1
2
3
4
History of Math
For the Liberal Arts
5
6
CHAPTER 2
7
8
Egyptian Mathematics
9
10
11
12
Lawrence Morales
13
14
15
Seattle Central
Community College
16
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 1
16
Table of Contents
17
PART 1: Introduction to the Egyptian Number System........................................................... 3
18
Introduction and Historical Background.................................................................................... 3
19
Egyptian Numbers and Base System........................................................................................... 4
20
Adding in the Egyptian Numeral System .................................................................................... 5
21
Egyptian Multiplication .............................................................................................................. 7
22
PART 2: Unit Fractions, Division, and the 2÷n Table............................................................ 13
23
Fractions and Division ............................................................................................................. 13
24
Egyptian Division...................................................................................................................... 19
25
The 2÷n Table ........................................................................................................................... 22
26
PART 3: Algebra and the Aha Problems ................................................................................. 27
27
Introduction............................................................................................................................... 27
28
Lines and Their Proportional Nature ....................................................................................... 27
29
Ideas Behind the Method of False Position .............................................................................. 28
30
The Egyptian Method of False Position.................................................................................... 31
31
PART 4: Homework Problems .................................................................................................. 39
32
Egyptian Numbers and Conversions......................................................................................... 39
33
Egyptian Multiplication ............................................................................................................ 39
34
Egyptian Fractions.................................................................................................................... 40
35
Powers of 2 ............................................................................................................................... 40
36
Unit Fractions........................................................................................................................... 41
37
Egyptian Division...................................................................................................................... 41
38
Egyptian 2/n Table.................................................................................................................... 42
39
Algebraic Practice .................................................................................................................... 42
40
Method of False Position .......................................................................................................... 42
41
Writing ...................................................................................................................................... 43
42
Endnotes....................................................................................................................................... 45
43
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 2
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
PART 1: Introduction to the Egyptian Number System
Introduction and Historical Background
Ancient Egyptian mathematics provides us with a rich source of material to explore.
We will look at their numeration system to compare not only their base system with
ours, but also their numeric symbols. We will study their algorithms for
multiplication and division, and even though they look very different than what we
are used to, we will see that they give us a lot of insight into what multiplication and division are
really all about. We will play with their “unit fractions” to see how they handled the
mathematical objects that so many of us seem to despise. (I don’t understand that…fractions are
your friends…fractions are your friends.) And we will look at Egyptian “algebra,” the methods
they used to solve “story problems.” Their methods, without the use of variables, will help us to
understand linear thinking and proportionality…topics that you have seen throughout your
mathematical experiences.
When people think about ancient Egypt, they inevitably think about
pyramids, pharaohs, the Nile, King Tut, and other such ideas. When
was the last time anyone ever mentioned their mathematics in the same
breath? Well, it’s about time they got their due credit, and we’ll try to
give them a little bit of that in this chapter.
Early in history, civilization reached a “high level” in Egypt.1 The Nile
River provided a fertile land with an agreeable climate. Egypt
experienced long periods of peace since the desert that surrounded it
provided a natural barrier to other groups that might attempt to
challenge them for dominance. These peaceful periods allowed the society to advance very
quickly.
By 3000 B.C.E., the Egyptians were using the natural seasons of the year to their advantage in
the area of agriculture. As the Nile flooded the lands, it provided very rich and fertile ground
with which to grow crops. They Egyptians made use of a complicated irrigation system in this
endeavor. It was important, therefore, to know when the rainy season would arrive. This meant
that it was vital that they be able to study the stars (astronomy) so that they could develop a
somewhat reliable calendar upon which they could depend for planning purposes. Also, Egypt
was made up of a large land area and as they grew and became more complex as a society, the
need to keep records of populations, taxes, armies, and other records pushed them towards the
development of a system of writing and numerals that could be used to keep track of such
information. In turn, the study of mathematics and the development of numbers became a
natural part of their historical evolution.
Most of the best information that we have about Egyptian Mathematics comes from the Rhind
Papyrus. Named after Henry Rhind, who bought it in the late 19th century, it is a collection of
work by a scribe named Ahmes who wrote it in about 1650 B.C.E.2 You can see a small portion
of it to the right3. In this text, Ahmes presents about 80 problems, some practical and some not.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 3
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
There are other smaller papyri that also give us information on their math, but not as extensively
as the Rhind papyrus does. Parts of the Moscow papyrus are pictured below. 4
Egyptian Numbers and Base System
The Egyptians used a base-ten (decimal) system, similar to our own. The main difference
between their system and ours is that they did not have a positional system. Recall that in a
positional base system, the position of a number symbol determines its value. For example, the
three in each of the following Hindu-Arabic numbers implies a different value due to the
difference in its position within the number:
23845 and 8923
In the first case, the 3 represents thousands, in the second case it represents ones.
The Egyptians did not use this kind of positional system, so the placement of a symbol did not
determine value. Instead, the number of symbols determined value. As such, we should know
what their symbols were. The hieroglyphic symbols
for their numbers are shown in the given figure.5
As you can see, they have a base-ten system since
unique symbols are created for each successive power
of ten. The symbol for 1 is a single stroke, perhaps a
stick. The symbol for 10 is perhaps the drawing of a
hobble for cattle6. The symbol for 100 is apparently a
coil of rope. The symbol for 1000 is thought to be a
lotus flower. 10,000 is represented by a bent finger.
100,000 is represented by a tadpole, and 1,000,000 is drawn as a man (or god) with his arms
raised in the air. To represent any number, these symbols were used in the appropriate quantities.
Example 1
To represent the number 3,265, we would need 5 single strokes, 6 “hobbles”,
2 coils of ropes, and 3 lotus flowers. The number would look like the
following:7
133
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 4
134
135
136
137
138
Example 2
139
140
141
142
143
144
Example 3
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
Here is the number 459,623.
Look at the representation for 5082. What do you notice?
Solution:
The Egyptians did not have a symbol for zero. There are zero hundreds…so
they simply did not draw/write any rope coils (the hundreds symbol) in this
number’s representation. ♦
Check Point A
Write the following number in Egyptian symbols: 15,603.
Solution
See endnotes for answer.8
Adding in the Egyptian Numeral System
Addition in this system is straightforward since all that needs to be done is to combine symbols,
literally. For each ten of a particular symbol, you exchange them for one symbol of the higher
place. For example, for every 10 single strokes that you see, you would erase them and replace
them with a “hobble.”
Example 4
Add
and
.
Solution:
We can see that there are a total of 7 single strokes, 5 hobbles, 3 coils, and 4
lotus flowers. Hence, the sum of these two numbers is:
174
175
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 5
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
These symbols simply state that 234 + 4123 = 4357. But note that if we were
going to add these two numbers using our “modern” algorithm, we would
place them vertically, make sure all the place values line up, and then add. It
would require that we knew that 4+3=7, 2+3 = 5, etc, and thus requires the
memorization of several addition combinations as well as a good knowledge
of the algorithm. With the Egyptian method, we simply gather together like
symbols and exchange groups of ten whenever necessary. The next example
shows when such an exchange is needed.♦
Example 5
Add
and
Solution:
In this example, we see that when we gather the single strokes together, we
have a total of 11. So we must exchange ten of those for one “hobble” and we
have one single stroke left over. That is,
is converted to
Proceeding from here we see we know have 7 + 1 + 1 = 9 “hobbles,” with no
need to carry. There are 8 + 4 =12 coils, so we need to covert that to 1 lotus
flower (for 10 coils) and 2 coils. Finally, we take the 5 + 1 + 1 = 7 coils. The
total then, is:
A visual representation of the gathering and carrying process is shown in the
picture below:
♦
204
205
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 6
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
Check Point B
Add the following two numbers without using the Hindu−Arabic numerals
and working only with Egyptian numeral symbols.
