Show that if G is a bipartite graph of odd order, then G is not Hamil-
tonian.

To show that a bipartite graph \(G\) of odd order is not Hamiltonian, we need to understand some key properties of bipartite graphs and Hamiltonian cycles. Let's break down the proof step by step.

### Definitions and Properties

1. **Bipartite Graph**: A graph \(G\) is bipartite if its vertex set \(V\) can be partitioned into two disjoint sets \(V_1\) and \(V_2\) such that every edge in \(G\) connects a vertex in \(V_1\) to a vertex in \(V_2\). That is, there are no edges within \(V_1\) or within \(V_2\).

2. **Hamiltonian Cycle**: A graph \(G\) is Hamiltonian if it contains a cycle that visits each vertex exactly once. Such a cycle is called a Hamiltonian cycle.

3. **Odd Order**: A graph \(G\) has odd order if it has an odd number of vertices, i.e., \(|V| = n\) where \(n\) is an odd number.

### Proof

To prove that a bipartite graph \(G\) of odd order is not Hamiltonian, assume for the sake of contradiction that \(G\) is Hamiltonian.

1. **Partitioning the Vertex Set**:
   
   Since \(G\) is bipartite, we can partition the vertex set \(V\) into two disjoint sets \(V_1\) and \(V_2\), such that every edge connects a vertex in \(V_1\) to a vertex in \(V_2\). 

2. **Number of Vertices**:
   
   Let \(|V_1| = a\) and \(|V_2| = b\). Since \(G\) has an odd number of vertices (\(n\)), we have:
   \[
   a + b = n \quad \text{and} \quad n \text{ is odd}.
   \]
   This implies that the sum of \(a\) and \(b\) is odd.

3. **Parity of \(a\) and \(b\)**:
   
   Since \(a + b\) is odd, it follows that one of \(a\) or \(b\) must be even, and the other must be odd. Without loss of generality, assume \(a\) is odd and \(b\) is even.

4. **Hamiltonian Cycle Requirements**:
   
   In a Hamiltonian cycle, vertices are visited in a cyclic order, and each vertex must alternate between \(V_1\) and \(V_2\). Hence, if a Hamiltonian cycle exists, the cycle would require an equal number of vertices from \(V_1\) and \(V_2\).

5. **Contradiction**:
   
   However, since one of \(a\) or \(b\) is odd and the other is even, it is impossible to have a Hamiltonian cycle that alternates between \(V_1\) and \(V_2\) with an equal number of vertices from each set. This is because an equal number of vertices from two sets implies both sets should have the same parity, which contradicts the fact that one is odd and the other is even.

Thus, we arrive at a contradiction. Therefore, our initial assumption that \(G\) is Hamiltonian must be false.

### Conclusion

Since the assumption leads to a contradiction, we conclude that a bipartite graph \(G\) of odd order cannot have a Hamiltonian cycle. Hence, such a graph is not Hamiltonian.

Would you like more details or have any questions?

Here are 5 questions you might want to ask next:
1. What are some examples of bipartite graphs that are Hamiltonian?
2. Can a bipartite graph with an even number of vertices be Hamiltonian?
3. How does the proof change for non-bipartite graphs of odd order?
4. What are some applications of bipartite graphs in real-world problems?
5. How can we use graph theory to solve problems in computer science?

Tip: When dealing with properties of bipartite graphs, always consider the implications of partitioning the vertex set and how this affects the graph's structure and cycles.

Explore the proof demonstrating why bipartite graphs of odd order cannot contain Hamiltonian cycles. This problem delves into fundamental concepts of graph theory, focusing on bipartite graphs and Hamiltonian cycles, with implications for mathematical proofs and graph structures.

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