The Ahmes Papyrus

The Ahmes was written in hieratic, and probably originated from the Middle Kingdom: 2000-1800 BC. It claims to be a ``thorough study of all things, insight into all that exists, knowledge of all obscure secrets." In fact, it is somewhat less. It is a collection of exercises, substantially rhetorical in form, designed primarily for students of mathematics. Included are exercises in

• fractions
• notation
• arithmetic
• algebra
• geometry
• mensuration

The practical mathematical tools for construction?

To illustrate the level and scope of Egyptian mathematics of this period, we select several of the problems and their solutions as found in the two papryi. For example, beer and bread problems are common in the Ahmes.

Problem 72. How many loaves of "strength" 45 are equivalent to 100 loaves of strength 10? Fact:

strength := $$1\over \mbox{\rm grain density}$$

Invoking the rule of three⁵, which was well known in the ancient world, we must solve the problem:

$${x\over 45}={{100}\over 10}$$

Answer: $x=100/10 \times 45 = 450$ loaves.

Problem 63. 700 loaves are to be divided among recipients where the amounts they are to receive are in the continued proportion

$${2\over3}:{1\over 2}:{1\over 3}:{1\over 4}$$

Solution. Add

$${2\over3}+{1\over 2}+{1\over 3}+{1\over 4}={7\over4}.$$

$$\begin{eqnarray*} {{700}\over 7/4}&=& 700\cdot{4\over7}\\ &=& 700({2\over7}+{2\... ... 28})\\ &=& 700({1\over 2}+{1\over14})\\ &=& 350+50\\ &=& 400 \end{eqnarray*}$$

The first value is 400. This is the base number. Now multiply each fraction by 400 to obtain the recipient's amount. Note the algorithm nature of this solution. It reveals no principles at all. Only when converting to modern notation and using modern symbols do we see that this is correct We have

$${{x_1}\over x_2}={{{2\over3}}\over {1\over2}},\quad {{x_2}\over x_3} ={{{1\over2}}\over {1\over3}},$$

etc. This will be the case if there is a base number $a$ such that

$$\begin{eqnarray*} x_1={2\over3}a\\ x_2={1\over2}a \\ x_3={1\over3}a\\ x_4={1\over4}a \end{eqnarray*}$$

Thus

$$x_1+x_2+x_3+x_4=({2\over3}+{1\over2}+{1\over3}+{1\over4})a=700$$

Now add the fractions to get ${7\over 4}$ and solve to get

$$a=400.$$

Now compute $x_1,\ x_2,\ x_3,\ x_4$.

The solution of linear algebra problems is present in the Ahmes. Equations of the modern form

$$x+ax=b\quad\mbox{\sf or} \quad x+ax+bx=x,$$

where $a,b,$ and $c$ are known are solved. The unknown, $x$, is called the heep. Note the rhetorical problem statement.

Problem 24. Find the heep if the heap and a seventh of the heep is 19. (Solve $x+x/7=19$.)

Method. Use the method of false position. Let $g$ be the guess. Substitute $g+ag=c$. Now solve $c\cdot y=b$. Answer: $x=g\cdot y$. Why?

Solution. Guess $g=7$.

$$7+1/7\cdot 7=8$$

$$19\div8=2+{3\over8}=2+{1\over4}+{1\over8}$$

Answer:

$$7\cdot(2+{3\over8})=7(2+{1\over4}+{1\over8})=16+{1\over2}+{1\over8}$$

Geometry and Mensuration Most geometry is related to mensuration. The Ahmes contains problems for the areas of

• isosceles triangles (correct)
• isosceles trapezoids (correct)
• quadrilaterals (incorrect)
• frustum (correct)
• circle (incorrect)
• curvilinear areas

In one problem the area for the quadrilateral was given by

$$A=({{b_1+b_2}\over 2})({{h_1+h_2}\over 2})$$

which of course is wrong in general but correct for rectangles. Yet the ``Rope stretchers" of ancient Egypt, that is the land surveyers, often had to deal with irregular quadrilaterals when measuring areas of land. This formula is quite accurate if the quadrilateral in question is nearly a rectangle.

The area for the triangle was given by replacement $b_2=0$ in the quadrilateral formula

$$A=({{b_1}\over 2})({{h_1+h_2}\over 2})$$

On Rigor

There is in Egyptian mathematics a search for relationships, but the Egyptians had only a vague distinction between the exact and the approximate. Formulas were not evident. Only solutions to specific problems were given, from which the student was left to generalize to other circumstances. Yet, as we shall see, several of the great Greek mathematicians, Pythagoras , Thales, and Eudoxus to name just three, went to Egypt to study. There must have been more there than some student exercises to consider!

Problem 79. This problem cites only ``seven houses, 49 cats, 343 mice, 2401 ears of spelt, 16,807 hekats."

Note the similarity to our familiar nursery rhyme:

As I was going to St. Ives,
I met a man with seven wives;
Every wife had seven sacks,
Every sack had seven cats,
Every cat had seven kits.
Kits, cats, sacks, and wives,
How many were going to St. Ives?

This rhyme asked for the very impractical sum of all and thus illustrates some knowledge and application of geometric progressions.

Problem 50. A circular field of diameter 9 has the same area as a square of side 8. This gives an effective $\pi=3\,{1\over6}$.

Problem 48 gives a hint of how this formula is constructed.

Trisect each side. Remove the corner triangles. The resulting octagonal figure approximates the circle. The area of the octagonal figure is:

$$9\times 9 -4({1\over 2}\cdot3\cdot3)=63\approx 64=8^2$$

Thus the number

$$4({8\over9})^2 = 3\,{{13}\over 81}$$

plays the role of $\pi$. That this octagonal figure, whose area is easily calculated, so accurately approximates the area of the circle is just plain good luck. Obtaining a better approximation to the area using finer divisions of a square and a similar argument is not easy.

Geometry and Mensuration

Problem 56 indicates an understanding of the idea of geometric similarity. This problem discusses the ratio

$${{\mbox{\rm rise}}\over \mbox{\rm run}}$$

The problem essentially asks to compute the $\cot\alpha$ for some angle $\alpha$. Such a formula would be need for building pyramids.

Note the obvious application to the construction of a pyramid for which the formula for the volume, $V=\frac{1}{3}b^2h$, was known. (How did they find that?)

Geometry and Mensuration

The are numerous myths about the presumed geometric relationship among the dimensions of the Great Pyramid. Here's one:

[perimeter of base]=
    [circumference of a circle of radius=height]

Such a formula would yield an effective $\pi=3{1\over7}$, not $\pi=3{1\over6}$, as already discussed.