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72 more on sets and logic

1. 1. More on the Language of SetsContinue from the previous Farmer Andy’s familysection with the given FarmerAndy’s family photo. We definethe following sets Andy Beth Cathy DanU = {Farmer Andy’s family} = {a, b, c, d}R = {member who is wearing something red} = {a, b, d}G = {member who is wearing something green} = {a, c}B = {member who is wearing something blue} = {b, c, d}F = {member who is a girl} = {b, c}M = {member who is a boy} = { a, d}
2. 2. More on the Language of SetsContinue from the previous Farmer Andy’s familysection with the given FarmerAndy’s family photo. We definethe following sets Andy Beth Cathy DanU = {Farmer Andy’s family} = {a, b, c, d}R = {member who is wearing something red} = {a, b, d}G = {member who is wearing something green} = {a, c}B = {member who is wearing something blue} = {b, c, d}F = {member who is a girl} = {b, c}M = {member who is a boy} = { a, d}Each description is called an attribute.
3. 3. More on the Language of SetsContinue from the previous Farmer Andy’s familysection with the given FarmerAndy’s family photo. We definethe following sets Andy Beth Cathy DanU = {Farmer Andy’s family} = {a, b, c, d}R = {member who is wearing something red} = {a, b, d}G = {member who is wearing something green} = {a, c}B = {member who is wearing something blue} = {b, c, d}F = {member who is a girl} = {b, c}M = {member who is a boy} = { a, d}Each description is called an attribute.Example A. Name another attribute in Farmer Andy’s familyphoto and list the respective subset.
4. 4. More on the Language of SetsContinue from the previous Farmer Andy’s familysection with the given FarmerAndy’s family photo. We definethe following sets Andy Beth Cathy DanU = {Farmer Andy’s family} = {a, b, c, d}R = {member who is wearing something red} = {a, b, d}G = {member who is wearing something green} = {a, c}B = {member who is wearing something blue} = {b, c, d}F = {member who is a girl} = {b, c}M = {member who is a boy} = { a, d}Each description is called an attribute.Example A. Name another attribute in Farmer Andy’s familyphoto and list the respective subset.For example, H = {member who has black hair} = {a, c}.
5. 5. More on the Language of SetsContinue from the previous Farmer Andy’s familysection with the given FarmerAndy’s family photo. We definethe following sets Andy Beth Cathy DanU = {Farmer Andy’s family} = {a, b, c, d}R = {member who is wearing something red} = {a, b, d}G = {member who is wearing something green} = {a, c}B = {member who is wearing something blue} = {b, c, d}F = {member who is a girl} = {b, c}M = {member who is a boy} = { a, d}Each description is called an attribute.Example A. Name another attribute in Farmer Andy’s familyphoto and list the respective subset.For example, H = {member who has black hair} = {a, c}.Note that H = G, i.e. different attributes may define the same set.
6. 6. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanNew attributes may be formed by linking other attributes usingthe three connective words “not”, “and” and “or”,
7. 7. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanNew attributes may be formed by linking other attributes usingthe three connective words “not”, “and” and “or”, and each ofthese words correspond to an operation of the correspondingset(s) involved.
8. 8. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanNew attributes may be formed by linking other attributes usingthe three connective words “not”, “and” and “or”, and each ofthese words correspond to an operation of the correspondingset(s) involved.The Complementary Operation and “Not”
9. 9. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanNew attributes may be formed by linking other attributes usingthe three connective words “not”, “and” and “or”, and each ofthese words correspond to an operation of the correspondingset(s) involved.The Complementary Operation and “Not”Given an universal set U and A U, the complement of A, ∩denoted as AC, consists of elements that are not in A,
10. 10. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanNew attributes may be formed by linking other attributes usingthe three connective words “not”, “and” and “or”, and each ofthese words correspond to an operation of the correspondingset(s) involved.The Complementary Operation and “Not”Given an universal set U and A U, the complement of A, ∩denoted as AC, consists of elements that are not in A, that is,AC = {x | x ϵ A}. (See diagram ) U C A A xϵ A The complement of A
11. 11. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanNew attributes may be formed by linking other attributes usingthe three connective words “not”, “and” and “or”, and each ofthese words correspond to an operation of the correspondingset(s) involved.The Complementary Operation and “Not”Given an universal set U and A U, the complement of A, ∩denoted as AC, consists of elements that are not in A, that is,AC = {x | x ϵ A}. (See diagram ) U ACFor example, using the sets above, ARC = xϵ A The complement of A
12. 12. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanNew attributes may be formed by linking other attributes usingthe three connective words “not”, “and” and “or”, and each ofthese words correspond to an operation of the correspondingset(s) involved.The Complementary Operation and “Not”Given an universal set U and A U, the complement of A, ∩denoted as AC, consists of elements that are not in A, that is,AC = {x | x ϵ A}. (See diagram ) U ACFor example, using the sets above, ARC = {x | x is not “wearing something red”} xϵ A The complement of A
