Proving an equivalence relation

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Homework Help Overview

The problem involves determining whether a defined relation R on the set of integers, where aRb if ab >= 0, qualifies as an equivalence relation. The original poster explores the properties of reflexivity, symmetry, and transitivity to assess this classification.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to demonstrate reflexivity by considering cases for positive, negative, and zero integers. They express uncertainty about the completeness of their reasoning.
  • For symmetry, the original poster initially believes the relation may not hold, but later questions their understanding of the implications of ab >= 0.
  • In discussing transitivity, the original poster presents a specific example to illustrate their reasoning but questions how to formally prove their assertion.
  • Participants raise concerns about the assumptions made in the original poster's reasoning, particularly regarding the implications of ab >= 0.

Discussion Status

Contextual Notes

Participants note that the original poster's approach may overlook certain cases, particularly when considering negative integers and the implications of the relation's definition. There is an emphasis on the need for careful consideration of the properties of the relation.

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Homework Statement


On the set of integers, define the relation R by: aRb if ab>=0.

Is R an equivalence relation?

Homework Equations





The Attempt at a Solution



R is an equivalence relation if it satisfies:

1) R is reflexive
Show that for all a∈Z, aRa.

Let a∈Z. Then if a is a negative integer, aa>=0. If a is a positive integer, aa>=0. And if a = 0, aa>=0.
Hence aRa

I feel like it is too simple.. lacking something??

2) R is symmetric
Show that for all a∈Z, aRb --> bRa

Let a∈Z, b∈Z such that aRb. By the definition of R, ab>=0.
This is not symmetric. Take a = -1, b = 2.
Then we have ab = -2 which is not >= 0.

3) R is transitive
if aRb and bRc implies aRc for all a,b,c ∈ Z

Let a, b, c ∈ Z s/t aRb, bRc --> aRc

Now I think this one is true.. but I'm not sure. But since aRb, and bRc, then you would always have ab or bc >=0 yea? so that means aRc must be true..
How would I prove it properly if it is correct?

Any help is appreciated! :) Thanks.
 
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1) is too simplistic because you didn't take account of the case a<0. 2) That (-1)R(2) is false doesn't prove anything. You want to prove if ab>=0 THEN ba>=0. For 3) suppose one of a,b and c is 0. Can you show with a simple example that it's not true?
 
1) I had that if it was a negative integer, which is the case of a<0 no?

The problem was I was doing them backwards, assuming the relation first rather than the property.

2)
Let a, b ∈ Z s/t ab>=0 --> a>=0, b>=0, so ab>=0 then ba>=0

So I think this is actually symmetric. I am not sure though, because I used the inequality to solve for a and b, but am I allowed to do that??

3) Let a, b, c ∈ Z s/t ab>=0 and bc>=0
So I should prove ac>=0
Let a = 1, b = 0, c = -1
Then we have ab>=0, bc>=0, but ac<0
so its false.

Thanks for the help!
 
Last edited:
Oh yeah, I see you did cover the negative case for 1). Sorry, somehow I didn't see that. But then for 2) ab>=0 doesn't imply a>=0 and b>=0. But it certainly does imply ba>=0. And yes, 3) is false. Just as you say.
 

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