# 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:

 $\displaystyle F_{-1}$ $=$ $\displaystyle F_1 - F_0$ $\displaystyle$ $=$ $\displaystyle 1 - 0$ $\displaystyle$ $=$ $\displaystyle 1$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{1 + 1} F_1$

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

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

 $\displaystyle F_{-2}$ $=$ $\displaystyle F_0 - F_{-1}$ $\displaystyle$ $=$ $\displaystyle 0 - 1$ $\displaystyle$ $=$ $\displaystyle -1$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{2 + 1} F_2$

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:

 $\displaystyle F_{- \left({k + 1}\right)}$ $=$ $\displaystyle F_{-\left({k - 1}\right)} - F_{-k}$ Definition of Fibonacci Number for Negative Index $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^k F_{k - 1} - \left({-1}\right)^{k + 1} F_k$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^k F_{k - 1} + \left({-1}\right)^k F_k$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^k \left({F_{k - 1} + F_k}\right)$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^k \left({F_{k + 1} }\right)$ Definition of Fibonacci Number $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{k + 2} \left({F_{k + 1} }\right)$

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:

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

$\blacksquare$