Relations and Function: CBSE Class 12 Maths NCERT Guide

Understanding Types of Relations

A relation R on a non-empty set A is simply a subset of the Cartesian product A × A. It defines a relationship between pairs of elements in A. For the CBSE Class 12 syllabus, we are particularly interested in three specific properties that a relation can have. Understanding these is the first step to mastering the chapter.

1. Reflexive: A relation R on set A is reflexive if every element of A is related to itself. In other words, for every a ∈ A, the pair (a, a) must be in R. For example, the relation "is equal to" on the set of integers is reflexive because every integer is equal to itself.

1. Symmetric: A relation R is symmetric if whenever (a, b) ∈ R, then (b, a) must also be in R. It's a two-way street. If 'a' is related to 'b', then 'b' must be related to 'a'. For example, the relation "is perpendicular to" on a set of lines is symmetric. If line L1 is perpendicular to L2, then L2 is also perpendicular to L1.

1. Transitive: A relation R is transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) must also be in R. This property creates a chain or a bridge. If 'a' is related to 'b' and 'b' is related to 'c', then 'a' must be related to 'c'. The relation "is less than" on integers is transitive: if a < b and b < c, then a < c.

A relation that possesses all three of these properties—reflexive, symmetric, and transitive—is called an Equivalence Relation.

Key Definitions: Types of Functions

One-one (Injective) Function

A function f: X → Y is defined to be one-one, or injective, if the images of distinct elements of X under f are distinct. Mathematically, for every x₁, x₂ ∈ X, if f(x₁) = f(x₂), then it must imply that x₁ = x₂. In simpler terms, no two inputs have the same output.

Onto (Surjective) Function

A function f: X → Y is said to be onto, or surjective, if every element of the codomain Y has at least one pre-image in the domain X. This means that for every y ∈ Y, there exists an element x ∈ X such that f(x) = y. In other words, the range of the function is equal to its codomain.

Bijective Function

A function is bijective if it is both one-one (injective) and onto (surjective). A bijective function creates a perfect pairing between the elements of the domain and the codomain. Bijective functions are important because they are invertible.

Worked Examples: Proving Relations and Classifying Functions

• Example 1: Proving an Equivalence Relation Question: Show that the relation R in the set Z of integers given by R = {(a, b) : 3 divides a - b} is an equivalence relation. Solution: To prove R is an equivalence relation, we must show it is reflexive, symmetric, and transitive. Step 1: Check for Reflexivity For any integer a ∈ Z, consider the pair (a, a). The difference is a - a = 0. Since 0 is divisible by 3 (as 0 = 3 × 0), the pair (a, a) ∈ R for all a ∈ Z. Therefore, R is reflexive. Step 2: Check for Symmetry Let (a, b) ∈ R. This means that (a - b) is divisible by 3. So, a - b = 3k for some integer k. Now, consider (b - a). We have b - a = -(a - b) = -3k = 3(-k). Since -k is also an integer, (b - a) is divisible by 3. Thus, (b, a) ∈ R. Therefore, R is symmetric. Step 3: Check for Transitivity Let (a, b) ∈ R and (b, c) ∈ R. This means (a - b) is divisible by 3 and (b - c) is divisible by 3. So, a - b = 3k₁ and b - c = 3k₂ for some integers k₁ and k₂. Let's add these two equations: (a - b) + (b - c) = 3k₁ + 3k₂. This simplifies to a - c = 3(k₁ + k₂). Since k₁ + k₂ is an integer, (a - c) is divisible by 3. Thus, (a, c) ∈ R. Therefore, R is transitive. Conclusion: Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
• Example 2: Classifying a Function Question: Check the injectivity and surjectivity of the function f: R → R given by f(x) = 2x³ + 1. Solution: Step 1: Check for Injectivity (One-one) Let x₁, x₂ be two arbitrary elements in the domain R such that f(x₁) = f(x₂). f(x₁) = f(x₂) ⇒ 2x₁³ + 1 = 2x₂³ + 1 ⇒ 2x₁³ = 2x₂³ ⇒ x₁³ = x₂³ ⇒ x₁ = x₂ (Taking the cube root on both sides is unique for real numbers) Since f(x₁) = f(x₂) implies x₁ = x₂, the function f is injective (one-one). Step 2: Check for Surjectivity (Onto) Let y be an arbitrary element in the codomain R. We need to find if there exists an x in the domain R such that f(x) = y. f(x) = y ⇒ 2x³ + 1 = y ⇒ 2x³ = y - 1 ⇒ x³ = (y - 1) / 2 ⇒ x = ³√((y - 1) / 2) For any real number y, the expression (y - 1) / 2 is also a real number. The cube root of any real number is a real and unique number. Therefore, for any y ∈ R (codomain), there exists an x = ³√((y - 1) / 2) ∈ R (domain). So, the function f is surjective (onto). * Conclusion: Since the function is both injective and surjective, it is a bijective function.

Exam Traps and Key Tips

For Equivalence Relations:

• Don't skip steps: To prove a relation is an equivalence relation, you must demonstrate all three properties: reflexive, symmetric, and transitive. Simply stating they hold is not enough; you must show the working for each.
• The 'for all' trap in reflexivity: The reflexive property (a,a) ∈ R must hold for every element 'a' in the given set. If it fails for even one element, the relation is not reflexive.

