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 2001, Lawrence Morales; MAT107 Chapter 2 - Page 1
16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 Table of Contents PART 1: Introduction to the Egyptian Number System... 3 Introduction and Historical Background... 3 Egyptian Numbers and Base System... 4 Adding in the Egyptian Numeral System... 5 Egyptian Multiplication... 7 PART 2: Unit Fractions, Division, and the 2 n Table... 13 Fractions and Division... 13 Egyptian Division... 19 The 2 n Table... 22 PART 3: Algebra and the Aha Problems... 27 Introduction... 27 Lines and Their Proportional Nature... 27 Ideas Behind the Method of False Position... 28 The Egyptian Method of False Position... 31 PART 4: Homework Problems... 39 Egyptian Numbers and Conversions... 39 Egyptian Multiplication... 39 Egyptian Fractions... 40 Powers of 2... 40 Unit Fractions... 41 Egyptian Division... 41 Egyptian 2/n Table... 43 Algebraic Practice... 43 Method of False Position... 43 Writing... 44 Endnotes... 46 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 19 th 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 right 3. 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 cattle 6. 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
134 135 136 137 138 139 140 141 142 143 144 Example 2 Example 3 Here is the number 459,623. Look at the representation for 5082. What do you notice? 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 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 4 168 169 170 171 172 173 174 175 Solution: Add and. 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: 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 Example 5 Solution: 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. Add and 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 Check Point B Add the following two numbers without using the Hindu Arabic numerals and working only with Egyptian numeral symbols. 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 Solution Egyptian Multiplication See the endnotes for final answer. 9 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. 3 5 7 2 3 9 3 2 1 3 1 0 7 1 0 7 1 4 0 0 8 5 3 2 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 Solution: Multiply 11 80. 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
288 289 290 291 292 293 294 295 296 297 298 299 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 80 \ 2 160 4 320 \ 8 640 Total 880 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
308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 Example 7 Solution: Multiply 35 15. We will provide a column with commentary that would not appear in an Egyptian papyrus, of course. Commentary \ 1 15 1 group of 15 is 15 \ 2 30 2 groups of 15 is 30 4 60 4 groups of 15 is 60 etc. 8 120 This line is not needed to produce 35 groups of 15 16 240 \ 32 480 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 Solution: Multiply 125 4. Commentary / 1 4 1 group of 4 2 8 2 groups of 4 is 8 / 4 16 4 groups of 4 is 16, etc / 8 32 Etc / 16 64 / 32 128 / 64 256 Sum 500 The final result is 500 when we add up all of the marked groups. 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. Keep doubling until the number in the left column exceeds the number of groups you need. 2. Do not include a number in the left column that is larger than the number of groups you need. 3. 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.) 328 329 330 331 332 333 334 335 336 337 Let s look at one last example. Example 9 Solution: Multiply 42 32. Commentary 1 32 1 group of 32 / 2 64 2 groups of 32 4 128 4 groups of 32 / 8 256 16 512 This line is not marked since 32 + 16 is 48, more than the 42 groups we need. / 32 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. 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 Check Point C Solution Check Point D Solution Multiply 43 12 using the Egyptian method. Check the endnotes for the solution. 10 Multiply 130 25 using the Egyptian method. 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: Example 10 Solution Write 75 as the sum of power of 2. 1 = 2 0 2 = 2 1 3 = 2 0 + 2 1 4 = 2 2 5 = 2 0 + 2 2 6 = 2 1 + 2 2 7 = 2 0 + 2 1 + 2 2 etc. This problem is similar to writing 75 in base 2. From Chapter 1, we can check that 75 10 = 1001011 2. This means we can write 75 as the sum of powers of 2 as follows: 6 3 1 0 75 = 2 + 2 + 2 + 2 You should check that this is correct and that you understand how 1001011 2 is 6 3 1 0 related to 2 + 2 + 2 + 2. 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
