solution:

(I) A = {x ∈ Z : 0 ≤ x ≤ 12}

In this way, A = {0, 1, 2, 3, … … , 12}

Presently R = {(a, b) : |a – b| is a different of 4}

R = {(4, 0), (0, 4), (5, 1), (1, 5), (6, 2), (2, 6), ….., (12, 9), (9, 12),…., (8, 0), (0, 8), …., (8, 4), (4,

8),….., (12, 12)}

Here, (x, x) = |4-4| = |8-8|= |12-12| = 0 : different of 4. R is reflexive.

|a – b| and |b – a| are different of 4. (a, b) ∈ R and (b, a) ∈ R. R is symmetric.

Furthermore, |a – b| and |b – c| then |a – c| are different of 4. (a, b) ∈ R and (b, c) ∈ R and (a, c) ∈ R is transitive.

Henceforth R is an identicalness connection.

(ii) Here, (a, a) = a = a.

(a, a) ∈ R . So R is reflexive.

a = b and b = a. (a, b) ∈ R and (b, a) ∈ R. R is symmetric.

Also, a = b and b = c then a = c. (a, b) ∈ R and (b, c) ∈ R and (a, c) ∈ R is transitive.

Henceforth R is an identicalness connection.

Presently set of all components identified with 1 for each situation is

(i) Required set = {1, 5, 9}

(ii) Required set = {1}