Equations In Greek Algebra example essay topic
The Pythagoreans studied Babylonian's algebra and used the Babylonian's method for solving equations in Greek algebra. Euclid recorded Pythagorean's findings. Euclid's solutions to problems were similar to Babylonian's solution to similar problems. Euclid rewrote Babylonian's problems so that the problems do not include fractions or irrational numbers. Plato and Euclid did not use fractions or irrational numbers, only Eratosthenes, Diophantus, and Archimedes. There is evidence that the Pythagoreans discovered irrational numbers, but since they did not believe them, they pretended that they did not discover it.
Greek algebra was an integration of geometry and algebra; it was referred to as geometric algebra. Early Greek algebra of the Pythagoreans and Euclid, Archimedes, and Apollonius was geometric because the Greeks had difficulties with irrational and even fractional numbers. Geometric algebra consists of "line segments, areas, and volumes [which] are strictly kept apart" (van der Waerden 74). The problems and solution in Greek geometric algebra was written in the rhetorical style, where every word is written out. In geometric algebra, the "fundamental relation between line segments or between areas is equality" (van der Waerden 76). There are three fundamental operations in this algebra: the sum of two line segments, the sum of two polygons, and the product of two line segments.
The first operation: The sum of two line segments a and b is a line segment c, which can be divided into two parts a' and b' that are equal to a and b respectively. Modern notation: a + b = c The second operation: The sum of two polygons A and B is a polygon C, which can be divided into two parts A' and B' that are equal to A and B. Modern notation: A + B = C The third operation: The product of two line segments a and b is a rectangle R contained by two line segments a' and b' that are equal to a and b. Modern notation: a x b = R When finding the product of line segments of a and b, the line segments are not always perpendicular so they need to be replaced with line segments a' and b,' which is perpendicular and forms a rectangle. The rectangle is "an object of algebraic operations" (van der Waerden 77).
Geometric Algebra Geometrical algebra is the area of mathematics that deals with geometric concepts and proofs in algebraic applications. Although Asia, Mesopotamia, and Egypt had the ideas of geometrical algebra, classical Greek mathematicians are normally credited with its development. The Greeks used this geometric algebra because they used a rhetorical style, which limited their ability to work out algebraic problems step by step. They would apply algebraic problems to geometric shapes-this technique also avoided the explanations of irrational numbers. The Greeks knew that the irrational numbers worked, but they did not believe in their existence. Therefore this method of geometric algebra was developed.
When we geometrically solve a quadratic equation, we are doing geometrical algebra. Greeks did algebra on a geometric basis. 1 For example: "O 2 1 The Greeks never solved algebraic equations in our symbolic style. They would work out several physical examples using geometry. With trial and error, the Greeks were able to show that their answers worked. The Greek period of Algebraic history runs up till the Romans conquered them around 400 ce.
Diophantus of Alexandria (Syncopated Style) Diophantus is a famous Greek mathematician, well known for his work in algebra. For several centuries, rhetorical algebra was the only algebra until Diophantus of Alexander pioneered syncopated algebra. He decided that rhetorical style was clumsy, so he started making abbreviations; and it was useful with powers up to the sixth and their reciprocals. Most people say it is not a whole lot better than the rhetorical style, but it gives a different perspective. Diophantus wrote a book called Arithmetica that tremendously influenced the number theory and algebra.
The word arithmetic comes from the Greek word arithmetike, which is composed of the Greek word for number (arith mos) and for science (techn e). The notation in Arithmetica helped move algebraic notation from rhetorical to syncopated, as well as influencing the evolution of some individual symbols. In Arithmetica "he gives an ingenious treatment of indeterminate equations usually two or more equations in several variables that have an infinite number of rational solutions" (NCTM, 240). His method is clever, however it lacks the development of a systematic method for finding general solutions. An example of Diophantus's syncopated algebra: or 2 x 3 + 8 x - (5 x 2 + 4) = 44 This is the syncopated style of writing equations. Syncopated style is where everything is abbreviated.
Rhetorical Algebra (Style) Rhetorical algebra can be traced back to both early Babylonian and Egyptian. This stage of algebra refers to the stage with when everything was written in words without the use of mathematical symbols. This type of algebra is also said to be a verbal or oral algebra since they wrote out everything they said. This was also known as prose algebra because of the written words. Since the people of this period were limited to writing material and their ability to write, most obtainable material were clay tablets. With the clay tablets they simply wrote out exactly what they said.
One of the earliest books written in rhetorical or prose algebra included the following problem: "ten and thing to be multiplied by thing less ten". If they had symbols as the symbols today the problem would be written as (x + 10) (x - 10) = x [sup 2]-100. Since, it is cumbersome to write out each step in words, rhetorical algebra ended as symbolic algebra evolved. Rhetorical algebra lasted until the 16th century when symbolic algebra became more renown. Difficulties are usually encountered in reading the problem in rhetorical algebra because of the non-existence of a standardized, efficient symbol. The rhetorical style used by the Greeks consisted of five simple steps: state the problem give the data give the answer explain the solution check the answer An Example: 1. state the problem Length, width.
I have multiplied length and width, thus obtaining area: 252. I have added length and width: 32. Required; length and width. 2. Give the data [Given] 32 the sum; x+y = k 252 the area. xy = P 3.