Proof. Theory graph. Please check. If graph $G = (V,E) $ where $|V| = n $ is connectivity then $ n-1 \le |E| $ My proof: The our thesis is: $ \forall G $ is connectivity $\Rightarrow$ $ n-1 \le |E| $ I prove that using 'reductio ad absurdum' $\neg (\forall G \mbox{ is connectivity }\Rightarrow n-1 \le |E| ) \iff |E| < n -1 \Rightarrow \exists G \mbox{ non-connectivity. }$ So, I proved that exists verticle $v$, such $deg(v) = 1$ It's obvious, so I don't present my proof. Because of the there is verticle $v$ such $deg(v) = 1$ removing one edge it can make that graph will become non-connectivity. This proof is OK? Why? Why not?

If you have learned about trees, it may be helpful to use them here. Note that a tree is a minimally connected graph. A tree $T$ has $V - 1$ edges, and every edge is a cut edge. That is, if you remove an edge from $T$, you disconnect the graph. Leveraging trees will make your proof pretty straight-forward.