384 385 386 387 388 389 390 391 392 393 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 40 394 Put another way: can be interpreted as asking, What number 2 Think About It 395 times 2 gives 40? Another important way to pose this question is 396 How many groups of 2 go into 40? Either way, the result is that How can we turn any 397 we can view any division problem as an equivalent problem of division problem into 398 multiplication! And, since we already know how the Egyptians did the equivalent 399 their multiplication, we can then do division as well. Or at least we problem of 400 should be able to. We just need to develop a way to convert a multiplication? 401 division problem to the equivalent of a multiplication problem. 402 403 Before we explore this question further, we first need to look at the way that Egyptians handled 404 their fractions. In general, the Egyptians only used fractions with numerator 1. 405 406 When a fraction has a 1 in the numerator, we will call it a unit fraction. (The word unit means 407 one.) 408 409 To write a unit fraction, the Egyptians would write the denominator in regular hieroglyphics and 410 place the symbol hieroglyphics: 411 412 above it. Thus, the fraction 12 1 looked like the following in 413 414 415 416 In most books, you will see unit fractions denoted by writing a horizontal bar above the denominator. Thus 12 1 would be represented by 12. In this document, you will see both notations. For 3 2, you will sometimes see the symbol 3, a three with two bars over it. 417 2001, Lawrence Morales; MAT107 Chapter 2 - Page 13
418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 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 = There is some evidence that a special symbol also existed for 3 4. 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 Notice in the last line that each of the two fractions have 1 in the numerator, and so each are considered unit fractions. 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: 438 439 440 441 442 443 444 445 446 447 There may be more than one way to write a unit fraction. For example: Another way to write the same fraction is this: 7 1 1 = + 24 6 8 7 1 1 = + 24 4 24 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 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 Bunt 12, 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. 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.) 2. Subtract this unit fraction from the given fraction to see what s left over. 3. 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. 4. 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
475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 Example 11 Solution Example 12 Solution 4 15 Write 4/15 as a sum of unit fractions: The quick way to do this is to think of 4 15 as 3 + 1 = 1 + 1 and you re done. 15 15 5 15 However, let s apply Sylvester s method and see how it works. 4 15 1 1 = 4 60 4 4 60 15 = + 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. 11 Write as a sum of unit fractions using the Sylvester Method: 18 Comments 11 The largest unit fraction less than this is 1/2 since 18/11 = 1.6; so we subtract 1/2 from 18 11/18 11 1 1 When we subtract we get a unit fraction, so we = 18 2 9 are done. 11 1 1 = + Add 1/2 to both sides of the previous equation. 18 2 9 11 2 9 18 = + Final answer. 2001, Lawrence Morales; MAT107 Chapter 2 - Page 16
490 491 Example 13 Write 7 6 as a sum of unit fractions using the Sylvester Method: 492 493 494 495 496 497 498 499 500 501 502 503 504 Solution 6 7 6 1 5 = 7 2 14 6 7 = 1 2 + 5 14 5 14 5 14 6 2 3 42 7 = + + 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. 5 The largest unit fraction less than this is 1/3 (since 14/5 = 2.8) so 14 we subtract it. This time, what s left is a unit fraction, so we are done. We only 1 1 need to combine this result with = 3 42 where we left off in step 3 above. First we rewrite 5/14 as a sum of unit fractions. 1 1 = + Add 1/3 to both sides of Step 5 3 42 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
505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 Check Point E Solution Check Point F Solution Write 7 as a sum of unit fractions. 10 See the endnotes for the answer. 13 Write 29 72 as a sum of unit fractions. See the endnotes for the answer. 14 2001, Lawrence Morales; MAT107 Chapter 2 - Page 18
520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 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 does 15. 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 Solution Find 19 8 using the Egyptian doubling method. 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. 2001, Lawrence Morales; MAT107 Chapter 2 - Page 19
