[proofplan] We encode the recurrence by multiplying the standard $2 \times 2$ matrices associated to the partial quotients. The product records the current and previous numerators and denominators, so the continued fraction identity follows by applying the corresponding fractional linear transformation to $\infty$, equivalently by reading off the first column. Positivity of the denominators follows directly from the recurrence and the convention $\mathbb{N}=\{1,2,3,\dots\}$, so each partial quotient $a_i$ with $i \geq 1$ is positive. The determinant identity follows because each matrix has determinant $-1$. [/proofplan]

[step:Encode the recurrence by a matrix product]The theorem statement defines the integer sequences $(p_n)_{n\geq -2}$ and $(q_n)_{n\geq -2}$ by the initial values \begin{align*} p_{-2}=0,\qquad p_{-1}=1,\qquad q_{-2}=1,\qquad q_{-1}=0, \end{align*} and by the recurrence relations \begin{align*} p_n=a_np_{n-1}+p_{n-2},\qquad q_n=a_nq_{n-1}+q_{n-2} \end{align*} for every integer $n\geq 0$. We will use these initial values and recurrence relations whenever we refer to the defining recurrences for the convergents. For each integer $i \geq 0$, define the matrix $A_i \in M_2(\mathbb{Z})$ by declaring its entries as follows: \begin{align*} (A_i)_{11}=a_i. \end{align*} \begin{align*} (A_i)_{12}=1. \end{align*} \begin{align*} (A_i)_{21}=1. \end{align*} \begin{align*} (A_i)_{22}=0. \end{align*} For each $n \geq 0$, define \begin{align*} P_n := A_0 A_1 \cdots A_n \in M_2(\mathbb{Z}). \end{align*} We prove by induction on $n \geq 0$ that the entries of $P_n$ are given by the following four identities: \begin{align*} (P_n)_{11}=p_n. \end{align*} \begin{align*} (P_n)_{12}=p_{n-1}. \end{align*} \begin{align*} (P_n)_{21}=q_n. \end{align*} \begin{align*} (P_n)_{22}=q_{n-1}. \end{align*} For $n = 0$, we have $P_0=A_0$. The entry identities are \begin{align*} (P_0)_{11}=a_0=p_0. \end{align*} \begin{align*} (P_0)_{12}=1=p_{-1}. \end{align*} \begin{align*} (P_0)_{21}=1=q_0. \end{align*} \begin{align*} (P_0)_{22}=0=q_{-1}. \end{align*} Here $p_0 = a_0p_{-1}+p_{-2}=a_0$, $q_0=a_0q_{-1}+q_{-2}=1$, $p_{-1}=1$, and $q_{-1}=0$. Assume the identity holds for some $n-1 \geq 0$. Since $P_n=P_{n-1}A_n$, ordinary matrix multiplication gives \begin{align*} (P_n)_{11}=p_{n-1}a_n+p_{n-2}=p_n. \end{align*} \begin{align*} (P_n)_{12}=p_{n-1}. \end{align*} \begin{align*} (P_n)_{21}=q_{n-1}a_n+q_{n-2}=q_n. \end{align*} \begin{align*} (P_n)_{22}=q_{n-1}. \end{align*} The two equalities involving $p_n$ and $q_n$ are exactly the defining recurrences. This completes the induction.[/step]

