Final answer:
To prove that there is an n-vertex tournament with in-degree equal to out-degree at every vertex if and only if n is odd, we can construct an odd-sized tournament with equal in-degrees and out-degrees, and also show that an even-sized tournament cannot have equal in-degrees and out-degrees.
Step-by-step explanation:
To prove that there is an n-vertex tournament with in-degree equal to out-degree at every vertex if and only if n is odd, we can follow these steps:
a) Construct an odd-sized tournament with equal in-degrees and out-degrees:
Let's consider a tournament with n vertices, where n is odd. We can assign a directed edge between every pair of distinct vertices. Since every vertex has (n-1) outgoing edges, the out-degree of each vertex is (n-1). Similarly, since every vertex has (n-1) incoming edges, the in-degree of each vertex is also (n-1).
b) Use induction to prove the statement:
We can use mathematical induction to prove that for any odd value of n, there exists an n-vertex tournament with equal in-degrees and out-degrees at every vertex.
c) Show that an even-sized tournament cannot have equal in-degrees and out-degrees:
For an even-sized tournament, let's assume that every vertex has equal in-degrees and out-degrees. However, the sum of in-degrees and out-degrees of all vertices should be even. Since the number of vertices is even, it implies that the sum of all in-degrees and out-degrees would be even, which contradicts the assumption.
d) Both a and c:
Therefore, the correct answer is 'd) Both a and c'.