History of Math for the Liberal Arts CHAPTER 3. Babylonian Mathematics. Lawrence Morales. Seattle Central Community College

1 3 4 5 6 History of Math for the Liberal Arts 7 8 9 CHAPTER 3 10 11 1 13 14 15 16 17 Babylonian Mathematics Lawrence Morales 18 19 0 Seattle Central Community College MAT107 Chapter 3, Lawrence Morales, 001; Page 1

1 3 4 5 6 7 8 9 30 31 3 33 34 35 36 37 38 39 40 41 4 43 44 45 46 47 48 49 50 51 Table of Contents Part 1: Introduction to Babylonian Numbers...4 Background & Historical Information...4 Early Mathematical Development...5 The Babylonian Writing System and Scribal Schools...5 Babylonian Number Symbols...7 Part : The Sexagesimal System...8 The Sexagesimal System as Used by the Babylonians...8 Difficulties With Babylonian Numbers...10 Babylonian Fractions...11 Going the Other Direction...14 Why Base 60?...18 Part 3: Babylonian Arithmetic...19 Addition...19 A Primer on Multiplication...1 The Babylonian Multiplication Table...8 Babylonian Multiplication with Tables...30 Babylonian Multiplication on Tablets...34 Babylonian Division...34 Division on Babylonian Tablets...37 Part 4: Babylonian Root Approximations...38 Babylonians and Square Roots...38 An Alternate Method of Estimating Roots...41 Plimpton 3...43 Part 5: Babylonian Algebra...44 Systems of Equations...44 Babylonian Quadratics...50 General Approaches to Quadratics...54 Part 6: Homework Problems...57 Conversions...57 A Babylonian Translation Problem...57 MAT107 Chapter 3, Lawrence Morales, 001; Page

5 53 54 55 56 57 58 59 60 61 6 63 64 65 66 67 68 Multiplication...58 Division...58 Tablet Problems...58 Root Approximations...59 The Alternate Method of Estimating Roots...60 The Babylonians and Pythagorean Triples...60 A Famous Babylonian Tablet...61 Babylonian Algebra Systems of Equations...61 Babylonian Algebra Solving Quadratic Equations...6 Practice and Application of the Quadratic Equation...63 New Quadratic Equations...63 Yet Another Quadratic Formula...64 Real Babylonian Algebra Problems From Tablets...65 Writing...66 Appendix: Blank Gelosia Grids...67 Specific Gelosia Grids...71 Part 6: Endnotes...73 MAT107 Chapter 3, Lawrence Morales, 001; Page 3

69 70 71 7 73 74 75 76 77 78 79 80 81 8 83 84 85 86 87 88 89 90 91 9 93 94 95 96 97 98 99 100 101 10 103 104 105 106 107 108 109 110 111 11 113 114 Part 1: Introduction to Babylonian Numbers The next civilization that we will explore in this course is ancient Mesopotamia. While this is its official, scholarly title, it is often referred to as Babylon. However, Babylon was only a small part of Mesopotamia. In this reading, I will use the terms interchangeably, despite the fact that they are not the same. As we did with the Egyptians, we will explore the symbols for their numbers, their base system, and their basic methods of doing arithmetic. We will need to come up with new methods of operations for multiplication and division, as the Babylonians did not give us as much information on how they did their calculations as the Egyptians did. We will also look at some of their algebraic techniques, particularly for solving quadratic equations. Background & Historical Information 1 The Babylonian civilization was made up of people who lived in the alluvial plain between the Tigris and Euphrates rivers. This is the area from Baghdad south to the Persian Gulf. The Greeks were originally those who called it Mesopotamia, which means, land between the rivers. (See the map of the region. ) Cities, writings, and metallurgy were necessary to form this civilization. 3 The cities had extensive irrigation systems, codes of law, postal services, and an administrative bureaucracy. Besides writing, they utilized other forms of technology such as wheeled vehicles, boats, plows, weaving, and brick towers known as ziggurats. To build tools (and weapons), they used copper smelting and bronze smelting techniques. Knowledge about their political and cultural history is somewhat limited 4 but is growing. There are many tablets that have been discovered with writings from this time period, but translation is slow and a relatively recent process. It wasn t until 1846 that the deciphering code was broken, which then allowed the translation process to begin. However, there are so many tablets that the task is still far from complete. 5 From what we do know, a series of invasions and sporadic warfare served to awaken and stimulate the culture. The geography of the land made cities of the river plain open to blatant attack from many groups. Many groups of invaders battled for control in the region for about three thousand years. One dynasty, the Amorite dynasty, achieved its greatest might under the rule of Hammurabi (circa 179-1750 B.C.E.). Hammurabi is now known for his code of law, The Hammurabi Code which was very logical but also very harsh. One of its injunctions is the familiar an eye for an eye and a tooth for a tooth. MAT107 Chapter 3, Lawrence Morales, 001; Page 4

