Egyptian algebra was rhetorical, that is, the problems and their solutions were expressed in words. There were symbols for addition (a pair of legs walking from left to right) and for subtraction (a pair of legs walking from right to left). Their algebra was limited almost entirely to linear equations in one variable, which were solved by the method of false position. This is most easily explained in modern terms. Given a formula of the form ax + x/b = c, where a, b and c are known, they would guess an answer for x, almost invariably the wrong one. Substituting this value into the left hand side of the equation they would get some value d (generally different to c) on the right hand side. Because the equation is linear, we can use proportions to deduce that the correct value of x is c/d times the false value.

Their knowledge of basic algebra was fairly advanced. They knew how to find the area of a rectangle, an isoscles triangle, and an isosceles trapezium. The area of a circle was taken as equal to that of a square on 8/9 of the diameter. The volumes of rectangular boxes and circular cylinders, all conceived of concretely as capacities of barns, were known. But perhaps the most remarkable result of Egyptian mensuration is to be found in Problem 14 of the Moscow Papyrus, where we find the equivalent, in modern notation, of the formula

V = (h/3)(a² + ab + b²)

for the volume V of the frustum of a square pyramid, where a and b are the sides of the square bases, and h is the height. Problem 56 of the Rhind Papyrus hints at the rudiments of trigonometry, and even a theory of similar triangles.

[Back: The Rhind Papyrus 2/n Table] [Next: Circle Formulae]