Fibonacci String Ends with ab or ba
Theorem
Let $S_n$ be a Fibonacci string of length $n$.
Then for $n \ge 3$, $S_n$ ends either with $\text{ba}$ or with $\text{ab}$.
Proof
The proof proceeds by induction.
For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition:
- $S_n$ ends either with $\text{ba}$ or with $\text{ab}$
We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these ends either with $\text{ba}$ or with $\text{ab}$.
Basis for the Induction
$P \left({3}\right)$ is the case:
- $S_3 = \text{ba}$
Thus $P \left({3}\right)$ is seen to hold.
$P \left({4}\right)$ is the case:
- $S_4 = \text{bab}$
Thus $P \left({4}\right)$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
- $S_k$ ends either with $\text{ba}$ or with $\text{ab}$
and:
- $S_{k - 1}$ ends either with $\text{ba}$ or with $\text{ab}$
from which it is to be shown that:
- $S_{k + 1}$ ends either with $\text{ba}$ or with $\text{ab}$
Induction Step
This is the induction step:
By definition of Fibonacci string:
- $S_{k + 1} = S_k S_{k - 1}$
So $S_{k + 1}$ end with $S_{k - 1}$.
By the induction hypothesis, $S_k$ ends either with $\text{ba}$ or with $\text{ab}$.
Thus $S_{k + 1}$ likewise ends either with $\text{ba}$ or with $\text{ab}$.
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- for all $n \in \Z$ such that $n \ge 3$, $S_n$ends either with $\text{ba}$ or with $\text{ab}$.