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Relations and Functions

NCERT Class 12 · Mathematics Based on NCERT Class 12 Mathematics textbook · Free CBSE study kit

Chapter Notes

RELATIONS AND FUNCTIONS — COMPREHENSIVE CHAPTER NOTES

INTRODUCTION TO RELATIONS

**Definition:** A **relation R from set A to set B** is an arbitrary subset of the Cartesian product A × B.

  • If (a, b) ∈ R, we say "a is related to b under R" and write **a R b**
  • A relation in a set A is a subset of A × A
  • Two extreme relations exist: empty relation φ and universal relation A × A
  • Functions are special kinds of relations (covered in section 1.3)
  • **Key Points:**

  • The mathematical definition of relation abstracts from real-world recognisable connections
  • Relations provide the foundational framework for understanding functions
  • Every function is a relation, but not every relation is a function
  • ---

    EMPTY AND UNIVERSAL RELATIONS

    **Definition 1: Empty Relation**

    A relation R in set A is **empty** if R = φ ⊂ A × A, meaning no element of A is related to any element of A.

    **Definition 2: Universal Relation**

    A relation R in set A is **universal** if R = A × A, meaning every element of A is related to every element of A.

    Both are called **trivial relations**.

    **Example:** Let A be the set of all students in a boys school.

  • R = {(a,b) : a is sister of b} is empty (boys school has no sisters among students)
  • R′ = {(a,b) : height difference < 3 meters} is universal (any two students differ by less than 3 meters)
  • ---

    REFLEXIVE RELATIONS

    **Definition:** A relation R in a set A is **reflexive** if **(a, a) ∈ R for every a ∈ A**.

    In other words, every element must be related to itself.

    **Testing for Reflexivity:**

  • Check if (a,a) is in the relation for ALL elements a ∈ A
  • If even one element fails this condition, R is not reflexive
  • **Examples:**

  • R = {(x,y) : x ≤ y} in ℝ is reflexive because x ≤ x for all x ∈ ℝ
  • R = {(x,y) : x = y} in any set is reflexive
  • R = {(x,y) : x > y} in ℤ is NOT reflexive because no integer is greater than itself
  • ---

    SYMMETRIC RELATIONS

    **Definition:** A relation R in a set A is **symmetric** if **(a₁, a₂) ∈ R implies (a₂, a₁) ∈ R for all a₁, a₂ ∈ A**.

    If one element is related to another, the reverse relation must also hold.

    **Testing for Symmetry:**

  • For every ordered pair (a,b) in R, check if (b,a) is also in R
  • If one pair lacks its reverse, R is not symmetric
  • **Examples:**

  • R = {(x,y) : x = y} is symmetric because if x = y then y = x
  • R = {(x,y) : x ≤ y} in ℝ is NOT symmetric (e.g., 2 ≤ 3 but 3 ≤ 2 is false)
  • R = {(L₁,L₂) : L₁ ⊥ L₂} (perpendicular lines) is symmetric (if L₁ ⊥ L₂ then L₂ ⊥ L₁)
  • ---

    TRANSITIVE RELATIONS

    **Definition:** A relation R in a set A is **transitive** if **(a₁, a₂) ∈ R and (a₂, a₃) ∈ R implies (a₁, a₃) ∈ R for all a₁, a₂, a₃ ∈ A**.

    The relation "chains" through intermediate elements.

    **Testing for Transitivity:**

  • Whenever you have a path a₁ → a₂ → a₃, check if a₁ → a₃ exists
  • If any path lacks its endpoint connection, R is not transitive
  • **Examples:**

  • R = {(x,y) : x ≤ y} in ℝ is transitive (if x ≤ y and y ≤ z then x ≤ z)
  • R = {(x,y) : x = y} is transitive (if x = y and y = z then x = z)
  • R = {(L₁,L₂) : L₁ ⊥ L₂} is NOT transitive (if L₁ ⊥ L₂ and L₂ ⊥ L₃, then L₁ ∥ L₃, not L₁ ⊥ L₃)
  • ---

    EQUIVALENCE RELATIONS

    **Definition 4:** A relation R in a set A is an **equivalence relation** if it is **simultaneously reflexive, symmetric, and transitive**.

