Class 12 Mathematics — Original Educational Content

Chapter 1: Relations and Functions 📐

Master the foundations of abstract mathematics — relations, equivalence classes, function types, composition, inverses, and binary operations.

📌 Relations & Equivalence | 🎯 Injective, Surjective, Bijective | 🧩 20 Practice Problems | 🎯 10-Question Quiz

Section 1

📌 Introduction to Relations

🔗 From Pairs to Patterns — The Language of Relationships

In everyday life, we constantly describe how things are connected: "Priya is taller than Ravi," "7 is greater than 3," "Mumbai is the capital of Maharashtra." Mathematics formalizes these connections using the powerful idea of a relation.

Before we can define a relation, we need to understand how to pair elements from two sets — enter the Cartesian Product! 🚀

Cartesian Product: Building All Possible Pairs

Given two non-empty sets A and B, the Cartesian product A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

A × B = {(a, b) : a ∈ A and b ∈ B}

Let A = {red, blue} and B = {1, 2, 3}. Then:

A × B = {(red,1), (red,2), (red,3), (blue,1), (blue,2), (blue,3)}

Notice: |A × B| = |A| × |B| = 2 × 3 = 6 ordered pairs.

Order matters! (red, 1) ≠ (1, red). The Cartesian product A × B is generally not the same as B × A.

What is a Relation?

A relation R from set A to set B is any subset of A × B. That is, R ⊆ A × B.

If (a, b) ∈ R, we write a R b and say "a is related to b under R."

A relation in a set A (or "on" A) means a relation from A to itself, i.e., R ⊆ A × A.

Let A = {1, 2, 3} and define a relation R by the rule "a is less than b." Then:

R = {(1,2), (1,3), (2,3)}

Here 1 R 2, 1 R 3, and 2 R 3 — but 3 R 1 is false (3 is not less than 1).

Domain, Codomain, and Range

For a relation R from A to B:

Domain of R = set of all first elements of the ordered pairs in R = {a : (a, b) ∈ R for some b ∈ B}
Codomain = the entire set B
Range of R = set of all second elements of the ordered pairs in R = {b : (a, b) ∈ R for some a ∈ A}

Range ⊆ Codomain, always. The range includes only those elements of B that are actually "hit" by R, while the codomain is the entire set B regardless of whether every element is used.

Two Extreme Relations

🕳️ Empty Relation (R = ∅)

No element of A is related to any element of A. R has zero ordered pairs.

Think of it as: "No student in the class has exactly 100 siblings" — the relation is empty because the condition is never satisfied.

🌐 Universal Relation (R = A × A)

Every element of A is related to every element of A (including itself). R contains ALL possible ordered pairs.

Think of it as: "Every student in the class breathes air" — the condition is satisfied by every pair, so all are related.

Both the empty relation ∅ and the universal relation A × A are trivially valid relations on A — they are subsets of A × A. Every other relation lies somewhere between these two extremes.

Section 2

🔄 Types of Relations

Not all relations behave the same way. We classify relations by three fundamental properties — each asking a simple question about the ordered pairs in R.

1. Reflexive Relations

A relation R on set A is reflexive if every element is related to itself:

∀ a ∈ A, (a, a) ∈ R

✅ Reflexive: R on ℤ defined by "a divides a." Since every integer divides itself (a = 1 × a), the pair (a, a) is always in R.

❌ Not Reflexive: R on ℝ defined by "a is strictly greater than b." We'd need a > a, which is impossible.

2. Symmetric Relations

A relation R on set A is symmetric if whenever a is related to b, then b is also related to a:

∀ a, b ∈ A, (a, b) ∈ R ⇒ (b, a) ∈ R

✅ Symmetric: R on a set of people defined by "a is a sibling of b." If Priya is Ravi's sibling, then Ravi is Priya's sibling.

❌ Not Symmetric: R on people defined by "a is the father of b." If Ramesh is the father of Suresh, Suresh is certainly not the father of Ramesh!

3. Transitive Relations

A relation R on set A is transitive if whenever a is related to b and b is related to c, then a is related to c:

∀ a, b, c ∈ A, (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R

✅ Transitive: "is less than" on ℝ. If a < b and b < c, then a < c.

