Assignment 1: Problem 3b

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Sep 15, 2010, 10:54:59 PM9/15/10

to CSC 2511 Fall 2010

{ (x, y) $\in \!\,$ R | abs(x - y) = 2 }

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Sep 15, 2010, 10:59:17 PM9/15/10

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Set A is symmetric if a, b $\in \!\,$ A and (a, b) $\in \!\,$ R, then (b, a) $\in \!\,$ R

A = { (x, y) $\in \!\,$ R | abs(x - y) = 2 } -> { (y, x) $\in \!\,$ R | abs(y - x) = 2 }

example

A = { (3, 1) $\in \!\,$ R | abs(3 - 1) = 2 } -> { (1, 3) $\in \!\,$ R | abs(1 - 3) = 2 }

2010/9/15 Aaron Albers <aaronc...@gmail.com>

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Sep 16, 2010, 1:18:02 PM9/16/10

to CSC 2511 Fall 2010

The binary relation R is not reflexive on N.

Proof by contradiction.

The binary relation R is reflexive on N if and only if a ~ a for all a ∈ N

( 1 ∈ N ) but ( 1 !~ 1 ) since ( | 1 - 1 | != 2 )

I think I am finally starting to get it.

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Sep 16, 2010, 1:25:07 PM9/16/10

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That one is symmetric only. you can show that it's not transitive or antisymmetric by a counterexample.

You already showed that it was symmetric.

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Sep 16, 2010, 1:32:06 PM9/16/10

to CSC 2511 Fall 2010

The binary relation R is symmetric on N.

Proof

The binary relation R is symmetric on N if and only if a,b ∈ N and a ~ b, then b ~ a.

If a,b ∈ N and | a - b | = 2, then | b - a | = 2

Just rehashing what I stated earlier. If anyone can expand on this go for it. I dont know how to be more specific.

On Wed, Sep 15, 2010 at 8:59 PM, Aaron Albers <aaronc...@gmail.com> wrote:

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