Relations and Functions

Introduction to Relations and its Types

Relations and Functions: Page 1 of 5

Introduction to Relations and its Types

Welcome to the fascinating world of Relations and Functions! This chapter builds upon what you learned in Class XI and lays the mathematical groundwork for calculus and other advanced topics. Think of it as learning the grammar of mathematics; understanding how different mathematical objects can be related to each other is fundamental.

So, why does this matter? In the real world, we are surrounded by relationships: a student is enrolled in a course, a city is the capital of a country, a number is greater than another number. In mathematics, we formalize these connections using the concept of relations. By classifying these relations into specific types—reflexive, symmetric, and transitive—we can uncover deep structural properties. This classification isn't just an academic exercise; it's the key to understanding concepts like equivalence classes, which are used in computer science, cryptography, and modern algebra.

In your CBSE board exams, the unit "Relations and Functions" carries significant weightage, typically around 8-10 marks. Questions testing the properties of relations are a staple, often appearing as 2, 3, or even 5-mark questions. Mastering this foundational topic will set you up for success in the entire calculus portion of your syllabus.

{{FORMULA: expr=(a, a) ∈ R; (a, b) ∈ R → (b, a) ∈ R; (a, b) ∈ R and (b, c) ∈ R → (a, c) ∈ R | symbols=R:Equivalence Relation, a,b,c:elements of Set A, →:implies}}

Definitions & Key Concepts

Let's start with a formal recap and introduce the new types of relations you'll be working with. A relation R from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B. If B = A, we say R is a relation on set A.

{{KEY: type=concept | title=The Golden Trio: RST | text=An Equivalence Relation is a special type of relationship that groups similar elements together. It must satisfy all three conditions: Reflexive, Symmetric, and Transitive. A simple mnemonic is RST. If a relation fails even one of these tests, it is not an equivalence relation.}}

Theorem Proof: Intersection of Equivalence Relations

Let's prove a fundamental property. This type of proof is excellent for building logical reasoning skills.

Theorem: If R and S are two equivalence relations on a set A, then their intersection R ∩ S is also an equivalence relation on A.

Proof:

1. Recall the Goal: To prove R ∩ S is an equivalence relation, we must show it is reflexive, symmetric, and transitive.
2. Check for Reflexivity:
   • Let a be an arbitrary element of A.
   • Since R is an equivalence relation, it is reflexive. Therefore, (a, a) ∈ R.
   • Similarly, since S is an equivalence relation, it is reflexive. Therefore, (a, a) ∈ S.
   • By the definition of intersection, if (a, a) is in both R and S, then (a, a) ∈ (R ∩ S).
   • Since this holds for every a ∈ A, R ∩ S is reflexive.
3. Check for Symmetry:
   • Let (a, b) ∈ (R ∩ S).
   • By definition of intersection, this means (a, b) ∈ R and (a, b) ∈ S.
   • Since R is symmetric, (a, b) ∈ R implies (b, a) ∈ R.
   • Since S is symmetric, (a, b) ∈ S implies (b, a) ∈ S.
   • Now, since (b, a) is in both R and S, we have (b, a) ∈ (R ∩ S).
   • Thus, (a, b) ∈ (R ∩ S) → (b, a) ∈ (R ∩ S). So, R ∩ S is symmetric.
4. Check for Transitivity:
   • Let (a, b) ∈ (R ∩ S) and (b, c) ∈ (R ∩ S).
   • This implies (a, b) ∈ R, (b, c) ∈ R, (a, b) ∈ S, and (b, c) ∈ S.
   • Since R is transitive, (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.
   • Since S is transitive, (a, b) ∈ S and (b, c) ∈ S implies (a, c) ∈ S.
   • Because (a, c) is in both R and S, we conclude that (a, c) ∈ (R ∩ S).
   • Thus, R ∩ S is transitive.
5. Conclusion: Since R ∩ S is reflexive, symmetric, and transitive, it is an equivalence relation.

Solved Examples

Let's solidify our understanding with examples ranging from simple checks to more complex, exam-style problems.

Example 1: Easy (Property Checking)

Given: A relation R on the set A = {1, 2, 3, 4} defined by R = {(1, 1), (2, 2), (4, 4), (1, 2), (2, 1), (2, 4)}.

To Find: Whether R is reflexive, symmetric, or transitive.

Approach: Check the definition of each property one by one against the elements of R.

Step-by-step Solution:

1. Reflexivity:

   • The set is A = {1, 2, 3, 4}. For R to be reflexive, it must contain (1, 1), (2, 2), (3, 3), and (4, 4).
   • We see (1, 1), (2, 2), and (4, 4) are in R.
   • However, (3, 3) ∉ R.
   • Therefore, R is not reflexive.

2. Symmetry:

   • We check each pair (a, b) and look for (b, a).
   • (1, 2) ∈ R. Is (2, 1) ∈ R? Yes.
   • (2, 4) ∈ R. Is (4, 2) ∈ R? No.
   • Since we found a counter-example, R is not symmetric.

3. Transitivity:

   • We look for chains (a, b) ∈ R and (b, c) ∈ R.
   • We have (1, 2) ∈ R and (2, 4) ∈ R.
   • For transitivity, we need (1, 4) to be in R.
   • Looking at the set R, (1, 4) ∉ R.
   • Therefore, R is not transitive.