❌ Not Transitive: "is perpendicular to" on lines in a plane. Line ℓ₁ ⊥ ℓ₂ and ℓ₂ ⊥ ℓ₃ does not imply ℓ₁ ⊥ ℓ₃ (in fact, ℓ₁ ∥ ℓ₃).

Quick Comparison Table

Relation (on appropriate set)	Reflexive	Symmetric	Transitive
"a = b" on ℤ	✓	✓	✓
"a < b" on ℝ	✗	✗	✓
"a ≤ b" on ℝ	✓	✗	✓
"a divides b" on ℤ⁺	✓	✗	✓
"a is sibling of b"	✗	✓	✓
"a ⊥ b" (lines)	✗	✓	✗

Consider the relation "is a friend of" on a social media platform (where you can be your own friend). Is it reflexive? If the platform auto-adds you as your own friend, yes. Symmetric? On Facebook, friendships are mutual — yes. Transitive? If A is friends with B and B is friends with C, is A necessarily friends with C? Not at all! So it's reflexive and symmetric, but not transitive.

A common error: "If R has no pairs of the form (a, b) and (b, c) where a ≠ c, then R is transitive." This is actually true — transitivity is satisfied vacuously! The condition "(a,b) ∈ R and (b,c) ∈ R" is never met, so the implication holds by default.

Section 3

⚖️ Equivalence Relations & Equivalence Classes

A relation R on a set A is called an equivalence relation if it is simultaneously:

Reflexive — every element is related to itself
Symmetric — if a is related to b, then b is related to a
Transitive — if a is related to b and b is related to c, then a is related to c

Worked Example: Even Difference

Define R on ℤ by: R = {(a, b) : 2 divides (a − b)}

In other words, a R b if and only if a − b is even (i.e., a and b have the same parity).

Checking all three properties:

✅ Reflexive

For any a ∈ ℤ: a − a = 0, and 2 | 0. So (a, a) ∈ R. ✓

✅ Symmetric

If (a, b) ∈ R, then 2 | (a − b), say a − b = 2k. Then b − a = −2k = 2(−k), so 2 | (b − a). Hence (b, a) ∈ R. ✓

✅ Transitive

If (a, b) ∈ R and (b, c) ∈ R, then a − b = 2m and b − c = 2n. Adding: a − c = 2m + 2n = 2(m + n), so 2 | (a − c). Hence (a, c) ∈ R. ✓

Since R is reflexive, symmetric, and transitive, R is an equivalence relation. ✨

Equivalence Classes

Given an equivalence relation R on A, the equivalence class of an element a ∈ A is:

[a] = {x ∈ A : (a, x) ∈ R}

It is the collection of all elements related to a.

For our even-difference relation on ℤ:

Equivalence Class [0] — The Even Integers

{…, −6, −4, −2, 0, 2, 4, 6, …}

All integers whose difference with 0 is even → all even numbers.

Equivalence Class [1] — The Odd Integers

{…, −5, −3, −1, 1, 3, 5, 7, …}

All integers whose difference with 1 is even → all odd numbers.

Notice: [0] ∪ [1] = ℤ (the full set) and [0] ∩ [1] = ∅ (no overlap). The equivalence classes form a partition of ℤ — they split the entire set into non-overlapping groups that together cover everything!

Example: Modular Arithmetic (mod 3)

Define R on ℤ by: a R b ⟺ 3 | (a − b) (i.e., a and b leave the same remainder when divided by 3).

This gives us three equivalence classes:

Class	Remainder	Elements
[0]	0	{…, −6, −3, 0, 3, 6, 9, …}
[1]	1	{…, −5, −2, 1, 4, 7, 10, …}
[2]	2	{…, −4, −1, 2, 5, 8, 11, …}

Equivalence relations are everywhere in daily life! "Born in the same month" partitions all people into 12 classes. "Living in the same city" partitions the population by city. "Having the same blood type" gives 8 classes (A+, A−, B+, B−, AB+, AB−, O+, O−). Any time you group things by a shared property, you're using an equivalence relation!

