The discussion revolves around a problem in graph theory concerning the degrees of vertices in a graph \( G \) of order \( 2n+1 \) where \( n \geq 2 \). Participants explore the conditions under which the graph contains a certain number of vertices with specified degrees, specifically focusing on proving that \( G \) contains at least \( n+1 \) vertices of degree \( n+2 \) or at least \( n+2 \) vertices of degree \( n+1 \).
Participants express uncertainty about the completeness of the problem statement, and there is no consensus on the best approach to proving the claim, indicating that multiple views and methods are being considered.
The discussion includes assumptions about the relationships between vertex degrees and the total number of vertices, which may not be fully resolved. The proof strategies suggested depend on specific interpretations of the problem conditions.
Try using an argument by contradiction. Suppose that there are $x$ vertices of degree $n+1$, and $y$ vertices of degree $n+2$, and suppose that the result is false. Then $x\leqslant n+1$ and $y\leqslant n$. But $x+y=2n+1$. It follows that we must have $x=n+1$ and $y=n.$ By counting the number of edges in the graph, show that this leads to a contradiction.Amer said:The degree of every vertex of a graph G of order \[2n+1 \geq 5\] is either n+1 or n+2. Prove that G contains at least n+1 vertices of degree n+2 or at least n+2 vertices of degree n+1