Fibonacci Number with Negative Index

Theorem
Let $F_n$ be the $n$th Fibonacci number.

Then:
 * $\forall n \in \Z_{> 0} : F_{-n} = \left({-1}\right)^{n + 1} F_n$

Proof
From the initial definition of Fibonacci numbers, we have:
 * $F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

By definition of the extension of the Fibonacci numbers to negative integers:
 * $F_n = F_{n + 2} - F_{n - 1}$

The proof proceeds by induction.

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

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

So $P(1)$ is seen to hold.

$P \left({2}\right)$ is the case:

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

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $F_{-\left({k - 1}\right)} = \left({-1}\right)^k F_{k - 1}$
 * $F_{-k} = \left({-1}\right)^{k + 1} F_k$

Then we need to show:
 * $F_{-\left({k + 1}\right)} = \left({-1}\right)^{k + 2} F_{k + 1}$

Induction Step
This is our induction step:

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

Therefore:
 * $\forall n \in \Z_{>0} : F_{-n} = \left({-1}\right)^{n + 1} F_n$