Consecutive Fibonacci Numbers are Coprime

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

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

where $\gcd \left\{{a,b}\right\}$ 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 initial definition of Fibonacci numbers, we have:
 * $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall n \ge 2: \gcd \left\{{F_n, F_{n+1}}\right\} = 1$

Basis for the Induction
$P(2)$ is the case $\gcd \left\{{F_2,F_3}\right\} = \gcd \left\{{2,3}\right\} = 1$, which holds.

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\displaystyle \forall n \ge 2: \gcd \left\{{F_k, F_{k+1}}\right\} = 1$

Then we need to show:
 * $\displaystyle \forall n \ge 2: \gcd \left\{{F_{k+1}, F_{k+2}}\right\} = 1$

Induction Step
This is our induction step:

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

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