Final Answer:

The relation R is not reflexive, not symmetric, and not transitive.

Example 2: Medium (Equivalence Relation on Integers)

Given: A relation R on the set of integers Z defined as R = {(a, b) : a - b is an even integer}.

To Find: Prove that R is an equivalence relation.

Approach: Systematically prove that the relation satisfies the three properties: reflexivity, symmetry, and transitivity.

Step-by-step Solution:

1. Reflexivity:

   • Let a ∈ Z.
   • Consider a - a = 0.
   • Since 0 is an even integer (0 = 2 × 0), (a, a) ∈ R for all a ∈ Z.
   • Thus, R is reflexive.

2. Symmetry:

   • Assume (a, b) ∈ R. This means a - b is an even integer.
   • So, a - b = 2k for some integer k.
   • Now consider b - a. We can write b - a = -(a - b) = -(2k) = 2(-k).
   • Since -k is also an integer, b - a is an even integer.
   • Therefore, (b, a) ∈ R.
   • Thus, R is symmetric.

3. Transitivity:

   • Assume (a, b) ∈ R and (b, c) ∈ R.
   • This means a - b is even, so a - b = 2k₁ for some integer k₁.
   • And b - c is even, so b - c = 2k₂ for some integer k₂.
   • We need to check a - c. Let's add the two equations:
   • (a - b) + (b - c) = 2k₁ + 2k₂
   • a - c = 2(k₁ + k₂)
   • Since k₁ + k₂ is an integer, a - c is an even integer.
   • Therefore, (a, c) ∈ R.
   • Thus, R is transitive.

Final Answer:

Since R is reflexive, symmetric, and transitive, it is an equivalence relation on the set of integers Z.

{{VISUAL: diagram: An arrow diagram on a set {a, b, c} showing a relation. An arrow from a to b represents (a, b). For reflexivity, there's a loop on each element (a to a). For symmetry, if there's an arrow from a to b, there's a return arrow from b to a. For transitivity, if there are arrows from a to b and b to c, there's a direct arrow from a to c.}}

Example 3: Hard (Divisibility Relation)

Given: A relation R on the set of natural numbers N where R = {(x, y) : x divides y}.

To Find: Check if R is reflexive, symmetric, and transitive.

Approach: Use the definition of divisibility (x divides y means y = kx for some integer k) to test the three properties.

Step-by-step Solution:

1. Reflexivity:

   • Let x ∈ N.
   • We need to check if (x, x) ∈ R, which means "does x divide x?".
   • Yes, any non-zero number divides itself (x = 1 × x).
   • So, (x, x) ∈ R for all x ∈ N.
   • Thus, R is reflexive.

2. Symmetry:

   • Assume (x, y) ∈ R. This means x divides y. So, y = kx for some k ∈ N.
   • We need to check if (y, x) ∈ R is always true, i.e., does y divide x?
   • Let's take a counter-example. Let x = 2 and y = 4.
   • (2, 4) ∈ R because 2 divides 4.
   • But does 4 divide 2? No.
   • So, (4, 2) ∉ R.
   • Since we found a case where (x, y) ∈ R but (y, x) ∉ R, the relation R is not symmetric.

3. Transitivity:

   • Assume (x, y) ∈ R and (y, z) ∈ R.
   • This means x divides y (y = k₁x) and y divides z (z = k₂y) for some k₁, k₂ ∈ N.
   • We need to check if x divides z.
   • Substitute the first equation into the second: z = k₂(k₁x) = (k₁k₂)x.
   • Since k₁ and k₂ are natural numbers, their product k₁k₂ is also a natural number.
   • Let k = k₁k₂. Then z = kx, which means x divides z.
   • So, (x, z) ∈ R.
   • Thus, R is transitive.

Final Answer:

The relation R is reflexive and transitive, but not symmetric.

Example 4: JEE-Tricky (Relation on a Cartesian Product)

Given: A relation R on the set A = N × N (pairs of natural numbers) is defined by (a, b) R (c, d) if and only if a + d = b + c.

To Find: Show that R is an equivalence relation.

Approach: The elements are ordered pairs. Apply the RST definitions carefully to these pairs. Rearranging the condition to a - b = c - d can simplify the proof.

Step-by-step Solution:

The condition a + d = b + c is equivalent to a - b = c - d. Let's use this form.

1. Reflexivity:

   • Let (a, b) ∈ N × N. We need to check if (a, b) R (a, b).
   • Using the condition, we check if a - b = a - b.
   • This is always true.
   • So, (a, b) R (a, b) for all (a, b) ∈ N × N. R is reflexive.

2. Symmetry:

   • Assume (a, b) R (c, d). This means a - b = c - d.
   • We need to check if (c, d) R (a, b), which means checking if c - d = a - b.
   • Since a - b = c - d implies c - d = a - b, the condition holds.
   • Thus, R is symmetric.

3. Transitivity:

   • Assume (a, b) R (c, d) and (c, d) R (e, f).
   • From the first relation, we have a - b = c - d ......(i)
   • From the second relation, we have c - d = e - f ......(ii)
   • From (i) and (ii), we can equate them: a - b = e - f.
   • This is exactly the condition for (a, b) R (e, f).
   • Thus, R is transitive.

Final Answer:

Since the relation R on N × N is reflexive, symmetric, and transitive, it is an equivalence relation.

