Question

In this problem, you will develop a new proof that every tree with two or more vertices has a leaf. Here is an outline for your proof.

a. First prove, using strong induction and the fact that every edge of a tree is a cut edge (Theorem 50.5), that a tree with n vertices has exactly n − 1 edges. Please note that our previous proof of this fact (Theorem 50.9) used the fact that trees have leaves; that is why we need an alternative proof.

Answer 1

The principle of strong induction, any tree with n vertices has exactly n - 1 edges for all n ≥ 2, without relying on the fact that trees have leaves.

Proof by Strong Induction:

Base Case: For a tree with 2 vertices, there is 1 edge (2 - 1 = 1). This is true as a tree with only two vertices forms a single edge between them.

Inductive Hypothesis: Assume that any tree with k vertices has k - 1 edges for all k ≥ 2.

Inductive Step: Let's consider a tree T with (k + 1) vertices.

Take any leaf vertex v in T. Removing this leaf vertex v (and the edge incident to it) from the tree leaves us with a tree T' of k vertices (as removing a leaf doesn't create any new disconnected components in a tree).

By the inductive hypothesis, the tree T' with k vertices has k - 1 edges. Now, when we add vertex v back to T', it becomes a tree with (k + 1) vertices and k edges (since we removed one edge by removing v earlier).

Thus, we've shown that if a tree with k vertices has k - 1 edges, then a tree with (k + 1) vertices has k edges. This completes the induction step.

Therefore, by the principle of strong induction, any tree with n vertices has exactly n - 1 edges for all n ≥ 2, without relying on the fact that trees have leaves.