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 .
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 .
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...
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.
Hello K A D C DilshanYou might find it helpful to study the examples and my replies in this posting.
Grandad
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 .