Periodic Continued Fractions

Proposition Suppose that \( \alpha \) has an eventually periodic continued fraction expansion. Then \( \alpha \) is a quadratic irrational.

Proof. We first show this when \( \alpha \) has a periodic continued fraction expansion. We then have a \( d \) such that

\[ \alpha = a _ { 0 } + \frac { 1 } { a _ { 1 } + \frac { 1 } { a _ { 2 } + \cdots + \frac { 1 } { a _ { d - 1 } + \frac { 1 } { \alpha } } } } \]

Since \( \alpha_0,\alpha_1, \cdots, \alpha_{d-1} \) are all integers

\[ \alpha=\frac{x\alpha+y}{z\alpha+w} \implies z\alpha^2+(w-x)\alpha-y=0 \]

Since \( \alpha \) is irrational, \( z \neq 0 \). Thus, \( \alpha \) is a quadratic irrational.

If \( \alpha=[\alpha_0,\alpha_1,\ldots,\alpha_m,\alpha_{m+1},\ldots,\alpha_{m+d-1},\alpha_{m+d},\ldots] \), then

\[ \beta=\frac{1}{\frac{1}{\frac{1}{\alpha-\alpha_{0}}-\alpha_{1}}-\cdots-\alpha_{m-2}}-\alpha_{m-1} \]

Clearly \( \beta \) has a periodic continued fraction expansion. So it is quadratic irrational.

Note,

\begin{align*} \beta=\frac{x\alpha+y}{z\alpha+w} &\implies a\left(\frac{x\alpha+y}{z\alpha+w}\right)^2+b\left(\frac{x\alpha+y}{z\alpha+w}\right)+c=0 \\ &\implies a'\alpha^2+b'\alpha+c'=0 \end{align*}

Since \( \alpha \) is irrational, \( a' \neq 0 \). Thus, \( \alpha \) is a quadratic irrational.

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