    **Fundamental Property:** An equivalence relation partitions a set into disjoint **equivalence classes**.

    **Equivalence Class:** For an element a ∈ A, the equivalence class [a] is the set of all elements related to a:

    **[a] = {x ∈ A : x R a}**

    **Properties of Equivalence Relations:**

  • All elements within an equivalence class are mutually related
  • No element from one class relates to elements of another class
  • All equivalence classes are pairwise disjoint
  • The union of all equivalence classes equals the original set A
  • **Example:** R = {(a,b) : 2 divides (a-b)} in ℤ

    *Proof of equivalence:*

  • **Reflexive:** 2 divides (a-a) = 0 for all a ∈ ℤ ✓
  • **Symmetric:** If 2|(a-b), then 2|-(a-b) = (b-a), so 2|(b-a) ✓
  • **Transitive:** If 2|(a-b) and 2|(b-c), then (a-b) and (b-c) are both even, so (a-c) = (a-b) + (b-c) is even, thus 2|(a-c) ✓
  • **Equivalence Classes:** [0] = {all even integers}, [1] = {all odd integers}

    **Worked Example:** In set A = {1,2,3,4,5,6,7}, define R = {(a,b) : both a and b are odd or both are even}

    *Testing:*

  • Reflexive: (a,a) ∈ R since a and a have same parity ✓
  • Symmetric: If a,b have same parity, so do b,a ✓
  • Transitive: If a,b have same parity AND b,c have same parity, then a,c have same parity ✓
  • **Equivalence classes:** [1] = {1,3,5,7} (all odd), [2] = {2,4,6} (all even)

    No element of {1,3,5,7} is related to any element of {2,4,6}.

    ---

    ONE-ONE FUNCTIONS (INJECTIVE)

    **Definition 5:** A function f : X → Y is **one-one (or injective)** if the images of distinct elements of X under f are distinct.

    **Formally:** For all x₁, x₂ ∈ X: **f(x₁) = f(x₂) ⟹ x₁ = x₂**

    **Alternative Testing Method:** If f(x₁) = f(x₂), you must be able to conclude x₁ = x₂.

    **Graphical Check:** A function is one-one if every horizontal line intersects its graph at most once.

    **Examples:**

  • f : ℝ → ℝ, f(x) = 2x is one-one [if 2x₁ = 2x₂ then x₁ = x₂]
  • f : ℕ → ℕ, f(x) = 2x is one-one [if 2x₁ = 2x₂ then x₁ = x₂]
  • f : ℝ → ℝ, f(x) = x² is NOT one-one [f(-1) = f(1) = 1 but -1 ≠ 1]
  • **Worked Example:** f : ℝ* → ℝ*, f(x) = 1/x

    Assume f(x₁) = f(x₂). Then 1/x₁ = 1/x₂ ⟹ x₁ = x₂. Therefore f is one-one. ✓

    ---

    ONTO FUNCTIONS (SURJECTIVE)

    **Definition 6:** A function f : X → Y is **onto (or surjective)** if every element of Y is the image of at least one element of X.

    **Formally:** For every y ∈ Y, **there exists x ∈ X such that f(x) = y**

    **Key Remark:** f is onto if and only if **Range of f = Co-domain Y**

    **Testing for Onto:**

  • Take an arbitrary element y from the co-domain
  • Try to find at least one x in the domain such that f(x) = y
  • If successful for all y, then f is onto
  • **Examples:**

  • f : ℝ → ℝ, f(x) = 2x is onto [for any y ∈ ℝ, choose x = y/2, then f(x) = y]
  • f : ℕ → ℕ, f(x) = 2x is NOT onto [odd numbers like 1,3,5 are never reached]
  • f : ℝ → ℝ, f(x) = x² is NOT onto [negative numbers like -1 cannot be images since x² ≥ 0]
  • **Worked Example:** f : ℕ → ℕ defined by f(1) = f(2) = 1 and f(x) = x-1 for x > 2

    For any y ∈ ℕ:

  • If y = 1: f(1) = 1 ✓
  • If y > 1: f(y+1) = (y+1) - 1 = y ✓
  • So every natural number is the image of some natural number. Therefore f is onto. ✓

    ---

    BIJECTIVE FUNCTIONS

    **Definition 7:** A function f : X → Y is **bijective (one-one and onto)** if it is both one-one AND onto.

    **Properties:**

  • Every element of X maps to a unique element of Y (one-one)
  • Every element of Y is mapped to by some element of X (onto)
  • There is a perfect "pairing" between X and Y
  • **Worked Example:** f : ℝ → ℝ, f(x) = 2x

    *One-one:* f(x₁) = f(x₂) ⟹ 2x₁ = 2x₂ ⟹ x₁ = x₂ ✓

    *Onto:* For any y ∈ ℝ, choose x = y/2 ∈ ℝ. Then f(y/2) = 2(y/2) = y ✓

    Therefore f is bijective.

