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Assign a grade of A(correct), C (partially correct), or F (failure) to each

A Transition to Advanced Mathematics | 7th Edition | ISBN: 9780495562023 | Authors: Douglas Smith, Maurice Eggen, Richard St. Andre ISBN: 9780495562023 335

Solution for problem 23 Chapter 6.2

A Transition to Advanced Mathematics | 7th Edition

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A Transition to Advanced Mathematics | 7th Edition | ISBN: 9780495562023 | Authors: Douglas Smith, Maurice Eggen, Richard St. Andre

A Transition to Advanced Mathematics | 7th Edition

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Problem 23

Assign a grade of A(correct), C (partially correct), or F (failure) to each. Justifyassignments of grades other than A.(a) Claim. If G is a group with identity e, then G is abelian.Proof. Let a and b be elements of G. Then ba.= e(ba)= (aa1)(ba)= a((a1b)a)= a((b1a)1a)= a(a(b1a)1)= (aa)(b1a)1= (aa)a1b= (aa)(bb1)a1b= a(ab)(b1a1)b= a(ab)(ab)1bT herefore and G is abelian. (b) Claim. If G is a group with elements x, y, and z, and if thenProof. If then implies that so Ifthen the inverse of z exists, and implies andHence in all cases, if then (c) Claim. The set of positive rationals with the operation of multiplicationis a group.Proof. The product of two positive rationals is a positive rational,so is closed under multiplication. Since for every1 is the identity. The inverse of the positive rational a is theb r , 1 r = r = r 1 ++xz = yz, x = y.xz x = y.z= yzz xz = yz(d) Claim. If m is prime, then has no divisors of zero.Proof. Suppose a is a divisor of zero in Thenand there exists in such that Then soThis contradicts the assumption that m is prime. (e) Claim. If m is prime, then has no divisors of zero.Proof. Suppose a is a divisor of zero in Thenand there exists in such that Then som divides ab. Since m is prime, m divides a or m divides b. But since aand b are elements of both are less than m. This is impossible. (f) Claim. For every natural number m, is a group.Proof. We know that is associative with identity element 1.Therefore, is associative with identity element 1. Itremains to show every element has an inverse. ForTherefore, and Therefore, everyelement of has an inverse. (g) Claim. If is a group, then m is prime.Proof. Assume that is a group. Suppose m is notprime. Let where r and s are integers greater than 1 and less thanm. Then Since r has an inverse t inThenThat is, This is impossible, becausez = e, xz = yz xe = ye, x = y. z = e,x = y.xz = yz,ab = ba= ba.= e(ba)= (aa1)(ba)= a((a1b)a)= a((b1a)1a)= a(a(b1a)1)= (aa)(b1a)1= (aa)a1b= (aa)(bb1)a1b= a(ab)(b1a1)b= a(ab)(ab)1bab = aeb{1, 1, i, i}{1, a, b}{1, 1}2x = 4 2x = 3 2x + 3 = 14 + x = 6 x + 7 = 3 3 + x = 1(8, +),6.2 Groups 291positive rational The rationals are associative under multiplicationbecause the reals are associative under multiplication.

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Chapter 6.2, Problem 23 is Solved
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Textbook: A Transition to Advanced Mathematics
Edition: 7
Author: Douglas Smith, Maurice Eggen, Richard St. Andre
ISBN: 9780495562023

Since the solution to 23 from 6.2 chapter was answered, more than 232 students have viewed the full step-by-step answer. A Transition to Advanced Mathematics was written by and is associated to the ISBN: 9780495562023. This textbook survival guide was created for the textbook: A Transition to Advanced Mathematics, edition: 7. The answer to “Assign a grade of A(correct), C (partially correct), or F (failure) to each. Justifyassignments of grades other than A.(a) Claim. If G is a group with identity e, then G is abelian.Proof. Let a and b be elements of G. Then ba.= e(ba)= (aa1)(ba)= a((a1b)a)= a((b1a)1a)= a(a(b1a)1)= (aa)(b1a)1= (aa)a1b= (aa)(bb1)a1b= a(ab)(b1a1)b= a(ab)(ab)1bT herefore and G is abelian. (b) Claim. If G is a group with elements x, y, and z, and if thenProof. If then implies that so Ifthen the inverse of z exists, and implies andHence in all cases, if then (c) Claim. The set of positive rationals with the operation of multiplicationis a group.Proof. The product of two positive rationals is a positive rational,so is closed under multiplication. Since for every1 is the identity. The inverse of the positive rational a is theb r , 1 r = r = r 1 ++xz = yz, x = y.xz x = y.z= yzz xz = yz(d) Claim. If m is prime, then has no divisors of zero.Proof. Suppose a is a divisor of zero in Thenand there exists in such that Then soThis contradicts the assumption that m is prime. (e) Claim. If m is prime, then has no divisors of zero.Proof. Suppose a is a divisor of zero in Thenand there exists in such that Then som divides ab. Since m is prime, m divides a or m divides b. But since aand b are elements of both are less than m. This is impossible. (f) Claim. For every natural number m, is a group.Proof. We know that is associative with identity element 1.Therefore, is associative with identity element 1. Itremains to show every element has an inverse. ForTherefore, and Therefore, everyelement of has an inverse. (g) Claim. If is a group, then m is prime.Proof. Assume that is a group. Suppose m is notprime. Let where r and s are integers greater than 1 and less thanm. Then Since r has an inverse t inThenThat is, This is impossible, becausez = e, xz = yz xe = ye, x = y. z = e,x = y.xz = yz,ab = ba= ba.= e(ba)= (aa1)(ba)= a((a1b)a)= a((b1a)1a)= a(a(b1a)1)= (aa)(b1a)1= (aa)a1b= (aa)(bb1)a1b= a(ab)(b1a1)b= a(ab)(ab)1bab = aeb{1, 1, i, i}{1, a, b}{1, 1}2x = 4 2x = 3 2x + 3 = 14 + x = 6 x + 7 = 3 3 + x = 1(8, +),6.2 Groups 291positive rational The rationals are associative under multiplicationbecause the reals are associative under multiplication.” is broken down into a number of easy to follow steps, and 412 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 39 chapters, and 619 solutions. The full step-by-step solution to problem: 23 from chapter: 6.2 was answered by , our top Math solution expert on 03/05/18, 08:54PM.

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Assign a grade of A(correct), C (partially correct), or F (failure) to each