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Can I get an example of a relation that is symmetric and transitive on the set $Z$

By Definition:

  1. $R$, a relation in a set $X$, is reflexive if and only if $∀x∈X$, $xRx$.
  2. $R$ is symmetric if and only if $∀x,y∈X, xRy⟹yRx$.
  3. $R$ is transitive if and only if $∀x,y,z∈X, xRy∧yRz⟹xRz$.

I know you can make a symmetric and transitive function that is not reflexive in general, i'm having trouble finding one on the set of integers.

And can we say anything in general about such relations?

JonathanZ
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Daniel Doyon
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    Hint: First, find a proof for "if R is symmetric and transitive, then it must be reflexive", and then try to find a sneaky, weird example for which your proof fails. – JonathanZ May 14 '23 at 21:33
  • Also, your comment that you can make one in general is a bit odd to me. I don't think I know of any example where the basic "trick" wouldn't also work for the integers. Maybe you could include one specific example in your question? – JonathanZ May 14 '23 at 21:42
  • You can always just bake it into the definition of the relation. xRy iff x ≠ y and x*y > 0 – Michael Carey May 15 '23 at 00:48
  • Wouldn't $R=${${(a,b)|a\neq b\land a,b\in\mathbb{Z}}$} work? Or am I missing something? – H. sapiens rex May 15 '23 at 02:13
  • That's not transitive. – JonathanZ May 15 '23 at 03:05
  • @JonathanZsupportsMonicaC indeed it's not! I realised a few minutes after posting, but by then I was already back at work lol. How about this instead? $aRb \Longleftrightarrow b=a+c, c \in \mathbb{Z} \land c \neq 0$? – H. sapiens rex May 15 '23 at 06:14
  • If all you are looking for is an example over the integers, perhaps the simplest to visualize is $aRb\iff a\neq 0\wedge b\neq 0$. Here, everything is related to everything else with the exception of zero. Nothing is related to zero and zero is not related to anything, in particular itself. Since zero is not related to itself, it fails to be reflexive. – JMoravitz May 15 '23 at 12:41
  • Indeed, if you were to go through the attempted fake-proof of trying to imply that symmetric and transitive implies reflexive, you will find that every example of a relation who is symmetric and transitive will be of the form that it is an equivalence relation with at least one equivalence class of size one and at least one of said size-one equivalence classes missing. – JMoravitz May 15 '23 at 12:46
  • That being said, two other incredibly simple examples: $aRb\iff a=b\wedge a\neq 0$ is the equivalent of the trivial equivalence relation "equals" but leaving out the equivalence class of zero, hence not reflexive... and then the other trivial example is the empty relation, the relation where there are no elements related to any others which is vacuously transitive and reflexive and obviously not reflexive either. – JMoravitz May 15 '23 at 12:49

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