[guided]The theorem statement defines the integer sequences $(p_n)_{n\geq -2}$ and $(q_n)_{n\geq -2}$ by \begin{align*} p_{-2}=0,\qquad p_{-1}=1,\qquad q_{-2}=1,\qquad q_{-1}=0, \end{align*} and by \begin{align*} p_n=a_np_{n-1}+p_{n-2},\qquad q_n=a_nq_{n-1}+q_{n-2} \end{align*} for every integer $n\geq 0$. The matrix method is a compact way to store these two second-order recurrences at once: one column records the current values, and the other column records the previous values. The recurrence has exactly the same shape as multiplication by the matrix $A_i \in M_2(\mathbb{Z})$ whose entries are \begin{align*} (A_i)_{11}=a_i. \end{align*} \begin{align*} (A_i)_{12}=1. \end{align*} \begin{align*} (A_i)_{21}=1. \end{align*} \begin{align*} (A_i)_{22}=0. \end{align*} Indeed, right multiplication by $A_i$ replaces a pair consisting of a current term and a previous term by the next term and the current term. This is why the product $P_n := A_0A_1\cdots A_n$ should contain both the $n$th and the $(n-1)$st convergent data. We prove the precise entry formula for $P_n$ by induction. The formula consists of the four identities \begin{align*} (P_n)_{11}=p_n. \end{align*} \begin{align*} (P_n)_{12}=p_{n-1}. \end{align*} \begin{align*} (P_n)_{21}=q_n. \end{align*} \begin{align*} (P_n)_{22}=q_{n-1}. \end{align*} For $n=0$, the initial recurrence gives \begin{align*} p_0=a_0p_{-1}+p_{-2}=a_0\cdot 1+0=a_0. \end{align*} \begin{align*} q_0=a_0q_{-1}+q_{-2}=a_0\cdot 0+1=1. \end{align*} Since $P_0=A_0$, its four entries are $(P_0)_{11}=a_0$, $(P_0)_{12}=1$, $(P_0)_{21}=1$, and $(P_0)_{22}=0$. These are exactly $p_0$, $p_{-1}$, $q_0$, and $q_{-1}$, respectively. Now assume the formula is known for $P_{n-1}$, where $n \geq 1$. Multiplying by the next matrix gives $P_n=P_{n-1}A_n$. By the row-column rule for matrix multiplication, \begin{align*} (P_n)_{11}=p_{n-1}a_n+p_{n-2}=p_n. \end{align*} \begin{align*} (P_n)_{12}=p_{n-1}. \end{align*} \begin{align*} (P_n)_{21}=q_{n-1}a_n+q_{n-2}=q_n. \end{align*} \begin{align*} (P_n)_{22}=q_{n-1}. \end{align*} The two entries in the first column are exactly $p_n$ and $q_n$ by the recurrence, while the second column shifts the previous values forward. Therefore the asserted entry formula holds for $P_n$.[/guided]

[step:Recover the finite continued fraction from the matrix product]For a real number $t > 0$ and an integer $i \geq 0$, define the fractional [linear map](/page/Linear%20Map) $T_i:(0,\infty)\to\mathbb{R}$ by \begin{align*} T_i(t)=a_i+\frac{1}{t}=\frac{a_i t+1}{t}. \end{align*} For every $i\geq 1$, the hypothesis $a_i\in\mathbb{N}$ gives $a_i>0$, so $T_i(t)>0$ whenever $t>0$. Thus the maps $T_1,\dots,T_{n-1}$ send positive inputs to positive outputs; no positivity of $T_0(t)$ is required because $T_0$ is applied only as the final map. For $n \geq 1$, the finite continued fraction satisfies \begin{align*} [a_0;a_1,\dots,a_n]=T_0\bigl(T_1(\cdots T_{n-1}(a_n)\cdots)\bigr). \end{align*} Since $a_n>0$ and each intermediate output under $T_i$ with $i\geq 1$ remains positive, all denominators occurring in this expression are positive, so the expression is defined. We next record the composition formula. For every $n\geq 0$ and every positive input $t>0$ for which the nested expression $T_0(T_1(\cdots T_n(t)\cdots))$ is defined, the denominator in the rational expression below is nonzero and \begin{align*} T_0\bigl(T_1(\cdots T_n(t)\cdots)\bigr)=\frac{p_n t+p_{n-1}}{q_n t+q_{n-1}}. \end{align*} This follows by induction on $n$. For $n=0$, the formula is the identity $T_0(t)=(p_0t+p_{-1})/(q_0t+q_{-1})$. For the induction step, assume \begin{align*} T_0\bigl(T_1(\cdots T_{n-1}(s)\cdots)\bigr)=\frac{p_{n-1}s+p_{n-2}}{q_{n-1}s+q_{n-2}} \end{align*} for every admissible positive input $s$. Substitute \begin{align*} s=T_n(t)=\frac{a_n t+1}{t}. \end{align*} Then \begin{align*} T_0\bigl(T_1(\cdots T_n(t)\cdots)\bigr) =\frac{p_{n-1}(a_n t+1)+p_{n-2}t}{q_{n-1}(a_n t+1)+q_{n-2}t} =\frac{(a_np_{n-1}+p_{n-2})t+p_{n-1}}{(a_nq_{n-1}+q_{n-2})t+q_{n-1}}. \end{align*} Using the defining recurrences for $p_n$ and $q_n$, this becomes \begin{align*} T_0\bigl(T_1(\cdots T_n(t)\cdots)\bigr)=\frac{p_n t+p_{n-1}}{q_n t+q_{n-1}}. \end{align*} For $n\geq 1$, apply this formula to the positive input $t=a_n$ after stopping the composition at $T_{n-1}$. This input is admissible because $a_n>0$ and, as checked above, each intermediate tail obtained by applying $T_i$ with $i\geq 1$ remains positive. Hence the denominator in the displayed rational expression is nonzero. The preceding formula with index $n-1$ gives \begin{align*} [a_0;a_1,\dots,a_n]=\frac{p_{n-1}a_n+p_{n-2}}{q_{n-1}a_n+q_{n-2}}. \end{align*} By the defining recurrences, the numerator is $p_n$ and the denominator is $q_n$. Hence \begin{align*} [a_0;a_1,\dots,a_n]=\frac{p_n}{q_n}. \end{align*} For $n=0$, the same identity is immediate: \begin{align*} [a_0]=a_0=\frac{p_0}{q_0}. \end{align*}[/step]

