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## #1 2013-04-11 13:06:37

mrpace
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### 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.
thanks.

## #2 2013-04-11 13:34:03

Agnishom
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### 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 2013-04-11 18:58:40

anonimnystefy
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### 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.

The limit operator is just an excuse for doing something you know you can't.
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