00:01
We are asked to prove a statement about ackermann's function.
00:06
So recall that ackermann's function is defined so that a of mn is equal to 2n if m equals 0.
00:32
0, if m is greater than or equal to 1, and n is equal to 0.
00:40
2, if m is greater than equal to 1, and n equals 1, and a of m minus 1, a of m n minus 1 if m is greater than or equal to 1 and n is greater than or equal to 2 and we are asked to prove in this case that a 1 n equals 2 to the n whenever n is greater than are equal to 1 and to do this we'll use induction so first of all we have that the base case if n is equal to 1 we have that a 1 1 1 1 well this is when n equals 1 and n equals 1 so this is going to be 2 by definition which is the same as 2 to the first and so the base case checks out now let's assume that a of 1 n equals i guess you should do k to make it easier to differentiate the a of 1 k equals 2 to the k for some k greater than or equal to 1 you want to show that a of 1 k plus 1 equals 2 to the k plus 1.
02:19
So we have that's a of 1, k plus 1 is equal to.
02:26
Well, we have that m is equal to 1, which is greater than or equal to 1.
02:31
And we have that k plus 1, since k is greater equal 1, is going to be greater than equal to 2...