1. **Problem:** Draw a simple, connected, undirected graph with 6 vertices each of degree 2, and prove it is not a tree. 2. **Step 1:** Construct the graph with vertices $v_1, v_2, v_3, v_4, v_5, v_6$ connected in a cycle: $$v_1 - v_2 - v_3 - v_4 - v_5 - v_6 - v_1$$ Each vertex has degree 2 because it connects to exactly two other vertices. 3. **Step 2:** Recall the properties of a tree: - It is connected. - It has no cycles. - It has exactly $n-1$ edges if there are $n$ vertices. 4. **Step 3:** Count edges in our graph: Since the graph is a cycle of 6 vertices, the number of edges is 6. 5. **Step 4:** Compare edges to vertices: For 6 vertices, a tree must have $6 - 1 = 5$ edges, but our graph has 6 edges. 6. **Step 5:** Since our graph has a cycle (closed loop) and more edges than a tree would have, it violates the no-cycle condition. 7. **Conclusion:** Therefore, the graph is not a tree because it contains a cycle and has more than $n-1$ edges. --- 8. **Problem:** Given $|A|=5$, $|B|=6$, and $|A \cup B|=8$, find $|A \cap B|$ using inclusion-exclusion. 9. **Step 1:** Apply the inclusion-exclusion principle: $$|A \cup B| = |A| + |B| - |A \cap B|$$ 10. **Step 2:** Substitute known values: $$8 = 5 + 6 - |A \cap B|$$ 11. **Step 3:** Solve for $|A \cap B|$: $$|A \cap B| = 5 + 6 - 8 = 3$$ 12. **Conclusion:** The intersection of $A$ and $B$ has 3 elements. --- 13. **Problem:** For universal set $U = \{1,2,3,4,5,6,7,8,9,10\}$, and sets $A = \{2,4,6,8,10\}$, $B = \{1,2,3,4,5\}$, find $(A' \cap B) \cup (A \cap B')$ and illustrate via Venn diagram. 14. **Step 1:** Find $A'$, the complement of $A$ in $U$: $$A' = U \setminus A = \{1,3,5,7,9\}$$ 15. **Step 2:** Find $B'$, the complement of $B$ in $U$: $$B' = U \setminus B = \{6,7,8,9,10\}$$ 16. **Step 3:** Compute $A' \cap B$: $$A' \cap B = \{1,3,5\}$$ 17. **Step 4:** Compute $A \cap B'$: $$A \cap B' = \{6,8,10\}$$ 18. **Step 5:** Compute union: $$(A' \cap B) \cup (A \cap B') = \{1,3,5,6,8,10\}$$ 19. **Step 6:** The Venn diagram would shade these elements outside the intersection but inside the union. --- 20. **Problem:** Prove or disprove: If $P(A) = P(B)$, then $A = B$. 21. **Step 1:** Recall $P(X)$ is the power set of $X$, which includes all subsets. 22. **Step 2:** If $P(A) = P(B)$, they have the exact same subsets, including singletons. 23. **Step 3:** Since singletons $\\{a\\}$ belong to $P(A)$ iff $a \in A$, equality of power sets means all elements correspond. 24. **Conclusion:** Therefore, if $P(A) = P(B)$, then $A = B$. The statement is true. --- 25. **Problem:** Prove $A \times (B \cap C) = (A \times B) \cap (A \times C)$ for sets $A,B,C$. 26. **Step 1:** Let $(a,x) \in A \times (B \cap C)$. 27. **Step 2:** Then $a \in A$ and $x \in B \cap C$, so $x \in B$ and $x \in C$. 28. **Step 3:** Therefore, $(a,x) \in A \times B$ and $(a,x) \in A \times C$. 29. **Step 4:** Hence, $(a,x) \in (A \times B) \cap (A \times C)$. 30. **Step 5:** Conversely, let $(a,x) \in (A \times B) \cap (A \times C)$. 31. **Step 6:** Then $(a,x)$ is in both $A \times B$ and $A \times C$, so $a \in A$, $x \in B$, and $x \in C$. 32. **Step 7:** Thus, $x \in B \cap C$ and $(a,x) \in A \times (B \cap C)$. 33. **Conclusion:** So, $A \times (B \cap C) = (A \times B) \cap (A \times C)$. --- 34. **Problem:** For $A = \{1,2,3\}$ and $B=\{x,y\}$, list all relations from $A$ to $B$, and represent one graphically and in matrix form. 35. **Step 1:** Relations are subsets of $A \times B$ which has $3 \times 2=6$ pairs. 36. **Step 2:** Possible pairs: $$(1,x), (1,y), (2,x), (2,y), (3,x), (3,y)$$ 37. **Step 3:** Number of possible relations: $2^6=64$. 38. **Step 4:** Example relation $R = \{(1,x), (2,y), (3,x)\}$. 39. **Step 5:** Graphically, vertices on left: elements of $A$, on right: elements of $B$, edges representing pairs of $R$. 40. **Step 6:** Matrix form rows: $A$, columns: $B$: $$\begin{bmatrix}1 & 0 \\ 0 & 1 \\ 1 & 0 \end{bmatrix}$$ Entries are 1 if pair included, 0 otherwise. --- 41. **Problem:** Prove or disprove: If a relation $R$ on $A$ is symmetric and transitive, then $R \cup I$ is an equivalence relation. 42. **Step 1:** $I$ is the identity relation; $R \cup I$ adds all pairs $(a,a)$. 43. **Step 2:** Symmetry and transitivity of $R$ extend to $R \cup I$. 44. **Step 3:** Addition of $I$ ensures reflexivity ($\forall a: (a,a) \in R \cup I$). 45. **Conclusion:** Hence, $R \cup I$ is reflexive, symmetric, and transitive, so it is an equivalence relation. --- 46. **Problem:** Given $f: \mathbb{R} \to \mathbb{R}$, $f(x) = |x|$, determine if $f$ is one-to-one, onto, or bijective. 47. **Step 1:** Check one-to-one (injective): $$f(x_1) = f(x_2) \implies |x_1| = |x_2|$$ 48. **Step 2:** This implies $x_1 = x_2$ or $x_1 = -x_2$, so not injective. 49. **Step 3:** Check onto (surjective): For any $y \in \mathbb{R}$, can we find $x$ with $f(x) = y$? 50. **Step 4:** Since $f(x) = |x| \geq 0$, negative $y$ have no pre-image. 51. **Step 5:** Hence $f$ is not onto. 52. **Step 6:** Therefore, $f$ is neither injective nor surjective, so not bijective. Final answers are derived with step-by-step explanations to aid learning.