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Chapter 9: Problem 11

Let \(n\) denote the number of vertices of a tree and \(e\) the number of edges. Verify that \(e=n-1\) for each tree. IMAGE IS NOT AVAILABLE TO COPY

Short Answer

Expert verified
By mathematical induction, we have shown that for any tree with \(n\) vertices, the number of edges \(e\) is equal to \(n-1\). This holds for the base case of a single vertex, and for a tree with \(k+1\) vertices, given our inductive hypothesis that it holds for a tree with \(k\) vertices.

Step by step solution

01

Base Case

First, let's consider the base case with a simple tree, where the number of vertices, \(n\), is equal to 1. In this case, there are no edges in the tree since there is only one vertex. Since \(n = 1\), \(n-1 = 0\), and thus \(e = n-1\), which means the property holds in the base case.
02

Inductive Hypothesis

Assume that the property holds for some tree with \(k\) vertices, such that the number of edges, \(e_k\), equals \(k-1\). This will be our inductive hypothesis.
03

Inductive Step

Now we need to prove that the property holds for a tree with \(k+1\) vertices. Let's consider a tree with \(k+1\) vertices. To create this tree, we need to add one vertex and an edge connecting it to a tree with \(k\) vertices, which we have assumed to hold the property according to the inductive hypothesis. Since there are \(k-1\) edges in the tree with \(k\) vertices, when adding one vertex and one edge, the number of edges \(e_{k+1}\) is equal to \((k-1) + 1 = k\). Thus, for a tree with \(k+1\) vertices, we have: \[e_{k+1} = k\] Since \(e_{k+1} = (k+1)-1 = k\), the property holds for a tree with \(k+1\) vertices as well.
04

Conclusion

By mathematical induction, we have shown that the property holds for a tree with any number of vertices \(n\). Therefore, we can conclude that for any tree, the number of edges \(e\) is equal to \(n-1\).

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