To solve the given problem using the contradiction method, we are asked to prove that for all $n = 2, 3, 4, \dots$, there exists an undirected graph $G = (V, E)$ where $|V| = n$, which:

1. Has $\frac{(n-1)(n-2)}{2}$ edges.
2. Is not connected.

Contradiction method steps:

1. Assume the negation of the statement: Assume that for all $n \geq 2$, any undirected graph $G = (V, E)$ with $|V| = n$ and $\frac{(n-1)(n-2)}{2}$ edges is connected.

2. Properties of a connected graph: A graph is connected if there is a path between every pair of vertices. For a graph with $n$ vertices, the maximum number of edges is $\frac{n(n-1)}{2}$, which corresponds to a complete graph $K_n$.

3. Edge count comparison: The given number of edges $\frac{(n-1)(n-2)}{2}$ can be simplified and compared to the total possible number of edges in a complete graph: $\frac{(n-1)(n-2)}{2} = \frac{n^2 - 3n + 2}{2}$ which is less than $\frac{n(n-1)}{2}$, the total number of edges in a complete graph. This implies that we have fewer edges than a complete graph, suggesting the possibility of disconnected vertices.

4. Construction of a disconnected graph: Consider a graph where one vertex $v_0$ is isolated (has no edges). The remaining $n-1$ vertices form a complete subgraph $K_{n-1}$. This subgraph will have exactly $\frac{(n-1)(n-2)}{2}$ edges. This construction satisfies both conditions:

   • The graph has $\frac{(n-1)(n-2)}{2}$ edges.
   • The graph is not connected because $v_0$ is isolated.

5. Contradiction: The assumption that a graph with $\frac{(n-1)(n-2)}{2}$ edges is always connected leads to a contradiction, as we have demonstrated a construction where such a graph is not connected.

Conclusion:

Thus, by contradiction, we have proven that for all $n \geq 2$, there exists an undirected graph $G = (V, E)$ with $|V| = n$ and $\frac{(n-1)(n-2)}{2}$ edges that is not connected.

Do you need further clarification or more details on any part of the proof?

Here are five related questions to explore further:

1. How does the edge count of a complete graph $K_n$ differ from the given formula?
2. Can you prove the same statement using an alternative method such as induction?
3. What is the significance of the condition $n \geq 2$?
4. How can the concept of graph connectivity be extended to weighted graphs?
5. What is the minimum number of edges required to ensure a graph is connected for any $n$?

Tip: When working with graph theory, always consider constructing specific cases of graphs to gain better insights into general properties.