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)}C_{2} 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 n_{i} vertices ===> n_{1} in 1^{st} component, n_{2} in 2^{nd} component.... n_{k} in k^{th} component

∴ n_{1}+n_{2}+....+n_{k} = 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 1^{st} component contains vertices ===> it contains ( n_{1} -1 ) edges

The 2^{nd} component contains vertices ===> it contains ( n_{2} -1 ) edges

.....

The k^{th} component contains vertices ===> it contains ( n_{k} -1 ) edges

total edges = edges in 1^{st} component + edges in 2^{nd} component + .......+ edges in k^{th} component

= ( n_{1} -1 ) + ( n_{2} -1 ) + .........+ ( n_{k} -1 )

= ( n_{1} + n_{2} + .........+ n_{k} ) - ( 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. )