# Relations on a set, check my answers?

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 $$(a,c)$$).

• It is not antisymmetric. – Jorge Fernández Hidalgo Jun 18 '14 at 14:36
• Note that $(4,2), (2,1) \in R$ but $(4,1) \notin R$, so $R$ is not transitive – DGRasines Jun 18 '14 at 14:38

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.