Relation between Adjacent Best Rational Approximations to Root 2

Theorem
Consider the Sequence of Best Rational Approximations to Square Root of 2:
 * $\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$

Let $\dfrac {p_n} {q_n}$ and $\dfrac {p_{n + 1} } {q_{n + 1} }$ be adjacent terms of $\sequence S$.

Then:
 * $\dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$

Proof
The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
 * $\dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown 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 the induction hypothesis:
 * $\dfrac {p_{k + 1} } {q_{k + 1} } = \dfrac {p_k + 2 q_k} {p_k + q_k}$

from which it is to be shown that:
 * $\dfrac {p_{k + 2} } {q_{k + 2} } = \dfrac {p_{k + 1} + 2 q_{k + 1} } {p_{k + 1} + q_{k + 1} }$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \Z_{>0}: \dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$