Definition: A graph $G$ is even if each of its vertices has even degree.
It is easy to show that $G$ is even if each of its blocks is even.
I was wondering if the converse is true. I started with assuming that this is not the case. Then there are at least two blocks that are not even.
This implies that at least two vertices in each of these blocks have odd degree.
How do I close this argument? Or is there a better proof or counter-example?
On
a proof by contradiction on a finite even graph G:
if a block in G has an odd degree vertex, it must have at least one more odd degree vertex (the sum of all degrees must be even). any vertex of even degree in G yet odd degree in it's block must be a cut vertex in G since that's the only way for its degree to be affected by getting rid of all the vertices unrelated to its block. not only it is a cut vertex, it must also have odd degree in another block (its sum of degrees in each of its blocks must be even since it's even in G). this means that each odd vertex in a block induces another block with at least one more odd vertex which is a cut vertex inducing another block with another odd cut vertex, going to infinity. our graph is finite, which leads to a contradiction.
Proof by contracdiction. Assume $G$ is even but not all blocks are. Then there are at least $2$ blocks ($A$ and $B$) with at least $2$ odd degree vertices each. As $A$ and $B$ are distinct blocks there has to exist a cut vertex $v$ in $G$ such that cutting $G$ at $v$ puts $A$ and $B$ into distinct connected components of $G$. As $A$ contains at least two vertices of odd degree, one of them is not equal to $v$. The degree of this vertex is the same before and after cutting at $v$. As we assumed $G$ to be even, this is a contradiction.