How to to turn round robin word problem into formal argument? A round-robin tournament consists of $n$ players and all possible games between any two players. Each game can result in win or loss of a player but no draws are allowed. A champion is a player $A$ such that for every other player $B$ either $A$ beats $B$ (in the game between A and B ) or there is a player $C$ such that $A$ beats $C$ who, in turn, beats $B$.
1.)Show that every tournament with at least two players has a champion. 
2.)Can you prove that there is more than one champion?
So part 1 seems obvious since there can be no draw between any two players and therefore one of the players must win resulting in a champion
For part 2, if $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$ in a tournament then all three would be considered champions.
Is there something I'm missing? I do not feel I am doing this right because the answers seem too straightforward. Also, how can I convey my answers properly? I have only ever worked with formal proofs of short propositions and never a word problem like this.
Any help would be much appreciated.
 A: There is a bit more subtelty to this problem because of the fact that this is not a transitive tournament i.e. just because $A$ beats $B$ and $B$ beats $C$, that doesn't mean that $A$ beats $C$. 
$\textbf{1)}$ Your justification is not enough. It is not an obvious fact because of the subtlety here: "or there is a player C such that A beats C who, in turn, beats B". A champion need not win every game (and in many cases will not have won every game, even if there is only one single champion!).
Proof by induction. Let $P(n)$ be the proposition that a tournament with $n$ players must have a champion.
Base Case: $(P(2))$ Suppose there is a tournament with two players. Since at least the players must play one another, and there can be no tie, then the winning player must be the champion, since he has beat every other player. So $P(2)$ is true.
Inductive Step: Suppose that $P(n)$ is true. Consider $P(n+1)$. For a tournament with $(n+1)$ players, select any $n$ players. By our hypothesis, this set of players must have a champion, lets call him $A$. Consider the other $n$ players besides $A$. By our hypothesis, this set of players must also have a champion, lets call him $B$. Since $A$ and $B$ play each other, then if $B$ beats $A$, then $B$ is the champion. Otherwise $A$ is the champion. So $P(n+1)$ is true. QED
$\textbf{2)}$ Your justification is enough. This is the simplest example. As mentioned in the comments, there need not be multiple champions, but there certainly could be.
