If a graph has less then $2n$ vertices, then it can't have $n$ spanning trees such that each pair is edge disjoint
This must apply for $n\geq 3$
I am not sure how to prove this.
Probably the best way is via contradiction: Prove that if is has less then $2n$ vertices, that it can have $n$ spanning trees such that each pair is edge disjoint, and find a contradiction.
The max number of nodes in a undirected graph is $a(a-1)/2$, and so in our case it is $2n(2n-1)/2=n(2n-1)$
We know that to be a spanning tree, you must use edges = number of vertices - 1. So in our case, spanning trees must use $2n-1$ edges.
Since I want $n$ spanning trees, that is also $n(2n-1)$ edges.
What I want to do is compare these two and arrive at a contraidction, but am stuck here.