# Prove that the set of integers has neither a greatest nor a least element

1. Dec 3, 2011

### snes_nerd

Prove that the set of integers has neither a greatest nor a least element.

I was given a hint: There are 2 different non existence results to prove, so prove them as separate propositions or claims. Divide into cases using the definition of the set of integers.

So I was kind of confused on the hint. The way I was going to solve this was to divide into 2 cases and use induction. One case would involve using induction to prove their is no greatest element and one case would involve using induction to prove their is no least element. Am I even on the right track here? If this is not the way to go then I dont really know how to go about it.

2. Dec 3, 2011

### HallsofIvy

I don't think it is two "cases" that are intended but two things you want to prove:
1) That the set of integers does not have a largest member.
2) That the set of integers does not have a smallest member.

I wouldn't use induction- those both "negative" statements- something is NOT true- and proof by contradiction is most natural for such statements.

Suppose the set of integers does has a largest, member, n. What can you say about n+1?

3. Dec 3, 2011

### snes_nerd

I can say that n+1 is an integer to and n+1 > n. So this would contradict the proposition that n is the greatest element. Is this the right idea?

4. Dec 3, 2011

### micromass

A lot here depends on what axioms you can accept and on what definitions you make. If you can only accept the Peano axioms, then showing that n<n+1 might not be trivial.

5. Dec 3, 2011

### micromass

Yes, that's ok.

6. Dec 3, 2011

### snes_nerd

Thanks everyone for the help. I didnt realize it was that simple and straightforward. So for proving their is an integer that is not the least element I go about it like this: Suppose there is an integer that has a least element n. Then n-1 is an integer to. But n-1 < n which is a contraction. Thus their is no integer that has a least element n.

7. Dec 3, 2011

Seems ok!