Discrete mathematics & numerical techniques, Study notes of Business Mathematics

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DISCREET MATHEMATICS AND

NUMERICAL TECHNIQUES

Contents

Section-I

UNIT 1 SET THEORY 3

UNIT 2 BOOLEAN ALGEBRA 41

Section-II

UNIT 3 FUNCTIONS 63

Section-III

UNIT 4 PERMUTATIONS, COMBINATIONS AND MATRICES 89

Unit 1 Set Theory Unit 2 Boolean Algebra

SECTION-I

Discrete Mathematics & Numerical Techniques

Notes

case letters. a, b, c, x, y, etc. The relationship of an object to a set, of which it is an element, is called a relation of belonging. Examples:

(vi) Set of all rational numbers: Any number of the form p / q , where p and q are integers and q ≠ 0 is called a rational numbers. The set of all rational numbers is denoted by Q. (vii) Set of all complex numbers: Any number of the form a + ib where i = − 1. Types of Sets

1. Singleton set: If a set consists of only one element, it is called a singleton set. For example: {1}, {0}, {a}, {b}, etc.
2. Finite set: A set consisting of a natural number of objects, i.e., in which the number of elements is finite, is called finite set. For example: A = {5, 7, 9, 11} and, B = {4, 8, 16, 32, 64, 91} Since A contains four elements and B contains six elements, so both are finite sets.
3. Infinite set: If number of elements in a set is infinite, the set is called infinite set. For Example Set of natural numbers, N = {1, 2, 3, 4, ...} is an infinite set.
4. Equal set: Two sets A and B consisting of the same elements are called equal set. For example: A = {1, 5, 9} B = {1, 5, 9} So here A = B.
5. Pair set: A set having two elements is called pair set. For example: {1, 2}, {0, 3}, (4, 9}, {3, 1}, etc.
6. Empty set: If a set consists of no elements, it is called the empty sector null set or void set and is represented by ϕ. Z = the set of integers = {0, 1, – 1, 2 – 2, –2, 3, –3, ⁄⁄.} N = the set of non negative integers of natural numbers = {0, 1, – 1, 2, 3⁄⁄.} Z+^ = the set of positive integers = {1, 2, 3,⁄⁄.} = {xtz | x > 0} Q = the set of rational numbers = {a, b|a, b ∈ z, b ≠ 0}

Set Theory

Notes

Q +^ = the set of positive rational numbers = {r ∈ q | r > 0} R = the set of real nos. R+^ = the set of positive real numbers R‰^ = the set of non zero real numbers C = the set of complex numbers = {x + iy|x, y ∈ R, i^2 = –1} C+^ = the set of non zero complex numbers. For each n ∈ Z+^ , Zn = {0, 1, 2, ⁄⁄⁄., n – 1} For real nos a, b with a > b, [a, b] = {x ε R|a ≤ x ≤ b} [a, b] = {x ε R|a < x < b} [a, b] = {x ε R|a ≤ x < b} [a, b] = {x ε R|a < x ≤ b} The first set is called a closed interval, the seemed set an open interval, and the other two sets help open intervals.

Here a point is to be noted that ϕ is a null set but {φ} is singleton set.

Example: Which of the following are valid set definitions? A = {4, 5, 8} B = {a, b, a, c} C = {a, b} D = {1, 2, 3, 4, 5, 6} S: A is a valid set, containing three elements 4, 5, and 8. B looks to be equally valid except for the occurring of two aÊs. We can certainly check for inclusion within the set and this is surely the most important element. Hence we might regard this as valid and equal to {a, b, c}. However, there are problems that original definition of B and remove one of the aÊs then we apparently have a ∈ B and a ∈ B. This conclusion is not allowed since it is inconsistent, hence we shall regard repetition within a set as referring to the same element and its duplication as being an oversight; the removal of duplicates forms the basis of several mathematical arguments later on. C is a valid set, containing two elements, a and b. D is a valid set, containing 6 elements 1, 2, 3, 4, 5 and 6.

