Classification of mathematical objects

[trees and plants]

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.

 

[fresh_42]

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.

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.

 

[trees and plants]

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.

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?

 

[fresh_42]

So a complete classification of some mathematical objects may sometimes be impossible? Because there are counterexamples?

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.

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. 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 can be applied to theorems, but theorems must not have a counterexample and must be proved?

Yes.

 

[trees and plants]

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.

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?

 

[fresh_42]

An example are finite simple groups. The theorem says:

If $G$ is a finite, simple group, then $G$ is one of the following groups:

• 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

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.

 

[trees and plants]

An example are finite simple groups. The theorem says:

If $G$ is a finite, simple group, then $G$ is one of the following groups:

• 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

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.

For them to show that there are no other possibilities they theorized there is one and reached contradiction?

 

[fresh_42]

For them to show that there are no other possibilities they theorized there is one and reached contradiction?

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 has
808 017 424 794 512 875 886 459 904 961 710 757 005 754 368 000 000 000 elements.

 

[Office_Shredder]

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.

 

[trees and plants]

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.

For the first and the second you wrote, to show that no more is of another kind, they theorized 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.

 

[Mark44]

For the first and the second you wrote, to show that no more is of another kind, they theorized another element and it lead to the already proven kinds?

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 third theorem, perhaps it can not be generalised? I think it can not.

Generalized in what way? Groups of larger size?

I have not tried to prove it but i think counterexamples perhaps can be shown that contradict more general cases.

You first need to state what you mean by "generalizations."

 

[trees and plants]

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."

One generalisation could be a group of 4n elements where n is natural number.

 

[Office_Shredder]

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.

 

[Mark44]

One generalisation could be a group of 4n elements where n is natural number.

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.

 

[trees and plants]

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.

I am sorry for that.

 

[trees and plants]

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.

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.

 

[Office_Shredder]

Why don't you post your proof then?

 

[trees and plants]

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?

 

[trees and plants]

For the the second statement, my proof is we have Z which is the set of the integer numbers. Z is equal to N union with -N union with {0}. So by having m belongs to Z and m=k, m=-k. m=0 for k>0 the proof is now easy i think. We apply for the first two equations similar procedure as i showed before for N and 0 belongs to Z so we get the proof of the statement.

 

[mathwonk]

i recommend you read a proof that classifies all compact topological surfaces. as i recall vaguely, they all arise, starting from the sphere, by adding on handles and crosscaps.

 

[pbuk]

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 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.

A dictionary that included every technical term from every field would be useless: almost any pronouncable combination of letters means something to someone!

 

[jbriggs444]

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?

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.

A good rule of thumb is that if you want to prove something about every integer, then mathematical induction is the appropriate tool.

Side note: It is very common for someone unfamiliar with mathematics to look at the word "induction" and immediately think "oh yes, I know what that is: 'deduction' is when you use a general principle and arrive at a conclusion about a specific case. 'induction' on the other hand is when you start with a bunch of specific cases and use them to guess at a general rule. Induction can never be used for a true proof because it is not 100% reliable".​

​

That is not what mathematical induction is. Mathematical induction is a valid tool for deductive reasoning. It is 100% reliable.​

​

An argument by mathematical induction takes a standard form:

1. Show that a particular property holds for the first natural number.
2. Show that whenever that property holds for a natural number n, it necessarily holds for n+1.
3. Conclude that the property holds for all natural numbers.

An inductive proof for the case at hand is simple. The property we will use is "is either odd or even". We will use your definition for odd: "Is equal to 2k-1 for some natural number k" and even: "Is equal to 2k for some natural number k"

1. The first natural number is either odd or even.

Proof: 1 is the first natural number. For natural number k=1, 1=2k-1

2. If n is a natural number that is either odd or even, n+1 is also either odd or even.

Proof:

Case 1: If n is odd then it is equal to 2k-1 for some natural number k. It follows that n+1 is equal to (2k-1)+1 = 2k. So n+1 is even.

Case 2: If n is even then it is equal to 2k-1 for some natural number k. It follows that n+1 is equal to 2k+1 = 2(k+1)-2. Since k+1 is a natural number, n+1 is odd.

3. By mathematical induction, we conclude that every natural number is either even or odd.

I leave it as an exercise for the interested reader (if any) to prove that no natural number is both even and odd.

 

[trees and plants]

I leave it as an exercise for the interested reader (if any) to prove that no natural number is both even and odd.

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.

 

[jbriggs444]

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.

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]

 

[trees and plants]

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]

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.

 

[trees and plants]

[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]

I have read in the past about Peano axioms.

 

[jbriggs444]

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.

Whoah. You start by using letters a through f for some reason. And then proceed to use k and l. It is going to be a tough slog to figure out where you are going with this pile of letters.

It looks like you confused yourself on line 1, writing "a=b" where you meant "a=c".

You start by supposing that a=2b-1 and c=2d.

One assumes that a, b, c and d are intended to be natural numbers. So at this point we know that a is odd and that c is even. Looking forward a bit it seems that you will be trying to prove that a is not equal to c. I should not have to guess. You should be providing some guidance as to how your proof is going to proceed.

Now you seem to be trying to split this into cases.

Case 1. If b = 2e and d = 2f [that is, if b and d are both even]
then if a=c then 2b-1 = 2d and then 2(d+b)=1

What you probably intended to write there is that 2(d-b) = 1.

No. I am not going to slog any further through this mess. The basic technique appears to be proof by contradiction: Assume the opposite of what is to be proved. Write down some complicated equations. Make some errors. Derive a contradiction. Declare success.

