1. **Problem 1: Construct the truth table for the biconditional statement:** "It is raining if and only if the ground is wet." The biconditional statement $p \leftrightarrow q$ is true when both $p$ and $q$ have the same truth value. | It is raining ($p$) | Ground is wet ($q$) | $p \leftrightarrow q$ | |---------------------|---------------------|-----------------------| | T | T | T | | T | F | F | | F | T | F | | F | F | T | 2. **Problem 2: Write the following sets:** (a) First ten composite numbers: - Tabular form: $\{4, 6, 8, 9, 10, 12, 14, 15, 16, 18\}$ - Descriptive form: The first ten composite numbers are 4, 6, 8, 9, 10, 12, 14, 15, 16, and 18. - Set builder form: $\{x \in \mathbb{N} \mid x \text{ is composite and } 4 \leq x \leq 18\}$ (b) Prime numbers between 10 and 50: - Tabular form: $\{11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$ - Descriptive form: The prime numbers between 10 and 50 are 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. - Set builder form: $\{x \in \mathbb{N} \mid 10 < x < 50 \text{ and } x \text{ is prime}\}$ (c) Factors of 50: - Tabular form: $\{1, 2, 5, 10, 25, 50\}$ - Descriptive form: The factors of 50 are 1, 2, 5, 10, 25, and 50. - Set builder form: $\{x \in \mathbb{N} \mid x \mid 50\}$ (where $x \mid 50$ means $x$ divides 50) 3. **Problem 3: Given sets** - $U = \{1, 2, 3, ..., 12\}$ - $A = \{1, 3, 5, 7, 9\}$ - $B = \{x \mid x = 2p, p \in P, x \leq 12\}$ where $P$ is the set of primes - $C = \{x \mid x^2 - 10x + 21 = 0\}$ Step 1: Find $B$. Primes $p$ up to 6 (since $2p \leq 12$): $2, 3, 5, 7, 11$ but $2p \leq 12$ limits $p$ to $2,3,5$. So $B = \{2\times2, 2\times3, 2\times5\} = \{4, 6, 10\}$ Step 2: Solve $C$. $$x^2 - 10x + 21 = 0$$ Using quadratic formula: $$x = \frac{10 \pm \sqrt{100 - 84}}{2} = \frac{10 \pm \sqrt{16}}{2} = \frac{10 \pm 4}{2}$$ So $x = 7$ or $x = 3$, thus $C = \{3, 7\}$ Step 3: Compute requested sets: - $A \cap B = \{1,3,5,7,9\} \cap \{4,6,10\} = \emptyset$ - $B \cup C = \{4,6,10\} \cup \{3,7\} = \{3,4,6,7,10\}$ - $A - C = \{1,3,5,7,9\} - \{3,7\} = \{1,5,9\}$ - $B^c = U - B = \{1,2,3,5,7,8,9,11,12\}$ (since $U=\{1..12\}$ and $B=\{4,6,10\}$) - $(A - C)^c = U - (A - C) = \{1..12\} - \{1,5,9\} = \{2,3,4,6,7,8,10,11,12\}$ 4. **Problem 4: Prove the set identity using membership table:** $$(A \cap B)^c = A^c \cup B^c$$ | $A$ | $B$ | $A \cap B$ | $(A \cap B)^c$ | $A^c$ | $B^c$ | $A^c \cup B^c$ | |-----|-----|------------|----------------|-------|-------|----------------| | 0 | 0 | 0 | 1 | 1 | 1 | 1 | | 0 | 1 | 0 | 1 | 1 | 0 | 1 | | 1 | 0 | 0 | 1 | 0 | 1 | 1 | | 1 | 1 | 1 | 0 | 0 | 0 | 0 | Since columns $(A \cap B)^c$ and $A^c \cup B^c$ are identical, the identity is proven. **Venn Diagram:** - Left side: Shade outside the intersection $A \cap B$. - Right side: Shade union of complements $A^c$ and $B^c$ (everything outside $A$ and outside $B$). This visually confirms the identity. Final answers are summarized above.