Key theorem: Every equivalence relation on A gives a partition of A, and conversely, every partition of A defines an equivalence relation (where two elements are related iff they belong to the same block of the partition).

Section 4

🎯 Types of Functions

A function f : X → Y assigns to each element x ∈ X exactly one element f(x) ∈ Y. While all functions share this basic property, they differ in how thoroughly they "cover" the codomain and whether they reuse outputs.

One-One (Injective) Functions

A function f : X → Y is one-one (or injective) if distinct inputs always produce distinct outputs:

f(x₁) = f(x₂) ⇒ x₁ = x₂

Equivalently: x₁ ≠ x₂ ⇒ f(x₁) ≠ f(x₂). No two different elements map to the same value.

123

→ f(x) = 2x + 1 → →

3579

↑ One-one: each element of X maps to a unique element of Y. Element 9 in Y is not hit.

Onto (Surjective) Functions

A function f : X → Y is onto (or surjective) if every element of Y is the image of at least one element in X:

∀ y ∈ Y, ∃ x ∈ X such that f(x) = y

In other words, the range of f equals the codomain Y.

✅ Onto: f : ℝ → ℝ defined by f(x) = x³. For any real number y, the cube root x = y^1/3 exists in ℝ, and f(y^1/3) = y.

❌ Not Onto: f : ℤ → ℤ defined by f(x) = 2x. The output is always even, so odd integers (like 3) have no pre-image. Range = even integers ≠ ℤ.

Bijective Functions (One-One AND Onto)

A function f : X → Y is bijective if it is both one-one and onto. Every element of Y is mapped to by exactly one element of X — a perfect pairing!

✅ Bijective: f : ℝ → ℝ defined by f(x) = 3x + 5.

One-one: 3x₁ + 5 = 3x₂ + 5 ⇒ 3x₁ = 3x₂ ⇒ x₁ = x₂. ✓

Onto: For any y ∈ ℝ, choose x = (y − 5)/3 ∈ ℝ. Then f(x) = 3 · (y−5)/3 + 5 = y. ✓

Counter-Examples to Watch For

Function	One-One?	Onto?	Why?
f(x) = x² on ℝ → ℝ	✗	✗	f(−2) = f(2) = 4; negative numbers not in range
f(x) = x² on ℝ⁺ → ℝ⁺	✓	✓	Restricted domain/codomain makes it bijective
f(x) = \|x\| on ℝ → ℝ	✗	✗	f(−3) = f(3); negatives not in range
f(x) = x³ on ℝ → ℝ	✓	✓	Strictly increasing, hits all reals

For finite sets, there is a beautiful result: if f : A → A (same finite set as domain and codomain), then f is one-one if and only if f is onto. So for finite sets, you only need to check one property! This fails for infinite sets — consider f : ℤ → ℤ, f(n) = 2n (one-one but not onto).

The number of bijections from a set with n elements to itself is n! (n factorial). So from {1, 2, 3} to itself, there are 3! = 6 bijections. These bijections are exactly the permutations of the set — a concept that connects algebra, combinatorics, and group theory!

Section 5

🔗 Composition of Functions

When two functions are applied one after the other, the result is a new function called their composition.

Let f : A → B and g : B → C. The composition of f and g, written g ∘ f (read "g composed with f" or "g of f"), is the function from A to C defined by:

(g ∘ f)(x) = g(f(x)) for all x ∈ A

First apply f, then apply g to the result.

The notation g ∘ f means "apply f first, then g." The function written on the right acts first! This is a very common source of errors.

Worked Example

Let f : ℝ → ℝ, f(x) = x² and g : ℝ → ℝ, g(x) = 2x + 1.

(g ∘ f)(x) = g(f(x)) = g(x²) = 2x² + 1

(f ∘ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)² = 4x² + 4x + 1

🚨 g ∘ f ≠ f ∘ g in general!

In the example above: g ∘ f gives 2x² + 1, while f ∘ g gives 4x² + 4x + 1. These are completely different functions.

Composition is NOT commutative. The order in which you compose matters. (However, composition IS associative: h ∘ (g ∘ f) = (h ∘ g) ∘ f.)

