# Fibonacci Number n+1 Minus Golden Mean by Fibonacci Number n

## Theorem

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

where:

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

## Proof 1

 $\ds F_{n + 1} - \phi F_n$ $=$ $\ds \dfrac 1 {\sqrt 5} \paren {\phi^{n + 1} - \hat \phi^{n + 1} } - \dfrac \phi {\sqrt 5} \paren {\phi^n - \hat \phi^n}$ Euler-Binet Formula $\ds$ $=$ $\ds \dfrac 1 {\sqrt 5} \paren {\phi^{n + 1} - \hat \phi^{n + 1} - \phi^{n + 1} + \phi \hat \phi^n}$ $\ds$ $=$ $\ds \hat \phi^n \dfrac {\phi - \hat \phi} {\sqrt 5}$ $\ds$ $=$ $\ds \hat \phi^n F_1$ Euler-Binet Formula $\ds$ $=$ $\ds \hat \phi^n$ Definition of Fibonacci Number: $F_1 = 1$

$\blacksquare$

## Proof 2

 $\ds F_n$ $=$ $\ds \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$ Euler-Binet Formula $\ds$ $=$ $\ds \dfrac {\phi^n - \hat \phi^n} {\frac {1 + \sqrt 5} 2 - \frac {1 - \sqrt 5} 2}$ $\ds$ $=$ $\ds \dfrac {\phi^n - \hat \phi^n} {\phi - \hat \phi}$ Definition 2 of Golden Mean

Thus from Recurrence Relation where n+1th Term is A by nth term + B to the n we have:

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

whence the result.

$\blacksquare$