Minimum number of edges for $n$ vertex graph to be connected I was reading about connectedness of a graph where I came across following points:


  
  The $n$ vertex graph with the maximal number of edges that is still disconnected is $K_{n-1}$ (complete graph with $n-1$ vertices) with an additional vertex, which has $\binom{n-1}{2}=\frac{(n-1)(n-2)}{2}$ edges.
  Adding any additional edge must connect the graph. So the minimum number of edges for $n$ vertex graph to be connected is $\frac{(n-1)(n-2)}{2}+1$.
  


I am ok with the first point but not with the second point. The minimum number of edges for $n$ vertex graph to be connected should be $n-1$. Then how does that second point makes sense? How does the minimum number of edges for $n$ vertex graph to be connected is $\frac{(n-1)(n-2)}{2}+1$? Was the text trying to say something else? Is $\frac{(n-1)(n-2)}{2}+1$ number characterizes/pertains to some other condition on the graph?
Likely they mean that if a graph has as at least $f(n) = {n-1 \choose 2}+1$ edges, then it is connected. For instance, $f(5) = 7$, and any 5-vertex graph is connected with 7 edges. However, as you observe, it would also be connected with a smaller number of edges (that is, with $n-1 = 5-1 = 4$ edges).