Babylonian Problems - NEW Pages

History of Mathematics in and for the Curriculum

‘Quadratic’ Problems: Babylon and Arabia

Many ancient cultures developed ways of dealing with problems involving areas. These were originally to do with measuring areas of land for reclaiming after a flood, for estimating the quantity of produce, or for tax purposes. There are large numbers of these problems from the Old Babylonian culture (1850-1600 BCE) that have been excavated and recently analysed, called ‘school problems’ which are the exercises used in training the scribes, or administrators. The techniques developed by the Babylonians and other cultures were incorporated and systematised into the new ‘algebra’ of al-Khowarizmi (c790 - c840 CE) and others in the 9th and 10th centuries.

Various types of problems became well known and standard methods were developed to deal with them1. Here, we look at ‘sum and product’ problems where the area and the semi-perimeter of a rectangular piece of land were given and the problem was to find the length and breadth of the rectangle; and ‘square and length equal a number’ problems where the length was some multiple of the side of the square.

1. Sum and Product problems This is a translation of a typical problem of this kind from a tablet found in Nippur, the capital of the Old Babylonian culture.

“Length, Width. I have raised, length and width. Surface: 252 I have put together length and width: 32 32 put together, 252 surface: 18 length, 14 width [Do it like this:] break off half of 32: this gives 16 raise 16 by itself: 256 leave out the surface: 256 − 252 = 4 find the side of this square 4: it is 2 put together 16 and 2: 18 length tear out 2 from 16: 14 width I have raised length and width. 252”

This type of problem was tackled by a method of ‘cut and paste’ geometry which was then developed into an algorithm (see my ‘Think of Two Numbers’ activity in Babylonian Classroom Activities 1), and in modern notation the problem was solved like this:

For e.g. Sum 7 and Product 12 Take half the sum S/2 7/2 Square it 49/4 Subtract the Product number 49/4 - 48/4 Find the square root of the result 1/2 This is the number you add to / subtract from 7/2 So we have (7/2 + 1/2) and (7/2 - 1/2) which are 4 and 3. To check, we add and multiply these two numbers together to get the original sum and product.

1 The problems chosen here were not the only problems where methods were developed. The MacTutor website gives more information and some good links. Don Allen’s website is a good one.

This algorithm works for any ‘sum and product’ of rational positive numbers.2 In ancient times, there were no negative numbers because in the solution process they were visualising areas, and negative areas are impossible.

See Babylonian Mathematics Classroom Activities 1 for variations and extensions. See NRICH http://nrich.maths.org/public/viewer.php?obj_id=6485

2. Square and length equal to a number. (The length was some multiple of the side of the square.)

The solution to this problem was based on the knowledge that, in our terms, (a + b)2 = a2 + 2ab + b2 or, any square can be seen as a symmetric diagram containing two smaller squares and two equal rectangles. For example, if a square added to six times its side is equal to 16, we make the diagram: The key to this solution is to realise that the area of the gnomon (the L- shaped part) is 16 and since the sides of the rectangles bordering the larger square are each 3, then the empty area must be 3x3 = 9. Thus the gnomon + the empty square is 16 + 9 = 25 and the side of the small square (the x2) must be 2. Looking at the algebra, x2 + 6x = 16 and comparing this with the diagram, we can see that the coefficient of the linear term must always be an even number, because there are two equal rectangles in the diagram.

Note that: 3 1. Even if the linear term is odd, we can always make it even; 3x can become 2ж цx . из 2шч 2. This is where the name ‘completing the square’ comes from. The Arab mathematicians in the 10th century CE made the correspondence between Euclid’s geometry and their newly invented algebra.

3. The right hand side of the equality does not always have to be a perfect square. The example often given to demonstrate Al-Khowarizmi’s solution is x2 + 10x = 39.

2 This algorithm was standard throughout the Mediaeval and Renaissance periods in Europe, right up to the 17th century. It was used by Cardano in 1545 when he discovered ‘Imaginary Numbers’.

‘A square and ten roots make thirty-nine.’ This is the first diagram Al-Khowarizmi shows in his book. He makes a central square for (x2) the unknown, and divides the ten roots into equal rectangles around the sides of the square. Here, the four blue rectangles and the green square together make 39. The next stage is to complete the large square by filling in the red squares in the corners. The short sides of the blue rectangles must be (5/2) units each. So each red square area is (5/2)2.

25 From the diagram, each of the red squares has area square units, and since there 4 are four of these, the total area of the red corners is 25. So (green square + blue rectangles) = 39, and the four red squares are 25, so the whole square is 39 + 25 = 64, and the length of the blue rectangle is 3 units.

The diagram above is reproduced in many history of mathematics books, and for some students the explanation can be quite baffling.

A simpler picture3 which was adopted by Abu Kamil later is as follows:

x2 + 10x = 39 Here, the gnomon (x2 + 10x) is used and it can be seen that the empty square is 25 square units. So the total area is 39 + 25 = 64. This makes the side of the large square 8 units and x, the unknown 3 units.

See NRICH http://nrich.maths.org/public/viewer.php?obj_id=6485 For my Development of Algebra Part 2

3 Katz uses this second diagram to demonstrate Al-Khowarizmi’s solution of x2 + 10 = 39, but Al Khowarizmi actually uses this diagram for a different problem.