Here is the question; I am unsure of how to continue with this proof and don't know if what I have so far is right. What I have so far is this;
Let a connected graph $G$, have two vertices of odd degree. Suppose an Euler trail does not have to begin at one of the odd vertices and end at the other. Then it must start at vertex $v$, with, $\deg(v)=2k$, for every $k\in \mathbb{N}$.
I am attempting to prove it by contradiction, but am stuck and don't know how to progress, and I don't know if I have done it right so far. For context I just started Specialist Mathematics 1&2 at the start of the year, so there is lots I don't know.
Help proving this would be much appreciated, thanks.
Assuming $G$ is undirected here. The proposition is every trail must start at an odd-degree vertice AND end at the other.
The negation then would be either we do not start at an odd-degree vertex or we do not end at one.
First part: Not starting at an odd-degree vertex.
Suppose we start at an even degree vertex, let the odd-degree vertex be $V$ and have degree $2n - 1, n \in \mathbb{N}$. Notice travelling to a vertex requires one edge and travelling out requires another distinct edge. Since $G$ is connected, we must travel each edge of $V$ exactly once.
Suppose we travel to $V$ from any other vertex using 1 edge, then we must travel outward using another edge. Repeating this process we exhaust $2(n-1)$ edges. Then since we must use the $2n-1$th edge, we travel to $V$ again but now we do not have any valid path outwards.
Second Part: Not ending at an odd degree vertex
Suppose $V'$ is the odd degree vertex with degree $2m-1, m \in \mathbb{N}$. Since $V'$ is not the ending vertex, and $G$ is connected, we must have traversed all edges of $V'$ exactly once. This means we must have used an edge to travel to $V'$ and then another to travel outwards. After exhausting $2(m-1)$ such edges, when we travel using the $2m-1$ th edge, we are stuck again. Thus, we did not have a valid path outwards, hence we could not have travelled every edge exactly once.
Since both parts of the negation are false, it implies the negation is false. Hence the proposition must be true.