Simple Infinite Continued Fraction Converges to Irrational Number

Theorem
The value of any simple infinite continued fraction is irrational.

Proof
Let $x = \left[{a_1, a_2, a_3, \ldots}\right]$ be a SICF.

Let $p_1, p_2, p_3, \ldots$ and $q_1, q_2, q_3, \ldots$ be its numerators and denominators.

Let $C_1, C_2, C_3, \ldots$ be the convergents of $\left[{a_1, a_2, a_3, \ldots}\right]$.

From corollary to the value of a SFCF, $C_k = \dfrac {p_k} {q_k}$.

From the corollary to the value of a SICF:
 * for each $n \ge 1$, $C_n < x < C_{n+1}$;
 * $\left|{x - \dfrac {p_n} {q_n}}\right| < \dfrac 1 {q_n q_{n+1}}$.

Suppose $x$ is rational. That is, let $x = r s$ where $r, s \in \Z$ such that $s > 0$.

Then:
 * $0 < \left|{\dfrac r s - \dfrac {p_n} {q_n}}\right| = \dfrac {\left|{r q_n - s p_n}\right|} {s q_n} < \dfrac 1 {q_n q_{n+1}}$.

(Note that $\dfrac r s \ne \dfrac {p_n} {q_n}$ or otherwise the continued fraction would be finite.)

So:
 * $0 < \left|{r q_n - s p_n}\right| < \dfrac s {q_{n+1}}$.

But the denominators of a SCF form a strictly increasing sequence of integers.

That means we can choose $n$ so that $q_{n+1} > s$.

But then $\left|{r q_n - s p_n}\right|$ would be an integer lying strictly between $0$ and $1$, which can't happen.

So no such integers $r, s$ exist, and so $x$ must be irrational.