How to prove that gcd of Fibonacci numbers set is $1$, I mean
$\gcd(a_{n+1},a_{n})=\gcd(a_{n},a_{n−1})$
because of Euclidean algorithm
$\gcd(a_{n+1},a_{n})=\gcd(a_{n},r)$
$r$ - the remainder after division of $a_{n+1}$ by $a_{n}$
but
$r=a_{n−1}$
it means that
$\gcd(a_{n+1},a_{n})=\gcd(a_{n},a_{n−1})$
And
$\gcd(a_{3},a_{2})=\gcd(a_{2},a_{1}) = 1$
it means that gcd of Fibonacci numbers set is $1$, right ?
GCD of all Fibonacci numbers is 1 simply because 1 is a Fibonacci number. There is nothing to prove here.
If you want to prove that consecutive Fibonacci numbers are relatively prime:
Suppose that there is some number $p$ dividing $a_{k}$ and $a_{k+1}$ for some $k$. It means that $p$ also divides their difference which is $a_{k+1}-a_k=a_{k-1}$. Repeat the same process and you’ll find out that $p$ also divides $a_{k-2}$, $a_{k-3}$, ..., $a_{1}=1$, which is possible only if $p=1$. So consecutive numbers must be coprime.