1. Chapter at a Glance
In Class 11 you learnt what relations and functions are. Now we classify them. Some relations behave like "equality": every element is related to itself, the relation works both ways, and it chains. These are equivalence relations, and they sort a set into non-overlapping groups (equivalence classes). Functions, too, come in special kinds: one-one (no two inputs share an output), onto (every element of the codomain is used) and bijective (both). These ideas are the base for inverse trigonometric functions and for invertible matrices later in the year.
- Empty and universal relations.
- Reflexive, symmetric and transitive relations.
- Equivalence relations and equivalence classes.
- One-one (injective), onto (surjective) and bijective functions.
- Testing functions on finite sets and on ℝ.
- Counting relations and functions.
2. Types of Relations
A relation R on a set A is a subset of A × A.
| Type | Condition | Example on a set of people |
|---|---|---|
| Empty | R = ∅ | "is 10 m taller than" in a class |
| Universal | R = A × A | "difference of heights is less than 3 m" |
| Reflexive | (a, a) ∈ R for every a ∈ A | "same age as" |
| Symmetric | (a, b) ∈ R ⇒ (b, a) ∈ R | "is a sibling of" |
| Transitive | (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R | "is taller than" |
A relation that is reflexive, symmetric and transitive. It partitions A into disjoint equivalence classes: [a] = {x ∈ A : (x, a) ∈ R}. Any two classes are either equal or disjoint, and together they cover A.
Show that R = {(a, b) : 2 divides a − b} on ℤ is an equivalence relation, and find its classes.
Solution: Reflexive: a − a = 0 is divisible by 2. Symmetric: if 2 | (a − b) then 2 | (b − a). Transitive: a − c = (a − b) + (b − c). Classes: [0] = even integers, [1] = odd integers.
R on ℝ: (a, b) ∈ R if a ≤ b². Is R reflexive, symmetric, transitive?
Solution: Not reflexive: ½ ≤ ¼ is false. Not symmetric: (1, 2) ∈ R (1 ≤ 4) but (2, 1) ∉ R (2 ≤ 1 false). Not transitive: (3, 2), (2, 1.5) ∈ R (3 ≤ 4, 2 ≤ 2.25) but (3, 1.5) ∉ R (3 ≤ 2.25 false).
On the set of all triangles, T₁ R T₂ if T₁ is congruent to T₂. Show R is an equivalence relation.
Solution: Every triangle is congruent to itself; congruence works both ways; if T₁ ≅ T₂ and T₂ ≅ T₃ then T₁ ≅ T₃. ✓
3. Types of Functions
- One-one (injective): f(x₁) = f(x₂) ⇒ x₁ = x₂ (distinct inputs, distinct outputs).
- Onto (surjective): every y ∈ Y has some x with f(x) = y (range = codomain).
- Bijective: both one-one and onto.
Show that f : ℝ → ℝ, f(x) = 2x + 3 is bijective.
Solution: One-one: 2x₁ + 3 = 2x₂ + 3 ⇒ x₁ = x₂. Onto: for any y, x = (y − 3)/2 ∈ ℝ gives f(x) = y. Hence bijective.
Is f : ℝ → ℝ, f(x) = x² one-one? onto?
Solution: Not one-one: f(−1) = f(1). Not onto: −1 has no pre-image (x² ≥ 0). But f : [0, ∞) → [0, ∞), f(x) = x² is bijective — the domain and codomain matter!
Is f : ℕ → ℕ, f(x) = 2x one-one and onto?
Solution: One-one (2x₁ = 2x₂ ⇒ x₁ = x₂). Not onto: odd numbers have no pre-image in ℕ.
If f : A → A with A finite, then f is one-one ⇔ f is onto. (This fails for infinite sets, as f(x) = 2x on ℕ shows.)
4. Counting
| From A (m elements) to B (n elements) | Number |
|---|---|
| Relations | 2^(mn) |
| Functions | nᵐ |
| One-one functions (m ≤ n) | ⁿPₘ |
| Bijections (m = n) | n! |
5. Solved Problems (Board Pattern)
Give an example of a relation on {1, 2, 3} that is reflexive and transitive but not symmetric.
Solution: R = {(1,1), (2,2), (3,3), (1,2)}. Reflexive ✓; transitive ✓ (only (1,2) chains with (2,2)); not symmetric since (2,1) ∉ R.
Show that f : ℝ → ℝ, f(x) = |x| is neither one-one nor onto.
Solution: f(−2) = f(2) → not one-one; −1 has no pre-image → not onto.
f : ℕ → ℕ, f(n) = (n + 1)/2 if n odd, n/2 if n even. Is f bijective?
Solution: f(1) = 1 and f(2) = 1 → not one-one. Onto: for any m, f(2m) = m. So f is onto but not one-one.
6. Master Formula Table
| Property | Test |
|---|---|
| Reflexive | ∀a: (a, a) ∈ R |
| Symmetric | (a, b) ∈ R ⇒ (b, a) ∈ R |
| Transitive | (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R |
| One-one | f(x₁) = f(x₂) ⇒ x₁ = x₂ |
| Onto | range = codomain |
| Relations A → B | 2^(mn) |
| One-one maps | ⁿPₘ |
| Bijections on n elements | n! |
7. Master Summary Table
| Concept | Key idea | Exam focus |
|---|---|---|
| Equivalence relation | R + S + T | Prove; find classes |
| Counterexamples | One failing case disproves | Show not reflexive/symmetric/transitive |
| One-one | No sharing of outputs | Algebraic proof |
| Onto | Solve y = f(x) for x in domain | Find pre-image |
| Bijective | Both | Needed for inverses |
8. Mind Map
- Reflexive, symmetric
- Transitive
- All three
- Classes partition
- One-one
- Onto
- Both
- Counting n!
