Question

If \( R \) is an equivalence relation on a set \( A \) then show that \( R^{-1} \) is also an equivalence relation on \( A \).

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Answer 1

\(\because R\) is an equivalance relation on set A.

If \(a \in A\) then \((a, a)\in R\)   (∵ R is reflexine)

⇒ \(R(a) = a\)

⇒ \(R^{-1}(a) = a \,\,\,\, \forall \,a\in A\) 

\(\therefore R^{-1}\) is reflexive.

Let \(a, b \in A \) such that \((a, b)\,\,\in R^{-1}\)

⇒ \(R^{-1}(a) = b\)

⇒ \(a = R(b)\)

⇒ \((b,a )\in R\)

⇒ \((a, b) \in R\)      (∵ R is symmetric)

⇒ \(R(a) = b\)

⇒  \(a = R^{-1}(b)\)

⇒ \((b,a)\in R^{-1}\,\,\forall \,a, b \in A\)

 \(\therefore R^{-1}\) is symmetric

Let \(a, b, c \in A\) such that \((a, b) \in R^{-1} \) & \((b, c)\in R^{-1}\)

Then \(R^{-1}(a) = b\) & \(R^{-1} (b) = c\)

⇒ \(a = R(b)\) & \(b = R(c)\)

⇒ \((b,a)\in R\) & \((c,b) \in R\)

⇒ \((a,b)\in R\) & \((b,c) \in R\)    (As \(R\) is symmetric)

⇒ \((a, c)\in R\)       (As \(R\) is transitive)

⇒ \((c, a) \in R\)       (As \(R\) is symmetric)

⇒ \(R(c) = a\)

⇒ \(R^{-1} (a) = c\)

⇒ \((a, c)\in R^{-1}\)

 \(\therefore R^{-1}\) is transitive 

 \(\therefore R\) is an equivalance relation on A.