Answer
## Conclusion
By mathematical induction, *every* round‐robin tournament on \(n\) teams admits an ordering \(T_1, T_2, \dots, T_n\) so that \(T_i\) beats \(T_{i+1}\) for \(i=1,2,\dots,n-1\). Equivalently, *any tournament has a directed Hamiltonian path.*
Work Step by Step
Below is a common induction argument showing that in any round‐robin tournament (i.e., a “tournament graph” on \(n\) teams), one can label the teams \(T_1, T_2, \dots, T_n\) so that \(T_1\) beats \(T_2\), \(T_2\) beats \(T_3\), …, \(T_{n-1}\) beats \(T_n\). In graph‐theory terms, *every tournament has a directed Hamiltonian path*.
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## Statement
> **Claim.** In a round‐robin tournament of \(n\) teams (where each pair of teams plays exactly once, and there is a directed edge from the winner to the loser), there is a way to relabel the teams as \(T_1, T_2, \dots, T_n\) so that
> \[
> T_1 \to T_2,\quad
> T_2 \to T_3,\quad
> \dots,\quad
> T_{n-1} \to T_n.
> \]
> Equivalently, we can order the teams in a sequence so that each team in the list beats the next team in the list.
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## Proof by Mathematical Induction
### Base Case
For \(n=1\) (just one team), the statement is trivial. There is only one team \(T_1\), and the condition holds vacuously.
For \(n=2\), label the two teams \(A\) and \(B\). Whichever team wins that game, call it \(T_1\); the other is \(T_2\). Then \(T_1\to T_2\). So the statement holds for \(n=2\) as well.
*(Either base case is sufficient; commonly, one starts at \(n=1\) or \(n=2\).)*
### Inductive Step
Assume the statement is true for all tournaments of size \(n\). We must prove it holds for size \(n+1\). So consider any round‐robin tournament with \(n+1\) teams. Call them \(A_1, A_2, \dots, A_{n+1}\).
1. **Remove one team:** Temporarily remove \(A_{n+1}\) from the tournament, leaving \(n\) teams \(A_1,\dots,A_n\). By the inductive hypothesis, we can label these \(n\) teams as \(T_1, T_2, \dots, T_n\) so that
\[
T_1 \to T_2,\quad T_2 \to T_3,\quad \dots,\quad T_{n-1} \to T_n.
\]
2. **Reinsert the removed team:** Now put \(A_{n+1}\) back into the tournament. We want to “insert” \(A_{n+1}\) into the existing chain \(T_1 \to T_2 \to \cdots \to T_n\) in a position that preserves the property.
- Compare \(A_{n+1}\) to \(T_1\).
- If \(A_{n+1}\) **beats** \(T_1\), then place \(A_{n+1}\) at the **front** of the sequence, calling it \(T_1\) and shifting all others by 1.
- Otherwise, compare \(A_{n+1}\) to \(T_2\). If \(A_{n+1}\) beats \(T_2\), insert \(A_{n+1}\) between \(T_1\) and \(T_2\).
- If it does not beat \(T_2\), compare to \(T_3\), and so on.
In general, you scan from \(T_1\) to \(T_n\). If \(A_{n+1}\) beats \(T_k\) but loses to \(T_{k+1}\), place \(A_{n+1}\) between \(T_k\) and \(T_{k+1}\). If \(A_{n+1}\) beats *all* of \(T_1,\dots,T_n\), then \(A_{n+1}\) goes at the front. If \(A_{n+1}\) loses to *all*, it goes at the end.
3. **Check the new chain:** By inserting \(A_{n+1}\) in exactly the spot where it transitions from “beats” to “loses,” we ensure the final chain
\[
T_1 \to T_2 \to \cdots \to T_k \to A_{n+1} \to T_{k+1} \to \cdots \to T_n
\]
maintains the property that each team in the list beats the next one. Thus we have found a labeling of size \(n+1\) with the desired property.
Since we have shown how to insert the \((n+1)\)st team without breaking the sequence, the inductive step is complete.