Properties of Composition

✅ Composition preserves Injectivity

If f : A → B and g : B → C are both one-one, then g ∘ f : A → C is also one-one.

Proof sketch: If g(f(x₁)) = g(f(x₂)), then f(x₁) = f(x₂) (since g is one-one), hence x₁ = x₂ (since f is one-one).

✅ Composition preserves Surjectivity

If f : A → B and g : B → C are both onto, then g ∘ f : A → C is also onto.

Proof sketch: For any c ∈ C, ∃ b ∈ B with g(b) = c (g onto). For that b, ∃ a ∈ A with f(a) = b (f onto). So g(f(a)) = c.

Combining the above: if both f and g are bijective, then g ∘ f is also bijective.

If g ∘ f is one-one, can we conclude that f is one-one? What about g? Hint: g ∘ f one-one guarantees f is one-one, but says nothing about g by itself.

Section 6

↩️ Invertible Functions

An invertible function is one whose action can be completely "undone." Just as subtraction undoes addition, an inverse function reverses the mapping.

A function f : X → Y is invertible if there exists a function g : Y → X such that:

g ∘ f = I_X and f ∘ g = I_Y

where I_X(x) = x and I_Y(y) = y are the identity functions. We write g = f⁻¹.

A function is invertible if and only if it is bijective. This is the central theorem connecting inverses to the one-one/onto classification. If f is not one-one, two inputs go to the same output, and the inverse can't decide which to pick. If f is not onto, some outputs have no source, leaving the inverse undefined there.

Finding the Inverse

To find f⁻¹, follow these steps:

Write y = f(x)
Solve for x in terms of y
Replace y with x in the final expression to get f⁻¹(x)

Find the inverse of f : ℝ → ℝ, f(x) = 3x + 7.

Step 1: y = 3x + 7

Step 2: y − 7 = 3x, so x = (y − 7)/3

Step 3: f⁻¹(x) = (x − 7)/3

Verification: f(f⁻¹(x)) = 3 · (x−7)/3 + 7 = x − 7 + 7 = x ✓

Inverse of a Composition

🔄 Shoe-Sock Principle

If f and g are both invertible, then: (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹

Just like putting on socks then shoes — to undo the process, you remove shoes first, then socks. The reverse order applies!

Let f(x) = 2x and g(x) = x + 5. Then g ∘ f(x) = 2x + 5.

Direct inverse: (g ∘ f)⁻¹(x) = (x − 5)/2

Via formula: f⁻¹(x) = x/2 and g⁻¹(x) = x − 5.

f⁻¹ ∘ g⁻¹(x) = f⁻¹(x − 5) = (x − 5)/2 ✓ They match!

For any bijective f, we have (f⁻¹)⁻¹ = f. The inverse of the inverse takes you right back to the original function.

Section 7

⊕ Binary Operations

A binary operation combines two elements of a set to produce another element of the same set. This is the foundation of algebraic structures like groups, rings, and fields.

A binary operation ∗ on a set A is a function ∗ : A × A → A. That is, for every pair (a, b) ∈ A × A, the result a ∗ b must also be in A.

The crucial requirement is closure: the operation must not take you outside the set.

✅ Binary operations on ℤ:

Addition (+): a + b ∈ ℤ for all a, b ∈ ℤ ✓
Multiplication (×): a × b ∈ ℤ for all a, b ∈ ℤ ✓
Subtraction (−): a − b ∈ ℤ for all a, b ∈ ℤ ✓

❌ Not a binary operation on ℤ:

Division (÷): 1 ÷ 2 = 0.5 ∉ ℤ. Fails closure!

Properties of Binary Operations

Property	Definition	Example
Commutative	a ∗ b = b ∗ a for all a, b	+ on ℤ: a + b = b + a ✓
Associative	(a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c	× on ℤ: (ab)c = a(bc) ✓

Subtraction on ℤ is not commutative (5 − 3 ≠ 3 − 5) and not associative ((8 − 3) − 2 = 3 but 8 − (3 − 2) = 7).

Identity Element

An element e ∈ A is called the identity element for ∗ if:

a ∗ e = e ∗ a = a for all a ∈ A

The identity element, if it exists, is unique.