13. 13. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanNew attributes may be formed by linking other attributes usingthe three connective words “not”, “and” and “or”, and each ofthese words correspond to an operation of the correspondingset(s) involved.The Complementary Operation and “Not”Given an universal set U and A U, the complement of A, ∩denoted as AC, consists of elements that are not in A, that is,AC = {x | x ϵ A}. (See diagram ) U ACFor example, using the sets above, ARC = {x | x is not “wearing something red”} xϵ A = {c}. The complement of A
14. 14. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanNew attributes may be formed by linking other attributes usingthe three connective words “not”, “and” and “or”, and each ofthese words correspond to an operation of the correspondingset(s) involved.The Complementary Operation and “Not”Given an universal set U and A U, the complement of A, ∩denoted as AC, consists of elements that are not in A, that is,AC = {x | x ϵ A}. (See diagram ) U ACFor example, using the sets above, ARC = {x | x is not “wearing something red”} xϵ A = {c}. The complement of A
15. 15. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “And”
16. 16. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “And”Given two set A and B, the intersection of A and B,denoted as A ∩ B consists of elements that common toboth A and B,
17. 17. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “And”Given two set A and B, the intersection of A and B,denoted as A ∩ B consists of elements that common toboth A and B, that is,A ∩ B = {x | x is a member of A and also a member of B} = {x | x ϵ A and x ϵ B }
18. 18. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “And”Given two set A and B, the intersection of A and B,denoted as A ∩ B consists of elements that common toboth A and B, that is,A ∩ B = {x | x is a member of A and also a member of B} = {x | x ϵ A and x ϵ B } (See diagram ). U A A∩B B The intersection A ∩ B
19. 19. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “And”Given two set A and B, the intersection of A and B,denoted as A ∩ B consists of elements that common toboth A and B, that is,A ∩ B = {x | x is a member of A and also a member of B} = {x | x ϵ A and x ϵ B } (See diagram ).For example, using the sets above, UR∩B= A A∩B B The intersection A ∩ B
20. 20. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “And”Given two set A and B, the intersection of A and B,denoted as A ∩ B consists of elements that common toboth A and B, that is,A ∩ B = {x | x is a member of A and also a member of B} = {x | x ϵ A and x ϵ B } (See diagram ).For example, using the sets above, UR ∩ B = {x | x is “wearing red” and is also “wearing blue”} A A∩B B The intersection A ∩ B
21. 21. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “And”Given two set A and B, the intersection of A and B,denoted as A ∩ B consists of elements that common toboth A and B, that is,A ∩ B = {x | x is a member of A and also a member of B} = {x | x ϵ A and x ϵ B } (See diagram ).For example, using the sets above, UR ∩ B = {x | x is “wearing red” and is also “wearing blue”} A A∩B B= {x | x is “wearing both red and blue”} The intersection A ∩ B
22. 22. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “And”Given two set A and B, the intersection of A and B,denoted as A ∩ B consists of elements that common toboth A and B, that is,A ∩ B = {x | x is a member of A and also a member of B} = {x | x ϵ A and x ϵ B } (See diagram ).For example, using the sets above, UR ∩ B = {x | x is “wearing red” and is also “wearing blue”} A A∩B B= {x | x is “wearing both red and blue”}= {b, d}. The intersection A ∩ B
23. 23. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “Or”
24. 24. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “Or”Given two set A and B, the union of A and B,denoted as A U B consists of elements that belong toeither A or B,
25. 25. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “Or”Given two set A and B, the union of A and B,denoted as A U B consists of elements that belong toeither A or B, that is,A U B = {x | x is a member of A or a member of B, or both} = {x | x ϵ A or x ϵ B }
26. 26. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “Or”Given two set A and B, the union of A and B,denoted as A U B consists of elements that belong toeither A or B, that is,A U B = {x | x is a member of A or a member of B, or both} = {x | x ϵ A or x ϵ B } (See diagram ). U A AUB B The union A U B
27. 27. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “Or”Given two set A and B, the union of A and B,denoted as A U B consists of elements that belong toeither A or B, that is,A U B = {x | x is a member of A or a member of B, or both} = {x | x ϵ A or x ϵ B } (See diagram ).For example, using the sets above, UGUF= A AUB B The union A U B
28. 28. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “Or”Given two set A and B, the union of A and B,denoted as A U B consists of elements that belong toeither A or B, that is,A U B = {x | x is a member of A or a member of B, or both} = {x | x ϵ A or x ϵ B } (See diagram ).For example, using the sets above, UGUF= {x | “x is wearing green” or “x is a girl”} A AUB B The union A U B
29. 29. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanThe Intersection Operation and “Or”Given two set A and B, the union of A and B,denoted as A U B consists of elements that belong toeither A or B, that is,A U B = {x | x is a member of A or a member of B, or both} = {x | x ϵ A or x ϵ B } (See diagram ).For example, using the sets above, UGUF= {x | “x is wearing green” or “x is a girl”} A AUB B= {a, c} U {b, c}= {a, b, c}. The union A U B
30. 30. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BC
31. 31. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.
