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Hello. When does someone know a classification of some kind of mathematical objects has been completed? For example of differentiable manifolds, or abelian groups or ordered fields? Thank you.
You know it when there is a theorem that proves it. Differential manifolds are too manifold to finally categorize them, Abelian groups and ordered fields are known.Hello. When does someone know a classification of some kind of mathematical objects has been completed? For example of differentiable manifolds, or abelian groups or ordered fields? Thank you.
So a complete classification of some mathematical objects may sometimes be impossible? Because there are counterexamples? The same is for generalisations of theorems or definitions of things in math? Or perhaps when having a definition of math object and someone changes it a little and gets another example then this may lead to a generalisation? The same can be applied to theorems, but theorems must not have a counterexample and must be proved?You know it when there is a theorem that proves it. Differential manifolds are too manifold to finally categorize them, Abelian groups and ordered fields are known.
You often need additional properties to get a chance for a classification, e.g. finite simple groups instead of all finite groups. Unclassified examples are nilpotent or solvable objects, groups as well as rings and algebras. There are too many of them, so any classification will probably - if at all - be done over additional constraints.
No, because there are too many of a kind. E.g. if you want to classify all stars, then you will have a problem. But you can introduce additional properties like the main sequence or certain sorts like pulsars, and concentrate on them. The same is true in mathematics. "Classify all nilpotent finite groups!" is not doable for there are too many of them. You will need additional properties as e.g. the nilpotency class (degree), or those with a given center.So a complete classification of some mathematical objects may sometimes be impossible? Because there are counterexamples?
Yes. In the example of nilpotent groups you could require a certain kind of automorphism group, anything that breaks down the task into portions that can be handled.The same is for generalisations of theorems or definitions of things in math? Or perhaps when having a definition of math object and someone changes it a little and gets another example then this may lead to a generalisation?
Yes.The same can be applied to theorems, but theorems must not have a counterexample and must be proved?
I do not understand how with a theorem a person shows that he has all types for classification and none is left out of it. Could you tell me?No, because there are too many of a kind. E.g. if you want to classify all stars, then you will have a problem. But you can introduce additional properties like the main sequence or certain sorts like pulsars, and concentrate on them. The same is true in mathematics. "Classify all nilpotent finite groups!" is not doable for there are too many of them. You will need additional properties as e.g. the nilpotency class (degree), or those with a given center.
Yes. In the example of nilpotent groups you could require a certain kind of automorphism group, anything that breaks down the task into portions that can be handled.
Yes.
For them to show that there are no other possibilities they theorized there is one and reached contradiction?An example are finite simple groups. The theorem says:
If ##G## is a finite, simple group, then ##G## is one of the following groups:
This is a complete classification for finite simple groups. The proof of the theorem has of course many steps and is historically spread over decades. You have to show that these groups are all finite and simple, and that there are no other possibilities.
- cyclic groups of prime order
- alternating groups ##A_n## with ##n>4##
- groups of Lie type over a finite field
- one of 26 sporadic (known) groups
They have found the 26 sporadic ones by assuming there is one, and finally that there is no room for a 27th. The proof is pretty complicated. The biggest of them hasFor them to show that there are no other possibilities they theorized there is one and reached contradiction?
For the first and the second you wrote, to show that no more is of another kind, they theorised another element and it lead to the already proven kinds? For the third theorem, perhaps it can not be generalised? I think it can not. I have not tried to prove it but i think counterexamples perhaps can be shown that contradict more general cases.Here are some simple classification theorems, of various levels of triviality.
Every integer is either even or odd.
Every integer is either zero, positive, or negative.
Every group of four elements is either ##\mathbb{Z}_4## or ##\mathbb{Z}_2 \times \mathbb{Z}_2##.
Every abelian group is either the direct sum of cyclic groups, or is infinite.
As @Office_Shredder mentioned, some of the statements he posted are very trivial. It's very easy to show directly that any integer is either evenly divisible by 2 or it is not. The same is true for showing that any integer falls into exactly one of the three possible divisions: negative, zero, or positive.For the first and the second you wrote, to show that no more is of another kind, they theorised another element and it lead to the already proven kinds?