 

[trees and plants]

Whoah. You start by using letters a through f for some reason. And then proceed to use k and l. It is going to be a tough slog to figure out where you are going with this pile of letters.

It looks like you confused yourself on line 1, writing "a=b" where you meant "a=c".

You start by supposing that a=2b-1 and c=2d.

One assumes that a, b, c and d are intended to be natural numbers. So at this point we know that a is odd and that c is even. Looking forward a bit it seems that you will be trying to prove that a is not equal to c. I should not have to guess. You should be providing some guidance as to how your proof is going to proceed.

Now you seem to be trying to split this into cases.

Case 1. If b = 2e and d = 2f [that is, if b and d are both even]
then if a=c then 2b-1 = 2d and then 2(d+b)=1

What you probably intended to write there is that 2(d-b) = 1.

No. I am not going to slog any further through this mess. The basic technique appears to be proof by contradiction: Assume the opposite of what is to be proved. Write down some complicated equations. Make some errors. Derive a contradiction. Declare success.

I am sorry for this. a is an odd number with b a natural number , c is an even number where d is a natural number, then if b is an even number, e is natural number and d is an even number, f is a natural number. I made a mistake it is not a=b, but a=c to continue for the contradiction. Then i make some substitutions then because b is even, e must be odd, and because d is even, f must be odd. I put k,l for natural numbers and for two odd numbers (2k-1)+(2l-1) we have that equation. If something else is needed tell me if you want.

 

[jbriggs444]

I am sorry for this. a is an odd number with b a natural number , c is an even number where d is a natural number, then if b is an even number, e is natural number and d is an even number, f is a natural number. I made a mistake it is not a=b, but a=c to continue for the contradiction. Then i make some substitutions then because b is even, e must be odd, and because d is even, f must be odd. I put k,l for natural numbers and for two odd numbers (2k-1)+(2l-1) we have that equation. If something else is needed tell me if you want.

*sigh*. I know that the proof is going to fail because your ultimate contradiction does not arise from an axiom. That's because you have not identified the axioms you are using.

Let's recap and actually declare our variables.

Let a be an odd natural number. Let b be a natural number so that a=2b-1. The existence of b is assured by the definition of oddness.

Let c be an even natural number. Let d be a natural number so that c=2d. The existence of d is assured by the definition of evenness.

We are trying to demonstrate that a and b must be distinct. We will proceed with a proof by contradiction.

Suppose that a=b.

Then it follows that 2b-1 = 2d. Or, equivalently, that 2(d-b) = 1. We wish to show that this is impossible.

Since all natural numbers are either even or odd, there are at most four (possibly overlapping) cases to consider:

1. b even, d even
2. b even, d odd
3. b odd, d even
4. b odd, d odd.

You are trying to reason through case 1.

Let e be a natural number so that 2b=e. The existence of e is assured by the fact that e is even.
Let f be a natural number so that 2d=f. The existence of f is assured by the fact that f is even.

You now assert that if b is even then 2b must be odd.
You now assert that if d is even then 2d must be odd.

Why must either assertion hold?

 

[trees and plants]

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) so

4(e-f)=1=>4((2(k-l))=1 =>8(k-l)=1 contradiction . This is with the corrections i made.

 

[jbriggs444]

2(2e-2f)=1=>4(e-f)=1

but this is a contradiction because e and f must be odd integers and

WHY must e and f be odd integers. You've asserted that they are but you have not proved it.

You'll also want to prove that there is no natural number that, when multiplied by 8, yields 1.

But if you could prove that, then you could likely prove more easily that there is no natural number that, when multiplied by 4 yields 1.

But if you could prove that, then you could likely prove more easily that there is no natural number that, when multiplied by 2 yields 1.

But if you could do that then all these equations would have been wasted.

 

[trees and plants]

WHY must e and f be odd integers. You've asserted that they are but you have not proved it.

If they are a) e=2k and f=2l-1 or b) e=2k and f=2l., where k,l are natural numbers then it still is a contradiction.

 

[jbriggs444]

If they are a) e=2k and f=2l-1 or b) e=2k and f=2l., where k,l are natural numbers then it still is a contradiction.

So you are retracting the claim that e and f are odd. That's fine.

Now prove the contradiction. Proof by personal incredulity does not count.

If you had a definition for multiplication by 2, a definition for addition and some axioms for the behavior of the successor function, that might be a place to start.

 

[trees and plants]

a) (2k-1)-(2l)=2(k-l)-1 so

4(e-f)=1=>4((2(k-l)-1)=1 ,4 times an integer can not equal 1 contradiction

b) (2k)-(2l)=2(k-l) so

4(e-f)=1=>4((2(k-l))=1 =>8(k-l)=1 ,contradiction

for k,l natural numbers a) is the case where e is odd and f is even, b) is the case where e is even and f is even. May i ask something if this is correct do you know a similar proof of the statement like the one i probably made?

 

[jbriggs444]

a) (2k-1)-(2l)=2(k-l)-1 so

4(e-f)=1=>4((2(k-l)-1)=1 ,4 times an integer can not equal 1 contradiction

b) (2k)-(2l)=2(k-l) so

4(e-f)=1=>4((2(k-l))=1 =>8(k-l)=1 ,contradiction

for k,l natural numbers a) is the case where e is odd and f is even, b) is the case where e is even and f is even. May i ask something if this is correct do you know a similar proof of the statement like the one i probably made?

You say that 4 times an integer can not equal 1. But saying it is not the same as proving it. Prove it.