Show, for $F(n)$ the $n$-th Fibonacci number, that $\gcd(F(n),F(n-1))=1$, and that the Euclidean algorithm calculation takes $n-2$ steps 
Let $F(n)$ be the $n^{th}$ Fibonacci term where $F(1) = F(2) = 1$. Show that $GCD(F(n), F(n - 1)) = 1, n >  1$ and that if one calculates $GCD(F(n), F(n - 1)) = 1$ with Euclides algorithm, its going to take $n - 2$ steps. 

My approach to this was that since the GCD was equivalent to 1, then i could attempt to prove this by induction. 
When $n = 2$, $GCD(F(2), F(2 - 1)) = 1$ satisfied the equation
When $n = k$, $GCD(F(k), F(k - 1)) = 1$ satisfied the equation
When $n = k + 1$, $GCD(F(k + 1), F(k)) = 1$ satisfied the equation
So the task here was to prove that the term in $n = k + 1$ results in a 1. 
I dont know if this approach is the best but am a little bit doubtful on the journey am taking. Could i use some help here please?
 A: You're on the right road.
I'll denote $\text{GCD}(a,b)$ by $a\wedge b$.
So assuming $F_{k+1}\wedge F_k=1$, you want to show that $F_{k+2}\wedge F_{k+1}=1$. A common technique is to consider an arbitrary common divisor of $F_{k+2}$ and $F_{k+1}$, let's call it $d$, and you show that $d\mid 1$. That would mean that $d=1$ and hence there's no common divisor other than $1$.
So let $d$ be a common divisor of $F_{k+2}$ and $F_{k+1}$. Then, you know that $d$ divides any combination of the form $aF_{k+2}+bF_{k+1}$ where $a,b\in\mathbb Z$. But you know that there's a relation between $F_{k+2}$ and $F_{k+1}$, namely $F_{k+2}=F_{k+1}+F_k$. So the simple combination you can think consider is $F_{k+2}-F_{k+1}=F_{k}$. This shows that $d$ divides $F_k$. Now, using the induction hypothesis, I let you conclude that $d=1$.
As for why the Euclidean algorithm takes $n-2$ steps, well use the fact that the Fibonacci sequence is increasing. So keep in mind that when you write $F_n=F_{n-1}+F_{n-2}$, you have $F_{n-2}<F_{n-1}$.
