The Algebra of the Egyptians and Babylonians, as it was practised from about 2000 B.C., was purely rhetorical; that is, all calculations were expressed in words. Rules and Methods were then used to solve particular problems with no attempt at generalisation. However we can now discern in these particular examples some basic ideas of what was to become the algebra of today.
The majority of the Egyptian problems were of the type which we would now solve using linear equations in one unknown. An example, from the Rhind Papyrus, states a problem:
"Two thirds added and one third taken away : 10 remains."
In our notation this would be:
(x + x) - (x + x) = 10
which leads to x = 10, and hence x = 9
The Egyptian "method" is given as :
"Make one tenth of this 10 : the result is 1 : remainder 9"
This suggests that they could see that the unknown was nine tenths of 10 (c.f. x = 10 above), so they had to find one tenth of 10 and subtract it from 10 to give the answer. Such a "method" applied only to the problem in question and we cannot tell whether or not they had a general method for solving similar problems. This is true for all Egyptian and Babylonian problems which we would now solve using algebra.
Both civilizations have left numerous problems such as that above, often stated in an abiguous way. Some, particularly Babylonian ones, involved problems we would now solve using simultaneous linear or quadratic equations. As an example, a Babylonian problem asks for :
"Two numbers whose sum is 13/2 and whose product is 15/2."
Our method would be to eliminate y from the equations
x + y = 13/2 (1)
xy = 15/2 (2)
and then solve a quadratic in x. The original did not state the equations in this way, but described a calculation which would correspond to the following steps:
(i) Divide equation (1) by 2 : (x + y)/2 = 13/4 (3)
(ii) Square equation (3) : ((x + y)/2) = 169/16 (4)
(iii) Subtract (2) from (4) : ((x + y)/2) - xy = 169/16 - 15/2 = 49/16 (5)
(iv) Take square roots (they used positive ones only!) of both sides of equation (5), making use of the fact that (x + y) - 4xy = (x - y) : (x - y)/2 = 7/4 (6)
(v) Add (3) and (6) to obtain x = 5, and then y = 3/2.
These examples illustrate the type of problem that was solved and the methods used, though the notation was not algebraic in the modern sense, but wordy and cumbersome. This was also the way they described "formulas" such as those for geometric areas and solids. One remarkable such example obtained by the Egyptians was for the volume of a truncated pyramid with square base. In our notation this would be:
V = h ( a + ab + b )
where "a" and "b" are the sides of the top and bottom squares, and "h" is the height.
[ Comments ] [ Module Leader ]
These pages are maintained by M.I.Woodcock.