# How can a bipartite graph be Eulerian?

From the way I understand it:

(1) a trail is Eulerian if it contains every edge exactly once.

(2) a graph has a closed Eulerian trail iff it is connected and every vertex has even degree

(3) a complete bipartite graph has two sets of vertices in which the vertices in each set never form an edge with each other, only with the vertices of the other set.

So by definition a bipartite graph has some edges that are not used (i.e. the edges between vertices of the same set). That would then mean that there are unused edges and so the graph cannot be Eulerian.

What am I missing here?

• There aren't edges between vertices in the same set that "aren't used". There aren't edges of this type at all. Dec 23, 2020 at 12:10
• Notice that even cycles $C_{2n}$ are bipartite and (obviously) Eulerian. Moreover, if your interested only in complete bipartite graphs $K_{p,q}$, then notice that such a graph has $p$ vertices of degree $q$ and $q$ vertices of degree $q$, whence, using (2), $K_{2n,2m}$ is both bipartite (by construction), connected, and has all vertices of even degree, thus is Eulerian. Apr 26, 2021 at 15:23

For example $$k_{2,2}$$ is Eulerian.