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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.
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.universe function said: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?fresh_42 said: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.universe function said: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?fresh_42 said: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?fresh_42 said: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 hasuniverse function said:For 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 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.Office_Shredder said: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.universe function said: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?
Generalized in what way? Groups of larger size?universe function said: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."universe function said: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.Mark44 said: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.universe function said:One generalisation could be a group of 4n elements where n is natural number.
I am sorry for that.Mark44 said: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.Office_Shredder said: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.mathwonk said: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.universe function said: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.jbriggs444 said: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.universe function said: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.jbriggs444 said: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]
I have read in the past about Peano axioms.jbriggs444 said:[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]
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.universe function said: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.
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 said: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.
*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.universe function said: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.
WHY must e and f be odd integers. You've asserted that they are but you have not proved it.universe function said:2(2e-2f)=1=>4(e-f)=1
but this is a contradiction because e and f must be odd integers and
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 said:WHY must e and f be odd integers. You've asserted that they are but you have not proved it.
So you are retracting the claim that e and f are odd. That's fine.universe function said: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.
You say that 4 times an integer can not equal 1. But saying it is not the same as proving it. Prove it.universe function said: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?
The purpose of classifying mathematical objects is to organize and categorize them based on their properties and characteristics. This helps to better understand and study these objects, as well as identify patterns and relationships between them.
Mathematical objects are classified based on their properties, such as shape, size, symmetry, and relationships with other objects. They can also be classified based on their purpose or function, such as numbers, shapes, or operations.
There are many different types of mathematical objects, including numbers, shapes, functions, equations, and graphs. These can be further divided into subcategories, such as natural numbers, polygons, linear functions, quadratic equations, and Cartesian coordinates.
A classification system for mathematical objects allows for easier organization and understanding of these objects. It also helps to identify commonalities and differences between objects, and can lead to new discoveries and advancements in mathematics.
Yes, mathematical objects can belong to multiple categories. For example, a square can be classified as a polygon, a quadrilateral, and a regular shape. This is because it possesses the properties and characteristics of all these categories.