The discussion focuses on proving properties of the degrees of vertices in a graph G of order \(2n+1 \geq 5\). It establishes that each vertex has a degree of either \(n+1\) or \(n+2\) and asserts that G must contain at least \(n+1\) vertices of degree \(n+2\) or at least \(n+2\) vertices of degree \(n+1\). A proof by contradiction is suggested, where the assumption of having \(x\) vertices of degree \(n+1\) and \(y\) vertices of degree \(n+2\) leads to a logical inconsistency when counting edges.
PREREQUISITESMathematicians, computer scientists, and students studying graph theory, particularly those interested in vertex properties and proof techniques.
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