Prove that if $R$ is symmetric, then $R^{-1}$ is symmetric, $R$ being a relation over $A$, and $\lnot(A = \varnothing)$.

This came as an exercise in my book.

I couldn't do anything - there is no (evident) explanation about how to prove things like that in my book (actually it seemed like a bonus question). Some Google searches like "prove if a relation is symmetric then the inverse is symmetric" seem to return other topics.

All I could do (nothing) was begin with:

Our hypothesis is $\forall a,b \in A [aRb \implies bRa]$. We want to show, based on it, that $a,b\in A[aR^{-1}b \implies aR^{-1}b]$... and I'm stuck.

Note that, I don't really want the solution to this exercise. I just want an explanation of a "general procedure" I should be taking when working with this kind of exercises, as in, "observe that you can do this and that to reach this, etc".

What do I mean by this "kind" of exercises? Well, like these:

  • If $R$ is antisymmetric, then $R^{-1}$ is antisymmetric.
  • If $R$ is reflexive, then $R \cap R^{-1}$ is reflexive.
  • If $R$ is transitive, then $R \cap R^{-1}$ is transitive.
  • Etc...

Think about the definition of $R^{-1}$

$aRb \implies bR^{-1}a$

If by symmetry of $R$ you know that $aRb \implies bRa$, what does $bRa$ imply about $R^{-1}$?

  • $\begingroup$ Hm..., so I could do something like... $$(aRb \implies bRa) \implies (bR^{-1}a \implies aR^{-1}b)$$ and then... I'm done since this proves that $R^{-1}$ is symmetric? $\endgroup$ – Zol Tun Kul Dec 6 '12 at 5:08
  • $\begingroup$ I do believe so. Not too bad, huh? $\endgroup$ – cheepychappy Dec 6 '12 at 5:11
  • $\begingroup$ Well that was silly... thanks a lot! I guess the other exercises are similar to this. $\endgroup$ – Zol Tun Kul Dec 6 '12 at 5:12
  • $\begingroup$ Indeed. Out of curiosity, what book are you working from? $\endgroup$ – cheepychappy Dec 6 '12 at 5:17
  • $\begingroup$ Well, it is in spanish, "Introduction to Discrete Mathematics" ("Introduccion a la Matematica Discreta") by Manuel Murillo Tsiji $\endgroup$ – Zol Tun Kul Dec 6 '12 at 5:20

You have a typo which may be preventing you from solving the problem. Instead of $a,b\in A[aR^{-1}b \implies aR^{-1}b]$ you want $a,b\in A[aR^{-1}b \implies bR^{-1}a]$ Now imagine you are given $c,d$ such that $cRd$. Now you know that $dRc$ because of symmetry. What does that tell you about $R^{-1}?$ What is your definition of $R^{-1}?$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.