32. 32. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”}
33. 33. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”} ={a}.
34. 34. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”} ={a}. ∩ Fb. R
35. 35. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”} ={a}. ∩ Fb. RR ∩ F is the “intersection of R with F”.
36. 36. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”} ={a}. ∩ Fb. RR ∩ F is the “intersection of R with F”.R ∩ F = {x | x is “wearing something red” and “is a girl”}
37. 37. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”} ={a}. ∩ Fb. RR ∩ F is the “intersection of R with F”.R ∩ F = {x | x is “wearing something red” and “is a girl”} = {b}.
38. 38. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”} ={a}. ∩ Fb. RR ∩ F is the “intersection of R with F”.R ∩ F = {x | x is “wearing something red” and “is a girl”} = {b}.c. G U M
39. 39. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”} ={a}. ∩ Fb. RR ∩ F is the “intersection of R with F”.R ∩ F = {x | x is “wearing something red” and “is a girl”} = {b}.c. G U MG U M is the “union of G with M”.
40. 40. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”} ={a}. ∩ Fb. RR ∩ F is the “intersection of R with F”.R ∩ F = {x | x is “wearing something red” and “is a girl”} = {b}.c. G U MG U M is the “union of G with M”. G U M= {x | x is “wearing something green” or “x is a boy”}.
41. 41. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanExample B. Using the above, name and describe each of thefollowing sets in the form of {x | x is ###} then list its elements.a. BCBC is the “complement of B”.BC = {x | x is not “wearing something blue”} ={a}. ∩ Fb. RR ∩ F is the “intersection of R with F”.R ∩ F = {x | x is “wearing something red” and “is a girl”} = {b}.c. G U MG U M is the “union of G with M”. G U M = {a, c, d}= {x | x is “wearing something green” or “x is a boy”}.
42. 42. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanMutually Exclusive Sets and Basic Counting
43. 43. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanMutually Exclusive Sets and Basic CountingGiven two sets A and B, we say A and Bare mutually exclusive if A ∩ B = Φ = { },i.e. A and B do not have any common items.
44. 44. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanMutually Exclusive Sets and Basic CountingGiven two sets A and B, we say A and B Uare mutually exclusive if A ∩ B = Φ = { }, A Bi.e. A and B do not have any common items. Mutually exclusive sets
45. 45. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanMutually Exclusive Sets and Basic CountingGiven two sets A and B, we say A and B Uare mutually exclusive if A ∩ B = Φ = { }, A Bi.e. A and B do not have any common items.Using the sets above, we have that Mutually exclusive setsBC = {not wearing anything blue} = {a}is mutually exclusive from the set of F = {is a girl} = {b, c},
46. 46. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanMutually Exclusive Sets and Basic CountingGiven two sets A and B, we say A and B Uare mutually exclusive if A ∩ B = Φ = { }, A Bi.e. A and B do not have any common items.Using the sets above, we have that Mutually exclusive setsBC = {not wearing anything blue} = {a}is mutually exclusive from the set of F = {is a girl} = {b, c},i.e. BC ∩ F = Φ.
47. 47. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanMutually Exclusive Sets and Basic CountingGiven two sets A and B, we say A and B Uare mutually exclusive if A ∩ B = Φ = { }, A Bi.e. A and B do not have any common items.Using the sets above, we have that Mutually exclusive setsBC = {not wearing anything blue} = {a}is mutually exclusive from the set of F = {is a girl} = {b, c},i.e. BC ∩ F = Φ. Any set B and its complement BC, by definition, are alwaysmutually exclusive, or that BC ∩ B = Φ.
48. 48. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {Is a girl} = {b, c}M = {Is a boy} = {a, d} Andy Beth Cathy DanMutually Exclusive Sets and Basic CountingGiven two sets A and B, we say A and B Uare mutually exclusive if A ∩ B = Φ = { }, A Bi.e. A and B do not have any common items.Using the sets above, we have that Mutually exclusive setsBC = {not wearing anything blue} = {a}is mutually exclusive from the set of F = {is a girl} = {b, c},i.e. BC ∩ F = Φ.Any set B and its complement BC, by definition, are alwaysmutually exclusive, or that BC ∩ B = Φ. Nothing can possess anattribute while not possessing the same attribute simultaneously.
49. 49. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanFor the following discussion, we assume all the sets are finite.
50. 50. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanFor the following discussion, we assume all the sets are finite.Recall o(B) is the number of elements in the set B.
