let g = (v, e) be a connected graph in which every vertex is a leaf. prove that g has exactly one edge
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To prove that a connected graph with every vertex being a leaf has exactly one edge, we can proceed with the following proof by contradiction:
Assume that the graph g has more than one edge. This means there exists at least one vertex v with degree greater than 1. Since g is connected, there must be a path between any two vertices in the graph.
Let's consider the case where we have at least two edges incident to vertex v. Since v is a leaf, each of these edges must connect to different vertices, say v1 and v2.
Now, we can construct a path between v1 and v2 by following the edges incident to v. This path will include vertices other than v, which contradicts our assumption that every vertex in g is a leaf.
Therefore, our assumption that g has more than one edge leads to a contradiction. Thus, g can have only one edge.
To prove that a connected graph with every vertex being a leaf has exactly one edge, we can use a proof by contradiction.
Assume that g has more than one edge. This means that there are at least two distinct edges, e1 and e2, in g.
Since g is connected, there exists a path between any two vertices in g. Let's consider the path from one of the endpoints of e1 to the other endpoint. Since every vertex in g is a leaf, this path can only consist of the two vertices connected by e1.
Similarly, consider the path from one of the endpoints of e2 to the other endpoint. Again, this path can only consist of the two vertices connected by e2.
Now, since every vertex in g is a leaf, these two paths cannot have any common vertices except for the endpoints of e1 and e2.
But this contradicts the fact that g is connected because there is no path connecting the endpoints of e1 and e2. Therefore, our assumption that g has more than one edge must be false.
Hence, we can conclude that a connected graph with every vertex being a leaf has exactly one edge.