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I've been struggling with identifying relations on a set, and was hoping someone could check my answers and make sure I'm on the right track.

Let $A = \{1,2,3,4\}$ and $R$ be a relation on the set $A$ defined by: $R = \{(1,1),(1,2),(1,4),(2,1),(2,2),(3,3),(4,2),(4,4)\}$

Determine whether $R$ is reflexive, irreflexive, symmetric, asymmetric, antisymmetric, or transitive. Here are the answers that I came up with - do the answers (and my explanations) make sense at all? Thanks!

Reflexive, because of $(1,1), (2,2), (3,3), (4,4)$ (if the pairs were graphed they would point back to themselves).

NOT irreflexive because of $(1,1), (2,2), (3,3), (4,4)$ (because there are pairs that point back to themselves).

NOT symmetric because there is no $(4,1)$ or $(2,4)$ (if the pairs were graphed they would all need to connect in both directions).

NOT asymmetric because there is $(1,2)$ and $(2,1)$ (no pairs should connect in both directions if they were graphed).

ANTISYMMETRIC because of $(1,2)$ and $(2,1)$ (both are in the set, and $1$ does not equal $2$).

TRANSITIVE because of $(1,2), (2,1), (1,1), (2,2)$ or there’s a $(1,4)$ and a $(4,2)$ and also a $(1,2)$ (if there’s an $(a,b)$ and a $(b,c)$ there has to be an $(a,c)$).

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    $\begingroup$ It is not antisymmetric. $\endgroup$ – Jorge Fernández Hidalgo Jun 18 '14 at 14:36
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    $\begingroup$ Note that $(4,2), (2,1) \in R$ but $(4,1) \notin R$, so $R$ is not transitive $\endgroup$ – DGRasines Jun 18 '14 at 14:38
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You've done a fine job, though you're off on antisymmetry and also transitivity.

For antisymmetry: The reasoning you give is precisely why it is NOT antisymmetric.

If it were antisymmetric, then $(1, 2) \in R$ and $(2, 1) \in R$ would imply $1 = 2$, which is absurd.

With respect to transitivity: We see that $(4, 2), (2, 1) \in R$, but $(4, 1)$ is not in $R$. So the relation is not transitive.

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  • $\begingroup$ Thank you - the explanations you gave made the concept a lot clearer! Those are the two properties I kind of expected to get wrong. $\endgroup$ – Tina Jun 18 '14 at 14:50
  • $\begingroup$ You're welcome, Tina! $\endgroup$ – Namaste Jun 18 '14 at 15:00
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All of it is correct, except that I think you meant to say the relation is NOT antisymmetric (your reasoning is correct, and I think you meant to conclude it is not antisymmetric). And since (2,1), (1,4) are in the relation, but (2,4) isn't in the relation, the relation is not transitive.

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