Let G(V,E) be an undirected graph.
if there exists a vertex called u that its degree is not even then it is connected to another vertex v that its degree is also not even.
I tried to prove this by contradiction but reached a closed end, may I get help with this?
Note: The sentence says there exists one so I can't choose v and u specifically.
I take it that "$u$ is connected to $v$" means that there is a path from $u$ to $v$. Let $C$ be the component of $G$ containing $u$, that is the largest connected subgraph of $G$ containing $u$. Note that the vertices $C$, other than $u$ itself, are precisely of those vertices $v$ such that there is a path from $u$ to $v$.
If the theorem is false, then $C$ is a graph with exactly one vertex of odd degree. What lemma have you learned that shows this is impossible?