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I am trying to brush up my graph theory skills. I have not done any in over 4 years and i am rusty...If someone could help me out with this simple proof i would appreciate it.

Prove that for any graph $G$ of order at least 2, the degree sequence has at least one pair of repeated entries.

So the degree sequence if a list of the degrees of each vertex (usually written in descending order).

I know that the sum of the degrees of the vertices of a graph is equal to $2|E|$ and that the number of vertices of odd degree is even.

If someone could help me out and point me in the right direction I would appreciate it.

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Yes, it just came to me after I posted the comment to the previous answer.

A graph with at least 2 vertices and no edges the 0 degree is repeated. A graph with 2 vertices and 1 edge the degree 1 is repeated. Each vertex has at least one entry in the degree sequence, so there are a total of $N$ entries. But each degree can have a maximum of degree = $N-1$. Therefore according to the pigeonhole principle at least one of the degrees has to repeat.

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  • $\begingroup$ And why can't the sequence 0,1,2,...,N-1 be a degree sequence? [signature removed by moderator] $\endgroup$
    – G. Paseman
    Aug 11 '10 at 17:12
  • $\begingroup$ If there is a vertex of degree N-1 in your graph on N vertices, every other vertex has degree at least 1. $\endgroup$ Sep 1 '10 at 22:18
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If there were no repeats, what would the degree sequence look like? Why could it not look like that?

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    $\begingroup$ I'd probably add such hints/questions as a comment rather than an answer--then if the OP figures it out from those he can post a full answer himself. $\endgroup$ Aug 11 '10 at 0:26
  • $\begingroup$ so the degree sequence would be unique, obviously, and it cannot look like that because...my mind is drawing a blank... $\endgroup$
    – gprime
    Aug 11 '10 at 1:35
  • $\begingroup$ Perhaps. For this problem, it felt more like an answer. [signature removed by moderator] $\endgroup$
    – G. Paseman
    Aug 11 '10 at 3:35
  • $\begingroup$ @Katie, @G. Paseman: Hints are acceptable as answers when they are hints because the answerer doesn't want to give away the problem $\endgroup$
    – Casebash
    Aug 11 '10 at 7:55

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