Example 5: NEET-Style (Conceptual/Combinatorial)

Given: A set A = {a, b, c}.

To Find: What is the total number of reflexive relations possible on set A?

Approach: Think about the structure of the Cartesian product A × A and which elements must be included for a relation to be reflexive. Then, consider the choices for the remaining elements.

Step-by-step Solution:

1. Total possible pairs: The set A has n = 3 elements. The Cartesian product A × A will have n × n = 3 × 3 = 9 ordered pairs. A × A = {(a, a), (a, b), (a, c), (b, a), (b, b), (b, c), (c, a), (c, b), (c, c)}

2. Condition for Reflexivity: For any relation R on A to be reflexive, it must contain all pairs of the form (x, x). These are the diagonal elements. In our case, the pairs (a, a), (b, b), and (c, c) MUST be in R. There is no choice for these 3 elements; they have to be included.

3. Choices for other pairs: The remaining pairs are the non-diagonal elements. There are n² - n = 9 - 3 = 6 such pairs. These pairs are: {(a, b), (a, c), (b, a), (b, c), (c, a), (c, b)}.

4. Counting the possibilities: For each of these 6 non-diagonal pairs, we have two choices: either include it in the relation R or not include it.

   • For (a, b): 2 choices (in or out)
   • For (a, c): 2 choices (in or out)
   • ...and so on for all 6 pairs.

5. Total Number of Relations: The total number of ways to form a reflexive relation is the product of the number of choices for each non-diagonal element. Total reflexive relations = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶. 2⁶ = 64.

   The general formula for a set with n elements is 2^(n² - n).

Final Answer:

The total number of reflexive relations possible on set A is 64.

Tips & Tricks (Shortcuts)

Common Mistakes / Sign-Errors

Concept-Trap Questions

1. Question: Consider the empty relation R = {} on a non-empty set A = {1, 2}. Is this relation reflexive, symmetric, or transitive?

   💡 Answer & Trap: The relation is not reflexive but it is symmetric and transitive. Trap: Students often think an empty relation has no properties. Reflexive: It fails because for A={1, 2}, (1, 1) and (2, 2) must be in R, but R is empty. Symmetric: The condition is "if (a, b) ∈ R, then (b, a) ∈ R". Since there is no (a, b) in R to begin with, the "if" part is false, making the whole statement vacuously true. Transitive: The condition is "if (a, b) ∈ R and (b, c) ∈ R, then...". Since we can't find such a pair, this is also vacuously true.

2. Question: A relation R on the set of all people is defined as R = {(a, b) : a and b have a common friend}. Is this relation transitive?

   💡 Answer & Trap: The relation is not transitive. Trap: It seems intuitive that it might be. Counter-example: Let person A and B be friends with X. So (A, B) ∈ R. Let person B and C be friends with Y. So (B, C) ∈ R. It is possible that A and C have no common friends at all (especially if X and Y are different people and don't know A or C). Therefore, (A, C) may not be in R.

3. Question: Let R be a relation on the set of lines L in a plane, defined by (L₁, L₂) ∈ R if line L₁ is perpendicular to line L₂. Check the properties of R.

   💡 Answer & Trap: The relation is symmetric, but not reflexive and not transitive. Trap: Students might confuse this with the "parallel lines" relation, which is an equivalence relation. Reflexive: A line L₁ cannot be perpendicular to itself. So (L₁, L₁) ∉ R. Not reflexive. Symmetric: If L₁ is perpendicular to L₂, then L₂ is perpendicular to L₁. So (L₁, L₂) ∈ R → (L₂, L₁) ∈ R. It is symmetric. Transitive: If L₁ is perpendicular to L₂, and L₂ is perpendicular to L₃, then L₁ is parallel to L₃ (or they are the same line). L₁ is not perpendicular to L₃. So (L₁, L₂) ∈ R and (L₂, L₃) ∈ R does not imply (L₁, L₃) ∈ R. Not transitive.

Mini Cheatsheet

Equivalence Relations

Equivalence Relations

Concept Introduction

Have you ever noticed how we group things in daily life? All triangles with the same angles are "similar". All people born in the same year belong to the same age group. All integers that leave the same remainder when divided by 3 (like 1, 4, 7, 10...) behave similarly in some contexts. This act of grouping based on a shared property is the core idea behind an equivalence relation.

An equivalence relation is a special type of relation that formalizes the idea of "sameness" or "equivalence". It's a relation that is reflexive (everything is related to itself), symmetric (if A is related to B, then B is related to A), and transitive (if A is related to B and B is related to C, then A is related to C). Think of the "=" sign; it perfectly embodies these properties.

In your CBSE board exams, questions on proving a given relation to be an equivalence relation are very common and often carry 3 to 5 marks. Mastering this concept is crucial as it builds the foundation for understanding partitions of sets and quotient structures, which are fundamental in higher mathematics.

{{FORMULA: expr=R is an Equivalence Relation ⇔ R is Reflexive, Symmetric, and Transitive | symbols=R:Relation}}

Definitions & Key Properties

To formally define an equivalence relation, we must first understand the three properties it must satisfy. Let R be a relation on a non-empty set A.

Theorem Proof: Intersection of Two Equivalence Relations

A very important property tested in exams is that the "equivalence" nature is preserved under intersection. Let's prove it.

