How to demonstrate that there would exist a vertex with degree less than $ 5$?

Question

Question:

                           Letting S be a planar graph. 
                Demonstrate that there would exist a vertex of S whose degree
                       would be no greater than 5, ie, $\leq 5$

I have just gotten started with Discrete Math Graph Theory and it would be a great help if someone could assist me in working out the steps to understand the problem and work through the answer with a detailed explanation? Thanks.

score 0 · Accepted Answer · answered Dec 17 '20 at 19:35

Assume that every vertex of a planar graph $G$ has degree $6$ or more. Thus by the handshaking lemma, we have $2m\geq 6n$ where $n$ and $m$ denote the order and size of $G$. Since $G$ is planar we know that $m\leq3n-6$ and we have $3n\leq m$. Thus $3n\leq 3n-6$ which is a contradiction.

score 0 · Answer 2 · answered Dec 17 '20 at 19:38

Suppose that every vertex has degree $\geq 6$. By the handshake lemma,

$$|E| = \frac{1}{2} \sum_{v \in V} d(v) \geq 3|V|$$

However, it is a well-known result that in a planar graph $|E| \leq 3v - 6$ (https://en.wikipedia.org/wiki/Planar_graph#, Theorem 1 under other planarity criteria)

How to demonstrate that there would exist a vertex with degree less than $ 5$?

2 Answers2

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