544 545 546 547 548 549 550 551 552 553 554 555 556 557 The following is an example that shows a slightly different approach, but retains the same spirit. Example 15 Solution Find 300 16 using the Egyptian doubling method. We will compute with groups of 16 until we get 300. In other words, how many groups of 16 go into 300? Comments 1 16 We start with one group of 16 \ 2 32 Double 1 to get 2 double 16 to get 32 4 64 etc. 8 128 etc. We stop doubling here because doing so again would yield 512, which is too much \ 16 256 (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. 16 1 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. 8 2 Double 1/16 to get 2/16 = 1/8 double 1 to get 2. \ 4 4 Double 1/8 to get 1/4 double 2 to get 4. \ 2 8 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. Check Point G Sum 300 Here s the sum we wanted. Answer 18 2 4 + + 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. Divide 39 12 using the Egyptian Method. See endnotes for solution. 16 2001, Lawrence Morales; MAT107 Chapter 2 - Page 20
558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 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 Solution Find 14 24. We will compute with groups of 24 until we get 14, just as before. Comments 1 24 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. 24 1 1/24 of the group has one (since a whole group has 24). Now we can start doubling until we get back to 14. \ 12 2 Double 1/24 to get 2/24 = 1/12 double 1 to get 2. \ 6 4 Double 1/12 to get 2/12 = 1/6 double 2 to get 4. \ 3 8 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 1 1 1 Answer The answer here is + +. 3 6 12 3+ 6+ 12 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 2 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 The 2 n Table \ 3 6 Sum 115 Answer 6+ 18+ 3 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. Comments 1 13 1 group has 13. Too big we only need 11 13 1 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 2/13 2 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 article 17 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
597 598 599 600 601 602 603 604 605 606 607 608 2 13 = 8 + 52 + 104. This means that 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. Example 18 Solution Find 20 3. We are seeking to find what number times 3 gives 20. Alternatively, how many groups of 3 go into 20? We will compute with groups of 3 until we get 20. Comments 1 3 One group has 3. \ 2 6 Double as usual. \ 4 12 More doubling. We have 6+12=18 we need 2 more 8 24 This line does not help us it would normally be omitted 3 1 1/3 of a group has 1. \ 2 + 6 2 Sum 20 Answer 6+ 2 + 6 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 + + 2001, Lawrence Morales; MAT107 Chapter 2 - Page 23
609 610 The Egyptian 2 n Table 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 2001, Lawrence Morales; MAT107 Chapter 2 - Page 24
611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630 631 632 633 Example 19 Solution Find 19 5. We will compute with groups of 5 until we get 19. In other words, how many groups of 5 go into 19? Comments \ 1 5 One group has 5 in it. \ 2 10 Double to get 10 we have 15 total in the right column we need four more. 5 1 One fifth of a group has 1 we double until we get four. 3+ 15 2 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 \ 2+ 6+ 10+ 30 4 Check Point H Solution Example 20 Solution Sum 19 Answer 3+ 2+ 6+ 10+ 30 Divide 59 7 using the 2 n Table. See the endnotes for the solution. 18 Divide 6+ 2+ 4 4+ 8 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. 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
634 635 636 637 638 639 640 641 642 643 644 645 646 647 648 649 650 Check Point I 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. 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? 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: rise y y y m = = = run x x x 2 1 2 1 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. -4-2 0 2 2 x -2 2001, Lawrence Morales; MAT107 Chapter 2 - Page 27
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: Ideas Behind the Method of False Position 1 x + 3= 21 5 1 x = 21 3 5 1 x = 18 5 x = 518 x = 90 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.) 1 x+ 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
y 10 8 6 1 y = x+ x 2 y = 6 722 723 724 725 726 727 728 729 730 731 732 733 734 735 736 737 738 739 740 741 742 743 744 745 746 747 4 (2,3) 2 x -1 0 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 Solution Solve with guess and check 2 2 40 3 x+ x=. 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. Our goal is here at the point (?,6) 2001, Lawrence Morales; MAT107 Chapter 2 - Page 29
