Download Algebra (AMS Chelsea Publishing) by Saunders Mac Lane, Garret Birkhoff PDF

By Saunders Mac Lane, Garret Birkhoff

This ebook provides sleek algebra from first rules and is offered to undergraduates or graduates. It combines general fabrics and valuable algebraic manipulations with common recommendations that make clear that means and significance.
This conceptual method of algebra starts off with an outline of algebraic constructions by way of axioms selected to fit the examples, for example, axioms for teams, earrings, fields, lattices, and vector areas. This axiomatic approach--emphasized through Hilbert and built in Germany by means of Noether, Artin, Van der Waerden, et al., within the 1920s--was popularized for the graduate point within the Nineteen Forties and Fifties to a point through the authors' ebook of A Survey of recent Algebra. the current e-book provides the advancements from that point to the 1st printing of this publication. This 3rd version comprises corrections made by means of the authors.

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Extra resources for Algebra (AMS Chelsea Publishing)

Example text

Since σ (2 c ) = 2 c+1 − 1 , we have 2 c+1 · b = (2 c+1 − 1 )σ (b). Note that 2 c+1 and 2 c+1 −1 are relatively prime. Thus 2 c+1 is a divisor of σ (b). Hence we can write σ (b) = d · 2 c+1 for some positive integer d . Equivalently, b = (2 c+1 − 1 )d . If we can show that d = 1 , then we have shown that n is of the form 2 c+1 · (2 c+1 − 1 ). Suppose, for the sake of contradiction, that d > 1 . Then the number b has at least 1, b, and d as distinct divisors. It follows that d · 2 c+1 = σ (b) ≥ b + d + 1 = (2 c+1 − 1 )d + d + 1 = 2 c+1 · d + 1 , a clear contradiction.

But now let’s return to a decade earlier when Euler became intrigued with the sum of the reciprocals of the squares problem. In 1731, Euler was able to equate the desired series with a sum involving a faster converging series. In this way, far fewer terms were required to evaluate the sum to any given degree of accuracy. In particular, Euler showed that ∞ n=1 1 = (log 2)2 + n2 ∞ n=1 1 2n−1 n2 . 644934, accurate to six decimal places. 64493406684822643647. n2 n=1 Even so, a closed form for the sum, if there even was one, seemed no closer now than when Jakob Bernoulli first considered the problem.

If there is such a beast, it must have at least eight prime factors (one of which is larger than ten million) and be at least 300 digits in length. So what do you make of this example due to the great French philosopher and mathematician Ren´e Descartes (1596–1650)? Let n = 32 · 72 · 112 · 132 · 22,021. All factors listed are relatively prime. Hence σ (n) = σ (32 ) · σ (72 ) · σ (112 ) · σ (132 ) · σ (22,021). But σ (32 ) = (33 − 1)/(3 − 1) = 13, σ (72 ) = (73 − 1)/(7 − 1) = 57, σ (112 ) = (113 − 1)/(11 − 1) = 133, and σ (132 ) = (133 − 1)/(13 − 1) = 183.

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