AJAY PATHAK (MATH)

Q.3 (a) Question: Define a complete graph and find the order and size of the graph K 2025 . Definition: Complete Graph A complete graph is a simple graph in which every pair of distinct vertices is connected by exactly one edge . A complete graph with n vertices is denoted by K n . Order of the Graph: The order of a graph is the number of vertices in the graph. Order of K 2025 = 2025 Size of the Graph: The size of a graph is the number of edges. A complete graph with n vertices has: n(n − 1)/2 edges For K 2025 : = 2025 × 2024 / 2 = 2025 × 1012 = 2049300 edges Final Answer: Order of K 2025 = 2025 Size of K 2025 = 2049300 Q.3 (b) Question: Show that the proposition [(p → q) ∧ (q → r)] → (p → r) is a tautology . Method: Trut...

Q.1 (a) Prove that: (A ∩ B) ∪ (A − B) = A A ∪ (B − A) = A ∪ B Solution: (i) To prove: (A ∩ B) ∪ (A − B) = A Let x ∈ (A ∩ B) ∪ (A − B) . Then either x ∈ A ∩ B or x ∈ A − B . In both cases, x ∈ A . Hence, (A ∩ B) ∪ (A − B) ⊆ A . Now let x ∈ A . Then either x ∈ B or x ∉ B . If x ∈ B , then x ∈ A ∩ B . If x ∉ B , then x ∈ A − B . Hence, x ∈ (A ∩ B) ∪ (A − B) . Therefore, A ⊆ (A ∩ B) ∪ (A − B) . Thus, (A ∩ B) ∪ (A − B) = A. (ii) To prove: A ∪ (B − A) = A ∪ B Let x ∈ A ∪ (B − A) . Then x ∈ A or x ∈ B − A . In both cases, x ∈ A ∪ B . Hence, A ∪ (B − A) ⊆ A ∪ B . Conversely, let x ∈ A ∪ B . If x ∈ A , then x ∈ A ∪ (B − A) . If x ∈ B and x ∉ A , then x ∈ B − A . Hence, A ∪ B ⊆ A ∪ (B − A) . Therefore, A ∪ (B − A) = A ∪ B. ✔ Proved ...

Recurrence Relations A recurrence relation defines a sequence in terms of its previous terms. Such relations occur frequently in mathematics, computer science, and discrete structures. Step Rule 1 Write the characteristic equation for the given recurrence relation. 2 Solve the characteristic equation and find all roots. 3 The order of the recurrence = highest index − lowest index. Nature of Roots Complementary Solution a n (h) Real and distinct λ₁ ≠ λ₂ ≠ … ≠ λₖ a n = C₁λ₁ n + C₂λ₂ n + … + Cₖλₖ n Real and equal λ₁ = λ₂ = … = λₖ a n = (C₁ + C₂n + C₃n² + …) λ₁ n Complex conjugate roots λ = α ± iβ a n = r n [ C₁ cos(nθ) + C₂ sin(nθ) ] where r = √(α² + β²),   θ = tan −1 (β/α) f(n) Assumed Particular Solution a n (p) Constant (C) P₀ Polynomial a + bn + cn² + … P₀ + P₁n + P₂n² + … a·b n , b ≠ λ P₀·b n a·b n , b = λ (with multiplicity m) P₀·n m b n Total ...

Matrix Representation of a Relation in Discrete Mathematics In Discrete Mathematics, relations can be represented in different ways such as ordered pairs, digraphs, and matrices. Among these, the matrix representation of a relation is very useful for performing operations like union, intersection, complement, and composition of relations. Matrix Representation of a Relation Let $A = \{a_1, a_2, a_3, \dots, a_n\}$ $B = \{b_1, b_2, b_3, \dots, b_m\}$ be two finite sets containing $n$ and $m$ elements respectively. Then the Cartesian product $A \times B$ contains $n \times m$ ordered pairs. Let $R$ be a relation from set $A$ to set $B$. Then, $$ R \subseteq A \times B $$ The matrix of relation $R$ , denoted by $M_R$, is an $n \times m$ matrix defined as follows: $$ M_R = [m_{ij}] $$ where $m_{ij} = 1$ if $(a_i, b_j) \in R$ $m_{ij} = 0$ if $(a_i, b_j) \notin R$ Question 1 Let $A = \{1,2,3,4\}$ $B = \{1,4,6,8,9,16\}$ A rel...

