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, $$C_k = \frac {p_k} {q_k}$$.

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

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

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

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

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

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.