By Bliss G.A.

**Read or Download A Necessary and Sufficient Condition for the Existence of a Stieltjes Integral (1917)(en)(5s) PDF**

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**Additional info for A Necessary and Sufficient Condition for the Existence of a Stieltjes Integral (1917)(en)(5s)**

**Sample text**

PROBLEMS: 1. 2. 3. 4. 5. 6. 7. Find all S-units in Z210. Find all S-units of the group ring Z2S3. How many S-units does the semigroup ring Z4S(3) have? Find those units, which are not S-units in Z24. Does M3×3 = {(aij ) / aij ∈ Z4 = {0, 1, 2, 3}}, the ring of 3 × 3 matrices have S-units? Justify your answer. Find all S-units in QS8, the group ring of the group S8 over the rational field Q. Find the S-units in ZS(7); the semigroup ring of the semigroup S(7) over the ring of integers Z. 45 8. 9. 10.

3: Let R be a ring. I a S-ideal of R; we say I is a Smarandache minimal ideal (S-minimal ideal) of R if we have a J ⊂ I where J is another S-ideal of R then J = I is the only ideal. 4: Let R be a S-ring and M be a S-ideal of R, we say M is a Smarandache maximal ideal (S-maximal ideal) of R if we have another S-ideal N such that M ⊂ N ⊂ R then the only possibility is M = N or N = R. 9: Let Z15 = {0, 1, 2, …, 13, 14} be the ring of integers modulo 15. Clearly I = {0, 3, 6, 9, 12} is a S-ideal of Z15 which is also a S-maximal ideal of Z15.

Zn has idempotents which are not S-idempotents when n = 2p, where p is a prime. Proof: Given Zn = {0, 1, 2, …, n –1} is the ring of integers modulo n and n = 2p, 2 where p is an odd prime. , p(p + 1) ≡ 0 (mod 2p). Now p ∈ Z2p is an idempotent but p is not a S-idempotent for there does not exist 52 a m ∈ Z2p \ {p, 0, 1} such that m ≡ p and mp ≡ m. But if p is a prime m = p is impossible. Thus in Z2p, when p is prime, p is an idempotent which is not a Sidempotent. 5: Let Z30 = {0, 1, 2, …, 29} be the ring of integers modulo 30.