[guided]We now connect the matrix identity to the actual continued fraction. For each integer $i\geq 0$, the map $T_i:(0,\infty)\to\mathbb{R}$ is defined by \begin{align*} T_i(t)=a_i+\frac{1}{t}=\frac{a_i t+1}{t}. \end{align*} The codomain is only $\mathbb{R}$ because $a_0$ may be negative. This causes no problem: $T_0$ is applied last, while the maps applied before it are $T_i$ with $i\geq 1$. For those indices, $a_i\in\mathbb{N}$ gives $a_i>0$, and hence $T_i(t)>0$ whenever $t>0$. For $n\geq 1$, the recursive definition of the finite continued fraction gives \begin{align*} [a_0;a_1,\dots,a_n]=T_0\bigl(T_1(\cdots T_{n-1}(a_n)\cdots)\bigr). \end{align*} The input $a_n$ is positive, and every intermediate map $T_i$ with $i\geq 1$ preserves positivity, so all reciprocal operations in the expression are defined. The product matrix $P_n$ represents the composition of these fractional linear maps. More precisely, for every $n\geq 0$ and every positive input $t>0$ for which the nested expression $T_0(T_1(\cdots T_n(t)\cdots))$ is defined, the denominator in the rational expression below is nonzero and \begin{align*} T_0\bigl(T_1(\cdots T_n(t)\cdots)\bigr)=\frac{p_n t+p_{n-1}}{q_n t+q_{n-1}}. \end{align*} This is proved by induction on $n$. For $n=0$, the formula is \begin{align*} T_0(t)=a_0+\frac{1}{t}=\frac{a_0t+1}{t}=\frac{p_0t+p_{-1}}{q_0t+q_{-1}}, \end{align*} using $p_0=a_0$, $p_{-1}=1$, $q_0=1$, and $q_{-1}=0$. Now assume the formula holds with index $n-1$. For an admissible positive input $t$, define \begin{align*} s=T_n(t)=\frac{a_nt+1}{t}. \end{align*} Substituting this value of $s$ into the induction hypothesis gives \begin{align*} T_0\bigl(T_1(\cdots T_n(t)\cdots)\bigr) =\frac{p_{n-1}s+p_{n-2}}{q_{n-1}s+q_{n-2}} =\frac{p_{n-1}\frac{a_nt+1}{t}+p_{n-2}}{q_{n-1}\frac{a_nt+1}{t}+q_{n-2}}. \end{align*} Multiplying numerator and denominator by the positive number $t$ gives \begin{align*} T_0\bigl(T_1(\cdots T_n(t)\cdots)\bigr) =\frac{(a_np_{n-1}+p_{n-2})t+p_{n-1}}{(a_nq_{n-1}+q_{n-2})t+q_{n-1}}. \end{align*} The recurrence relations identify $a_np_{n-1}+p_{n-2}$ with $p_n$ and $a_nq_{n-1}+q_{n-2}$ with $q_n$, so \begin{align*} T_0\bigl(T_1(\cdots T_n(t)\cdots)\bigr)=\frac{p_nt+p_{n-1}}{q_nt+q_{n-1}}. \end{align*} Now apply this composition formula with index $n-1$ and input $t=a_n$. The input is admissible because $a_n>0$, and each map $T_i$ with $i\geq 1$ preserves positivity on positive inputs. Therefore the nested continued fraction tail is defined, and the denominator in the rational expression is nonzero. We obtain \begin{align*} [a_0;a_1,\dots,a_n]=\frac{p_{n-1}a_n+p_{n-2}}{q_{n-1}a_n+q_{n-2}}. \end{align*} By the defining recurrences for the convergents, the numerator is $p_n$ and the denominator is $q_n$. Therefore \begin{align*} [a_0;a_1,\dots,a_n]=\frac{p_n}{q_n}. \end{align*} For $n=0$, the identity is \begin{align*} [a_0]=a_0=\frac{p_0}{q_0}, \end{align*} since $p_0=a_0$ and $q_0=1$.[/guided]

