Definition talk:Inverse Relation

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Actually, $R$ does not have to be a relation - in fact, this is the way it is most commonly addressed in literature - to have converse defined for any set, not just relations. In fact it is quite useful this way (for example, we could have that $R$ is a relation $\iff$ the converse of the converse $= R$. Same for domain and range. Would you approve this change? -Andrew Salmon 23:39, 12 September 2011 (CDT)

Not sure I do, because I don't understand your first sentence (and I'm not sure about the "most commonly addressed in literature" either, I have a colossal pile of books and this is new to me). If you have another definition for the "converse of a set", then go for it, but this page describes the inverse of a relation (which is called by some, confusingly or not, the "converse"). --prime mover 00:19, 13 September 2011 (CDT)
Well, no, it's the same thing, but without the condition that $R$ is a relation. I'm saying that the definition works for any class - not just classes of ordered pairs. -Andrew Salmon 00:48, 13 September 2011 (CDT)
What! How can it? It's defined in the sense of ordered pairs. Are you sure you mean the same as what's written here? Write what you think you mean in mathematical language, because I think what you're thinking of is something different. --prime mover 01:33, 13 September 2011 (CDT)
Sorry - looking back on it, my response was worded terribly. It's exactly the same definition: The converse of $R = \{ < x,y > | < y , x > \in R \}$. But it's just without the proviso that $R$ is a relation (a class that contains ordered pairs only). -Andrew Salmon 01:45, 13 September 2011 (CDT)
But as a (binary) relation is (by definition) a subset of the cartesian product of two sets, then it is by definition a set of ordered pairs taken from a pair of sets. Hence a collection of ordered pairs taken from a pair of sets is by definition a relation. There is no difference, so there is no need to go any further with this.
Also note that there is already a page which proves $\left({R^{-1} }\right)^{-1} = R$ so your initial point is covered. --prime mover 03:15, 13 September 2011 (CDT)