[proofplan] We construct the expansion by iterating the continued fraction algorithm: take the integer part, invert the nonzero fractional part, and repeat. Irrationality ensures that the algorithm never terminates, and the finite convergents obtained from this algorithm converge back to the original number. For uniqueness, the integer part of an infinite simple continued fraction is recovered by taking the floor; after subtracting it and inverting, the same argument recovers each later coefficient inductively. [/proofplan]

[step:Construct the complete quotients by the continued fraction algorithm]Define the initial complete quotient $\theta_0 := \theta$. Since $\theta_0$ is irrational, it is not an integer. Define \begin{align*} a_0 := \lfloor \theta_0 \rfloor \in \mathbb{Z}, \end{align*} and define the next complete quotient \begin{align*} \theta_1 := \frac{1}{\theta_0 - a_0}. \end{align*} Because $a_0 \leq \theta_0 < a_0 + 1$ and $\theta_0 \notin \mathbb{Z}$, we have $0 < \theta_0 - a_0 < 1$, hence $\theta_1 > 1$. Inductively, suppose $\theta_n$ has been defined and is irrational. Define \begin{align*} a_n := \lfloor \theta_n \rfloor \end{align*} and \begin{align*} \theta_{n+1} := \frac{1}{\theta_n - a_n}. \end{align*} For $n \geq 1$, the inequality $\theta_n > 1$ gives $a_n \in \mathbb{N}$. Since $\theta_n$ is irrational, $\theta_n - a_n$ is a nonzero irrational number, so $\theta_{n+1}$ is irrational. Also $0 < \theta_n - a_n < 1$, so $\theta_{n+1} > 1$. Thus the construction gives a sequence $(a_n)_{n=0}^{\infty}$ with $a_0 \in \mathbb{Z}$ and $a_n \in \mathbb{N}$ for every $n \geq 1$.[/step]

[guided]The continued fraction algorithm is forced by the elementary decomposition of a real number into its integer part and fractional part. We begin with \begin{align*} \theta_0 := \theta. \end{align*} Since $\theta$ is irrational, $\theta_0$ is not an integer. Therefore its fractional part is nonzero. Define \begin{align*} a_0 := \lfloor \theta_0 \rfloor \in \mathbb{Z}. \end{align*} The defining property of the floor gives \begin{align*} a_0 \leq \theta_0 < a_0 + 1. \end{align*} Because $\theta_0$ is not an integer, equality cannot occur on the left after subtracting $a_0$, so \begin{align*} 0 < \theta_0 - a_0 < 1. \end{align*} This makes the reciprocal well-defined and positive, so we define \begin{align*} \theta_1 := \frac{1}{\theta_0 - a_0}. \end{align*} The last inequality gives $\theta_1 > 1$. Now repeat this construction. Suppose $\theta_n$ is already defined and irrational. Put \begin{align*} a_n := \lfloor \theta_n \rfloor, \qquad \theta_{n+1} := \frac{1}{\theta_n - a_n}. \end{align*} Again $\theta_n$ is not an integer, so \begin{align*} 0 < \theta_n - a_n < 1. \end{align*} Thus $\theta_{n+1}$ is defined and satisfies $\theta_{n+1} > 1$. Also $\theta_{n+1}$ is irrational: if $\theta_{n+1}$ were rational, then \begin{align*} \theta_n = a_n + \frac{1}{\theta_{n+1}} \end{align*} would be rational, contradicting the inductive hypothesis. Hence the algorithm never terminates. Since $\theta_n > 1$ for every $n \geq 1$, its floor is a positive integer, so $a_n \in \mathbb{N}$ for every $n \geq 1$.[/guided]

[step:Express the original number through every finite prefix and remaining complete quotient] For each $m \geq 0$, the defining relation \begin{align*} \theta_n = a_n + \frac{1}{\theta_{n+1}} \end{align*} holds for $0 \leq n \leq m$. Substituting these identities successively gives \begin{align*} \theta = [a_0; a_1, \dots, a_m, \theta_{m+1}]. \end{align*} Here $[a_0; a_1, \dots, a_m, x]$ denotes the finite expression obtained by replacing the final tail after $a_m$ with the positive real number $x$. [/step]

[step:Show that the finite convergents converge to $\theta$] For $m \geq 0$, define integers $p_m$ and $q_m$ by \begin{align*} p_{-2} := 0,\quad p_{-1} := 1,\quad p_m := a_m p_{m-1} + p_{m-2}, \end{align*} and \begin{align*} q_{-2} := 1,\quad q_{-1} := 0,\quad q_m := a_m q_{m-1} + q_{m-2}. \end{align*} Then $q_m \geq 1$ for every $m \geq 0$. Moreover, for $m \geq 1$, the recurrence and the positivity of $a_{m+1}$ give \begin{align*} q_{m+1} = a_{m+1}q_m + q_{m-1} \geq q_m + q_{m-1}. \end{align*} Since $q_0 = 1$ and $q_1 = a_1 \geq 1$, this implies by induction that $(q_m)_{m=0}^{\infty}$ is unbounded, hence $q_m \to \infty$. A direct induction on $m$ gives \begin{align*} [a_0; a_1, \dots, a_m] = \frac{p_m}{q_m} \end{align*} and, for every $x > 0$, \begin{align*} [a_0; a_1, \dots, a_m, x] = \frac{x p_m + p_{m-1}}{x q_m + q_{m-1}}. \end{align*} The same recurrence gives the determinant identity \begin{align*} p_m q_{m-1} - p_{m-1} q_m = (-1)^{m-1}. \end{align*} Using $x = \theta_{m+1} > 1$, subtracting the two rational expressions over the common denominator $q_m(xq_m + q_{m-1})$ gives \begin{align*} \left|\theta - [a_0; a_1, \dots, a_m]\right| = \frac{|p_{m-1}q_m - p_m q_{m-1}|}{q_m(xq_m + q_{m-1})}. \end{align*} By the determinant identity, this becomes \begin{align*} \left|\theta - [a_0; a_1, \dots, a_m]\right| = \frac{1}{q_m(xq_m + q_{m-1})}. \end{align*} Since $x > 1$ and $q_{m-1} \geq 0$, we have $xq_m + q_{m-1} \geq q_m$, so \begin{align*} \left|\theta - [a_0; a_1, \dots, a_m]\right| \leq \frac{1}{q_m^2}. \end{align*} Because $q_m \to \infty$, it follows that \begin{align*} \lim_{m \to \infty} [a_0; a_1, \dots, a_m] = \theta. \end{align*} Thus the constructed sequence is an infinite simple continued fraction expansion of $\theta$. [/step]

