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With reference to the scenario, prepare a report which examines the set theory and functions applicable to software engineering and analyze mathematical structures of objects using graph theory. Section 1 Perform algebraic set operations in a formulated mathematical problem. Determine the cardinality of a given bag (multi set). Determine the inverse of a function using appropriate mathematical techniques. Formulate corresponding proof principles to prove properties about defined sets. Section 2 Model contextualized problems using trees, both quantitatively and qualitatively. Use Dijkstra’s algorithm to find a shortest path spanning tree in a graph. Assess whether an Euler Ian and Hamiltonian circuit exists in an undirected graph. Construct a proof of the Five Color Theorem. Introduction As we are working as a mathematical analyst for the Infrastructure and Planning division within the Nepal Railways, the department helps the Government of Nepal by providing statistics in planning and developing infrastructures. Here, we are supposed to contribute by examining set theory and functions for collecting and analyzing socio-demographic data and developing methodologies for reliable data collection. In this part, we are going to prepare a report which examines the set theory and functions applicable to software engineering and analyze mathematical structures of objects using graph theory. Firstly, we will collect data with respect to the scenario and then we will perform algebraic set operations in a formulated mathematical problem, determining the cardinality of multi-set, determining the inverse of a function using appropriate mathematical techniques and formulate corresponding proof principles to prove properties about defined sets. Moreover, I will be modeling the contextualized problems using trees with both quantitatively and qualitatively, finding shortest path of spanning tree in a graph using Dijkstra’s algorithm, Assess whether an Euler Ian and Hamiltonian circuit exists in an undirected graph and also discussed about an undirected graph. Perform algebraic set operations in a formulated mathematical problem. Set
(Mathsisfun.com,, n.d.)The collection of well-defined distinct objects, unordered discrete objects of the same type is known as a set. The word well-defined refers to a specific property which makes it easy to identify whether the given objects belongs to the set or not. The word distinct means that the objects of a set must be all different. A set is said to contain its elements. We write a ∈ A to denote that a is an element of the set A. The notation a ∉ A denotes that a is not an element of the set A. For example:- i).collection of all vowels in english alphabet= {a, e,i,o,u} ii).Collection of all odd numbers= {1, 3, 5, 7, 9, 11……..} Types of set Finite set A set consisting of a natural number of objects, i.e. in which number element is finite is said to be a finite set. For example:- i). A = { 5, 7, 9, 11} and B = { 4 , 8 , 16, 32, 64, 128} Obviously, A, B contain a finite number of elements, i.e. 4 objects in A and 6 in B. Thus, they are finite sets. ii). A is the set of natural numbers less than 6 .A ={1,2,3,4,5} Since, set A has 5 elements.so,the element can be counted.Thus,set A is finite set. Infinite set The number of element cannot be counted or the set of all natural numbers is called infinite set. For example:- i). Set of all points in a plane is an infinite set.
We know that the universal set contains all the elements of the given sets; thus, the universal set of A,B and C is given by: U={1,2,3,4,5,6,8,9} 6.subset If A and B are two sets, and every element of set A is also an element of set B, then A is called a subset of B and we write it as A ⊆ B or B ⊇ A.The symbole of subset is ‘⊆’ For example i).Let A = {2, 4, 6} , B = {6, 4, 8, 2} Here A is a subset of B Since, all the elements of set A are contained in set B. ii).Let X = {Toyota}, Y ={All brands of car} Set B cover all brands of car;Maruti,Suzuki,Toyota,Hyundai,Tata motors etc,Moreover x is a set of Toyota.then all the elements of set X contained in set Y.Hence,X is a subject of Y. Suset classifies as Types of Subsets Subsets are classified as Proper Subset Improper Subsets Proper Subset:
If A and B are two sets, then A is called the proper subset of B if A ⊆ B but B ⊇ A i.e., A ≠ B. The symbol ‘⊂’ is used to denote proper subset. Symbolically, we write A ⊂ B. For example; A = {1, 2, 3, 4} Here n(A) = 4 B = {1, 2, 3, 4, 5} Here n(B) = 5 We observe that, all the elements of A are present in B but the element ‘5’ of B is not present in A.So, we say that A is a proper subset of B.Symbolically, we write it as A ⊂ B. Improper set A subset which contains all the elements of the original set is called an improper subset. It is denoted by ⊆. For example Let,P ={2,4,6} Then, the subsets of P are; {}, {2}, {4}, {6}, {2,4}, {4,6}, {2,6} and {2,4,6}. Where, {}, {2}, {4}, {6}, {2,4}, {4,6}, {2,6} are the proper subsets and {2,4,6} is the improper subsets. Therefore, we can write {2,4,6} ⊆ P. 7.Power set In set theory, the power set (or power set) of a Set A is defined as the set of all subsets of the Set A including the Set itself and the null or empty set. It is denoted by P(A). Basically, this set is the combination of all subsets including null set, of a given set. If the given set has n elements, then its Power Set will contain 2n^ elements. It also represents the cardinality of the power set. For example
For example i). Given U = {1, 2, 3, 4, 5, 6, 7, 8, 10} X = {1, 2, 6, 7} and Y = {1, 3, 4, 5, 8} X ∪ Y = {1, 2, 3, 4, 5, 6, 7, 8} ii). If set P = {2, 3, 4, 5, 6, 7}, set Q = {0, 3, 6, 9, 12} and set R = {2, 4, 6, 8}. PUQUR={0,2,3,4,5,6,7,8,9,12} Joint set Two sets are said to be joint sets when they have at least one common element. For example M={5,6,7},N = {4,6,7,9} are joint set Disjoint set A pair of sets which does not have any common element are called disjoint sets.