Methods of Representing a Set

The most common method of describing the sets is as follows: z Roster method or listing method or Tabular method. z Set Builder method or property method or Rule method.

1. Roster Method: In this method, a set is described by listing all its elements, separating them by commas and enclosing them within brackets (only brackets). Example: (i) If A is the set of even natural numbers less than 8. A = {2, 4, 6} (ii) If B is the set of odd and numbers less than 17. B = {1, 3, 5, 7, 9, 11, 13, 15}

Set Theory

Notes

Example 2: Let P = {a, b, c, d, e, f} and Q = {a, b, c, d, g, h, i, j}, then P ∪ Q = {a, b, c, d, e, f, g, h, i, j}

5. Disjoint sets: Let A, Be Ω. The set A and B are called disjoint, or mutually disjoint, when A ∩ B = φ i.e. two sets are called disjoint if there intersection is a null set. Example 1: Let A = {1, 2, 3, 4} and B = {5, 6, 7, 8} A ∩ B = φ Since no element is common in set A and B. Example 2: Let A = {a, b, c} and B = {e, f, g} So A ∩ B = φ Since no element in A and B is common.
6. Difference of two sets: The difference of two sets A and B is the set of all elements which are in A but not in B and is denoted by A – B. Or symbolically: A – B = {x| x ∈A; x ∉ B}. Similarly: The difference of two sets B and is the set of all elements which are in B but not in A and is denoted by B – A. Or Symbolically: B – A = {x|x ∈ B; x ∉ A} Example 1: Let A = {1, 2, 3, 4} and B = {2, 4, 6, 8}, then find A – B and B – A? Solution: A – B = {1, 3} and B – A = {6, 8} Example 2: Let P = {a, b, c, d} and Q = {c, d, e, f, g, h}, then find P – Q and Q – P? Solution: P – Q = {a, b}. Q – P = {e, f, g, h}.
7. Symmetric difference of two sets: The symmetric difference of two sets. A and B is the set of the elements which are in the union of (A – B) and (B – A) and is represented by Δ. So, Δ (A, B) = (A –B) ∪ (B – A) Example: Let A = {1, 2, 3, 4, 5, 6} and B = {4, 5, 6, 7, 8, 9, 10} then, find symmetric difference of sets A and B. Solution: Here A – B = {1, 2, 3} B – A = {7, 8, 9, 10} So, Δ (A, B) = (A – B) ╦∪ (B – A) = {1, 2, 3, 7, 8, 9, 10}
8. Super set: If A and B are two sets and A is the subset of B then B is a super set of A. It can be written as, B ⊃ A Example: A = {1,2,3}, B = {1, 2, 3, 4} since A ⊂ B then here, we can say that, B ⊃ A
9. Universal set: Any set which is super set of all the sets under consideration is known as the universal set and is denoted by Ω or S. Example: Let us take some sets A = {1, 2, 3}, B = {4, 5, 6}, C = {6, 7, 8, 9}

Discrete Mathematics & Numerical Techniques

Notes

D = {10, 11, 12}, E = {9, 10, 11} F = {6, 7, 9}, then Universal set, S = {1, 2, 3, 4, 5, 6 7, 8, 9, 10, 11, or any super set of S}

10. Complement of A set: Let U be the universal set and A ⊂ U. Then, the complement of A is the set of those elements of U which are not in A. The complement of A is denoted by Ac^. A c^ = U – A = {x : x ∪ ∈ and x ∉ A} = {x : x ∉ A} Complement of a set is set itself. i.e. (A c^ ) c^ = A

Some Symbols

11. Partition of a set: Give a set A and on index set I, Let φ ≠ A (^) i ≤ A for each i ∈ I. Then {A (^) i} (^) i ∈I is a partition of A if (a) A = ∪i ∈I and (b) A (^) i ∩ Aj = φ, for all for i, j ∈ I where I ≠ j. Each subset A (^) i is called a cell or block of the partition. Subsets A, B and C such as, A = {1, 5, 9}, B = {3, 7}, C = {11} Clearly, (i) A, B, C are non empty. (ii) A ∪ B ∪ C = {1, 3, 5, 7, 9, 11} = S. (iii) A ∩ B = φ, B ∪ C = φ and C ∩ A = φ Hence, the set {1, 5, 9}, {3, 7} and {11} is a partition of the set S.

