# Proving an equivalence relation

1. Jan 24, 2010

### missavvy

1. The problem statement, all variables and given/known data
On the set of integers, define the relation R by: aRb if ab>=0.

Is R an equivalence relation?

2. Relevant equations

3. 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.

2. Jan 24, 2010

### Dick

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?

3. Jan 24, 2010

### missavvy

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. Im 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: Jan 24, 2010
4. Jan 24, 2010

### Dick

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.