**Definition:** A **relation R from set A to set B** is an arbitrary subset of the Cartesian product A × B.
**Key Points:**
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**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.
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**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:**
**Examples:**
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**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:**
**Examples:**
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**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:**
**Examples:**
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**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:**
**Example:** R = {(a,b) : 2 divides (a-b)} in ℤ
*Proof of equivalence:*
**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:*
**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}.
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**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:**
**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. ✓
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**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:**
**Examples:**
**Worked Example:** f : ℕ → ℕ defined by f(1) = f(2) = 1 and f(x) = x-1 for x > 2
For any y ∈ ℕ:
So every natural number is the image of some natural number. Therefore f is onto. ✓
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**Definition 7:** A function f : X → Y is **bijective (one-one and onto)** if it is both one-one AND onto.
**Properties:**
**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.
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**Important Theorem:** For a finite set X:
This means for finite sets, one-one ⟺ onto ⟺ bijective.
**Why Different for Infinite Sets:**
**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.
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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
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For a relation R in set A:
For a function f : X → Y:
Q1. Let A = {1, 2, 3}. Which of the following is the empty relation on 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?
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?
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?
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.
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?
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:
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?
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?
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.
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.
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.
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.
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