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

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Discussion Overview

The discussion revolves around proving that the set of integers has neither a greatest nor a least element. Participants explore different approaches to establish these two non-existence results, considering the definitions and axioms related to integers.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests dividing the proof into two separate claims: one for the absence of a greatest member and another for the absence of a least member, using induction.
  • Another participant argues against using induction for these "negative" statements, proposing proof by contradiction instead.
  • A participant proposes that if there exists a largest member n, then n+1 must also be an integer and greater than n, leading to a contradiction.
  • Concerns are raised about the reliance on axioms, particularly the Peano axioms, which may complicate the proof of n < n+1.
  • One participant outlines a proof for the absence of a least element by assuming there is a least element n and showing that n-1 is also an integer, leading to a contradiction.

Areas of Agreement / Disagreement

Participants generally agree on the structure of the proofs but differ on the methods to use, particularly regarding the appropriateness of induction versus proof by contradiction. The discussion remains unresolved on the best approach to take.

Contextual Notes

Participants express uncertainty regarding the axioms they can accept and how these might affect the proofs. There is also a mention of the potential complexity introduced by the Peano axioms.

snes_nerd
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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 don't really know how to go about it.
 
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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?
 
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?
 
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.
 
snes_nerd said:
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?

Yes, that's ok.
 
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.
 
snes_nerd said:
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.

Seems ok! :smile:
 

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