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I have seen some proofs on this website for this problem using degree counting, but I was wondering if we could use induction?

My proof is as follows:

Base Case: $n = 2$ vertices Here, the number of leaves is 2 and the maximum degree $k$ in the tree is 1. Thus, the claim holds true.

Inductive Hypothesis: Assume the claim holds for trees of size $n$ vertices.

Inductive Step: WTS Claim holds for trees of size $n+1$ vertices

Suppose we have some tree $T$ on $n$ vertices with max degree $k$. By IH, $T$ has at least $k$ leaves. Suppose we add a leaf to $T$ to create $T'$, some tree with $n+1$ vertices. We now have two cases:

(1) The max degree of $T'$ is still $k$, in which case we are done. (2) The max degree of $T'$ increases. The max degree can only increase by 1 because we are only adding one vertex/edge to $T$. So, the claim still holds true and we are done.

Yep, that's my proof. I have a bad feeling that I am overlooking some facts when I came up with this. Could someone let me know? This is not a homework problem, just preparing for an exam. Thanks!

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  • $\begingroup$ In th einduction step, it would be more stringent to start with an arbitrary (!) tree with $n+1$ vertices and then transfrom this to a tree with $n$ vertices by removing a leaf $\endgroup$ – Hagen von Eitzen Feb 14 '19 at 19:22
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As specified in the comment, you need to reverse your induction, going from any tree on $n+1$ to a tree on $n$ vertices.

Otherwise you need to argue that you cover all possible $n+1$ trees by your construction.

Here you are only showing that from a $n$-tree satisfying the hypothesis, you can build a ($n+1$)-tree also satisfying the hypothesis. Your are not prooving that any $(n+1)$-tree satisfies the hypothesis.

The general reasoning stays the same in this reverse construction.

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