Operation	Set	Identity
+ (addition)	ℤ	0 (since a + 0 = 0 + a = a)
× (multiplication)	ℚ*	1 (since a × 1 = 1 × a = a)
∪ (union)	Power set 𝒫(S)	∅ (since A ∪ ∅ = A)
∩ (intersection)	Power set 𝒫(S)	S (since A ∩ S = A)

Inverse Element

Given a binary operation ∗ on A with identity e, an element b ∈ A is called the inverse of a ∈ A if:

a ∗ b = b ∗ a = e

Addition on ℤ (identity = 0): The inverse of a is −a, since a + (−a) = (−a) + a = 0.

Multiplication on ℚ* (identity = 1): The inverse of a (≠ 0) is 1/a, since a × (1/a) = (1/a) × a = 1.

Does every element of ℤ have a multiplicative inverse within ℤ? Consider: what integer b satisfies 3 × b = 1? Since b = 1/3 ∉ ℤ, the answer is no — most integers lack multiplicative inverses in ℤ. (Only 1 and −1 have multiplicative inverses in ℤ.)

Section 8

🧩 Practice Problems — Set 1

Test your understanding with these problems. Try solving each one before revealing the answer!

PROBLEM 1

Let A = {1, 2, 3} and R = {(1,1), (2,2), (3,3), (1,2), (2,3)}. Determine whether R is reflexive, symmetric, and/or transitive.

Reflexive: ✅ Yes — (1,1), (2,2), (3,3) are all present.
Symmetric: ❌ No — (1,2) ∈ R but (2,1) ∉ R.
Transitive: ❌ No — (1,2) ∈ R and (2,3) ∈ R, but (1,3) ∉ R.

PROBLEM 2

Show that the relation R = {(a, b) : a ≤ b} on ℝ is reflexive and transitive, but not symmetric.

Reflexive: For every a ∈ ℝ, a ≤ a is true. So (a, a) ∈ R. ✅
Transitive: If a ≤ b and b ≤ c, then a ≤ c. So (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R. ✅
Symmetric: Consider a = 1, b = 2. We have 1 ≤ 2, so (1,2) ∈ R. But 2 ≤ 1 is false, so (2,1) ∉ R. ❌

PROBLEM 3

Is the function f : ℝ → ℝ defined by f(x) = x³ one-one? Justify your answer.

Yes, f is one-one.
Suppose f(x₁) = f(x₂), i.e., x₁³ = x₂³.
Taking cube roots (which is well-defined and order-preserving on ℝ): x₁ = x₂.
Since equal outputs force equal inputs, f is injective. ✅

PROBLEM 4

Let f(x) = |x| and g(x) = 2x − 3. Find (g ∘ f)(x).

(g ∘ f)(x) = g(f(x)) = g(|x|) = 2|x| − 3.

For example: (g ∘ f)(−4) = 2|−4| − 3 = 2(4) − 3 = 5.
And: (g ∘ f)(3) = 2|3| − 3 = 6 − 3 = 3.

PROBLEM 5

A binary operation ∗ on ℤ is defined by a ∗ b = a + b − ab. Is ∗ commutative?

Yes, ∗ is commutative.
a ∗ b = a + b − ab
b ∗ a = b + a − ba = a + b − ab (since addition and multiplication of integers are commutative)
Therefore a ∗ b = b ∗ a for all a, b ∈ ℤ. ✅

PROBLEM 6

Find the inverse of the function f : ℝ → ℝ defined by f(x) = 5x + 2.

Let y = 5x + 2.
Solving for x: y − 2 = 5x, so x = (y − 2)/5.
Therefore f⁻¹(x) = (x − 2)/5.

Verification: f(f⁻¹(x)) = 5 · (x − 2)/5 + 2 = (x − 2) + 2 = x ✅

PROBLEM 7

Let R be a relation on ℤ defined by R = {(a, b) : a − b is even}. Show that R is an equivalence relation.