9. Tips and Tricks
- For "onto", solve y = f(x) for x and check that x lies in the domain for every y.
- A strictly increasing or decreasing function on an interval is one-one.
- Horizontal line test: a function ℝ → ℝ is one-one if every horizontal line meets its graph at most once.
- Concluding "reflexive" from (a, a) for some a — it must hold for every a.
- Ignoring the codomain when testing onto.
- Thinking a relation that is not symmetric must be anti-symmetric.
10. Important One-liners
- An equivalence relation is reflexive, symmetric and transitive.
- Equivalence classes are pairwise disjoint and cover the set.
- The empty relation on a non-empty set is symmetric and transitive but not reflexive.
- The universal relation is an equivalence relation.
- On a finite set A, f : A → A is one-one iff it is onto.
11. Exam-pattern Questions with Detailed Answers
The number of equivalence relations on {1, 2} is:
- 1
- 2
- 3
- 4
Show answer
{(1,1),(2,2)} and the universal relation — 2.
Answer: (B)
Is R = {(1,1), (2,2), (3,3), (1,2), (2,3)} on {1, 2, 3} transitive?
Show answer
(1,2), (2,3) ∈ R but (1,3) ∉ R → not transitive.
Answer: No
How many bijections are there from {a, b, c, d} to itself?
Show answer
4! = 24.
Answer: 24
Show that R = {(a, b) : |a − b| is even} on {1, 2, 3, 4, 5} is an equivalence relation, and list the classes.
Show answer
|a − a| = 0 even; |a − b| = |b − a|; if a − b and b − c are even, so is their sum a − c. Classes {1, 3, 5} and {2, 4}.
Answer: Classes {1, 3, 5}, {2, 4}
Show that f : ℝ − {0} → ℝ − {0}, f(x) = 1/x is bijective.
Show answer
1/x₁ = 1/x₂ ⇒ x₁ = x₂. For y ≠ 0, x = 1/y is in the domain and f(x) = y. Bijective.
Answer: Proved
Check whether f : ℝ → ℝ, f(x) = x³ is one-one and onto.
Show answer
x₁³ = x₂³ ⇒ x₁ = x₂ (one-one). For any y, x = ∛y works (onto). Bijective.
Answer: Bijective
Let L be the set of lines in a plane and R = {(L₁, L₂) : L₁ ⊥ L₂}. Show R is symmetric but neither reflexive nor transitive.
Show answer
No line is perpendicular to itself (not reflexive). L₁ ⊥ L₂ ⇒ L₂ ⊥ L₁ (symmetric). L₁ ⊥ L₂, L₂ ⊥ L₃ ⇒ L₁ ∥ L₃, not ⊥ (not transitive).
Answer: Proved
Is f : ℤ → ℤ, f(x) = x² one-one? onto?
Show answer
f(−1) = f(1) → not one-one; 2 has no integer square root → not onto.
Answer: Neither
On ℕ × ℕ define (a, b) R (c, d) if a + d = b + c. Show R is an equivalence relation.
Show answer
Reflexive: a + b = b + a. Symmetric: a + d = b + c ⇒ c + b = d + a. Transitive: a + d = b + c and c + f = d + e ⇒ adding, a + f = b + e. ∎
Answer: Proved
Show that f : ℝ → (−1, 1), f(x) = x/(1 + |x|), is one-one and onto.
Show answer
For x ≥ 0, f(x) = x/(1 + x) ∈ [0, 1); for x < 0, f(x) = x/(1 − x) ∈ (−1, 0). f preserves sign, and each piece is strictly increasing, so f is one-one. Onto: for 0 ≤ y < 1, x = y/(1 − y) ≥ 0; for −1 < y < 0, x = y/(1 + y) < 0. Bijective.
Answer: Proved
Let A = ℝ − {3} and B = ℝ − {1}. Show that f : A → B, f(x) = (x − 2)/(x − 3) is bijective.
Show answer
One-one: (x₁ − 2)/(x₁ − 3) = (x₂ − 2)/(x₂ − 3) ⇒ −3x₁ − 2x₂ = −2x₁ − 3x₂ ⇒ x₁ = x₂. Onto: y = (x − 2)/(x − 3) ⇒ x = (3y − 2)/(y − 1), defined for y ≠ 1 and never equal to 3. Bijective.
Answer: Proved
Find the number of relations on {1, 2, 3} containing (1, 2) and (1, 3) that are reflexive and symmetric but not transitive.
Show answer
Must contain (1,1), (2,2), (3,3), (1,2), (2,1), (1,3), (3,1). Then (2,1), (1,3) need (2,3) for transitivity. Without (2,3) and (3,2) it is not transitive — 1 relation. Adding (2,3) forces (3,2) (symmetry) and becomes transitive. So exactly 1.
Answer: 1
12. Quick Revision Summary
- Relation types: empty, universal, reflexive, symmetric, transitive.
- Equivalence = all three; classes partition the set.
- One-one: f(x₁) = f(x₂) ⇒ x₁ = x₂; onto: range = codomain; bijective: both.
- Domain and codomain change the answer (x² on ℝ vs on [0, ∞)).
- Counting: 2^(mn) relations, nᵐ functions, ⁿPₘ one-one, n! bijections.