[proofplan] The irrationality of $\theta$ supplies the infinite simple continued fraction setting; the coprimality argument itself uses only the standard recurrence and the positivity of the partial quotients after $a_0$. We first verify the initial convergent and prove by induction that all denominators $q_n$ are positive. We then prove the standard determinant identity relating two consecutive convergents: \begin{align*} p_nq_{n-1}-p_{n-1}q_n = (-1)^{n-1} \end{align*} for every $n \ge 1$. Once this identity is known, any common divisor of $p_n$ and $q_n$ must divide the integer $(-1)^{n-1}$, hence must be $1$. [/proofplan]

[step:Verify the initial convergent is already in lowest terms] From the recurrence at $n=0$, using $p_{-1}=1$, $p_{-2}=0$, $q_{-1}=0$, and $q_{-2}=1$, we obtain \begin{align*} p_0 = a_0p_{-1}+p_{-2}=a_0. \end{align*} The same initial values give \begin{align*} q_0 = a_0q_{-1}+q_{-2}=1. \end{align*} Therefore \begin{align*} \gcd(p_0,q_0)=\gcd(a_0,1)=1. \end{align*} [/step]

[step:Prove all denominators are positive] For a simple continued fraction, the partial quotients satisfy $a_m \in \mathbb{N}$ for every integer $m \ge 1$, so $a_m \ge 1$. We prove $q_n>0$ for every integer $n \ge 0$ by induction. The initial value is \begin{align*} q_0=1>0. \end{align*} For $n=1$, the recurrence and $q_{-1}=0$ give \begin{align*} q_1=a_1q_0+q_{-1}=a_1>0. \end{align*} Now let $n \ge 2$ and assume $q_{n-1}>0$ and $q_{n-2}>0$. Since $a_n \ge 1$, the recurrence gives \begin{align*} q_n=a_nq_{n-1}+q_{n-2}>0. \end{align*} By induction, $q_n>0$ for every integer $n \ge 0$. [/step]

[step:Prove the determinant identity for consecutive convergents]For each integer $n \ge 1$, define the determinant-like integer \begin{align*} D_n := p_nq_{n-1}-p_{n-1}q_n. \end{align*} We prove by induction that \begin{align*} D_n = (-1)^{n-1} \end{align*} for every $n \ge 1$. For $n=1$, the recurrence gives \begin{align*} p_1 = a_1p_0+p_{-1}=a_1a_0+1. \end{align*} It also gives \begin{align*} q_1 = a_1q_0+q_{-1}=a_1. \end{align*} Hence \begin{align*} D_1=(a_1a_0+1)\cdot 1-a_0a_1=1=(-1)^0. \end{align*} Now let $n \ge 2$ and assume \begin{align*} D_{n-1}=p_{n-1}q_{n-2}-p_{n-2}q_{n-1}=(-1)^{n-2}. \end{align*} Using the recurrence for $p_n$ and $q_n$, we compute \begin{align*} D_n=(a_np_{n-1}+p_{n-2})q_{n-1}-p_{n-1}(a_nq_{n-1}+q_{n-2}). \end{align*} Expanding and cancelling the two equal terms $a_np_{n-1}q_{n-1}$ and $-a_np_{n-1}q_{n-1}$ gives \begin{align*} D_n=p_{n-2}q_{n-1}-p_{n-1}q_{n-2}. \end{align*} By the definition of $D_{n-1}$, this is $-D_{n-1}$. Therefore the induction hypothesis gives \begin{align*} D_n=-D_{n-1}=-(-1)^{n-2}=(-1)^{n-1}. \end{align*} By induction, the determinant identity holds for every $n \ge 1$.[/step]

[guided]The purpose of this step is to prove an arithmetic invariant of the recurrence. Define, for each integer $n \ge 1$, the integer \begin{align*} D_n := p_nq_{n-1}-p_{n-1}q_n. \end{align*} This quantity measures the cross-difference of two consecutive convergents. We will show that it is always equal to either $1$ or $-1$: \begin{align*} D_n = (-1)^{n-1}. \end{align*} We start with $n=1$. From the recurrence and the initial values, \begin{align*} p_0 = a_0p_{-1}+p_{-2}=a_0. \end{align*} The denominator recurrence gives \begin{align*} q_0 = a_0q_{-1}+q_{-2}=1. \end{align*} Applying the recurrence once more gives \begin{align*} p_1 = a_1p_0+p_{-1}=a_1a_0+1. \end{align*} Similarly, \begin{align*} q_1 = a_1q_0+q_{-1}=a_1. \end{align*} Therefore \begin{align*} D_1=(a_1a_0+1)\cdot 1-a_0a_1=1=(-1)^0. \end{align*} This proves the base case. Now suppose $n \ge 2$ and assume the identity has already been proved at the previous index: \begin{align*} D_{n-1}=p_{n-1}q_{n-2}-p_{n-2}q_{n-1}=(-1)^{n-2}. \end{align*} We compute $D_n$ directly from the recurrence. The key point is that the terms involving $a_n$ cancel. Substituting $p_n=a_np_{n-1}+p_{n-2}$ and $q_n=a_nq_{n-1}+q_{n-2}$ into the definition of $D_n$ gives \begin{align*} D_n=(a_np_{n-1}+p_{n-2})q_{n-1}-p_{n-1}(a_nq_{n-1}+q_{n-2}). \end{align*} Expanding the products gives \begin{align*} D_n=a_np_{n-1}q_{n-1}+p_{n-2}q_{n-1}-a_np_{n-1}q_{n-1}-p_{n-1}q_{n-2}. \end{align*} Cancelling the two opposite terms involving $a_n$ leaves \begin{align*} D_n=p_{n-2}q_{n-1}-p_{n-1}q_{n-2}. \end{align*} The last expression is the negative of $D_{n-1}$, since \begin{align*} D_{n-1}=p_{n-1}q_{n-2}-p_{n-2}q_{n-1}. \end{align*} Thus the induction hypothesis gives \begin{align*} D_n=-D_{n-1}=-(-1)^{n-2}=(-1)^{n-1}. \end{align*} This completes the induction and proves the determinant identity for every $n \ge 1$.[/guided]

[step:Use the determinant identity to rule out nontrivial common divisors] Fix an integer $n \ge 1$. Let $d:=\gcd(p_n,q_n)$, so $d$ is a positive integer satisfying $d \mid p_n$ and $d \mid q_n$. Since $d \mid p_n$, the integer $d$ divides $p_nq_{n-1}$. Since $d \mid q_n$, the integer $d$ divides $p_{n-1}q_n$. The elementary divisibility rule $d\mid x$ and $d\mid y$ imply $d\mid x-y$ now gives \begin{align*} d \mid \bigl(p_nq_{n-1}-p_{n-1}q_n\bigr). \end{align*} By the determinant identity proved above, \begin{align*} p_nq_{n-1}-p_{n-1}q_n = (-1)^{n-1}. \end{align*} Hence $d \mid (-1)^{n-1}$. Since $d$ is positive, this forces $d=1$. Thus \begin{align*} \gcd(p_n,q_n)=1 \end{align*} for every $n \ge 1$. [/step]

[step:Combine positivity and coprimality] The first step proved $\gcd(p_0,q_0)=1$. The positivity step proved $q_n>0$ for every integer $n \ge 0$. The previous step proved $\gcd(p_n,q_n)=1$ for every integer $n \ge 1$. Therefore, for every integer $n \ge 0$, the convergent \begin{align*} \frac{p_n}{q_n} \end{align*} has positive denominator and is in lowest terms. [/step]