Discrete Mathematics & Numerical Techniques

Notes

Example: P is a mustiest that has as its elements the types of computer equipment needed by one department of a university where the multiplicities are the number of items required. Q is the same type of multi-set but for a different department. P = {107 PCs, 44 routers, 6 servers} Q = {14 PCs, 6 routers, 2 Macs} Problem 1: What combination of P and Q represents that the university must buy if there is no sharing of equipment? Solution: Since no equipment is shared, each department needs to have its own version of the equipment. Thus, the combination of equipment needed is: P + Q = {121 PCÊs, 50 routers, 6 servers, 2 Macs} Problem 2: What combination of P and Q represents, what must be bought if the two departments share? Solution: If the two departments share, then the number of each item is simply the maximum of the needs of both departments. Thus, the combination of equipment needed is: P ∪ Q = {107 PCÊs, 44 routers, 6 × servers, 2 Macs} Problem 3: What combination of P and Q represents the actual equipment that both departments share? Solution: The combination of equipment shared by the two departments is simply the overlap in the equipment needed by each department. Thus, the combination of equipment shared is: P ∪ Q = {14 PCÊs, 6 routers}. Solved Examples Example 1: Find the smallest set X such as, X ∪ {1, 2} = {1, 2, 3, 5, 9} Solution: Since X ∪ {1, 2} = {1, 2, 3, 5, 9} So, Smallest set X contains all the elements except {1, 2} So, X = {3, 5, 9} Example 2: Let A = {1, 2}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6} Find. A ∪ B, B ∩ C, A – B, B – A, A ∩ (B ∪ C), A ∪ (B ∩ C), and (A ∪ B ∪ C) Solution: Here given, A = {1, 2, 3}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6} A ∪ B = {1, 2, 3, 4, 6, 8} B ∩ C = {4, 6} A – B = {1, 3} B – A = {4, 6, 8} A ∩ (B ∪ C) = {1, 2, 3} ∩ {2, 4, 6, 8} ∪ {3, 4, 5, 6} = {1, 2, 3} ∩ {2, 3, 4, 5, 6, 8} = {2, 3} A ∪ (B ∩ C) = {1, 2, 3} ∪ {2, 4, 6, 8} ∩ {3, 4, 5, 6} = {1, 2, 3} ∪ {4, 6} = {1, 2, 3, 4, 6}

Set Theory

Notes

A ∪ B ∪ C = {1, 2, 3} ∪ {2, 4, 6, 8} ∪ {3, 4, 5, 6}

Example 3: If U = {1, 2, 3, 4, 5, 6, 7, 8, 9} A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}

Solution: Here given U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6} Now, A′ = U – A A′ = {5, 6, 7, 8, 9} A ∪ B = {1, 2, 3, 4, 6, 8} (A ∪ B)′ = ∪ – (A ∪ B) = {5, 7, 9} (A ∩ C) = {3, 4} (A ∩ C)′ = U – (A ∩ C) = {1, 2, 5, 6, 7, 8, 9} B – C = {2, 8} (B – C)′ = U – (B – C) = {1, 3, 4, 5, 6, 7, 9} Example 4: Let U = R the set of all real numbers. If A = {x : x ∈ R, 0 < x < 2}, B = {x : x ∈ R, 1 < x 3} then, Find A′, B′, A ∪ B, A ∩ B. Solution: A′ = R – A = {x : x ∈ R and x ∉ A} = {x : (x ∈ R and x > 2) or (x ∈ R and x ª 0)} = {x : x ∈ R and x > R} ∪ {x : x ∈ R and x ≤ 0} B′ = {x : x ∈ R and x < 1} ∪ {x : x ∈ R and x > 3} Similarly: A ∪ B = {x : x ∈ R and 0 < x < 3} A ∩ B = {x : x ∈ R and 1 < x < 2} A – B = {x : x ∈ R and 0 < =1} Example 5 : If A = {2, 3, 4, 8, 10}, B = {1, 3, 4, 10, 12} and C = {4, 5, 6, 12, 14, 16} Find (A ∪ B) ∩ (A ∪ C) and (A ∩ B) ∪ (A ∩ C)? Solution: Here given A = {2, 3, 4, 8, 10}, B = {1, 3, 4, 10, 12} and C = {4, 5, 6, 12, 14, 16} then, A ∪ B = {1, 2, 3, 4, 8, 10, 12} A ∪ C = {2, 3, 4, 5, 6, 8, 10, 12, 14, 16} (A ∪ B) ∩ (A ∪ C) = {2, 3, 4, 8, 10, 12} A ∩ B = {3, 4, 10} A ∩ C = {4} (A ∩ B) ∪ (A ∩ C) = {3, 4, 10}