Solution
See the endnotes for final answer.9
Egyptian Multiplication
One of the most interesting aspects of Egyptian mathematics is their system of multiplication.
While some people might say that it is cumbersome (and it may be), when properly understood,
it beautifully illustrates what multiplication is really about. You may recall that multiplication is
often called “repeated addition.” For example, when we say that 4×5 = 20, what we are saying is
that 4 groups of 5, when added together, give 20. That is:
5 + 5 + 5 + 5 = 20
4 groups of 5
This is fine when we have smaller numbers. However, for problems like
357×239, who wants to add a string of 357 instances of 239? Yikes. To avoid
this tedious work, we have a multiplication algorithm that makes things easier
for us. In the United States, the algorithm would look something like what you
see to the right.
Of course, when we do this, we do so almost without thinking. We
probably have no idea what this algorithm has to do with repeated
addition. And if we want to see what the link is between these two
mathematical operations, we would really have to think about it for a
while.
However, the Egyptian method of multiplication makes it much easier
to see the link between multiplication and addition. Indeed, it is
relatively easy to see multiplication as repeated in the Egyptian
method.
1
7
8
×
3
0
1
5
3
2
2
7
4
3
5
3
1
1
0
2
7
9
3
0
0
3
Think About It
Why does the modern
method of
multiplication, as
shown above, actually
work?
The method the Egyptians use to multiply is often called the method of doubling. The method
involves adding together appropriate amounts that have been obtained by doubling. The method
is best described with a detailed example and explanation, followed by other examples where the
notation is more compact and “Egyptian.” We’ll start by doing 11×80.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 7
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
Example 6
Multiply 11×80.
Solution:
Keep in mind that what this means is that we want to add up 11 groups of 80.
That is, we want:
80+80+80+80+80+80+80+80+80+80+80=?
11 groups of 80
To simplify this process we will use the method of doubling, only written in a
more modern form. (We’ll show the Egyptian method later.) This method
starts with the obvious statement that:
1 group of 80 is 80
We now double that to get:
2 groups of 80 is 160
But 2 groups of 80 are not enough…we want 11 of them, after all. So double
again, since it’s easy to do, and we get:
4 groups of 80 is 320
Note that I double 2 to get 4, and I double 160 to get 320. 4 groups are still
not enough, so we double again, getting:
8 groups of 80 is 640
And again…
16 groups of 80 is 1280
Here we stop because we have 16 groups of 80, which is more than we need.
We never got 11 of them. So we simply assemble all the appropriate groups
so that we have 11 of them. Note that:
8 groups + 2 groups + 1 group = 11 groups. Therefore:
1 group gives 80
2 groups give 160
8 groups give 640
We add them all up and we have 80 + 160 + 640 = 880, which can easily be
checked.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 8
288
289
290
291
Now, the Egyptians would not have written all of that out. They had a more
compact notation for this that looked something like what follows. For now,
let’s just use Hindu-Arabic numbers to see how they would do this problem.
\ 1
\ 2
4
\ 8
80
160
320
640
Total 880
292
293
294
295
296
297
298
299
Note that the first line is the obvious statement that one group of 80 is 80. The
second line doubles the first and says 2 groups of 80 are 160, and so on. The
backward slash marks are used to indicate which groups will eventually be
added together to get the final desired sum.
Now let’s see what this might have looked line in hieroglyphic form:
\
\
This is like an
equals sign.
\
300
301
302
303
304
305
306
307
Now that you know how they wrote their numbers, you should be able to see
that these two tables are representing the same problem. The little symbol that
looks like a TV set means that the answer follows…it’s sort of like our
modern equal sign. Let’s try to do a couple of examples using this compact
notation. ♦
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 9
308
309
310
311
312
313
314
Example 7
Multiply 35×15.
Solution:
We will provide a column with commentary that would not appear in an
Egyptian papyrus, of course.
\ 1
\ 2
4
8
16
\ 32
315
316
317
318
319
320
321
15
30
60
120
240
480
Commentary
1 group of 15 is 15
2 groups of 15 is 30
4 groups of 15 is 60 etc.
This line is not needed to produce 35 groups of 15
⇔
⇔
⇔
⇔
⇔
⇔
Sum 525 ⇔ The final result is 525 when we add 15+30+480
which corresponds to 1+2+32 = 35 groups of 15.♦
Example 8
Multiply 125×4.
Solution:
/ 1
2
/ 4
/ 8
/ 16
/ 32
/ 64
4
8
16
32
64
128
256
Commentary
1
group of 4
⇔
⇔ 2 groups of 4 is 8
⇔ 4 groups of 4 is 16, etc
Etc…
Sum 500 ⇔ The final result is 500 when we add up all of the
marked groups.9
322
323
324
325
326
327
328
This is the basic
format that the
Egyptians would use,
except they would use
their own symbols.
Related homework
problems should take
on this basic format.
This would not appear in the Egyptian scrolls…it is
provided here only as an explanation of what they
were doing.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 10
Strategies to Use in the Doubling Method
1.
2.
3.
328
329
330
331
332
333
334
Keep doubling until the number in the left column exceeds the number of groups you
need.
Do not include a number in the left column that is larger than the number of groups you
need.
When determining what rows to mark, start with the last row and continue to add rows
above it until you have the total number of groups required. (For example, in the last
problem, we wanted 125 groups of 4. So we start with the last row, giving us 64 groups.
We move to next row up and add 64+32=96. This is still not enough. 96+16=112.
Continuing we have 112+8=120. Again, 120+4=124. Now, 124+2=126, which is too
much, so we skip the row with 2 groups and move to the first row, 124+1=125.)
Let’s look at one last example.
Example 9
Multiply 42×32.
Solution:
1
/ 2
4
/ 8
16
/ 32
32
64
128
256
512
Commentary
⇔ 1 group of 32
⇔ 2 groups of 32
⇔ 4 groups of 32
⇔ This line is not marked since 32 + 16 is 48, more than
the 42 groups we need.
1024
⇔ If we double again we get 64 groups, which is more
than the 42 we need, so we stop here.
Sum 1344
We add up the rows of 32,8,and 2 groups for a total of
32.
335
336
337
Let’s see what this would look like in hieroglyphics.
/
/
/
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 11
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
Check Point C
Multiply 43×12 using the Egyptian method.
Solution
Check the endnotes for the solution.10
Check Point D
Multiply 130×25 using the Egyptian method.
Solution
Check the endnotes for the solution.11
Why does the method of doubling work? Well, it can be shown that any positive integer can
be written as the sum of powers of 2. For example:
1 = 20
2 = 21
3 = 20 + 21
4 = 22
5 = 20 + 22
6 = 21 + 22
7 = 20 + 21 + 22 etc.
Example 10
Write 75 as the sum of power of 2.
Solution
This problem is similar to writing 75 in base 2. From Chapter 1, we can check
that 7510 = 10010112. This means we can write 75 as the sum of powers of 2
as follows:
75 = 26 + 23 + 21 + 20
You should check that this is correct and that you understand how 10010112 is
related to 26 + 23 + 21 + 20 . ♦
Powers of 2 are the only ones for which it is possible to write all of the positive integers. (Try
writing 2 as a single power of 3 and you will see what I mean….powers of three fail very
quickly!) This is probably not the reason why the Egyptians used this method, however…they
were not interested so much in why something worked as much as they were in the fact that it
gave correct results. They most likely used this method because doubling is a relatively easy
process to carry out and they probably viewed multiplication for what it really is: repeated
addition.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 12
384
385
386
387
388
389
390
391
PART 2: Unit Fractions, Division, and the 2÷n Table
Fractions and Division
Having just finished looking at Egyptian multiplication, we can ask “What about division?”
