A B is shaded A B A B

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1 NION: Let and be subsets of a universal set. The union of sets and is the set of all elements in that belong to or to or to both, and is denoted. Symbolically: = {x x or x } EMMPLE: Let = {a, b, c, d, e, f, g} = {a, c, e, g}, = {d, e, f, g} Then = {x x or x } ={a, c, d, e, f, g} VENN DIGRM FOR NION: is shaded REMRK:. = that is union is commutative you can prove this very easily only by using definition.. and The above remark of subset is easily seen by the definition of union. MEMERSHIP TLE FOR NION: REMRK: This membership table is similar to the truth table for logical connective, disjunction ( ). INTERSECTION: Let and subsets of a universal set. The intersection of sets and is the set of all elements in that belong to both and and is denoted. Symbolically: = {x x and x } EXMPLE: Let = {a, b, c, d, e, f, g} Page of

2 = {a, c, e, g}, = {d, e, f, g} Then = {e, g} is shaded VENN DIGRM FOR INTERSECTION: REMRK:. =. and. If = φ, then & are called disjoint sets. MEMERSHIP TLE FOR INTERSECTION: REMRK: This membership table is similar to the truth table for logical connective, conjunction ( ). DIFFERENCE: Let and be subsets of a universal set. The difference of and (or relative complement of in ) is the set of all elements in that belong to but not to, and is denoted or \. Symbolically: = {x x and x } EXMPLE: Let = {a, b, c, d, e, f, g} = {a, c, e, g}, = {d, e, f, g} Then = {a, c} VENN DIGRM FOR SET DIFFERENCE: Page of

3 - is shaded REMRK:. that is Set difference is not commutative..., and are mutually disjoint sets. MEMERSHIP TLE FOR SET DIFFERENCE: REMRK: The membership table is similar to the truth table for ~ (p q). COMPLEMENT: Let be a subset of universal set. The complement of is the set of all element in that do not belong to, and is denoted Ν, or c Symbolically: c = {x x } EXMPLE: Let = {a, b, c, d, e, f, g] = {a, c, e, g} Then c = {b, d, f} VENN DIGRM FOR COMPLEMENT: c c is shaded REMRK :. c =. c = φ Page of

4 . c = MEMERSHIP TLE FOR COMPLEMENT: 0 0 c REMRK This membership table is similar to the truth table for logical connective negation (~) EXERCISE: Let = {,,,, 0}, X = {,,,, 5} Y = {y y = x, x X}, Z = {z z 9 z + = 0} Enumerate: ()X Y () Y Z () X Z ()Y c (5) X c Z c (6) (X Z) c Firstly we enumerate the given sets. Given = {,,,, 0}, X = {,,,, 5} Y = {y y = x, x X} = {,, 6, 8, 0} Z = {z z 9 z + = 0} = {, 7} () X Y = {,,,, 5} {,, 6, 8, 0} = {, } () Y Z = {,, 6, 8, 0} {, 7} = {,, 6, 7, 8, 0} () X Z = {,,,, 5} {, 7} = {,,, 5} () Y c = Y = {,,,, 0} {,, 6, 8, 0} = {,, 5, 7, 9 (5) X c Z c = {6, 7, 8, 9, 0} {,,, 5, 6, 8, 9, 0} = {7} (6) (X Z)c = (X Z) = {,,,, 0} {,,, 5} = {, 6, 7, 8, 9, 0} NOTE (X Z)c Xc - Zc EXERCISE: Given the following universal set and its two subsets P and Q, where = {x x Z,0 x 0} P = {x x is a prime number} Q = {x x < 70} (i) Draw a Venn diagram for the above (ii) List the elements in Pc Q SOLTION: Page of

5 First we write the sets in Tabular form. = {x x Z, 0 x 0} Since it is the set of integers that are greater then or equal 0 and less or equal to 0. So we have = {0,,,,, 0} P = {x x is a prime number} It is the set of prime numbers between 0 and 0. Remember Prime numbers are those numbers which have only two distinct divisors. P = {,, 5, 7} Q = {x x < 70} The set Q contains the elements between 0 and 0 which has their square less or equal to 70. Q= {0,,,,, 5, 6, 7, 8} Thus we write the sets in Tabular form. VENN DIGRM:,,5,7P Q (i) P c Q =? 0,,,6,8 9,0 P c = P = {0,,,,, 0}- {,, 5, 7} = {0,,, 6, 8, 9, 0} and P c Q = {0,,, 6, 8, 9, 0} {0,,,,, 5, 6, 7, 8} = {0,,, 6, 8} EXERCISE: Let = {,,,, 5}, C = {, } and and are non empty sets. Find in each of the following: (i) =, = φ and = {} (ii) and = {, 5} (iii) = {}, = {,, } and C = {,,} (iv) and are disjoint, and C are disjoint, and the union of and is the set {, }. (i) =, = φ and = {} SOLTION Since = = {,,,, 5} and = φ, Therefore = c = {} c = {,,, 5} Page 5 of

6 (i) and = {, 5} also C = {, } SOLTION When, then = = {, 5} lso being a proper subset of implies = {} or = {5} (iii) = {}, = {,, }and C = {,,} lso C = {, } SOLTION C Since we have in the intersection of and as well as in C so we place in common part shared by the three sets in the Venn diagram. Now since is in the union of and C it means that may be in C or may be in, but cannot be in because if is in the then it must be in but is not there, thus we place in the part of C which is not shared by any other set. Same is the reason for and we place it in the set which is not shared by any other set. Now will be in, cannot be in because = {}, and is not in C. So = {, } and = {, } (i) = φ, C = φ, = {, }. lso C = {, } SOLTION EXERCISE: C, 5 = {} se a Venn diagram to represent the following: (i) ( ) C c (ii) c ( C) (iii) ( ) C (iv) ( c ) C c Page 6 of

7 5 6 C 7 8 () ( ) C c C 8 ( ) C c is shaded (ii) c ( C) is shaded C 8 (iii) ( ) C 5 6 C 7 8 ( ) C is shaded Page 7 of

8 (iii) ( c ) C c is shaded C 8 PROVING SET IDENTITIES Y VENN DIGRMS: Prove the following using Venn Diagrams: (i) ( ) = (ii) ( ) c = c c (iii) = c SOLTION (i) - ( ) = (a) = {, } = {, } ={ } (b) is shaded = {, } = { } ( ) = { } ( ) is shaded Page 8 of

9 (c) is shaded = {, } = {, } = {} RESLT: ( ) = SOLTION (ii) ( ) c = c c (a) (b) ( ) c Page 9 of

10 (c) c is shaded. (d) (e) c is shaded. c c is shaded. Now diagrams (b) and (e) are same hence RESLT: ( ) c = c c Page 0 of

11 SOLTION (iii) = c (a) (b) is shaded. (c) c is shaded. From diagrams (a) and (b) we can say c is shaded RESLT: = c PROVING SET IDENTITIES Y MEMERSHIP TLE: Prove the following using Membership Table: (i) ( ) = Page of

12 SOLTION (i) (ii) (iii) ( ) c = c c = c ( ) = - -(-) Since the last two columns of the above table are same hence the corresponding set expressions are same. That is ( ) = SOLTION (ii) ( ) c = c c ( ) c c c c c Since the fourth and last columns of the above table are same hence the corresponding set expressions are same. That is ( ) c = c c SOLTION (iii) c c Page of

Exclusive Disjunction

Exclusive Disjunction Recall A statement is a declarative sentence that is either true or false, but not both. If we have a declarative sentence s, p: s is true, and q: s is false, can we rewrite s is