prove that a connected graph with $n$ vertices has at least $n-1$ edges Show that every connected graph with $n$ vertices has at least $n − 1$ edges.
How can I prove this?  Conceptually, I understand that the following graph has 3 vertices, and two edges:
a-----b-----c
with $a$, $b$ and $c$ being vertices, and $\{a,b\}$, $\{b,c\}$ being edges.
Is there some way to prove this logically?
--UPDATE--
Does this look correct?  Any advice on how to improve this proof would be appreciated.  Thank you.


We can use Disjoint Sets here. Disjoint Set Data Structure aka Union-Find Data Structure is quite popular in Computer Science (Cycle Detection etc.).
Disjoint Sets of Vertices are very handy in defining connected components in a graph. The number of Disjoint Sets (DS) equals the number of connected components. $v,u \in V$ are members of same DS if there is a path from $u$ to $v$.
To construct DSs of a graph, we start with $|V|$ Sets with each $v\in V$ representing a DS. Then we add the edges information. For each $\{v,u\} \in E$, we find the DSs to which $v$ and $u$ belong and union the two sets if they lie in different DSs. So for each edge, there is either $1$ or $0$ union operation.
In this setting, since the graph is connected, we must have a single DS at the end. To merge $|V|$ DSs to one, we must have $|V|-1$ Union operations. Hence at least $|V|-1$ edges are required.
Generalisation :
If a graph has $k$ connected components, we must have $|V|-k$ union operations. Hence, $|E| \geq |V|-k$ or $|E|+k \geq |V|$
