[proofplan] We prove the identity directly from the standard [convergent](/page/Convergent) recurrence and its initial values; no irrationality or positivity property is used beyond that recurrence data. First we compute the determinant for the initial adjacent pair $n=1$. Then we show that passing from the pair indexed by $n-1$ to the pair indexed by $n$ changes the determinant by a factor of $-1$. Iterating this sign reversal from the base value gives the formula $(-1)^{n-1}$ for every $n \geq 1$. [/proofplan]

[step:Compute the first adjacent determinant]Define, for each integer $n \geq 1$, the integer \begin{align*} D_n := \det((q_{n-1},p_{n-1}), (q_n,p_n)) = q_{n-1}p_n - p_{n-1}q_n. \end{align*} The standard initial values for the [convergent](/page/Convergent) recurrence are $p_{-2}=0$, $p_{-1}=1$, $q_{-2}=1$, and $q_{-1}=0$. Using the recurrence at $n=0$ gives \begin{align*} p_0 = a_0p_{-1} + p_{-2} = a_0. \end{align*} The same recurrence gives \begin{align*} q_0 = a_0q_{-1} + q_{-2} = 1. \end{align*} Using the recurrence at $n=1$ gives \begin{align*} p_1 = a_1p_0 + p_{-1} = a_1a_0 + 1. \end{align*} The same recurrence gives \begin{align*} q_1 = a_1q_0 + q_{-1} = a_1. \end{align*} Therefore \begin{align*} D_1 = q_0p_1 - p_0q_1 = 1 \cdot (a_1a_0 + 1) - a_0a_1 = 1 = (-1)^0. \end{align*} Thus the desired identity holds for $n=1$.[/step]

[guided]We first isolate the determinant we want to compute. For every integer $n \geq 1$, define \begin{align*} D_n := \det((q_{n-1},p_{n-1}), (q_n,p_n)). \end{align*} By the definition of the determinant on $\mathbb{Z}^2$, this means \begin{align*} D_n = q_{n-1}p_n - p_{n-1}q_n. \end{align*} The base case is $n=1$, so we need the first two convergents. The standard [convergent](/page/Convergent) recurrence starts from the initial values $p_{-2}=0$, $p_{-1}=1$, $q_{-2}=1$, and $q_{-1}=0$. Therefore \begin{align*} p_0 = a_0p_{-1} + p_{-2} = a_0 \cdot 1 + 0 = a_0. \end{align*} The denominator recurrence gives \begin{align*} q_0 = a_0q_{-1} + q_{-2} = a_0 \cdot 0 + 1 = 1. \end{align*} Applying the recurrence once more gives \begin{align*} p_1 = a_1p_0 + p_{-1} = a_1a_0 + 1. \end{align*} The corresponding denominator is \begin{align*} q_1 = a_1q_0 + q_{-1} = a_1. \end{align*} Substituting these four values into the determinant formula yields \begin{align*} D_1 = q_0p_1 - p_0q_1 = 1 \cdot (a_1a_0 + 1) - a_0a_1 = 1. \end{align*} Since $1 = (-1)^0$, the formula holds for the first adjacent pair.[/guided]

[step:Show that each adjacent determinant is the negative of the previous one]Let $n \geq 2$ be an integer. Using the recurrence relations $p_n = a_np_{n-1} + p_{n-2}$ and $q_n = a_nq_{n-1} + q_{n-2}$, we compute \begin{align*} D_n = q_{n-1}p_n - p_{n-1}q_n. \end{align*} Substituting the two recurrence relations gives \begin{align*} D_n = q_{n-1}(a_np_{n-1} + p_{n-2}) - p_{n-1}(a_nq_{n-1} + q_{n-2}). \end{align*} Expanding both products and cancelling the equal terms $a_nq_{n-1}p_{n-1}$ and $a_np_{n-1}q_{n-1}$ gives \begin{align*} D_n = q_{n-1}p_{n-2} - p_{n-1}q_{n-2}. \end{align*} Reordering the two products inside the determinant expression yields \begin{align*} D_n = -(q_{n-2}p_{n-1} - p_{n-2}q_{n-1}) = -D_{n-1}. \end{align*} Hence $D_n = -D_{n-1}$ for every integer $n \geq 2$.[/step]

[guided]Fix an integer $n \geq 2$. The goal of this step is to compare the determinant built from the adjacent pair indexed by $n$ with the determinant built from the adjacent pair indexed by $n-1$. By definition, \begin{align*} D_n = q_{n-1}p_n - p_{n-1}q_n. \end{align*} The recurrence relations for convergents apply at this index and give $p_n = a_np_{n-1} + p_{n-2}$ and $q_n = a_nq_{n-1} + q_{n-2}$. Substituting both formulas into the determinant gives \begin{align*} D_n = q_{n-1}(a_np_{n-1} + p_{n-2}) - p_{n-1}(a_nq_{n-1} + q_{n-2}). \end{align*} Now expand the two products: \begin{align*} D_n = a_nq_{n-1}p_{n-1} + q_{n-1}p_{n-2} - a_np_{n-1}q_{n-1} - p_{n-1}q_{n-2}. \end{align*} The first and third terms cancel because multiplication in $\mathbb{Z}$ is commutative. Thus \begin{align*} D_n = q_{n-1}p_{n-2} - p_{n-1}q_{n-2}. \end{align*} This is almost $D_{n-1}$, but with the two determinant columns reversed. Since reversing the two columns of a $2 \times 2$ determinant changes its sign, \begin{align*} q_{n-1}p_{n-2} - p_{n-1}q_{n-2} = -(q_{n-2}p_{n-1} - p_{n-2}q_{n-1}). \end{align*} The expression in parentheses is exactly $D_{n-1}$ by the definition of $D_{n-1}$. Therefore \begin{align*} D_n = -D_{n-1}. \end{align*}[/guided]

[step:Iterate the sign reversal to obtain the closed formula] We prove by induction on $n \geq 1$ that $D_n = (-1)^{n-1}$. The base case $n=1$ was proved above. Suppose that $n \geq 2$ and that $D_{n-1} = (-1)^{n-2}$. From the recurrence identity $D_n = -D_{n-1}$, \begin{align*} D_n = -D_{n-1} = -(-1)^{n-2} = (-1)^{n-1}. \end{align*} By induction, $D_n = (-1)^{n-1}$ for every integer $n \geq 1$. Since $D_n$ is exactly \begin{align*} \det((q_{n-1},p_{n-1}), (q_n,p_n)), \end{align*} the theorem follows. [/step]