Theorem: If R and S are two equivalence relations on a set A, then their intersection, R ∩ S, is also an equivalence relation on A.

Proof:

1. State the Goal: We are given that R and S are equivalence relations on set A. This means both R and S are reflexive, symmetric, and transitive. We need to prove that the relation T = R ∩ S is also reflexive, symmetric, and transitive.

2. Check for Reflexivity:

   • Let a be any arbitrary element of set A (a ∈ A).
   • Since R is an equivalence relation, it is reflexive. Therefore, (a, a) ∈ R.
   • Since S is an equivalence relation, it is also reflexive. Therefore, (a, a) ∈ S.
   • By the definition of intersection, if an element is in R and in S, it must be in their intersection R ∩ S.
   • Thus, (a, a) ∈ R ∩ S for all a ∈ A.
   • Conclusion: R ∩ S is reflexive.

3. Check for Symmetry:

   • Let (a, b) be an arbitrary element of R ∩ S. This means (a, b) ∈ R and (a, b) ∈ S.
   • Since R is symmetric, if (a, b) ∈ R, then (b, a) ∈ R.
   • Since S is symmetric, if (a, b) ∈ S, then (b, a) ∈ S.
   • Now we have (b, a) ∈ R and (b, a) ∈ S.
   • By the definition of intersection, this implies (b, a) ∈ R ∩ S.
   • Conclusion: R ∩ S is symmetric.

4. Check for Transitivity:

   • Let (a, b) ∈ R ∩ S and (b, c) ∈ R ∩ S.
   • From (a, b) ∈ R ∩ S, we get (a, b) ∈ R and (a, b) ∈ S.
   • From (b, c) ∈ R ∩ S, we get (b, c) ∈ R and (b, c) ∈ S.
   • Since R is transitive, (a, b) ∈ R and (b, c) ∈ R implies (a, c) ∈ R.
   • Since S is transitive, (a, b) ∈ S and (b, c) ∈ S implies (a, c) ∈ S.
   • Now we have (a, c) ∈ R and (a, c) ∈ S.
   • By the definition of intersection, this implies (a, c) ∈ R ∩ S.
   • Conclusion: R ∩ S is transitive.

5. Final Conclusion: Since R ∩ S has been proven to be reflexive, symmetric, and transitive, it is an equivalence relation.

Solved Examples

Here are some graded examples, from basic to advanced, to solidify your understanding.

Example 1: Easy (Set of Integers)

Given: Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}. To Find: Determine if R is an equivalence relation on A. Approach: Check the three properties (Reflexive, Symmetric, Transitive) one by one.

Solution:

1. Reflexivity:

   • The set is A = {1, 2, 3}. We need to check if (1, 1), (2, 2), and (3, 3) are in R.
   • Looking at the relation, R contains (1, 1), (2, 2), and (3, 3).
   • So, R is reflexive.

2. Symmetry:

   • We check every pair (a, b) in R for its reverse (b, a).
   • (1, 1), (2, 2), (3, 3) are their own reverses.
   • For (1, 2) ∈ R, we check for (2, 1). Yes, (2, 1) ∈ R.
   • For (2, 1) ∈ R, we check for (1, 2). Yes, (1, 2) ∈ R.
   • So, R is symmetric.

3. Transitivity:

   • We check for pairs (a, b) and (b, c) in R.
   • Consider (1, 2) ∈ R and (2, 1) ∈ R. Here, a=1, b=2, c=1.
   • We must check if (a, c) = (1, 1) is in R.
   • Yes, (1, 1) ∈ R.
   • Consider (2, 1) ∈ R and (1, 2) ∈ R. Here, a=2, b=1, c=2.
   • We must check if (a, c) = (2, 2) is in R.
   • Yes, (2, 2) ∈ R.
   • After checking all such combinations, we find the condition holds.
   • So, R is transitive.

Final Answer: <br>

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

Example 2: Medium (Relation on Z)

Given: A relation R on the set of integers Z, defined as R = {(a, b) | 2 divides (a - b)}. To Find: Show that R is an equivalence relation. Approach: Prove reflexivity, symmetry, and transitivity using the properties of integers.

Solution:

1. Reflexivity:

   • Let a be any integer (a ∈ Z).
   • Consider a - a = 0.
   • Since 2 divides 0 (as 0 = 2 × 0), the condition is satisfied.
   • Therefore, (a, a) ∈ R for all a ∈ Z.
   • R is reflexive.

2. Symmetry:

   • Assume (a, b) ∈ R. By definition, this means 2 divides (a - b).
   • This implies a - b = 2k for some integer k.
   • Now consider b - a. We can write b - a = -(a - b) = -(2k) = 2(-k).
   • Since k is an integer, -k is also an integer.
   • This shows that 2 divides (b - a).
   • Therefore, (b, a) ∈ R.
   • R is symmetric.

3. Transitivity:

   • Assume (a, b) ∈ R and (b, c) ∈ R.
   • (a, b) ∈ R means 2 divides (a - b), so a - b = 2k₁ for some integer k₁.
   • (b, c) ∈ R means 2 divides (b - c), so b - c = 2k₂ for some integer k₂.
   • We need to check (a - c). Let's add the two equations:
   • (a - b) + (b - c) = 2k₁ + 2k₂
   • a - c = 2(k₁ + k₂)
   • Since k₁ and k₂ are integers, their sum (k₁ + k₂) is also an integer.
   • This shows that 2 divides (a - c).
   • Therefore, (a, c) ∈ R.
   • R is transitive.