Set Theory

Notes

8. Law of inclusion: (a) A (A ∪ B) and B (A ∪ B) (b) (A ∩ B) A) and (A ∩) B
9. Law of difference: (a) A – φ = A (b) A – A = φ
10. Absorption Laws: (a) A ∪ (A ∩ B) = A (b) A ∩ (A ∪ B) = A Proof of operation on sets:
11. Properties of subset, (a) To prove A A = A Let, x A A ⇒ x A or x A ⇒ x ∈ A i.e., A A A ...(1) Conversely, Let, x ∈ A ⇒ x ∈ A or x ∈ A ⇒ x ∈ A A i.e., A A A ...(2) By equations (1) and (2) we get, A A = A (b) To prove A ∩ A = A Here, x ∈ A ⇒ x ∈ A and x ∈ A ⇒ x ∈ A i .e. A ∩ A A ...(3) Conversely, x ∈ A ⇒ x ∈ A and x ∈ A ⇒ x ∈ A ∩ A A A ∩ A ...(4) ⇒ A ∩ A = A (c) To prove A B = A, If B A ∴ x ∈ B ⇒ x ∈ A ...(5) Now x ∈ A B ⇒ x ∈ A or x ∈ B ⇒ x ∈ A by equation (5) i.e., A B A ...(6) Conversely, x ∈ A ⇒ x ∈ A or x ∈ B

Discrete Mathematics & Numerical Techniques

Notes

⇒ x ∈ A ∪ B A A ∪ B ...(7) So by equations (6) and (7) we get, A B = A Now let A B = A i.e., x ∈ A or x ∈ B ⇒ x ∈ A x ∈ B ⇒ x ∈ A Hence, B ⊆ A i.e., A ∪ B = A ⇒ B ⊆ A or A ∪ B = A, Iff B ⊆ A (d) To prove A ∩ B = A, Iff A ⊆ B Let, A ⊆ B ∴ x ∈A ⇒ x ∈B ...(8) Now x ∈A ∩ B ⇒ x ∈A and x ∈B, ⇒ x ∈A i.e., A ∩ B ⊆ A ...(9) and x ∈A ⇒ x ∈ A and x ∈B by equation (8) ⇒ x ∈ A ∩ B A ⊆ A ∩ B ...(10) By equations (9) and (10) we get, A ∩ B = A ...(11) Now let A ∩ B = A i.e., x ∈A → x ∈A and x ∈B ∴ x ∈ A ⇒ x ∈ B i.e., A ⊆ B ...(12) By using equations (11) and (12) we get, A ∩ B = A iff A ⊆ B Properties of Intersection operation on sets:

1. Intersection operation is commutative i.e., A ∩ B = B ∩ A Proof: Let x ∈ A ∩ B by the definition of intersection: x ∈ A ∩ B ⇒ x ∈ A and x ∈ B ⇒ x ∈ B and x ∈ A ⇒ x ∈ B ∩ A

Discrete Mathematics & Numerical Techniques

Notes

⇒ b ∈ A or b ∈ B or b ∈ C ⇒ b ∈ (A ∪ B) or b ∈ C ⇒ b ∈ (A ∪ B) or b ∈ C ⇒ b ∈ (A ∪ B) ∪ C ⇒ A ∪ (B ∪ C) ⊆ (A ∪ B)∪ C ...(2) By using equations (1) and (2) we get, A ∪ (B ∪ C) = (A ∪ B) ∪ C Properties of Union operation:

