I will now demonstarte that a tournament T (a complete digraph) is transitive if an only if it contains no cycles. Let T be transitive. This means that for any three vertics of a tournament, a, b and c, with a > (directed to) b, and b > c, that a > c. Now suppose T has a cycle, C: v1 > v2 > v3 > ... vn > v1. Since v1 > v2, and v2 > v3, then v1 > v3 and the cycle can be reduced to v1 > v3 > ... vn > v1. Continuing in this manner, C is reduced to v1 > v(n-1) > vn > v1. But this is a contradiction of the tournament's transitivitiy, since v1 > v(n-1) > vn suggests that v1 > vn is a directed edge of T. Yet the cycle suggests that vn > v1. Hence T can contain no cycles.
I will leave it as an exercise to the reader to prove the converse. If no takers, I'll follow up later.
I will now demonstarte that a tournament T (a complete digraph) is transitive if an only if it contains no cycles. Let T be transitive. This means that for any three vertics of a tournament, a, b and c, with a > (directed to) b, and b > c, that a > c. Now suppose T has a cycle, C: v1 > v2 > v3 > ... vn > v1. Since v1 > v2, and v2 > v3, then v1 > v3 and the cycle can be reduced to v1 > v3 > ... vn > v1. Continuing in this manner, C is reduced to v1 > v(n-1) > vn > v1. But this is a contradiction of the tournament's transitivitiy, since v1 > v(n-1) > vn suggests that v1 > vn is a directed edge of T. Yet the cycle suggests that vn > v1. Hence T can contain no cycles.
I will leave it as an exercise to the reader to prove the converse. If no takers, I'll follow up later.
My shed must be T, cos it doesn't contain a cycle??
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