For Classifying Functions:

• Proving vs. Disproving: To prove a function is one-one or onto, you must use general variables (like x₁, x₂, y). To disprove it, a single counterexample is sufficient and is often the quickest method. For f(x) = x², showing f(-2) = 4 and f(2) = 4 is enough to prove it's not one-one.
• Domain and Codomain are key: The same function definition, like f(x) = x², can be injective/surjective or not, depending on the specified domain and codomain. For example, f: [0, ∞) → [0, ∞) defined by f(x) = x² is bijective, whereas f: R → R is not.

Practice Questions with Solutions

• Q: Let A = {1, 2, 3}. Let R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}. Determine whether the relation R on A is reflexive, symmetric, and transitive. A: Step 1: Check Reflexivity. A relation on A is reflexive if (a, a) ∈ R for all a ∈ A. Here, A = {1, 2, 3}. We have (1, 1) ∈ R, (2, 2) ∈ R, and (3, 3) ∈ R. So, R is reflexive. Step 2: Check Symmetry. A relation is symmetric if (a, b) ∈ R implies (b, a) ∈ R. We check the pairs other than reflexive ones. We have (1, 2) ∈ R. Is (2, 1) ∈ R? Yes, it is. Since for every (a,b) in R, (b,a) is also in R, the relation is symmetric. Step 3: Check Transitivity. A relation is transitive if (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R. Let's check. We have (1, 2) ∈ R and (2, 1) ∈ R. For transitivity, (1, 1) should be in R. Yes, (1, 1) ∈ R. Let's try another pair. We have (2, 1) ∈ R and (1, 2) ∈ R. This implies (2, 2) should be in R. Yes, (2, 2) ∈ R. At first glance it seems transitive, but wait. If we take (a,b) = (1,2) and (b,c) = (2,2), then (a,c) = (1,2) which is in R. Let's take (a,b) = (1,1) and (b,c) = (1,2), then (a,c) = (1,2) which is in R. After checking all combinations, the relation holds for transitivity. Final answer: The relation R is reflexive, symmetric, and transitive. Therefore, it is an equivalence relation.
• Q: Show that the relation R in the set L of all lines in a plane, defined as R = {(L₁, L₂) : L₁ is parallel to L₂}, is an equivalence relation. A: Step 1: Check Reflexivity. Any line L₁ is parallel to itself (by convention). Thus, (L₁, L₁) ∈ R for all L₁ ∈ L. So, R is reflexive. Step 2: Check Symmetry. Let (L₁, L₂) ∈ R. This means L₁ is parallel to L₂. If L₁ is parallel to L₂, then L₂ is also parallel to L₁. So, (L₂, L₁) ∈ R. Therefore, R is symmetric. Step 3: Check Transitivity. Let (L₁, L₂) ∈ R and (L₂, L₃) ∈ R. This means L₁ is parallel to L₂ and L₂ is parallel to L₃. From elementary geometry, we know that two lines parallel to the same line are parallel to each other. Thus, L₁ is parallel to L₃. So, (L₁, L₃) ∈ R. Therefore, R is transitive. Final answer: Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
• Q: Check the injectivity and surjectivity of the function f: R → R given by f(x) = 5x - 3. A: Step 1: Check Injectivity (One-one). Let f(x₁) = f(x₂), where x₁, x₂ ∈ R. This gives 5x₁ - 3 = 5x₂ - 3. Adding 3 to both sides gives 5x₁ = 5x₂. Dividing by 5 gives x₁ = x₂. Since f(x₁) = f(x₂) ⇒ x₁ = x₂, the function is injective. Step 2: Check Surjectivity (Onto). Let y be an arbitrary element in the codomain R. We need to find an x in the domain such that f(x) = y. So, 5x - 3 = y. Solving for x, we get 5x = y + 3, which means x = (y + 3)/5. For any real number y, (y + 3)/5 is also a well-defined real number. So, for every y in the codomain, there exists an x in the domain. Therefore, the function is surjective. Final answer: The function f(x) = 5x - 3 is both injective and surjective, hence it is a bijective function.
• Q: Is the function f: N → N (natural numbers to natural numbers) defined by f(x) = x² one-one and/or onto? A: Step 1: Check Injectivity (One-one). Let x₁, x₂ ∈ N such that f(x₁) = f(x₂). This implies x₁² = x₂². Since x₁ and x₂ are natural numbers (and thus positive), taking the square root gives x₁ = x₂. (We don't have the ± issue as we do in real numbers). Therefore, the function is injective. Step 2: Check Surjectivity (Onto). Let y be an element in the codomain N. We need to see if there is an x ∈ N such that f(x) = y. This means x² = y, or x = √y. Let's pick an element from the codomain N, for example, y = 3. Then x = √3, which is not a natural number. Since not all elements in the codomain N (like 2, 3, 5, 6, 7...) have a pre-image in the domain N, the function is not surjective. Final answer: The function f(x) = x² from N to N is one-one (injective) but not onto (surjective).

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