Final answer:
To prove that there is an n-vertex tournament with indegree equal to outdegree at every vertex iff n is odd, we use a contrapositive argument. By assuming no tournament exists with these properties and using contradiction, we can show that if n is odd, such a tournament must exist.
Step-by-step explanation:
To prove that there is an n-vertex tournament with in-degree equal to outdegree at every vertex if n is odd, we can use a contrapositive argument. The contrapositive statement is: If there is no n-vertex tournament with indegree equal to outdegree at every vertex, then n is even.
We can prove this by contradiction. Assume that there is no n-vertex tournament with in-degree equal to outdegree at every vertex, but n is odd. We can then construct a tournament with n vertices and count the total number of edges entering and leaving each vertex.
Since the indegree and outdegree at every vertex is not equal, there exists at least one vertex with a different number of edges entering and leaving. Let's call this vertex v. Without loss of generality, assume that the in-degree of v is greater than the outdegree.
Now, consider the edges pointing into vertex v. Since the in-degree is greater than the outdegree, there must be at least one incoming edge that is not leaving vertex v. This creates a contradiction, because the total number of edges entering and leaving vertex v should be equal.
Therefore, our assumption that there is no n-vertex tournament with in-degree equal to outdegree at every vertex, but n is odd, is false. This means that if n is odd, there must exist an n-vertex tournament with in-degree equal to an outdegree at every vertex.