By Asger Aaboe

ISBN-10: 0883856131

ISBN-13: 9780883856130

Professor Aaboe offers the following the reader a sense for the universality of significant arithmetic, placing each one selected subject into its right environment, hence bringing out the continuity and cumulative nature of mathematical wisdom. the fabric he selects is mathematically ordinary, but shows the intensity that's attribute of really nice proposal styles in every age. The good fortune of this exposition is because of the author's new angle to his topic. He correctly refrains from trying a basic survey of arithmetic in antiquity, yet selects, as an alternative, a couple of consultant goods that he can deal with intimately. He describes Babylonian arithmetic as printed from cuneiform texts found just recently, in addition to extra favourite issues built through the Greeks. even supposing each one bankruptcy could be learn as a separate unit, there are numerous connecting threads. Aaboe remains as with reference to the unique texts as is cozy for a latest reader, and the bibliography permits the pupil to delve extra deeply into any point of historic arithmetic that catches his or her fancy.

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Professor Aaboe provides the following the reader a sense for the universality of significant arithmetic, placing each one selected subject into its right surroundings, therefore bringing out the continuity and cumulative nature of mathematical wisdom. the fabric he selects is mathematically user-friendly, but indicates the intensity that's attribute of really nice notion styles in every age.

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**Example text**

The Diagonal of a Square. 6b is a copy and Figure 1 . 6a (the photograph is full size). We find three numbers: a = 30, b = 1,24,51,10, c = 42,25,35. We observe first that c = a-b if we insert semicolons in the proper places, for to multiply by 30 is essentially the same as to divide by 2. If a is taken to mean the side of a square, as suggested by the figure, and c the diagonal, then, by Pythagoras’ theorem, c2 = 2a2,and c = a d , so b ought to be an approximation of 6, that is, if it is interpreted as 1;24,51,10.

1) I have added the area and two thirds of the side of my square and it is 0;35. Two thirds of 1, the coefficient, is 0;40. Half of this, 0;20, you multiply by 0;20 (and the result) 0;6,40 you add to 0;35, and (the result) 0;41,40 has 0;SO as its square root. 0;20, which you multiplied by itself, you subtract from 0;SO and 0;30 is the (side of) the square. This example states and solves the quadratic equation x2 + $X = 0;35. Q is written with a special sign, and the first step is to convert to its sexagesimal equivalent 0;40.

We find three numbers: a = 30, b = 1,24,51,10, c = 42,25,35. We observe first that c = a-b if we insert semicolons in the proper places, for to multiply by 30 is essentially the same as to divide by 2. If a is taken to mean the side of a square, as suggested by the figure, and c the diagonal, then, by Pythagoras’ theorem, c2 = 2a2,and c = a d , so b ought to be an approximation of 6, that is, if it is interpreted as 1;24,51,10. This is indeed correct; for ( 1;24,51,10)2 = 1;59,59,59,38,1,40 is very close to 2.

### Episodes from the Early History of Mathematics by Asger Aaboe

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