51. 51. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanFor the following discussion, we assume all the sets are finite.Recall o(B) is the number of elements in the set B.The set U = {a, b, c, d} can be split into two non-overlapping sets,
52. 52. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanFor the following discussion, we assume all the sets are finite.Recall o(B) is the number of elements in the set B.The set U = {a, b, c, d} can be split into two non-overlapping sets,R = {wearing red} = {a, b. d} with o(R) = 3, andRC = {not wearing any red} = {c} with o(RC) = 1.
53. 53. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanFor the following discussion, we assume all the sets are finite.Recall o(B) is the number of elements in the set B.The set U = {a, b, c, d} can be split into two non-overlapping sets,R = {wearing red} = {a, b. d} with o(R) = 3, andRC = {not wearing any red} = {c} with o(RC) = 1. U RC: R: d c b a
54. 54. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanFor the following discussion, we assume all the sets are finite.Recall o(B) is the number of elements in the set B.The set U = {a, b, c, d} can be split into two non-overlapping sets,R = {wearing red} = {a, b. d} with o(R) = 3, andRC = {not wearing any red} = {c} with o(RC) = 1.Each member of Farmer Andy’s family is in precisely one ofthese two sets. U RC: R: d c b a
55. 55. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanFor the following discussion, we assume all the sets are finite.Recall o(B) is the number of elements in the set B.The set U = {a, b, c, d} can be split into two non-overlapping sets,R = {wearing red} = {a, b. d} with o(R) = 3, andRC = {not wearing any red} = {c} with o(RC) = 1.Each member of Farmer Andy’s family is in precisely one ofthese two sets. U RC: R:The Complementary Counting Principle d bGiven U be the universal set and that A U, ∩ c athen o(A) + o(AC) = o(U),or that o(A) = o(U) – o(AC).
56. 56. More on the Language of SetsU = {Farmer Andy’s family} = {a, b, c, d}R = {wearing something red} = {a, b, d} Farmer Andy’s familyG = {wearing something green} = {a, c}B = {wearing something blue} = {b, c, d}F = {is a girl} = {b, c}M = {is a boy} = {a, d} Andy Beth Cathy DanFor the following discussion, we assume all the sets are finite.Recall o(B) is the number of elements in the set B.The set U = {a, b, c, d} can be split into two non-overlapping sets,R = {wearing red} = {a, b. d} with o(R) = 3, andRC = {not wearing any red} = {c} with o(RC) = 1.Each member of Farmer Andy’s family is in precisely one ofthese two sets. U RC: R:The Complementary Counting Principle d bGiven U be the universal set and that A U, ∩ c athen o(A) + o(AC) = o(U),or that o(A) = o(U) – o(AC). o(R) + o(RC) = o(U) = 4
57. 57. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors? Adult Child Teen Male 5 17 18 Female 15 26 19
58. 58. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer Female 15 26 19
59. 59. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer or we may observe that Female 15 26 19a non–minor is an adult.
60. 60. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer or we may observe that Female 15 26 19a non–minor is an adult. Since there are 5 + 15 = 20 adults outof the 100 viewers,
61. 61. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer or we may observe that Female 15 26 19a non–minor is an adult. Since there are 5 + 15 = 20 adults outof the 100 viewers, we see that there are 100 – 20 = 80 minors.
62. 62. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer or we may observe that Female 15 26 19a non–minor is an adult. Since there are 5 + 15 = 20 adults outof the 100 viewers, we see that there are 100 – 20 = 80 minors.Counting the Union of Two Sets
63. 63. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer or we may observe that Female 15 26 19a non–minor is an adult. Since there are 5 + 15 = 20 adults outof the 100 viewers, we see that there are 100 – 20 = 80 minors.Counting the Union of Two SetsRecall that G = {wearing something green} = {a, c} Uso o(G) = 2, G: a c
64. 64. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer or we may observe that Female 15 26 19a non–minor is an adult. Since there are 5 + 15 = 20 adults outof the 100 viewers, we see that there are 100 – 20 = 80 minors.Counting the Union of Two SetsRecall that G = {wearing something green} = {a, c} Uso o(G) = 2, F = {is a girl} = {b, c} so o(F) = 2, G: F: a c b d
65. 65. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer or we may observe that Female 15 26 19a non–minor is an adult. Since there are 5 + 15 = 20 adults outof the 100 viewers, we see that there are 100 – 20 = 80 minors.Counting the Union of Two SetsRecall that G = {wearing something green} = {a, c} Uso o(G) = 2, F = {is a girl} = {b, c} so o(F) = 2, G: F:and G U F = {a, b, c} with o(G U F) = 3. a c b d
66. 66. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer or we may observe that Female 15 26 19a non–minor is an adult. Since there are 5 + 15 = 20 adults outof the 100 viewers, we see that there are 100 – 20 = 80 minors.Counting the Union of Two SetsRecall that G = {wearing something green} = {a, c} Uso o(G) = 2, F = {is a girl} = {b, c} so o(F) = 2, G: F:and G U F = {a, b, c} with o(G U F) = 3.But note that o(G U F) = 3 ≠ o(G) + o(F) = 4. a c b d G U F = {a, b, c}
67. 67. More on the Language of SetsSometime it’s easier to count the complement of a set.Example C. The Following is a break down of 100 audiencesat a showing of “Mary Porter”. Minors are teens and children.How many in the audience are minors?We may add up the numbers of the Adult Child Teenteens and children to obtain the Male 5 17 18answer or we may observe that Female 15 26 19a non–minor is an adult. Since there are 5 + 15 = 20 adults outof the 100 viewers, we see that there are 100 – 20 = 80 minors.Counting the Union of Two SetsRecall that G = {wearing something green} = {a, c} Uso o(G) = 2, F = {is a girl} = {b, c} so o(F) = 2, G: F:and G U F = {a, b, c} with o(G U F) = 3.But note that o(G U F) = 3 ≠ o(G) + o(F) = 4. a c b dThe sum o(G) + o(F) is higher because their G U F = {a, b, c}common element “c” is counted twice as
