prove that for every connected graph G, if G has no cycles then for every pair of vertices a,b in G, there is only one path from a to b in G.

the contrapositive of this would be easier to prove but i'm not exactly sure how to do that..

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- May 5th 2010, 12:22 PMmathh18contrapositive proof
prove that for every connected graph G, if G has no cycles then for every pair of vertices a,b in G, there is only one path from a to b in G.

the contrapositive of this would be easier to prove but i'm not exactly sure how to do that.. - May 5th 2010, 12:30 PMPlato
- May 5th 2010, 12:37 PMmathh18
a cycle is a circuit with the only repeated nodes at the beginning and end.

every path from a to b is also a path from b to a

if there are two paths from a to b, then...there is more than one cycle? - May 5th 2010, 12:44 PMPlato
- May 5th 2010, 12:59 PMmathh18
I don't know I'm confused. Does it mean there is one cycle?

- May 5th 2010, 01:01 PMPlato
- May 5th 2010, 01:07 PMmathh18
That's what I'm confused on(Worried)