Indeed, this is a good question. Recall that multiplication and division are inverse operations of
each other. For example:
40
= 20 is true because 40 = 2 × 20
2
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
40
can be interpreted as asking, “What number
2
times 2 gives 40?” Another important way to pose this question is
“How many groups of 2 go into 40?” Either way, the result is that
we can view any division problem as an equivalent problem of
multiplication! And, since we already know how the Egyptians did
their multiplication, we can then do division as well. Or at least we
should be able to. We just need to develop a way to convert a
division problem to the equivalent of a multiplication problem.
Put another way:
Think About It
How can we turn any
division problem into
the “equivalent”
problem of
multiplication?
Before we explore this question further, we first need to look at the way that Egyptians handled
their fractions. In general, the Egyptians only used fractions with numerator 1.
When a fraction has a 1 in the numerator, we will call it a unit fraction. (The word “unit” means
one.)
To write a unit fraction, the Egyptians would write the denominator in regular hieroglyphics and
1
looked like the following in
place the symbol
above it. Thus, the fraction
12
hieroglyphics:
In most books, you will see unit fractions denoted by writing a horizontal bar above the
1
denominator. Thus
would be represented by 12 . In this document, you will see both
12
2
notations. For , you will sometimes see the symbol 3 , a three with two bars over it.
3
417
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 13
418
419
420
The Egyptian had special symbols only for a limited number of fractions. Here are three special
symbols that they used:
1
=
2
1
=×
4
2
=
3
421
3
.
4
422
There is some evidence that a special symbol also existed for
423
424
425
426
427
428
Besides the three special symbols given above, all other Egyptian fractions were written as the
sum of unit fractions. This may seem odd to us, but it is a fact that we must deal with if we are
going to understand the Egyptian mathematical system. For example, there was no special
symbol for 5/6 so it had to be written as the sum of unit fractions. Here’s one way to do that:
5 2 3
= +
6 6 6
1 1
= +
3 2
429
430
Notice in the last line that each of the two fractions have 1 in the numerator, and so each are
431
considered unit fractions.
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
You will also see 5/6 written in the following way using our alternate notation for unit fractions:
5
= 3+ 2
6
Of course, the Egyptians would have written it with hieroglyphics as:
There may be more than one way to write a unit fraction. For example:
7 1 1
= +
24 6 8
Another way to write the same fraction is this:
7 1 1
= +
24 4 24
447
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 14
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
The big question remains: How do you know how to take a fraction and break it into unit
fractions?
J.J. Sylvester (1814-1897) provided us with a system to just this very thing. We summarize the
method here, as described by Bunt12, and then try to carefully explain it with some examples.
This is not how the Egyptians found their unit fraction “decompositions.” Instead this is a much
more modern method developed to help us to try to understand how such fractions can be
manipulated.
Sylvester’s Method of Writing a Given Fraction as the Sum of Unit Fractions
1.
2.
3.
4.
Find the largest unit fraction (the one with the smallest denominator) that is less than
the given fraction. (Hint: To find the largest unit fraction less than the given fraction,
divide the denominator by the numerator and take the next integer greater than the
quotient for the new denominator.)
Subtract this unit fraction from the given fraction to see what’s left over.
If what is left over is a unit fraction, you’re done. If not, continue this process by
finding the largest unit fraction that goes into what’s left over.
Continue until you end with a unit fraction.
As you read through the following examples, you may want or need to refer back to the box
above to make sure you follow what is happening. It may take a bit of practice but after a while it
becomes rather routine. Also, you may find it helpful to use a calculator with the ability to
handle fractions to do some of the intermediate calculations.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 15
475
476
477
478
Example 11
Write 4/15 as a sum of unit fractions:
Solution
4
3 1 1 1
as
+ = + and you’re done.
15 15 15 5 15
However, let’s apply Sylvester’s method and see how it works.
479
The quick way to do this is to think of
480
481
4
15
4 1 1
− =
15 4 60
4
= 4 + 60
15
482
483
484
Example 12
485
486
487
Comments
The quotient of the denominator and numerator
is 15/4 = 3.75. We take the next highest integer,
⇔ which is 4, and that tells us that ¼ is the largest
fraction that is smaller than 4/15. (See Step 1
above)
We subtract 1/4 from 4/15 to see what’s left over.
⇔ Since 1/60 is a unit fraction, we are done. (See
Steps 2 and 3 above.)
Add 1/4 to both sides of the equation above to
get this. This is our final answer. Note that this is
⇔ not the same combination of unit fractions that
we got before. ♦
Write
11
as a sum of unit fractions using the Sylvester Method:
18
Solution
11
18
11 1 1
− =
18 2 9
11 1 1
= +
18 2 9
11
=2+9
18
Comments
The largest unit fraction less than this is 1/2
⇔ since 18/11 = 1.6; so we subtract 1/2 from
11/18
When we subtract we get a unit fraction, so we
⇔
are done.
⇔ Add 1/2 to both sides of the previous equation.
⇔ Final answer.♦
488
489
490
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 16
490
Example 13
491
492
493
494
Write
6
as a sum of unit fractions using the Sylvester Method:
7
Solution
6
7
⇔
6 1 5
− =
7 2 14
⇔
6 1 5
= +
7 2 14
⇔
6
= 2 + 3 + 42
7
495
496
497
498
499
500
501
502
503
504
Comments
The largest unit fraction less than
this is 1/2. So we subtract 1/2
from 6/7
When we subtract, we don’t get a
unit fraction. So we will need to
continue. Let’s first write 6/7 =
1/2 + 5/14, by adding 1/2 to both
sides of the previous equation.
We will now continue the process
on 5/14, since it is not a unit
fraction. We don’t have to worry
about the 1/2 since it’s already a
unit fraction.
The largest unit fraction less than
this is 1/3 (since 14/5 = 2.8) so
we subtract it.
This time, what’s left is a unit
fraction, so we are done. We only
need to combine this result with
where we left off in step 3 above.
First we rewrite 5/14 as a sum of
unit fractions.
5
14
⇔
5 1 1
− =
14 3 42
⇔
5 1 1
= +
14 3 42
⇔ Add 1/3 to both sides of Step 5
⇔
Substitute the equation in Step 6
into Step 3
This combination of unit fractions for 6/7 is not obvious nor is it easy to get to otherwise,
so the Sylvester Method is helpful here.♦
Caution: Don’t get the wrong impression. The Egyptians did not use this method or even one like
it. (Sylvester lived in the nineteenth century, after all.) In all likelihood, Egyptian mathematicians
found unit “decompositions” for a variety of fractions and then recorded them in “tables” for
general use thereafter. As we shall see, their 2÷n Table was a powerful tool for division.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 17
505
Check Point E
506
507
508
509
510
511
512
Write
Solution
See the endnotes for the answer.13
Check Point F
513
514
515
516
517
518
519
520
7
as a sum of unit fractions.
10
Write
29
as a sum of unit fractions.
72
Solution
See the endnotes for the answer.14
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 18
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
Egyptian Division
Let’s go back to the question of division. Suppose we want to divide 19 by 8, as Problem 24 of
the Rhind Papyrus does15. We need to keep in mind that determining the value of 19÷8 is
equivalent to determining how many times 8 goes into 19. So we can see that 19 = 8 + 8 + 3.