68. 68. More on the Language of SetsThe Formula for Counting the Union (A U B)Given two sets and A and B,o(A U B) = o(A) + o(B) – o(A ∩ B)
69. 69. More on the Language of SetsThe Formula for Counting the Union (A U B)Given two sets and A and B,o(A U B) = o(A) + o(B) – o(A ∩ B)Example D. a. Given two squares each with an area of 10 cm 2,find the total area they cover if we overlay them as shown. 10 cm2 10 cm2 5 cm2 A B AUB
70. 70. More on the Language of SetsThe Formula for Counting the Union (A U B)Given two sets and A and B,o(A U B) = o(A) + o(B) – o(A ∩ B)Example D. a. Given two squares each with an area of 10 cm 2,find the total area they cover if we overlay them as shown. 10 cm2 10 cm2 5 cm2 A B AUBThe area the joint–squares covered is10 + 10 – (the overlapping portion)
71. 71. More on the Language of SetsThe Formula for Counting the Union (A U B)Given two sets and A and B,o(A U B) = o(A) + o(B) – o(A ∩ B)Example D. a. Given two squares each with an area of 10 cm 2,find the total area they cover if we overlay them as shown. 10 cm2 10 cm2 5 cm2 A B AUBThe area the joint–squares covered is10 + 10 – (the overlapping portion) total area of= 10 + 10 – 5 = 15 cm2, A U B = 15 cm 2
72. 72. More on the Language of SetsThe Formula for Counting the Union (A U B)Given two sets and A and B,o(A U B) = o(A) + o(B) – o(A ∩ B)Example D. a. Given two squares each with an area of 10 cm 2,find the total area they cover if we overlay them as shown. 10 cm2 10 cm2 5 cm2 A B AUBThe area the joint–squares covered is10 + 10 – (the overlapping portion) total area of= 10 + 10 – 5 = 15 cm2, A U B = 15 cm 2or that + is 5 + 10 = 15 cm2
73. 73. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test.
74. 74. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test.
75. 75. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test.
76. 76. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.
77. 77. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.How many students have passed at least one required test?
78. 78. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.How many students have passed at least one required test?Let A = {students who passed algebra},B = {students who passed biology},
79. 79. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.How many students have passed at least one required test?Let A = {students who passed algebra},B = {students who passed biology}, thenA ∩ B = {students who passed both tests} and
80. 80. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.How many students have passed at least one required test?Let A = {students who passed algebra},B = {students who passed biology}, thenA ∩ B = {students who passed both tests} andA U B = {students who passed at least one test}.
81. 81. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.How many students have passed at least one required test?Let A = {students who passed algebra},B = {students who passed biology}, thenA ∩ B = {students who passed both tests} andA U B = {students who passed at least one test}. o(A)=10 o(A ∩ B)=5 o(B)=10
82. 82. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.How many students have passed at least one required test?Let A = {students who passed algebra},B = {students who passed biology}, thenA ∩ B = {students who passed both tests} andA U B = {students who passed at least one test}.Hence o(A U B) = o(A) + o(B) – o(A ∩ B) = o(A)=10 o(A ∩ B)=5 o(B)=10
83. 83. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.How many students have passed at least one required test?Let A = {students who passed algebra},B = {students who passed biology}, thenA ∩ B = {students who passed both tests} andA U B = {students who passed at least one test}.Hence o(A U B) = o(A) + o(B) – o(A ∩ B) = 10 + 10 – 5 = 15 o(A)=10 o(A ∩ B)=5 o(B)=10
84. 84. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.How many students have passed at least one required test?Let A = {students who passed algebra},B = {students who passed biology}, thenA ∩ B = {students who passed both tests} andA U B = {students who passed at least one test}.Hence o(A U B) = o(A) + o(B) – o(A ∩ B) = 10 + 10 – 5 = 15or that there are 15 peoplewho passed at least one test. o(A)=10 o(A ∩ B)=5 o(B)=10 o(A U B) = o(A) + o(B) – o(A ∩ B) = 15
85. 85. More on the Language of Setsb. To graduate from a school, the student must pass thealgebra test and the biology test. In a class, there are tenstudents that have passed the algebra test. There are also tenstudents that have passed the biology test. In these groups,five students passed both the algebra and the biology tests.How many students have passed at least one required test?Let A = {students who passed algebra},B = {students who passed biology}, thenA ∩ B = {students who passed both tests} andA U B = {students who passed at least one test}.Hence o(A U B) = o(A) + o(B) – o(A ∩ B) = 10 + 10 – 5 = 15or that there are 15 peoplewho passed at least one test. o(A)=10 o(A ∩ B)=5 o(B)=10We may use the countingformula for the union todeduce the order of A. o(A U B) = o(A) + o(B) – o(A ∩ B) = 15
86. 86. More on the Language of SetsExample E. As in example D, suppose there are 14 studentsthat have passed the biology test and in this group,5 students passed both the algebra and the biology.There are 20 students that have passed at least one test.How may students have passed only algebra and need to passbiology? How many students have passed the algebra test?
