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#1 2013-04-10 15:06:37

Registered: 2012-08-16
Posts: 74

Show that ~ is an equivalence relation

so yea i'm not really understanding what an equivalence relation is even. Can anyone do this problem?

Let ~ be the relation defined on Z by
m~n <--> 2 devides m+n

show that ~ is an equivalence relation
describe the partition of Z determined by the equivalence classes of ~

any help is much appreciated.


#2 2013-04-10 15:34:03

Real Member
From: Riemann Sphere
Registered: 2011-01-29
Posts: 24,850

Re: Show that ~ is an equivalence relation

1. m~m since 2|m+m = 2m (Thats reflexivity)
2. Let us take for granted m~n which means 2|m+n.
    Again, n~m, means 2|n + m. Now m+n=n+m
    Therefore, IF m~n, THEN n~m (Thats symmetry)
3. Let us take for granted m~n and n~o
    Therefore, 2|m+n and 2|n+o. Let m + n = 2k and n + o = 2j, Now m + o = 2k + 2j - 2n = 2(k+j -n)
    Thus 2|m+o; Thus, m~o (Thats Transitivity)

From 1, 2 and 3 ~ is symmetrical, transitive and reflexive. Therefore ~ is an equivalence relation


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#3 2013-04-10 20:58:40

Real Member
From: Harlan's World
Registered: 2011-05-23
Posts: 16,037

Re: Show that ~ is an equivalence relation

And, you will notice that x~y if and only if x and y are of the same parity, so the classes of equivalence in Z are the even integers and the odd integers.

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