That is, 8 divides into 19 two times with a remainder of 3. (We can do this quickly in our head
to see that 19÷8 = 2+3/8.) What we are doing here is trying to determine how many groups of 8
we need to get 19. So what the Egyptians do is work with groups of 8 and they keep doubling
until they get close to 19. To make up for any remainder, they then resort to taking halves or
other fractional parts until they get 19 exactly. Let’s do an example so we can see how the
process works.
Example 14
Find 19÷8 using the Egyptian doubling method.
Solution
We will compute with groups of 8 and keep doubling until the right column
adds up to 19. Then the corresponding numbers in the marked left column
give us our desired result. Remember that we want to determine the
following: how many groups of 8 go into 19?
1
8
⇔
2
16
⇔
2
4
⇔
\
4
2
⇔
\
8
1
⇔
\
Sum 19
Answer
2+ 4+8
⇔
Comments
We have one group of 8, for a total of 8…we want 19, so
we need to double since 8 is not enough.
Double 1 to get 2…double 8 to get 16…still not 19. If we
double again, we’ll get 32 in the right column…that’s
too much…we only need 19. We have 16, which means
we need 3 more. So, watch this little trick. We switch to
taking half of each column instead of doubling. (Think
about it…it’s basically the same process.)
If one group has 8, then one half a group has 4. Too
much, since we only need 3. So take half again.
Half of a half is one fourth…half of 4 is 2. So now we
have 16 and 2 for a total of 18. We only need one more.
Talk half again.
Half of one fourth is one eighth…half of 2 is1, which is
what we need to finish the process.
We now add up all of the right hand columns so that our
total is 19 and then added up the corresponding left
hand columns to get our final answer. We get 2 + 1/4 +
1/8, which is2+3/8, the value of 19÷8. 9
542
543
544
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 19
544
545
546
547
548
549
550
551
552
The following is an example that shows a slightly different approach, but retains the same spirit.
Example 15
Find 300÷16 using the Egyptian doubling method.
Solution
We will compute with groups of 16 until we get 300. In other words, how
many groups of 16 go into 300?
1
2
4
8
16
32
64
128
⇔
⇔
⇔
⇔
\ 16
256
⇔
16
1
⇔
8
2
⇔
\
4
4
⇔
\
2
8
⇔
\
Sum 300
Answer
18 + 2 + 4
553
554
555
556
557
⇔
Comments
We start with one group of 16
Double 1 to get 2…double 16 to get 32
etc.
etc.
We stop doubling here because doing so
again would yield 512, which is too much
(we only need 300). We have 256 and then
add 32 to get 288. We need 12 more, but the
smallest we have in the right column is 16. So
we switch to taking fractional pieces.
One group has 16, so 1/16 of a group has 1.
Now we go back to doubling until we can get
the 12 more than we need. Note that this is
different that the previous example. In this
example, we go back to the beginning and
generate 1 in the right column and then
revert to doubling after this point.
Double 1/16 to get 2/16 = 1/8…double 1 to
get 2.
Double 1/8 to get 1/4…double 2 to get 4.
Double 1/4 to get 1/2…double 4 to get 8. We
stop doubling here because doubling 8 gives
us 16, which is more than the 12 we need. We
just need to pick up 12 more from the right
hand column. Note that the 1/4 row and the
1/2 row add give us 12.
Here’s the sum we wanted.
Now we add up the left hand columns that
are marked off and get 16+2+1/4+1/2 =
18+3/4. This is the correct answer. ♦
Check Point G
Divide 39÷12 using the Egyptian Method. See endnotes for solution.16
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 20
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
If you do this enough times the process becomes faster and more intuitive. And it’s easy to see
why it works! You’re simply adding up the appropriate number of groups by doubling and
perhaps taking a fractional piece here and there. Now, think about the division algorithm that you
learned as a child. Can you explain why it works? Probably not because it’s not as intuitive!
What we see here are algorithms that the Egyptians developed and used that really use the basic
ideas of multiplication and division. Both of these methods are said to be additive because they
both boil down to the simpler task of addition. Our modern algorithms, while perhaps a bit more
compact, are not as clearly grounded in these ideas, at least not on the surface. Some, or many, of
you will feel that the modern methods are “better.” I would challenge you to consider why you
think that’s true! Let’s continue with more examples.
Example 16
Find 14÷24.
Solution
We will compute with groups of 24 until we get 14, just as before.
1
24
24
1
\
12
2
\
6
4
\
3
8
Comments
I group of 24 gives 24. We only want a total of
14…this is already too many! So immediately turn
⇔ to fractions. We can either take a half of this group
or 1/24 of the group. It’s often easier to do the
latter.
1/24 of the group has one (since a whole group has
⇔ 24). Now we can start doubling until we get back
to 14.
⇔ Double 1/24 to get 2/24 = 1/12…double 1 to get 2.
⇔ Double 1/12 to get 2/12 = 1/6…double 2 to get 4.
Double 1/6 to get 2/6 = 1/3…double 4 to get 8. We
stop doubling here because doing so would give us
⇔
16 and we only need 14. We now just add the
appropriate numbers in the right hand column.
Sum 14
Answer
3 + 6 + 12
⇔ The answer here is
1 1 1
+ + .♦
3 6 12
575
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 21
576
577
578
579
580
581
Here’s one last example without comments…can you follow each step?
Example 17
Find 115÷18.
Solution
1
18
\
2
36
\
4
72
\ 18
1
9
2
3
6
\
Sum 115
Answer 6 +18 + 3 9
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
The 2÷n Table
The examples we saw in the previous section don’t take into account a variety of problems that
one could encounter doing division. For example, let’s try to divide 11÷13.
1
13
13
1
2/13
2
Comments
⇔ 1 group has 13. Too big…we only need 11
⇔ 1/13 of a group will have 1. Now start to double.
The problem here is that 2/13 is not a unit fraction
and it’s not one of the three exceptions that
Egyptians had for unit fractions. So, they did not
⇔
recognize this entry in their calculations. To
continue they would need to know how to write
2/13 as a sum of unit fractions.
In order to get out of this jam, the Egyptians had the 2÷n Table (see forthcoming page for a
complete table and the internet article17 for more information) to which they could refer. It was
like a “cheat-sheet” they could use whenever they needed it. Note that it’s only used for fractions
with 2 in the numerator…these fractions will often appear when you try to double a unit fraction
and the result does not reduce to another unit fraction. If we look at the table, we see that
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 22
597
598
599
600
601
602
603
604
605
606
607
2 1 1
1
= + +
. This allows us to continue in the
13 8 52 104
division process where we got stuck above. Let’s do an example of this to see how this table is
used.
2 ÷ 13 = 8 + 52 + 104 . This means that
Example 18
Find 20÷3. We are seeking to find what number times 3 gives 20.
Alternatively, how many groups of 3 go into 20?
Solution
We will compute with groups of 3 until we get 20.
\
\
\
1
2
4
3
6
12
⇔
⇔
⇔
8
24
⇔
3
1
⇔
2 +6
2
⇔
Sum 20
Answer
6 + 2 +6
Comments
One group has 3.
Double as usual.
More doubling. We have 6+12=18…we need 2 more
This line does not help us…it would normally be
omitted
1/3 of a group has 1.
Double 1/3 and we get 2/3…we go to the table and see
that 2/3 = 1/2 + 1/6. Double 1 to 2, which is what we
need.