[step:Recover the first coefficient from the finite convergents of any infinite simple continued fraction] Suppose \begin{align*} \theta = [a_0; a_1, a_2, \dots] \end{align*} is an infinite simple continued fraction with $a_0 \in \mathbb{Z}$ and $a_n \in \mathbb{N}$ for $n \geq 1$. By the meaning of this equality, the finite convergents \begin{align*} \gamma_m := [a_0; a_1, \dots, a_m] \end{align*} are defined for $m \geq 0$ and satisfy $\lim_{m \to \infty}\gamma_m = \theta$. For $m \geq 1$, define the finite tail value \begin{align*} \beta_m := [a_1; a_2, \dots, a_m]. \end{align*} Since $a_1 \in \mathbb{N}$ and every later partial quotient is positive, the recursive finite expression gives $\beta_m > 1$ for every $m \geq 2$, while $\beta_1 = a_1 \geq 1$. Hence \begin{align*} 0 < \frac{1}{\beta_m} \leq 1 \end{align*} for every $m \geq 1$. The finite convergent satisfies \begin{align*} \gamma_m = a_0 + \frac{1}{\beta_m}, \end{align*} so \begin{align*} a_0 < \gamma_m \leq a_0 + 1 \end{align*} for every $m \geq 1$. Passing to the limit in these two inequalities gives \begin{align*} a_0 \leq \theta \leq a_0 + 1. \end{align*} Since $\theta$ is irrational, it is not equal to either integer endpoint $a_0$ or $a_0 + 1$. Therefore \begin{align*} a_0 < \theta < a_0 + 1, \end{align*} and the defining property of the floor gives $a_0 = \lfloor \theta \rfloor$. [/step]

[step:Apply continuity of the inverse map to shift the expansion and prove uniqueness] Suppose \begin{align*} \theta = [a_0; a_1, a_2, \dots] = [b_0; b_1, b_2, \dots] \end{align*} are two infinite simple continued fraction expansions of the same irrational number, with $a_0,b_0 \in \mathbb{Z}$ and $a_n,b_n \in \mathbb{N}$ for $n \geq 1$. By the previous step, \begin{align*} a_0 = \lfloor \theta \rfloor = b_0. \end{align*} Let $c_0 := a_0 = b_0$. Since $\theta$ is irrational, $\theta \neq c_0$, so define \begin{align*} \theta_1 := \frac{1}{\theta - c_0}. \end{align*} For $m \geq 1$, define the finite convergents of the first expansion and its shifted tail by \begin{align*} \gamma_m := [c_0; a_1, \dots, a_m] \end{align*} and \begin{align*} \alpha_m := [a_1; a_2, \dots, a_m]. \end{align*} Then $\gamma_m = c_0 + 1/\alpha_m$, so \begin{align*} \alpha_m = \frac{1}{\gamma_m - c_0}. \end{align*} Because $\gamma_m \to \theta$ and $\theta - c_0 \neq 0$, the reciprocal map $x \mapsto 1/(x-c_0)$ is continuous at $\theta$, and therefore \begin{align*} \lim_{m \to \infty} \alpha_m = \frac{1}{\theta - c_0} = \theta_1. \end{align*} Thus $\theta_1 = [a_1; a_2, a_3, \dots]$. Applying the same argument to the finite convergents $[c_0; b_1, \dots, b_m]$ gives \begin{align*} \theta_1 = [b_1; b_2, b_3, \dots]. \end{align*} The number $\theta_1$ is irrational, because if it were rational then $\theta = c_0 + 1/\theta_1$ would be rational. Applying the previous step to the two expansions of $\theta_1$ gives $a_1 = b_1$. Repeating this argument inductively, suppose $a_j = b_j$ for every $0 \leq j \leq n-1$. Let $c_j := a_j = b_j$ for $0 \leq j \leq n-1$. Define the common $n$-th complete quotient recursively by \begin{align*} \theta_{j+1} := \frac{1}{\theta_j - c_j} \end{align*} for $0 \leq j \leq n-1$, with $\theta_0 := \theta$. At each stage, $\theta_j$ is irrational, because otherwise reversing the finitely many rational operations $x \mapsto c_i + 1/x$ would make $\theta$ rational. The same finite-convergent continuity argument applied after the common prefix $c_0,\dots,c_{n-1}$ shows that \begin{align*} \theta_n = [a_n; a_{n+1}, a_{n+2}, \dots] = [b_n; b_{n+1}, b_{n+2}, \dots]. \end{align*} Applying the previous step to the two expansions of the irrational number $\theta_n$ gives \begin{align*} a_n = \lfloor \theta_n \rfloor = b_n. \end{align*} By induction, $a_n = b_n$ for every $n \geq 0$. Hence the infinite simple continued fraction expansion of $\theta$ is unique. [/step]