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For an example relation $R$ on $\{1,2\}$ of $\{(1,1), (2,2)\}$.

This relation is reflexive, but is it also symmetric and transitive?

It appears to be symmetric because $(1,1)$ is present as is $(1,1)$, and $(2,2)$ is present as is $(2,2)$, in other words, if $x = 1$ and $y = 1$, then $(x,y)$ is present and $(y,x)$ is present.

Following from that, it also appears to be transitive because for $x = 1, y = 1$ and $z = 1$, $(x,y)$ is present, $(y,z)$ is present, and so is $(x,z)$.

Is there some restriction on symmetric and transitive relations in that the elements of the ordered pair cannot be equal?

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If a relation is reflexive, symmetric and transitive it is an equivalence relation. This means that it splits the base set into disjoint subsets (equivalence classes) in which every element is related to itself and every other element in the class to which it belongs. Can you see equivalence classes in this case? – Mark Bennet Jul 15 '11 at 22:43

No, you're only considering the diagonal of the set, which is always an equivalence relation. But what if you took $R=\{(1,1),(2,2),(2,1)\}$? It's still a valid relation, it's reflexive on $\{1,2\}$ but it's not symmetric since $(1,2)\not\in R$. The point is you can have more than just pairs of form $(x,x)$ in your relation.

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Thank you. I understand now, I think my mind got too focused in on the specific instances (only pairs of the form (x,x). – Tinker Jul 15 '11 at 22:45
@Tinker, no problem, the set you mention is actually the diagonal relation, or the equality relation. – yunone Jul 15 '11 at 22:48
@Tinker Would not the current relation i.e. {(1,1),(2,2)} be symmetric. You said that, "what if you took 2,1".. well, why shall i take 2,1? I thought i got it pretty straigth in the early classes, but i would really appreciate that what am i missing in this question. Would not the answer to the curretn question posted be 'Yes'? – Kraken Sep 16 '11 at 16:14

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