Note that 20/3 is 6 and 2/3. This is the same as
6 + 2 +6 ♦
608
609
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 23
The Egyptian 2÷n Table
609
610
2÷3 = 2+ 6
2 ÷ 53 = 30 + 318 + 795
2 ÷ 5 = 3 + 15
2 ÷ 55 = 30 + 330
2 ÷ 7 = 4 + 28
2 ÷ 57 = 38 + 114
2 ÷ 9 = 6 + 18
2 ÷ 59 = 36 + 236 + 531
2 ÷ 11 = 6 + 66
2 ÷ 61 = 40 + 244 + 488 + 610
2 ÷ 13 = 8 + 52 + 104
2 ÷ 63 = 42 + 126
2 ÷ 15 = 10 + 30
2 ÷ 65 = 39 + 195
2 ÷ 17 = 12 + 51 + 68
2 ÷ 67 = 40 + 335 + 536
2 ÷ 19 = 12 + 76 + 114
2 ÷ 69 = 46 + 138
2 ÷ 21 = 14 + 42
2 ÷ 71 = 40 + 568 + 710
2 ÷ 23 = 12 + 276
2 ÷ 73 = 60 + 219 + 292 + 365
2 ÷ 25 = 15 + 75
2 ÷ 75 = 50 + 150
2 ÷ 27 = 18 + 54
2 ÷ 77 = 44 + 308
2 ÷ 29 = 24 + 58 + 174 + 232
2 ÷ 79 = 60 + 237 + 316 + 790
2 ÷ 31 = 20 + 124 + 155
2 ÷ 81 = 54 + 162
2 ÷ 33 = 22 + 66
2 ÷ 83 = 60 + 332 + 415 + 498
2 ÷ 35 = 30 + 42
2 ÷ 85 = 51 + 255
2 ÷ 37 = 24 + 111 + 296
2 ÷ 87 = 58 + 174
2 ÷ 39 = 26 + 78
2 ÷ 89 = 60 + 356 + 534 + 890
2 ÷ 41 = 24 + 246 + 328
2 ÷ 91 = 70 + 130
2 ÷ 43 = 42 + 86 + 129 + 301
2 ÷ 93 = 62 + 186
2 ÷ 45 = 30 + 90
2 ÷ 95 = 60 + 380 + 570
2 ÷ 47 = 30 + 141 + 470
2 ÷ 97 = 56 + 679 + 776
2 ÷ 49 = 28 + 196
2 ÷ 99 = 66 + 198
2 ÷ 51 = 102 + 34
2 ÷ 101 = 101 + 202 + 303 + 606
611
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 24
611
612
613
614
615
616
617
618
Example 19
Find 19÷5.
Solution
We will compute with groups of 5 until we get 19. In other words, how many
groups of 5 go into 19?
\
1
5
⇔
\
2
10
⇔
5
1
⇔
3 + 15
2
⇔
2 + 6 + 10 + 30
4
⇔
Sum 19
Answer
3 + 2 + 6 + 10 + 30
⇔
\
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
Comments
One group has 5 in it.
Double to get 10…we have 15
total in the right column…we
need four more.
One fifth of a group has 1…we
double until we get four.
Double 1/5 to get 2/5…look up
2/5 in the table to get the correct
unit fraction combination of
1/3+1/15
Double 1/3 to get 1/2+1/6 from
the table…also double 1/15 to
get 1/10+1/30 from the table. All
these go in the left column. In the
right column, doubling 2 to get 4
is the easy part. Note again that
all the left hand column data is
coming from the 2÷n table.
Believe it or not, this is 3 + 4/5,
just as it should be. ♦
Check Point H
Divide 59÷7 using the 2÷n Table.
Solution
See the endnotes for the solution.18
Example 20
Divide 6 + 2 + 4 ÷ 4 + 8
Solution
This problem is asking that we solve a problem where two fractions are being
divided by each other. The process is exactly the same as in the previous
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 25
634
635
636
637
problems…we just need to be careful about keeping the details in order. For
this problem we are going to compute with groups of 4 + 8 until we get
6 + 2 + 4 . That is, how many groups of 4 + 8 go into 6 + 2 + 4 ?
1
4 +8
⇔
2
2+ 4
⇔
4
1+ 2
⇔
8
3
⇔
\ 16
6
⇔
\
Sum 6 + 2 + 4
Answer 18
Comments
We are computing with fractional
groups of 4 + 8 . We want to keep
doubling until we get to 6 + 2 + 4
Double 1/4 gives 1/2…double 1/8
gives ¼
Double 1/2 gives 1…double 1/4
gives 1/2
Double 1 gives 2…double 1/2
gives 1…this is a total of 3.
Double 8 gives 16. Double 3 gives
6. We are looking for 6 + 2 + 4 so
we note that the two columns
marked gives us precisely that.
⇔ Our answer is 18.
638
3 3
This problem is equivalent to computing 6 ÷ . Can you verify that 18 is the
4 8
answer using your typical division algorithm for fractions? ♦
639
640
641
642
643
644
645
646
647
648
649
650
651
Check Point I
Divide 10 + 3 ÷ 2 + 3 + 4 using the 2÷n Table. See the endnotes for the
solution.19
This concludes our study of Egyptian arithmetic. With this information, we can now proceed to
Part 3 where we will look at Egyptian algebraic methods.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 26
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
PART 3: Algebra and the Aha Problems
Introduction
Having finished looking at various forms of arithmetic in ancient Egypt, we now move on to
algebraic problems. In the Rhind Papyrus, there are a variety of math problems, some of which
could be considered “algebraic.” The reason for this is that, when translated into modern
mathematical notation, they are not simple arithmetic, but resemble what we would call algebraic
problems. The focus of our explorations here will be on the “Aha” problems that require what is
called the method of false position. Essentially, what this method does is make an educated
guess about a solution to an equation, check the guess, and then adjust the guess to produce the
correct answer. A quick check then assures that the process has worked. In order to prepare
ourselves for this, we would be wise to first review our own knowledge of lines, linear equation,
and their proportional nature.
Lines and Their Proportional Nature
Recall that the equation of a line can be given in the following form:
y = mx + b
where m is the slope of the line and b is the y-intercept. The slope of a line is a measure of the
change in y divided by the change in x. You are probably familiar with the following ways to
express slope:
m=
rise ∆y y2 − y1
=
=
run ∆x x2 − x1
As an example, consider the line y=3x+2. This line has slope of 3 = 3/1, which means that when
we move three units in the positive y direction, we move one unit in the positive x direction. The
y-intercept is 2, which means the line crosses
y
the y axis at the point (0,2). A graph of this line
y = 3x + 2
is shown here.
4
From the graph, we are reminded that the rate
of change of a line is constant. That is, between
any two points on a line, the slope is always
the same.
2
x
-4
-2
0
-2
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 27
2
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
For us, solving linear equations is rather easy. Of course, we have hundreds of years of
experience, convenient notation, and many other historical developments to help us out. For
example, let’s solve the following linear equation for x:
1
x + 3 = 21
5
1
x = 21 − 3
5
1
x = 18
5
x = 5 ⋅18
x = 90
Ideas Behind the Method of False Position
The Egyptians, however, did not have the luxury of symbols for variables, or algebraic equations
as shown here. To solve these equations, they used the method of false position. Before we show
the Egyptian version of this method, let’s explore it from a more modern point of view. This will
allow us to see what they were doing and better understand it when we examine their method in
more detail.