115 116 117 118 119 10 11 1 13 14 15 16 17 18 19 130 131 13 133 134 135 136 137 138 139 140 141 14 143 144 145 146 147 148 149 150 151 15 153 154 155 Early Mathematical Development In terms of mathematics, influences of imperialism and trade over long distances served as the catalysts of mathematical development 6. These were both relatively new to human society at the time and encouraged the establishment of news schools and temple scribes who could record and manage the new collections of information needed to support these endeavors. However, everyday needs such as religion, commerce, and agriculture were even stronger influences on the development of mathematics. 7 Grain supplies had to be tracked and distributed among an increasingly large population. Daily business transactions and the use of wills encouraged the creation of numerical tables. The building of dams, irrigation canals, granaries and other buildings necessitated calculations be made while the religious practices of the time were heavily reliant on having a dependable calendar. This means keeping careful records of astronomical data. In the following table, you can see a general breakdown of periods of history in the context of their mathematics. Period Mathematical Development 5000 3000BCE Development of number concepts in prehistoric Middle East 3000 000BCE Evolution of the sexagesimal place value system in southern Iraq 000 1600BCE Arithmetic in Old Babylonian scribal schools 1600 0BCE Old Babylonian mathematics As the table indicates, the Babylonians had what is called a sexagesimal place system, which means they had a base 60 system. Also, notice that there is an entry in the table dealing with scribal schools. We ll briefly look at the role of such schools in the development of Babylonian mathematics. The Babylonian Writing System and Scribal Schools There are about 400,000 Babylonian tablets that have survived to the present day. Many of them are hidden in the library collections of older universities and have not been looked at in many years. (In the summer of 1999, Dr. Eleanor Robson discovered several at Catholic University of America while we were attending an NSF Institute on the Use of History of Math in the Classroom.) Of these, about 400 are related to mathematics. Most are what are called Old Babylonian, (1800 to 1600 B.C.E.) and they contain less information than the Egyptian mathematics sources that we have available to us. However, from what we do have, it appears that the Babylonians were more advanced in mathematics than the Egyptians were. The tablets come in a variety of shapes and sizes: Large multi columned tablets: to 6 columns, printed on both sides. Large student teacher tablets: The teacher would use large script while student writing would be smaller and to the right. Small one-column tablets. Round buns : to 4 lines, usually used for student teacher exercises. MAT107 Chapter 3, Lawrence Morales, 001; Page 5

156 157 158 159 160 161 16 163 164 165 166 167 168 169 170 171 17 173 174 175 176 177 178 179 A bun tablet, which is a school exercise in repeated multiplication and division. 8 A meteorological calculation From the list of types of tables above you see a hint that a big source of tablets, math, and the start of the sexagesimal system comes from school tablets. These were tablets used in scribal schools that were used to train young boys/men to do the work of scribes. The curriculum of the school was progressive 9 : 1. Learn the basics of writing letters, names, and forms.. Learn lists of trees, reeds, vessels, animals, stones, plants, and other nouns by theme. 3. Learn how to represent meteorological data, weights, multiplication, and reciprocals. 4. Learn how to write contracts and proverbs. Shortly after 3000 B.C.E., the Babylonians developed a system of pictographs (sort of like hieroglyphics) to represent their numbers. These are called cuneiform. 10 While the Egyptians used a form of ink to write on their papyrus, the Babylonians used a reed and later a stylus with a triangular head to make pictographic impressions into clay tablets. These tablets were then baked so they could become hardened to preserve the writing. The limitation of the writing stylus that they used prevented them from having a wide variety of symbols that they could use to represent numbers. In particular, the creation of curved lines was pretty much impossible so they resorted to a series of vertical, horizontal, and oblique marks, as you can see above. As time passed, they also found a way to draw a wedge that looks like an angled bracket opening to the right. (Holding the stylus so that its sides were inclined on the tablet could do this.) MAT107 Chapter 3, Lawrence Morales, 001; Page 6

