Statement 1 :- A simple graph is necessarily connected if |E| > $\frac{(n-1)(n-2)}{2}$
( Proof already given by Rishav Kumar Singh, credit goes to him )
Let we have n vertices in a graph.
If we put 1 vertex aside we have n-1 vertices.
with n-1 vertices max edges we could have when graph is complete is (n-1)C2 which is $\frac{(n-1)(n-2)}{2}$.
We can see that with $\frac{(n-1)(n-2)}{2}$ much edges we have one vertex left aside which makes graph disconnected.
So to connect that vertex we need one more edge. So total number of edges > $\frac{(n-1)(n-2)}{2}$
Statement 2:- A simple graph with n vertices and k components has at least n-k edges.
let take in each component we have ni vertices ===> n1 in 1st component, n2 in 2nd component.... nk in kth component
∴ n1+n2+....+nk = n
note that minimum edges occurs in a tree ( if the tree has P vertices then it contains P-1 edges ) ==> Assume each component is Tree
The 1st component contains vertices ===> it contains ( n1 -1 ) edges
The 2nd component contains vertices ===> it contains ( n2 -1 ) edges
.....
The kth component contains vertices ===> it contains ( nk -1 ) edges
total edges = edges in 1st component + edges in 2nd component + .......+ edges in kth component
= ( n1 -1 ) + ( n2 -1 ) + .........+ ( nk -1 )
= ( n1 + n2 + .........+ nk ) - ( 1+1+1+....k times)
= n - k
these no.of edges atleast possible in the graph which have k components ( if any component is not tree then no.of edges should be increase. )