00:01
Okay, so here for part a, we let s be the set of positive integers where n is greater than equal to two for which the inequalities hold.
00:08
And then since we have that the square root of b squared minus two times squared of ab plus square of a squared of a squared is greater than zero, which then is going to imply that the square root of b minus the square root of a, all squared, is then greater than zero.
00:25
And then it follows that we have that a plus b over two is greater than the square root of a b.
00:35
So we get that a1 is greater than b1.
00:40
And now we have that a2 is then going to be equal to a1 plus b1 all over two, which is going to be less than a1.
00:51
And then we get that b2, which is going to be equal to the square root.
00:56
Of a1 times b1 is going to be less than b1.
01:02
And then by this argument, we get that a2 is going to be less than b2, because a2 is equal to a1 plus b1, all divided by 2, which is less than the square root of a1 times b1, which is going to be equal to b2.
01:21
And so we get that a1 is greater than a2, which is greater than b2, which is less than b1.
01:30
So we have that 2 is an element of s.
01:33
So we assume that let's say k is an element of s.
01:36
Then we have that a sub k plus 1 is going to be equal to a subk plus b subk, all divided by 2...