Prove that graph isn't Eulerian Let $G$ be a regular graph with even number of vertices and odd number of edges. Show that $G$ isnt an Eulerian graph.
I'm not sure if my solution is correct/misses something:
$|V(G)| = 2k%$ and $|E(G)| = \frac{l \cdot 2k}{2} = k \cdot l$, where $l$ is the degree of every vertice. If $E$ is odd then both $k$ and $l$ must be odd, so if $l$ is odd then the graph isn't Eulerian. What happens if the graph isn't regular, can I somehow explain why the graph could be Eulerian (other than showing an example?)
 A: 
See the picture for an example.
A: For an example, take a hexagon $ABCDEF$ with an inscribed triangle $ACE$.
A: ETA: It is not clear on whether or not the reader is insisting that the graph $H=(V,E)$, besides satisfying $|V|$ even and $|E|$ odd, is regular.
If $H$ is a regular graph with an even number of vertices and an odd number of edges, then $H$ is not Eulerian. Indeed, let $d$ be the degree of every vertex in $H$. Then the number $|E|$ of edges in $H$ satisfies $|E|=\frac{d|V|}{2}$; for $|E|$ to be odd with $|V|$ even, it follows that $d$ must be odd, and $|V|$ must not be a multiple of $4$. This was already noted in the OP and comments.
This is not true however, if the condition that $H$ is regular is removed.There is, in particular, a graph $H$ on an even number of vertices and with a odd number of edges [but not regular] that is Eulerian. In fact, there is a graph $H$ $8$ vertices where $7$ of the vertices have degree $4$ and the $1$ remaining vertex has degree $2$, and has $(7 \times 4 +2)/2 = 15$ edges [so $8$ vertices and $15$ edges total]. As every vertex has even degree in $H$, this graph $H$ is Eulerian.
To construct $H$, first let $H'$ be a complete bipartite graph on $7$ vertices, where one side $V_1$ has $4$ vertices and the other side $V_2$ has $3$ vertices [so $H'$ has exactly $4+3=7$ vertices]. Then every vertex on $V_1$ has degree exactly $3$ in $H'$ and every vertex in $V_2$ has degree exactly $4$ vertices. Then from $H'$ construct $H$ as follows:

*

*First, put an edge between a pair of vertices in $V_1$. So now in $V_1$ there are $2$ vertices $u_1$ and $u_2$ of degree $4$ and $2$ vertices of degree $3$ in this graph, and every vertex in $V_2$ still has degree exactly $4$.


*Now add an additional vertex $v$, and the edges $u_1v$ and $u_2v$. The resulting graph is $H$. Every vertex in $V_1$ has degree exactly $4$, and so does every vertex in $V_2$. The remaining vertex $v$ has degree $2$. The number of edges in $H$ is $15$.
