# Taking the negative of a continued fraction

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 $y*z$?

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I fixed your LaTeX. Note that you should include entire formulas in between dollar signs, not parts of them: write $z= \langle a_0;a_1,a_2,a_3,\ldots\rangle$. Also, due to html you should not use < and >, rather use $\lt$ and $\gt$. – t.b. Nov 15 '11 at 8:26
Maybe you should give a definition of the continued fraction representation. – Phira Nov 15 '11 at 8:31
Phira is probably hinting to you that "continued fraction" is commonly interpreted to mean "regular continued fraction" and that means that $a_1,a_2,\dots$ must all be positive, so your fraction for $-z$ doesn't qualify. – Gerry Myerson Nov 15 '11 at 12:04

No, your formula for $-z$ is incorrect. Simply write your proposed formula and add it to the continued fraction of $z$; it must evaluate to $0$, which it does not. 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 these notes by van der Poorten.
If continued fraction $$a_0+\frac{1}{\displaystyle a_1+\frac{1}{a_2+\ddots}}$$ converges to $z$, where all $a_k$ and $z$ are complex numbers, then continued fraction $$-a_0+\frac{1}{\displaystyle -a_1+\frac{1}{-a_2+\ddots}}$$ converges to $-z$.