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Given set $A={1, 2, 3}$ consider the following relation on $A$

$R=\{(1, 1), (2, 1), (3, 3), (3, 2)\}$

Which one of the following statements are true

  1. $R$ is antisymmetric and transitive
  2. $R$ is antisymmetric but not transitive
  3. $R$ is not transitive and not antisymmetric
  4. $R$ is reflexive and antisymmetric

I have 4 as the answer, because:

  • $R$ is reflexive: $(1, 1), (3, 3)$
  • $R$ is antisymmetric: $(2, 1), (3, 2)$ are in $R$ but not $(1, 2), (2, 3)$

But there may also be a case for transitivity because of $(2, 1), (3, 2)$

I'm at a bit of a loss really

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$R$ is not reflexive since $(2,2) \notin R$. It is not transitive since $(3,2), (2,1) \in R$, but $(3,1)\notin R$. –  Prahlad Vaidyanathan Sep 2 '13 at 18:06
Also, I think you are confusing antisymmetry "$(x,y), (y,x)\in R \Longrightarrow x=y$" with asymmetry "$(x,y)\in R \Longrightarrow (y,x)\notin R$". That said, it is still true that $R$ is antisymmetric. –  walcher Sep 2 '13 at 18:11

1 Answer 1

The relation is not reflexive because $(2,2)\notin R$. Remember that the definition of reflexivity is that $(a,a)\in R$ for all $a\in A$.

Also, the relation is not transitive, because $(3,2)\in R$ and $(2,1)\in R$, but $(3,1)\notin R$.

Finally, a relation is antisymmetric if the conditions $(a,b)\in R$ and $(b,a)\in R$ imply $a=b$. Is this true in the $R$ you've given?

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