87. 87. More on the Language of SetsExample E. As in example D, suppose there are 14 studentsthat have passed the biology test and in this group,5 students passed both the algebra and the biology.There are 20 students that have passed at least one test.How may students have passed only algebra and need to passbiology? How many students have passed the algebra test?Again A = {passed algebra},B = {passed biology},A ∩ B = {passed both tests} andA U B = {passed at least one test}.We’ll use the diagram method.
88. 88. More on the Language of SetsExample E. As in example D, suppose there are 14 studentsthat have passed the biology test and in this group,5 students passed both the algebra and the biology.There are 20 students that have passed at least one test.How may students have passed only algebra and need to passbiology? How many students have passed the algebra test?Again A = {passed algebra}, AB = {passed biology},A ∩ B = {passed both tests} andA U B = {passed at least one test}.We’ll use the diagram method.
89. 89. More on the Language of SetsExample E. As in example D, suppose there are 14 studentsthat have passed the biology test and in this group,5 students passed both the algebra and the biology.There are 20 students that have passed at least one test.How may students have passed only algebra and need to passbiology? How many students have passed the algebra test?Again A = {passed algebra}, A BB = {passed biology},A ∩ B = {passed both tests} andA U B = {passed at least one test}.We’ll use the diagram method.
90. 90. More on the Language of SetsExample E. As in example D, suppose there are 14 studentsthat have passed the biology test and in this group,5 students passed both the algebra and the biology.There are 20 students that have passed at least one test.How may students have passed only algebra and need to passbiology? How many students have passed the algebra test?Again A = {passed algebra}, A BB = {passed biology},A ∩ B = {passed both tests} andA U B = {passed at least one test}.We’ll use the diagram method.
91. 91. More on the Language of SetsExample E. As in example D, suppose there are 14 studentsthat have passed the biology test and in this group,5 students passed both the algebra and the biology.There are 20 students that have passed at least one test.How may students have passed only algebra and need to passbiology? How many students have passed the algebra test?Again A = {passed algebra}, A BB = {passed biology},A ∩ B = {passed both tests} andA U B = {passed at least one test}.We’ll use the diagram method.Imagine the 20 students of AUB are total number of students:standing in two overlapping squares o(A U B) = 20as shown.
92. 92. More on the Language of SetsExample E. As in example D, suppose there are 14 studentsthat have passed the biology test and in this group,5 students passed both the algebra and the biology.There are 20 students that have passed at least one test.How may students have passed only algebra and need to passbiology? How many students have passed the algebra test?Again A = {passed algebra}, A BB = {passed biology},A ∩ B = {passed both tests} and 14A U B = {passed at least one test}.We’ll use the diagram method.Imagine the 20 students of AUB are total number of students:standing in two overlapping squares o(A U B) = 20as shown. There are 14 studentsinside B.
93. 93. More on the Language of SetsExample E. As in example D, suppose there are 14 studentsthat have passed the biology test and in this group,5 students passed both the algebra and the biology.There are 20 students that have passed at least one test.How may students have passed only algebra and need to passbiology? How many students have passed the algebra test?Again A = {passed algebra}, A BB = {passed biology},A ∩ B = {passed both tests} and 6 14A U B = {passed at least one test}.We’ll use the diagram method.Imagine the 20 students of AUB are total number of students:standing in two overlapping squares o(A U B) = 20as shown. There are 14 studentsinside B. So there are 6 outside who have passed only algebra
94. 94. More on the Language of SetsExample E. As in example D, suppose there are 14 studentsthat have passed the biology test and in this group,5 students passed both the algebra and the biology.There are 20 students that have passed at least one test.How may students have passed only algebra and need to passbiology? How many students have passed the algebra test?Again A = {passed algebra}, A BB = {passed biology},A ∩ B = {passed both tests} and 6 5 14A U B = {passed at least one test}.We’ll use the diagram method.Imagine the 20 students of AUB are total number of students:standing in two overlapping squares o(A U B) = 20as shown. There are 14 studentsinside B. So there are 6 outside who have passed only algebraand need to pass biology and o(A) = 6 + 5 = 11.
