Download A History of Algebra: From al-Khwārizmī to Emmy Noether by Prof. Dr. Bartel Leenert van der Waerden (auth.) PDF

By Prof. Dr. Bartel Leenert van der Waerden (auth.)

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Extra resources for A History of Algebra: From al-Khwārizmī to Emmy Noether

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Then it follows that x = u - v satisfies the required equation In Cardano's geometrical terminology the reduction of (U-v)3 to is very cumbersome, but the fundamental idea is the same. It is easy to determine u and v from the conditions (6) and (7). From (7) one finds uv=2 hence Now the difTerence and the product of the two cubes u 3 and v3 are known, and one finds u3 =V 108 + 1O v3 =V108-1O, so u and v are cube roots of known numbers, and we have x=YV108 + 10 -YV108-10. e. one-third the coefficient of x).

182 and 183 deserve special attention, because they involve irreducible mixed cubic and biquadratic equations. In modern notation, these four equations can be written as (1) =n (2) dx+cx +bx +ax =n (3) 2 3 4 dx +cx 2 +ax 4 =n+bx 3 dx +ax 4 =n+cx 2 +bx 3 . (4) Dardi presents rules for the solution of these equations. However, as he himself admits, his rules are valid only in the special cases considered, not in general. In aIl cases, he first instructs us to divide all coefficients by a, so that, for instance, Equation (1) is reduced to the simpler form (1 ') The solution of (1') is given as (5) x=V(c/W+n-c/b.

V'g, 52 Chapter 2. Algebra in Italy Dardi's problem was: How can Iwrite the solution (11) in a form like (9), in which not special numbers like 5 and g (equal to 18 or 28) occur, but only expressions which can be calculated from the coefficients (13)? Now let us try to find out how Dardi solved his problem. Let's consider the three terms of (11) separately. The first term 5 was obtained by halving the 10 given in the Problem P, and b=20 was found by doubling this term. So, the first term in (11) can only be generalized to b/4.

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