Math Problem Statement

If U = {1,2,3,4,5,6}, A = {1,2,3,4}, B = {3,5,6}, C = {3,4,5,6}, find out (i) (A∪B)' (ii) A'∩B' (iii) What do you notice (iv) Prove that (A∩B)' = A'∪B' (v) Find AΔB.

Solution

Let's analyze and solve the given problem step-by-step:

Provided Information

  1. Universal Set, UU: {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}
  2. Set AA: {1,2,3,4}\{1, 2, 3, 4\}
  3. Set BB: {3,5,6}\{3, 5, 6\}
  4. Set CC: {3,4,5,6}\{3, 4, 5, 6\}

The questions to solve are:

  1. (i)(i) Find (AB)(A \cup B)'.
  2. (ii)(ii) Find ABA' \cap B'.
  3. (iii)(iii) What do you notice?
  4. (iv)(iv) Prove that (AB)=AB(A \cap B)' = A' \cup B'.
  5. (v)(v) Find AΔBA \Delta B.

Solutions

  1. (i) (AB)(A \cup B)':

    • ABA \cup B represents the union of sets AA and BB, which includes all elements in either AA or BB. AB={1,2,3,4,5,6}A \cup B = \{1, 2, 3, 4, 5, 6\}
    • The complement, (AB)(A \cup B)', includes all elements in UU but not in ABA \cup B. (AB)=U(AB)={}(A \cup B)' = U - (A \cup B) = \{\}
  2. (ii) ABA' \cap B':

    • AA' is the complement of AA, meaning all elements in UU but not in AA. A={5,6}A' = \{5, 6\}
    • BB' is the complement of BB, meaning all elements in UU but not in BB. B={1,2,4}B' = \{1, 2, 4\}
    • The intersection, ABA' \cap B', includes all elements that are in both AA' and BB'. AB={}A' \cap B' = \{\}
  3. (iii) What do you notice?

    • Notice that (AB)=AB(A \cup B)' = A' \cap B'. This is a standard property in set theory, known as De Morgan's Law.
  4. (iv) Prove that (AB)=AB(A \cap B)' = A' \cup B':

    • By De Morgan's Law, the complement of the intersection of two sets is equal to the union of their complements: (AB)=AB(A \cap B)' = A' \cup B'
    • Let's verify:
      • AB={3}A \cap B = \{3\}
      • A={5,6}A' = \{5, 6\} and B={1,2,4}B' = \{1, 2, 4\}
      • AB={1,2,4,5,6}A' \cup B' = \{1, 2, 4, 5, 6\}
      • (AB)=U{3}={1,2,4,5,6}(A \cap B)' = U - \{3\} = \{1, 2, 4, 5, 6\}
      • This shows that (AB)=AB(A \cap B)' = A' \cup B'.
  5. (v) Find AΔBA \Delta B:

    • The symmetric difference AΔBA \Delta B includes elements in ABA \cup B but not in ABA \cap B. AΔB=(AB)(AB)A \Delta B = (A \cup B) - (A \cap B)
    • AB={1,2,3,4,5,6}A \cup B = \{1, 2, 3, 4, 5, 6\}
    • AB={3}A \cap B = \{3\}
    • Therefore: AΔB={1,2,4,5,6}A \Delta B = \{1, 2, 4, 5, 6\}

Would you like me to expand further or provide more details? Here are some related questions for deeper understanding:

  1. What is the relation between union and intersection of sets?
  2. How can we represent De Morgan’s laws using Venn diagrams?
  3. Can you prove that (AΔB)=AB(A \Delta B)' = A' \cap B'?
  4. What are the applications of symmetric difference in probability?
  5. How does the complement operation affect the cardinality of a set?

Tip: Always try using Venn diagrams to visualize set operations like union, intersection, and complements for a clearer understanding.

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Math Problem Analysis

Mathematical Concepts

Set Theory
Union
Intersection
Complement
De Morgan's Law
Symmetric Difference

Formulas

A ∪ B = {x | x ∈ A or x ∈ B}
A ∩ B = {x | x ∈ A and x ∈ B}
(A ∪ B)' = U - (A ∪ B)
(A ∩ B)' = U - (A ∩ B)
A Δ B = (A ∪ B) - (A ∩ B)

Theorems

De Morgan's Law: (A ∪ B)' = A' ∩ B' and (A ∩ B)' = A' ∪ B'

Suitable Grade Level

Grades 9-12