00:01
Alrighty, so i wrote the well -ordering property of z down first.
00:06
So the well -ordering property of z says that if a is a non -empty subset of z plus the positive integers, there is some element m in a such that m is less than or equal to a for all a in a.
00:20
And we're going to prove this using induction and also using contradiction.
00:25
So let's begin the proof.
00:27
We're going to suppose a non -empty subset a of z plus has no minimal element.
01:09
Then it is a fact that 1 is not in a because 1 is less than or equal to a for all a in z plus.
01:43
So basically what we're saying here is that if we're supposing that a has no minimal element, then 1 can't be an element of a because 1 is the minimal element of all positive integers.
01:55
And so we're going to use this as a base case for our induction.
01:59
So we use this observation as the basis of the following induction argument.
02:21
Okay, so let it be true that there is some k in z plus such that 1, 2, dot, dot, dot, all the way up through k are not in a.
03:10
So the question is, is k plus 1 in a? the answer is no, because since 1, 2, all the way up through k is not in a, then if k plus 1 was in a, then k plus 1 would be a minimal element...