180 181 18 183 Babylonian Number Symbols The symbols that we will be using (and which they used) are the following: Symbol Name Decimal Value or Wedge 1 184 185 186 187 188 189 190 or Corner 10 Empty Placeholder None. It specifies when a place is empty and was not used until 300 B.C.E. As you can see, there are only two main symbols for numbers in the Babylonian system, the for ones and the for tens. Compare this with the Egyptian system and even our own system that needs many more symbols to represent some arbitrary number. The Babylonians would use just these two symbols within a base sixty system to represent any number they wanted. In the picture below 11, you can see the symbols for various numbers: 1 = = 3 = 4 = 5 = 6 = 7 = 8 = 9 = 10 = 11 = 1 = 191 19 193 194 195 196 15 = 0 = 30 = 40 = 50 = They also has certain special symbols for some common fractions which we will not use much in this text, but they are interesting to note nonetheless. 60 = Think About It In the list of symbols above, the symbol for 1 and for 60 is the same? Why would that be true? 197 198 199 00 01 Note the absence of a symbol for zero or for the equivalent of a decimal point (some symbol that separates whole numbers from fractional numbers). This will cause some problems and ambiguity in interpreting the values of Babylonian numbers, but for now it s just interesting to note the limited number of symbols they had available to them to write numbers. MAT107 Chapter 3, Lawrence Morales, 001; Page 7

0 03 04 05 06 07 08 09 10 11 1 13 14 15 16 17 18 19 0 1 3 4 5 6 7 8 9 30 31 3 33 34 35 36 37 38 39 40 41 Part : The Sexagesimal System In the first topic of this course, we learned about different base systems. In particular, we looked at the Mayan base 0 system and how to convert between their system and our own. The process of converting between base 60 and base 10 is essentially the same as it was with the Mayans, except that we replace the base of 0 with the base of 60. Let s first start by converting from base sixty to base ten, which is generally easier. The Sexagesimal System as Used by the Babylonians Unlike the Egyptians system, the sexagesimal system is a positional system. That means that the position of a symbol will be important in determining the value of that symbol. In the sexagesimal system, every new place is a sequential power of 60. Let s look at a table to compare our decimal system to the Babylonian s. Decimal (Base 10) System Sexagesimal (Base 60) System 10 0 1 60 0 1 10 1 10 60 1 60 10 100 60 3,600 10 3 1000 60 3 16,000 Etc Etc Etc Etc As you can see, both systems start off with the ones place, but then they diverge pretty quickly from there. The sexagesimal system place values increase much more quickly, for obvious reasons. We will need a standard way to represent a sexagesimal number so that it s easy for us to see what number is being discussed. Let s take the following example. Example 1 Convert 1,40,59 60 to base 10. (The subscript, once again, gives the base in use.) This number may look odd to you. The commas appear to be in the wrong places. However, this is a sexagesimal number, not a decimal number, so you can t read it the way you are used to. This number has the following meaning. 1,40,59 = 59 in the ones place 40 in the sixties place 1 in the thirty six hundreds place Note that the commas serve to separate the places from each other, which is a different role than they play in the decimal system. Because we have 60 as our base, each place value can have as many as 59 in it before we have to carry to MAT107 Chapter 3, Lawrence Morales, 001; Page 8

