bk_{2j}} \geq \frac{1}{bk_{2j}} \quad (**) \end{align}

• So from $(*)$ and $(**)$ we have that for all $j$:

(4)

\begin{align} \quad \frac{1}{bk_{2j}} < \frac{1}{k_{2j+1}k_{2j}} \end{align}

• So $\displaystyle{\frac{1}{b} < \frac{1}{k_{2j+1}}}$ for all $j$ which implies that $b > k_{2j+1}$ for all $j$. This is a contradiction since $b$ is fixed and $k_{2j+1} \to \infty$ as $j \to \infty$. Hence the assumption that $x$ is rational is false. So $x$ is irrational, that is, $\langle a_0; a_1, a_2, ... \rangle$ is an irrational number. $\blacksquare$

Now suppose that $x$ is an irrational number. We can construct an infinite simple continued fraction for $\epsilon_0$ as follows.

Let $x_0 = \left \lfloor \epsilon_0 \right \rfloor$. Then $0 < x - \left \lfloor x \right \rfloor < 1$. So $\epsilon_1 = \frac{1}{x - \left \lfloor x \right \rfloor} > 1$. Then let $x_1 = \left \lfloor \epsilon_1 \right \rfloor$. Then $0 < \epsilon_1 - \left \lfloor \epsilon_1 \right \rfloor < 1$ so $\epsilon_2 = \frac{1}{\epsilon_1 - \left \lfloor \epsilon_1 \right \rfloor} > 1$. So let $x_2 = \left \lfloor \epsilon_2 \right \rfloor$. We continue this process indefinitely to obtain an infinite simple continued fraction $\langle x_0; x_1, x_2, ... \rangle$ for $\epsilon_0$. Hence:

Proposition 2: Every irrational number can be represented by an infinite simple continued fraction.

For an example of this process, consider $\epsilon_0 = \pi$. We have that:

(5)

\begin{align} \quad x_0 = \left \lfloor \epsilon_0 \right \rfloor = \left \lfloor \pi \right \rfloor = 3 \\ \end{align}

Therefore $\epsilon_1 = \frac{1}{\epsilon_0 - \left \lfloor \epsilon_0 \right \rfloor} = \frac{1}{\pi - 3} \approx 7.0625...$. So:

(6)

\begin{align} \quad x_1 = \left \lfloor \epsilon_1 \right \rfloor = 7 \end{align}

Therefore $\epsilon_2 = \frac{1}{\epsilon_1 - \left \lfloor \epsilon_1 \right \rfloor} \approx 15.99...$. So:

(7)

\begin{align} \quad x_2 = \left \lfloor \epsilon_2 \right \rfloor = 15 \end{align}

So the first few terms of an infinite simple continued fraction for $\pi$ are $\pi = \langle 3; 7, 15, ... \rangle$.