Download Communications in Algebra, volume 25, number 12 by Taft E.J. (ed.) PDF

By Taft E.J. (ed.)

Communications in Algebra supplies the reader entry to the competitively speedy booklet of vital articles of well timed and enduring curiosity that experience made this magazine the finest foreign discussion board for the trade of keystone algebraic rules. moreover, all parts of algebraic study are coated, together with classical quantity thought. No own or institutional arithmetic library can manage to pay for to be with no this constantly more desirable, undeniably influential, on-going presentation of present pursuits and actions.

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Extra resources for Communications in Algebra, volume 25, number 12

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59. (i) Find all solutions of 34x − 62y = 8 with x, y ≥ 0. (ii) Find all solutions of 62y − 34x = 8 with x, y ≥ 0. 60. Find all solutions: (i) 242x + 1870y = 66, (ii) 327x + 870y = 66 (iii) 327x + 870y = 56. 61. Find d = (3731, 1894) and write d = 3731r + 1894s where (i) r > 0 and s < 0; (ii) r < 0 and s > 0. 62. Decide if each of the following has a solution or not. If so, find the solution with the smallest possible x ≥ 0: (i) 133x + 203y = 38, (ii) 133x + 203y = 40, (iii) 133x + 203y = 42, (iv) 133x + 203y = 44.

You might guess that every fifth Fibonacci number thereafter gains another digit, and that is the case. We leave the verification as an exercise, below. Now if F5d+2 has at least d + 1 decimal digits, then any number a with d digits satisfies a < F5d+2 . So from Lam´e’s theorem we get: Corollary 15. If a < b and a has d digits, then N(a, b) ≤ (5d + 2) − 3 < 5d. The corollary shows how efficient Euclid’s Algorithm is. Even on the worst possible examples, Euclid’s Algorithm takes less than 5d steps, where d is the number of decimal digits of the smaller of the two numbers being computed.

Based on the roles of a, b, q and r in long division, we call a the divisor, b the dividend, q the quotient, and r the remainder. Proof. We prove the Division Theorem by well-ordering. Let S = {b − ax|x is a nonnegative integer and b − ax ≥ 0}. Then S is a set of nonnegative integers and is non-empty because b = b − a · 0 is in S . So by well-ordering, S has a least element r. Clearly r = b − aq for some integer q ≥ 0. We must show that 0 ≤ r < a. Since r is in S , r ≥ 0. Is r < a? If not, then r − a ≥ 0, and r − a = b − qa − a = b − (q + 1)a ≥ 0.

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