Proof. Let the simple continued fraction of $\alpha$ be periodic, so that
$$\alpha=\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{N-1}, \overline{a_{N}, a_{N+1}, \ldots, a_{N+k}}\right]$$

Now let
$$\beta=\left[\overline{a_{N} ; a_{N+1}, \ldots, a_{N+k}}\right]$$

Then
$$\beta=\left[a_{N} ; a_{N+1}, \ldots, a_{N+k}, \beta\right]$$
and from Theorem 10.9, it follows that
$$\beta=\frac{\beta p_{k}+p_{k-1}}{\beta q_{k}+q_{k-1}}$$
where $p_{k} / q_{k}$ and $p_{k-1} / q_{k-1}$ are convergents of $\left[a_{N} ; a_{N+1}, \ldots, a_{N+k}\right]$. Since the simple continued fraction of $\beta$ is infinite, $\beta$ is irrational, and from (10.13) we have
$$q_{k} \beta^{2}+\left(q_{k-1}-p_{k}\right) \beta-p_{k-1}=0$$
so that $\beta$ is a quadratic irrational. Now note that
$$\alpha=\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{N-1}, \beta\right]$$
so that from Theorem 10.9 we have
$$\alpha=\frac{\beta p_{N-1}+p_{N-2}}{\beta q_{N-1}+q_{N-2}}$$
where $p_{N-1} / q_{N-1}$ and $p_{N-2} / q_{N-2}$ are convergents of $\left[a_{0} ; a_{1}, a_{2}, \ldots, a_{N-1}\right]$. Since $\beta$ is a quadratic irrational, Lemma 10.2 tells us that $\alpha$ is also a quadratic irrational (we know that $\alpha$ is irrational because it has an infinite simple continued fraction expansion).