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Suppose first that a simple continued fraction is eventually periodic. Its periodic tail $\xi$ is fixed by the Möbius transformation represented by the continued-fraction matrix of one full period:ThusThe infinite tail is not rational, since a rational simple continued fraction terminates. Hence $\xi$ is a quadratic irrational, and the finite initial block expresses the original number as another rational Möbius transform of $\xi$. Therefore every eventually periodic simple continued fraction represents a quadratic irrational.

For the purely periodic fraction $x=[7,7,7,\dots ]$,so $x^2-7x-1=0$. Positivity selects

Apply the continued-fraction algorithm to $\sqrt{23}$:The complete quotient then repeats, givingThe convergents through one term before the final $8$ areThe last of these satisfies the Pell equationso a positive solution is

Setand, for $n\ge 1$, defineWriting $M(a)=\left(\begin{array}{cc}a& 1\\ 1& 0\end{array}\right)$, these recurrences are equivalent to the continued-fraction matrix identityThe associated Möbius transformation sends a positive tail $\beta$ to the continued-fraction tail formula

Since $\det M(a_j)=-1$, taking determinants givesMoreover,whereasTheir signs are opposite, so ${\theta }_n$ lies between the two adjacent convergents. For $n=0$, the second endpoint is interpreted as $p_{-1}/q_{-1}=\infty$.

The continued-fraction matrix for the original finite continued fraction isEach $M(a_j)$ is a symmetric matrix, so reversing the order gives the transpose:Reading off the columns proves the reversal identity for a finite continued fraction:These are respectively the $n$th and $(n-1)$th convergents of the reversed continued fraction.