1. Problem 1 asks to prove the equivalence of three statements about a graph $G$ with $v-1$ edges: (a) $G$ is connected, (b) $G$ is acyclic, and (c) $G$ is a tree. 2. Recall the definitions: A graph is a tree if it is connected and acyclic. The given $G$ has $v-1$ edges where $v$ is the number of vertices. 3. Step 1: Show (a) and (b) imply (c). - If $G$ is connected and acyclic, then by definition $G$ is a tree. - Also, a tree with $v$ vertices always has $v-1$ edges, matching the given. 4. Step 2: Show (c) implies (a) and (b). - If $G$ is a tree, then by definition it is connected and acyclic. 5. Step 3: Show that (a) implies (b) and (c). - Given $G$ is connected with $v-1$ edges. - Assume $G$ has a cycle. Removing an edge from this cycle reduces edges to $v-2$, but graph remains connected. - This contradicts the minimality of $v-1$ edges required for connectivity. - Therefore, $G$ is acyclic. - Hence, $G$ is a tree. 6. Step 4: Show (b) implies (a) and (c). - If $G$ is acyclic with $v-1$ edges, suppose $G$ is not connected. - Then $G$ has at least two components, each acyclic. - Total number of edges in disconnected components is less than $v-1$, contradiction. - So $G$ is connected and hence a tree. 7. Conclusion: The three statements (a), (b), and (c) are equivalent for a graph $G$ with $v-1$ edges. 8. Problem 2 asks: Show that if every vertex degree in $G$ is even, then $G$ has no cut edge. 9. A cut edge (bridge) is an edge whose removal increases the number of connected components. 10. Assume to the contrary there is a cut edge $e$. 11. Removing $e$ breaks $G$ into two components, say $G_1$ and $G_2$. 12. Sum of degrees in $G_1$ and $G_2$ respectively would be even since all vertices have even degree. 13. But removing $e$ reduces the degree of exactly two vertices by 1, which are endpoints of $e$, making their degrees odd in respective components. 14. This contradicts the assumption that all vertex degrees are even. 15. Therefore, no cut edge exists in $G$ if all degrees are even. Final answers: - The three statements about $G$ with $v-1$ edges are equivalent: connected, acyclic, and tree. - If each vertex degree in $G$ is even, $G$ has no cut edge.