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## From Irrational to Rational

I find the relationship between the Golden Ratio and the Fibonacci numbers amazing.

Here is a derivation:

Consider the following  recursive relationship , where $A$ and $B$ are positive integers:

$x_n=A x_{n-1} +Bx_{n-2}$

We can represent this in matrix form:

$\left (\begin{array}{ll} 0 &1\\B&A\end{array}\right) \left ( \begin{array}{l}x_{n-2}\\x_{n-1}\end{array}\right ) =\left ( \begin{array}{l} x_{n-1}\\x_{n}\end{array}\right )$

Let

$R =\left (\begin{array}{ll} 0 &1\\B&A\end{array}\right)$

The characteristic polynomial of $R$ is

$\lambda^2 - A\lambda -B =0$

Now by the choice of $A$ and $B$ the characteristic polynomial has distinct and real roots: $\alpha$ and $\beta$ . This follows because the discriminant $\forall A, B \in \mathrm{N}, A^2 +4 B >0$

$R$ can be represented as:

$R =\frac{1}{\beta - \alpha} \left (\begin{array}{ll} 1& 1\\\alpha&\beta\end{array}\right ) \left (\begin{array}{ll} \alpha & 0 \\ 0&\beta\end{array}\right ) \left ( \begin{array}{ll}\beta&-1\\-\alpha&1\end{array}\right )$

where the scalar is the determinant of the first matrix, the first matrix holds the eigenvectors of $R$ , the central diagonal matrix the eigenvalues of $R$ and the third matrix (when scalar is included) is the inverse of the first matirx, i.e. given the distinct real eigenvalues:

$R\Sigma = \Sigma\Lambda \Longleftrightarrow R=\Sigma\Lambda\Sigma^{-1}$

Consequently,

$R \vec{x_n} =R^2 \vec{x_{n-1}}= R^n \vec{x_1}=\vec{x_{n+1}}\Longleftrightarrow \Sigma \Lambda^n\Sigma^{-1} \vec{x_1} =\vec{x_{n+1}}$

where $\vec{x_n}=\left (\begin{array}{l} x_{n-1}\\x_{n}\end{array}\right )$

Let $x_0=0$ and $x_1 =1$ then

$\Sigma^{-1} \vec{x_1} =\frac{1}{\beta-\alpha} \left (\begin{array}{l} -1 \\ 1\end{array}\right )$

$\Sigma\Lambda^n =\left ( \begin{array}{ll} \alpha^n &\beta^n \\ \alpha^{n+1}&\beta^{n+1}\end{array}\right )$

Therefore,

$\left ( \begin{array}{l} x_n\\x_{n+1}\end{array}\right ) = \left ( \begin{array}{l} \frac{\beta^n-\alpha^n}{\beta-\alpha}\\\frac{\beta^{n+1}-\alpha^{n+1}}{\beta-\alpha}\end{array}\right )$

Simplifying,

$x_n = \frac{\beta^n-\alpha^n}{\beta-\alpha}$

For the Fibonacci series $A =B=1$ and $\beta =\frac{1 +\sqrt{5}}{2}=\phi$ and $\alpha =\frac{1-\sqrt{5}}{2}=1-\phi=-\frac{1}{\phi}$ and  hence $\beta -\alpha =\sqrt{5}$

Therefore,

$F_n =\frac{\phi^n -(1-\phi)^n}{\sqrt{5}}=\frac{\phi^n -(-\phi^{-1})^n}{\sqrt{5}}$

An integer on the left hand side and an expression of irrationals on the right!