If I have a continued fraction for an irrational number $z= \langle a_0;a_1,a_2,a_3,\ldots\rangle$ it seems that $(-1)*z = \langle-a_0;-a_1,-a_2,-a_3,\ldots\rangle$. Is this true?
In general, if you have the continued fraction representation for $y$ and $z$ can you say something about the continued fraction representation of $y*z$?
Best Answer
While evaluating the continued fraction of the negative coefficients of the continued fraction expansion of $z$ does indeed evaluate to $-z$, your formula $[-a_0; -a_1, \dots, ]$ is not regarded as "the continued fraction" of $-z$, which is usually defined using the Euclidean algorithm resulting in non-negative coefficients after the first. In general, for $z \in \mathbb{R}$ and $z= [a_1; a_2, a_3 \dots]$, then \begin{align} -z= [-a_1-1;1,a_2-1, a_3, \dots], \end{align} where the terms in the ellipses are identical. For example, $\frac{4}{3} = [1,3]$, while $-\frac{4}{3} = [-2,1,2]$. To address your second question, there are formulas to compute the continued fraction expansion of $\frac{az+b}{cz+d}$, where $a, b, c, d \in \mathbb{Z}$, relying only on the continued fraction expansion of $z$ and certain $2 \times 2$ matrices defined using the coefficients. See An Introduction to Continued Fractions by van der Poorten.