This section exhibits some algebra with regular continued fractions.
We find two cases, depending on whether the first quotient is even or odd. First, the even case: \[ \newcommand{\sslash}{\mathbin{/\mkern-6mu/}} \newcommand{\tslash}{\mathbin{/\mkern-5mu/}} \newcommand{\uslash}{\mathbin{/\mkern-4mu/}} \begin{array}{ll} 2 \sslash 2a, b, c, \ldots \sslash & = \frac{2}{2a + \frac{1}{b + \cdots}} \\ & = \frac{1}{a + \frac{1}{2b + \frac{2}{\cdots}}} \\ & = \sslash a, 2b + 2\sslash c , d, \ldots \sslash \sslash \end{array} \] Then the odd case: \[ \begin{array}{ll} 2 \sslash 2a + 1, b, c, \ldots \sslash & = \frac{2}{2a + 1 + \frac{1}{b + \uslash c, d, \ldots \uslash}} \\ & = \frac{1}{a + \frac{1}{2} + \frac{1}{2b + 2\uslash c, d, \ldots \uslash}} \\ & = \frac{1}{a + \frac{(b + 1) + \uslash c, d, \ldots \uslash } {2b + 2\uslash c, d, \ldots \uslash}} \\ & = \sslash a, 1, 1 + 2\sslash b - 1, c , d, \ldots \sslash \sslash \end{array} \]
Given \( A = A_0 + \sslash A_1, A_2, \ldots \sslash \) compute \[ B = B_0 + \sslash B_1, B_2, \ldots \sslash = 1 / A \] The result always has the form \[ B_0 + \sslash B_1, \ldots, B_m, A_4, A_5, \ldots \sslash \] but various cases arise, depending on \( A_0 \) and \( A_1 \).
This case reduces easily. \[ B = A_1 + \sslash A_2, A_3, \ldots \sslash = 1 / \sslash A_1, A2, \ldots \sslash \]
In this case, \( A_0 \) is absorbed into the continued fraction. \[ B = \sslash A_0, A_1, A_2, \ldots \sslash = 1 / (A_0 + \sslash A_1, A2, \ldots \sslash) \] The interesting cases arise when \( A_0 < 0 \).
Here, \( A < -1 \) so \( -1 < 1/A < 0 \) making \( B_0 = -1 \). \[\begin{array}{ll} B & = \frac{1}{A_0 + \tslash A_1, A_2, \ldots \tslash} \\ & = -1 + \frac{1} {\frac{A_0 + 1 + \uslash A_1, A_2, \ldots \uslash -1} {A_0 + 1 + \uslash A_1, A_2, \ldots \uslash}} \\ & = -1 + \frac{1}{1 + \frac{1}{(-A_0 - 2) + (1 - \uslash A_1, A_2, \ldots \uslash ) }} \\ & = -1 + \sslash 1, -A_0 - 2, 1, A_1 - 1, A_2, A_3, \ldots \sslash \end{array} \]
Here, \( A = -1 + f \) where \( 0 < f \leq 1/2 \) so \( -2 \leq 1/A < -1 \) making \( B_0 = -2 \). \[ \begin{array}{ll} B & = \frac{1}{-1 + \frac{1} { A_1 + \uslash A_2, A_3, \ldots \uslash}} \\ & = -1 + \frac{1} {-A_1 + 1 - \uslash A_2, A_3, \ldots \uslash} \\ & = -2 + \frac{1} {\frac{ A_1 - 1 + \uslash A_2, A_3, \ldots \uslash} {A_1 - 2 + \uslash A_2, A_3, \ldots \uslash}} \\ & = -2 + \frac{1} {1 + \frac{1}{A_1 - 2 + \uslash A_2, A_3, \ldots \uslash}} \\ & = -2 + \sslash 1, A_1 - 2, A_2, A_3, \ldots \sslash \end{array} \]
Here, \( A = -1 + g \) where \( 1/2 < g \leq 1 \) so \( 1/A < -2 \) making \( B_0 < -2 \). \[ \begin{array}{ll} B & = \frac{1}{-1 + \frac{1} { 1 + \uslash A_2, A_3, \ldots \uslash }} \\ & = \frac{1 + \tslash A_2, A_3, \ldots \tslash} { - \tslash A_2, A_3, \ldots \tslash } \\ & = -1 - A_2 - \sslash A_3, A_4, \ldots \sslash \\ & = -A_2 - 2 + \sslash 1, A_3 - 1, A_4, \ldots \sslash \end{array} \]