# Discrete Mathematics

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{a}, {b}, {c}, the subsets of A having two elements are {a, b}, {a, c}, {b, c}, and the subset having three elements is {a, b, c}, i.e., the set A itself and the set which is having no element of A, i.e., φ. Therefore all subsets of A are. φ, {a}, {b}, {c}, {a, b}, {b, c}, (a, c}, and A. 36 DISCRETE MATHEMATICS PROBLEM 2.4 1. Let A = {a, b, c, d}, list all elements of the power set P(A). 2. How many elements are there in the power set of a set having the following number of objects: (a) Nine;

lattices and the partial ordering relations on L and S corresponding to the operations of meet and join be ≤ and ≤′ respectively. If f : L → S is a homomorphism, then for a, b ∈ L such that a ≤ b, then f(a) ≤′ f(b). 98 DISCRETE MATHEMATICS Proof. We have, a ≤ b ⇒ ⇒ ⇒ ⇒ This means that a ≤ b ⇒ a*b=a f (a * b) = f (a) f (a) ∧ f (b) = f (a) f (a) ≤′ f (b). f(a) ≤′ f(b) if f is a homomorphism. Definition 4.12.2. Two lattices (L, *, ⊕) and (S, ∧, ∨) are said to be isomorphic, if there is a

the identity in L with respect to ⊕ and c ≠ 1. This leads to contradiction. So, 1 is the complement of 0. In a similar manner, we can show that the 0 is the complement of 1. Definition 4.14.2. A lattice (L, *, ⊕, 0, 1) is said to be a complemented lattice if every element of L has at least one complement. Example 4.14.1. In the adjacent figures, the diagram of some lattices are given and the complements of some elements are given on next page. ORDERED SETS AND LATTICES 103 1 1 1 b a a b

5.1.1. The set B = {0, 1} with two binary operations + and · and a unary operation′ defined on B by the following operation tables is a Boolean Algebra. For it satisfies the following properties: + 0 1 0 0 1 1 1 1 0 1 a a′ 0 0 0 0 1 1 0 1 1 0 Table 1 Table 2 (1) Closure Property. Since every entry of the operation tables is in B, the set B is closed under all three operations. (2) Commutative Property. Since Tables 1 and 2 are symmetric, both operations + and · are

GH MN y PQJK L1O TM P N0Q L0O TM 1 P N Q = = = LM kx OP . NyQ LM k ⋅1 OP = LM k OP N 0 Q N0Q LM k 0 OP = LM 0 OP . N 1 Q N1Q Therefore the standard matrix for T is A = LM k N0 0 1 OP . Q 1 (— x, y) 2 h (2x, y) (x, y) 1 K=— 2 (Compression) Initial figure K=2 (Expansion) Fig. 6.6 Similarly, the standard matrix for an expansion or compression in the y-direction is LM 1 N0 0 k OP Q (x, 2y) 3 (x, — y) 4 (x, y) Initial figure K=2 (Expansion) 3 K= — 4 (Compression) Fig. 6.7