# Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less

## Theorem

Let $n \in \Z$.

Then:

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

where:

$F_n$ denotes the $n$th Fibonacci number
$\phi$ denotes the golden mean.

## Proof

### Positive Index

First the result is proved for positive integers.

The proof proceeds by induction.

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:

 $\displaystyle F_0 \times \phi + F_{-1}$ $=$ $\displaystyle F_0 \times \phi + \left({-1}\right)^0 F_1$ Fibonacci Number with Negative Index $\displaystyle$ $=$ $\displaystyle F_0 \times \phi + 1$ Definition of Fibonacci Number $F_1 = 1$ $\displaystyle$ $=$ $\displaystyle 0 \times \phi + 1$ Definition of Fibonacci Number $F_0 = 0$ $\displaystyle$ $=$ $\displaystyle 1$ $\displaystyle$ $=$ $\displaystyle \phi^0$

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

#### Basis for the Induction

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

 $\displaystyle F_1 \times \phi + F_0$ $=$ $\displaystyle F_1 \times \phi$ Definition of Fibonacci Number $F_0 = 0$ $\displaystyle$ $=$ $\displaystyle 1 \times \phi$ Definition of Fibonacci Number $F_1 = 1$ $\displaystyle$ $=$ $\displaystyle \phi^1$

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:

$\phi^{k + 1} = F_{k + 1} \phi + F_k$

#### Induction Step

This is the induction step:

 $\displaystyle F_{k + 1} \phi + F_k$ $=$ $\displaystyle \left({F_k + F_{k - 1} }\right) \phi + F_k$ Definition of Fibonacci Number $\displaystyle$ $=$ $\displaystyle F_k \left({1 + \phi}\right) + F_{k - 1} \phi$ $\displaystyle$ $=$ $\displaystyle F_k \phi^2 + F_{k - 1} \phi$ Square of Golden Mean equals One plus Golden Mean $\displaystyle$ $=$ $\displaystyle \phi \left({F_k \phi + F_{k - 1} }\right)$ $\displaystyle$ $=$ $\displaystyle \phi \left({\phi^n}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \phi^{n + 1}$

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

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

$\Box$

### Negative Index

Then the result is extended to negative integers.

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:

 $\displaystyle F_0 \times \phi + F_{-1}$ $=$ $\displaystyle F_0 \times \phi + \left({-1}\right)^0 F_1$ Fibonacci Number with Negative Index $\displaystyle$ $=$ $\displaystyle F_0 \times \phi + 1$ Definition of Fibonacci Number $F_1 = 1$ $\displaystyle$ $=$ $\displaystyle 0 \times \phi + 1$ Definition of Fibonacci Number $F_0 = 0$ $\displaystyle$ $=$ $\displaystyle 1$ $\displaystyle$ $=$ $\displaystyle \phi^0$

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

#### Basis for the Induction

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

 $\displaystyle F_{-1} \times \phi + F_{-2}$ $=$ $\displaystyle \left({-1}\right)^0 F_1 \times \phi + \left({-1}\right)^{-1} F_2$ Fibonacci Number with Negative Index $\displaystyle$ $=$ $\displaystyle 1 \times \phi - F_2$ Definition of Fibonacci Number $F_1 = 1$ $\displaystyle$ $=$ $\displaystyle \phi - 1$ Definition of Fibonacci Number: $F_2 = 1$ $\displaystyle$ $=$ $\displaystyle \dfrac 1 \phi$ Definition 3 of Golden Mean

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:

 $\displaystyle F_{-\left({k + 1}\right)} \phi + F_{-\left({k + 1}\right) - 1}$ $=$ $\displaystyle \left({-1}\right)^{-\left({k + 1}\right) + 1} F_{k + 1} \phi + \left({-1}\right)^{-\left({k + 1}\right)} F_{\left({k + 1}\right) + 1}$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{k + 2} F_{k + 1} \phi - \left({-1}\right)^{k + 2} F_{k + 2}$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{k + 2} \left({F_{k + 1} \phi - \left({F_{k + 1} + F_k}\right)}\right)$ Definition of Fibonacci Number $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{k + 2} \left({F_{k + 1} \left({\phi - 1}\right) - F_k}\right)$ $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{k + 2} \left({F_{k + 1} \phi^{-1} - F_k}\right)$ Definition 3 of Golden Mean $\displaystyle$ $=$ $\displaystyle \left({-1}\right)^{k + 2} F_{k + 1} \phi^{-1} + \left({-1}\right)^{k + 1} F_k$ $\displaystyle$ $=$ $\displaystyle \frac 1 \phi \left({F_{-k - 1} + F_{-k} \phi}\right)$ Fibonacci Number with Negative Index $\displaystyle$ $=$ $\displaystyle \frac 1 \phi \left({\phi^{-k} }\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \phi^{-\left({k + 1}\right)}$

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

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

$\Box$

Hence the result is seen to show for both positive and negative integers.

$\blacksquare$