Question

Theorem: Let G be a graph with the following properties: (1) G is connected. (2) G has n vertices, where n ≥ 1. (3) G has n-1 edges. Use mathematical induction by the number of vertices n to prove that G is cycle-free, meaning G does not contain any simple circuit. (Hint: The handshaking theorem, 2m = ∑v∈V deg(v), can be useful for the inductive step of the proof. Here, m is the number of edges in G, deg(v) is the degree of vertex v, and V is the set of all vertices of G.)

          Theorem: Let G be a graph with the following properties: (1) G is connected. (2) G has n vertices, where n ≥ 1. (3) G has n-1 edges. Use mathematical induction by the number of vertices n to prove that G is cycle-free, meaning G does not contain any simple circuit. (Hint: The handshaking theorem, 2m = ∑v∈V deg(v), can be useful for the inductive step of the proof. Here, m is the number of edges in G, deg(v) is the degree of vertex v, and V is the set of all vertices of G.)

Added by Samuel A.

Question

Please give Ace some feedback