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**Abstract Algebra: Relations; Find a relation that is symmetric, etc**

## Homework Statement

Find a relation that is symmetric and transitive but not reflexive.

## Homework Equations

None, other than my chosen condition on the relation, namely: xy > |x + y|.

## The Attempt at a Solution

I'm sure this is on here every semester, but if possible I'd like some feedback on my attempt. This is a product relation in Z x Z.

Call the set O.

O = {(x,y)∈ Z

^{2}: xy > |x + y|}

**Symmetric:**

Since in the integers xy = yx and |x + y| = |y + x|, (y,x)∈ O (it is symmetric).

**Transitive:**

This means, if (a,b)∈ O and (b,c)∈ O then (a,c)∈ O, that is, if ab > |a + b| and bc > |b + c|, then ac > |a + c|.

First of all, none of a, b or c can be zero, otherwise somewhere along here you'd have 0 > a positive number, because the absolute values are taken.

Second, either a AND b are positive or a AND b are negative, or you'd get a negative number > a positive number.

In the first case, if a and b are positive, then by the assumption that bc > |b + c|, c has to be positive. In that case, it is true that ac > |a + c| (if a were 1 and c were 2 this would be false, but we're already assuming that ab > |a + b| and bc > |b+ c|).

**HOWEVER, I'm not exactly clear on how I would give a good reason for the above.**

The similar reasoning is done if a and b are negative. The relation seems to be

**transitive**. I need a little help in demonstrating this, assuming I'm right.

**Reflexive:**

It is NOT reflexive for the following reason: (1,1)∉ O. (1)(1) = 1, but |1 + 1| = 2, and 1 > 2 is nonsense.

What do you think? Thanks for any help.

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