Simple Infinite Continued Fraction Converges to Irrational Number

Theorem
The value of any simple infinite continued fraction in $\R$ is irrational.

Proof
Let $\left[{a_0, a_1, a_2, \ldots}\right]$ be a simple infinite continued fraction.

Note that by Simple Infinite Continued Fraction Converges, a simple infinite continued fraction is indeed convergent, say to $x \in \R$.

Let $p_0, p_1, \ldots$ and $q_0, q_1, \ldots$ be its numerators and denominators.

Let $C_0, C_1, \ldots$ be its convergents, so that $C_n = p_n/q_n$ for $n \geq 0$.

For all $n \geq 0$, from Bound for Difference of Limit of Simple Infinite Continued Fraction with Convergent:
 * $\left\vert{x - \dfrac {p_n} {q_n} }\right\vert < \dfrac 1 {q_n q_{n + 1} }$

Suppose $x$ is rational.

That is, let $x = \dfrac r s$ 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 Denominators of Simple Continued Fraction are Strictly Increasing.

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 cannot happen.

So no such integers $r, s$ exist.

Thus $x$ must be irrational.