Question

Prove the Well ordering property of Z by induction and prove the minimal element is unique

It seems so obvious lol but I don't really know how to go about it

Answer 1

n < n+1

basis step: 1 < 2, is true

assume it is true for some k in Z

k < k+1 ; add 1 to each side ?

k+1 < (k+1)+1

Answer 2

I think I have to use sets though..
The books defines it as:
If \(A\) is any nonempty subset of \(\mathbb{Z}^+\), there is some element \(m \in A\) such that \(m < a\), for all \(a \in A\) (\(m\) is called a minimal element of \(A\)).

So my basis step was \(1 \in A\), then obviously a has a least element since \(1 \leq n\) for every \(n \in \mathbb{Z}^+\)

Then assume \(m \in A\), either m is a least element of A, or there is an integer n so that \(1 \leq n < m\)
And similarly the same would go for n, so you always have a least element

but I'm not sure if this is a good way to prove it

Answer 3

ive seen proofs that show that if p <= q, and q <= p, then p=q which is the standard practice of showing that two values p and q are the same.

but i still get a headache trying to read thru them :/

Answer 4

i never really know how much of the wheel has to be reinvented during "proofs" either

Answer 5

But it seems so weird to prove that it's unique since Z contains no duplicate elements so any element of a subset should be unique right?

Answer 6

wait scratch that brainfart

Answer 7

For proving it's unique:
Let \(m, n \in A\)
Assume \(n \in A\) is a least element, then \(n \leq a, \quad \forall a\in A \)
So \(n \leq m\)

Similary, \(m \leq n\)
so \(m = n\) and it's unique

Answer 8

That would work right :D

Answer 9

looks familiar yes :)

Answer 10

Is the induction proof above (that it has a least element) good or not? I think it is but I'm not that great with proofs :s