An Egyptian Maths Problem

To get an idea of just how dexterous the Egyptians were with unit fractions, we just have to look at Problem 31 from the Rhind papyrus. In the text the problem, the working and the answer are given. Here we shall just give the problem (translated into English) and its solution, and leave the reader to work out the intermediate steps!

The problem: A quantity, its 2/3 its 1/2 and its 1/7, added together, become 33. What is the quantity?

The solution:

A Babylonian Quadratic Equation

The following is an example of quadratic problem (translated into English and with each sexagesimal place separated by a comma, and the integer and fractional part separated by a semi-colon):

I have subtracted the side of my square from the area: 14, 30. You write down 1, the coefficient. You break off half of 1. 0; 30 and 0; 30 you multiply. You add 0; 15 to 14, 30. Result 14, 30; 15. This is the square of side 29; 30. You add 0; 30, which you multiplied, to 29; 30. Result 30, the side of the square.

Put into decimals and modern algebraic notation, mathematically the problem is equivalent to solving the equation:

x2 – x = 870 [(14 x 60) + 30 = 870]

by the method we call ‘completing the square’.

However, while such a transformation can help a modern reader to get to grips with the mathematical problem being solved, it also strips out all contextual information about Babylonian mathematical practice. For if we look again at the original problem we see that it is much more than just a bare equation. It contains (what seems to us) the rather strange idea of subtracting the side of a square from its area, it demonstrates an algorithmic procedure, it provides evidence of the sexagesimal system, and it shows that the scribe must have had a way of knowing that the square root of 14, 30; 15 (870¼ ) is 29; 30 (29½), since it seems unlikely that he could do that kind of calculation in his head.

Content last updated: 13/01/2005

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