MHB Degrees of Vertices I: Is it Possible?

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In a graph with vertex set V = {v1, v2, v3, v4, v5}, the proposed degrees of the vertices are 3, 6, 2, 1, and 5. The calculation shows that 2E equals 17, leading to E being 8.5, which is not a whole number. Since the number of edges in a graph must be an integer, it is impossible for the vertices to have these specified degrees. Therefore, a graph with these vertex degrees cannot exist.
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Let G be a graph with vertex set V = {v1, v2, v3, v4, v5}.

Is it possible for the degrees of the vertices to be 3, 6, 2, 1, 5, respectively? Why or why not?

2E = deg v1 + deg v2 + deg v3 + deg v4 + deg v5

2E = 3 + 6 + 2 + 1 + 5

2E = 17

E = 8.5

Is this correct to say that yes it is possible for the degrees of the vertices to be 3, 6, 2, 1, 5 because you end up with the number of edges being a decimal number?

Is it correct to say no that its not possible for the degrees of the vertices to be 3, 6, 2, 1, 5 because you end up with the number of edges being a decimal number?
 
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Joystar1977 said:
Let G be a graph with vertex set V = {v1, v2, v3, v4, v5}.

Is it possible for the degrees of the vertices to be 3, 6, 2, 1, 5, respectively? Why or why not?

2E = deg v1 + deg v2 + deg v3 + deg v4 + deg v5

2E = 3 + 6 + 2 + 1 + 5

2E = 17

E = 8.5

Is this correct to say that yes it is possible for the degrees of the vertices to be 3, 6, 2, 1, 5 because you end up with the number of edges being a decimal number?

Is it correct to say no that its not possible for the degrees of the vertices to be 3, 6, 2, 1, 5 because you end up with the number of edges being a decimal number?
The number of edges in a graph cannot be other than a whole number. Thus there is no graph possible which has vertices of degrees 3, 6, 2, 1, 5.
 
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