Let A = {1, 2, 3, ..., 12} and R = {(a, b): |a − b| is divisible by 4}. Show that R is an equivalence relation on A. Find the set of all elements related to 3, and write all the equivalence classes.

1. 1 · Write what is given and what is to be found. Given: A = {1, ..., 12} and R = {(a, b): |a − b| divisible by 4}. To show: R is reflexive, symmetric and transitive. To find: [3] and all the classes.
2. 2 · Write the rule. R is an equivalence relation if it is reflexive, symmetric and transitive.
3. 3 · Reflexive. For every a ∈ A, |a − a| = 0 = 4 × 0, so (a, a) ∈ R. Hence R is reflexive.
4. 4 · Symmetric. If (a, b) ∈ R, then |a − b| is divisible by 4. Since |b − a| = |a − b|, (b, a) ∈ R. Hence R is symmetric.
5. 5 · Transitive.
   • Let (a, b), (b, c) ∈ R. Then a − b = 4k and b − c = 4m for some integers k and m (each may be negative).
   • Add: a − c = 4(k + m), so |a − c| is divisible by 4.
   • So (a, c) ∈ R. Hence R is transitive.
6. 6 · Find the elements related to 3. |a − 3| must be 0, 4 or 8, since the largest gap in A is 11. So a = 3, 7 or 11: [3] = {3, 7, 11}.
7. 7 · Write all the classes. [1] = {1, 5, 9}, [2] = {2, 6, 10}, [3] = {3, 7, 11}, [4] = {4, 8, 12}.
8. 8 · Check and box the answer. R is an equivalence relation, and [3] = {3, 7, 11}. The 4 classes have 3 elements each, 12 in all, with no overlap.

Scoring tip: Write 'Hence R is reflexive' (and so on) after each part. Examiners give a mark for each property clearly concluded.