    ---

    FINITE SET PROPERTY

    **Important Theorem:** For a finite set X:

  • A one-one function f : X → X is necessarily **onto**
  • An onto function f : X → X is necessarily **one-one**
  • This means for finite sets, one-one ⟺ onto ⟺ bijective.

    **Why Different for Infinite Sets:**

  • f : ℕ → ℕ, f(x) = 2x is one-one but NOT onto (Example 8)
  • f : ℕ → ℕ with f(1)=f(2)=1, f(x)=x-1 (x>2) is onto but NOT one-one (Example 10)
  • **Proof Sketch for Finite Set:** If f : X → X (|X| = n) is one-one, then n distinct domain elements map to n distinct range elements. Since co-domain has only n elements, range must equal co-domain, so f is onto.

    ---

    COMMON MISTAKES TO AVOID

    1. **Confusing Reflexivity:** R is reflexive only if (a,a) ∈ R for EVERY a ∈ A, not just some

    2. **Symmetry Error:** Check both directions—if (a,b) ∈ R then (b,a) must be in R

    3. **Transitivity Chain:** Requires: if (a,b) and (b,c) both in R, then (a,c) must be in R

    4. **One-One vs Onto:** One-one means no two different inputs give same output; onto means every output is reached

    5. **Domain vs Co-domain:** Range (actual outputs) may differ from co-domain (permitted outputs); a function is onto only when these equal

    6. **Equivalence Class Partition:** Equivalence classes are pairwise disjoint AND their union is the entire set

    ---

    EXAM QUICK-CHECK CHECKLIST

    For a relation R in set A:

  • ✓ Reflexive? Test: (a,a) ∈ R for ALL a
  • ✓ Symmetric? Test: (a,b) ∈ R ⟹ (b,a) ∈ R
  • ✓ Transitive? Test: (a,b) ∈ R and (b,c) ∈ R ⟹ (a,c) ∈ R
  • ✓ Equivalence? If all three above properties hold
  • For a function f : X → Y:

  • ✓ One-one? Test: f(x₁) = f(x₂) ⟹ x₁ = x₂
  • ✓ Onto? Test: For every y in Y, ∃x in X such that f(x) = y
  • ✓ Bijective? Both one-one AND onto simultaneously
  • MCQs — 10 Questions with Answers

    Q1. Let A = {1, 2, 3}. Which of the following is the empty relation on A?

    • A. {(a, b) : a − b = 10} ✓
    • B. {(a, b) : a < b}
    • C. {(a, b) : |a − b| ≥ 0}
    • D. {(a, a) : a ∈ A}

    Answer: A — No pair (a, b) in A × A satisfies a − b = 10 since the maximum difference is 3 − 1 = 2, making the relation empty; all other options contain at least one pair.

    Q2. Which property does the relation R = {(a, b) ∈ Z : a = b} satisfy?

    • A. Only reflexive
    • B. Only symmetric
    • C. Reflexive and symmetric, but not transitive
    • D. Reflexive, symmetric, and transitive (equivalence relation) ✓

    Answer: D — Equality is reflexive (a = a), symmetric (a = b ⟹ b = a), and transitive (a = b and b = c ⟹ a = c), making it an equivalence relation.

    Q3. Let R be a relation in set {1, 2, 3, 4} defined as R = {(1,2), (2,1), (2,3), (3,2)}. Is R symmetric? Is R transitive?

    • A. Symmetric: Yes, Transitive: Yes
    • B. Symmetric: Yes, Transitive: No ✓
    • C. Symmetric: No, Transitive: Yes
    • D. Symmetric: No, Transitive: No

    Answer: B — R is symmetric because (1,2) ∈ R and (2,1) ∈ R, (2,3) ∈ R and (3,2) ∈ R; R is not transitive because (1,2) ∈ R and (2,3) ∈ R but (1,3) ∉ R.