4 43 44 45 46 47 48 49 50 57 58 59 60 61 6 63 64 65 66 67 68 69 70 71 the next place up. In the sexagesimal system, no place can have more than 59 in it, just as in the decimal system no place can have more than 9 in it before we carry up to the next place. If we want to convert this number to base 10, we would to the following: 1 0 ( ) ( ) ( ) ( 1 60 ) ( 40 60) ( 59 1) 1,40,59 = 1 60 + 40 60 + 59 60 = + + = 75600 + 400 + 59 = 78059 Hence, 1,40,59 in base 60 is equal to 78,059 in base 10. 51 Of course, 1,40,59 is a modern 5 way of representing sexagesimal 53 numbers. We ll call it modern 1 thirty-six hundreds 40 sixties 59 ones 54 sexagesimal notation for 1 times 3600 40 times 60 59 times 55 1 representing Babylonian numbers. 56 It is used for our own convenience. The Babylonians, on the other hand, would write this number as shown here. Note some key points: (1) The ones places starts on the right, as ours does. () Each place value is separated by a space to give the reader a clue about how many symbols are in each position. (3) The symbols generally go to the right of the symbols. (4) There is no zero symbol. To represent 40, four corners (each of which has a value of 10) are recorded and no ones symbols are recorded (like the Egyptians). The four corners in the sixties represent a total of,400 in base 10 because of their position 40 sixties is,400. Example Write 13,7,34 in Babylonian notation and determine its decimal value. In Babylonian notation, it looks like this: 7 73 74 75 76 77 Its decimal value is: 13 60 + 7 60 + 34 1 = 46800 + 40 + 34 ( ) ( ) ( ) = 4754 MAT107 Chapter 3, Lawrence Morales, 001; Page 9

78 79 80 81 Check Point A What is the decimal value of the following Babylonian number? Write it in modern notation. Be careful to note where the spaces are to distinguish place values. 8 83 84 85 86 87 88 89 90 91 9 93 94 95 96 97 98 99 300 301 30 303 304 305 306 307 308 309 310 311 31 313 314 315 316 317 318 319 30 31 See endnote to check your answer. 1 Check Point B Write 1,13,5,11 in Babylonian notation and then find the decimal value of the number by converting to base 10. See endnote to check your answer. 13 Difficulties With Babylonian Numbers As you can imagine, these numbers can be very hard to read if you re not carefully trained. The spaces between positions may not be large and may make distinguishing the positions difficult. As you can see in the picture, the clay tables that the Babylonians used were usually not much larger than the palm of your hand and could have a lot of writing on them. So reading them is not easy. (This table is only about 6cm long, or about.5 inches, and look at the detail on it!) Another major difficulty with reading the numbers accurately is the fact that the Babylonians did not generally have a symbol for zero. Therefore, if a place has zero in it, that place is just skipped and the reader of the tablet has to determine this fact from the context of the rest of the tablet. For example the following number could have many different values: It could be 34 60 + 1, the most straightforward value. It could be 34 60 + 1 if the 60 s place is empty. It could be 34 60 + 60 if the ones place is empty. It could be 34 60 3 + 60 if the 60 and ones places are empty. And so on. MAT107 Chapter 3, Lawrence Morales, 001; Page 10

3 33 34 35 36 37 38 39 This is certainly a drawback to their system, at least from our own modern viewpoint. In our modern notation, we will go ahead and use our symbol for zero if there is something missing from a place value. Hence, the number 18,0,3 means that there are no 60 s. In about 300 B.C.E. (relatively late), they did start using the symbol as a placeholder to indicate that a position was blank. However, as far as we know, they did not view this as a number as we do the number zero. Hence, to write the number 18,0,3 in Babylonian notation, they would mark the following into a clay tablet: 330 331 33 333 334 335 336 337 338 339 340 341 34 343 344 345 346 347 348 349 350 351 35 Babylonian Fractions What about non whole numbers? We know that in the decimal system, 39.7 means thirty nine whole parts plus seven tenths (fractional parts). The decimal point tells us this. The Babylonians, however, did not have a symbol that accomplishes the same thing. This meant that the tablet reader had to determine the value of the number from the context of the tablet. In modern sexagesimal notation we will use the semicolon to separate whole numbers from fractions. For example, we might see a number that looks like this in modern works that study Babylonian mathematics. 3,16;30 This is how this number might appear on a Babylonian tablet: There is no marker to tell you where the whole numbers end and where the fraction part starts. To help us keep track of what is what, we ll use the non standard notation of drawing a dotted line where the whole numbers and fractions are split. Thus, we will write the number above as: 353 354 355 MAT107 Chapter 3, Lawrence Morales, 001; Page 11

