Consecutive Fibonacci Numbers are Coprime

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

Then:
 * $\forall n \ge 2: \gcd \set {F_n, F_{n + 1} } = 1$

where $\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.

That is, a Fibonacci number and the one next to it are coprime.

Proof
From the definition of Fibonacci numbers:
 * $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $\gcd \set {F_n, F_{n + 1} } = 1$

Basis for the Induction
$\map P 2$ is the case:
 * $\gcd \set {F_2, F_3} = \gcd \set {2, 3} = 1$

Thus $\map P 2$ is seen to hold.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\gcd \set {F_k, F_{k + 1} } = 1$

Then we need to show:
 * $\gcd \set {F_{k + 1}, F_{k + 2} } = 1$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \ge 2: \gcd \set {F_n, F_{n + 1} } = 1$