Question

let u be the universal set, where u = {1,2,3,4,5,6}. let sets a and b be subsets of u, where set a = {5,6} and set b = {3,4,5}. in this problem you will find (a∪b) ∩ b by performing a series of steps using set order of operations. it is typically easier to complete each set operation one at a time and write down the result before moving on to complete the next operation. the operations are laid out below to lead you through completing the operations one at a time. b = { } a∪b = { } (a∪b) = { } (a∪b) ∩ b = { }

Explanation:

Step1: Find complement of B ($B'$)

The complement of a set \( B \) with respect to the universal set \( U \) is the set of all elements in \( U \) that are not in \( B \). Given \( U = \{1, 2, 3, 4, 5, 6\} \) and \( B = \{3, 4, 5\} \), we remove the elements of \( B \) from \( U \). So \( B' = U - B = \{1, 2, 6\} \).

Step2: Find union of A and \( B' \) (\( A \cup B' \))

The union of two sets \( A \) and \( B' \) is the set of all elements that are in \( A \) or in \( B' \). Given \( A = \{5, 6\} \) and \( B' = \{1, 2, 6\} \), we combine the elements (removing duplicates). So \( A \cup B' = \{1, 2, 5, 6\} \).

Step3: Find complement of \( A \cup B' \) (\( (A \cup B')' \))

The complement of \( A \cup B' \) with respect to \( U \) is the set of all elements in \( U \) that are not in \( A \cup B' \). Given \( U = \{1, 2, 3, 4, 5, 6\} \) and \( A \cup B' = \{1, 2, 5, 6\} \), we remove the elements of \( A \cup B' \) from \( U \). So \( (A \cup B')' = \{3, 4\} \).

Step4: Find intersection of \( (A \cup B')' \) and B (\( (A \cup B')' \cap B \))

The intersection of two sets \( (A \cup B')' \) and \( B \) is the set of all elements that are in both \( (A \cup B')' \) and \( B \). Given \( (A \cup B')' = \{3, 4\} \) and \( B = \{3, 4, 5\} \), we find the common elements. So \( (A \cup B')' \cap B = \{3, 4\} \).

Answer:

s (for each step):

• \( B' = \{1, 2, 6\} \)
• \( A \cup B' = \{1, 2, 5, 6\} \)
• \( (A \cup B')' = \{3, 4\} \)
• \( (A \cup B')' \cap B = \{3, 4\} \)