Consider the relation $R \subseteq \Bbb Z \times \Bbb Z$ given as $$R = \{(x,y) \in \Bbb Z \times \Bbb Z \mid xy >0 \textrm{ or } x=y=0\}.$$ Prove that R is an equivalence and write down its equivalence classes.
Can anybody tell me how do I approach a task like this in general?
$[0] = \{0, 1, 4, 9\}$ for $xy>0$ would be my approach.
But then, what about $[1]$ for example?
I would like to see an example, let's say with $xy > 1$ or $xy>2$, as I am honestly a little bit lost (first time doing that kind of stuff) and my literature hasn't any kind of example for that, other than some theoretically stuff.
Thanks in advance everyone!
To get you started: you are correct that $R$ must be symmetric, transitive, and reflexive. Prove these one at a time. Here's symmetry, for example:
Suppose that $(x,y) \in R$. Then (by definition of $R$), $xy>0$ or $x=y=0$. In the first case, since $xy=yx$ for all real numbers $x,y$, then $yx>0$. In the second case, $y=x=0$. Therefore $yx>0$ or $y=x=0$, so $(y,x) \in R$.
Now follow the same kind of reasoning for transitivity and reflexivity.
Once you've established that, figure out the classes. Take $x=0$, for example. For what $y$ do we have $(x,y)\in R$? That would require $xy>0$ or $x=y=0$. But $xy>0$ is impossible since $x=0$, so we must have $y=0$. In other words, $[0] = \{0\}$, which is to say that $0$ is the only element in its class. What happens if you do the same thing with $x=1$? What other number remain after this?
Finally, to give you an idea of why this might not work with other relations, let's change $xy>0$ to $xy>1$: let $S = \{(x,y) \in \Bbb Z \times \Bbb Z \mid xy >1 \textrm{ or } x=y=0\}.$
If this were an equivalence relation, it would be reflexive: that is, for every $x\in \Bbb Z$, we would have $x^2 > 1$ or $x=0$. This fails for $x=1$. (Exercise: is this relation transitive? symmetric?)