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Let \( A = \{1, 4, 7, 8\} \), \( B = \{4, 6, 8, 9\} \), and \( C = \{3, 4, 5, 7\} \) be three subsets of a universal set \( U = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \). Find:

(a) \( A \cap (B - C) \)

(b) \( A - (B - C) \)

(c) \( A \cup B^c \)

(d) \( A^c \cap (B^c - C^c) \)

(e) \( A \cup (B \cap C^c) \)

(f) [tex]\( (A - C) \cup (C - A) \)[/tex]

Sagot :

GinnyAnswer
Let's solve each part of the question step-by-step:

(a) \( A \cap (B - C) \)

First, we need to find the difference between sets \( B \) and \( C \):
[tex]\[ B - C = \{ x \mid x \in B \text{ and } x \notin C \} = \{6, 8, 9\} \][/tex]

Now, we find the intersection of \( A \) with \( B - C \):
[tex]\[ A \cap (B - C) = \{ x \mid x \in A \text{ and } x \in (B - C) \} = \{8\} \][/tex]

Thus, the answer is:
[tex]\[ \boxed{\{8\}} \][/tex]

(b) \( A - (B - C) \)

Using the result from part (a) for \( B - C = \{6, 8, 9\} \), we find the difference between sets \( A \) and \( B - C \):
[tex]\[ A - (B - C) = \{ x \mid x \in A \text{ and } x \notin (B - C) \} = \{1, 4, 7\} \][/tex]

Thus, the answer is:
[tex]\[ \boxed{\{1, 4, 7\}} \][/tex]

(d) \( A^c \cap (B^c - C^c) \)

First, we find the complements of \( A \), \( B \), and \( C \) in the universal set \( U \):
[tex]\[ A^c = U - A = \{2, 3, 5, 6, 9\} \][/tex]
[tex]\[ B^c = U - B = \{1, 2, 3, 5, 7\} \][/tex]
[tex]\[ C^c = U - C = \{1, 2, 6, 8, 9\} \][/tex]

Next, we find the difference between \( B^c \) and \( C^c \):
[tex]\[ B^c - C^c = \{ x \mid x \in B^c \text{ and } x \notin C^c \} = \{3, 5\} \][/tex]

Now, we find the intersection of \( A^c \) with \( B^c - C^c \):
[tex]\[ A^c \cap (B^c - C^c) = \{ x \mid x \in A^c \text{ and } x \in (B^c - C^c) \} = \{3, 5\} \][/tex]

Thus, the answer is:
[tex]\[ \boxed{\{3, 5\}} \][/tex]

(e) \( A \cup (B \cap C^c) \)

Using the complement of \( C \) found previously \( C^c = \{1, 2, 6, 8, 9\} \), we find the intersection of \( B \) with \( C^c \):
[tex]\[ B \cap C^c = \{ x \mid x \in B \text{ and } x \in C^c \} = \{6, 8, 9\} \][/tex]

Now, we find the union of \( A \) with \( B \cap C^c \):
[tex]\[ A \cup (B \cap C^c) = \{ x \mid x \in A \text{ or } x \in (B \cap C^c) \} = \{1, 4, 6, 7, 8, 9\} \][/tex]

Thus, the answer is:
[tex]\[ \boxed{\{1, 4, 6, 7, 8, 9\}} \][/tex]

(c) \( A \cup B^c \)

Using the complement of \( B \) found previously \( B^c = \{1, 2, 3, 5, 7\} \), we find the union of \( A \) with \( B^c \):
[tex]\[ A \cup B^c = \{ x \mid x \in A \text{ or } x \in B^c \} = \{1, 2, 3, 4, 5, 7, 8\} \][/tex]

Thus, the answer is:
[tex]\[ \boxed{\{1, 2, 3, 4, 5, 7, 8\}} \][/tex]

(f) \( (A - C) \cup (C - A) \)

First, we find the difference between sets \( A \) and \( C \):
[tex]\[ A - C = \{ x \mid x \in A \text{ and } x \notin C \} = \{1, 8\} \][/tex]

Next, we find the difference between sets \( C \) and \( A \):
[tex]\[ C - A = \{ x \mid x \in C \text{ and } x \notin A \} = \{3, 5\} \][/tex]

Now, we find the union of \( A - C \) and \( C - A \):
[tex]\[ (A - C) \cup (C - A) = \{1, 8\} \cup \{3, 5\} = \{1, 3, 5, 8\} \][/tex]

Thus, the answer is:
[tex]\[ \boxed{\{1, 3, 5, 8\}} \][/tex]
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