**Theorem 3.14:** In a tree T, there is one and only one simple chain joining any two vertices x
and y.

**Pf**: Two vertices, x and y are joined in T since T is connected. The shortest chains between x
and y must be simple, if not the section between repeated vertices could be removed to obtain
a shorter chain joining x and y. Suppose then that there are two simple chains joining x to y.
Let x_{p} be the first vertex at which the two chains differ, and let x_{q} be the next vertex at which
they agree. Then following the first chain from x_{p} to x_{q} and then the second chain from x_{q} back to x_{p} forms a circuit. > <.

**Theorem 3.10**: If T is a tree with n vertices and e edges then n = e + 1.

**Pf**: By induction on n.

If n = 1, there is only 1 vertex and no edges, so 1 = 0 + 1.

Assume true for all trees with fewer than k vertices (strong form of induction). Remove an
edge from a tree G with k vertices. This will give us two disconnected graphs, both of which
are trees (this follows from Thm 3.14). Since they each have fewer than k vertices, we can
apply the induction hypothesis to each, and in the first tree we have n_{1} = e_{1} + 1 and in the
second we have n_{2} = e_{2} + 1. Thus, in G we have k = n_{1} + n_{2} = e_{1} + e_{2} + 2 = e + 1, since the
number of edges in G is e = e_{1} + e_{2} + 1.

**Theorem 3.12**: In any graph G with e edges, the sum of the degrees of all the vertices = 2e.

**Def**: *Leaf (pendant vertex)*

**Theorem 3.13**: If T is a tree with more than 1 vertex, there are at least 2 pendant vertices.

**Pf**: Since T is connected, every vertex has degree at least 1. The sum of the degrees of all the
vertices = 2e = 2(n-1) = 2n - 2. If n-1 of the vertices of T had degree at least 2 we would get
a contradiction.

**Theorem**: If T is a tree with n vertices, then the chromatic polynomial, P(T,x) = x(x-1)^{n-1}.

**Pf**: By induction on n.

If n = 1 then the tree is a single isolated vertex with chromatic polynomial x = x(x-1)^{0}.

Assume the statement is true for all trees with k or fewer vertices (strong form of induction).
Let G be a tree with k + 1 vertices. Since G has at least two pendant vertices, let ß be an
edge to one of these leaves. By the fundamental reduction theorem,

**Theorem 3.15:** A graph G is connected if and only if it has a spanning tree.

**Pf**: <= Since any two vertices are joined in the spanning tree, they are joined in G, so G is
connected.

=> If G has no circuits then it is a tree (and so, is its own spanning tree). If it has a circuit, remove one edge from the circuit. The result is still connected. Repeat this until there are no circuits left. This gives a graph having all the vertices of G and no circuits, i.e., it is a spanning tree for G.

**Theorem 3.11**: Suppose that G is a graph with n vertices and e edges. Then G is a tree if and
only if G is connected and n = e + 1.

**Pf**: => Follows from the definition of a tree and Theorem 3.10.

<= Since G is connected it has a spanning tree (Thm. 3.15), say T. T has n vertices and n-1 = e edges since it is a tree. Since G also has e edges, we must have G = T.

**Theorem 3.16**: Suppose that G is a graph with n vertices and e edges. Then G is a tree if and
only if G has no circuits and n = e + 1.

**Pf**: => Follows from the definition of a tree and Theorem 3.10.

<= If G is not connected, consider its connected components K_{1}, ..., K_{p} with p > 1. Since each
of these is connected and has no circuits, each is a tree. By counting the edges in all of these
components, we get n_{1} - 1 + n_{2} - 1 + ... + n_{p} - 1 = n - p which can not equal e = n - 1 if p > 1.