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I tried to prove that if $R$ is transitive then its inverse is transitive as well. $$\begin{align} & a{{R}^{-1}}b\,\,,\,\,b{{R}^{-1}}c \\ & \Rightarrow bRa\,\,,\,\,cRb \\ & \Rightarrow cRa \\ & \Rightarrow a{{R}^{-1}}c \\ \end{align}$$

is this a correct proof (or am I completely wrong and it's not even true?)

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up vote 3 down vote accepted

A note: What you call "complementary" is conventionally called the inverse relation of $R$. (It's interesting that you got the notation right.) Wikipedia says that alternative terms are converse or transpose relation. I am not sure if "complementary" is used by anyone.

True statement and correct proof.

This statement from wikipedia article on inverse relation is relevant:

If a relation is reflexive, irreflexive, symmetric, antisymmetric, asymmetric, transitive, total, trichotomous, a partial order, total order, strict weak order, total preorder (weak order), or an equivalence relation, its inverse is too.

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Great I confused myself and thought I might be wrong :) Thanks – Jason Jul 30 '11 at 21:04
You are welcome, @Jason. When in doubt, check Wikipedia :) – Srivatsan Jul 30 '11 at 21:06
The complement of a relation $R$ is defined as the relation $S$ (on the same set) which holds iff $R$ does not hold. But I have never heard the term in the wild. – André Nicolas Jul 31 '11 at 1:39
@AndreN Oh yes, thanks for pointing it out. But I sure hope OP did not have this complement in mind. – Srivatsan Jul 31 '11 at 1:42
@Srivatsan Narayanan: Luckily not, the OP's proof could not be written by someone who was thinking complement. – André Nicolas Jul 31 '11 at 2:01

Yes, that proof looks fine to me.

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And Thanks to you 2 :) – Jason Jul 30 '11 at 21:04

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