For example Let set A={2,5,6} and set B={4,7,8} Therefore, A ∩ B = {2,5,6} ∩ {4,7,8} As we can see, A and B do not have any common element. So, A ∩ B = {} Hence, proved A and B are disjoint. 2.Intersection of set Intersection of two given sets is the largest set which contains all the elements that are common to both sets (Mathstopia.net, n.d.). To find the intersection of two given sets A and B is a set which consists of all the elements which are common to both A and B. The symbol for denoting intersection of sets is ‘ꓵ’ For example i).
ii). Let the set of natural numbers be the universal set and A is a set of even natural numbers, then A' {x: x is a set of odd natural numbers} 4.Set Difference If A and B are two sets, then their difference is given by A - B or B - A. For example Let, A = {2, 3, 4} and B = {4, 5, 6} A - B means elements of A which are not the elements of B. A - B = {2, 3} As Nepal is multi-religious country.for this government took a small survey on particular religion.The surey showed 60% people are hindu,20% Buddhist and 20% nor hindu neither Buddhist. No. of hindu n(A) = 60% No. of Buddhist n(B) = 20 % No. of nor hindu neither buddhist( A ∪B)I^ = 20% Above venn diagram shows the given data. Where ‘x’ is ( A ∩B).
Here n(U) = 100 % We know that, n(U) = n(A) + n(B) - ( A ∩B)+ n ( A ∪B)I 100 = 60 + 20 + x + 20 x = 100 – 60 – 20 – 20 x = 0 For Union For any two set A and B, the union of A and B written as AUB is the set of all elements which are members of the set A or the set B or both, Symbolically it is written as: A U B = { x:x ∈ A or X ∈ B }. From above question n(A∪B) = n(A) + n(B) – n (A∩B) n(A∪B) = 60 + 20 –x n(A∪B) = 80 – x n(A∪B) = 80 – 3 * 0 n(A∪B) = 80 Hence, union of set is 80. For Intersection The intersection of two set A and B denoted A intersection B is the set of elements which belongs to both A and B, it is written as: A intersection B = { x:x ∈ A and x ∈ B}. n (A∩B) = n(A) + n(B) - (A∪B) n (A∩B) = 60 + 20 – 80 n (A∩B) = 60 + 20 – 80 n (A∩B) = 0
Let S be a finite set. The cardinality |S| of S is the number of elements in S. That is, if: S∼N<n Where, ∼ denotes set equivalence N<n is the set of all-natural numbers less than n Then we define: |S|=n Cardinality of Infinite Set Let S be an infinite set. The cardinality |S| of S can be indicated as: |S|=∞ However, it needs to be noted that this just means that the cardinality of S cannot be assigned a number n ∈ N. It means that |S| is at least ℵ0 (aleph null). Example- Let A= {a, b, c, d, e, f} and B= {b, c, d, g, h} What is the cardinality of A? A ∪ B, A-B, A ꓵ B? The cardinality of A is 6, since there are 6 elements in the set. The cardinality of A ∪ B is 8, since A ∪ B= {a, b, c, d, e, f, g, h}, which contains 8 elements. The cardinality of A-B is 3, since A-B= {a, e, f}, which contains 3 elements. The cardinality of A ꓵ B is 3, since A ꓵ B= {b, c, d}, which contains 3 elements.
