00:02
We're asked to prove that a of m plus 1 n is greater than or equal to a of mn, whenever m and n are non -negative integers, where a is the aclymine function.
00:15
Call that the acriman function, a of mn, is defined to be 2n when m is equal to 0, 0 when n is greater than or equal to 1 and n is equal to 0 2 when n is greater than equal to 1 and n is equal to 1 and a of m minus 1 a of m n minus 1 when m is greater than are equal to 1 and n is greater than or equal to 2 so to prove the statement first, let pm be the statement a of m plus 1n is greater than or equal to a of m, m, n for all non -negative integers n.
01:21
In fact, we can make this a strict inequality.
01:24
I don't know why they made it that way.
01:27
The problem statement, but yeah, for all n greater than are equal to zero.
01:42
So we'll use induction to prove this.
01:49
So for the basis, we have, well i suppose first, let's consider if n is equal to 0, then we have no matter what m is, a of m plus 1, n, well, m plus 1 has to be greater than or equal to 1, and so it follows that this is going to be if n is 0, it is greater than or equal to.
02:25
Okay, this is greater than or equal to then.
02:32
This is going to be a of m plus 1, 0, which is equal to 0, which is equal to.
02:43
And then we have, depending on what m is, a of m 0, is going to be 2 times 0, which is equal to 0, if m is equal to 0, and 0, if m is greater than or equal to 1.
03:08
And as we see in another case, it's going to be 0 for all m.
03:13
And so we've shown that a of m plus 1, 0, is greater than or equal to a of m 0.
03:23
In fact, they're equal.
03:25
And so we've proven that part.
03:28
Now, from here on out, we're just going to consider n greater than or equal to 1, the rest of the entire proof.
03:38
Now, for the base case, we have that, suppose the m is equal to 0, well, we'll be.
03:45
We have a of 0 plus 1, which is 1, and then n.
04:02
We have a previous exercise, this is 2 to the n.
04:07
And we have that a of 0, n, since n is 0, this is simply going to be 2 times n.
04:17
And we know for a fact that 2 to the n is greater than or equal to 2 times n, which is a of 0 n.
04:39
So it follows that p0 is true.
04:49
Now let's consider an m equals 1.
04:53
We have that a of 1 plus 1 is 2n.
04:59
And this is by the recursive definition, a of 1, a of 1, and minus 1.
05:20
And recall from a previous problem, we have that this is the same as 2 .2.
05:28
To the a of 1 n minus 1.
05:38
My mistake, this should be a of 1, a of 2, n minus 1.
05:46
And instead of writing it like this, i think i'm going to write it out more explicitly.
05:51
So this is equal to a of 1, a of again, a of 1, a of 2, n minus 2...