Periodic Continued Fractions

Proposition Suppose that $α$ has an eventually periodic continued fraction expansion. Then $α$ is a quadratic irrational.

Proof. We first show this when $α$ has a periodic continued fraction expansion. We then have a $d$ such that

α = a_{0} + \frac{1}{a_{1} + \frac{1}{a_{2} + \dots + \frac{1}{a_{d - 1} + \frac{1}{α}}}}

Since $α_{0}, α_{1}, \dots, α_{d - 1}$ are all integers

α = \frac{x α + y}{z α + w} ⟹ z α^{2} + (w - x) α - y = 0

Since $α$ is irrational, $z \neq 0$ . Thus, $α$ is a quadratic irrational.

If $α = [α_{0}, α_{1}, \dots, α_{m}, α_{m + 1}, \dots, α_{m + d - 1}, α_{m + d}, \dots]$ , then

β = \frac{1}{\frac{1}{\frac{1}{α - α_{0}} - α_{1}} - \dots - α_{m - 2}} - α_{m - 1}

Clearly $β$ has a periodic continued fraction expansion. So it is quadratic irrational.

Note,

\begin{aligned} β = \frac{x α + y}{z α + w} & ⟹ a {(\frac{x α + y}{z α + w})}^{2} + b (\frac{x α + y}{z α + w}) + c = 0 \\ ⟹ a^{'} α^{2} + b^{'} α + c^{'} = 0 \end{aligned}

Since $α$ is irrational, $a^{'} \neq 0$ . Thus, $α$ is a quadratic irrational.

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