95. 95. More on the Language of SetsOr we may solve this algebraically by setting,o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5in the union–formula o(A U B) = o(A) + o(B) – o(A ∩ B),
96. 96. More on the Language of SetsOr we may solve this algebraically by setting,o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5in the union–formula o(A U B) = o(A) + o(B) – o(A ∩ B),we have that20 = o(A) + 14 – 5, or11 = o(A)
97. 97. More on the Language of SetsOr we may solve this algebraically by setting,o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5in the union–formula o(A U B) = o(A) + o(B) – o(A ∩ B),we have that20 = o(A) + 14 – 5, or11 = o(A)Hence there are 11 students that passed algebra and out ofwhich 9 still have to pass the biology test.
98. 98. More on the Language of SetsOr we may solve this algebraically by setting,o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5in the union–formula o(A U B) = o(A) + o(B) – o(A ∩ B),we have that20 = o(A) + 14 – 5, or11 = o(A)Hence there are 11 students that passed algebra and out ofwhich 9 still have to pass the biology test.Finally we note that if A and B are mutually exclusive, theno(A U B) = o(A) + o(B) since o(A ∩ B) = o(Φ) = 0.
99. 99. More on the Language of SetsOr we may solve this algebraically by setting,o(A U B) = 20, o(B) = 14, o(A ∩ B) = 5in the union–formula o(A U B) = o(A) + o(B) – o(A ∩ B),we have that20 = o(A) + 14 – 5, or11 = o(A)Hence there are 11 students that passed algebra and out ofwhich 9 still have to pass the biology test.Finally we note that if A and B are mutually exclusive, theno(A U B) = o(A) + o(B) since o(A ∩ B) = o(Φ) = 0. It A and B does not overlap A U B, then A B Area of A U B = Area(A) + Area(B) or o(A U B) = o(A) + o(B) AUB
100. 100. More on the Language of SetsIf I and J are two sets of intervals, then I U J is the set obtainedby gluing the two intervals together, and that I ∩ J is theoverlapping part of the two intervals.
101. 101. More on the Language of SetsIf I and J are two sets of intervals, then I U J is the set obtainedby gluing the two intervals together, and that I ∩ J is theoverlapping part of the two intervals.Example E. Given intervals I, J , and K, perform the followingset operation. Give the answers in intervals and in inequalities. –4 –1I: 0 J = {x > –2} K= {–3 < x ≤ 1}a. K U J
102. 102. More on the Language of SetsIf I and J are two sets of intervals, then I U J is the set obtainedby gluing the two intervals together, and that I ∩ J is theoverlapping part of the two intervals.Example E. Given intervals I, J , and K, perform the followingset operation. Give the answers in intervals and in inequalities. –4 –1I: 0 J = {x > –2} K= {–3 < x ≤ 1}a. K U J –3 0 K
103. 103. More on the Language of SetsIf I and J are two sets of intervals, then I U J is the set obtainedby gluing the two intervals together, and that I ∩ J is theoverlapping part of the two intervals.Example E. Given intervals I, J , and K, perform the followingset operation. Give the answers in intervals and in inequalities. –4 –1I: 0 J = {x > –2} K= {–3 < x ≤ 1}a. K U J –3 –2 1 J 0 K
104. 104. More on the Language of SetsIf I and J are two sets of intervals, then I U J is the set obtainedby gluing the two intervals together, and that I ∩ J is theoverlapping part of the two intervals.Example E. Given intervals I, J , and K, perform the followingset operation. Give the answers in intervals and in inequalities. –4 –1I: 0 J = {x > –2} K= {–3 < x ≤ 1}a. K U J –3 –2 1 J –3 K 0 so K U J is 0K U J = {–3 < x} or (–3, ∞).
