[proofplan] We first recall the standard recursive definition of convergent numerators and denominators for a [simple continued fraction](/page/Simple%20Continued%20Fraction): the integrality condition $a_0 \in \mathbb{Z}$ and positivity condition $a_i \in \mathbb{N}$ for $i \geq 1$ are part of that definition, while the algebra below only uses the displayed recurrence relations. We then prove the matrix identity by induction on $n$: the base case is the first convergent, and the inductive step is right multiplication by $M(a_n)$. The determinant identity follows by taking determinants of the product formula and computing that each matrix $M(a_i)$ has determinant $-1$. [/proofplan]

[step:Recall the convergent recurrence supplied by the continued fraction definition] For each integer $k \geq 0$, let $C_k \in \mathbb{R}$ denote the finite [simple continued fraction](/page/Simple%20Continued%20Fraction) value $[a_0;a_1,\dots,a_k]$. The standard definition of the $k$th [convergent](/page/Convergent) writes \begin{align*} C_k=\frac{p_k}{q_k}, \end{align*} where the continuants $p_k$ and $q_k$ are defined by the initial values $p_{-2}=0$, $p_{-1}=1$, $q_{-2}=1$, $q_{-1}=0$ and, for every integer $k \geq 0$, by the recurrence relations \begin{align*} p_k=a_kp_{k-1}+p_{k-2}, \qquad q_k=a_kq_{k-1}+q_{k-2}. \end{align*} The conditions $a_0 \in \mathbb{Z}$ and $a_i \in \mathbb{N}$ for $i \geq 1$ are the standard hypotheses under which the displayed expression is a simple continued fraction; the matrix computation below only uses the real scalars $a_i$ and the recurrence relations just stated. [/step]

[step:Verify the matrix identity for the first convergent] Define the auxiliary map $A: \mathbb{R}^4 \to \mathbb{R}^{2 \times 2}$ by declaring $A(r,s,t,u)$ to be the real matrix whose first row is $(r,s)$ and whose second row is $(t,u)$. With the map $M: \mathbb{R} \to \mathbb{R}^{2 \times 2}$ from the theorem statement, we have $M(a_i)=A(a_i,1,1,0)$ for every allowed index $i$. For $n = 0$, the recurrence relations give \begin{align*} p_0 = a_0p_{-1} + p_{-2} = a_0. \end{align*} They also give \begin{align*} q_0 = a_0q_{-1} + q_{-2} = 1. \end{align*} Therefore \begin{align*} M(a_0)=A(a_0,1,1,0)=A(p_0,p_{-1},q_0,q_{-1}). \end{align*} This proves the formula for $n = 0$. [/step]

[step:Propagate the matrix formula through the recurrence relations]Assume that $n \geq 1$ and that the matrix identity holds for $n-1$, namely \begin{align*} M(a_0)M(a_1)\cdots M(a_{n-1})=A(p_{n-1},p_{n-2},q_{n-1},q_{n-2}). \end{align*} Multiplying both sides on the right by $M(a_n)=A(a_n,1,1,0)$ gives \begin{align*} M(a_0)M(a_1)\cdots M(a_n)=A(a_n p_{n-1}+p_{n-2},p_{n-1},a_n q_{n-1}+q_{n-2},q_{n-1}). \end{align*} By the defining recurrence relations, \begin{align*} a_n p_{n-1} + p_{n-2} = p_n. \end{align*} Also, \begin{align*} a_n q_{n-1} + q_{n-2} = q_n. \end{align*} Hence \begin{align*} M(a_0)M(a_1)\cdots M(a_n)=A(p_n,p_{n-1},q_n,q_{n-1}). \end{align*} By induction, the matrix identity holds for every allowed integer $n$.[/step]

[guided]The induction hypothesis says that, after multiplying the first $n$ matrices up to $M(a_{n-1})$, the two columns store two consecutive convergents: \begin{align*} M(a_0)M(a_1)\cdots M(a_{n-1})=A(p_{n-1},p_{n-2},q_{n-1},q_{n-2}). \end{align*} The next partial quotient $a_n$ should transform the pair of previous numerators and denominators according to the recurrence relations. This is exactly what right multiplication by $M(a_n)=A(a_n,1,1,0)$ does. The ordinary rule for multiplying $2 \times 2$ matrices gives \begin{align*} A(p_{n-1},p_{n-2},q_{n-1},q_{n-2})A(a_n,1,1,0)=A(p_{n-1}a_n+p_{n-2},p_{n-1},q_{n-1}a_n+q_{n-2},q_{n-1}). \end{align*} Thus \begin{align*} M(a_0)M(a_1)\cdots M(a_n)=A(a_n p_{n-1}+p_{n-2},p_{n-1},a_n q_{n-1}+q_{n-2},q_{n-1}). \end{align*} The first column is now the new numerator-denominator pair because the convergent recurrences define \begin{align*} p_n = a_n p_{n-1} + p_{n-2}. \end{align*} They also define \begin{align*} q_n = a_n q_{n-1} + q_{n-2}. \end{align*} The second column is the old numerator-denominator pair $(p_{n-1}, q_{n-1})$. Therefore \begin{align*} M(a_0)M(a_1)\cdots M(a_n)=A(p_n,p_{n-1},q_n,q_{n-1}). \end{align*} Since the formula was already verified for $n=0$, this proves the formula for every allowed $n$ by induction.[/guided]

[step:Take determinants to obtain the alternating determinant identity] For every integer $i$ with $0 \leq i \leq n$, \begin{align*} \det M(a_i)=\det A(a_i,1,1,0)=a_i \cdot 0 - 1 \cdot 1=-1. \end{align*} Taking determinants in the matrix identity gives \begin{align*} \det\left(M(a_0)M(a_1)\cdots M(a_n)\right)=\det A(p_n,p_{n-1},q_n,q_{n-1}). \end{align*} We compute the left-hand side directly for this product. For each integer $j$ with $0 \leq j \leq n$, define $B_j:=M(a_0)M(a_1)\cdots M(a_j)$. We have $\det B_0=\det M(a_0)=-1$. If $1 \leq j \leq n$ and $B_{j-1}=A(r,s,t,u)$ for some scalars $r,s,t,u$, then \begin{align*} B_j=A(r,s,t,u)A(a_j,1,1,0)=A(ra_j+s,r,ta_j+u,t). \end{align*} Therefore \begin{align*} \det B_j=(ra_j+s)t-r(ta_j+u)=st-ru=-(ru-st)=-\det B_{j-1}. \end{align*} By induction on $j$, this gives \begin{align*} \det\left(M(a_0)M(a_1)\cdots M(a_n)\right)=\det B_n=(-1)^{n+1}. \end{align*} The determinant of the right-hand side is \begin{align*} \det A(p_n,p_{n-1},q_n,q_{n-1})=p_n q_{n-1} - p_{n-1} q_n. \end{align*} Thus \begin{align*} p_n q_{n-1} - p_{n-1} q_n = (-1)^{n+1}. \end{align*} Since $(-1)^{n+1} = (-1)^{n-1}$ for every integer $n$, we obtain \begin{align*} p_n q_{n-1} - p_{n-1} q_n = (-1)^{n-1}. \end{align*} This is the claimed determinant identity. [/step]