Example- What is the cardinality of W = the set of English names of the days of the week? The cardinality of this set is 7, since there are 7 days a week. Sometimes we may be interested in the cardinality of the union or intersection of sets, but not know the actual elements of each set. This is common in surveying. Formulate corresponding proof principles to prove properties about defined sets. Boolean Algebra of sets A set is a collection of objects, called members or elements. The members of a set can be physical objects, such as people, stars, or red roses, or they can be abstract objects, such as ideas, numbers, or even other sets. Boolean algebra is often referred to as the algebra of logic, because the English mathematician George Boole, who is largely responsible for its beginnings, was the first to apply algebraic techniques to logical methodology? Boole showed that logical propositions and their connectives could be expressed in the language of set theory. Thus, Boolean algebra is also the algebra of sets. Algebra, in general, is the language of mathematics, together with the rules for manipulating that language. Beginning with the members of a specific set (called the universal set), together with one or more binary operations defined on that set, procedures are derived for manipulating the members of the set using the defined operations, and combinations of those operations. Both the language and the rules of manipulation vary, depending on the properties of elements in the universal set. For instance, the algebra of real numbers differs from the algebra of complex numbers, because real numbers and complex numbers are defined differently, leading to differing definitions for the binary operations of addition and multiplication, and resulting in different rules for manipulating the two types of numbers. Boolean algebra consists of the rules for manipulating the subsets of any universal set, independent of the particular properties associated with individual members of that set. It depends, instead, on the properties of sets. The universal set may be any set, including the set of real numbers or the set of complex numbers, because the elements of interest, in Boolean
Now to prove different law, we have: Idempotent law A∪A = A and A∩A = A For A∪A = A, Here, A= {a, b, c, d} Now, A∪A = {a, b, c, d} ∪ {a, b, c, d} = {a, b, c, d } Hence, A∪A = A. For A∩A = A, Here, A= {a, b, c, d} Now, A ꓵ A = {a, b, c, d} ꓵ { a, b, c, d } = {a, b, c, d } Hence, A∩A = A. So, idempotent law is proved. Associative law A∪(B∪C) = (A∪B) ∪C and A∩(B∩C) = (A∩B) ∩ C For A∪(B∪C) = (A∪B) ∪C LHS, A∪(B∪C)
= {a, b, c, d} ∪ ({c, d, e, f} ∪ {c, d, g, h}) = {a, b, c, d} ∪ {c, d, e, f, g, h} = {a, b, c, d, e, f, g, h } Now, RHS (A∪B) ∪C = ({a, b, c, d} ∪ {c, d, e, f}) ∪ {c, d, g, h} = {a, b, c, d, e, f} ∪ {c, d, g, h} = {a, b, c, d, e, f, g, h } Here, LHS=RHS For A∩(B∩C) = (A∩B) ∩ C LHS, A∩(B∩C) = {a, b, c, d} ꓵ ({c, d, e, f} ꓵ { c, d, g, h }) = {a, b, c, d} ꓵ { c, d } = {c, d } RHS, ( A∩B) ∩ C = ({a, b, c, d} ∩ {c, d, e, f}) ∩ {c, d, g, h } = {c, d} ∩ {c, d, g, h } = {c, d } Hence, LHS=RHS So, associative property is proved.
So, commutative law is proved. Distributive law A∪(B∩C) = (A∪B) ∩ (A∪C) and A∩(B∪C) = (A∩B) ∪ (A∩C) For A∪(B∩C) = (A∪B) ∩ (A∪C), LHS, A∪( B∩C ) = {a, b, c, d} ∪ ({c, d, e, f} ꓵ { c, d, g, h } = {a, b, c, d} ∪ {c, d} = {a, b, c, d } RHS, (A∪B) ∩ (A∪C) = ({a, b, c, d} ∪ {c, d, e, f}) ꓵ ({a, b, c, d} ∪ {c, d, g, h}) = {a, b, c, d, e, f} ꓵ { a, b, c, d, g, h } = {a, b, c, d } Hence LHS=RHS. For A∩(B∪C) = (A∩B) ∪ (A∩C) LHS, A∩(B∪C) = {a, b, c, d} ꓵ ({c, d, e, f} ∪ {c, d, g, h}) = {a, b, c, d} ꓵ { c, d, e, f, g, h } = {c, d } RHS, (A∩B) ∪ (A∩C)
= ({a, b, c, d} ꓵ {c, d, e, f}) ∪ ({a, b, c, d} ꓵ { c, d, g, h } = {c, d} ∪ {c, d} = {c, d } Hence, LHS=RHS. So, distributive law is proved. Complement law A∪ A´ = U and A∩ A´ = ∅ and A´ = A ∪= {a, b, c, d, e, f, g, h, i } A= {a, b, c, d} B= {c, d, e, f} C= {c, d, g, h} A´ = {e, f, g, h, i } For A∪ A´ = U, Here, U= {a, b, c, d, e, f, g, h, i} Now, A∪ A´ = {a, b, c, d} ∪ {e, f, g, h, i} = {a, b, c, d, e, f, g, h, i } Hence, A∪ A´ = U For A∩ A´ = ∅, = {a, b, c, d} ꓵ { e, f, g, h, i } = ∅ Hence, A∩ A´ = ∅.
