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Prove that if a non-empty relation $R$ on $A$ is transitive and irreflexive, then it is asymmetric.

I assume that I need to prove this one by contradiction, but I'm having a hard time wrapping my head around it. If a relation is transitive and irreflexive wouldn't it also be symmetric?

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Suppose for a contradiction that we can find $x,y \in A$ such that $(x,y) \in R$ and $(y,x) \in R$. Then by transitivity, $(x,x) \in R$, contradicting irreflexivity.

Edit. By the way, if this still seems at all mysterious, try visualizing it as follows. Consider two points $x$ and $y$, and an arrow going $x \rightarrow y$ and another $y \rightarrow x$. Use transitivity to deduce that there must be a loop at $x$. But irreflexivity basically says: "no loops."

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    $\begingroup$ Yes this makes sense, thank you for the clarity. $\endgroup$ – Joe Caruso Oct 14 '13 at 19:43

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