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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.

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  • $\begingroup$ math.stackexchange.com/questions/1255057/… may be relevant. $\endgroup$ – Gerry Myerson Feb 15 at 3:06
  • $\begingroup$ You say: "The max number of nodes in an undirected graph is $a(a-1)/2$." What is $a$? And what is a "node"? Are "nodes" edges or vertices or something else? $\endgroup$ – bof Feb 15 at 3:19
  • $\begingroup$ $a$ is the number of nodes. Nodes and vertices are used interchangably here $\endgroup$ – K Split X Feb 15 at 3:33
  • $\begingroup$ Note that the title says less than $2n$ vertices (well, actually, it says "less then", and it should say "fewer than", but never mind), while the reasoning allows as many as $2n$ vertices. So, which is it? Are $2n$ vertices allowed, or not? $\endgroup$ – Gerry Myerson Feb 15 at 9:01
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Let $|V|=v < 2n$ be the number of vertices of your graph. Assume that it has $n$ spanning trees, pairwise edge disjoint.

Any spanning tree must have $v-1$ edges. If all $n$ trees are pairwise edge disjoint, then you graph must have at least $n(v-1)$ edges. $$ |E| \geq n(v-1)$$

However, using $n >\frac{v}{2}$

$$ |E| \geq n(v-1) > \frac{v(v-1)}{2}=e_\max$$ With $e_\max$ the maximum number of edge of a graph on $v$ vertices ($K_v$). Therefore this is a contradiction.

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