Equivalence relations

J

Jhony

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Hey i have a question that i got stuck on for long time

E is equivalence relation on group A. B is another group and we know ∀ X ∈ A/E. |X| = |B|
Prove .|A/E|·|B| = |A|

And do we use Axiom Of Choice to solve this?

How do i do that?


Thanks Everybody :)
 
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We'll want to see at least some of the work you tried, so we can get a sense of what you have learned and see whether you made any mistakes we can correct. What theorems have you learned related to equivalence classes and associated concepts?
 
QUESTION: Do you mean by group an algebraic structure or is that word a miss-translation of collection or set?
 
Dr.Peterson said:
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Welcome to FreeMathHelp.com! Please take the time to read the following before you make your first post. It will help you to get your math questions answered promptly and in the most helpful manner. A summary is available here, but please read the complete guidelines below at your earliest...
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We'll want to see at least some of the work you tried, so we can get a sense of what you have learned and see whether you made any mistakes we can correct. What theorems have you learned related to equivalence classes and associated concepts?
Click to expand...

I proved that |A/E|= |A| i found a one to one function from A/E to A. But the question is to find function for this equivalence
|A/E|·|B| = |A|. Can you help me with an idea?

Thanks
 
Jhony said:
Its a question from the set theory
Click to expand...
So are you saying that the word is not "group"? If you translated from another language, please say so, and perhaps show us the original so we can check on it.

Jhony said:
I proved that |A/E|= |A| i found a one to one function from A/E to A. But the question is to find function for this equivalence
|A/E|·|B| = |A|. Can you help me with an idea?
Click to expand...
That doesn't sound right to me; maybe you didn't mean quite what you said. Please show us the function you used.
 
Dr.Peterson said:
So are you saying that the word is not "group"? If you translated from another language, please say so, and perhaps show us the original so we can check on it.


That doesn't sound right to me; maybe you didn't mean quite what you said. Please show us the function you used.
Click to expand...


I meant <= and not = sorry. This is what i could prove
 
That doesn't really change what I asked for. I'd like to see what you have done, and specifically your one-to-one function, as a basis for suggesting other things. In particular, I want to see what definitions you were given, and so on.
 
I'll try to make the image almost readable:

1585592851689.png

But I still can't tell what you are saying. Where did R come from? What is A'? Where are you using the definition of A/E?

What theorems have you learned in this topic?

Most important, you still haven't shown us the original problem and why you used the word "group".
 
Jhony said:
I proved that |A/E|= |A| i found a one to one function from A/E to A. But the question is to find function for this equivalence |A/E|·|B| = |A|.
Click to expand...
Please give a complete answer to Prof. Peterson's repeated questions along with the definitions.
For example: it is common to see \(\#(A)\text{ or }\|A\|\) used for the cordiality of a set \(A\) I assume that is what you mean by \(|A|\), right?
The symbol \(A/E\) usually stands for the collection so equivalence classes of \(A\) induced by \(E\).
Thus it makes sense to have \(|A/E|\le |A|\). What is meant by "find function for this equivalence \(|A/E|\cdot |B| = |A|~?\)"
Please answer.
 
pka said:
Please give a complete answer to Prof. Peterson's repeated questions along with the definitions.
For example: it is common to see \(\#(A)\text{ or }\|A\|\) used for the cordiality of a set \(A\) I assume that is what you mean by \(|A|\), right?
The symbol \(A/E\) usually stands for the collection so equivalence classes of \(A\) induced by \(E\).
Thus it makes sense to have \(|A/E|\le |A|\). What is meant by "find function for this equivalence \(|A/E|\cdot |B| = |A|~?\)"
Please answer.
Click to expand...

The question is:
E is equivalence relation on set A. B is another set and we know ∀ X ∈ A/E. |X| = |B|
Prove .|A/E|·|B| = |A|

Now i know in order to prove this equation i need to find an equivalence function (1 to 1 and onto). So im not sure what function i can define in order to prove it.

Thanks again
 
Last edited:
Jhony said:
The question is: E is equivalence relation on set A. B is another set and we know ∀ X ∈ A/E. |X| = |B|
Prove .|A/E|·|B| = |A| Now i know in order to prove this equation i need to find an equivalence function (1 to 1 and onto). So im not sure what function i can define in order to prove it.
Click to expand...
I am still concerned about this given: \((\forall X\in A/E)[|X|=|B|]\).
Strictly speaking that says that every equivalence class in \(A\) with respect to \(E\) has the same cardinality as set \(B\).
Among other things that would mean that each equivalence class in \(A/E\) has the same cardinality.
Because the collection of equivalence classes form a partition of the set. Thus \(\bigcup\limits_{X \in A/E} X = A\)
What can you do with those facts?
 
pka said:
I am still concerned about this given: \((\forall X\in A/E)[|X|=|B|]\).
Strictly speaking that says that every equivalence class in \(A\) with respect to \(E\) has the same cardinality as set \(B\).
Among other things that would mean that each equivalence class in \(A/E\) has the same cardinality.
Because the collection of equivalence classes form a partition of the set. Thus \(\bigcup\limits_{X \in A/E} X = A\)
What can you do with those facts?
Click to expand...


I understood it as each X which is an equivalence class has the cardinal as B.
So in faxt there is a 1 to 1 function from X to B that im not sure how to define.
 
Jhony said:
I understood it as each X which is an equivalence class has the cardinal as B.
So in faxt there is a 1 to 1 function from X to B that im not sure how to define.
Click to expand...
Start simple. Suppose \(A=\{0,1,2,3,4,5,6,7,8\}\) and \(B=\{a,s,d\}\) define \((x,y)\in E\iff x\cong y\mod 3\).
How many classes are there? What is the cardinally of each class?
Can you construct the question out of those? See if you can extend to \(A=\mathbb{Z}\); same \(E\) but different \(B\).
In general: \(\forall X\in A/E)(\exists {f_X}:~X\overset {1 - 1} \longleftrightarrow B)\), Now that is just a suggestion.
 
pka said:
Start simple. Suppose \(A=\{0,1,2,3,4,5,6,7,8\}\) and \(B=\{a,s,d\}\) define \((x,y)\in E\iff x\cong y\mod 3\).
How many classes are there? What is the cardinally of each class?
Can you construct the question out of those? See if you can extend to \(A=\mathbb{Z}\); same \(E\) but different \(B\).
In general: \(\forall X\in A/E)(\exists {f_X}:~X\overset {1 - 1} \longleftrightarrow B)\), Now that is just a suggestion.
Click to expand...

Hey
I thought that since X is an equivalence class with number of items which the cardinal is equal to cardinal B. To difine
f: X---->B by f(Xn,m)=Bn,m . while n define 1 specific equivalence class (when we have X1, X2....) and m is one item in this specific equivalence class
Then i get a 1 to 1 function.

Im just not sure about the formal writing of the function

What do you think?
 
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