Finding GCD with Fibonacci: Base Case

RoboNerd
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Homework Statement



Suppose that m divisions are required to find gcd(a,b). Prove by induction that for m >= 1, a >= F(m+2) and b>= F(m+1) where F(n) is the Fibonacci sequence.

Hint: to find gcd(a,b), after the first division the algorithm computes gcd(b,r).

Homework Equations



Fibonacci Sequence: [/B]F(n) = F(n-1) + F(n-2)

Euclidean algorithm for finding the GCD of two numbers.

The Attempt at a Solution


[/B]
Hi everyone,

I know that the fibonacci sequence [F(n) = F(n-1) + F(n-2)] looks a lot like the expression
num = q*x + r, which is the expression for a number being divided by x with a quotient q and remainder r.

I know that I am supposed to prove by induction that for m >= 1, a >= F(m+2) and b>= F(m+1) where F(n) is the Fibonacci sequence, but I have no idea how to get started even with the base case or how to prove it.

Any guidance would be greatly appreciated. Thanks
 
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RoboNerd said:

Homework Statement



Suppose that m divisions are required to find gcd(a,b). Prove by induction that for m >= 1, a >= F(m+2) and b>= F(m+1) where F(n) is the Fibonacci sequence.

Hint: to find gcd(a,b), after the first division the algorithm computes gcd(b,r).

Homework Equations



Fibonacci Sequence: [/B]F(n) = F(n-1) + F(n-2)

Euclidean algorithm for finding the GCD of two numbers.

The Attempt at a Solution


[/B]
Hi everyone,

I know that the fibonacci sequence [F(n) = F(n-1) + F(n-2)] looks a lot like the expression
num = q*x + r, which is the expression for a number being divided by x with a quotient q and remainder r.

I know that I am supposed to prove by induction that for m >= 1, a >= F(m+2) and b>= F(m+1) where F(n) is the Fibonacci sequence, but I have no idea how to get started even with the base case or how to prove it.

Any guidance would be greatly appreciated. Thanks
For a base case, chose the smallest (distinct) positive integers. How many divisions does it take to find their gcd? What are the corresponding fibonacci numbers?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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