Final Answer: <br>

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

Example 3: Hard (CBSE Favorite - Geometry)

Given: Let L be the set of all lines in a plane and R be the relation in L defined as R = {(L₁, L₂) | L₁ is parallel to L₂}. To Find: Show that R is an equivalence relation. Approach: Use the geometric properties of parallel lines to verify the three conditions.

Solution:

1. Reflexivity:

   • Let L₁ be any line in L.
   • We know that every line is parallel to itself.
   • Therefore, (L₁, L₁) ∈ R for all L₁ ∈ L.
   • R is reflexive.

2. Symmetry:

   • Assume (L₁, L₂) ∈ R. This means L₁ is parallel to L₂.
   • By the geometric definition of parallelism, if L₁ is parallel to L₂, then L₂ is also parallel to L₁.
   • Therefore, (L₂, L₁) ∈ R.
   • R is symmetric.

3. Transitivity:

   • Assume (L₁, L₂) ∈ R and (L₂, L₃) ∈ R.
   • This means L₁ is parallel to L₂, and L₂ is parallel to L₃.
   • From geometry, we know that two lines parallel to the same line are parallel to each other.
   • Therefore, L₁ is parallel to L₃.
   • This implies (L₁, L₃) ∈ R.
   • R is transitive.

Final Answer: <br>

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

Example 4: JEE-Tricky (Modular Arithmetic)

Given: A relation R on the set of integers Z defined by a R b if and only if a ≡ b (mod 5). (This means 5 divides a - b). To Find: Prove R is an equivalence relation and find the equivalence class of 2, i.e., [2]. Approach: This is identical in structure to Example 2, just with a different divisor. We'll prove the three properties and then find all integers x such that x - 2 is a multiple of 5.

Solution:

1. Reflexivity:

   • For any a ∈ Z, a - a = 0.
   • Since 5 divides 0 (0 = 5 × 0), we have a ≡ a (mod 5).
   • So, (a, a) ∈ R. R is reflexive.

2. Symmetry:

   • Assume (a, b) ∈ R, so a ≡ b (mod 5).
   • This means a - b = 5k for some integer k.
   • Then b - a = -5k = 5(-k).
   • Since -k is an integer, 5 divides b - a.
   • So, b ≡ a (mod 5), which means (b, a) ∈ R. R is symmetric.

3. Transitivity:

   • Assume (a, b) ∈ R and (b, c) ∈ R.
   • a - b = 5k₁ and b - c = 5k₂.
   • Adding them: (a - b) + (b - c) = 5k₁ + 5k₂
   • a - c = 5(k₁ + k₂)
   • So, 5 divides a - c, which means a ≡ c (mod 5).
   • Therefore, (a, c) ∈ R. R is transitive.

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

4. Finding the Equivalence Class [2]:
   • [2] = {x ∈ Z | (x, 2) ∈ R}
   • [2] = {x ∈ Z | x ≡ 2 (mod 5)}
   • [2] = {x ∈ Z | x - 2 = 5k \text{ for some integer } k}
   • [2] = {x ∈ Z | x = 5k + 2 \text{ for some integer } k}
   • Let's list some values by substituting k = ..., -2, -1, 0, 1, 2, ...
   • k=-2: x = 5(-2)+2 = -8
   • k=-1: x = 5(-1)+2 = -3
   • k=0: x = 5(0)+2 = 2
   • k=1: x = 5(1)+2 = 7
   • k=2: x = 5(2)+2 = 12

Final Answer: <br>

The relation R is an equivalence relation. The equivalence class of 2 is [2] = {..., -8, -3, 2, 7, 12, ...}.

Example 5: CBSE-style (Triangles)

Given: Let T be the set of all triangles in a plane. A relation R is defined on T as R = {(T₁, T₂) : T₁ is congruent to T₂}. To Find: Show that R is an equivalence relation. Approach: Use the properties of triangle congruence from Class 9 and 10 geometry.

Solution:

1. Reflexivity:

   • Let T₁ be any triangle in T.
   • Every triangle is congruent to itself (by SAS, SSS, or ASA congruence rule, where sides and angles are matched with themselves).
   • Therefore, (T₁, T₁) ∈ R.
   • R is reflexive.

2. Symmetry:

   • Assume (T₁, T₂) ∈ R. This means T₁ is congruent to T₂ (T₁ ≅ T₂).
   • The property of congruence is symmetric. If T₁ is congruent to T₂, then T₂ is congruent to T₁ (T₂ ≅ T₁).
   • Therefore, (T₂, T₁) ∈ R.
   • R is symmetric.

3. Transitivity:

   • Assume (T₁, T₂) ∈ R and (T₂, T₃) ∈ R.
   • This means T₁ is congruent to T₂ (T₁ ≅ T₂) and T₂ is congruent to T₃ (T₂ ≅ T₃).
   • From geometry, if one figure is congruent to a second, and the second is congruent to a third, then the first is congruent to the third.
   • Therefore, T₁ is congruent to T₃ (T₁ ≅ T₃).
   • This implies (T₁, T₃) ∈ R.
   • R is transitive.

