Answered step by step
Verified Expert Solution
Question
1 Approved Answer
Suppose we want to prove that any connected graph G = (V, E) with |V| = |E| + 1 is a tree, i.e. the
Suppose we want to prove that any connected graph G = (V, E) with |V| = |E| + 1 is a tree, i.e. the implication (v) (i) in Theorem 5.1.2. What is wrong with the following proof? We already assume that the considered graph is connected, so all we need to prove is that it has no cycle. We proceed by induction on the number of vertices. For V| = 1, we have a single vertex and no edge, and the statement holds. So assume the implication holds for any graph G = (V, E) on n vertices. We want to prove it also for a graph G' = (V', E') arising from G by adding a new vertex. In order that the assumption |V'| = |E|+ 1 holds for G', we must also add one new edge, and because we assume G' is connected, this new edge must connect the new vertex to some vertex in V. Hence the new vertex has degree 1 and so it cannot be contained in a cycle. And because G has no cycle (by the inductive hypothesis), we get that neither does G' have a cycle, which finishes the induction step.
Step by Step Solution
★★★★★
3.43 Rating (153 Votes )
There are 3 Steps involved in it
Step: 1
For this we have to prove the followingtheorem Let G be a ...Get Instant Access to Expert-Tailored Solutions
See step-by-step solutions with expert insights and AI powered tools for academic success
Step: 2
Step: 3
Ace Your Homework with AI
Get the answers you need in no time with our AI-driven, step-by-step assistance
Get Started