Show that R is an equivalence relation, and find all equivalence classes.(a) R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 5), (5, 1), (3, 5), (5, 3), (1, 3), (3, 1)} on {1, 2, 3, 4, 5}(b) R = {(x, y) : 3 divides (x − y)} on R(c) R = {(x, y) : 3 divides (x + 2y)} on Z(d) R = {(x, y) : x and y are both even} on {1, 2, 3, 4, 5}

(a) To show that R is an equivalence relation, we need to show that it is reflexive, symmetric, and transitive.

Reflexivity: For all a in {1, 2, 3, 4, 5}, we have (a, a) in R. So, R is reflexive.

Symmetry: For all (a, b) in R, we also have (b, a) in R. So, R is symmetric.

Transitivity: For all (a, b) and (b, c) in R, we also have (a, c) in R. For example, (1, 5) and (5, 3) are in R, and so is (1, 3). So, R is transitive.

Therefore, R is an equivalence relation.

The equivalence classes are: [1] = {1, 3, 5}, [2] = {2}, [3] = {1, 3, 5}, [4] = {4}, [5] = {1, 3, 5}.

(b) R is an equivalence relation because it is reflexive (x-x=0 is divisible by 3 for all x), symmetric (if x-y is divisible by 3, then y-x is also divisible by 3), and transitive (if x-y and y-z are divisible by 3, then x-z is also divisible by 3). The equivalence classes are infinite and cannot be listed.

(c) R is not an equivalence relation because it is not symmetric. For example, if x=3 and y=1, then 3 divides (3+21)=5, so (3,1) is in R. But 3 does not divide (1+23)=7, so (1,3) is not in R.

(d) R is an equivalence relation because it is reflexive (every even number is even), symmetric (if both x and y are even, then both y and x are even), and transitive (if x and y are even, and y and z are even, then x and z are even). The equivalence classes are [1] = {1}, [2] = {2, 4}, [3] = {3}, [4] = {2, 4}, [5] = {5}.