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

An equivalence relation on a set A is a relation that is reflexive, symmetric, and transitive. Let's denote this relation as R. The inverse of a relation R, denoted as R^(-1), is defined as follows: for any (a, b) ∈ R, (b, a) ∈ R^(-1).

To prove that if R is an equivalence relation on a set A, then R^(-1) is also an equivalence relation on A, we need to show that R^(-1) is reflexive, symmetric, and transitive.

1. Reflexivity: A relation R is reflexive if for every element a in A, (a, a) ∈ R. Since R is an equivalence relation, it is reflexive. Therefore, for every a in A, (a, a) ∈ R, and hence (a, a) ∈ R^(-1). So, R^(-1) is reflexive.

2. Symmetry: A relation R is symmetric if for every (a, b) ∈ R, (b, a) ∈ R. Since R is an equivalence relation, it is symmetric. Therefore, for every (a, b) ∈ R, (b, a) ∈ R. But (b, a) ∈ R is exactly the definition of R^(-1). Hence, for every (a, b) ∈ R^(-1), (b, a) ∈ R^(-1). So, R^(-1) is symmetric.

3. Transitivity: A relation R is transitive if for every (a, b) ∈ R and (b, c) ∈ R, (a, c) ∈ R. Since R is an equivalence relation, it is transitive. Therefore, for every (a, b) ∈ R and (b, c) ∈ R, (a, c) ∈ R. By the definition of R^(-1), this means that for every (b, a) ∈ R^(-1) and (c, b) ∈ R^(-1), (c, a) ∈ R^(-1). So, R^(-1) is transitive.

In conclusion, R^(-1) is reflexive, symmetric, and transitive, and hence it is an equivalence relation on A.

Here is a table summarizing the properties:

Property	R	R^(-1)
Reflexivity	For every a in A, (a, a) ∈ R	For every a in A, (a, a) ∈ R^(-1)
Symmetry	For every (a, b) ∈ R, (b, a) ∈ R	For every (a, b) ∈ R^(-1), (b, a) ∈ R^(-1)
Transitivity	For every (a, b) ∈ R and (b, c) ∈ R, (a, c) ∈ R	For every (b, a) ∈ R^(-1) and (c, b) ∈ R^(-1), (c, a) ∈ R^(-1)