Let’s start with the following equation that we will solve in a slightly different way than what we
are used to. (Before we begin, why don’t you solve this for x using standard algebra.)
x+
1
x=6
2
Suppose we do not know how to solve this equation with algebraic steps. One useful approach
in the absence of a specific algebraic strategy is to guess. Let’s pick a guess for the solution of
this equation. Because of the fraction, we may as well choose a guess that will help us rid
ourselves of the fraction. The simplest guess for x that would do this is 2. By choosing 2 as our
guess and substituting it into the equation, we get the following on the left side of the equation:
1
2 + (2) = 2 + 1 = 3
2
Of course, we don’t want the left side to be 3, we want it to be 6 (which is the right side of the
original equation). So x = 2 is definitely not the solution. But let’s look at the graph of this line.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 28
10
y
8
y = x+
6
y=6
1
x
2
4
(2,3)
2
x
-1
722
723
724
725
726
727
728
729
730
731
732
733
739
740
741
742
743
744
745
746
747
1
2
3
4
5
In this picture the point (2,3), corresponding to our guess, has been marked with the black dot.
Because the rate of change for this line is constant, we can draw some useful conclusions. The
value x = 2 gets us to y = 3. But our goal is to get to y = 6. This means that we have to go twice
as far in the y direction than we’ve already gone. But, to go twice as far in the y direction, we
would have to go twice as far in the x direction as well. This statement is making use of the
proportional nature of lines! Thus, if x = 2 takes us to y = 3, then going twice as far to x = 4 will
take us to y = 6. When we check x = 4, in the equation, we see that it does indeed work. Here,
because we need to multiply by 2, we call 2 the proportionality factor.
Example 21
2
Solve with guess and check… x + 2 x = 40 .
3
734
735
736
737
738
0
Our goal is here at the
point (?,6)
Solution
We start with our guess of x = 3. This will help get rid of the fraction. Check
the guess and get
2
(3) + 2(3) = 2 + 6 = 8
3
Since our result is 8 and we actually want it to be 40, we see that we off by a
factor of 5. That is, 40÷8=5. (The proportionality factor is 5.) Therefore, our
guess of x = 3 is also too small by a factor of 5. The correct answer should be
3×5=15. Checking we see:
2
(15) + 2(15) = 10 + 30 = 40
3
Let’s look at a graph of what is happening here.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 29
40
y=
y
(15,40)
2
x + 2x
3
20
(3,8)
x
0
748
749
750
751
752
753
754
755
756
0
5
10
15
Our first guess takes us to the point (3,8). This is one fifth of the way to our
goal of 40, so we must move five times as far in each direction to get there.
The small rectangle is one-fifth the size of the large rectangle. In order to get
to our goal of 40, we must take five times as many steps as we did with our
guess, as shown below.
40
y=
y
(15,40)
2
x + 2x
3
20
(3,8)
x
0
757
758
759
760
761
762
763
0
5
10
15
This takes us to x = 15, which is easily checked to be the correct solution of
this equation? ♦
Think About It
Can we use this same
kind of guess and
check method with
quadratic equations?
Why or why not?
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 30
763
Example 22
1
Solve with guess and check…. x + x = 14
5
764
765
766
767
Solution
We start with a guess of x = 5. Checking our guess we get:
1
(5) + 5 = 1 + 5 = 6
5
Our goal is 14 and our guess gave us 6. So we’re off by a factor of a little
more than 2. To determine the exact factor, we only need to divide: 14÷6 =
7÷3 = 7/3. (7/3 is the proportionality factor.) So just as before, we multiply
our original guess by 7/3 to get the correct answer: 5×(7/3)=35/3. You should
check this with regular algebra to verify that it is correct. ♦
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
The Egyptian Method of False Position
This, then, is the idea behind the method of false position. Note that it depends on a fundamental
property of lines…proportionality. Let’s explore how the Egyptians would present such a
solution.
Example 23
A quantity and its seventh, added together, give 19. What is the number?
This problem is typical of those found in the Rhind Papyrus.
Note that this can be translated into the following algebraic equation (which
we write here only for our own clarification…the Egyptians did not have
variables.)
1
x + x = 19
7
789
790
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 31
790
791
792
793
794
795
796
797
798
799
800
Solution
This is Problem 24 from the Rhind Papyrus and requires the method of false
position in an Egyptian context. Text that the scribe might write will be bold
and underlined.
Assume 7
\
1
\
7
Comments
Start with a guess that is easy to check. 7 will take
care of the fraction so this is the scribe’s first guess.
One quantity or group is seven, since we’re
7
⇔
assuming 7 as the answer.
Since one-seventh of the quantity is added to the
1
⇔ quantity, the scribe needs to compute what oneseventh of 7 is…it’s 1.
Add up the quantity and its seventh. The 7 and the 1
Sum 8 ⇔ in the right hand column add up to 8. But we need
19.
Note that in this first step, the scribe is essentially
taking an educated guess and then plugging that
into the “equation” in front of him. The steps above
do just that.
As many times as 8 must be multiplied to give 19, so 7 multiplied that
many times will give the required number.
Comments
Here, the scribe is basically doing what we did
above. He sees that his answer, 8, is too small. He
wants 19. If he can determine by what factor he’s off
(the number of times 8 must be multiplied to give
19), then he can multiply that by 7 (the guess) to get
the correct answer. To compute the appropriate
proportionality factor, he uses basic division, shown
below…
801
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 32
1
8
⇔
2
16
⇔
2
4
⇔
\
4
2
⇔
\
8
1
⇔
\
Sum 19
2+4+8
⇔
\
1
2+4+8
⇔
\
2
4+2+4
⇔
\
4
9+2
⇔
Sum 16 + 2 + 8
⇔
We need to know the value of 19÷8, so…compute
with groups of 8 until we get 19…one group has 8
Double each column. If we double again it’s too
much so we switch to fractions as in the last section.
We only need three more to get to 19.
One half of a group has 4 in it…that’s too much.
Reduce again…
One fourth of a group has 2 in it. 16 + 2 is 18…only
one more needed…so…
One eighth of a group has 1 in it. We now just
mark off the columns we need.
The answer is 2 + 4 + 8 . This is 19÷8, which is the
factor we needed.
At this point, we now need to multiply our original
guess of 7 by this number to get the actual result.
That is, we want to compute 7 × (2 + 4 + 8) . In other
words, we need seven groups of 2 + 4 + 8 . At this
point, we switch to a multiplication problem
One group has 2 + 4 + 8
Double 1 to get 2 in the left column…In the right
column, double 2 to get 4, double 1/4 to get 1/2, and
double 1/8 to get 1/4 for a total of 4 + 2 + 4 .
Double 2 to get 4 in the left column…In the right
column, double 4 to get 8, double 1/2 to get 1, and
double 1/4 to get 1/2. This gives us a total of 8 + 1
+ 1/2 = 9 1/2. Now just mark off the rows that add
up to 7, since we need 7 groups of 2 + 4 + 8 .
This sum comes from the left hand columns. The
2,4,and 9 give 15, the two halves give 1, for a total
of 16 so far. The two 1/4’s give 1/2 and the 1/8 is
remaining. This is a total of 16 + 1/2 + 1/8, as
indicated.
802
803
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 33
803
804
805
806
807
808
809
810
811
812
813
814
Do it thus. The quantity is
\
1
16 + 2 + 8
\
7
2+4+8
Comments
This last step is a check of the process. This line
⇔
represents “the quantity”
⇔ This is one seventh of the quantity
When we add these together, we get 19. The scribe
Total 19 ⇔ has thus checked that the solution obtained above
actually works, as it should. ♦
It may be advantageous to see what this might look like from start to finish on a
papyrus without the commentary, as the scribe might record it. (Of course, this is
an English translation of what would be written on a papyrus.)
