Explanation Given Information: Definition of gcd ( a 1 , a 2 , ....... a n ) . For n ≥ 3 , the greatest common divisor of n nonzero integers a 1 , a 2 , ........ a n can be defined inductively by gcd ( a 1 a 2 ........ a n ) = gcd ( a 1 , g c d ( a 2 ........ a n ) ) . Proof: Let g = gcd ( a 1 a 2 ........ a n ) . We will prove by induction on n that g divides each a i . If n = 3 , we know g = gcd ( a 1 a 2 ........ a n ) divides both a 1 and gcd ( a 2 , a 3 ) . Since gcd ( a 2 , a 3 ) divides a 2 and a 3 , by definition, we conclude that g divides each of a 1 , a 2 , and a 3 . Now assume that k ≥ 3 and that the result is true for n = k

Discrete Mathematics with Graph Theory (Classic Version) (3rd Edition) (Pearson Modern Classics for Advanced Mathematics Series)

3rd Edition

ISBN: 9780134689555

Author: Edgar Goodaire, Michael Parmenter

Publisher: PEARSON

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Set Theory

Every field in the present world involves a set. The theory of sets was developed by German mathematician Georg Cantor while working on the “problems of trigonometric series”.

Question

Chapter 5.1, Problem 23E

To determine

To prove: That gcd(a1,a2,.......an), as defined in Exercise 22, satisfies the properties given in Definition 4.2.14.

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Chapter 5.1, Problem 22E

Chapter 5.1, Problem 24E

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Chapter 5 Solutions

Discrete Mathematics with Graph Theory (Classic Version) (3rd Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 5.1 - True/False Questions The statement i=1n(2i1)=n2...Ch. 5.1 - Prob. 2TFQ Ch. 5.1 - Prob. 3TFQ Ch. 5.1 - Prob. 4TFQ Ch. 5.1 - Prob. 5TFQ Ch. 5.1 - Prob. 6TFQ Ch. 5.1 - Prob. 7TFQ Ch. 5.1 - Prob. 8TFQ Ch. 5.1 - Prob. 9TFQ Ch. 5.1 - Prob. 10TFQ

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To prove: That gcd ( a 1 , a 2 , ....... a n ) , as defined in Exercise 22, satisfies the properties given in Definition 4.2.14.

Solution Summary: The author explains that mathrmgcd(a_1, a.......n) satisfies the properties given in Definition

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To prove: That gcd(a1,a2,.......an), as defined in Exercise 22, satisfies the properties given in Definition 4.2.14.

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Chapter 5 Solutions