1. Problem: Classify the following sets and name their types: $A=\{1,2,3\}$, $B=\{3,6,9,12,\ldots\}$, $C=\varnothing$. 1.1 Step: The formula or idea used is definition based: a set is finite if it has a countable number of elements listed, and infinite if it continues without end. 1.2 Work: $A$ lists three elements so $A$ is a finite set. 1.3 Work: $B$ lists multiples of 3 in natural numbers so it is infinite. 1.4 Work: $C$ equals $\varnothing$ so it is the empty set. 1.5 Answer: $A$ is finite, $B$ is infinite, $C$ is empty. 2. Problem: Write the set of English vowels in three forms: descriptive, tabular and set-builder. 2.1 Step: Formula used is representation forms of a set. 2.2 Work: Descriptive form: the set of vowels in English. 2.3 Work: Tabular (roster) form: $V=\{a,e,i,o,u\}$. 2.4 Work: Set-builder form: $V=\{x\mid x\text{ is a vowel in the English alphabet}\}$. 2.5 Answer: Descriptive: the set of vowels. Tabular: $V=\{a,e,i,o,u\}$. Set-builder: $V=\{x\mid x\text{ is a vowel}\}$. 3. Problem: Given universe $U=\{1,2,3,4,5,6\}$, $A=\{1,2,3\}$ and $B=\{3,4\}$ compute $A\cup B$, $A\cap B$, $A\setminus B$, and complement of $A$. 3.1 Step: Use definitions: $A\cup B=\{x\mid x\in A \text{ or } x\in B\}$, $A\cap B=\{x\mid x\in A \text{ and } x\in B\}$, $A\setminus B=\{x\mid x\in A \text{ and } x\notin B\}$, $A^{\complement}=U\setminus A$. 3.2 Work: $A\cup B=\{1,2,3,4\}$. 3.3 Work: $A\cap B=\{3\}$. 3.4 Work: $A\setminus B=\{1,2\}$. 3.5 Work: $A^{\complement}=\{4,5,6\}$. 3.6 Answer: $A\cup B=\{1,2,3,4\}$, $A\cap B=\{3\}$, $A\setminus B=\{1,2\}$, $A^{\complement}=\{4,5,6\}$. 4. Problem: Show commutative and associative properties for union and intersection with an example. 4.1 Step: Rules: Commutative: $A\cup B=B\cup A$ and $A\cap B=B\cap A$. Associative: $(A\cup B)\cup C=A\cup(B\cup C)$ and similarly for $\cap$. 4.2 Work: Example sets $A=\{1,2\}$, $B=\{2,3\}$, $C=\{3,4\}$. 4.3 Verify commutative union: $A\cup B=\{1,2,3\}=B\cup A$. 4.4 Verify commutative intersection: $A\cap B=\{2\}=B\cap A$. 4.5 Verify associative union: $(A\cup B)\cup C=\{1,2,3,4\}=A\cup(B\cup C)$. 4.6 Answer: The equalities hold by element membership checks. 5. Problem: State and verify De Morgan's laws for two sets. 5.1 Step: Laws: $\bigl(A\cup B\bigr)^{\complement}=A^{\complement}\cap B^{\complement}$ and $\bigl(A\cap B\bigr)^{\complement}=A^{\complement}\cup B^{\complement}$. 5.2 Work: Use $U=\{1,2,3,4\}$, $A=\{1,2\}$, $B=\{2,3\}$. 5.3 Compute left side: $A\cup B=\{1,2,3\}$ so $(A\cup B)^{\complement}=\{4\}$. 5.4 Compute right side: $A^{\complement}=\{3,4\}$ and $B^{\complement}=\{1,4\}$ so $A^{\complement}\cap B^{\complement}=\{4\}$. 5.5 Answer: Both sides equal $\{4\}$ so the first De Morgan law is verified; the second is similar. 6. Problem: Venn diagram with two sets: In a class of 30 students 18 like Math, 12 like Science and 5 like both. How many like Math or Science and how many like neither? 