Denominators of Simple Continued Fraction are Strictly Increasing

Theorem

Let $N \in \N \cup \set \infty$ be an extended natural number.

Let $\sqbrk {a_0, a_1, a_2, \ldots}$ be a simple continued fraction in $\R$ of length $N$.

Let $q_0, q_1, q_2, \ldots$ be its denominators.

Then with the possible exception of $q_0 = q_1$, the sequence $\sequence {q_n}$ is strictly increasing.

Proof

By definition of simple continued fraction, all partial quotients of $\sqbrk {a_0, a_1, a_2, \ldots}$ are strictly positive integers, with the possible exception of $a_0$.

So:

$q_1 = a_1 \ge 1 = q_0$
$q_2 = a_2 a_1 + 1 \ge a_1 + 1 > a_1 = q_1$

hence:

$1 = q_0 \le q_1 < q_2$.

Suppose $q_k > q_{k-1} \ge 1$ for some $k \ge 2$.

Then $q_{k+1} = a_{k+1} q_k + q_{k-1} \ge q_k + q_{k-1} \ge q_{k} + 1 > q_k$.

So, by induction, $\sequence {q_n}$ is strictly increasing except when possibly $q_0 = q_1 = 1$.

$\blacksquare$