Reflexive: For any a ∈ ℤ, a − a = 0, which is even. So (a, a) ∈ R. ✅

Symmetric: If (a, b) ∈ R, then a − b is even, say a − b = 2k. Then b − a = −2k = 2(−k), which is also even. So (b, a) ∈ R. ✅

Transitive: If (a, b) ∈ R and (b, c) ∈ R, then a − b = 2m and b − c = 2n for some integers m, n. Adding: a − c = (a − b) + (b − c) = 2m + 2n = 2(m + n), which is even. So (a, c) ∈ R. ✅

Since R is reflexive, symmetric, and transitive, R is an equivalence relation.

PROBLEM 8

How many bijections exist from the set {a, b, c, d} to itself?

A bijection from a set to itself is a permutation.
For a set with 4 elements, the number of permutations is 4! = 4 × 3 × 2 × 1 = 24.

Reasoning: The first element has 4 choices for its image, the second has 3 remaining choices, the third has 2, and the last has 1.

PROBLEM 9

Is division (÷) a binary operation on ℚ* (non-zero rationals)?

Yes, ÷ is a binary operation on ℚ*.
For any a, b ∈ ℚ* (both non-zero rationals), a ÷ b = a/b.
Since a ≠ 0 and b ≠ 0, the result a/b is a non-zero rational number, i.e., a/b ∈ ℚ*.
Therefore, closure is satisfied. ✅

Note: Division is NOT a binary operation on ℤ (1 ÷ 2 = 0.5 ∉ ℤ) or on ℚ (0 ÷ 0 is undefined).

PROBLEM 10

Find all equivalence classes of the relation R = {(a, b) : 3 | (a − b)} on ℤ.

Two integers are related iff they give the same remainder when divided by 3. There are exactly 3 equivalence classes:

[0] = {…, −6, −3, 0, 3, 6, 9, …} — multiples of 3 (remainder 0)
[1] = {…, −5, −2, 1, 4, 7, 10, …} — remainder 1 when divided by 3
[2] = {…, −4, −1, 2, 5, 8, 11, …} — remainder 2 when divided by 3

These three classes are disjoint and their union is all of ℤ, forming a partition.

Section 9

🧩 Practice Problems — Set 2 (Challenging)

These problems are more involved and test deeper understanding. Work through each carefully!

PROBLEM 1

Show that the relation R on ℝ defined by R = {(a, b) : a ≤ b²} is neither reflexive, nor symmetric, nor transitive.

Not Reflexive: Take a = 1/2. We need 1/2 ≤ (1/2)² = 1/4. But 1/2 > 1/4. So (1/2, 1/2) ∉ R. ❌

Not Symmetric: Take a = 1, b = 2. We have 1 ≤ 2² = 4 ✓, so (1, 2) ∈ R. But 2 ≤ 1² = 1? No, 2 > 1. So (2, 1) ∉ R. ❌

Not Transitive: Take a = 2, b = −2, c = 1/2. We have 2 ≤ (−2)² = 4 ✓ and −2 ≤ (1/2)² = 1/4? Hmm, −2 ≤ 1/4 ✓. But 2 ≤ (1/2)² = 1/4? No. So (a,b) ∈ R and (b,c) ∈ R but (a,c) ∉ R. ❌

PROBLEM 2

Let f(x) = x² and g(x) = x + 1 (both ℝ → ℝ). Find f ∘ g and g ∘ f. Are they equal?

f ∘ g(x) = f(g(x)) = f(x + 1) = (x + 1)² = x² + 2x + 1

g ∘ f(x) = g(f(x)) = g(x²) = x² + 1

Are they equal? No! For example, at x = 1:
f ∘ g(1) = (1+1)² = 4, but g ∘ f(1) = 1² + 1 = 2.
This confirms composition is not commutative in general.

PROBLEM 3

Show that f : ℕ → ℕ defined by f(x) = 2x is one-one but not onto.

One-one: Suppose f(x₁) = f(x₂), i.e., 2x₁ = 2x₂. Dividing by 2: x₁ = x₂. ✅

Not onto: Consider 3 ∈ ℕ (codomain). Is there any x ∈ ℕ with f(x) = 3? We'd need 2x = 3, giving x = 3/2, which is not a natural number. So 3 has no pre-image, and f is not onto. ❌

The range of f is {2, 4, 6, 8, …} — only even natural numbers, which is a proper subset of ℕ.

