Proving if a graph is connected

Question

How can I prove this theorem:
For a graph with n vertices, if the degree of each vertex is at least n/2, then G is connected.

Answer 1

If every vertex has degree at least $\frac{n}{2}$, then every connected component has at least $\frac{n}{2}+1$ vertices. Clearly, there can only be one component now.

Answer 2

I'm not a topologist, but this clearly depends on the definition of the degree of a vertex.

If degree refers solely to the number of edges meeting at a vertex and there can be multiple edges between two vertices, then the theorem doesn't hold for $n>3$. We can simply divide the vertices into pairs (plus one group of $3$ if $n$ is odd), then draw $\geq \frac{n}{2}$ edges between each pair while keeping the pairs disconnected from each other. (This requires at least $2$ pairs, so at least $4$ vertices.)

If degree refers to the number of distinct vertices a vertex is directly connected to, or multiple edges between two vertices are not allowed, then see Klaus' answer.

Answer 3

To prove that the graph is connected it is sufficient to show that taken any 2 vertices in graph G say u and v, there exists a path between them. But it can be clearly seen that the degree sum of 2 vertices is greater n which implies that there is at least one common neighbor for both u and v. Hence there always exists a path between any 2 vertices of the graph. Hence the graph is connected.