DM _CH(2) - HackMD

The Islamic University of Gaza Discrete Mathematics Lab - 2023/2024 Discussion || ch(2) Eng: Amal I. Mahfouz , Eng: hashem hjazy **** section 2.1 * 2.1.2: Use set builder notation to give a description of each of these sets. a) {0,3,6,9,12} { X ∣ X = 3n, n ∈ Z, 0 ≤ n ≤ 4} b) {−3,−2,−1,0,1,2,3} { X | -3 <= X <= 3, x ∈ Z } c) {m,n,o,p} {X ∣ X ∈ {m, n, o, p } } *2.1.7: Determine whether each of these pairs of sets are equal. Note : sets do not contain duplicate elements. a) {1,3,3,3,5,5,5,5,5},{5,3,1} Equal b) {{1}},{1,{1}} Not Equal, the first set has one element while the second set has two elements. c) ∅,{∅} Not Equal, the first set has no elements, the second set has one element (∅) *2.1.9: For each of the following sets, determine whether 2 is an element of that set. a) x ∈ R|xisaninteger greater than 1 2 is an element of this set. ex : set={2,3,4,5,etc.} b) x ∈ R|xisthe square of an integer 2 is not an element of this set. ex : set={0,1,4,9,16,etc.} c) {2,{2}} 2 is an element of this set. ex : This set contains two elements: the number 2 and the set {2} d) {{2},{{2}}} 2 is not an element of this set. ex: The number 2 itself is not directly listed as an element of this set e) {{2},{2,{2}}} 2 is not an element of this set. f) {{{2}}} 2 is not an element of this set. *2.1.11: Determine whether each of these statements is true or false. a) 0 ∈∅ False (0 is not an element of the empty set.) b) ∅ ∈ {0} False (The set {0} only contains 0, not the empty set ∅.) c) {0} ⊂ ∅ False d) ∅ ⊂ {0} True (The empty set ∅ is a subset of every set) e) {0} ∈ {0} False (the set {0} is an element of the set {0}? No , because the latter is contain element 0 not element {0}) f) {0} ⊂ {0} False g) {∅} ⊆ {∅} True *2.1.21: What is the cardinality of each of these sets? a) {a} 1 b) {{a}} 1 c) {a,{a}} 2 d) {a,{a},{a,{a}}} 3 --- --- section 2.2 *2.2.3: Let A ={1,2,3,4,5} and B ={0,3,6}.Find a) A∪B. A ∪ B = {1,2,3,4,5} ∪ {0,3,6} = {0,1,2,3,4,5,6} b) A∩B. A ∩ B = {1,2,3,4,5} ∩ {0,3,6} = {3} c) A−B. A − B = {1,2,3,4,5} − {0,3,6} = {1,2,4,5} d) B−A. B − A= {0,3,6} − {1,2,3,4,5} = {0,6} *2.2.7: Prove the domination laws in Table 1 by showing that: to solve this question , let us remember : Universal Set (U): Contains all possible elements. Empty Set (∅): Contains no elements. Set A: Any arbitrary set. a) A ∪ U = U. the universal set contains all elements , so that the union of any set with the universal set is the universal set if x ∈ A , then x is in A ∪ U ; because A ⊆ U b) A ∩ ∅ = ∅. the intersection of any set with the empty set is the empty set ; because there are no common elements in ∅. *2.2.14:Find the sets A and B if A−B ={1,5,7,8}, B−A = {2, 10},and A ∩ B ={3,6,9}. A = {1,5,7,8,3,6,9} B = {2,10,3,6,9} 2.2.21: Show that if A and B are sets, then a) A − B = A ∩ 'B. A − B = { x:x ∈ A and x (not ∈) B}. the set of elements that are in 𝐴 but not in 𝐵 A ∩ 'B = {x:x ∈ A and x (not ∈ B)}. the set of elements that are in both 𝐴 and complement 𝐵 Since both sets contain the same elements (those that are in 𝐴 but not in 𝐵) ; Thus, the statement is true: A - B = A ∩ 'B these two sets are not the same, 𝐴 − 𝐵 contains elements in 𝐴 but not in 𝐵, while 𝐴 ∩ 𝐵 contains elements that are in both 𝐴 and 𝐵. These are two different operations, so 𝐴 − 𝐵 ≠ 𝐴 ∩𝐵 b) (A ∩ B) ∪(A ∩ 'B) = A. A ∩ 'B = {x:x∈A and x (not ∈)B} the set of elements that are in both 𝐴 and complement 𝐵 A ∩ B = {x:x ∈ A and x ∈ B}. the set of elements that are in both 𝐴 and 𝐵 The union of these two sets,(A ∩ B) ∪(A ∩ 'B) includes all the elements that are either in both A and 𝐵, or in A but not in 𝐵. This means all the elements that are in 𝐴 in either case are included in the union.