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Graph theory proof related to trees

Yuuki

Member
Jun 7, 2013
43
Prove that a graph on v vertices that has no cycle is connected iff it has precisely v-1 edges.

Necessary Condition:
A connected graph with no cycles is a tree. Therefore, it has v-1 edges.

Sufficient Condition:
I need help with this.
How can I use "a connected graph has no cycles iff it has exactly v-1 edges"?

Also, is my proof for the necessary condition correct?

edit:
I was able to come up with one.

Suppose G is disconnected.
Let the components be G1 and G2, each having n and m vertices respectively.
Since both are connected and have no cycles, they have n-1 edges and m-1 edges.
Hence, G has only (n-1)+(m-1)=v-2 edges.
Therefore, if G has v-1 edges and has no cycles, G must be connected.
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Prove that a graph on v vertices that has no cycle is connected iff it has precisely v-1 edges.

Necessary Condition:
A connected graph with no cycles is a tree. Therefore, it has v-1 edges.

Sufficient Condition:
I need help with this.
How can I use "a connected graph has no cycles iff it has exactly v-1 edges"?

Also, is my proof for the necessary condition correct?

edit:
I was able to come up with one.

Suppose G is disconnected.
Let the components be G1 and G2, each having n and m vertices respectively.
Since both are connected and have no cycles, they have n-1 edges and m-1 edges.
Hence, G has only (n-1)+(m-1)=v-2 edges.
Therefore, if G has v-1 edges and has no cycles, G must be connected.
Hi Yuuki, :)

I think you'll have to revise what is meant by Necessity and Sufficiency in mathematics. In you problem you have to show that,

"A graph with \(v\) vertices that has no cycles is connected iff it has precisely \(v-1\) edges"

Here the condition, "a connected graph with \(v\) vertices and no cycles" is sufficient for that graph to have \(v-1\) edges. You have proved this using the definition of Trees.

On the other hand "a connected graph with \(v\) vertices and no cycles" is also necessary for that graph to have \(v-1\) edges. Hence we use the phrase, "necessary and sufficient" for "iff" statements. For this you may assume that, "\(G\) is a disconnected graph with no cycles which has \(v\) vertices and \(v-1\) edges". This will give you a contradiction.