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**mrpace****Member**- Registered: 2012-08-16
- Posts: 69

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.

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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

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

I'm not crazy, my mother had me tested.

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,954

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.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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