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 varied lengths and diversified degrees of mathematical sophistication, starting from a conventional summary algebra direction to 1 with a extra utilized taste. The publication is prepared round issues: mathematics and congruence. every one subject is built first for the integers, then for polynomials, and at last for jewelry and teams, so scholars can see the place many summary techniques come from, why they're vital, and the way they relate to 1 another.

New Features:

- A groups-first alternative that allows those that are looking to conceal teams earlier than jewelry to take action easily.

- Proofs for newbies within the early chapters, that are damaged into steps, every one of that's defined and proved in detail.

- within the middle direction (chapters 1-8), there are 35% extra examples and thirteen% extra routines.

**Read Online or Download Abstract Algebra: An Introduction PDF**

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**Extra info for Abstract Algebra: An Introduction**

**Sample text**

15. If (a, n) 16. If [a ] = = 1, prove that there is an integer bsuch that ab = 1 (mod n). , prove that (a, n) = 1. Show by example that the converse may be false. 17. Prove that 10" = (-lf (mod 11) for every positive n. 18. /:- 2891235306. ] 19. Prove or disprove: If [a] 20. , then (a, n) = (b, n). a2 = b2 (mod n), then a = b(mod n) or a = -b (mod n). (b) Do part (a) when n is prime. Ol'ia:PKI. - ... it. 32 Chap ter 2 Congruence inland Modular Arithmetic 21. (a) Show that 10" = 1 (mod 9) for every positive n.

B = c (mod n). Therefore, by symmetry and transitivity, a= c (mod n). Hence, [a] = [c] n [c] is nonempty. 3. 5 Let n > 1 be an integer and (1) consider congruence modulo n. If a is any integer and r is the remainder when a is divided by n, then [a] = [r]. (2) There are exactly n distinct congruences classes, namely, [OJ, [1], [ 2], [n - 1], o o • I Proof•(l) Let aEZ By Thus the DivisionAlgorithm, a= nq + r, with Os r < n. a - r= qn, so that a = r (mod n). 3, [a] = [r]. (2) If [a] is any congruence class, then (1) shows 0 s r < n.

For example, we know that equality is reflexive: a = a for every integer a; symmetric: if a=b, then b =a; transitive: if a= b and b = c, then a= c. We now see that congruence modulo n is also reflexive, symmetric, and transitive. 1 Let n be a positive integer. For all a, b, cEZ, (1) a= a (mod n); (2) if a = b (mod n), then b =a(mod n); (3) if a = b (mod n) and b = c (mod n), then a = c (mod n). Proof ... (1) To prove that a= a(modn), we must show thatn I (a - a). But a - a = 0 andn I 0 (see Example 2 on page 9).