# 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:

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

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

#### Basis for the Induction

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

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

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:

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

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 $\map P n$ 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 $\map P n$ be the proposition:

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

$\map P 0$ is the case:

 $\ds F_0 \times \phi + F_{-1}$ $=$ $\ds F_0 \times \phi + \paren {-1}^0 F_1$ Fibonacci Number with Negative Index $\ds$ $=$ $\ds F_0 \times \phi + 1$ Definition of Fibonacci Number $F_1 = 1$ $\ds$ $=$ $\ds 0 \times \phi + 1$ Definition of Fibonacci Number $F_0 = 0$ $\ds$ $=$ $\ds 1$ $\ds$ $=$ $\ds \phi^0$

Thus $\map P 0$ is seen to hold.

#### Basis for the Induction

$\map P 1$ is the case:

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

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

#### Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

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

from which it is to be shown that:

$\phi^{-\paren {k + 1} } = F_{-\paren {k + 1} } \phi + F_{-\paren {k + 1} - 1}$

#### Induction Step

This is the induction step:

 $\ds F_{-\paren {k + 1} } \phi + F_{-\paren {k + 1} - 1}$ $=$ $\ds \paren {-1}^{-\paren {k + 1} + 1} F_{k + 1} \phi + \paren {-1}^{-\paren {k + 1} } F_{\paren {k + 1} + 1}$ $\ds$ $=$ $\ds \paren {-1}^{k + 2} F_{k + 1} \phi - \paren {-1}^{k + 2} F_{k + 2}$ $\ds$ $=$ $\ds \paren {-1}^{k + 2} \paren {F_{k + 1} \phi - \paren {F_{k + 1} + F_k} }$ Definition of Fibonacci Number $\ds$ $=$ $\ds \paren {-1}^{k + 2} \paren {F_{k + 1} \paren {\phi - 1} - F_k}$ $\ds$ $=$ $\ds \paren {-1}^{k + 2} \paren {F_{k + 1} \phi^{-1} - F_k}$ Definition 3 of Golden Mean $\ds$ $=$ $\ds \paren {-1}^{k + 2} F_{k + 1} \phi^{-1} + \paren {-1}^{k + 1} F_k$ $\ds$ $=$ $\ds \frac 1 \phi \paren {F_{-k - 1} + F_{-k} \phi}$ Fibonacci Number with Negative Index $\ds$ $=$ $\ds \frac 1 \phi \paren {\phi^{-k} }$ Induction Hypothesis $\ds$ $=$ $\ds \phi^{-\paren {k + 1} }$

So $\map P k \implies \map P {k + 1}$ 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.

$\blacksquare$