# 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)

- What! How can it? It's

- 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)

- 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)