Fibonacci Number with Negative Index

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

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

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

Then:
 * $F_0 = F_2 - F_1 = 0$

Proof by induction:

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

Basis for the Induction
$P \left({1}\right)$ is the case $F_{-1} = F_1$, which holds because $F_{-1} = F_1 - F_0 = 1$.

$P \left({2}\right)$ is the case $F_{-2} = -F_2$, which also holds because $F_{-2} = F_0 - F_{-1} = -1$.

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:
 * $\displaystyle F_{-k+1} = \left({-1}\right)^{k} F_{k-1}$
 * $\displaystyle F_{-k} = \left({-1}\right)^{k+1} F_{k}$

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

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