Equivalence Relations

Equivalence relations are the high-value idea in Relations and Functions — CBSE loves a 3–5 mark question asking you to prove a relation is an equivalence relation and then find an equivalence class. Master the definition and the partition picture and these are safe marks.

Definition

A relation R on A is an equivalence relation if it is simultaneously reflexive, symmetric AND transitive. All three must hold.

Equivalence classes

For a ∈ A, the equivalence class [a] = { x ∈ A : xRa }. Key facts:

• Every element lies in exactly one class (a ∈ [a] by reflexivity).
• Two classes are either identical or disjoint: [a] = [b] if aRb, else [a] ∩ [b] = ∅.
• The classes partition A — they cover A with no overlap.

Worked template

Show R = {(a, b) : a − b is divisible by n} on ℤ is an equivalence relation:

• Reflexive: a − a = 0 is divisible by n ✓
• Symmetric: if n | (a − b) then n | (b − a) ✓
• Transitive: n | (a − b) and n | (b − c) ⇒ n | (a − c) ✓
  This gives the classes [0], [1], …, [n−1] (residues mod n).

Exam Tricks & Tips

• 🎯 Always check ALL THREE properties explicitly; missing one loses the "equivalence" mark.
• 🎯 "Divisible by n", "same remainder", "a − b even", "parallel to" are classic equivalence relations.
• 🎯 The number of distinct equivalence classes = number of blocks in the partition.
• 🎯 "aRb if a − b is a multiple of 4" on {1..12}: classes are {1,5,9},{2,6,10},{3,7,11},{4,8,12}.
• 🎯 To find [a], collect every x satisfying the rule with a — don't forget a itself.
• ❌ Common mistake: writing that "≥" is an equivalence relation. It is reflexive and transitive but NOT symmetric, so it fails.

Expected exam pattern

A 3–4 mark proof ("prove R is an equivalence relation") often followed by "find the equivalence class of a given element" for 1–2 marks.

Quick recap

Equivalence = reflexive + symmetric + transitive. It splits A into disjoint equivalence classes that partition the set. Prove all three, then read off the classes.

Cover the answer, recall, then check. 11 cards on equivalence relations and classes.

Q1. Define an equivalence relation.
A1. A relation that is reflexive, symmetric, AND transitive — all three together.

Q2. What is the equivalence class [a]?
A2. [a] = { x ∈ A : xRa } — all elements related to a.

Q3. Can two equivalence classes partly overlap?
A3. No. They are either identical or completely disjoint.

Q4. What do the equivalence classes form?
A4. A partition of A — they cover A with no overlaps.

Q5. Is "a − b divisible by n" on ℤ an equivalence relation?
A5. Yes. Reflexive, symmetric and transitive; classes are the residues 0…n−1 mod n.

Q6. Why is [a] never empty?
A6. Because a ∈ [a] by reflexivity (aRa always holds).

Q7. Is "≥" on ℝ an equivalence relation?
A7. No — it fails symmetry (3 ≥ 2 but 2 ≱ 3).

Q8. For "a − b multiple of 4" on {1,…,12}, give [2].
A8. {2, 6, 10}.

Q9. Is "line L₁ is parallel to line L₂" an equivalence relation?
A9. Yes — reflexive (a line is parallel to itself), symmetric and transitive.

Q10. If aRb in an equivalence relation, what is true of [a] and [b]?
A10. [a] = [b] — they are the same class.

Q11. Number of equivalence classes equals what?
A11. The number of blocks in the induced partition of A.

When you sort a pile of coins by denomination, you never compare a coin to itself and you happily group all ₹5 coins together as "the same for our purpose". That instinct — treating different objects as equivalent under some criterion — is captured exactly by an equivalence relation.

Definition: A relation R on a set A is an equivalence relation if it is simultaneously reflexive, symmetric and transitive. Equivalence relations generalise the idea of equality: things related by R are "the same" with respect to the property R encodes.

