(1) Show that if $G$ is simple and $\delta$ > $\frac{n-2}{2}$ then $G$ is connected.
(2) For $n$ even, find a disconnected $\frac{n-2}{2}$-regular simple graph.
($\delta$ - minimum degree of a vertex in a graph, $n$ - vertices)
How do I approach this problem?
For (1), assume that $G$ is disconnected. Then $G$ has at least two connected components. Let $C$ be the smallest connected component of $G$; that is, $C$ contains the minimum # of vertices. It is easy to see that $|C| \leq \frac{n}{2}$. Let $u$ be a vertex in $C$. Since $\delta(u) > \frac{n}{2} -1 \geq |C| - 1$, $u$ must connect to some vertex outside $C$, violating the fact that $C$ is a connected component.
For (2), let the vertices be $v_1, v_2, \cdots, v_n$. Construct two disconnected cliques such that the first clique involves the first $\frac{n}{2}$ vertices and the other clique involves the remaining vertices.