Prove that, if G is a bipartite graph with an odd number of vertices, then G is non-Hamiltonian 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.


We would like to show that a bipartite graph with an odd number of vertices does not have a Hamilton circuit.
Let's suppose we have a bipartite graph, G=(V,E) where V can be partitioned into disjoint sets, V1 and V2, where no edges connect any respective vertices in V1 and the same applies to those in V2. Assume that we have a Hamilton circuit. This circuit must begin and end at the same vertex, and traverse all of the vertices. It follows that the vertices sequence for the circuit would start at s1 which is in V1, and then the second would be s2 in v2, then s3 in v3, s4 in v4, ..., sn in v1. We can see that the cardinality of V1, |V1|, is equal to that of V2, |V2|. Because of this, we can see that the number of vertices in this graph G is even (2m where m is either |V1| or |V2|, since |V1| = |V2|). It now follows that if a bipartite graph has such a hamilton circuit, it cannot have an odd number of vertices. The circuit will always, like shown, have even number
