I need to prove that if we have a tree (undirected connected graph with no cycles) with $k$ leaves, then for every vertices in the tree: $\deg(v)\leq k$.
Is this possible to prove by induction? Or any other way? I tried to do it in different ways.
First I tried to prove it by induction on the number of vertices in a tree. Let tree $G$ have $k$ leaves and $n$ vertices. Base of induction: $n=2$ If $n=2$ then there is one edge that connects the two nodes. So every node has a degree of one meaning there are two leaves in the tree and every vertices has a degree that is less then the number of leaves. We assume by induction: We have a tree $G$ with $n$ vertices and $k$ leaves, every node has a degree that is not bigger than $k$.
Induction step: lets add a new vertices to the graph. We can add it to a leaf or a vertices that isn't a leaf. If we add it to a leaf then the number of leaves in the tree doesn't change, the only degree that changes is the leaf's degree it becomes $2$. The number of leaves in a tree is at least $2$. So the degree of the once leaf is not bigger than the number of leaves.if we add it to a node that isn't a leaf the number of leaves become bigger by one. The degree to the vertice is bigger by one. Before adding the leaf the degree was. $\deg(v)\leq k$ after adding the leaf it becomes $\deg(v)+1\leq k+1$ which is still not bigger then then the number of leaves. I don't feel like this induction is right.
