00:01
We're asked to determine the number of divisions is by the greatest common divisor of the fibonacci numbers on their consecutive, and the verified answer with mathematical and done.
00:19
So recall the definition of the fibonacci numbers, we have that f of 0 equals 0, f of 1 is equal 1, you have it f of n plus 2 is going to be equal to fm plus 1 plus f of n, or any n greater than an equality, 0, 1, i want to prove that the number of divisions by the clicking algorithm for ffn and ffm plus 1 is n minus 1, when n is greater than or equal to 2.
01:03
So to do this, let pn be this statement.
01:18
Now, the basis step, i guess before we actually do the proof by induction.
01:26
First, let's consider the other cases.
01:29
So we have that the number of divisions for f of 0 will n -z.
01:36
Equal to 0, then we have that f0 is already 0, and so the euclidian algorithm requires no divisions.
01:54
The greatest common divisor of the 0 is equal to, now that is equal to 1, then is equal to 1, then we have f of 1 equal 1, f2 is 1, and so we have the euclidean algorithm that requires exactly 1 in addition in this case, since f1 is visible by a and now we'll move on to the base case.
02:54
So now if n is equal to 2.
03:02
We're going to execute the euclidean algorithm.
03:21
So first of all, you have f of 1 is equal to 1, and f of 2 is equal to 1.
03:41
And then therefore we have that f of 2 is 1, and then therefore we have that f of 2 is 1, and so you have that f of 3 which is 2 is equal to 1.
03:59
Times 1 plus 1 which is going to be f of 2 times 1 plus 1 so it requires 1 which is equal to 2 minus 1 divisions i'm sorry in mistake here this should be f of 2 2 times 1 plus 0 which is f2 plus 2 plus 0 if you gave remainder of 0 that's mean 1 or 2 minus 1 but 8 now the inductive step...