To prove that for all the integers a , b a n d c , gcd ( a , gcd ( b , c ) ) = gcd ( gcd ( a , b ) , c ) and hence to show that the gcd operator is associative.

Question

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Answer 1

Question

Chapter 31.1, Problem 10E

Program Plan Intro

To prove that for all the integers a,b and c,

gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.

Expert Solution & Answer

Explanation of Solution

Assume

d = gcd(a,b,c), a=dp, b=dp and c=dr.

Claim gcd(a,gcd(b,c)) = d

Let e= gcd (b,c)

b = es,c = et

Since d|b and d|c, thus d|e.

Let e = dm, thus

b=(dm)s=sq,c=(dm)t=dr.

Assume k = gcd(p,m),

k|p,k|m,⇒dk|dp,dk|dm,⇒dk|dp,dk|(dm)s,dk|(dm)t,⇒dk|a,dk|b,dk|c.

Since d = gcd (a,b,c), thus k=1.

gcd (a,gcd (b,c))=gcd (a,e) =gcd (dp,dm) = d*gcd (p,m) = d*k =d

Hence,

gcd (a,gcd (b,c))=d=gcd (gcd (a,b),c).

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Answer 2

Question

Chapter 31.1, Problem 10E

Program Plan Intro

To prove that for all the integers a,b and c,

gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.

Expert Solution & Answer

Explanation of Solution

Assume

d = gcd(a,b,c), a=dp, b=dp and c=dr.

Claim gcd(a,gcd(b,c)) = d

Let e= gcd (b,c)

b = es,c = et

Since d|b and d|c, thus d|e.

Let e = dm, thus

b=(dm)s=sq,c=(dm)t=dr.

Assume k = gcd(p,m),

k|p,k|m,⇒dk|dp,dk|dm,⇒dk|dp,dk|(dm)s,dk|(dm)t,⇒dk|a,dk|b,dk|c.

Since d = gcd (a,b,c), thus k=1.

gcd (a,gcd (b,c))=gcd (a,e) =gcd (dp,dm) = d*gcd (p,m) = d*k =d

Hence,

gcd (a,gcd (b,c))=d=gcd (gcd (a,b),c).

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Answer 3

Question

Chapter 31.1, Problem 10E

Program Plan Intro

To prove that for all the integers a,b and c,

gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.

Expert Solution & Answer

Explanation of Solution

Assume

d = gcd(a,b,c), a=dp, b=dp and c=dr.

Claim gcd(a,gcd(b,c)) = d

Let e= gcd (b,c)

b = es,c = et

Since d|b and d|c, thus d|e.

Let e = dm, thus

b=(dm)s=sq,c=(dm)t=dr.

Assume k = gcd(p,m),

k|p,k|m,⇒dk|dp,dk|dm,⇒dk|dp,dk|(dm)s,dk|(dm)t,⇒dk|a,dk|b,dk|c.

Since d = gcd (a,b,c), thus k=1.

gcd (a,gcd (b,c))=gcd (a,e) =gcd (dp,dm) = d*gcd (p,m) = d*k =d

Hence,

gcd (a,gcd (b,c))=d=gcd (gcd (a,b),c).

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Answer 4

Question

Chapter 31.1, Problem 10E

Program Plan Intro

To prove that for all the integers a,b and c,

gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.

Expert Solution & Answer

Explanation of Solution

Assume

d = gcd(a,b,c), a=dp, b=dp and c=dr.

Claim gcd(a,gcd(b,c)) = d

Let e= gcd (b,c)

b = es,c = et

Since d|b and d|c, thus d|e.

Let e = dm, thus

b=(dm)s=sq,c=(dm)t=dr.

Assume k = gcd(p,m),

k|p,k|m,⇒dk|dp,dk|dm,⇒dk|dp,dk|(dm)s,dk|(dm)t,⇒dk|a,dk|b,dk|c.

Since d = gcd (a,b,c), thus k=1.

gcd (a,gcd (b,c))=gcd (a,e) =gcd (dp,dm) = d*gcd (p,m) = d*k =d

Hence,

gcd (a,gcd (b,c))=d=gcd (gcd (a,b),c).