Types of Relations in Discrete Mathematics – Definitions, Examples & Solved Problems Relations are a fundamental concept in Discrete Mathematics and Graph Theory . They are widely used in equivalence relations, partial order relations, posets, lattices, and many real-life applications. Questions based on relations are frequently asked in GTU and engineering mathematics examinations . In this article, we explain the types of relations in a clear and exam-oriented manner with suitable examples and fully solved problems. Question 1 Explain types of a relation with a suitable example. Solution Definition of Relation Let A and B be two non-empty sets. A relation R from set A to set B is defined as a subset of the Cartesian product A × B . Mathematically, $$ R \subseteq A \times B $$ Types of Relations 1. Reflexive Relation A relation R on a set A is said to be reflexive if every element of A is related to itself. $$ (a,a) \in R \quad \text{for ...

Solution: Triangle Inequality Statement: For all real numbers x and y, |x + y| ≤ |x| + |y| (a) Verification Case (i): x = 2, y = 3 |x + y| = |2 + 3| = |5| = 5 |x| + |y| = |2| + |3| = 2 + 3 = 5 ∴ |x + y| = |x| + |y| Case (ii): x = −2, y = −3 |x + y| = |−2 − 3| = |−5| = 5 |x| + |y| = |−2| + |−3| = 2 + 3 = 5 ∴ |x + y| = |x| + |y| Case (iii): x = −2, y = 3 |x + y| = |−2 + 3| = |1| = 1 |x| + |y| = |−2| + |3| = 2 + 3 = 5 ∴ |x + y| ≤ |x| + |y| (b) Proof for all real numbers We prove the inequality by considering different cases. Case 1: x ≥ 0, y ≥ 0 |x + y| = x + y = |x| + |y| Hence, |x + y| ≤ |x| + |y|. Case 2: x ≤ 0, y ≤ 0 |x + y| = −(x + y) = −x − y = |x| + |y| Hence, |x + y| ≤ |x| + |y|. Case 3: x and y have opposite signs In this case, the sum x + y is reduced in magnitude. Therefore, |x + y| < |x| + |y| Hence, the inequality holds. Therefore, the Triangle Inequality holds for all real numbers x and ...

Max and Min of Two Numbers Using Absolute Value Maximum and Minimum of Two Numbers Using Absolute Value The maximum and minimum of two real numbers a and b can be expressed using the absolute value function. These formulas are very useful in mathematical analysis and proofs. 1. Formula for Maximum max(a, b) = (a + b + |a − b|) / 2 Proof Case 1: a ≥ b a − b ≥ 0 ⇒ |a − b| = a − b (a + b + |a − b|) / 2 = (a + b + (a − b)) / 2 = a Hence, max(a, b) = a . Case 2: b > a a − b < 0 ⇒ |a − b| = b − a (a + b + |a − b|) / 2 = (a + b + (b − a)) / 2 = b Hence, max(a, b) = b . 2. Formula for Minimum min(a, b) = (a + b − |a − b|) / 2 Proof Case 1: a ≤ b a − b ≤ 0 ⇒ |a − b| = b − a (a + b − |a − b|) / 2 = (a + b − (b − a)) / 2 = a Hence, min(a, b) = a . Case 2: b < a a − b > 0 ⇒ |a − b| = a − b (a + b − |a − b|) / 2 = (a + b − (a − b)) / 2 = b Hence, min(a, b) = b . Importa...