Proving an equivalence relation

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The relation R defined on the integers by aRb if ab >= 0 is analyzed for equivalence properties. It is established that R is reflexive since for any integer a, aa >= 0 holds true. However, R is not symmetric, as demonstrated by the counterexample of a = -1 and b = 2, where ab < 0. Furthermore, R is not transitive, illustrated by the case where a = 1, b = 0, and c = -1, leading to ac < 0 despite ab and bc being non-negative. Overall, R fails to meet the criteria for being an equivalence 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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