Simple Infinite Continued Fraction Converges to Irrational Number

Theorem

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

Proof

Let $\sqbrk {a_0, a_1, a_2, \ldots}$ 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 $n \ge 0$.

For all $n \ge 0$, from Accuracy of Convergents of Convergent Simple Infinite Continued Fraction:

$\size {x - \dfrac {p_n} {q_n} } < \dfrac 1 {q_n q_{n + 1} }$

Suppose $x$ is rational.

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

Then:

$0 < \size {\dfrac r s - \dfrac {p_n} {q_n} } = \dfrac {\size {r q_n - s p_n} } {s q_n} < \dfrac 1 {q_n q_{n + 1} }$

We note that:

$\dfrac r s \ne \dfrac {p_n} {q_n}$

because otherwise the continued fraction would be finite.

So:

$0 < \size {r q_n - s p_n} < \dfrac s {q_{n + 1} }$

But Denominators of Simple Continued Fraction are Strictly Increasing.

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

But then $\size {r q_n - s p_n}$ 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.

$\blacksquare$