356 357 358 359 360 361 36 363 364 365 366 367 368 369 370 371 However, assuming we know that a group of symbols is not in a Decimal Sexagesimal whole number place, how do we interpret their values? In our (Base 10) (Base 60) decimal system, once you move to the right of the decimal point, every place is one tenth of the previous place. In the sexagesimal 1 s 1 s system, each place to the right of the sexagesimal point is 1 1 one sixtieth of the previous place. Here is a table that illustrates 10 60 this. 1 1 1 1 = = The fractional place values in the sexagesimal system get 10 100 60 3600 smaller much more quickly than they do in the decimal system again, for obvious reasons. 1 1 1 1 = = 3 3 10 1000 60 16000 So now we can interpret the 30 after the dotted line. It means that we have 30 sixtieths, which is the same as 30/60. Therefore, the decimal value of the number is as follows: 1 3,16;30 = ( 3 60) + ( 16 1) + 30 60 1 = 1380 + 16 + 37 373 374 375 Example 3 = 1396.5 Find the value of the following Babylonian number: 376 377 378 379 380 381 38 383 384 385 386 387 388 Example 4 In modern notation, this would be represented by 6,10;40. Its decimal value is: 1 ( 6 60) + ( 10 1) + 40 = 60 = 360 + 10 + 3 370.666 Note that 40/60 has been reduced to /3. Find the value of the following Babylonian number: MAT107 Chapter 3, Lawrence Morales, 001; Page 1

389 390 391 39 393 394 395 396 397 398 399 400 Check Point C In modern notation, this number would be represented by 10;10,. Here is its decimal value: 1 1 10 + 10 + = 60 60 1 1 = 10 + + 6 1800 10.167 Write the following Babylonian number in modern notation and find its decimal value: 401 40 403 404 405 406 407 408 409 410 Check Point D See the endnote to check your answer. 14 Write the following Babylonian number in modern notation and find its decimal value: 411 41 413 414 415 416 See the endnote to check your answer. 15 MAT107 Chapter 3, Lawrence Morales, 001; Page 13

417 418 419 40 41 4 43 44 45 46 47 48 49 430 431 43 433 434 435 436 437 Going the Other Direction. Now suppose we have a decimal number that we want to convert to sexagesimal notation. You will see that the process is once again Power of 60 Value the same that we used to convert from base ten to base twenty (or 60 1 60 any other base). Let s start with a simple example and work up from there. We will need to know what the powers of 60 are to do 60 3600 this so here is a quick little table to help us. 60 3 16,000 60 4 1,960,000 Example 5 Convert the decimal value of 85 to modern and Babylonian sexagesimal notation. Since there are more than 59 ones, we must go to the 60 s place. There is one 60 in 85 since 85 = (60 1) + 5. We see there are 5 ones left over. Therefore, 85 = 1,5. The Babylonian representation would therefore be: 438 439 440 441 44 443 444 445 446 447 448 449 450 451 45 453 454 455 456 457 458 459 460 Example 6 Alternatively, we could use our methods from Chapter 1. If we divide 85 by 60, we get 1.416666.. So there is one in the sixties place. Subtracting that one from the quotient gives 0.416666. which we can then multiply by 60 to get 5. (Review Chapter 1 for a discussion of this general method.) Convert the decimal value of 8,000 to modern sexagesimal notation. For this number, it s easiest to ask, What is the highest power of 60 that will go into 8,000? From the table we see that 60 =3600 is the highest power of 60 that will go into 8,000. 3600 goes into 8000 a total of times, with remainder 800. We now move to the 60 s place. 60 goes into 800 a total of 13 times with a remainder of 0. Since 0 is less than 59 we have 0 ones. Therefore, 8,000 =,13,0. MAT107 Chapter 3, Lawrence Morales, 001; Page 14

