Periodic Continued Fractions are Quadratic Irrationals

Theorem 1: Let $\theta \in \mathbb{R} \setminus \mathbb{Q}$ with $\theta = \langle a_0; a_1, a_2, ... \rangle$. If $\theta$'s infinite continued fraction is periodic then $\theta$ is a quadratic irrational.

• Proof: Since $\theta$ is periodic the continued fraction of $\theta$ has a repeating part, say:

(1)

\begin{align} \theta = \langle a_0; a_1, a_2, ..., a_r, \overline{a_{r+1}, a_{r+2}, ..., a_n} \rangle \end{align}

• Let $\epsilon = \langle \overline{a_{r+1}; a_{r+2}, ..., a_n} \rangle$. Then:

(2)

\begin{align} \quad \theta &= \langle a_0; a_1, a_2, ..., a_r, \epsilon \rangle \\ &= \frac{\epsilon h_r + h_{r-1}}{\epsilon k_r + k_{r-1}} \quad (*) \end{align}

• Now also observe that:

(3)

\begin{align} \quad \theta &= \langle a_0; a_1, a_2, ..., a_r, a_{r+1}, a_{r+2}, ..., a_n, \epsilon \rangle \\ &= \frac{\epsilon h_n + h_{n-1}}{\epsilon k_n + k_{n-1}} \quad (**) \end{align}

• The equation at $(*)$ implies that:

(4)

\begin{align} \quad \theta [\epsilon k_r + k_{r-1}] &= \epsilon h_r + h_{r-1} \\ \theta \epsilon k_r + \theta k_{r-1} &= \epsilon h_r + h_{r-1} \\ \theta \epsilon k_r - \epsilon h_r &= h_{r-1} - \theta k_{r-1} \\ \epsilon ( \theta k_r - h_r) &= h_{r-1} - \theta k_{r-1} \\ \epsilon &= \frac{h_{r-1} - \theta k_{r-1}}{\theta k_r - h_r} \quad (\dagger) \end{align}

• While the equation at $(*)$ implies similarly that:

(5)

\begin{align} \quad \epsilon &= \frac{h_{n-1} - \theta k_{n-1}}{\theta k_n - h_n} \quad (\dagger \dagger) \end{align}

• So from $(\dagger)$ and $(\dagger \dagger)$ we have that:

(6)

\begin{align} \frac{h_{r-1} - \theta k_{r-1}}{\theta k_r - h_r} = \frac{h_{n-1} - \theta k_{n-1}}{\theta k_n - h_n} \end{align}

• Cross multiplying these equations and bringing all the terms to one side gives us:

(7)

\begin{align} [h_{r-1} - \theta k_{r-1}][\theta k_n - h_n] - [h_{n-1} - \theta k_{n-1}][\theta k_r - h_r] = 0 \end{align}

• So let $f(x) = [h_{r-1} - x k_{r-1}][x k_n - h_n] - [h_{n-1} - x k_{n-1}][x k_r - h_r]$. Then $f$ is a quadratic polynomial since the $x^2$ terms do not cancel each other out, and $f(\theta) = 0$. $\blacksquare$