Value of Finite Continued Fraction equals Numerator Divided by Denominator

Theorem
Let $F$ be a field.

Let $\tuple {a_0, a_1, \ldots, a_n}$ be a finite continued fraction of length $n \ge 0$.

Let $p_n$ and $q_n$ be its $n$th numerator and denominator.

Then the value $\sqbrk {a_0, a_1, \ldots, a_n}$ equals $\dfrac {p_n} {q_n}$.

Proof
We will use a proof by induction on the length $n$.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
 * $\sqbrk {a_0, a_1, \ldots, a_n} = \dfrac {p_n} {q_n}$

Basis for the Induction
$\map P 0$ is the case:
 * $\sqbrk {a_0} = \dfrac {a_0} 1 = \dfrac {p_0} {q_0}$

This holds for any continued fraction.

$\map P 1$ is the case:

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\sqbrk {a_0, a_1, \ldots, a_k} = \dfrac {p_k} {q_k}$

Then we need to show:
 * $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \dfrac {p_{k + 1} } {q_{k + 1} }$

Induction Step
This is our induction step:

Consider the continued fraction:
 * $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} }$

The numerators are:
 * $p_0, p_1, \ldots, p_k, p_{k + 1}$

and the denominators are:
 * $q_0, q_1, \ldots, q_k, q_{k + 1}$

By definition of value of a finite continued fraction:
 * $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \sqbrk {a_0, a_1, \ldots, a_{k - 1}, a_k'}$

where $a_k' = a_k + \dfrac 1 {a_{k + 1} }$

Consider the.

Take the continued fraction:
 * $\sqbrk {a_0, a_1, \ldots, a_{k - 1}, a_k'}$

Its numerators are:
 * $p_0, p_1, \ldots, p_{k-1}$ and $p_k'$

where $p_k' = \paren {a_k + \dfrac 1 {a_{k + 1} } } p_{k - 1} + p_{k - 2}$ by definition.

Its denominators are:
 * $q_0, q_1, \ldots, p_{k - 1}$ and $q_k'$

where $q_k' = \paren {a_k + \dfrac 1 {a_{k + 1} } } q_{k - 1} + q_{k - 2}$ by definition.

As it has just $k$ partial quotients, the induction hypothesis tells us that its value is:
 * $\dfrac {p_k'} {q_k'}$

So:

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

Therefore:
 * $\sqbrk {a_0, a_1, \ldots, a_k, a_{k + 1} } = \dfrac {p_{k + 1} } {q_{k + 1} }$