# Thread: Can someone show me how to prove this as an equivalence relation or not?

1. ## Can someone show me how to prove this as an equivalence relation or not?

a) Consider the relation R defined on Z(the integers) by x~y when xy is greater than or equal to 0. Is R an equivalence relation?
b) Would your answer to (a) change if R was defined on Z by x~y when xy is greater than 0?

2. ## Re: Can someone show me how to prove this as an equivalence relation or not?

Originally Posted by allstar2
a) Consider the relation R defined on Z(the integers) by x~y when xy is greater than or equal to 0.
Is R an equivalence relation?
b) Would your answer to (a) change if R was defined on Z by x~y when xy is greater than 0?
$\forall x,~x\cdot x\ge 0$

If $x\cdot y\ge 0$ then $y\cdot x\ge ?$

If $x\cdot y\ge 0~\&~y\cdot z\ge 0$ then $x\cdot z\ge ?$

3. ## Re: Can someone show me how to prove this as an equivalence relation or not?

I often think about equivalence relations in terms of the following fact. Let X, Y be sets and f a function from X to Y. Define a binary relation R on X as follows: R(x, y) iff f(x) = f(y). Then R is an equivalence relation. This can be straightforwardly proved by definition.

Considering (a), can we find a function f such that x ~ y iff f(x) = f(y)? The fact that x ~ y, i.e., xy ≥ 0, approximately means that x and y have the same sign, i.e., x and y are both positive or both negative. I said "approximately" because there are problems when x or y is zero. Can a positive number be related to 0? What about a negative number? What does this say about transitivity?

Concerning (b), x ~ y does mean that x and y have the same sign. Let

$\mathop{\text{sgn}}(x)=\begin{cases}1&x>0\\-1&x<0\end{cases}$

Then x ~ y iff sgn(x) = sgn(y). Unfortunately, we can't conclude that ~ is an equivalence relation using the fact above because sgn is not a function from Z: it is not defined on 0. Which property of an equivalence relation does this break?