Question

Prove that if R is an equivalence relation on a set A, then R ^-1 is an equivalence relation on A.

Answer 1

solution:

given data:

Given that R is an equivalence relation on A, so it is reflexive, symmetric and transitive.

So, for $(a,a)\epsilon R\Rightarrow (a,a)\epsilon R^{-1}$

hence, $R^{-1}$ is reflexive

Also, R is symmetric, so, $for (a,b)\epsilon R\Rightarrow (b,a)\epsilon R$

Then, in $R^{-1}$, $if (b,a)\epsilon R^{-1}$ then $(a,b)\epsilon R^{-1}$

which shows $R^{-1}$ is symmetric.

Also, R is transitive, so $if (a,b)\epsilon R an d(b,c)\epsilon R\Rightarrow (a,c)\epsilon R$

Then, in $R^{-1}$, $if (b,a)\epsilon R^{-1} and(c,b)\epsilon R^{-1}\Rightarrow (c,a)\epsilon R^{-1}$

which means $R^{-1}$ is transitive

hence, we can conclude that $R^{-1}$ is equivalence relation on A

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