Properties of Convergents of Continued Fractions

Theorem
Let $\left[{a_1, a_2, a_3, \ldots}\right]$ be a simple continued fraction.

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 $\left[{a_1, a_2, a_3, \ldots}\right]$.

Then the following results apply:

Difference between Adjacent Convergents
$p_k q_{k-1} - p_{k-1} q_k = \left({-1}\right)^k$, that is:
 * $\displaystyle C_k - C_{k-1} = \frac {p_k} {q_k} - \frac {p_{k-1}} {q_{k-1}} = \frac {\left({-1}\right)^k} {q_k q_{k-1}}$

for $k \ge 2$;

Difference between Adjacent Convergents But One
$p_k q_{k-2} - p_{k-2} q_k = \left({-1}\right)^{k-1} a_k$, that is:
 * $\displaystyle C_k - C_{k-2} = \frac {p_k} {q_k} - \frac {p_{k-2}} {q_{k-2}} = \frac {\left({-1}\right)^{k-1} a_k} {q_k q_{k-2}}$

for $k \ge 3$;

Convergents are Rationals in Canonical Form

 * $p_k$ and $q_k$ are coprime for $k \ge 1$;
 * $q_k > 0$ for $k \ge 1$.

Proof of Difference between Adjacent Convergents
Proof by induction:

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

Basis for the Induction

 * $P(2)$ is the case where $p_1 = a_1, p_2 = a_1 a_2 + 1, q_1 = 1, q_2 = a_2$.

So $p_2 q_1 - p_1 q_2 = \left({a_1 a_2 + 1}\right) \times 1 - a_1 a_2 = 1 = \left({-1}\right)^2$.

So $P(2)$ holds. 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:


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

Then we need to show:


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

Induction Step
This is our induction step:

Consider $\left[{a_1, a_2, a_3, \ldots, a_k, a_{k+1}}\right]$.

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:
 * $\forall n \ge 2: p_n q_{n-1} - p_{n-1} q_n = \left({-1}\right)^n$

Proof of Difference between Adjacent Convergents But One
This is a simple consequence of Difference between Adjacent Convergents above.

Proof that Convergents are Rationals in Canonical Form
This also follows directly from Difference between Adjacent Convergents above.

Let $d = \gcd \left\{{p_k, q_k}\right\}$.

Then $p_k q_{k-1} - p_{k-1} q_k$ is a multiple of $d$ from Common Divisor Divides Integer Combination.

So $d \backslash \left({-1}\right)^k$ and so $d = 1$.

Also note that $q_1 = 1$, and $a_k > 0$ for all $k \ge 2$.

So it follows that $q_k > 0$ for all $k \ge 1$ from definition of denominator.

Hence the result.