Why all three properties are essential

Drop any one property and "sameness" breaks:

• Without reflexivity, an object need not equal itself — absurd.
• Without symmetry, "a is same as b" would not guarantee "b is same as a".
• Without transitivity, sameness would not pass along a chain.

The canonical example is congruence modulo n: a R b ⟺ n divides (a − b), written a ≡ b (mod n). It is reflexive (n | 0), symmetric (n | (a−b) ⟹ n | (b−a)) and transitive (n | (a−b), n | (b−c) ⟹ n | (a−c)). So "same remainder on dividing by n" is a genuine equivalence relation.

Equivalence classes — the real payoff

Definition: For a ∈ A, the equivalence class [a] = { x ∈ A : (x, a) ∈ R } is the set of all elements equivalent to a.

Two fundamental facts (the Partition Theorem):

1. Every element lies in exactly one class ([a] is never empty since a ∈ [a]).
2. Two classes are either identical or disjoint: [a] = [b] if aRb, else [a] ∩ [b] = φ.

Hence an equivalence relation partitions A into non-overlapping classes whose union is all of A. Conversely, every partition defines an equivalence relation. Equivalence relations and partitions are two faces of one idea.

For congruence mod 3 on integers, the classes are exactly three: [0] = {…,−3,0,3,6,…}, [1] = {…,−2,1,4,…}, [2] = {…,−1,2,5,…}. These are the residue classes.

Worked example

Question: On the set A of all lines in a plane, define L₁ R L₂ ⟺ L₁ is parallel to L₂. Show R is an equivalence relation and describe [L].

Solution:
Step 1 — Reflexive: every line is parallel to itself ⟹ (L, L) ∈ R. ✓
Step 2 — Symmetric: L₁ ∥ L₂ ⟹ L₂ ∥ L₁. ✓
Step 3 — Transitive: L₁ ∥ L₂ and L₂ ∥ L₃ ⟹ L₁ ∥ L₃. ✓
Conclusion: R is an equivalence relation. The class [L] is the set of all lines parallel to L, i.e. all lines sharing L's direction (same slope). Each "direction" is one equivalence class.

Exam application

A recurring NCERT question: R on integers with a R b ⟺ a − b is even. Show it is equivalence and find the classes — there are exactly two (evens and odds). Another favourite: R = {(a,b) : a and b have the same remainder mod 5}. Master the template — prove R, S, T, then list classes.

Shortcut: If a relation is defined by "a and b have the same value of f( )" (same remainder, same length, same colour), it is automatically an equivalence relation, because equality of f-values is reflexive, symmetric and transitive for free.

Key formulas / definitions

• Equivalence relation: reflexive AND symmetric AND transitive.
• Equivalence class of a: [a] = {x ∈ A : (x, a) ∈ R}.
• Partition: equivalence classes are disjoint and their union is A.
• Property: [a] = [b] iff (a, b) ∈ R; otherwise [a] ∩ [b] = ∅.

An equivalence relation divides a set into non-overlapping equivalence classes that together cover the whole set.

Worked Example

Problem: On the set of integers Z, define a R b if and only if (a − b) is divisible by 3. Show that R is an equivalence relation.

Solution:
Step 1 — Test reflexivity. For any integer a, a − a = 0, which is divisible by 3 (0 = 3 × 0). So a R a holds for all a — reflexive.

Step 2 — Test symmetry. Suppose a R b, so (a − b) = 3k for some integer k. Then (b − a) = −3k = 3(−k), also divisible by 3. So b R a — symmetric.

Step 3 — Test transitivity. Suppose a R b and b R c, so (a − b) = 3m and (b − c) = 3n. Adding: (a − c) = 3m + 3n = 3(m + n), divisible by 3. So a R c — transitive.

Step 4 — Conclude. Since R is reflexive, symmetric, and transitive, it is an equivalence relation. (It partitions Z into three classes: numbers leaving remainder 0, 1, or 2 on division by 3.)

Answer: R is reflexive, symmetric, and transitive, so it is an equivalence relation on Z.