CBSE Class 12 Maths Chapter 1: Relations and Functions Revision Notes

Key Definitions Glossary

Relation (R)

A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B. Mathematically, R ⊆ A × B.

Reflexive Relation

A relation R on set A is reflexive if every element of A is related to itself. Formally, (a, a) ∈ R for every a ∈ A.

Symmetric Relation

A relation R on set A is symmetric if (a, b) ∈ R implies (b, a) ∈ R for all a, b ∈ A.

Transitive Relation

A relation R on set A is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R for all a, b, c ∈ A.

Equivalence Relation

A relation R on set A that is simultaneously reflexive, symmetric, and transitive.

Injective Function (One-to-One)

A function f: X → Y is injective if distinct elements in the domain have distinct images in the codomain. Formally, f(x1) = f(x2) ⇒ x1 = x2.

Surjective Function (Onto)

A function f: X → Y is surjective if every element in the codomain Y has at least one pre-image in the domain X. Range of f = Codomain Y.

Bijective Function

A function f: X → Y that is both injective (one-to-one) and surjective (onto).

Core Conceptual Intuition: Proving Bijectivity and Equivalence

To score full marks on board questions, you must follow a highly structured approach for proofs. When proving that a relation is an equivalence relation, you must explicitly write down three distinct steps proving Reflexivity, Symmetry, and Transitivity. If even one property fails, the relation is not an equivalence relation.

Similarly, for verifying if a function is bijective, you must separately prove it is injective (One-to-One) and surjective (Onto).

• For Injectivity, assume $f(x_1) = f(x_2)$ and use algebraic simplification to show that this holds true if and only if $x_1 = x_2$. If you find any counterexample where $x_1 \neq x_2$ but $f(x_1) = f(x_2)$, the function is Many-to-One.
• For Surjectivity, let $y \in Y$ (codomain). Express $x$ in terms of $y$ (i.e., $x = g(y)$) and check if this $x$ always belongs to the domain $X$ for every $y$ in the codomain. If any $y$ is left without a valid pre-image in $X$, the function is Into.

Comparison of Function Classifications

Must-Remember Formulas & Properties

• The total number of relations from a set A with n elements to itself is 2^(n^2).
• The number of reflexive relations on a set with n elements is 2^(n^2 - n).
• The number of symmetric relations on a set with n elements is 2^(n(n+1)/2).
• An empty relation on a non-empty set is symmetric and transitive, but never reflexive (unless the set itself is empty).
• An identity relation on any set is always an equivalence relation.
• The intersection of two equivalence relations on a set is also an equivalence relation, but their union may not be.
• For a function f: A → B, where n(A) = m and n(B) = n, the number of injective functions is nPm (if n >= m) and 0 (if n < m).
• The total number of bijective functions from set A to B (where n(A) = n(B) = n) is n! (n factorial).

Solved Mini-Examples for Revision

• {"title":"Example 1: Checking Equivalence Relations","content":"Let R be a relation on the set of all integers Z defined by R = {(a, b) : a - b is divisible by 3}. Show that R is an equivalence relation.\n\nProof:\n1. Reflexive: For any a ∈ Z, a - a = 0. Since 0 is divisible by 3, (a, a) ∈ R. Hence, R is reflexive.\n2. Symmetric: Let (a, b) ∈ R. Then a - b = 3k for some k ∈ Z. This implies b - a = -3k = 3(-k). Since -k ∈ Z, b - a is divisible by 3. Thus, (b, a) ∈ R. Hence, R is symmetric.\n3. Transitive: Let (a, b) ∈ R and (b, c) ∈ R. Then a - b = 3k and b - c = 3m for some k, m ∈ Z. Adding these equations, (a - b) + (b - c) = 3k + 3m => a - c = 3(k + m). Since k + m ∈ Z, a - c is divisible by 3. Thus, (a, c) ∈ R. Hence, R is transitive.\n\nSince R is reflexive, symmetric, and transitive, it is an equivalence relation."}
• {"title":"Example 2: Proving Bijectivity","content":"Show that the function f: R → R defined by f(x) = 2x + 5 is bijective.\n\nProof:\n1. Injectivity: Let f(x1) = f(x2) for x1, x2 ∈ R.\n => 2x1 + 5 = 2x2 + 5\n => 2x1 = 2x2\n => x1 = x2.\n Thus, f is injective.\n2. Surjectivity: Let y ∈ R (codomain). Set y = f(x) => y = 2x + 5.\n => x = (y - 5)/2.\n Since y ∈ R, (y - 5)/2 is also a real number (x ∈ R). \n Now, f(x) = f((y - 5)/2) = 2((y - 5)/2) + 5 = y - 5 + 5 = y.\n Thus, every element in the codomain has a pre-image in the domain. Hence, f is surjective.\n\nSince f is both injective and surjective, it is a bijective function."}

Board Exam Traps and Marking Cues

• The Domain & Codomain Trap: Always read the sets defining the function carefully. For example, $f: \mathbb{N} \to \mathbb{N}$ defined by $f(x) = 2x$ is injective but NOT surjective (since odd numbers in the codomain have no pre-images in natural numbers). However, if the function is defined as $f: \mathbb{R} \to \mathbb{R}$, it is still injective but still not surjective (since the range is only positive or depending on the exact polynomial). Always verify the domains!
• Avoid Short-cut Explanations: In Board Exams, writing 'clearly onto' or 'by observation it is one-to-one' will cost you marks. Write down the algebraic proofs step-by-step to secure full method marks.

Quick Revision Self-Assessment

• What is the identity relation on a set A? The identity relation I_A on set A is defined as I_A = {(a, a) : a ∈ A}. Every element is related to only itself.
• If set A has 3 elements, how many bijective functions can be defined from A to itself? The number of bijective functions from A to itself is n! = 3! = 3 × 2 × 1 = 6 functions.
• Can a relation be symmetric and transitive but not reflexive? Give an example. Yes. Let A = {1, 2}. Consider relation R = {(1, 1)}. R is symmetric and transitive, but it is not reflexive because (2, 2) ∉ R.
• Is the constant function f: R → R defined by f(x) = c surjective? No. The range of f is only {c}, which is a proper subset of the codomain R (unless R itself only contains c, which is not the case). Therefore, it is into, not surjective.

Frequently Asked Questions