Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm going through a proof of the following theorem:

A connected graph $ G$ is Eulerian iff $d(x)$ is even $ \forall x \in G $.

The proof of the harder direction is roughly as follows:

  1. Use induction on $ e(G) $. Done if $ e(G) = 0 $.
  2. Given connected $G$ with $ e(G) > 0 $ and the relevant condition on the degrees of the vertices, suppose $G$ not Eulerian and let $C$ be a largest circuit in $G$ with no edges repeated. Note that $ e(C) > 0 $ as $G$ is not a tree.
  3. Let $H$ be a component of $ G - E(C) $ with $e(H) > 0$. Then $H$ is connected and $ d_H(x) $ is even for all $ x \in H $.
  4. etc

My problem is understanding why $d_H(x)$ is even for all x.

My thoughts:

We start with a connected graph in which every vertex has even degree. We remove a circuit C (and I see that $d_C(x) $ is even for all $ x \in C$). We then take a component $H$ of the remaining graph $ G - E(C) $ that has an edge. Suppose $H$ had a vertex $y$ with odd degree (in $H$). Then if we add $C$ back in to the graph, we see that $y$ must be adjacent to an odd number of vertices in $C$. This means that each of these vertices in $C$ must be adjacent to an odd number of vertices in $ G - E(C) - y $.

I'm stuck at this point. Am I overcomplicating things? I feel like this is something that should be obvious.

Thanks in advance.

share|cite|improve this question
up vote 1 down vote accepted

As you noted, all vertices in the circuit $C$ have even degree. Thus, you are taking the degrees of $G$ (all of which are even) and decreasing them all by an even number (possibly 0) by removing the edges of $C$. The result is that every vertex in $G - E(C)$ has even degree ("even minus even is even"). That's all there is to it.

share|cite|improve this answer
Thank you, I understand fully. – TRY Jul 20 '11 at 17:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.