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Answer 5

Chapter 31.1, Problem 10E

Program Plan Intro

To prove that for all the integers a,b and c,

gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.

Expert Solution & Answer

Explanation of Solution

Assume

d = gcd(a,b,c), a=dp, b=dp and c=dr.

Claim gcd(a,gcd(b,c)) = d

Let e= gcd (b,c)

b = es,c = et

Since d|b and d|c, thus d|e.

Let e = dm, thus

b=(dm)s=sq,c=(dm)t=dr.

Assume k = gcd(p,m),

k|p,k|m,⇒dk|dp,dk|dm,⇒dk|dp,dk|(dm)s,dk|(dm)t,⇒dk|a,dk|b,dk|c.

Since d = gcd (a,b,c), thus k=1.

gcd (a,gcd (b,c))=gcd (a,e) =gcd (dp,dm) = d*gcd (p,m) = d*k =d

Hence,

gcd (a,gcd (b,c))=d=gcd (gcd (a,b),c).

Answer 6

Chapter 31.1, Problem 10E

Program Plan Intro

To prove that for all the integers a,b and c,

gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.

Expert Solution & Answer

Explanation of Solution

Assume

d = gcd(a,b,c), a=dp, b=dp and c=dr.

Claim gcd(a,gcd(b,c)) = d

Let e= gcd (b,c)

b = es,c = et

Since d|b and d|c, thus d|e.

Let e = dm, thus

b=(dm)s=sq,c=(dm)t=dr.

Assume k = gcd(p,m),

k|p,k|m,⇒dk|dp,dk|dm,⇒dk|dp,dk|(dm)s,dk|(dm)t,⇒dk|a,dk|b,dk|c.

Since d = gcd (a,b,c), thus k=1.

gcd (a,gcd (b,c))=gcd (a,e) =gcd (dp,dm) = d*gcd (p,m) = d*k =d

Hence,

gcd (a,gcd (b,c))=d=gcd (gcd (a,b),c).

Answer 7

Chapter 31.1, Problem 10E

Program Plan Intro

To prove that for all the integers a,b and c,

gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.

Answer 8

Expert Solution & Answer

Explanation of Solution

Assume

d = gcd(a,b,c), a=dp, b=dp and c=dr.

Claim gcd(a,gcd(b,c)) = d

Let e= gcd (b,c)

b = es,c = et

Since d|b and d|c, thus d|e.

Let e = dm, thus

b=(dm)s=sq,c=(dm)t=dr.

Assume k = gcd(p,m),

k|p,k|m,⇒dk|dp,dk|dm,⇒dk|dp,dk|(dm)s,dk|(dm)t,⇒dk|a,dk|b,dk|c.

Since d = gcd (a,b,c), thus k=1.

gcd (a,gcd (b,c))=gcd (a,e) =gcd (dp,dm) = d*gcd (p,m) = d*k =d

Hence,

gcd (a,gcd (b,c))=d=gcd (gcd (a,b),c).

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Ch. 31.1 - Prob. 1E Ch. 31.1 - Prob. 2E Ch. 31.1 - Prob. 3E Ch. 31.1 - Prob. 4E Ch. 31.1 - Prob. 5E Ch. 31.1 - Prob. 6E Ch. 31.1 - Prob. 7E Ch. 31.1 - Prob. 8E Ch. 31.1 - Prob. 9E Ch. 31.1 - Prob. 10E

To prove that for all the integers a , b a n d c , gcd ( a , gcd ( b , c ) ) = gcd ( gcd ( a , b ) , c ) and hence to show that the gcd operator is associative.

Concept explainers

To prove that for all the integers a,b and c,

gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.

Explanation of Solution

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Chapter 31 Solutions

To prove that for all the integers a , b a n d c , gcd ( a , gcd ( b , c ) ) = gcd ( gcd ( a , b ) , c ) and hence to show that the gcd operator is associative.

Concept explainers

To prove that for all the integers a,b and c, gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.

Explanation of Solution

Want to see more full solutions like this?

Chapter 31 Solutions

To prove that for all the integers a,b and c,

gcd(a,gcd(b,c)) = gcd(gcd(a,b),c) and hence to show that the gcd operator is associative.