461 46 463 464 465 466 467 468 469 470 471 47 473 474 475 476 477 478 479 480 481 48 483 484 485 486 487 Check Point E Of course, we can check this simply by starting at,13,0 and converting back to decimal to make sure that we get 8,000. (You should do so now on your own.) Once again, we could use our alternative method from Chapter 1, in which case we would dived 8000 by 3600 = 60. Continuing the process will produce the same result as the one obtained above. Convert the decimal value of 300,000 to modern sexagesimal notation. See the endnotes to check your answer. 16 We have one more conversion issue to deal with, and that is decimals. How do we convert the decimal value of 1/3 into Babylonian notation? How about 0.4? Well, the value 1/3 is not too hard. We can see that 1/3 is the same as 0/60, so in modern sexagesimal notation, 1/3 = 0;0. But what about 0.4? In order to see how to convert this, let s convert 1/3 in a slightly different way: 1 1 3 60 0 = = = 0; 0 3 1 60 60 With this method, we see that multiplying 1/3 by 60 gives us the desired result. So this is the technique we ll use, only we ll streamline it a bit. MAT107 Chapter 3, Lawrence Morales, 001; Page 15

488 489 490 491 49 493 494 495 496 497 498 499 500 501 50 503 504 505 506 Example 7 Convert 0.4 to modern sexagesimal notation. We ll display our steps in a table with comments to help explain the process: 60 0.4=14.4 Therefore 0.4 = 14 +? 60 60 0.4 60=4 4 Therefore we have 60 14 4 Finally, 0.4 = + = 0;14; 4 60 60 Check: 14 4 0;14, 4 = + 60 60 6 = 5 = 0.4 Comments Since we get 14.4, this means we have 14.4. But we only want whole values 60 in the sixtieth s place. So we take 14 60 and then have some undetermined number left 1 over in the 60 place. Since we have 0.4 left over (after removing the 14), we multiply that by 60. This tells us how many we have in the 1 60 place. We combine our results to get the final answer. By carefully adding fractions we can get to 6/5, which when converted to a decimal is 0.4, confirming our work. As a result of this example, we now have a method for converting a fraction decimal to sexagesimal notation. (1) Multiply the decimal by 60. The whole number part of the product is the number of 1/60 s in the number. () Multiply the left over, fractional part of the product in (1) to get the number of 1/60 s. (3) Continue this process until multiplying by 60 gives you a whole number. MAT107 Chapter 3, Lawrence Morales, 001; Page 16

507 508 509 510 511 51 513 Example 8 Ones Convert 5.66 to modern sexagesimal notation. We can ignore the 5 since it is the whole number part. We only want to start multiplying 60 by the fractional part of this number. 1 60 1 60 1 60 3 514 515 516 517 518 519 50 51 5 53 54 55 56 57 58 59 530 531 53 533 534 535 536 537 538 539 5 + 0.66 60 = 15.96 0.96 60 = 57.6 0.6 60 = 36 5 15 57 36 We can now assemble all these pieces to see that 5.66 = 5;15,57,36. If you don t believe it, here s a check: 15 57 36 633 5 + + + = = 5.66 3 60 60 60 500 Check Point F Convert 75.10 to modern sexagesimal notation. See the endnotes to check your answer. 17 Check Point G Convert 13.55 to modern sexagesimal notation. See the endnotes to check your answer. 18 MAT107 Chapter 3, Lawrence Morales, 001; Page 17

540 541 54 543 544 545 546 547 548 549 550 551 55 553 554 555 556 557 558 559 Why Base 60? It is uncertain why the Babylonians chose 60 as their base. Theon of Think About It Alexandria (600 C.E.) made a comment that 60 was one of the smaller numbers that has a large number of divisors and that 60 was chosen for How many divisors this reason. Others who have studied the Babylonians think there was a does 60 have and what more natural origin to the system. For example, the Babylonian took the are they? Why is the year to have 360 days. Since 360 is a multiple of 60, they chose this as concept of having their base. (But why not some other multiple such as 0?) Dr. Robson has more divisors helpful? suggested that it was invented to shortcut the complicated procedures needed for multiplication and division in particular. 19 Others have suggested that the base evolved from two or more groups. Perhaps one set of people had base 10 and another had base 6 and they merged the two. None of these are definitive and are all theoretical. It has yet to be determined for certain why 60 was chosen as their base. We saw earlier that their writing system had some disadvantages. The base system also has some disadvantages. For example, even certain small numbers can require numerous marks in the clay. For example, 999 only requires three symbols in our decimal system. How many does it require in the sexagesimal system? 0 MAT107 Chapter 3, Lawrence Morales, 001; Page 18

