By Prof. Dr. Bartel Leenert van der Waerden (auth.)
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There are lots of bits and items of folklore in arithmetic which are handed down from consultant to pupil, or from collaborator to collaborator, yet that are too fuzzy and non-rigorous to be mentioned within the formal literature. characteristically, it used to be an issue of good fortune and placement as to who discovered such folklore arithmetic.
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Extra info for A History of Algebra: From al-Khwārizmī to Emmy Noether
Chapters 8-11 contain problems of concern to merchants. A remarkable playful problem is the "problem of the 30 birds". A man buys 30 birds: patridges, doves, and sparrows. A patridge costs 3 silver co ins, a dove 2, and a sparrow l He pays with 30 coins. How many patridges, doves, and sparrows does he buy? The problem is, to solve the pair of equations x+y+z=30 3x+2y+tz= 30 in positive integers x, y, z. The only solution is x = 3, y = 5, z = 22. This problem is a variant of the "problem of 100 birds", which is found in Chinese, Indian, and Arabic sources.
7 -VB). It seems that Antonio Mazzinghi was the first to introduce, besides the traditional name "cosa" for an unknown quantity, a special name for another unknown. One of his problems reads: to find two numbers such that their sum is 18 and the sum of their squares 27. He now assurnes the first number to be "una cosa meno la radice d'alchuna quantita", and the second "una chosa piu la radice d'alchuna quantita". That is, he supposes the two numbers to have the form x-VY and x+Vy· I feel we cannot but admire the mathematical ability of Maestro Antonio.
From D one draws perpendiculars DZ and DE to BZ and Be. Omar now proves that DZ=x solves the equation (11). In modern terminology, let x=DZ and y=DE be the coordinates of D. The equation of the parabola is (12) x 2 = yc, or, in Omar's own words: "The square of DZ will be equal to the product of BZ and AB". The equation of the circle is (13) which Omar writes as a proportion "BE is to ED as ED is to EC". Just so, (12) is written as a proportion: "AB is to BE as BE is to ED". From these two proportions Omar concludes that EB=x is a solution, and the only solution of his problem.