# Fibonacci Number in terms of Smaller Fibonacci Numbers

## Theorem

Let $F_k$ be the $k$th Fibonacci number.

Then:

$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$

### Negative Indices

Let $n \in \Z_{< 0}$ be a negative integer.

Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).

$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$

continues to hold, whether $m$ or $n$ are positive or negative.

## Proof 1

From the initial definition of Fibonacci numbers, we have:

$F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

Proof by induction:

For all $n \in \Z_{>0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \forall m \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$

### Basis for the Induction

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

 $\displaystyle F_{m + 1}$ $=$ $\displaystyle F_{m - 1} + F_m$ Definition of Fibonacci Number $\displaystyle$ $=$ $\displaystyle F_{m - 1} \times 1 + F_m \times 1$ $\displaystyle$ $=$ $\displaystyle F_{m - 1} F_1 + F_m F_2$ Definition of Fibonacci Number $\displaystyle$ $=$ $\displaystyle F_{m - 1} F_n + F_m F_{n + 1}$ for $n = 1$

and so $P \left({1}\right)$ is seen to hold.

$P \left({2}\right)$ is the case:

 $\displaystyle F_{m + 2}$ $=$ $\displaystyle F_{m + 1} + F_m$ Definition of Fibonacci Number $\displaystyle$ $=$ $\displaystyle F_{m - 1} + F_m + F_m$ Definition of Fibonacci Number $\displaystyle$ $=$ $\displaystyle F_{m - 1} \times 1 + F_m \times 2$ $\displaystyle$ $=$ $\displaystyle F_{m - 1} F_2 + F_m F_3$ Definition of Fibonacci Number $\displaystyle$ $=$ $\displaystyle F_{m - 1} F_n + F_m F_{n + 1}$ for $n = 2$

and so $P \left({2}\right)$ is seen to hold.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ and $P \left({k-1}\right)$ are true, where $k > 1$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle F_{m + k} = F_{m - 1} F_k + F_m F_{k + 1}$

and:

$\displaystyle F_{m + k - 1} = F_{m - 1} F_{k - 1} + F_m F_k$

from which it is to be shown:

$\displaystyle F_{m + k + 1} = F_{m - 1} F_{k + 1} + F_m F_{k + 2}$

### Induction Step

This is our induction step:

 $\displaystyle F_{m + k + 1}$ $=$ $\displaystyle F_{m + k} + F_{m + k - 1}$ Definition of Fibonacci Number $\displaystyle$ $=$ $\displaystyle F_{m - 1} F_k + F_m F_{k + 1} + F_{m - 1} F_{k - 1} + F_m F_k$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle F_{m - 1} \left({F_k + F_{k - 1} }\right) + F_m \left({F_{k + 1} + F_k}\right)$ $\displaystyle$ $=$ $\displaystyle F_{m - 1} F_{k + 1} + F_m F_{k + 2}$ Definition of Fibonacci Number

So $P \left({k}\right) \land P \left({k - 1}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall m, n \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$

$\blacksquare$

## Proof 2

 $\displaystyle$  $\displaystyle F_{m - 1} F_n + F_m F_{n + 1}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\phi^{m - 1} - \hat \phi^{m - 1} } {\sqrt 5} \dfrac {\phi^n - \hat \phi^n} {\sqrt 5} + \dfrac {\phi^m - \hat \phi^m} {\sqrt 5} \dfrac {\phi^{n + 1} - \hat \phi^{n + 1} } {\sqrt 5}$ Euler-Binet Formula $\displaystyle$ $=$ $\displaystyle \dfrac {\phi^{m + n - 1} - \phi^{m - 1} \hat \phi^n - \phi^n \hat \phi^{m - 1} + \hat \phi^{m + n - 1} + \phi^{m + n + 1} - \phi^m \hat \phi^{n + 1} - \phi^{n + 1} \hat \phi^m + \hat \phi^{m + n + 1} } 5$ $\displaystyle$ $=$ $\displaystyle \dfrac {\phi^{m + n - 1} \left({1 + \phi^2}\right) + \hat \phi^{m + n - 1} \left({1 + \hat \phi^2}\right) - \phi^{m - 1} \hat \phi^n \left({1 + \phi \hat \phi}\right) -\phi^n \hat \phi^{m - 1} \left({1 + \phi \hat \phi}\right)} 5$ $\displaystyle$ $=$ $\displaystyle \dfrac {\phi^{m + n - 1} \left({1 + \phi^2}\right) + \hat \phi^{m + n - 1} \left({1 + \hat \phi^2}\right)} 5$ as $\phi \hat \phi = -1$ $\displaystyle$ $=$ $\displaystyle \dfrac {\phi^{m + n - 1} \left({2 + \phi}\right) + \hat \phi^{m + n - 1} \left({2 + \hat \phi}\right)} 5$ as both $\phi$ and $\hat \phi$ satisfy $x^2 = x + 1$ $\displaystyle$ $=$ $\displaystyle \dfrac {\phi^{m + n - 1} \left({2 + \dfrac {1 + \sqrt 5} 2}\right) + \hat \phi^{m + n - 1} \left({2 + \dfrac {1 - \sqrt 5} 2}\right) } 5$ Definition of $\phi$ and $\hat \phi$ $\displaystyle$ $=$ $\displaystyle \dfrac {\phi^{m + n - 1} \left({\dfrac {5 + \sqrt 5} 2}\right) + \hat \phi^{m + n - 1} \left({\dfrac{ 5 - \sqrt 5} 2}\right)} 5$ $\displaystyle$ $=$ $\displaystyle \dfrac {\phi^{m + n - 1} \left({\dfrac{1 + \sqrt 5} 2}\right) - \hat \phi^{m + n - 1} \left({\dfrac {1 - \sqrt 5} 2}\right)} {\sqrt 5}$ dividing numerator and denominator by $\sqrt 5$ $\displaystyle$ $=$ $\displaystyle \dfrac {\phi^{m + n} - \hat \phi^{m + n} } {\sqrt 5}$ Definition of $\phi$ and $\hat \phi$ $\displaystyle$ $=$ $\displaystyle F_{m + n}$ Euler-Binet Formula

$\blacksquare$