A quantity and its seventh, added together, give
19. What is the number?
Assume 7
\
\
1
7
7
1
Sum 8
815
816
817
818
819
As many times as 8 must be multiplied to give
19, so many times 7 will give the required
number.
\
\
\
1
2
2
4
8
8
16
4
2
1
Sum 19
Answer 2 + 4 + 8
820
\ 1
\ 2
\ 4
2+4+8
4+2+4
9+2
The quantity is 16 + 2 + 8
821
822
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 34
This is how the
scribe would do
a problem with
method of false
position…refer
to this for
homework.
822
823
Do it thus. The quantity is
\
\
1
7
16 + 2 + 8
2+4+8
Total 19
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
This example is not an easy one to follow. You really do have to think about what
is going on. Please review it once more if you are unsure and then proceed to the
following new example. ♦
Example 24
Here’s another problem from the Rhind Papyrus:
A quantity and its fourth added give 15. What is the quantity?
Solution
We are going to try to emulate the approach of the scribe as in solution to the
Rhind Papyrus, Problem 24. Comments will be less extensive.
Assume 4
\
1
4
\
4
1
Comments
Our guess comes from the
⇔ “fourth” in the stated problem.
Here we compute the quantity…
…and it’s fourth, which is 1. We
⇔ total them to get a preliminary
total of…
Sum 5
841
842
843
844
As many times as 5 must be multiplied to give 15, so many times 4 will
give the required number.
\ 1
5
\ 2
10
Comments
We are dividing 15 by 5 to find
⇔ the proportionality factor. We
want a total of 15
The proportionality factor is 3,
Sum 15
⇔ gotten by adding up the rows
Answer 3
that total 15.
845
1
3
Comments
⇔ Now multiply the original guess of
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 35
4 by 3, the proportionality factor.
6
⇔ Double
Here are the 4 groups of 3 that we
12
⇔
want.
This is the answer that we will
Sum 12 ⇔
present and check in the next step.
2
\ 4
846
847
848
Do it thus. The quantity is
\
\
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
Comments
12
⇔ One quantity is 12
3
⇔ 1/4 of the quantity is 3
Sum 15 ⇔ For a total of 15
1
4
This concludes the problem. ♦
Check Point J
A quantity and its two fifth gives 28. What is the quantity?
Solution
See the endnote for the answer. 20
This last example will incorporate the 2÷n table that we discussed earlier.
Example 25
A quantity and its fifth are added to give 19. What is the quantity?
Translate the problem as an Egyptian scribe might.
Solution
Assume 5
\
\
1
5
Comments
⇔ Here is a check of the initial guess
5
1
Sum 6
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 36
869
870
871
872
As many times as 6 must be multiplied to give 19, so must 5 [the guess] be
multiplied to give the required number.
\
1
6
\
\
2
6
12
1
Comments
Here we divide 19 by 6 to get the
⇔
proportionality factor.
Sum 19
1
Answer ⇔ The proportionality factor is 3 6
3+ 6
873
874
\ 1
3+ 6
⇔
2
6+3
⇔
12 + 2 + 6
⇔
\ 4
Sum
15 + 2 + 3
875
876
877
⇔
Comments
We now multiply the original guess of 5 by
the proportionality factor. We want 5 groups
of 3 + 1/6
Double 3 + 1/6 and you get 6 + 1/3
Double 6 1/3 and get 12 1/2 1/6, which we
read from the 2÷n Table
5 groups of 3 1/6 give 15 + 1/2 + 1/3. This is
the final answer, which we now check.
Do it thus. The quantity is
\
\
1
5
15 + 2 + 3
3+ 6
Sum 19
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
♦
Conclusion
So there we have it. Egyptian algebra. Of course, linear equations are not hard. You’ve seem
them many times and done them many times. But throw in the more abstract idea of
proportionality, blend in the method of false position, and stir in Egyptian algorithms for
multiplication and division and you have a challenging task at hand. But don’t be discouraged. It
will probably take you at least a couple of solid readings to get the feel, as well as having to ask
several questions and get help.
One more note as we end: We need to remember that the Egyptians did mathematics primarily
because it helped them solve some kind of problem that was at hand. They were not overly
concerned with why a certain process worked. Their primary concern was that it worked and
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 37
893
894
895
896
897
898
899
900
901
provided a solution. Many of the tools that we have used in this chapter to study the Egyptians
(linear equations and their graphs, the Sylvester method, etc) are modern lenses through which
we are studying Egyptian mathematics. The fact that the Egyptians lived thousands of years
before these tools arrived on the scene is more evidence of how much credit they deserve for
getting as far as they did in their development of mathematics.
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 38
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
PART 4: Homework Problems
Egyptian Numbers and Conversions
Write the following numbers in Egyptian hieroglyphics:
1)
5,425
2) 32,520
3)
850,633
4) 2,304,540
5)
1,238,110
6) 300,400
7)
3,000,003
8) 398,442
After converting to Egyptian hieroglyphics, add the groups of numbers using the visual method
shown in Example 5.
9)
7,321 + 8,2,44
11) 468 + 543 + 3,681
10) 24,634 + 95,858
12) 1245 + 326 + 8,774
Egyptian Multiplication
Use the Egyptian Method of multiplication to do the following multiplication problems. Do not
change the order of multiplication. 30×55 means 30 groups of 55 so the first line would be 1
55. Use base-10 numbers throughout, including your answers. See Example 8 for comments on
the write up.
13) 28×24
14) 117×30
15) 35×55
16) 240×15
17) 64×22
18) 22×67
19) 300×15
20) 15×300
Use the Egyptian Method of multiplication to do the following multiplication problems. Use
Egyptian symbols throughout, including in your answer. See Example 9 for a model for these
problems.
21) 22×12
22) 120×8
23) 33×15
24) 230×11
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 39
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
25) Consider the two problems 93×12 and 12×93. Do each problem using the Egyptian method,
first by adding up 93 groups of 12 and then adding 12 groups of 93. (Important: the order
made a difference to the Egyptians…they viewed them as separate problems.) Use your
resulting work to discuss if and when there is an advantage to the order in which an
Egyptian scribe might compute a multiplication problem. After doing both computations
using a modern algorithm, compare the two methods…does picking the “right” order
generally make modern multiplication easier? Why or why not? (The point here is to think
and write about how the order of multiplication affects each of the methods.) Clearly write
your observations and results…your score will depend on it in part.
Egyptian Fractions
What is the largest unit fraction that is less than the given fraction?
26)
2
5
27)
23
98
28)
79
1004
29)
131
980
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
Do the following subtractions (in preparations for the next problems) any way you
want:
30)
4 1
−
5 2
31)
2 1
−
35 18
Powers of 2
Write each of the following positive integers as single powers of 2. (Hint: You can use methods
from Chapter 1 to convert between bases to do this if you want to.) See Example 10 for help
Please give some indication of how you got your sum and show your work.
32) 9
33) 25
34) 80
35) 300
36) 1000
37) 3500
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 40
984
985
986
987
988
989
990
991
992
993
Unit Fractions
994
38)
9
20
39)
23
25
40)
13
36
41)
41
140
42)
335
336
43)
25
28
44)
3
17
45)
91
120
Write the following fractions as the sum of unit fractions by using the Sylvester Method
described in this chapter. (See Example 11 to Example 13.) Show all steps used, including the
subtraction of unit fractions in each step. You can do your work in Hindu-Arabic numbers
symbols, but write your final answer in Egyptian number symbols. No credit will be given for
methods that do not use the Sylvester Method. (Hint: to find the largest unit fraction less than
the given fraction, divide the denominator by the numerator and take the next integer greater
than the quotient for the new denominator.)
