Difference between Adjacent Convergents of Simple Continued Fraction

Theorem
Let $F$ be a field, such as the field of real numbers $\R$.

Let $n \in \N \cup \{\infty\}$ be an extended natural number.

Let $\left[{a_0, a_1, a_2, \ldots}\right]$ be a continued fraction in $F$ of length $n$.

Let $p_0, p_1, p_2, \ldots$ and $q_0, q_1, q_2, \ldots$ be its numerators and denominators.

Let $C_0, C_1, C_2, \ldots$ be the convergents of $\left[{a_0, a_1, a_2, \ldots}\right]$.

Then for $k \geq 1$:
 * $p_k q_{k-1} - p_{k-1} q_k = \left({-1}\right)^{k+1}$

That is:
 * $C_k - C_{k - 1} = \dfrac {p_k} {q_k} - \dfrac {p_{k - 1} } {q_{k - 1} } = \dfrac {\left({-1}\right)^{k+1}} {q_k q_{k - 1} }$

Proof
Proof by induction:

For all $n \in \Z: n \ge 2$, let $P \left({n}\right)$ be the proposition:
 * $p_n q_{n - 1} - p_{n - 1} q_n = \left({-1}\right)^{n+1}$

Basis for the Induction
$P(1)$ is the case:
 * $p_0 = a_0, p_1 = a_0 a_1 + 1, q_0 = 1, q_1 = a_1$

So:
 * $p_1 q_0 - p_0 q_1 = \left({a_0 a_1 + 1}\right) \times 1 - a_0 a_1 = 1 = \left({-1}\right)^2$

So $P(2)$ is seen to hold.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:


 * $p_k q_{k-1} - p_{k-1} q_k = \left({-1}\right)^{k+1}$

Then we need to show:


 * $p_{k + 1} q_k - p_k q_{k + 1} = \left({-1}\right)^{k + 2}$

Induction Step
This is our induction step:

Consider $\left[{a_0, a_1, a_2, \ldots, a_k, a_{k + 1}}\right]$.

Its final numerator and denominator are by definition:
 * $p_{k + 1} = a_{k + 1} p_k + p_{k - 1}, q_{k + 1} = a_{k + 1} q_k + q_{k - 1}$

Therefore:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \ge 1: p_n q_{n-1} - p_{n-1} q_n = \left({-1}\right)^{n+1}$