105. 105. More on the Language of SetsIf I and J are two sets of intervals, then I U J is the set obtainedby gluing the two intervals together, and that I ∩ J is theoverlapping part of the two intervals.Example E. Given intervals I, J , and K, perform the followingset operation. Give the answers in intervals and in inequalities. –4 –1I: 0 J = {x > –2} K= {–3 < x ≤ 1}a. K U J –3 –2 1 J –3 K 0 so K U J is 0K U J = {–3 < x} or (–3, ∞).b. I ∩ K
106. 106. More on the Language of SetsIf I and J are two sets of intervals, then I U J is the set obtainedby gluing the two intervals together, and that I ∩ J is theoverlapping part of the two intervals.Example E. Given intervals I, J , and K, perform the followingset operation. Give the answers in intervals and in inequalities. –4 –1I: 0 J = {x > –2} K= {–3 < x ≤ 1}a. K U J –3 –2 1 J –3 K 0 so K U J is 0K U J = {–3 < x} or (–3, ∞).b. I ∩ K –4 I
107. 107. More on the Language of SetsIf I and J are two sets of intervals, then I U J is the set obtainedby gluing the two intervals together, and that I ∩ J is theoverlapping part of the two intervals.Example E. Given intervals I, J , and K, perform the followingset operation. Give the answers in intervals and in inequalities. –4 –1I: 0 J = {x > –2} K= {–3 < x ≤ 1}a. K U J –3 –2 1 J –3 K 0 so K U J is 0K U J = {–3 < x} or (–3, ∞).b. I ∩ K –4 –3 –1 K 1 0 I
108. 108. More on the Language of SetsIf I and J are two sets of intervals, then I U J is the set obtainedby gluing the two intervals together, and that I ∩ J is theoverlapping part of the two intervals.Example E. Given intervals I, J , and K, perform the followingset operation. Give the answers in intervals and in inequalities. –4 –1I: 0 J = {x > –2} K= {–3 < x ≤ 1}a. K U J –3 –2 1 J –3 K 0 so K U J is 0K U J = {–3 < x} or (–3, ∞).b. I ∩ K –4 –3 –1 K 1 0 IThe common or overlapping portion is shown.I ∩ K = {–3 < x < –1} or (–3, –1).
109. 109. More on the Language of SetsExercise A. From the photo of Farmer Andy’s family in theirnew dresses, we define the following setsU = { the family} Farmer Andy’s family in new dressesR = {red}G = {green}B = {blue}F = {female}M = {male} Andy Beth Cathy DanT = {has triangular decoration} C = {has circular decoration}Exercise A.1.a. List each set defined above and pick out all the mutuallyexclusive pairs by inspection.List the elements of the following sets. Name and describeeach set in the form of {x | x is ###}.b. FC c. MC d. CC e. TC f. GC g. UCh. T U F i. F ∩ M j. B U C k. C U FC l. GC ∩ T k. TC ∩ CC
110. 110. More on the Language of Sets2. From Example C. we have the table as shown.Let A = {Adult} ,T = {Teen}, Adult Child TeenC = {Child}, F = {Female}, Male 5 17 18M = {Male}, Female 15 26 19Which pair of sets are mutually exclusively?Name and describe each of the following sets in the form of{x | x is ###} and find its order.b. TC c. CC d. MC e. FC f. ACg. A U F h. T ∩ C i. F ∩ MC j. AC U C k. CC U FC3. Given intervals I, J , and K, do the set operation. Draw.Put the answer in the interval and the inequality notations. –8 –2 0 J: x < –3½ K: –5 ≤ x < 2I:a. K U J b. K ∩ I c. I U J d. K U Ie. I ∩ J f. K ∩ J g. (K ∩ J) U I h. (K U J) ∩ I
111. 111. More on the Language of Sets4. Given following shapes and their areas, find the total area ifwe glue them together as shown.a. 10 cm2 20 cm2 5 cm2 A B AUBb. 8 12 5 A B AUBc. Each overlap is 1 A 20 A B 20 C B AUBUC C 20
112. 112. More on the Language of Sets5. Given following shapes and their two areas and we glue themtogether as shown, find the missing area.a. Area of A U B = 31 A 20 cm2 5 cm2 Area of A = ? Bb. 8 B Area of A U B = 19 5 A Area of B = ?c. A A B Area of C B A U B U C = 45 C Each overlap is 1 Area of each = ? A =B=C
113. 113. More on the Language of Sets7. Farmer Andy produces nuts and sells them in three types ofbags. These are: peanuts only, cashews only, andmixed peanuts–cashews. We bought many bags of each type.Draw a diagram that reflects each of the following questions.Answer the question if possible. If not, explain why not.a. Out of the bags we bought, 30 contain peanut and12 contain cashews only. How many bags we bought in total?How many peanuts only bags did we buy?How many mixed begs did we buy?b. Out of the bags we bought, 30 contain peanut only and12 contain cashews only. How many bags we bought in total?c. Out of the bags we bought, 30 contain peanut, 12 containcashews and there are 6 of mixed type. How many bags didwe buy in total and how may of each type did we buy?d. Out of the 40 bags we bought, 30 contain peanut and 12 areof mixed type. How many bags of each type did we buy?
114. 114. More on the Language of Sets8. Everyone in an extended family of 100 people have eithergenetic traits A or B, or both. Answer each of the followingquestions with the given information. Define and draw each set.a. There are 85 people who have trait A. Out of the people withtrait A , 27 people have both traits A an d B. How many people ofeach type did we have?b. There are 85 people who have trait A and there 27 peoplewith B. How many people of each type did we have?c. Is it possible to have 85 people who don’t have trait A andthere 15 people with both traits A, B? What may weconclude?d. Is it possible to have 85 people who don’t have trait B andthere 16 people with both traits A, B? What may weconclude?