995
996
997
998
999
1000
1001
1002
1003
1004
1005
1006
1007
1008
Egyptian Division
Divide the following using the Egyptian Method of Division. You may work in Hindu-Arabic
numerals but express your answer in both Hindu-Arabic and Egyptian numerals.
46)
360
24
47)
1011
48)
468
26
49) 943 ÷ 23
1012
1013
1014
50)
2088 ÷ 36
51) 35926 ÷ 142
1015
52)
7
12
53)
1009
3676
24
1010
11
16
1016
1017
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 41
1018
1019
1020
1021
Egyptian 2/n Table
1022
54)
11
17
55)
23
30
1024
56)
17
90
57)
41
150
1025
1026
58)
(109 + 3) ÷ ( 6 + 2 + 3)
59)
( 22 + 3 + 7 ) ÷ ( 2 + 2 + 6 + 7 )
60)
( 4 + 3 + 10 + 30) ÷ (1 + 5)
61) 17 + 3 + 5 + 15 ÷ 2 + 5
Use the Egyptian Method of Division and the 2÷n Table to calculate the following.
1023
1027
1028
1029
1030
1031
1032
1033
1034
1035
1036
1037
1038
1039
1040
1041
1042
1043
1044
1045
1046
1047
1048
1049
1050
1051
1052
1053
1054
1055
1056
1057
1058
1059
1060
(
) (
)
Algebraic Practice
For each of the following problems, translate the sentence into an algebraic equation using the
variable x and then solve the equation for x. You can use modern algebraic methods to solve
these problems.
62) A quantity and its 1/3, when added, give 220. What is the number?
63) A quantity and its 1/7, when added, give 264. What is the number?
64) From the Rhind Papyrus: A quantity, its 2/3, and its 1/7, added together, become 38. What
is the quantity?
Method of False Position
For the following problems, use the method of false position to solve the following problem. Your
answer must be written in the form a scribe would have, including all intermediate computations
to get to a final answer. No credit will be given for another kind of solution. See Example 23 for
a guide on how to write up the problem as a scribe would. Please show a check of your answer
using “regular” algebra.
65) A quantity and its 1/6 added together become 63. What is the quantity?
66) A quantity and its 1/2 added together become 16. What is the quantity?
67) A quantity and its 1/5 added together become 35. What is the quantity?
68) A quantity and its 1/8 added together become 100. What is the quantity?
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 42
1060
1061
1062
1063
1064
1065
1066
1067
1068
1069
1070
1071
1072
1073
69) Rhind Papyrus Problem 28, when translated into modern English, says the following:
Think of a number, and add 2/3 of this number to itself. From this sum subtract 1/3 its
value and say what your number is. Suppose the answer is 10. Then take away 1/10 of this
10 giving 9. Then 9 was the number first thought of.”
The scribe, Ahmes, then gave the following proof:
If the original number was 9 [method of false position], then 2/3 is 6, which added makes
15. then 1/3 of 15 is 5 which on subtraction leaves 10. That is how you do it.
If you study this carefully you find that what Ahmes was really doing was proving the
following algebraic identity:
2n  1 
2n  1  
2n  1 
2n  

 n + 3  − 3  n + 3  − 10   n + 3  − 3  n + 3   = n


1074
1075
1076
1077
1078
1079
1080
1081
1082
1083
1084
1085
1086
1087
1088
1089
1090
1091
1092
1093
1094
1095
1096
1097
1098
1099
1100
1101
1102
1103
1104
1105
a.)
Use n = 9 and verify that the left and right side of this identity actually work. Show all
steps.
b.)
Without using any particular value for n, algebraically simplify the left side of the identity
above and show that, when completely reduced, it is equal to the right side. All work
should be shown and should be in terms of the variable n. Hint: Notice that the quantity in
square brackets appears several times…if you simplify it once you can use that throughout
the expression.
Writing
Write a short essay on the given topic. It should not be more than one page and if you can type it
(double−spaced), I would appreciate it. If you cannot type it, your writing must be legible.
Attention to grammar is important, although it does not have to be perfect grammatically…I just
want to be able to understand it.
70) Explain how writing a positive integer as a sum of powers of 2 is essentially the same thing
as converting a base 10 number to base 2. Use a specific example as part of your
explanation.
71) Suppose you were an elementary school teacher who was teaching multiplication for the
first time. You have decided to teach both the method of doubling and the modern
algorithm for multiplication to your students. Which one would you teach first and why?
Be specific with your reasons. Carefully think them through before responding.
72) Investigate via the library or the Internet some interesting aspect of Egyptian civilization
that we did not touch on in this chapter and present it in a short essay. You must provide
reference(s) of the materials you used (either the publishing information or Internet
address).
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 43
1105
1106
Blank Page
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 44
1106
1107
1108
Endnotes
1
2
http://www-groups.dcs.st-andrews.ac.uk/~history/HistTopics/Egyptian_mathematics.html
Bunt, Jones, Bedient; The Historical Roots of Elementary Mathematics, page 6.
3
http://www-groups.dcs.st-and.ac.uk/~history/Diagrams/Rhind_papyrus.jpeg
4
http://www-groups.dcs.st-and.ac.uk/~history/Diagrams/Moscow_papyrus.jpeg
5
http://www.eyelid.co.uk/numbers.htm All Egyptian numbers like these in this chapter are taken from this web site.
6
http://www.eyelid.co.uk/numbers.htm
7
http://www.eyelid.co.uk/calc.htm
Solution to Check Point A
8
9
Solution to Check Point B
10
Solution to Check Point C
\
\
1
2
4
8
16
32
\
\
12
24
48
96
192
384
Total 516
11
Solution to Check Point D
\
\
1
2
4
8
16
32
64
128
25
50
100
200
400
800
1600
3200
3250
12
13
Bunt, page 17
Solution to Check Point E
7 1 1
= +
10 2 5
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 45
14
Solution to Check Point F
29 1 1
1
= + +
72 3 15 360
29/72
29/72 − 1/3 = 5/72
29/72 = 1/3 + 5/72
72/29 = 2.48… so subtract 1/3
5/72
5/72 − 1/15 = 1/360
5/72 = 1/15 + 1/360
72/5 = 14.4…so subtract 1/15
29/72 = 1/3 + 1/15 + 1/360
15
16
Bunt, page 18.
Solution to Check Point G
\
\
\
\
17
18
1
2
12
6
12
24
1
2
http://www.seanet.com/~ksbrown/rhind.htm
Solution to Check Point H
1
2
4
\
8
\
7
\ 4 + 28
7
14
28
56
1
Total 3 + 6 + 12
2
Total 8 + 4 + 7 + 28
19
Solution to Check Point I
\
1
2+3+ 4
2
4+2+6+2=5+6
4
10 + 3
Total 10 + 3
Answer 4
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 46
20
Solution to Check Point J
Assume 5
\
\
1
5
2
3 + 15
Sum 7
As many times as 7 must be multiplied to give 28, so many times 5 will give the required number.
\
1
2
7
14
4
28
Sum 28
Answer 4
Multiply 4 by 5…
\
1
2
4
5
10
20
Sum 20
Do it thus. The quantity is…
\
1
\
5
3 + 15
20
4
8
Sum 28
©2001, Lawrence Morales; MAT107 Chapter 2 - Page 47