Consecutive Fibonacci Numbers are Coprime

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

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

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 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 $P \left({n}\right)$ be the proposition:
 * $\forall n \ge 2: \gcd \left\{{F_n, F_{n + 1}}\right\} = 1$

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

Thus $P(2)$ is seen to hold.

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:
 * $\forall n \ge 2: \gcd \left\{ {F_k, F_{k + 1} }\right\} = 1$

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

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