PROBLEM 4

Let A = {1, 2, 3}. How many equivalence relations on A contain both (1, 2) and (2, 1)?

Any equivalence relation must be reflexive, so it must contain {(1,1), (2,2), (3,3)}. It must contain (1,2) and (2,1) (given). By symmetry and the required pairs, we have the base: {(1,1), (2,2), (3,3), (1,2), (2,1)}.

Now, should 3 be in the same class as 1 and 2?
Option 1: No — classes are {1, 2} and {3}. Relation = {(1,1), (2,2), (3,3), (1,2), (2,1)}. ✅ (Check: transitive since no chain leads to new pairs.)
Option 2: Yes — all in one class {1, 2, 3}. Relation = A × A (universal). ✅

Answer: 2 equivalence relations.

PROBLEM 5

Define ∗ on ℚ by a ∗ b = (a + b)/2. Is ∗ associative?

Check associativity: Compare (a ∗ b) ∗ c with a ∗ (b ∗ c).

(a ∗ b) ∗ c = ((a + b)/2) ∗ c = ((a + b)/2 + c) / 2 = (a + b + 2c) / 4

a ∗ (b ∗ c) = a ∗ ((b + c)/2) = (a + (b + c)/2) / 2 = (2a + b + c) / 4

Since (a + b + 2c)/4 ≠ (2a + b + c)/4 in general (e.g., a = 0, b = 0, c = 4 gives 2 vs 1), ∗ is NOT associative. ❌

PROBLEM 6

Show that f : ℝ \ {−1} → ℝ defined by f(x) = (x − 1)/(x + 1) is one-one but not onto (codomain ℝ).

One-one: Suppose f(x₁) = f(x₂):
(x₁ − 1)/(x₁ + 1) = (x₂ − 1)/(x₂ + 1)
Cross-multiply: (x₁ − 1)(x₂ + 1) = (x₂ − 1)(x₁ + 1)
Expand: x₁x₂ + x₁ − x₂ − 1 = x₁x₂ + x₂ − x₁ − 1
Simplify: 2x₁ = 2x₂, so x₁ = x₂. ✅

Not onto: Is there x such that f(x) = 1?
(x − 1)/(x + 1) = 1 ⇒ x − 1 = x + 1 ⇒ −1 = 1, which is impossible.
So 1 is not in the range of f, hence f is not onto. ❌

PROBLEM 7

If f is bijective, show that (f⁻¹)⁻¹ = f.

Since f is bijective, f⁻¹ exists and satisfies:
f ∘ f⁻¹ = I_Y and f⁻¹ ∘ f = I_X

The inverse of f⁻¹ is the function g such that g ∘ f⁻¹ = I_Y and f⁻¹ ∘ g = I_X.

From the equations above, g = f satisfies both conditions! Therefore (f⁻¹)⁻¹ = f. ✅

Intuition: Undoing an undo brings you back to where you started.

PROBLEM 8

Let A = {1, 2, 3, 4, 5, 6, 7} and R = {(a, b) : both a and b are odd, or both are even}. Find the equivalence classes.

This relation groups elements by parity (odd/even).

Equivalence class of odd numbers:
[1] = [3] = [5] = [7] = {1, 3, 5, 7}

Equivalence class of even numbers:
[2] = [4] = [6] = {2, 4, 6}

There are exactly 2 equivalence classes: {1, 3, 5, 7} and {2, 4, 6}. They are disjoint and their union is A.

PROBLEM 9

For the operation ∗ on ℤ⁺ defined by a ∗ b = LCM(a, b), find the identity element.

We need e ∈ ℤ⁺ such that LCM(a, e) = a for all a ∈ ℤ⁺.

Recall that LCM(a, 1) = a for every positive integer a (since 1 divides every integer, the LCM is just a).

Also, LCM(1, a) = a. ✅

Therefore, the identity element is e = 1.

PROBLEM 10

If f : A → B is bijective and g : B → C is bijective, show that g ∘ f : A → C is bijective.

