[Math] Prove that trees have at least two vertices of degree one

Question

Prove that every tree with $n\geq 2$ vertices has at least two vertices of degree one.

What I tried:

Suppose that there are fewer than two vertices of degree one. So we can split into two cases.

Case one: there is no vertex of degree one. Since we know that every tree has $n-1$ edges then the total degree of any tree have to be $2(n-1)$. But for this case since no vertex have degree $1$ then every vertex have at least a degree of $2$ and since there are $n$ vertices, the total degree is $\geq 2n$ which is a contradiction.

Case two: there is only one vertex of degree one. Similarly the total degree of any tree have to be $2(n-1)$. Then there are $(n-1)$ vertices with which have degree of $\geq 2$ while only one vertex with degree of one. Thus summing up to find the total of the vertices, we have that the total degree of vertices is $\geq 2(n-1)+1=2n-1$ which is also a contradiction.

This thus proves the statement for both cases.

Is my proof correct? Could anyone explain better?

Answer 1

Your proof is correct. Although, you could tighten it up a bit. Notice that your argument in case two is general enough to cover case one too. The heart of your argument could be boiled down to:

Every tree has $n-1$ edges, so the the sum of the degrees of all the vertices of any tree have to be $2(n-1).$ But if there are fewer than two vertices of degree one, then the sum of the degrees of all the vertices must be at least $2(n-1)+1,$ which is a contradiction.