560 561 56 563 564 565 566 567 568 569 570 571 57 573 574 575 576 577 Part 3: Babylonian Arithmetic Addition We now return to the questions of Babylonian mathematics and examine their basic arithmetic techniques. We will want to look at addition, multiplication, division, and computing the values of square roots. Recall from our studies of the Egyptians that addition was as straightforward as gathering like symbols and then carrying when we had more than 10 in a certain place value. In Babylonian addition, we will use the same idea we combine like symbols and when we have more than 59 in a place, we carry a ones symbol up to the next place. One example will hopefully be enough to illustrate this. Just keep in mind that the idea is exactly the same as what we did with Egyptian math. Example 9 Add the following two Babylonian numbers: 578 579 580 581 58 583 584 585 We ll assume that the right columns are the ones. When we gather up all the symbols in each place (like we did with the Mayans), we get the following. 586 587 588 589 590 591 59 We need to keep in mind that 10 s make one. Also, six s make 60 since each one of them represents 10. When we look at the ones place, we see that we have a total of 73. So we take 60 of them and carry to the 60 s place. Note that six s make one in the 60 s place. (Why?) MAT107 Chapter 3, Lawrence Morales, 001; Page 19

593 594 595 596 597 598 599 600 601 60 That leaves one and three s in the ones place (for a total of 13). Now we look at the 60 s place in the middle. There are more than 10 s here so we need to convert ten of them to a since every 10 s makes a. (Why does it get converted to a which stays in its current place rather than moving up to the next place?) 603 604 605 606 607 608 This leaves s and 6 s in the 60 s place, for total of 6 60 s. There is nothing to carry from the 60 s to the 60 s since 6 is less than 60. The 60 s place has five s in it for a total of 50, so it s okay as well. So we have our final result. 609 610 611 61 613 614 615 616 Thus, 30,16,40 + 0,9,3 = 50,6,13 60 You can check this by converting the original numbers into base 10 and then adding and comparing to the answer above. MAT107 Chapter 3, Lawrence Morales, 001; Page 0

617 618 619 Check Point H Add the following Babylonian numbers. 60 61 6 63 64 65 66 67 68 69 630 631 63 633 634 635 636 637 See the endnotes to check your answer. 1 A Primer on Multiplication When we get to this topic, the method of the Babylonians changes from the method of the Egyptians. With the Egyptians, we saw that they used the method of doubling to do their multiplication and that this was a great illustration of multiplication as repeated addition. We do not know any specific multiplication process that the Babylonians used. Instead, they appeared to use multiplication tables instead. These tables had the values for numerous products that students and scribes could look up if they had not already memorized them. Here is a picture of a typical multiplication table. While it is hard to read, it gives you a sense of what they looked like. 638 639 MAT107 Chapter 3, Lawrence Morales, 001; Page 1

640 641 64 643 644 645 646 647 648 649 650 651 65 653 654 655 656 657 658 It would be very difficult to try to reproduce these tables and then actually use them, but there is still some value in trying to learn how to multiply in a base other than 10, a task that we have yet to undertake. So let s do so now. We start by examining our own (U.S.) multiplication algorithm. (If you did not learn arithmetic in the U.S., you may have a different method of multiplication.) Let s start with a simple example: 56 38. To do this problem, we would proceed as follows 5 6 3 8 4 4 8 This line represents 8 56 1 6 8 0 This line represents 30 56 1 8 This line is the sum of the products This process is a very clever one indeed, when you examine it in more detail. (I ll bet very few of you have ever thought about how or why this method works. It just does. ) We ll do that on the next page, where we can see all the steps and reasoning together. MAT107 Chapter 3, Lawrence Morales, 001; Page

