Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Negative Index

Theorem
Let $n \in \Z_{\le 0}$.

Then:


 * $\phi^n = F_n \phi + F_{n - 1}$

where:
 * $F_n$ denotes the $n$th Fibonacci number as extended to negative indices
 * $\phi$ denotes the golden mean.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\le 0}$, let $P \left({n}\right)$ be the proposition:
 * $\phi^n = F_n \phi + F_{n - 1}$

This can equivalently be expressed as:

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\phi^{-n} = F_{-n} \phi + F_{-n - 1}$

$P \left({0}\right)$ is the case:

Thus $P \left({0}\right)$ is seen to hold.

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $\phi^{-k} = F_{-k} \phi + F_{-k - 1}$

from which it is to be shown that:
 * $\phi^{-\left({k + 1}\right)} = F_{-\left({k + 1}\right)} \phi + F_{-\left({k + 1}\right) - 1}$

Induction Step
This is the induction step:

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


 * $\forall n \in \Z_{\le 0}: \phi^n = F_n \phi + F_{n - 1}$