MHB Graph theory proof related to trees

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A graph with v vertices that has no cycles is connected if and only if it has exactly v-1 edges. The necessary condition is established by proving that a connected graph with no cycles is a tree, which inherently has v-1 edges. The sufficient condition requires demonstrating that if a graph has v-1 edges and no cycles, it must be connected. A proof approach involves assuming a disconnected graph and showing that it leads to a contradiction regarding the number of edges. Thus, the relationship between cycles, connectivity, and edge count in graphs is clarified.
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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.
 
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Yuuki said:
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.
 

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