Fibonacci String Begins with ba
Theorem
Let $S_n$ be a Fibonacci string of length $n$.
Then for $n \ge 3$, $S_n$ begins with $\text {ba}$.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
- $S_n$ begins with $\text {ba}$
We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these begin with $\text{ba}$.
Basis for the Induction
$\map P 3$ is the case:
- $S_3 = \text {ba}$
Thus $\map P 3$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $S_k$ begins with $\text {ba}$
from which it is to be shown that:
- $S_{k + 1}$ begins with $\text {ba}$
Induction Step
This is the induction step:
By definition of Fibonacci string:
- $S_{k + 1} = S_k S_{k - 1}$
So $S_{k + 1}$ begins with $S_k$.
By the induction hypothesis, $S_k$ begins with $\text {ba}$.
Thus $S_{k + 1}$ likewise begins with $\text {ba}$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- for all $n \in \Z$ such that $n \ge 3$, $S_n$ begins with $\text {ba}$.