Let $G$ be a graph with $n$ vertices and $e$ edges such that $e>\binom{n-1}{2}$. Then $G$ must be connected. As usual, hints would be greatly appreciated. If it where$\binom{n}{2}$ then wouldn't everything be adjacent to itself? I'll try to answer the question myself at the end if I'm able to.
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2http://math.stackexchange.com/questions/34212/for-a-graph-g-if-m-binomn-12-then-g-is-connected?rq=1 – Daniel Montealegre Sep 11 '13 at 01:41
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Yep. My bad. Thanks for the link – Brent J Sep 11 '13 at 03:31
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If $G$ is disconnected, the complement of $G$ contains a complete bipartite graph $K_{m,n-m}$ subgraph for some $m \in \{1,2,\ldots,n-1\}$ (the vertices of one component of $G$ forms one part, the remaining vertices are in the other part). Thus, there are at least $m(n-m)$ non-edges in $G$.
Since $K_{m,n-m}$ has $m(n-m)$ edges, the number of edges in $G$ is at most \begin{align*} \binom{n}{2}-m(n-m) &= \binom{n-1}{2}+n-1-m(n-m) \\ &\leq \binom{n-1}{2} \end{align*} since the parabola $f_n(m):=n-1-m(n-m)\leq 0$ for all $m \in \{1,2,\ldots,n-1\}$. (This can be proved by showing that $1$ and $n-1$ are roots of $f_n(m)-1$ and showing $f_n(m)-1<0$ when $1<m<n-1$.)
Hence a disconnected graph has at most $\binom{n-1}{2}$ edges.
Rebecca J. Stones
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