Final Answer: <br>

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

{{KEY: type=concept | title=Equivalence Classes Partition a Set | text=An equivalence relation on a set A carves up or 'partitions' the set into a collection of non-empty, disjoint subsets called equivalence classes. The union of all these classes gives back the original set A. For example, the relation a ≡ b (mod 2) on integers partitions Z into two classes: [0] (the set of all even integers) and [1] (the set of all odd integers).}}

{{VISUAL: diagram: A large circle labeled 'Set A' is shown completely filled by three smaller, non-overlapping, irregularly shaped regions labeled '[a]', '[b]', and '[c]'. This illustrates that equivalence classes are disjoint and their union is the entire set.}}

Tips & Tricks (Shortcuts)

Common Mistakes / Sign-Errors

Concept-Trap Questions

Here are a few tricky questions designed to test the nuances of the definitions.

Trap 1: Let R be a relation on the set A = {1, 2, 3} defined as R = {(1, 1), (2, 2)}. Is R an equivalence relation?

💡 Answer & Trap: No. The trap is that it seems to satisfy all conditions for the elements it contains. However, it fails reflexivity. Reflexivity: For R to be reflexive on A, it must contain (a, a) for ALL a ∈ A. Here, (3, 3) is missing from R. Therefore, R is not reflexive and cannot be an equivalence relation.

Trap 2: Is the relation R = {(a, b) | a ≤ b} on the set of integers Z an equivalence relation?

💡 Answer & Trap: No. It is reflexive (a ≤ a) and transitive (if a ≤ b and b ≤ c, then a ≤ c). The trap is symmetry. Symmetry: Let's take a = 2 and b = 5. We have (2, 5) ∈ R because 2 ≤ 5. But (5, 2) ∉ R because 5 is not less than or equal to 2. Since it is not symmetric, it is not an equivalence relation.

Trap 3: If R and S are two equivalence relations on a set A, is their union, R ∪ S, necessarily an equivalence relation?

💡 Answer & Trap: No. The trap is assuming that properties that hold for intersection also hold for union. Reflexivity and symmetry will hold, but transitivity often fails. Counterexample: Let A = {1, 2, 3}. Let R = {(1,1), (2,2), (3,3), (1,2), (2,1)} (Equivalence relation with classes {1,2}, {3}). Let S = {(1,1), (2,2), (3,3), (2,3), (3,2)} (Equivalence relation with classes {1}, {2,3}). Then R ∪ S = {(1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2)}. Here, (1, 2) ∈ R ∪ S and (2, 3) ∈ R ∪ S. For transitivity, we need (1, 3) to be in R ∪ S. But it is not. Therefore, R ∪ S is not transitive and not an equivalence relation.

Mini Cheatsheet

Use this quick summary table for last-minute revision. Let R be a relation on a set A.

Types of Functions

Types of Functions

Welcome back! In our last session, we built the foundation of what relations and functions are. Think of a function as a well-behaved machine: you put in an input (from the domain), and it gives you exactly one output (in the co-domain).

Now, we'll dive deeper. Not all machines work the same way. Some give a unique output for every unique input. Others might give the same output for different inputs. Some machines can produce every possible output we care about, while others have a limited range. Understanding these behaviours is crucial for advanced topics like inverse functions and calculus. This classification helps us predict a function's properties and its suitability for various mathematical operations.

In your CBSE board exams, questions on identifying function types (one-one, onto) are very common, typically appearing as 2, 3, or even 4-mark questions. Mastering this section is key to scoring well in this chapter, which holds an overall weightage of around 8 marks.

{{FORMULA: expr=f(x₁) = f(x₂) ⇒ x₁ = x₂ | symbols=f: a function, x₁ and x₂: elements from the domain}}

Definitions & Key Concepts

Let's classify functions based on how their inputs and outputs are mapped. A function f is defined from a set A (the domain) to a set B (the co-domain), written as f: A → B.

Proof of a Key Theorem: Composition of Injective Functions

A very important property in functions is how they behave under composition. Let's prove a fundamental theorem that you'll often use.

Theorem: If f: A → B and g: B → C are both one-one functions, then their composition gof: A → C is also a one-one function.

Proof:

1. State the Goal: To prove that the function gof is one-one (injective), we must show that for any two elements x₁ and x₂ in the domain A, if gof(x₁) = gof(x₂), then it must imply that x₁ = x₂.

2. Assumption: Let's take two arbitrary elements x₁, x₂ ∈ A and assume that their images under gof are equal. gof(x₁) = gof(x₂)

3. Apply Definition of Composition: The composition gof(x) is defined as g(f(x)). Applying this to our assumption: g(f(x₁)) = g(f(x₂))

4. Use the Property of g: We are given that the function g: B → C is one-one. The definition of a one-one function states that if the outputs are equal, the inputs must be equal. Here, f(x₁) and f(x₂) are the inputs to g. Since g(f(x₁)) = g(f(x₂)), it must be that f(x₁) = f(x₂).

5. Use the Property of f: We are also given that the function f: A → B is one-one. We just established that f(x₁) = f(x₂). Applying the one-one property to f, this implies that x₁ = x₂.

6. Conclusion: We started with the assumption gof(x₁) = gof(x₂) and, through logical steps using the given properties of f and g, we have successfully shown that x₁ = x₂.

