Can somebody clarify me how equivalence classes are obtained ?

And how can I answer this type of Qn ?

Qn:- Let S={1,2,3,4,5} have a partition consisting of the sets {1,3,5} and {2,4} .Show that this partition determines an equivalence relation .

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- Apr 9th 2009, 07:31 PMK A D C Dilshan[SOLVED] Equivalance Relations
Can somebody clarify me how equivalence classes are obtained ?

And how can I answer this type of Qn ?

Qn:- Let S={1,2,3,4,5} have a partition consisting of the sets {1,3,5} and {2,4} .Show that this partition determines an equivalence relation . - Apr 9th 2009, 10:23 PMdr3amrunn3r
An equivalence relation always forms a partition on the set in which it is defined. Each cell, or element, of the partition is an equivalence class.

For example, in your set:

{1,2,3,4,5} you have the partition

{1,3,5}

{2,4}

A good choice would be the equivalence relation:

element a of A is related to element b of A, if:

a is of the same parity as b

notice {1, 3, 5} are all odd,

{2,4} are both even.

I'll leave it up to you to confirm that this equivalence relation is in fact an equivalence relation:

It's reflexive, a R a

Symmetric, a R b implies b R a

Transitive.. a R b and b R c implies a R c....

a R b is the conventional notation used to denote element a is Related to b... - Apr 9th 2009, 10:35 PMdr3amrunn3r
Also. Given an equivalence relation, an equivalence class is formally defined and denoted as the following:

Given an equivalence relation defined on some set A

[a] : {b ϵ A | b R a}

That is, the set of all the elements of A which are related to a.

As an example, consider the equivalence class of nonnegative integers congruent to 1 modulo 2:

[1(mod 2)] = {1, 3, 5, 7, 9, 11, etc} essentially, all the odd integers.

similarly the equivalence class of integers congruent to 0 mod 2:

[0 (mod 2)] = {0, 2, 4, 6, 8, 10, etch} all the even nonnegative integers.

The relation a R b if a is congruent to b modulo 2 is another equivalence relation that might help you with your question. The only equivalence classes of this relation are [0] and [1]. Together, these two equivalence classes partition all of the integers in the set on which this relation is defined. - Apr 9th 2009, 10:36 PMdr3amrunn3r
But I wouldn't directly copy this. I'm a bit rusty on my notation.

- Apr 10th 2009, 12:46 AMGrandadEquivalence Relations
Hello K A D C DilshanYou might find it helpful to study the examples and my replies in this posting.

Grandad - Apr 10th 2009, 02:10 AMK A D C DilshanThank you !!!!!
Actually dr3amrunn3r I had this idea before .

Your definition of equivalence classes is not correct I think .

I think it should be corrected as follows

[a] = {x : (a,x) ∊ R} Isn`t it ?

so [1(mod2)] and [2(mod2)] are the partitions as I thought as we can not take 0 .

But I had no experience on building relations using partitions .

I think a is congruent to b modulo 2 is something similar to a≣b(mod2) i.e 2 | (a-b)

Thank you very much for confirming the solution .

If I have done any mistake please tell me .