A graph G has m vertices of odd degree and n | vedclass

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Question

(B) According to the Handshaking Lemma,the sum of the degrees of all vertices in a graph is equal to twice the number of edges,which is an even number

Vedclass · Accepted Answer

$m$ is an even number

Vedclass · Answer

(B) According to the Handshaking Lemma,the sum of the degrees of all vertices in a graph is equal to twice the number of edges,which is an even number.Let $V_{odd}$ be the set of vertices with odd degree and $V_{even}$ be the set of vertices with even degree.The sum of degrees is $\sum_{v \in V_{odd}} 	ext{deg}(v) + \sum_{v \in V_{even}} 	ext{deg}(v) = 2|E|$.Since $\sum_{v \in V_{even}} 	ext{deg}(v)$ is always even,the sum $\sum_{v \in V_{odd}} 	ext{deg}(v)$ must also be even for the total sum to be even.For the sum of $m$ odd numbers to be even,$m$ must be an even number.Therefore,the number of vertices of odd degree,$m$,is always even.

$(i) \{ P,R,I,N,C,A,L\} $	$(a) \{ x:x \text{ is a positive integer and is a divisor of } 18\} $
$(ii) \{ 0\} $	$(b) \{ x:x \text{ is an integer and } x^2 - 9 = 0\} $
$(iii) \{ 1,2,3,6,9,18\} $	$(c) \{ x:x \text{ is an integer and } x + 1 = 1\} $
$(iv) \{ 3, -3\} $	$(d) \{ x:x \text{ is a letter of the word } PRINCIPAL\} $

$A$ graph $G$ has $m$ vertices of odd degree and $n$ vertices of even degree. Then which of the following statements is necessarily true?

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