By Thomas W. Hungerford
Summary ALGEBRA: AN creation is meant for a primary undergraduate path in sleek summary algebra. Its versatile layout makes it appropriate for classes of assorted lengths and various degrees of mathematical sophistication, starting from a conventional summary algebra direction to at least one with a extra utilized style. The booklet is prepared round topics: mathematics and congruence. every one topic is constructed first for the integers, then for polynomials, and at last for jewelry and teams, so scholars can see the place many summary recommendations come from, why they're vital, and the way they relate to 1 another.
- A groups-first alternative that permits those that are looking to conceal teams ahead of earrings to take action easily.
- Proofs for newbies within the early chapters, that are damaged into steps, every one of that is defined and proved in detail.
- within the middle direction (chapters 1-8), there are 35% extra examples and thirteen% extra workouts.
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Extra resources for Abstract Algebra: An Introduction
Justify your answer. ] 16. Prove that (a, b) =1 if and only if there is no primepsuch thatpI aand p I b. 17. If pis prime and (a, 18. b) =p, then (a2, il) =? Prove or disprove each of the following statements: (a) If pis prime andpI (a2 + b2) andpI (c2 + Jl-), then p I (a2 - c2). (b) If pis prime andpI(a2 + b 2) andpI (c2 + t:P), then p I (a2 + c2). (c) If pis prime andpI aandpI (a2 + il), thenpI b. f B. 19. ' P'i [If, where p1, Pz, ... , Pk are distinct positive primes and each r1, s1 2: 0. Prove that aI b if and only if r1 s s1 for every i.
9. If a I c and b I c, must ab divide c? Justify your answer. 10. If (a, 0)= 1, what can a possibly be? 11. If n E Z, what are the possible values of (a) (n, n + 2) (b) (n,n + 6) 12. Suppose that (a, b)= 1and(a, c) = 1. Are any of the following statements false? Justify your answers. (a) (ab, a) =1 (b) (b, c) =1 (c) (ab, c) =1 13. Suppose that a, b, q, and r are integers such that a= bq+ r. Prove each of the following statements. (a) Every common divisor c of a and bis also a common divisor of b and r.
Neither of these proofs will use the definition of congruence. 1 ). d. KlUOlld,, or�:iawtdlioriaj*t. it. 3. First, assume that a"" c (modn). ::[c]. To do this, letbE[a]. Then by definitionb= a(modn). Since a= c (modn), wehaveb = c(modn)bytransitivity. :: [c]. :: [a]. Therefore, [a]= [c]. [a] = [c]. Since a=a (mod n) by reflexivity, aE[c]. By the definition of [c] , we see that Conversely, assume that we have aE[a] and, hence, a= c (modn). • A and Care two sets, there are usually three possibilities: Either A and Care dis A = C, or A n C is nonempty but A * C.
Abstract Algebra: An Introduction by Thomas W. Hungerford