Let $G(V, E)$ be an undirected final and simple graph. $\bar{G}(V, E')$ a simple graph on the same vertices, while $e \in E'$ iff $e \notin E$
Prove that if G is not a connected graph, then $\bar{G}$ is a connected graph.
This is my idea:
Let $v_1, v_2$ $(v_1 \neq v_2)$ be vertices in $V$, assume there isn't such a route $v_1 \overset{e_1}{\rightarrow} ... \overset{e_m}{\rightarrow} v_2$, while $e_i \in E'$. therefore in particular $(v_1, v_2) \notin E'$, therefore $(v_1, v_2) \in E$. this is true for every $(v_1, v_2)$ in contradiction to G not being connected.
Is this ok? I feel like there is something wrong here...
*I translated it so if anything is not clear please tell me.
Your proof is not correct.
In proof by contradiction you started with letting there be two vertices $v_1$ and $v_2$ such that there is no path connecting them in $\bar{G}$. Which shows $v_1$ and $v_2$ are adjusten in $G$.
This does not prove that every pair $v_1$ and $v_2$ are adjacent in $G$.