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Relations and Functions 2 Relations and Functions 2.1 INTRODUCTION The word relation implies an association of two objects, two persons, etc., according to some property possessed by them, e.g. (i) A is the son of B. The relation R here is “is the son of ”. (ii) Delhi is the capital of INDIA. The relation R here is “is the capital of ”. In Mathematics we can study relationship between a real number and a number which is smaller than it, between a program and a variable it uses. Relationship between elements of sets is represented using a structure called relation. 2.2 RELATION OR BINARY RELATION Let A and B be two non-empty sets. A binary relation or simply a relation from A to B is a sub- set of A × B. Given x ∈A and y ∈ B, we write x R y if (x, y) ∈ R and x R y if (x, y) ∉ R. If R is relation from A to A, then R is said to be relation on A. A binary relation on a set A is a subset of A × A. This definition needs some explanation, since the connection between it and the informal idea of a relation might not be obvious. Note: The word ‘binary’ refers to the fact that the relation is between two elements of A. As this is the only kind of relation we will be studying, we will omit the word ‘binary’ in what follows. The easiest way to do this is by means of an example. Let A = {1, 3, 5} be a set. Then A × A = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)}. Unit2.indd 29 9/15/2009 3:48:24 PM Relations and Functions • 31 3. If x > y. The relation x R3 y means the ordered pairs in which the first element is greater than the second. These are (3, 1), (5, 1), and (5, 3). ∴ R3 = {(3, 1), (5, 1), (5, 3)}, a subset of A × A and so is a relation of A. ABU 1 1 3 3 5 5 Hence, in all the above cases, we have R1 ⊆ A × A, R2 ⊆ A × A, and R3 ⊆ A × A. Thus a relation R on any set is any subset of the Cartesian product A × A. 2.3 DOMAIN AND RANGE OF RELATION For a relation R, domain is a set of elements in A which are related to some element in B. Similarly range of R is a set of elements in B which are related to some element in A. Example 1: If A = {a, b, c}, B = {x, y, z}; and R = {(a, x), (a, z), (b, y), (c, x)}. Find the domain and range of R. Solution Domain of R = the set of first components of the ordered pairs in R = d(R) = {a, b, c} Range of R = the set of second components of the ordered pairs in R = r(R) = {x, y, z}. Example 2: Find the number of relations from A = {a, b, c} to B = {x, y, z}. Solution There are 3 × 3 = 9 elements in A × B and hence there are m = 29 = 512 subsets of A × B. Thus there are m = 512 relations from A × B Unit2.indd 31 9/15/2009 3:48:25 PM 34 • A Textbook of Discrete Mathematics Let A = {1, 2, 3} and B = {5, 6} R = {(1, 5), (1, 6), (2, 6), (3, 6)} 56 1 11 2 01 3 01 2.7 COMPOSITION OF RELATIONS Let A, B, C be three sets and a relation R1 from the set A to B be a R1 b where a ∈ A and b ∈ B, and the relation R2 from B to C be b R2 c where b ∈ B and c ∈ C. Then the composition of the relation R1 and R2 denoted by R1 o R2 or R1R2 is a relation from A to C defined by R1 o R2 = {(a, c): b ∈ B is an element such that aRb and bRc}, where a ∈ A and c ∈ C. Computing the composite of two relations requires that we find elements that are the sec- ond element of ordered pairs in first relation and the first element of ordered pairs in the second relation. Example 5: Let A = {1, 3, 5}, Let R be the relation such that xRy, iff y = x + 2 and S be the relation such that xRy iff x < y. (a) Find RS. (b) Find SR. (c) Illustrate RS and SR via a diagram. (d ) Is the relation (set) RS equal to the relation SR? Why? Solution R = {(1, 3), (3, 5)} S = {(1, 3), (1, 5), (3, 5)}, (a) RS = {(1, 5)} (b) SR = {(1, 5)} AAA RRS 1 1 1 1 (c) 3 3 3 3 5 5 5 5 (d ) ∴ The relations RS = SR Relations are same as the elements of the set. Unit2.indd 34 9/15/2009 3:48:27 PM Relations and Functions • 35 2.8 TYPES OF RELATIONS OR PROPERTIES OF RELATION There are several properties that are used to classify relations on a set. (i) Reflexive: R is reflexive if a R a ∀ a ∈ A. (ii) Symmetric: R is symmetric if a R b ⇒ b R a. (iii) Asymmetric: If R is not symmetric, it is called Asymmetric relation. (iv) Anti-symmetric: R is anti-symmetric if a R b and b R a ⇒ a = b; i.e. if (a, b) ∈ R then (b, a) ∉ R unless a = b (v) Transitive: R is transitive if a R b and b R c ⇒ a R c. ∀ a, b, c. These properties are only defined for a relation on a set. Example 6: If A = {a, b, c, d } and R = {(a, a), (b, b), (c, c), (d, d )}. Prove that R is reflexive. Solution a ∈ A, (a, a) ∈ R, b ∈ A, (b, b) ∈ R, c ∈ A, (c, c) ∈ R, d ∈ A, (d, d ) ∈ R. ∴ a R a ∀ a ∈ A, hence R is reflexive. Example 7: Prove that the relation R = {(1, 1),(2, 1), (2, 2),(2, 3),(2, 4),(3, 1),(3, 2),(3, 3),(3, 4)} on set A = {1, 2, 3, 4} is neither reflexive, nor symmetric or anti-symmetric but transitive. Solution A = {1, 2, 3, 4} (4, 4) ∉ A, hence R is not reflexive. (2, 1) ∈ R and (1, 2) ∉ R, hence R is not symmetric. (2, 3) ∈ R and (3, 2) ∈ R, but 2 ≠ 3, hence R is not anti-symmetric. (a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R ∀ a, b, c ∈ A, hence transitive. 2.8.1 Universal Relation A relation R in a set A is called universal relation if R = A × A. e.g.: If A = {x, y, z}. Then R = {(x, x), (x, y), (x, z), ( y, x), ( y, y), ( y, z), (z, x), (z, y), (z, z)} is a universal relation in A. 2.8.2 Identity Relation A relation in a set A is called identity relation if R = {(x, x): x ∈ A} e.g.: If A {x, y, z}. Then R = {(x, x), ( y, y), (z, z)} is an identity relation in A. Unit2.indd 35 9/15/2009 3:48:27 PM 36 • A Textbook of Discrete Mathematics 2.8.3 Equivalence Relation A relation R on a set A is equivalence relation if it is reflexive, symmetric and transitive. Two elements that are related by equivalence relation are called equivalent. 2.8.4 Equivalence Class Equivalence class of an element a ∈A, denoted by [a], having relation R on A, is the set of ele- ments of A to which a is related i.e., [a] = {x: (a, x) ∈ R}. Example 8: Let R be the relation on the set N of positive integers defined by R= {(a, b): a + b is even}. Prove that R is an equivalent relation. Solution R is reflexive:∀ a ∈ N, a + a is even, hence a R a ∀ a. R is symmetric: If a + b is even then b + a is even, hence a R b ⇒ b R a. R is transitive: Let a, b, c have the same parity i.e., these are either odd or even. We see if a R b and b R c then a R c. Hence R is equivalence relation. Example 9: Let R be the relation on the set Z of integers defined by R = {(x, y): x ∈ Z, y ∈ Z, (x - y) is divisible by 6}. Prove that R is an equivalent relation. Solution R is an equivalence relation if it is reflexive, symmetric and transitive. (a) Reflexivity: a - a = 0 is divisible by 6, hence a R a and R is reflexive. (b) Symmetry: a - b = - (b - a), if a - b is divisible by 6, (b - a) is also divisible by 6, hence a R b =b R a and R is symmetric. (c) Transitivity: If a - b and b - c are divisible by 6, let a - b = 6m, m ∈ Z and b - c = 6n, n ∈ Z so that (m + n) ∈Z and (a - c) = 6(m + n), divisible by 6, accord- ingly if a R b and b R c then a R c, so that R is transitive. Example 10: If A is a set of all integers. Then prove that R = {(a, b): a, b ∈ A, (a - b) is an even integer} is an equivalence relation on A. Solution 0 = a – a is an even integer. So (a, a) ∈ R If (a, b) ∈ R, a - b is an even integer.
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