[step:Prove positivity of the denominators]We prove $q_n>0$ for every $n\geq 0$ by induction. The initial values are \begin{align*} q_0 = 1 > 0, \qquad q_1 = a_1 q_0 + q_{-1} = a_1 > 0. \end{align*} Now let $n \geq 2$ and assume $q_{n-1}>0$ and $q_{n-2}>0$. Since $a_n\in\mathbb{N}=\{1,2,3,\dots\}$, we have $a_n\geq 1$, and hence \begin{align*} q_n = a_n q_{n-1}+q_{n-2} > 0. \end{align*} Thus $q_n>0$ for all $n\geq 0$.[/step]

[guided]We prove the denominator positivity by induction because the recurrence for $q_n$ only involves the two preceding denominators. The initial values are \begin{align*} q_0=1>0. \end{align*} Also, \begin{align*} q_1=a_1q_0+q_{-1}=a_1\cdot 1+0=a_1>0, \end{align*} because $a_1\in\mathbb{N}$. Now let $n\geq 2$ and assume the induction hypotheses $q_{n-1}>0$ and $q_{n-2}>0$. Since $a_n\in\mathbb{N}=\{1,2,3,\dots\}$, we have $a_n\geq 1$. Therefore both summands in the recurrence \begin{align*} q_n=a_nq_{n-1}+q_{n-2} \end{align*} are nonnegative, and in fact $q_{n-2}>0$. Hence \begin{align*} q_n=a_nq_{n-1}+q_{n-2}>0. \end{align*} By induction, $q_n>0$ for every $n\geq 0$.[/guided]

[step:Take determinants to obtain the alternating identity]For every $i\geq 0$, the determinant formula for a $2\times 2$ matrix gives \begin{align*} \det A_i=a_i\cdot 0-1\cdot 1=-1. \end{align*} Therefore, for every $n\geq 0$, \begin{align*} \det P_n=\prod_{i=0}^n \det A_i=(-1)^{n+1}. \end{align*} Using the entry formula for $P_n$ from the first step, the determinant formula gives \begin{align*} \det P_n=p_n q_{n-1}-p_{n-1}q_n. \end{align*} Thus, for every $n\geq 1$, \begin{align*} p_n q_{n-1}-p_{n-1}q_n=(-1)^{n+1}=(-1)^{n-1}, \end{align*} because the exponents $n+1$ and $n-1$ differ by $2$. This proves the determinant identity and completes the proof.[/step]

[guided]The determinant identity is the invariant carried by the matrix product. For each $i\geq 0$, the determinant formula for a $2\times 2$ matrix gives \begin{align*} \det A_i=a_i\cdot 0-1\cdot 1=-1. \end{align*} The determinant is multiplicative for square matrices, so for \begin{align*} P_n=A_0A_1\cdots A_n \end{align*} we obtain \begin{align*} \det P_n=\prod_{i=0}^n \det A_i=(-1)^{n+1}. \end{align*} There are $n+1$ factors because the product runs from $0$ through $n$. The first step identified the entries of $P_n$ as \begin{align*} (P_n)_{11}=p_n, \end{align*} \begin{align*} (P_n)_{12}=p_{n-1}, \end{align*} \begin{align*} (P_n)_{21}=q_n, \end{align*} \begin{align*} (P_n)_{22}=q_{n-1}. \end{align*} Applying the $2\times 2$ determinant formula to this entry description gives \begin{align*} \det P_n=p_nq_{n-1}-p_{n-1}q_n. \end{align*} Combining the two computations of $\det P_n$, we get \begin{align*} p_nq_{n-1}-p_{n-1}q_n=(-1)^{n+1}. \end{align*} For $n\geq 1$, the integers $n+1$ and $n-1$ differ by $2$, so the two powers of $-1$ agree: \begin{align*} (-1)^{n+1}=(-1)^{n-1}. \end{align*} Therefore \begin{align*} p_nq_{n-1}-p_{n-1}q_n=(-1)^{n-1}, \end{align*} which is the claimed alternating determinant identity.[/guided]