Can someone help to verify the following two properties, perhaps by indicating what properties of eulerian graphs is used in them?
Q1: Let $G$ be a connected graph containing a Eulerian circuit. If $G$ is bipartite then it has an even number of edges.
Q2: Let $G$ be as in Q1. For edges $e$ and $f$ sharing a vertex, $G$ has an Euler circuit in which $e$ and $f$ appear consecutively.
All help appreciated!
For Q1, consider the partition of the vertices of the graph $G$ into $S_1$ and $S_2$. Let the circuit in question start in $S_1$. No edges $e = (v_i,v_j)$ exist where $v_i$ and $v_j$ are both in $S_1$ or $S_2$.
A Eulerian circuit cannot go from a member of $S_1$ to another member of $S_1$, as no such edges exist, so it must alternate between $S_1$ and $S_2$. As a result, if we label the n-th edge $e_n$, for even $n$, it will land in $S_1$, and for odd $n$ it will land in $S_2$.
However, it is also a circuit, and so it must return to $S_1$, meaning that the final edge will have an even index, also showing that the total number of edges is even.
For Q2, it is not necessarily true. Consider the following graph:
It is not possible to traverse the sequence $BAC$ without cutting off the vertices $E$ and $D$. Also answered here: If $G$ is an Eulerian graph with edges $a$, $b$ sharing a vertex $v$, is it true that $G$ has an Eulerian trail in which $a$, $b$ are consecutive?