Show that each of the relation R in the set A = { x ∈Z: 0≤x≤12}, A={x} 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

A = { x ∈ Z : 0 ≤ x ≤ 12} = {0,1,2,3,4,5,6,7,8,9,10,11,12}

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

For any element a ∈A, we have (a, a) ∈ R as |a - a = 0|is a multiple of 4.

∴R is reflexive.

Now, let (a, b) ∈ R ⇒ |a - b| is a multiple of 4.

⇒ |-(a - b)| = ⇒ |b - a| is a multiple of 4.

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b), (b, c) ∈ R.

⇒ |(a - b)| is a multiple of 4 and |(b - c)| is a multiple of 4.

⇒ (a - b) is a multiple of 4 and (b - c) is a multiple of 4.

⇒ (a - c) = (a – b) + (b – c) is a multiple of 4.

⇒ |a - c| is a multiple of 4.

⇒ (a, c) ∈R

∴ R is transitive.

Hence, R is an equivalence relation.

The set of elements related to 1 is {1, 5, 9} since

|1 - 1| = 0 is a multiple of 4,

|5 - 1| = 4 is a multiple of 4, and

|9 - 1| = 8 is a multiple of 4.

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

For any element a ∈A, we have (a, a) ∈ R, since a = a.

∴R is reflexive.

Now, let (a, b) ∈ R.

⇒ a = b

⇒ b = a

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

⇒ a = b and b = c

⇒ a = c

⇒ (a, c) ∈ R

∴ R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.