a)Let graph $T=(V,E,f)$ where $|V|=n>1$
Prove that those statements are equivalents:
- T is a tree;
- For each $v$ $\in$ V there's only a path from $u$ to $v$.
b) Let G a connected graph whose vertexes are all even. Prove that for each edge $e \in E(G)$ the graph you obtain deleting that edge keeps being connected.
c) Prove that if a graph has no loops, then it has at least a 1-degree vertex or it is an empty graph.
a) $=>$ for definition of tree, what about $<=$?
b) A graph is connected if for each vertex $v,z \in V$ there is a path from $v$ to $z$. If all the vertexes are even, then each vertex is connected with other 2 ones, then removing 1 edge won't compromise the connectednessof the whole graph.
c) if the graph has no loops it could be an empty graph or a tree. If it's a tree then it has at least 1-degree vertex. (Because every tree has at least a leaf and a leaf is 1-degre vertex)