    Q4. The relation R = {(a, b) ∈ Z : 2 divides (a + b)} on set Z. Which statement is NOT correct?

    • A. R is reflexive because 2 divides 2a for all a ∈ Z
    • B. R is symmetric because if 2 divides (a + b), then 2 divides (b + a)
    • C. R is transitive because if 2 divides (a + b) and 2 divides (b + c), then 2 divides (a + c) ✓
    • D. R is an equivalence relation

    Answer: C — R is not transitive: if a = 1, b = 1, then a + b = 2 (divisible by 2); if b = 1, c = 2, then b + c = 3 (not divisible by 2), showing transitivity fails.

    Q5. Assertion (A): The relation 'is parallel to' on the set of all lines in a plane is an equivalence relation. Reason (R): A relation is an equivalence relation if it is reflexive, symmetric, and transitive.

    • A. Both A and R are true, and R is the correct explanation of A
    • B. Both A and R are true, but R is NOT the correct explanation of A
    • C. A is false, but R is true ✓
    • D. Both A and R are false

    Answer: C — R is true (definition of equivalence relation), but A is false because 'parallel to' is not reflexive (a line is not parallel to itself by standard definition, it coincides with itself).

    Q6. Let T be the set of all triangles in a plane and R be 'is similar to' on T. Verify: R is reflexive, R is symmetric, and R is transitive. Which of the following is true?

    • A. Only reflexive
    • B. Reflexive and symmetric, but not transitive
    • C. Reflexive, symmetric, and transitive ✓
    • D. None of the above

    Answer: C — Reflexive: every triangle is similar to itself; symmetric: T₁ ~ T₂ ⟹ T₂ ~ T₁; transitive: T₁ ~ T₂ and T₂ ~ T₃ ⟹ T₁ ~ T₃; hence R is an equivalence relation.

    Q7. For the relation R in Z given by R = {(a, b) : 3 divides (a − b)}, the equivalence class of 2 is:

    • A. {..., −4, −1, 2, 5, 8, ...} ✓
    • B. {..., −3, 0, 3, 6, 9, ...}
    • C. {..., −2, 1, 4, 7, 10, ...}
    • D. {2}

    Answer: A — [2] contains all integers b such that 3 divides (b − 2), i.e., b ≡ 2 (mod 3), giving {..., −4, −1, 2, 5, 8, ...}.

    Q8. Which of the following relations on the set {1, 2, 3, 4} is NOT a function from the set to itself?

    • A. {(1, 2), (2, 3), (3, 4), (4, 1)}
    • B. {(1, 1), (2, 2), (3, 3), (4, 4)}
    • C. {(1, 2), (2, 1), (3, 4), (4, 3)}
    • D. {(1, 2), (2, 1), (2, 3), (3, 4), (4, 1)} ✓

    Answer: D — Option D fails the function property because the element 2 maps to both 1 and 3 (not single-valued); options A, B, C each assign exactly one output to each input.

    Q9. Consider R = {(a, b) ∈ N × N : a divides b}. Check each property: (i) reflexive? (ii) symmetric? (iii) transitive?

    • A. (i) Yes, (ii) Yes, (iii) Yes
    • B. (i) Yes, (ii) No, (iii) Yes ✓
    • C. (i) Yes, (ii) Yes, (iii) No
    • D. (i) No, (ii) Yes, (iii) Yes

    Answer: B — Reflexive: a | a ✓; symmetric: if a | b, does b | a? Only if a = b, so ✗; transitive: if a | b and b | c, then a | c ✓.

    Q10. HOTS: A relation R on set A = {1, 2, 3, 4, 5} is defined by R = {(a, b) : both a and b are odd}. Show that R is reflexive and symmetric. Is it transitive? If not, find a counterexample and explain why it fails.