7. Final Statement: Therefore, by the definition of an injective function, the composite function gof: A → C is one-one.

This proof demonstrates how the properties of individual functions carry over to their compositions, a fundamental concept in higher mathematics.

Solved Examples

Let's solidify our understanding with some examples, ranging from simple to more complex.

Example 1 (Easy)

Given: A function f: R → R defined by f(x) = 3x - 5. To Find: Check if f is one-one and onto. Approach: Use the algebraic definitions for injectivity (f(x₁) = f(x₂) ⇒ x₁ = x₂) and surjectivity (express x in terms of y to check if a pre-image exists for every y).

Step-by-step Solution:

1. Check for One-one (Injectivity): Let x₁, x₂ be two arbitrary elements in the domain R. Assume f(x₁) = f(x₂). 3x₁ - 5 = 3x₂ - 5 3x₁ = 3x₂ x₁ = x₂ Since f(x₁) = f(x₂) ⇒ x₁ = x₂, the function f is one-one.

2. Check for Onto (Surjectivity): Let y be an arbitrary element in the co-domain R. We need to find if there exists an x in the domain R such that f(x) = y. y = 3x - 5 Let's solve for x: y + 5 = 3x x = (y + 5) / 3 For any real number y, the value (y + 5) / 3 is also a real number. This means for any y in the co-domain, we can find a corresponding x in the domain. Therefore, the function f is onto.

3. Conclusion: Since f is both one-one and onto, it is a bijective function.

Final Answer: The function f(x) = 3x - 5 is one-one and onto, hence it is bijective.

Example 2 (Medium)

Given: A function f: R → R defined by f(x) = x² + 1. To Find: Check if f is injective and surjective. Approach: Test injectivity with counter-examples and find the range of the function to check for surjectivity against the co-domain R.

Step-by-step Solution:

1. Check for One-one (Injectivity): Let's test with two different inputs. Consider x₁ = -2 and x₂ = 2. f(-2) = (-2)² + 1 = 4 + 1 = 5 f(2) = (2)² + 1 = 4 + 1 = 5 Here, x₁ ≠ x₂ but f(x₁) = f(x₂). Since different inputs lead to the same output, the function f is not one-one (it is many-one).

2. Check for Onto (Surjectivity): Let y be an arbitrary element in the co-domain R. f(x) = y means x² + 1 = y. x² = y - 1 x = ±√(y - 1) For x to be a real number (as the domain is R), the term inside the square root must be non-negative. y - 1 ≥ 0, which means y ≥ 1. This shows that f(x) can only produce values greater than or equal to 1. The range of f is [1, ∞). The co-domain is R (all real numbers). Since Range [1, ∞) is a proper subset of the Co-domain R, the function f is not onto. For example, there is no x ∈ R such that f(x) = 0.

Final Answer: The function f(x) = x² + 1 from R → R is neither one-one nor onto.

{{KEY: type=concept | title=Domain and Co-domain are Critical | text=The properties of a function depend heavily on its specified domain and co-domain. The function f(x) = x² is not one-one on R, but it is one-one on [0, ∞). It is not onto for co-domain R, but it is onto for co-domain [0, ∞). Always check the sets A and B in f: A → B!}}

Example 3 (Hard - CBSE Style)

Given: The function f: N → Y defined by f(x) = x², where Y is the set of all perfect squares in N. (Y = {1, 4, 9, 16, ...}). To Find: Show that f is a bijective function. Approach: Prove injectivity and surjectivity separately, paying close attention to the specified domain (N) and co-domain (Y).

Step-by-step Solution:

1. Check for One-one (Injectivity): Let x₁, x₂ ∈ N (the set of natural numbers). Assume f(x₁) = f(x₂). x₁² = x₂² x₁² - x₂² = 0 (x₁ - x₂)(x₁ + x₂) = 0 This gives two possibilities: x₁ - x₂ = 0 or x₁ + x₂ = 0. This means x₁ = x₂ or x₁ = -x₂. Since x₁ and x₂ belong to N (natural numbers), they are positive. Therefore, x₁ + x₂ cannot be zero. The only possibility is x₁ = x₂. Thus, f is one-one.

2. Check for Onto (Surjectivity): The co-domain is Y, the set of all perfect squares ({1, 4, 9, ...}). Let y be an arbitrary element in the co-domain Y. By the definition of set Y, y is a perfect square. This means there must exist a natural number k such that y = k². We need to find an x in the domain N such that f(x) = y. f(x) = x² = y = k² x = k Since k is a natural number, this x is in the domain N. So for every y in Y, there exists an x = √y in N which is its pre-image. Therefore, f is onto.

3. Conclusion: Since the function f is both one-one and onto, it is bijective.

Final Answer: The function f: N → Y where f(x) = x² and Y is the set of perfect squares is proven to be one-one and onto, hence it is bijective.

Example 4 (JEE-Tricky)

Given: A function f: R - {2} → R - {1} defined by f(x) = (x-1)/(x-2). To Find: Determine if f is bijective. Approach: Prove injectivity algebraically. For surjectivity, express x in terms of y and check if a valid x exists for every y in the co-domain.

