Euler-Binet Formula/Proof 1

Proof
Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition:
 * $F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$

Basis for the Induction
$\map P 0$ is true, as this just says:
 * $\dfrac {\phi^0 - \hat \phi^0} {\sqrt 5} = \dfrac {1 - 1} {\sqrt 5} = 0 = F_0$

$\map P 1$ is the case:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P j$ is true for all $0 \le j \le k + 1$, then it logically follows that $\map P {k + 2}$ is true.

So this is our induction hypothesis:
 * $\forall 0 \le j \le k + 1: F_j = \dfrac {\phi^j - \hat \phi^j} {\sqrt 5}$

Then we need to show:
 * $F_{k + 2} = \dfrac {\phi^{k + 2} - \hat \phi^{k + 2} } {\sqrt 5}$

Induction Step
This is our induction step:

We observe that we have the following two identities:
 * $\phi^2 = \paren {\dfrac {1 + \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 + 2 \sqrt 5} = \dfrac {3 + \sqrt 5} 2 = 1 + \phi$
 * $\hat \phi^2 = \paren {\dfrac {1 - \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 - 2 \sqrt 5} = \dfrac {3 - \sqrt 5} 2 = 1 + \hat \phi$

This can also be deduced from the definition of the golden mean: the fact that $\phi$ and $\hat \phi$ are the solutions to the quadratic equation $x^2 = x + 1$.

Thus:

The result follows by the Second Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$