One-one: Suppose (g ∘ f)(x₁) = (g ∘ f)(x₂), i.e., g(f(x₁)) = g(f(x₂)).
Since g is one-one: f(x₁) = f(x₂).
Since f is one-one: x₁ = x₂.
Therefore g ∘ f is one-one. ✅

Onto: Let c ∈ C (arbitrary).
Since g is onto: ∃ b ∈ B such that g(b) = c.
Since f is onto: ∃ a ∈ A such that f(a) = b.
Then (g ∘ f)(a) = g(f(a)) = g(b) = c.
Therefore g ∘ f is onto. ✅

Since g ∘ f is both one-one and onto, it is bijective.

Section 10

🎯 Quick Quiz — Test Yourself! (10 MCQs)

Select the correct answer for each question. You can't change your answer once selected!

QUESTION 1 OF 10

A relation that is reflexive, symmetric, and transitive is called:

AEmpty relation

BUniversal relation

CEquivalence relation

DPartial order relation

QUESTION 2 OF 10

The function f : ℝ → ℝ defined by f(x) = x² is:

AOne-one only

BOnto only

CBoth one-one and onto

DNeither one-one nor onto

QUESTION 3 OF 10

The number of bijections from {1, 2, 3} to itself is:

D27

QUESTION 4 OF 10

If f(x) = 2x + 3, then f⁻¹(x) equals:

A(x − 3) / 2

B(x + 3) / 2

C2x − 3

D3 − 2x

QUESTION 5 OF 10

The notation g ∘ f means:

AApply g first, then f

BApply f first, then g

CApply both simultaneously

DMultiply f and g

QUESTION 6 OF 10

Which of the following is a binary operation on ℤ?

ADivision (÷)

BAddition (+)

CSquare root (√)

DLogarithm (log)

QUESTION 7 OF 10

The identity element for addition (+) on ℤ is:

B−1

DDoes not exist

QUESTION 8 OF 10

The relation R = {(1,1), (2,2)} on A = {1, 2, 3} is:

AReflexive

BSymmetric only

CTransitive only

DNot reflexive

QUESTION 9 OF 10

If a function is both one-one and onto, it is called:

AInjective

BSurjective

CBijective

DReflexive

QUESTION 10 OF 10

The number of equivalence classes of the relation "congruent modulo 5" on ℤ is:

DInfinite

Your Quiz Results

Section 11

📋 Chapter Summary

🔑 Key Definitions at a Glance

Relations

A relation R from A to B is a subset of A × B. A relation on A means R ⊆ A × A.
Reflexive: (a, a) ∈ R for all a ∈ A.
Symmetric: (a, b) ∈ R ⇒ (b, a) ∈ R.
Transitive: (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R.
Equivalence relation: Reflexive + Symmetric + Transitive.
Equivalence class of a: [a] = {x ∈ A : (a, x) ∈ R}. Equivalence classes form a partition of A.

Functions

One-one (Injective): f(x₁) = f(x₂) ⇒ x₁ = x₂.
Onto (Surjective): For every y in codomain, ∃ x with f(x) = y.
Bijective: One-one AND onto.
For finite sets: f : A → A is one-one ⟺ f is onto.
Number of bijections from an n-element set to itself = n!

Composition

(g ∘ f)(x) = g(f(x)) — apply f first, then g.
Composition is associative but not commutative.
Composition preserves injectivity, surjectivity, and bijectivity.

Invertible Functions

f is invertible ⟺ f is bijective.
f⁻¹(y) = x where f(x) = y.
(g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹ (reverse order).
(f⁻¹)⁻¹ = f.

Binary Operations

A binary operation ∗ on A is a function ∗ : A × A → A (closure is essential).
Commutative: a ∗ b = b ∗ a.
Associative: (a ∗ b) ∗ c = a ∗ (b ∗ c).
Identity element e: a ∗ e = e ∗ a = a for all a.
Inverse of a: a ∗ b = b ∗ a = e.

This chapter forms the gateway to abstract algebra — one of the most beautiful branches of mathematics. The concepts of equivalence relations, bijections, and binary operations are the building blocks of groups, rings, fields, and beyond. What you've learned here will serve you throughout higher mathematics! 🎓

🎉 Congratulations on completing Chapter 1!

Relations and Functions — Class 12 Mathematics | EduArtha

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