So I have this exercise in my discrete math course that I don't understand:

Put $A = \{1,2,3\}$ and form the relation R on A by putting $R = \{(1,1),(2,2),(1,2),(2,3),(3,1)\}.$

Investigate if $R$ is: reflexive, symmetric, antisymmetric. If the relation has a property, give proof for it and if the relation doesn't have the property, prove it.

My attempt:

Reflexive: Yes, since $\{1,2,3\} = \{1,2,3\}.$

Symmetric: No, because in R every sub-pair of elements has max 2 elements and thus doesn't contain $\{1,2,3\}.$

Anti-symmetric: No ... but don't know why.

There is no hindsight but I know I'm wrong, can someone please help me out?

Thanks in advance,

  • $\begingroup$ the relation is neither reflexive nor symmetric nor transitive,that's just a binary relation $\endgroup$ – user715522 Feb 11 '20 at 19:15
  • $\begingroup$ $R$ being reflexive means for each $a\in A,$ $(a,a)\in R.$ Is $(3,3)\in R?$ $\endgroup$ – Thomas Andrews Feb 11 '20 at 19:19
  • $\begingroup$ "Yes, since $\{1,2,3\}=\{1,2,3\}$" The domain being the same as the codomain is irrelevant here. To be reflexive requires that for every element $x$ in the domain $(x,x)$ is an element of the relation. Here, $(1,1)$ and $(2,2)$ are both elements of the relation however $(3,3)$ is not. As such it is not true that every element $x$ satisfies that $(x,x)$ is in the relation and so it is not reflexive. $\endgroup$ – JMoravitz Feb 11 '20 at 19:20
  • $\begingroup$ I recommend thinking about problems like these from a graph-theoretical point of view. See this answer of mine for instance on some of the interpretations. It is plain to see in your example that all arrows (not loops) are single-sided and so the relation is antisymmetric. $\endgroup$ – JMoravitz Feb 11 '20 at 19:22
  • $\begingroup$ Can someone please explain why this relation is anti-symmetric? Anti-symmetry means aRb ^bRa if and only if a=b. How does it apply to this case? $\endgroup$ – Jean Doe Feb 15 '20 at 16:38

Not reflexive as $(3,3) \notin R$.

Not symmetric as $(1,2) \in R$ but $(2,1) \notin R$.

The relation is anti symmetric.

Not transitive as $(1,2),(2,3) \in R$ but $(1,3) \notin R$.

  • $\begingroup$ @JMoravitz Thanks for noting the typo! $\endgroup$ – mathcounterexamples.net Feb 11 '20 at 19:24
  • $\begingroup$ Thank you for answering. But could you please explain why it is anti-symmetric? And also, why would (1,3) have to be in R for it to be transitive? $\endgroup$ – Jean Doe Feb 11 '20 at 19:46
  • 1
    $\begingroup$ Regarding transitive, please refer to the definition. Regarding anti symmetric, notice that for any $a\neq b$, you never have both $(a,b) \in R$ and $(b,a) \in R$. $\endgroup$ – mathcounterexamples.net Feb 11 '20 at 19:51
  • $\begingroup$ I still do not understand why this relation is anti-symmetric. Anti-symmetry means aRb ^bRa if and only if a=b. How does it apply to this case? $\endgroup$ – Jean Doe Feb 15 '20 at 16:37
  • $\begingroup$ @JeanDoe For what integers $a,b$ do you both have $(a,b)\in R$ and $(b,a) \in R$? $\endgroup$ – mathcounterexamples.net Feb 15 '20 at 19:40

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