1. Commutative Law: To prove that A ∪ B = B ∪ A Proof: Let x ∈ A ∪ B ∴ x ∈ A ∪ B ⇒ x ∈ A or x ∈ B ⇒ x ∈ B or x ∈ A ⇒ x ∈ B ∪ A ∴ A ∪ B ⊆ B ∪ A ...(1) Conversely let, y ∈ B ∪ A But y ∈ B ∪ A ⇒ y ∈ B or y ∈ A ⇒ y ∈ A or y ∈ B ⇒ y ∈ A ∪ B ∴ B ∪ A ⊆ A ∪ B ...(2) By using equation (1) and (2) we have equations: A ∪ B = B ∪ A
2. Associative Law: To prove that, A ∪ (B ∪ C) = (A ∪ B) ∪ C Proof: Let, x ∈ A ∪ (B ∪ C) Hence, x ∈ A ∪ B ∪ C ⇒ x ∈ A or x ∈ B or x ⇒ (x ∈ A or x ∈ B) or x ∈ C ⇒ x ∈ (A ∪ B) or x ∈ C ⇒ x ∈ (A ∪ B) ∪ C ∴ A ∪ (B ∪ C ) ⊆ (A ∪ B) ∪ C ...(1) Conversely, let x ∈ (A ∪ B) ∪ C, Here, y ∈ (A ∪ B) ∪ C ⇒ (x ∈ A or x ∈ B) or x ∈ C

Set Theory

Notes

⇒ x ∈ A or x ∈ B or x ∈ C ⇒ x ∈ A or (x ∈ B or x ∈ C) ⇒ x ∈ A or x ∈ (B ∪ C) ⇒ x ∈ A ∪ (B ∪ C) ⇒ (A ∪ B) ∪ C ⊆ A ∪ (B ∪ C) ...(2) By using equations (1) and (2) we get, A ∪ (B ∪ C) = (A ∪ B) ∪ C

3. Distributive laws: For three given sets A, B and C to prove that, (i) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) (ii) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) Proof: Let a ∈ A ∩ (B ∪ C), Here a ∈ A ∩ (B ∪ C) ⇒ a ∈ A and a ∈ B ∪ C ⇒ a ∈ A and (a ∈ B or a ∈ C) ⇒ (a ∈ A and x ∈ B) or (a ∈ A) or a ∈ C) ⇒ a ∈ A ∩ B or a ∈ A ∩ C ⇒ a ∈ (A ∩ B) ∪ (A ∩ C) Hence, A ∪ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) ...(1) Conversely let b ∈ (A ∩ B) ∪ (A ∩ C) Here, b ∈ (A ∩ B) ∪ (A ∩ C) ⇒ b ∈ A ∩ B or b ∈ A ∩ C ⇒ (b ∈ A and b ∈ B) or (b ∈ A and b ∈ C) ⇒ (b ∈ A and b ∈ B) or (b ∈ C) ⇒ (b ∈ A and b ∈ B ∪ C) ⇒ (b ∈ A ∩ (B ∪ C) ∴ (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C) ...(2) By using equations (1) and (2) we get, Hence, A ∪ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) Number of elements in the union of two or more sets: Let A, B and C be three finite sets and let n(A), n(B), n(C) respectively denote the number of elements in these sets. Then we see that, n ( A ∪ B) = n (A) + n (B) – n (A ∩ B) ...(1) In case A and B are disjoint sets then, A ∩ B = and n (A ∩ B) = n (φ) = 0 i.e., for disjoint sets A and B n(A ∪ B) = n(A) + n(B) Similarly we can show that, n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C)– n(A ∩ C) + n(A ∩ B ∩ C) ...(2) Example 1: In a group of 1000 people, there are 750 who can speak Spanish and 400 who can speak Bengali. How many people can speak Spanish only? How many