Step-by-step Solution:

1. Check for One-one (Injectivity): Let x₁, x₂ ∈ R - {2}. Assume f(x₁) = f(x₂). (x₁-1)/(x₁-2) = (x₂-1)/(x₂-2) (x₁-1)(x₂-2) = (x₂-1)(x₁-2) x₁x₂ - 2x₁ - x₂ + 2 = x₁x₂ - 2x₂ - x₁ + 2 -2x₁ - x₂ = -2x₂ - x₁ -2x₁ + x₁ = -2x₂ + x₂ -x₁ = -x₂ x₁ = x₂ Hence, f is one-one.

2. Check for Onto (Surjectivity): Let y be an arbitrary element in the co-domain R - {1}. We need to find x such that f(x) = y. y = (x-1)/(x-2) y(x-2) = x-1 yx - 2y = x - 1 yx - x = 2y - 1 x(y - 1) = 2y - 1 x = (2y - 1) / (y - 1) Now, does this x exist for every y in the co-domain R - {1}? Yes, this expression is a valid real number for all y except y = 1. Since the co-domain is explicitly R - {1}, a valid x can be found for every y. Also, we must ensure the x we find is in the domain R - {2}. Can x ever be 2? If x = 2, then 2 = (2y - 1) / (y - 1) ⇒ 2y - 2 = 2y - 1 ⇒ -2 = -1, which is impossible. So x can never be 2. Thus, for every y ∈ R - {1}, there exists an x ∈ R - {2}. f is onto.

Final Answer: The function f(x) = (x-1)/(x-2) is both one-one and onto, therefore it is bijective.

Example 5 (NEET-Style / Conceptual MCQ)

Given: The greatest integer function f: R → Z defined by f(x) = [x]. To Find: Classify the function (one-one, onto, bijective, neither). Approach: Use counter-examples to quickly check the definitions of injectivity and surjectivity.

Step-by-step Solution:

1. Check for One-one (Injectivity): The greatest integer function [x] gives the greatest integer less than or equal to x. Let's take x₁ = 2.1 and x₂ = 2.8. f(2.1) = [2.1] = 2 f(2.8) = [2.8] = 2 Here, x₁ ≠ x₂ but f(x₁) = f(x₂). Therefore, the function is not one-one (it is many-one).

2. Check for Onto (Surjectivity): The domain is R (all real numbers). The co-domain is Z (all integers). The range of the function f(x) = [x] is the set of all integers. For any integer z, we can always find a real number x such that [x] = z. For example, we can simply choose x = z. f(z) = [z] = z. So, every integer z in the co-domain has a pre-image (in fact, infinitely many pre-images, like the entire interval [z, z+1)). Since Range (Z) = Co-domain (Z), the function is onto.

Final Answer: The function f(x) = [x] from R → Z is onto but not one-one.

{{VISUAL: diagram: A graph of a parabola y=x² is shown. A horizontal line y=4 intersects the graph at two points, x=-2 and x=2. A caption reads: "Horizontal Line Test: A line intersects the graph more than once, so the function is many-to-one." Another graph of a line y=2x is shown. Any horizontal line intersects it only once, with a caption: "Horizontal line intersects at most once, so the function is one-to-one."}}

Tips & Tricks (Shortcuts)

Common Mistakes / Sign-Errors

Concept-Trap Questions

These questions look simple but are designed to catch common misunderstandings.

1. A function f: {a, b, c} → {1, 2, 3} is defined by the mapping f = {(a, 2), (b, 3), (c, 2)}. Is this function one-one? Is it onto?

   💡 Answer & Trap: The function is not one-one (it is many-one). The trap is seeing three inputs and three outputs and assuming a perfect match. Here, both a and c map to the same output 2. The function is also not onto. The trap is again the number of elements. The element 1 in the co-domain {1, 2, 3} has no pre-image in the domain {a, b, c}. The range is {2, 3}, which is a subset of the co-domain.

2. Is the function f: R → R defined by f(x) = x³ - 6x² + 11x - 6 one-one?

   💡 Answer & Trap: No, it is not one-one. The trap is thinking that all cubic (odd-degree) polynomials are strictly monotonic and therefore one-one, like f(x)=x³. However, this specific cubic can be factored as f(x) = (x-1)(x-2)(x-3). We can see that f(1) = 0, f(2) = 0, and f(3) = 0. Since multiple inputs (1, 2, 3) give the same output (0), it is a many-one function. Always be careful with polynomials that aren't simple monomials.

3. Let f: N → Z be defined by f(n) = n/2 if n is even, and f(n) = -(n-1)/2 if n is odd. Is this function bijective?

   💡 Answer & Trap: Yes, this function is bijective. The trap lies in the piecewise definition which can be confusing. One-one: If n is even, f(n) is a positive integer or zero (1, 2, 3, ...). If n is odd, f(n) is a negative integer (-0, -1, -2, ...). Since the outputs for even and odd inputs fall into disjoint sets of integers, there's no overlap. Within each case, the function is clearly one-one. So, f is one-one. Onto: Let's check if we can get any integer y. If y ≥ 1, we need n/2 = y, so n = 2y (which is an even natural number). If y ≤ 0, we need -(n-1)/2 = y, so (n-1)/2 = -y, n-1 = -2y, n = 1-2y (which is an odd natural number for y ≤ 0). Every integer y in the co-domain Z has a unique pre-image in N. So it is onto. Since it is both one-one and onto, it is bijective.

Mini Cheatsheet