# Fibonacci Number less than Golden Section to Power less One

## Theorem

For all $n \in \N_{> 0}$:

$F_n \le \phi^{n - 1}$

where:

$F_n$ is the $n$th Fibonacci number
$\phi$ is the golden section: $\phi = \dfrac {1 + \sqrt 5} 2$

## Proof

The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$F_n \le \phi^{n - 1}$

### Basis for the Induction

$\map P 1$ is true, as this just says:

$F_1 = 1 = \phi^0 = \phi^{1 - 1}$

It is also necessary to demonstrate $\map P 2$ is true:

$F_2 = 1 \le \dfrac {1 + \sqrt 5} 2 = \phi = \phi^{2 - 1}$

This is the basis for the induction.

### Induction Hypothesis

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

So this is the induction hypothesis:

$F_k \le \phi^{k - 1}$

from which it is to be shown that:

$F_{k + 1} \le \phi^k$

### Induction Step

This is the induction step:

 $\ds F_{k + 1}$ $=$ $\ds F_{k - 1} + F_k$ Definition of Fibonacci Numbers $\ds$ $\le$ $\ds \phi^{k - 2} + \phi^{k - 1}$ Induction Hypothesis $\ds$ $=$ $\ds \phi^{k - 2} \paren {1 + \phi}$ $\ds$ $=$ $\ds \phi^{k - 2} \paren {\phi^2}$ Square of Golden Mean equals One plus Golden Mean $\ds$ $=$ $\ds \phi^k$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction.

Therefore:

$\forall n \in \N_{> 0}: F_n \le \phi^{n - 1}$

$\blacksquare$