40 y (15,40) 2 y = x+ 2x 3 20 748 749 750 751 752 753 754 755 756 (3,8) 0 x 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. 2 y y = x+ 2x 40 (15,40) 3 20 757 758 759 760 761 762 (3,8) 0 x 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
763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 Example 22 Solution Solve with guess and check. 1 14 5 x+ x= 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. 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 2001, Lawrence Morales; MAT107 Chapter 2 - Page 31
790 791 792 793 794 795 796 797 798 799 800 801 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 \ 7 1 Sum 8 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 assuming 7 as the answer. Since one-seventh of the quantity is added to the 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 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 2001, Lawrence Morales; MAT107 Chapter 2 - Page 32
802 1 8 We need to know the value of 19 8, so compute with groups of 8 until we get 19 one group has 8 \ 2 16 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. 2 4 One half of a group has 4 in it that s too much. Reduce again \ 4 2 One fourth of a group has 2 in it. 16 + 2 is 18 only one more needed so \ 8 1 One eighth of a group has 1 in it. We now just mark off the columns we need. Sum 19 The answer is + 4 8. This is 19 8, which is the 2 + 4 + 8 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 \ 1 2 + 4 + 8 One group has 2 + 4 + 8 \ 2 4 + 2 + 4 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 \ 4 9 + 2 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 Sum 16 + 2 + 8 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. 2001, Lawrence Morales; MAT107 Chapter 2 - Page 33
803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 818 819 820 821 Do it thus. The quantity is Comments \ 1 16 + 2 + 8 This last step is a check of the process. This line represents the quantity \ 7 2 + 4 + 8 This is one seventh of the quantity Total 19 When we add these together, we get 19. The scribe 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 As many times as 8 must be multiplied to give 19, so many times 7 will give the required number. 1 8 \ 2 16 2 4 \ 4 2 \ 8 1 \ 1 2 + 4 + 8 \ 2 4 + 2 + 4 \ 4 9 + 2 Sum 19 Answer 2 + 4 + 8 The quantity is 16 + 2 + 8 This is how the scribe would do a problem with method of false position refer to this for homework. 2001, Lawrence Morales; MAT107 Chapter 2 - Page 34
822 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 839 840 841 842 843 844 845 Example 24 Solution Do it thus. The quantity is \ 1 16 + 2 + 8 \ 7 2 + 4 + 8 Total 19 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. Here s another problem from the Rhind Papyrus: A quantity and its fourth added give 15. What is the quantity? 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 Sum 5 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 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 Answer 3 gotten by adding up the rows that total 15. Comments 1 3 Now multiply the original guess of 2001, Lawrence Morales; MAT107 Chapter 2 - Page 35
846 847 848 849 850 851 852 853 854 855 856 857 858 859 860 861 862 863 864 865 866 867 868 Check Point J Solution 4 by 3, the proportionality factor. 2 6 Double \ 4 12 Here are the 4 groups of 3 that we want. Sum 12 This is the answer that we will present and check in the next step. Do it thus. The quantity is Comments \ 1 12 One quantity is 12 \ 4 3 1/4 of the quantity is 3 Sum 15 For a total of 15 This concludes the problem. A quantity and its two fifth gives 28. What is the quantity? See the endnote for the answer. 20 This last example will incorporate the 2 n table that we discussed earlier. Example 25 Solution A quantity and its fifth are added to give 19. What is the quantity? Translate the problem as an Egyptian scribe might. Assume 5 Comments \ 1 5 Here is a check of the initial guess \ 5 1 Sum 6 2001, Lawrence Morales; MAT107 Chapter 2 - Page 36
869 870 871 872 873 874 875 876 877 878 879 880 881 882 883 884 885 886 887 888 889 890 891 892 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 12 \ 6 1 Sum 19 Answer 3 + 6 Comments Here we divide 19 by 6 to get the proportionality factor. The proportionality factor is 1 3 6 Comments \ 1 3 + 6 We now multiply the original guess of 5 by the proportionality factor. We want 5 groups of 3 + 1/6 2 6 + 3 Double 3 + 1/6 and you get 6 + 1/3 \ 4 12 + 2 + 6 Double 6 1/3 and get 12 1/2 1/6, which we read from the 2 n Table Sum 5 groups of 3 1/6 give 15 + 1/2 + 1/3. This is 15 + 2 + 3 the final answer, which we now check. Conclusion Do it thus. The quantity is \ 1 15 + 2 + 3 \ 5 3 + 6 Sum 19 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