Diophantus of Alexandria: A Study in the History of Greek AlgebraUniversity Press, 1910 - 387 páginas |
Dentro del libro
Resultados 1-5 de 32
Página 106
... perpendicular sides vanishing . G. Vacca ( in Bibliotheca Mathematica , 113. 1901 , pp . 358-9 ) points out that Fermat seems to have been anticipated , in the matter of these theorems , by Albert Girard , who has the following note on ...
... perpendicular sides vanishing . G. Vacca ( in Bibliotheca Mathematica , 113. 1901 , pp . 358-9 ) points out that Fermat seems to have been anticipated , in the matter of these theorems , by Albert Girard , who has the following note on ...
Página 204
... perpendicular sides ; therefore the square of the hypotenuse = x2 + 8x + 25 = a square . Again , the area is 3x + 6 ; and , as this is to be 6 , it must be six times a certain square ; that is , 2x + 6 divided by 6 must be a square ...
... perpendicular sides ; therefore the square of the hypotenuse = x2 + 8x + 25 = a square . Again , the area is 3x + 6 ; and , as this is to be 6 , it must be six times a certain square ; that is , 2x + 6 divided by 6 must be a square ...
Página 218
... perpendicular . This is easy , and the triangle will be similar to ( 17 , 15 , 8 ) ; the three triangles will then be formed ( 1 ) from 17 + 4.8 , 2.17-4.8 or 49 , 2 , ( 2 ) from 6.8 , 17-2.8 or 48 , 1 , ( 3 ) from 4.8 + 15 , 4.8-2.15 ...
... perpendicular . This is easy , and the triangle will be similar to ( 17 , 15 , 8 ) ; the three triangles will then be formed ( 1 ) from 17 + 4.8 , 2.17-4.8 or 49 , 2 , ( 2 ) from 6.8 , 17-2.8 or 48 , 1 , ( 3 ) from 4.8 + 15 , 4.8-2.15 ...
Página 219
... perpendicular in one is five times the product of the hypotenuse and one perpendicular in the other1 . four times the third , and we can also in the same way find three right - angled triangles of the same area ; we can also construct ...
... perpendicular in one is five times the product of the hypotenuse and one perpendicular in the other1 . four times the third , and we can also in the same way find three right - angled triangles of the same area ; we can also construct ...
Página 226
... perpendicular = 6r , base = x2 - 9 . 1 = 3 Thus + 9- ( x - 9 ) = 18 should be a cube , but it is not . Now 182.3 ; therefore we must replace 3 by m , where 2. m2 is a cube ; and m = 2 . We form , therefore , a right - angled triangle ...
... perpendicular = 6r , base = x2 - 9 . 1 = 3 Thus + 9- ( x - 9 ) = 18 should be a cube , but it is not . Now 182.3 ; therefore we must replace 3 by m , where 2. m2 is a cube ; and m = 2 . We form , therefore , a right - angled triangle ...
Otras ediciones - Ver todo
Diophantus of Alexandria: A Study in the History of Greek Algebra Thomas L. Heath Vista previa restringida - 1910 |
Diophantus of Alexandria: A Study in the History of Greek Algebra Sir Thomas Little Heath,Leonhard Euler Vista de fragmentos - 1964 |
Términos y frases comunes
absolute term added Algebra arithm Assume auxiliary triangle Bachet Books coefficient commentary cube difference Dioph Diophantus double-equation equal Euler expression factors find a right-angled find three numbers find three squares find two numbers four numbers fraction Frénicle given number given ratio gives a square Greek hypotenuse Iamblichus indeterminate indeterminate equations integral Lagrange Lemma letter Maximus Planudes method minus multiplied Nesselmann obtain Oeuvres de Fermat polygonal number Porisms prime number problem proposition quadratic rational numbers required numbers required squares right-angled triangle satisfy the conditions solution square number substitute subtract Suppose Tannery Theon of Smyrna theorem third number triple-equation unknown quantity whence whole numbers Xylander καὶ