The sum of the values of the degree π(V) taken over all the vertices v of a graph .G= (V, E), is equal to twice the number of edges i.e.
β π(π£) = 2|E| vβV.
Write a proof to validate the equation.
To prove : Sum of degrees of the vertices of any graph is equal to twice of number of edges in the graph.
Proof:
Let "G = (V , E)" be a "(n, m)" graph where n represent number of vertices and m represent number of edges
So, "O(V)=n \\ ;\\ \\ O(E)=m"
Let "v\\in V" be any arbitrary vertex of G
Suppose "\\sigma(V)\\geq1" (degree of V)
then there exist an edge say 'e' incident at v
Let "w\\in V" be such that "e=\\{v,w\\}" (w is the other end point for edge 'e')
Thus the edge 'e' contributes one to the degree of vertex v and one to the degree of vertex w.
Thus each edge contributes two to the degree of vertices of V.
Hence, the sum of degrees of all the vertices is twice the number of edges.
[Basically the fact we used in this proof is that each edge in a graph has two end , so each edge gives two degrees to the vertices (one for each vertex)of graph]
No edge "\\implies" No degree
One edge "\\implies" 2 Degree
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