Diophantos of Alexandria a Study in the History of Greek Algebra
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In our eReader you can find the full English version of the book. Read Diophantos of Alexandria a Study in the History of Greek Algebra Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book Diophantos of Alexandria a Study in the History of Greek Algebra
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To quickly assess the difficulty of the text, read a short excerpt:
(1751), pp. 49 sqq. Commentationes arithmeticae, I. Pp. 62-72).
BOOK VI The first equation then gives 54 15 - or 1 5; 2 36 = a square.
This equation we cannot solve because \ 5 is not tfie sum of two squares*. Therefore we must change the assumed triangle. Now (with reference to the triangle 3, 4, 5) I5w 2 = the continued product of a square less than the area, the hypotenuse, and one perpendicular ; while 36 = the continued product of the area, the perpen- dicular, and the difference between t...he hypotenuse and the perpendicular. Therefore we have to find a right-angled triangle (h, p, b, say) and a square (w 2 ) less than 6 such that m^hp ^pb . P(]i p} is a square. If we form the triangle from two numbers X lt X 3 and suppose that p=2. X^X^ and if we then divide throughout by (X l X^f which is equal to // -/, we must find a square, 5 s [= m-l(X^ - XJf\ such that z*/ip ^pb . P is a square. Ttie problem can be solved if X^, X^ are "similar plane numbers*? Form the auxiliary triangle from similar plane numbers accordingly, say 4, i.
What to read after Diophantos of Alexandria a Study in the History of Greek Algebra?
You can find similar books in the "Read Also" column, or choose other free books by Thomas Little Heath to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
(1751), pp. 49 sqq. Commentationes arithmeticae, I. Pp. 62-72).
BOOK VI The first equation then gives 54 15 - or 1 5; 2 36 = a square.
This equation we cannot solve because \ 5 is not tfie sum of two squares*. Therefore we must change the assumed triangle. Now (with reference to the triangle 3, 4, 5) I5w 2 = the continued product of a square less than the area, the hypotenuse, and one perpendicular ; while 36 = the continued product of the area, the perpen- dicular, and the difference between t...he hypotenuse and the perpendicular. Therefore we have to find a right-angled triangle (h, p, b, say) and a square (w 2 ) less than 6 such that m^hp ^pb . P(]i p} is a square. If we form the triangle from two numbers X lt X 3 and suppose that p=2. X^X^ and if we then divide throughout by (X l X^f which is equal to // -/, we must find a square, 5 s [= m-l(X^ - XJf\ such that z*/ip ^pb . P is a square. Ttie problem can be solved if X^, X^ are "similar plane numbers*? Form the auxiliary triangle from similar plane numbers accordingly, say 4, i.
What to read after Diophantos of Alexandria a Study in the History of Greek Algebra?
You can find similar books in the "Read Also" column, or choose other free books by Thomas Little Heath to read onlineMoreLess
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