Question

Show that the relation R in the set A = [x ∈ Z : 0 ≤ x ≤ 12}, given by

(i) R = {(a, b): | a – b | is a multiple of 4}

(ii) R = {(a, b): a = b}

is an equivalence relation. Find the set of all elements related to 1 in each case.

Answer 1

The set A = {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2, ………, 12}
(i) R = {(a, b): | a – b | is a multiple of 4}
|a – b | = 4k or b = a + 4k.
∴ R = {(1, 5), (1, 9), (2, 6), (2,10), (3, 7), (3,11), (4, 8), (4, 12), (5, 9), (6, 10), (7, 11), (8, 12), (0, 0), (1, 1),
(2, 2), ……… (12, 12), (0, 4), (4, 0), (0, 8), (8, 0), (0, 12), (12, 0), (5,1), (9,1), (6, 2), (10, 2), (7, 3), (11, 3), (8, 4), (12,4), (9, 5), (10,6), (11, 7), (12, 8)}
(a) a – a = 0 = 4k, where k = 0 ⇒ (a, a) ∈ R.
∴ R is reflexive.
(b) If | a – b | = 4k, then | b – a | = 4k i.e., (a, b) and (b, a) both belong to R. ∴ R is symmetric.
(c) a – c = a – b + b – c.
When a – b and b – c both are multiples of 4, then a – c is also a multiple of

This shows if (a, b), (b, c) ∈ R, then (a, c) also ∈ R.
∴ R is an equivalence relation.
The set of all elements related to 1 is {(1, 5), (1, 9), (5, 1), (9, 1)}
(ii) R – {(a, b) : a = b} = {(0, 0), (1, 1), (2, 2), ……. ,(12, 12)}
(a) a = a ⇒ (a, a) ∈ R ∴R is reflexive.
(b) Again if (a, b) ∈ R, then (b, a) also ∈R.
Since a – b and (a, b) ∈ R ⇒ R is symmetric.
(c) If (a, b) ∈R, then (b, c) ∈ R ⇒ a = b = c
∴ a = c ⇒ (a, c) ∈ R. Hence R is transitive.
Set related to 1 is {1}