Classification of mathematical objects

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Discussion Overview

The discussion revolves around the classification of mathematical objects, such as differentiable manifolds, abelian groups, and ordered fields. Participants explore the conditions under which classifications can be considered complete, the challenges posed by counterexamples, and the necessity of additional properties for successful classification.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation

Main Points Raised

  • Some participants suggest that a classification is considered complete when there is a theorem proving it, while others argue that certain objects, like differentiable manifolds, may never be fully categorized due to their complexity.
  • It is proposed that additional properties are often required to classify mathematical objects effectively, such as focusing on finite simple groups rather than all finite groups.
  • Participants discuss the implications of counterexamples on the completeness of classifications and the potential for generalizations in mathematical definitions and theorems.
  • An example of a complete classification is provided for finite simple groups, detailing the types included and the historical development of the proof.
  • Some participants express uncertainty about how one can demonstrate that all types for classification are accounted for without leaving any out.
  • Trivial classification theorems are presented, with discussions on their generalizability and the potential for counterexamples in broader cases.
  • There is a suggestion that classifications of specific groups cannot necessarily be generalized to all groups, indicating a limitation in the scope of certain classification theorems.

Areas of Agreement / Disagreement

Participants express a mix of agreement and disagreement regarding the feasibility of complete classifications, the role of additional properties, and the implications of counterexamples. The discussion remains unresolved on several points, particularly concerning the generalizability of specific classification theorems.

Contextual Notes

Limitations include the dependence on definitions of mathematical objects and the unresolved nature of generalizations proposed by participants. The complexity of proofs and the historical context of classification theorems are acknowledged but not fully explored.

  • #61
trees and plants said:
1) ##m-1\geq1 \Rightarrow m\geq2 \Rightarrow m\in S## contradiction because ##m-1\leq m ## and m is the smallest element of S.

2) (i)##m## is of the form ##2k## ##\Rightarrow## ##m-1## is an odd,

or (ii) ##m## is of the form ##2k-1## ##\Rightarrow## ##m-1## is an even, but we said that m-1 is not a natural number , contradiction.

Both cases show that there is no such ##m\in S##

I don't get this at all. ##m-1\notin S## means that ##m-## can be written as ##2k## or ##2k+1##. You need to use that.
 
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  • #62
<sigh>
Writing random statements hoping you will accidentally prove this is not a good strategy.

<double sigh>
If m is the smallest number that is neither even nor odd, is m-2 even? Is it odd? Is it neither?
 
  • #63
Office_Shredder said:
I don't get this at all. ##m-1\notin S## means that ##m-## can be written as ##2k## or ##2k+1#. You need to use that.
I am sorry for the mistakes before.
a) ##m-1\notin S##
(i) if ##m-1=2k## ##\Rightarrow## ##m=2k+1## ##m## is odd contradiction because we said that ##m\in S##

(ii) if ##m-1=2k-1## ##\Rightarrow## ##m=2k-2## ##\Rightarrow## ##m## is even, contradiction because we said that ##m\in S##

(b) if it is neither of these two forms then it is a contradiction because then ##m-1\leq m## but we know that ##m## is the smallest element of ##S##.
 
  • #64
Is this effort correct?
 
  • #65
So, i had to suppose that there is a set ##S## that contains the rest of the natural numbers and prove that it is empty.

Thank you for your help and time.
 
  • #66
If m-1=2k-1 then m=2k, not 2k-2. But now what you have looks a lot more like a correct proof.

I think you are still not handling the m-1 not a natural number case. That just means m is 1 or 0 depending on your definition of natural numbers. You need to confirm that is either even or odd also.
 
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  • #67
Office_Shredder said:
I think you are still not handling the m-1 not a natural number case.
This proof requires careful examination of the edge cases. Also, as originally posed (by you) it was a statement about integers, not natural numbers, and that will make zero an edge case as well.
 
  • #68
Vanadium 50 said:
This proof requires careful examination of the edge cases. Also, as originally posed (by you) it was a statement about integers, not natural numbers, and that will make zero an edge case as well.
Is the proof for integers analogous to the one i made? Is there anything else needed?
 
  • #69
trees and plants said:
Is the proof for integers analogous to the one i made?
You tell me. You won't get very far asking other people to prove things. You want to get better, you need to work yourself.
 

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