Discrete Mathematics with Graph Theory (3rd Edition) 124

# Discrete Mathematics with Graph Theory (3rd Edition) 124 -...

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122 Solutions to Exercises 23. Let g = gcd(al, ... ,an), as defined in Exercise 22. We first prove by induction on n that g divides each ai. If n = 3, we know g = gcd(al, gcd(a2' a3)) divides both al and gcd(a2' a3)' Since gcd(a2' a3) divides a2 and a3, by definition (see Definition 4.2.4), we conclude that g divides each of al, a2 and a3. Now assume that k ~ 3 and that the result is true for n = k. Given k + 1 nonzero integers al,"" ak+l, we know that g = gcd(al, gcd(a2,"" ak+l) divides both al and gcd(a2,"" ak+l). By the induction hypothesis, gcd(a2' ... ,ak+l) divides each of a2, . .. ,ak+l, hence, g divides each of al, . .. , ak+l. By the Principle of Mathematical Induction, the result is proved. Now let c be any common divisor of al, ... ,an' Since gcd(a2' ... ,an) is an integral linear combina- tion of a2, ... ,an by Exercise 22, c is a divisor of gcd(a2' ... ,an)' Since c also divides al and g is an integral linear combination of al and gcd(a2' ... ,an), c is a divisor of g. Certainly then c ::;
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## This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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