Let A = {x ∈ Z: 0 ≤ x ≤ 12}. Show that R = {(a, b): a, b ∈ A, |a − b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2].

We know that a relation R is an equivalence relation, if it is reflexive, symmetric and transitive.

For reflexive: |a − a| = 0 ⟹ (a, a) ∈ R, ∀a ∈ A

For symmetric: (a, b) ∈ R ⟹ |a − b| = 4k = |b − a|, k ∈ Z ⟹ (b, a) ∈ R, ∀(a, b) ∈ R

For transitive: Let (a, b) ∈ R and (b, c) ∈ R
Thus, we have (a, b) ∈ R ⟹ |a − b| = 4k₁, k₁ ∈ Z
and (b, c) ∈ R ⟹ |b − c| = 4k₂, k₂ ∈ Z
Since a, b and c are integers, we have |a − c| = |a − b + b − c| = |a − b| ± |b − c| = 4(k₁ ± k₂) = 4m, m ∈ Z ⟹ (a, c) ∈ R

So, we have shown that the relation R is reflexive, symmetric and transitive. Therefore, the relation is an equivalence relation.

Let x be the element of A such that (x, 1) ∈ R
|x − 1| is a multiple of 4 ⟹ |x − 1| = 0, 4, 8, 12
x − 1 = 0, 4, 8, 12
x = 1, 5, 9 (As 13 ∉ set of A)
Set of elements related to 1 = {1, 5, 9}

Now, let us find the elements in the equivalence class [2].
Let y be the element of A such that (y, 2) ∈ R
|y − 2| is a multiple of 4 ⟹ |y − 2| = 0, 4, 8, 12
y − 2 = 0, 4, 8, 12
y = 2, 6, 10 (As 14 ∉ set of A)
Thus, the elements in the equivalence class [2] = {2, 6, 10}.