6.1 Step: Use formula $|A\cup B|=|A|+|B|-|A\cap B|$ and $\text{neither}=\text{total}-|A\cup B|$. 6.2 Work: $|A\cup B|=18+12-5=25$. 6.3 Work: Neither $=30-25=5$. 6.4 Answer: 25 students like Math or Science, 5 like neither. 7. Problem: Venn diagram with three sets: In a survey 40 students, 20 like A, 18 like B, 15 like C, pairwise overlaps $|A\cap B|=6$, $|A\cap C|=5$, $|B\cap C|=4$, and all three $|A\cap B\cap C|=2$. How many like at least one subject? 7.1 Step: Use inclusion-exclusion for three sets: $|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|$. 7.2 Work: Sum singles $=20+18+15=53$. Subtract pairs $=53-(6+5+4)=38$. Add triple $=38+2=40$. 7.3 Answer: $|A\cup B\cup C|=40$, so all students like at least one subject in this data. 8. Problem: Difference and complement: Let $U=\{1,2,3,4,5,6,7,8,9,10\}$, $A\text{ is even numbers}$, $B\text{ is primes }\{2,3,5,7\}$. Find $A\setminus B$ and $A^{\complement}$. 8.1 Step: Definitions used earlier for difference and complement. 8.2 Work: $A=\{2,4,6,8,10\}$. 8.3 Work: $A\setminus B=\{4,6,8,10\}$ because 2 is prime and removed. 8.4 Work: $A^{\complement}=\{1,3,5,7,9\}$. 8.5 Answer: $A\setminus B=\{4,6,8,10\}$ and $A^{\complement}=\{1,3,5,7,9\}$. 9. Problem: Use set-builder notation to describe solution sets: Solve $x^2<9$ for integers and write solution in set-builder form. 9.1 Step: Solve inequality and then express as set-builder. 9.2 Work: Integers satisfying $x^2<9$ are $x\in\{-2,-1,0,1,2\}$. 9.3 Work: Set-builder form: $S=\{x\in\mathbb{Z}\mid x^2<9\}$. 9.4 Answer: $S=\{-2,-1,0,1,2\}$ and $S=\{x\in\mathbb{Z}\mid x^2<9\}$. 10. Problem: Word problem with sets and inclusion-exclusion: In a club of 50 members 30 play football, 20 play cricket and 10 play both. How many play neither? 10.1 Step: Use $|F\cup C|=|F|+|C|-|F\cap C|$. 10.2 Work: $|F\cup C|=30+20-10=40$. 10.3 Work: Neither $=50-40=10$. 10.4 Answer: 10 members play neither sport. 11. Problem: Prove distributive law $A\cap( B\cup C)=(A\cap B)\cup( A\cap C)$ by element argument. 11.1 Step: Use element method: show $x$ is in left side iff $x$ is in right side. 11.2 Work: Suppose $x\in A\cap(B\cup C)$. Then $x\in A$ and $x\in(B\cup C)$. So $x\in A$ and $(x\in B$ or $x\in C)$. Thus $(x\in A\cap B)$ or $(x\in A\cap C)$. Hence $x\in(A\cap B)\cup(A\cap C)$. 11.3 Work: Reverse direction similar, so sets are equal. 11.4 Answer: Distributive law holds by element-wise proof. 12. Problem: Determine if the pairs are disjoint and explain: $X=\{1,3,5\}$ and $Y=\{2,4,6\}$; $P=\{1,2\}$ and $Q=\{2,3\}$. 12.1 Step: Definition: Two sets are disjoint if their intersection is $\varnothing$. 12.2 Work: $X\cap Y=\varnothing$ so $X$ and $Y$ are disjoint. 12.3 Work: $P\cap Q=\{2\}\neq\varnothing$ so $P$ and $Q$ are not disjoint. 12.4 Answer: $X$ and $Y$ are disjoint; $P$ and $Q$ are not disjoint.