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Graph Theory Proof

  1. Jan 5, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that all vertices of a complete graph Kn have deg(v) = (n-1)

    2. Relevant equations
    ∑ deg(v) = 2|E|

    |E| = ½(n)(n-1) for Kn

    3. The attempt at a solution
    I may have over thought this but this was my initial path at a formal proof.

    Using the degree sum formula above and the formula for |E| for Kn,

    ∑ deg(v) = 2|E| = n(n-1)

    Kn has n vertices so,

    deg(v1) + deg(v2) + . . . . + deg(vn) (1/n) = (n-1)

    ∴ Each vertex , v, has degree (n-1)


    EDIT: The degree sum formula was mistakenly labeled as the Handshaking lemma.
     
    Last edited: Jan 5, 2015
  2. jcsd
  3. Jan 5, 2015 #2
    Interesting approach... it feels like you are using more advanced results to try to prove a simpler result, though - and that simpler result was perhaps used to prove the more complex results.

    "A complete graph is a simple undirected graph in which every pair of distinct vertices is connected by a unique edge"
    ... that should be enough foundation to build your proof on.
     
  4. Jan 5, 2015 #3

    Mark44

    Staff: Mentor

    The above is incorrect. The 1/n factor is multiplying only the last term, deg(vn. I imagine that you meant this:
    [deg(v1) + deg(v2) + . . . . + deg(vn)] (1/n) = (n-1)

    In any case, what's your justification for dividing by n? It seems to me that this gives you the average number of vertex degrees.
    In a complete graph with n vertices, each vertex must be connected to each other vertex. Thus, each vertex is connected to n - 1 vertices, so for each vertex v, deg(v) = n - 1. I'm not sure if you need to go into any more detail than this.
     
  5. Jan 5, 2015 #4
    Ah ha! I just saw the degree sum formula and the edge formula for a complete graph and tried connecting the two to get the result I needed. I should have just went back to the definition of a complete graph. Derp!
     
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