The Arabic world and the cubic

 

The Arabs made a great contribution to the development of mathematics, is said to be their role as intermediary agent between Indian and Greek mathematics and the modern world. They had access to the great Indian and Greek mathematicians works in the original language and translated them to Arabic.

Al-Khwarizmi's (c. 780-c. 850) use of geometrical justifications of algebraic manipulations together with the fact that the Elements existed in two distinct translations from Greek into Arabic by his contemporary at the House of Wisdom, al-Hajjaj ibn Yusuf ibn Matar, suggest a line of descent from Euclid. 

On the other hand, because his treatment of practical geometry so closely followed that of the Hebrew text, Mishnat ha Middot, which dated from around A.D. 150, the evidence of Semitic ancestry exists. Al-Khwarizmi's concern with practical algebra and his treatment of equations through the second degree betray a vestige of the Babylonian line, while his totally rhetorical style points to a remote Hindu ancestor and a lack of contact with later Greek texts, particularly the Arithmetica of Diophantus. 

In fact, since the first known Arabic translation of the Arithmetica was not completed by Qusta ibn Luqa until the middle of the ninth century or later, we can be fairly certain that the more theoretical ideas of Diophantus had not yet entered the environment of, and so had not come into competition with, Arabic mathematics.

The Arabs did not make full use of the advances of the Hindus so they had neither negative quantities nor abbreviations for their unknowns.

However Muhammad ibn Musa Al-Khwarizmi gave a classification of different types of quadratics (although only numerical examples of each). The different types arise since al'Khwarizmi had no zero or negatives. He has six chapters each devoted to a different type of equation, the equations being made up of three types of quantities namely: roots, squares of roots and numbers i.e. x, x and numbers.

Squares equal to roots.
Squares equal to numbers.
Roots equal to numbers.
Square and roots equal to numbers, e.g. x2 + 10x = 39.
Squares and numbers equal to roots, e.g. x2+ 21 = 10x.
Roots and numbers equal to squares, e.g. 3x + 4 = x2.

al'Khwarizmi gives the rule for solving each type of equation, essentially the familiar quadratic formula given for a numerical example in each case, and then a geometrical proof for each example utilizing the method of completing the square.

The Arabs solved some cubics algebraically, and gave a geometrical explanation. This was done, for example, by Tabit ibn Qorra (836-901) and by al-hasan ibn al-haitham (965-1039).

`Umar al-Khayyami (Omar Khayyam) (c. 1048-c. 1125) went further and used conical sections to to solve general equations of third degree.

Khayyami used pure algebra to solve quadraatics, but he verified the results using constructions. He solved particular cubics also using constructions. He did admit that he couldnīt find an algebraic formula for the general cubic, but he invited other mathematicians to look for one. He also classified cubic equations, and he described those which he can solve and those which he couldnīt .

Khayyam believed that cubic equations could only be solved  geometrically, by using conic sections. His method, used in Algebra (c 1079) is based on a geometrical construction where the solution is found at the intersection between a parabola and a semicircle.

He solves the equation: x3 + Bx = C

by writing it as

x3 +p2x = p2q

 

where B = p2

 

and C = p2 q

A parabola with equation

x2 = p y and

a circle with equation

x2 + y2 = q x is constructed

A positive solution of the equation is found in the intersection between the curves. The following example shows the solution of x3 + 4x = 16

The equation x3 + 4*x = 16 rewrites to  

x3 + 22 *x = 22 * 4.

The equation for the parabola will then be
                       x2 = 2y

and for the circle x2 + y2 = 4x  which gives us         x2 - 4x + y2 = 0       which leads to x2 -4x + 4 +y2 =4 and hence the equation of the circle  becomes; 
                      (x - 2)2 +y2 =22

The circle cuts the parabola in the points (0,0) and (2,2).

Only positive answers were of course regarded so the answer is x = 2.

 

 

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