Prove theat in any graph G the number of points of odd degree is even.?

Answer 1

Every edge of a graph has two "ends", so if n is the number of edges of a graph then the sum of the degrees of all its vertices is 2n, which is an even number. Let r be the number of vertices of odd degree and s be the number of vertices of even degree.
Then when you add up the degrees of all the vertices, you are adding r odd numbers and s even numbers and the sum is 2n (an even number).
The sum of the s even numbers will always be even, and so the sum of the r odd numbers must also be even in order to get an even sum. But then r cannot be odd since the sum of an odd number of odd numbers is odd. Therefore r is even.

Answer 2

The sum of the degrees of all the vertices must be even because it double counts each edge. If an odd number of vertices had odd degree, the sum of the degrees of all the vertices would be odd, which is impossible. Therefore, there must be an even number of vertices with odd degree.

Answer 3

Since the sum of degrees of the vertices is even, the sum of the degrees of those vertices which have odd degree is even since it is equal to the difference between the sum of degrees of all vertices and the sum of even-degree vertices.
But for the sum of odd numbers to be even there has to be an even number of these odd numbers which represent the degree of odd-degree vertices.

Answer 4

The network has two vertices of odd degree and all other vertices of even degree- possible answers: defiantly a tree, defiantly not a tree, may or may not; must have exactly three edges, must have at least four edges, could have three or more edges

Answer 5

Each edge has two sides to it, from where it started to where it ended. So if you add up the total you should get an even sum. (<<<< Adding up all the edges). But if you add by vertex: you get an even contribution from all the vertices of even order, and so you have to have an even contribution from all the vertices of odd order. So there must be an even number of them. QED

Answer 6

of direction you propose any appropriate graph G, in spite of the undeniable fact that it nevertheless helps to assert it. coach by technique of induction on the style of issues. obviously if the graph has 2 factors this holds. Now assume the hypothesis holds for all graphs with ok factors. evaluate a level p in a graph with ok+a million factors. Whichever factors it connects to switches the parity from how they were contained in the ok-length graph, e.g., if a level had unusual degree earlier, by technique of connecting to p, now it has even degree, and vice versa. assume the degree of p is even. Then there are a good style of parity switches. obviously this keeps the style of issues contained in the ok-length graph with unusual degree even, and p has even degree, so it does no longer upload to their huge style contained in the ok + a million length graph, the count number remains even. Now assume that p has unusual degree. Then there are an unusual style of parity switches, so the style of issues contained in the ok-length which have unusual degree has develop into unusual, yet we've now a sparkling unusual aspect p to characteristic, an unusual huge style plus one is even, so there are a good style of issues with unusual degree contained in the ok + a million length graph.