Explanation Given: d 1 , d 2 ,.. ., d n arenpositive integers with sum 2n- 2. Proof: LetP(n) be “there exists a tree that hasnvertices and the degrees of the vertices ared 1 , d 2 ,.. ., d n ” Forn = 2 The sum of the two integersd 1 , d 2 has to be equal to 2n − 2 = 4 − 2 = 2. Since both integers have to be positive, they then both have to be equal to 1. d 1 , d 2 , = 1 The tree of height 1 with an edge between a root and one child is then a tree with two vertices and the degree of the two vertices is 1 each. Thus,P(1) is true. Let P(k) be true, thus there exists a tree that haskvertices and the degrees of the vertices ared 1 , d 2 ,.. ., d k . We need to prove that P(k + 1 ) is true. Let k be greater than 1. Let d 1 , d 2 ,.. ., d k , d k + 1 equal to 1 (because if allk+ 1 integers are at least 2, then their sum is 2(k + 1) =2k+ 2 ) Let d k + 1 = 1 At least one of the other positive integers then has to be greater than 1 (because if allk+ 1 integers are 1, then their sum isk+ 1, which is smaller than2k...

Discrete Mathematics and Its Applications ( 8th International Edition ) ISBN:9781260091991

8th Edition

ISBN: 9781259731709

Author: ROSEN

Publisher: MCG

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Chapter 11, Problem 6SE

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To prove:

there is a tree that hasnvertices such that the degrees of these vertices are d₁ , d₂ ,.. ., d_n.

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Chapter 11, Problem 5SE

Chapter 11, Problem 7SE

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Chapter 11 Solutions

Discrete Mathematics and Its Applications ( 8th International Edition ) ISBN:9781260091991

Ch. 11.1 - Prob. 1E Ch. 11.1 - Vhich of these graphs are trees?Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Prob. 6E Ch. 11.1 - Prob. 7E Ch. 11.1 - Prob. 8E Ch. 11.1 - Prob. 9E Ch. 11.1 - Prob. 10E

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To prove: there is a tree that hasnvertices such that the degrees of these vertices are d 1 , d 2 ,.. ., d n .

Solution Summary: The author proves that P(k) is true, since both integers have to be positive, and the degree of the two vertices is 1 each.

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To prove:

there is a tree that hasnvertices such that the degrees of these vertices are d₁ , d₂ ,.. ., d_n.

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Chapter 11 Solutions

To prove: there is a tree that hasnvertices such that the degrees of these vertices are d 1 , d 2 ,.. ., d n .

Solution Summary: The author proves that P(k) is true, since both integers have to be positive, and the degree of the two vertices is 1 each.

Videos

To prove: there is a tree that hasnvertices such that the degrees of these vertices are d1 , d2 ,.. ., dn.

Want to see the full answer?

Chapter 11 Solutions

To prove:

there is a tree that hasnvertices such that the degrees of these vertices are d₁ , d₂ ,.. ., d_n.