Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**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.

Offline

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

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

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

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

Offline

**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 15,955

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.

Here lies the reader who will never open this book. He is forever dead.

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.

Offline

Pages: **1**