Let graph $G$ have 2 edge-disjoint spanning trees.
This means none of the edges are common in the 2 spanning trees.
What is the least amount of vertices $k$, that $G$ can have?
Give an example of such a graph that has 2 edge-disjoint spanning trees but only $k$ vertices.
Would the answer just be this:
The two spanning trees are $(1,2)$ and $(2,1)$, and they are edge disjoint.
What about if an artibary amount of vertices $k'$ is given? Then what would the graph look like such that $k' \geq k$ and we still only have 2 edge-disjoint spanning trees?
Two questions^
Best Answer
Well as for your example 12 and 21 are the same edge I believe so that won't work. [I assume we are talking simple undirected graphs here.]
Actually there is a simple graph on 4 vertices with 2 edge-disjoint spanning trees...in particular $K_4$ (where the vertex-set is $\{1,2,3,4\}$) has one spanning tree with the edges $12,23,34$ and another where the edges are $31,14,42$.
And 4 is the smallest such integer for simple graphs. A graph on $k$ vertices would have to have at least $2k-2$ edges $(k-1$ edges for each spanning tree on $k$ vertices) to have two edge-disjoint spanning trees, and this is impossible for simple graphs on 3 or fewer vertices [make sure you see why].