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Continuing with my studies in Introduction to Graph Theory 5th Edition by Robin J Wilson, one of the exercises asked to prove that, if $G$ is a bipartite graph with an odd number of vertices, then $G$ is non-Hamiltonian. This is what I've come up with. Is it strong enough?

Let graph $G$ be a bipartite graph with an odd number of vertices and G be Hamiltonian, meaning that there is a directed cycle that includes every vertex of $G$ (Wilson 48). As such, there exists a cycle in G would of odd length. However, by Theorem 2.1, a graph $G$ is bipartite if and only if every cycle of $G$ has even length (Wilson 33). Proven by contradiction, if $G$ is a bipartite graph with an odd number of vertices, then $G$ is non-Hamiltonian.

As an example, the picture below has 13 vertices so it must be non-Hamiltonian.

enter image description here

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    $\begingroup$ Yes, this is correct. $\endgroup$ – Brian M. Scott Dec 14 '16 at 21:41
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    $\begingroup$ So the picture I just added in, as an example, has 13 vertices so that must be non-Hamiltonian as I just proved. $\endgroup$ – Learner Dec 14 '16 at 21:53
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Yes, your proof is quite correct.

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$ – I am Back Dec 15 '16 at 16:53
  • $\begingroup$ @THELONEWOLF. the OP's question is whether or not their proof is correct. It is correct. I don't think there is a more appropriate answer than this, and I would rather this question not be added to the Unanswered Question queue. $\endgroup$ – Mike Pierce Dec 15 '16 at 17:03
  • $\begingroup$ I have talked more than 10 experienced users on the site and everybody agree on one thing which is that avoid answers/comment which are unnecessary. I think that this answer is unnecessary, it is a good comment but a bad answer. Regards $\endgroup$ – I am Back Dec 15 '16 at 17:07

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