    • A. R is reflexive, symmetric, and transitive; no counterexample needed
    • B. R is reflexive and symmetric but not transitive; (1,1), (1,3), (3,5) shows 1 R 3 and 3 R 5 but 1 ∉ R is odd alone
    • C. R is reflexive and symmetric but not transitive; (1,3) ∈ R and (3,5) ∈ R but (1,5) ∉ R (both 1 and 5 are odd, so transitive holds) ✓
    • D. R is not reflexive because 2 ∈ A and 2 is not odd

    Answer: C — Reflexive: (a, a) ∈ R only if a is odd (holds for odd a); symmetric: if both a, b are odd then both b, a are odd ✓; transitive: (1,3) ∈ R (both odd) and (3,5) ∈ R (both odd) imply (1,5) ∈ R (both odd) ✓, so actually R IS transitive.

    Flashcards

    What is a relation R from set A to set B mathematically?

    Any arbitrary subset of A × B; written as R ⊆ A × B, where (a, b) ∈ R means a is related to b.

    Define reflexive relation with an example.

    A relation R in set A is reflexive if (a, a) ∈ R for every a ∈ A; example: equality relation {(a, a) : a ∈ A}.

    What makes a relation symmetric? Give one example.

    A relation is symmetric if (a₁, a₂) ∈ R implies (a₂, a₁) ∈ R; example: congruence of triangles where T₁ ≅ T₂ implies T₂ ≅ T₁.

    Define transitive relation and give a counterexample.

    Transitive: (a₁, a₂) and (a₂, a₃) ∈ R implies (a₁, a₃) ∈ R; counterexample: perpendicularity of lines (L₁ ⊥ L₂ and L₂ ⊥ L₃ does not imply L₁ ⊥ L₃).

    What is an equivalence relation? Name all three required properties.

    A relation that is simultaneously reflexive, symmetric, and transitive; these three properties together define an equivalence relation.

    Show that 'congruence of triangles' is an equivalence relation.

    Reflexive: every triangle is congruent to itself; symmetric: T₁ ≅ T₂ ⟹ T₂ ≅ T₁; transitive: T₁ ≅ T₂ and T₂ ≅ T₃ ⟹ T₁ ≅ T₃.

    For R = {(a, b) ∈ Z : 2 divides (a − b)}, prove R is reflexive.

    For any a ∈ Z, 2 divides (a − a) = 0, so (a, a) ∈ R for all a; hence R is reflexive.

    What are equivalence classes? Explain using even and odd integers.

    Equivalence classes are disjoint subsets created by an equivalence relation; for integers: [0] = all even integers, [1] = all odd integers, partitioning Z completely.

    Identify which property the relation 'perpendicularity of lines' satisfies.

    Only symmetric: L₁ ⊥ L₂ ⟹ L₂ ⊥ L₁; not reflexive (no line ⊥ itself) and not transitive (L₁ ⊥ L₂ and L₂ ⊥ L₃ means L₁ ∥ L₃, not L₁ ⊥ L₃).

    What conditions must equivalence classes satisfy for any equivalence relation?

    All elements within each class are related; no element from one class relates to any element of another; classes are disjoint and their union equals the entire set.

    Important Board Questions

    Define an empty relation and a universal relation. Give one example of each on the set {1, 2, 3}. [2 marks]

    Empty relation has no pairs: state R = φ and verify no (a, b) satisfies the condition. Universal relation: R = A × A with a simple example like |a − b| ≥ 0 or a ∈ A, b ∈ A.

    Prove that the relation R = {(a, b) ∈ Z : 2 divides (a − b)} is an equivalence relation. Then identify the two equivalence classes and describe what integers they contain. [5 marks]

    For reflexivity, show 2 | (a − a); for symmetry, if 2 | (a − b) then 2 | (b − a); for transitivity, sum two even differences. Equivalence classes: [0] = all even integers, [1] = all odd integers; verify they partition Z.

    State the definition of reflexive, symmetric, and transitive relations. Using the relation 'perpendicularity of lines' in a plane, determine which properties it satisfies and which it does not. Justify each answer with a reason or counterexample. [6 marks]

    Define all three properties precisely. For perpendicularity: not reflexive (a line cannot be ⊥ to itself), symmetric (if L₁ ⊥ L₂ then L₂ ⊥ L₁), and not transitive (if L₁ ⊥ L₂ and L₂ ⊥ L₃, then L₁ ∥ L₃, not L₁ ⊥ L₃); use counterexample with three lines to show failure of transitivity.

    Next chapterInverse Trigonometric Functions →

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