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3.27. Let R be a ring, possibly non commutative, in which xy = 0 implies x = 0 or y = 0. If a, b R and an = bn and am = fm for two relatively prime positive integers m and n, prove that a = b. Solution: (m, n) = 1, implies that, there exists u, k = Z such that mu+nk = 1. Since m, n ≥ 0 either u ≥ 0 or k ≥ 0. Assume that u > 0. Then munk = 1 where k > 0. Hence mu = nk + 1. Now, let

3.27. Let R be a ring, possibly non commutative, in which xy = 0 implies x = 0 or y = 0. If a, b R and an = bn and am = fm for two relatively prime positive integers m and n, prove that a = b. Solution: (m, n) = 1, implies that, there exists u, k = Z such that mu+nk = 1. Since m, n ≥ 0 either u ≥ 0 or k ≥ 0. Assume that u > 0. Then munk = 1 where k > 0. Hence mu = nk + 1. Now, let

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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3.27. Let R be a ring, possibly non commutative, in which xy 0 implies x = 0 or y = 0. If a,b ≤ R and an = bn and am a" = fm for two relatively prime positive integers m and n, prove that a = = b. Solution: (m, n) = 1, implies that, there exists u, k € Z such that mu+nk = 1. Since m, n ≥ 0 either u ≥ 0 or k ≥ 0. Assume that u > 0. Then mu - nk = 1 where k > 0. Hence mu = nk + 1. Now, let mu a = (am)u = (fm)u = fmu ank+1 =amu = a fnk+1 = aank = a(an)k = a(bn)k = abnk = Hence bb nk abrk implies that (b − a)bnk = 0. Then either bnk = 0 or b = a. The first possibility fnk = 0 is impossible if b ‡ 0. One can see this easily that, if t is the smallest positive integer such that bt-1 0, but bt = 0, then 0 = b = bt-¹.b 0. This implies that the assumption bt = 0 is impossible, whenever b 0. Hence b = Request explain the highlighted portion If u or k >=0, and we assume u>0, how do we say then k>0??
Transcribed Image Text:3.27. Let R be a ring, possibly non commutative, in which xy 0 implies x = 0 or y = 0. If a,b ≤ R and an = bn and am a" = fm for two relatively prime positive integers m and n, prove that a = = b. Solution: (m, n) = 1, implies that, there exists u, k € Z such that mu+nk = 1. Since m, n ≥ 0 either u ≥ 0 or k ≥ 0. Assume that u > 0. Then mu - nk = 1 where k > 0. Hence mu = nk + 1. Now, let mu a = (am)u = (fm)u = fmu ank+1 =amu = a fnk+1 = aank = a(an)k = a(bn)k = abnk = Hence bb nk abrk implies that (b − a)bnk = 0. Then either bnk = 0 or b = a. The first possibility fnk = 0 is impossible if b ‡ 0. One can see this easily that, if t is the smallest positive integer such that bt-1 0, but bt = 0, then 0 = b = bt-¹.b 0. This implies that the assumption bt = 0 is impossible, whenever b 0. Hence b = Request explain the highlighted portion If u or k >=0, and we assume u>0, how do we say then k>0??
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