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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!

Théophile
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Prometheus
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  • First, what is the task? The statement just asks us to "consider" something. Are you trying to prove that this is (or isn't) an equivalence relation, for example? – Théophile Dec 15 '20 at 19:51
  • Second, suppose this is an equivalence relation. In that case (and only in that case) it makes sense to group elements into equivalence classes, such as what you have done with $[0]$. Careful, though: $0$ should not be in the same set as $1$ (or $4$, etc.), since it is not true that $0\cdot1 > 0$ or that $0=1=0$. It is true that $1$ and $4$ are in the same set, since $1\cdot4 > 0$. With that in mind, $[1] = [4]$; it doesn't matter which representative you pick. – Théophile Dec 15 '20 at 19:53
  • Updated the question. Sorry for that. The Task is to proof that R is an equivalence and write down its equivalence classes – Prometheus Dec 15 '20 at 19:53
  • Great, that's much clearer now. What is your definition of an equivalence relation? – Théophile Dec 15 '20 at 19:54
  • The definition of an equivalence is that R must have symmetry, be transitive and reflexive. But with {0, 1,4,9} itself thats not the case, or? – Prometheus Dec 15 '20 at 19:58

1 Answers1

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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?)

Théophile
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  • Thank you. So i have [0] = {0} [1] = { 1,2,3,4,5,6,7,8,9... ℤ} [2] = { 2,4,6,8,9,10 .....} (and so on). Now if you don't mind. How would i approach symmetry now for that proof? Odd[1] R = {1,3,5,7,9} with [2] R = {2,4,6,8,9,10} => {(1,2), (3,4), (5,6)} etc. How should i show (b,a) in R than for the final symmetry proof? Mirror [2]R with Odd[1] R? Is odd[1] even allowed? – Prometheus Dec 15 '20 at 20:40
  • @Prometheus Okay, so you found that $[1] = {1,2,3,\ldots}$, the set of all positive integers. A key idea is that if you take any element from this, its class will be the same; this is exactly why these elements are said to be equivalent. So $[1] = [2] = [3] = \ldots = {1,2,3,\ldots}$. To put it another way, once you've found $[1]$, you don't need to determine $[2]$ because you've already found it; and moreover, it especially doesn't make sense to find that $[2]$ is something else entirely. – Théophile Dec 16 '20 at 01:46
  • @Prometheus After finding $[0]$ and $[1]$, don't forget about negative numbers. You can see that $[-1] = {-1,-2,-3,\ldots}$, and this covers all integers. The "moral" of this equivalence relation is that we can separate all integers into three classes: zero, positive, or negative. – Théophile Dec 16 '20 at 01:47