Here is the question I am trying to prove by contradiction as my professor recommended but in attempting to prove I cannot find a contradiction. Can someone point me in the right direction of what I am not considering? Thank you.
Let $G$ be a graph of order $n$. Prove that is the minimum degree $\delta(G)\ge\frac{n-1}{2}$ then $G$ is connected.
My proof: Let $G$ be a graph of order $n$. Assume, to the contrary that G is disconnected. Then $G$ has at least two components. Now assume, without loss of generality, that there are just two components $G_1$ and $G_2$. Let $n_1$ and $n_2$ be the order of $G_1$ and $G_2$ respectively. Then $n_1 \le n_2$ and $n_1 \le \frac{n}{2}$. Every vertex of $G_1$ has degree at most $n_1 - 1 \le \frac{n}{2}-1=\frac{n-2}{2}$ thus $\delta(G_1)\le \frac{n-2}{2}\le\frac{n}{2}$. It then follows that $\delta(G)\le\delta(G_1)\le\frac{n-2}{2}\le\frac{n}{2}$. (Here is where I have trouble proceeding, I am missing the contradiction.)