1. Stating the problem: Identify which sets among A, B, C, D, E, F, G, H are equal. Step 1: Solve equations for A and B. - For A: $x^2 - 4x + 3=0$. Factor: $(x-3)(x-1)=0$, so $A = \{1,3\}$. - For B: $x^2 - 3x + 2=0$. Factor: $(x-2)(x-1)=0$, so $B = \{1,2\}$. Step 2: List given sets explicitly where possible. - $C = \{x \in \mathbb{N} : x < 3\} = \{1,2\}$. - $E = \{1,2\}$. - $F = \{1,2,1\} = \{1,2\}$ (no repeats in sets). - $G = \{3,1\} = \{1,3\}$. - $D = \{x \in \mathbb{N} : x \text{ is odd}, x < 5\} = \{1,3\}$. - $H = \{1,1,3\} = \{1,3\}$. Step 3: Compare sets for equality. - $A = \{1,3\}$; equal to $G$, $D$, and $H$. - $B = \{1,2\}$; equal to $C$, $E$, and $F$. --- 2. List elements and find equal sets. - Universal set $U = \{a,b,c,...,y,z\}$. - $A = \{x \mid x \text{ is a vowel}\} = \{a,e,i,o,u\}$. - $C = \{x \mid x \text{ precedes } f \text{ in alphabet}\} = \{a,b,c,d,e\}$. - $B = \{x \mid x \text{ is in } "little"\} = \{l,i,t,e\}$ (distinct letters). - $D = \{x \mid x \text{ is in } "title"\} = \{t,i,l,e\}$. Check equalities: - $B = \{l,i,t,e\}$ and $D = \{t,i,l,e\}$ are equal (order and repetition irrelevant). - No other sets are equal. --- 3. Condition testing for $X$: - Given: $A=\{1,2,...,9\}$, $B=\{2,4,6,8\}$, $C=\{1,3,5,7,9\}$, $D=\{3,4,5\}$, $E=\{3,5\}$. (a) $X$ and $B$ are disjoint: $X \cap B = \emptyset$. So $X$ cannot contain $2,4,6,8$. Possible candidates: $C$, $D$, $E$ (check if these have no elements of $B$). - $C=\{1,3,5,7,9\}$ no overlap with $B$, so $C$ can be $X$. - $D=\{3,4,5\}$ contains $4 \in B$, so no. - $E=\{3,5\}$ no overlap with $B$, so yes. (b) $X \subseteq D$ but $X \nsubseteq B$. Since $B = \{2,4,6,8\}$, none of $D$ fully inside $B$. So any subset of $D$ that is NOT contained in $B$ satisfies. Possible $X$: $D$ itself or subsets like $\{3\}, \{5\}, \{3,5\}$. (c) $X \subseteq A$ but $X \nsubseteq C$. Since $C = \{1,3,5,7,9\}$, any subset of $A$ that is not a subset of $C$ qualifies. For example $B$ since it contains 2 which is not in $C$. (d) $X \subseteq C$ but $X \nsubseteq A$. Since $C \subseteq A$, every subset of $C$ must be subset of $A$. No such $X$ exists. --- 4. Universal set $U=\{1,...,9\}$, sets: $A=\{1,2,5,6\}$, $B=\{2,5,7\}$, $C=\{1,3,5,7,9\}$. (a) Compute intersections: - $A \cap B = \{2,5\}$. - $A \cap C = \{1,5\}$. (b) Unions: - $A \cup B = \{1,2,5,6,7\}$. - $B \cup C = \{1,2,3,5,7,9\}$. (c) Complements (relative to $U$): - $A^C = U \setminus A = \{3,4,7,8,9\}$. - $C^C = U \setminus C = \{2,4,6,8\}$. (d) Differences: - $A \setminus B = A \cap B^C = \{1,6\}$. - $A \setminus C = \{2,6\}$. (e) Symmetric differences: - $A \oplus B = (A \setminus B) \cup (B \setminus A) = \{1,6,7\}$. - $A \oplus C = \{2,3,6,7,9\}$. (f) Set operations: - $(A \cup C) \setminus B = \{1,2,3,4,5,6,7,9\} \setminus \{2,5,7\} = \{1,3,4,6,9\}$. - $B \oplus C = \{2,3,6,9\}$. - $(B \oplus C) \setminus A = \{2,3,6,9\} \setminus \{1,2,5,6\} = \{3,9\}$. --- 5. Given $A \setminus B = A \cap B^C$. We want $A \cup B$ as a formula using intersection and complement only. By De Morgan's law: $$ A \cup B = (A^C \cap B^C)^C $$ --- 6. Shade sets from Venn diagram: (a) $A \setminus (B \cup C) = A \cap (B \cup C)^C$. (b) $A^C \cap (B \cup C)$. (c) $A^C \cap (C \setminus B) = A^C \cap C \cap B^C$. --- 7. Write duals (interchanging union and intersection, empty set and universal set): (a) Original: $A = (B^C \cap A) \cup (A \cap B)$. Dual: $A = (B^C \cup A) \cap (A \cup B)$. (b) Original: $ (A \cap B) \cup (A^C \cap B) \cup (A \cap B^C) \cup (A^C \cap B^C) = U$. Dual: $(A \cup B) \cap (A^C \cup B) \cap (A \cup B^C) \cap (A^C \cup B^C) = \emptyset$. --- 8. Finite or infinite? (a) Lines parallel to $x$-axis: Infinite (infinite number of such lines). (b) Letters in English alphabet: Finite (26 letters). (c) Integers multiples of 5: Infinite. (d) Animals living on earth: Infinite (due to biodiversity). Final answers are integrated above stepwise.