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The question I'm tackling right now is this:

Give an example of a relation R on a set S that is not reflexive, transitive and not symmetric.

My answer:

Let S = {1,2,3} and let R = {(1,1), (2,2), (1,2)}. Then R is irreflexive since (s,s) is not in R for every element s of S and R is not symmetric since (1,2) is in R but (2,1) is not in R.

I dunno how to answer the transitive part. Could you please assist me in this question?

Your help would be greatly appreciated.

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  • $\begingroup$ What about an empty relation? $\endgroup$
    – dtldarek
    Dec 2, 2014 at 0:17

1 Answer 1

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To show that it is transitive, we have to show that if $(a,b)\in R$ and $(b,c)\in R$, then $(a,c)\in R$. In your example, $R = \{(1,1), (2,2), (1,2)\}$. So there are only two cases to check: $$(1,1)\in R\mbox{ and }(1,2)\in R,\mbox{ then }(1,2)\in R,$$ $$(1,2)\in R\mbox{ and }(2,2)\in R,\mbox{ then }(1,2)\in R.$$ Therefore, $R$ is transitive.

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  • $\begingroup$ What about the "3" in set S? @Paul $\endgroup$
    – apple
    Dec 2, 2014 at 0:08
  • $\begingroup$ To show $R$ is transitive, we just have to look at the elements in $R$. So $3$ in $S$ is irrelevant in this case. However, to show that $R$ is reflexive, you have to show that $(a,a)\in R$ for all $a\in S$. So $R$ is not reflexive because $(3,3)\not\in R$. $\endgroup$
    – Paul
    Dec 2, 2014 at 0:10
  • $\begingroup$ I get it now. Was my answer about not reflexive and antisymmetric correct? @Paul $\endgroup$
    – apple
    Dec 2, 2014 at 0:17
  • $\begingroup$ Yes, it's correct. You can upvote or accept the answer if it is helpful to you. $\endgroup$
    – Paul
    Dec 2, 2014 at 0:18
  • $\begingroup$ It is helpful without a doubt. Cheers @Paul $\endgroup$
    – apple
    Dec 2, 2014 at 0:39

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