A tournament is a complete graph where each edge has an orientation. A proper edge coloring is an edge coloring such that if $a, b, c$ are three distinct vertices of the graph, and if the edge between $a$ and $b$ points to $b$, and the edge between $b$ and $c$ points to $c$, then these two edges are of different colors.

The edge chromatic number of a complete oriented graph is the smallest number of colors needed to obtain a proper edge coloring. What is the smallest edge chromatic number possible for a tournament?
Let $k$ be the unique integer such that $2^{k-1}<n \leqslant 2^{k}$. Suppose the vertices of the graph are numbered from 0 to $n-1$, and the edges always point to the vertex with the smallest number. We color the edge between vertices numbered $a$ and $b$ with the color $1 \leqslant i \leqslant k$ if the highest digit in the binary representation of $a$ and $b$ that differs is the $k$-th. This coloring with $k$ colors is a proper coloring.

Let us show that no complete directed graph with $n$ vertices can be colored with $k-1$ colors. Indeed, for the color $1 \leqslant i \leqslant k-1$, we can define the set $A_{i}$ of vertices that are the destination of an edge of color $i$, and $B_{i}$ the others. Given two vertices $x$ and $y$, there is at least one $i$ (the color of the edge between $x$ and $y$) such that $x, y$ are not in the same part of the partition $A_{i} \cup B_{i}$. Thus, the set of $i$ for which $x \in A_{i}$ is different for each vertex $x$, so there are at most $2^{k-1}$ vertices, which is absurd because $2^{k-1}<n$, hence the result.