How could I go about proving that for any connected graph with n vertices, the minimum amount of edges is n-1?
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Check out www.esi2.us.es/~mbilbao/pdffiles/DiestelGT.pdf for corollary 1.5.3. – Moritz Oct 12 '15 at 06:20
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Related: Proving that each graph contains a spanning tree?. Note that a graph must be connected to have a spanning tree, and that the number of edges in a spanning tree is always one less than the number of vertices. – hardmath Oct 12 '15 at 15:57
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You could go about it by searching. – Oct 14 '15 at 05:47
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The real reason is that any connected graph contains a spanning tree, and a spanning tree always has exactly $(n-1)$ edges.
This is not a proof, but it points you in the direction of a proof: Consider the following procedure:
<0> Initialize $U = \phi$, the empty set.
<1> Choose an arbitrary vertex $v$ and add it to $U$.
<2> $G$ is connected, so there is an edge from a vertex in $U$ to some vertex $z \notin U$. Add $z$ to $U$.
<3> Repeat step <2> until $U=V$. Note that this takes exactly $(n-1)$ steps.
Note that this is an algorithm for constructing a spanning tree.
Consider $E(U)$, the set of edges between vertices in $U$. Now, since we repeat step <2> $(n-1)$ times, and each time we add at least one edge to $E(U)$, it follows that by the end of the procedure $E(U)$ contains at least $(n-1)$ edges.
Zur Luria
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