Proving that a relation is or is not an equivalence relation

425 Views Asked by At

So I have relation $R$ that is defined by $aRb$ if $ab>0, a,b \in Z-${$0$}

and I get that it's only an equivalence relation if

$aRa$

$aRb \rightarrow bRa$

$aRb, bRc \rightarrow aRc$

but I'm not sure how I can prove that it is or is not an equivalence relation. What steps are sensible to take, and how can I define the equivalent classes of the equivalence?

1

There are 1 best solutions below

6
On BEST ANSWER

To verify whether $R$ is an equivalence relation, you check each property in turn.

Let's start with the reflexive property and take $a \in \mathbb{Z} - \{0\}$. Is it true that $aa > 0$? Of course; hence $aRa$ and $R$ is reflexive.

For symmetry, you ask, "if $ab > 0$ is it true that $ba > 0$?" I'm sure you can answer that.

For transitivity, you ask, "if $ab > 0$ and $bc > 0$ is it true that $ac > 0$?" This takes a moment's thought to decide. If $a$ and $b$ have the same sign, and $b$ and $c$ have the same sign (so that their products are positive) what can we say of the signs of $a$ and $c$?

Reformulated this way, the answer should be easy to get to.

As for the equivalence classes of $R$, if you followed the argument above, you probably noticed that the sign of a nonzero integer is really all that matters in deciding whether it is related to another nonzero integer. This means that there are very few equivalence classes for $R$. Can you see them?