Generalized in what way? Groups of larger size?For the third theorem, perhaps it can not be generalised? I think it can not.
You first need to state what you mean by "generalizations."I have not tried to prove it but i think counterexamples perhaps can be shown that contradict more general cases.
One generalisation could be a group of 4n elements where n is natural number.As @Office_Shredder mentioned, some of the statements he posted are very trivial. It's very easy to show directly that any integer is either evenly divisible by 2 or it is not. The same is true for showing that any integer falls into exactly one of the three possible divisions: negative, zero, or positive.
Generalized in what way? Groups of larger size?
You first need to state what you mean by "generalizations."
This is now getting ridiculous. Office_Shredder simply gave some examples of very simple classifications. It was not meant to be an invitation to extend any of them to be more general.One generalisation could be a group of 4n elements where n is natural number.
I am sorry for that.This is now getting ridiculous. Office_Shredder simply gave some examples of very simple classifications. It was not meant to be an invitation to extend any of them to be more general.
Yes, i understand now that the first one is trivial but i am not sure if my proof is correct, because i have a problem sometimes when i am trying to prove a statement in math until now.Sure. In the same way that the classification of simple finite groups can't be generalized to a classification of all groups. But if you're wondering what the proof of a classification theorem looks like, proving those would be a good place to start. The first only requires knowing how remainders work when doing integer division, or perhaps is trivial depending on your definition of an odd integer.
No the arrogance is yours. The dictionary was not created for you, it was created for everyone. And when the majority of people type 'crosscaps' they have made a mistake. All you have to do is add the word to your local configuration of the dictionary.The programmer who did this is not an imbecile, rather he is probably very bright, but perhaps arrogant. I.e. he thinks he knows what I mean to say better than I do. This is annoying.
No. It is a handwave, not a proof. What is missing is a demonstration that {1,2} union {3,4} union {5,6} and so on is equal to the set of all natural numbers. As it is, we have a handwave rather than a proof.Well, we have n=2k-1 for an odd number and n=2k for an even number, where k is a natural number. For k=1 they give 1,2 for k=2 they give 3,4, for k=3 they give 5,6 so by continuing for all k as natural numbers we have N, which is the set of all natural numbers. Is this correct?
Thank you. I knew about mathematical induction, i was not sure if i should use it. If we have n=2k-1 for odds and n=2k for evens then a number can not be even and odd because 2k≠2k-1 because 0≠1. If we assume they are equal it leads to a contradiction.I leave it as an exercise for the interested reader (if any) to prove that no natural number is both even and odd.
No. That proof does not work.Thank you. I knew about mathematical induction, i was not sure if i should use it. If we have n=2k-1 for odds and n=2k for evens then a number can not be even and odd because 2k≠2k-1 because 0≠1. If we assume they are equal it leads to a contradiction.
Let us try. So, Suppose we have a=2b-1, c=2d, (i) if b=2e, d=2f, then if a=b=>2b-1=2d=>2(d+b)=1=>2(2e+2f)=1=>4(e+f)=1 but this is a contradiction because e and f must be odd integers and 2k-1+2l-1=2((k+l)-1) so 4(e+f)=1=>4((2(k+l)-1)=1 which means that 4 times an odd number is equal to 1, contradiction. Similar proofs are for the cases where b is an odd number and d an even , and b is an even number and d is an odd.No. That proof does not work.
You have not exhibited a guarantee that a "k" that exists to demonstrate that n=2k is even is the same "k" that exists to demonstrate that n=2k-1 is odd. [Which indicates that you also have some opportunities for better education about variable naming, scopes and quantifiers].
Let me provide you with an axiom to make your work easier: 1 is not even. Or, if you prefer: 0 is not odd.
[It is not easy to prove that 1 is not even. Especially if you do not know what axioms or lemmas you have to work with]