1. Consider the problem involving set operations and with conditions: Given: - Universal set $U = \{ x \mid x \in \mathbb{N}, x \leq 9 \}$ - Sets: $A = \{7, 10, 13, 16, 19\}$ (though 10 and above not in $U$ - context needed) We focus on the operations: 2. Find $(A - B)'$ where $B = \{1,2,3,4\}$ and $A = \{7,10,13,16,19\}$: - $A-B = A$ because $B$ elements not in $A$ - Hence $(A-B)' = U \setminus (A-B)$ but since $U$ is $\{1..9\}$ and $A$ contains elements not in $U$, $A-B$ in $U$ is $\{7\}$ and $(A-B)'=U \setminus \{7\} = \{1,2,3,4,5,6,8,9\}$. 3. For $U = \{ x \mid x \in \mathbb{N}, x \leq 8 \}$, and sets - $A=\{1,2,3,4\}, B=\{3,4,5,6\}, C=\{5,6,7,8\}$, Find: (1) $(A \cup B \cup C)' = U \setminus (A \cup B \cup C)$ - $A \cup B \cup C = \{1,2,3,4,5,6,7,8\} = U$ - So $(A \cup B \cup C)' = \emptyset$ (2) $(A \cap B')' \cap A'$: - $B' = U \setminus B = \{1,2,7,8\}$ - $A \cap B' = \{1,2,3,4\} \cap \{1,2,7,8\} = \{1,2\}$ - $(A \cap B')' = U \setminus \{1,2\} = \{3,4,5,6,7,8\}$ - $A' = U \setminus A = \{5,6,7,8\}$ - Intersection: $(A \cap B')' \cap A' = \{3,4,5,6,7,8\} \cap \{5,6,7,8\} = \{5,6,7,8\}$ (3) $(A \Delta B) \Delta C'$: - $A \Delta B = (A \setminus B) \cup (B \setminus A) = \{1,2\} \cup \{5,6\} = \{1,2,5,6\}$ - $C' = U \setminus C = \{1,2,3,4\}$ - Finally $(A \Delta B) \Delta C' = \{1,2,5,6\} \Delta \{1,2,3,4\} = \{3,4,5,6\}$ 4. Given $A' = \{4,5,6,7\}$, $U = \{ x \mid x \in \mathbb{N}, x < 10 \}$, and - $A = \{x \mid x \text{ even, } 1 \leq x \leq 10\} = \{2,4,6,8,10\}$ - $B = \{x \mid x \text{ multiple of 3, } 1 \leq x \leq 10\} = \{3,6,9\}$ Calculate $A \Delta B$ and $A \times (B \Delta A)$: - $B \Delta A = (B \setminus A) \cup (A \setminus B) = \{3,9\} \cup \{2,4,8,10\} = \{2,3,4,8,9,10\}$ - $A \times (B \Delta A)$ is the Cartesian product $\{ (a,b) \mid a \in A, b \in B \Delta A \}$. 5. Given $n(A \setminus B) = 6$, $n(B \setminus A) = 7$, and $n(A) = 8$ Find: (a) $n(B)$: - From $n(A) = n(A \cap B) + n(A \setminus B)$, so $n(A \cap B) = 8 - 6 = 2$ - Since $n(B) = n(B \setminus A) + n(A \cap B) = 7 + 2 = 9$ (b) $n(A \cap B) = 2$ (from above) (c) $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 8 + 9 - 2 = 15$ Summary answers: - $(A - B)' = \{1,2,3,4,5,6,8,9\}$ - $(A \cup B \cup C)' = \emptyset$ - $(A \cap B')' \cap A' = \{5,6,7,8\}$ - $(A \Delta B) \Delta C' = \{3,4,5,6\}$ - $A \Delta B = \{2,3,4,8,9,10\}$ - $n(B) = 9$ - $n(A \cap B) = 2$ - $n(A \cup B) = 15$