659 660 56 38 = (50 + 6) (30 + 8) 5 6 3 8 4 8 Multiplying the 6 and 8 is the equivalent of multiplying 6 ones by 8 ones, giving 48. 56 38 = (50 + 6) (30 + 8) 5 6 3 8 4 8 4 0 0 Multiplying the 5 and 8 is the same as multiplying 5 tens by 8 ones, which is 400. Note that when you up the first two lines, 48+400=448, which is what you get if you carry the 4 in the first step and then add it to the 40 you get from 8 times 5. 5 6 3 8 4 4 8 56 38 = (50 + 6) (30 + 8) 56 38 = (50 + 6) (30 + 8) 5 6 3 8 4 4 8 1 8 0 1 5 0 0 5 6 3 8 4 4 8 1 6 8 0 Multiplying the 3 and 6 is the same as multiplying 3 tens and 6 ones, giving 180. Multiplying the 3 and 5 is the same as multiplying 3 tens and 5 tens, giving 50 30=1500. Adding the last two computations gives 1680. 56 38 = (50 + 6) (30 + 8) =18 5 6 3 8 4 4 8 1 6 8 0 1 8 Add everything up and get,18. 661 66 MAT107 Chapter 3, Lawrence Morales, 001; Page 3

663 664 665 666 667 668 669 670 671 67 673 674 675 676 677 678 679 680 681 68 683 684 685 686 687 688 689 690 691 69 693 694 695 696 697 698 699 700 701 70 What is important to note in the previous page is that the algorithm that we use is essentially based on multiplying each digit in the first number by every digit in the second number AND making sure that when doing so, the values of the digits based on their places are properly accounted for. In our modern algorithm, this is accomplished by placing zeros in the appropriate places to shift numbers to the left. 5 6 3 8 4 4 8 1 6 8 0 1 8 Trying to use this same algorithm in base 60 would be possible, but because the base system is different, it becomes rather confusing. Instead, we ll use a different method that is equivalent but will help us to keep track of the place values. It s a method that actually comes from the Middle Ages, but we ll adapt it here for our own purposes. It s called the gelosia method. (In Chapter 6, Logs and Cubic, we ll give more background information on this method.) Before we start multiplying in base 60 with this new method, we ll first show how it works in base 10. Let s go back to the problem of multiplying 56 38. This zero serves to make sure that we take into account that the 5 and 3 are in the tens place In the gelosia method, we begin by placing the numbers to be multiplied on the outer edges of a grid, as shown below: 5 6 Note that there are essentially four blocks, each of which has a diagonal line running through it. The four blocks correspond to the four different multiplications that need to be done: 5 3, 6 3, 5 8, and 6 8. Where two numbers intersect, we place their product, using the diagonal line to separate the ones place from the tens place. Filling in this grid gives the following: 3 8 5 6 1 1 5 8 3 4 4 0 8 8 MAT107 Chapter 3, Lawrence Morales, 001; Page 4

703 704 705 706 707 708 709 710 711 71 713 714 715 716 717 718 719 70 71 7 73 74 75 76 77 78 79 730 731 73 733 734 735 736 737 738 739 740 741 74 743 744 745 746 747 748 Notice that there is no need to carry anything, yet. We are simply doing all the individual multiplications and recording the results. Now the diagonal lines play their most important role. We are going to use them as addition guides. We start at the lower right hand corner of the grid. All digits that lie within the same diagonal trough are added together. If the result is ten or more, we carry up to the next diagonal. For this grid, we would get the following: 5 6 1 1 5 8 3 4 4 0 8 8 5 1 6 1 1 1 5 8 3 4 4 1 0 8 8 8 In the first diagonal we have an 8, so it gets copied down. The first diagonal represents the ones place. In the second diagonal we have 8 + 4 + 0 = 1, so we carry 1 to the next diagonal up and write a in the diagonal s total box. (Note the one has been carried up and is written in a smaller font size above for distinction.) The second diagonal represents the tens place. In the third diagonal we have 4 + 5 + 1 + 1 = 11, so we carry 1 to the next diagonal up and write a 1 in the diagonal s total box. The third diagonal represents the hundreds place. In the fourth diagonal we have 1 + 1 =, which gives us a, which we write in the diagonal s total box. The fourth diagonal represents the thousands Thousands Hundreds place. Tens Now, the answer is easily read from the upper left corner around the edge to the lower right corner as 18. Why does this work? Well, note that when we multiply 6 8 and get 48 (lower right corner), this represents 4 tens and 8 ones. The 8 goes in the first diagonal and the 4 goes up one diagonal into the tens diagonal. When we do Carried up from the previous diagonal 5 6 1 1 5 8 3 4 4 1 0 8 8 8 Ones MAT107 Chapter 3, Lawrence Morales, 001; Page 5

749 750 751 75 753 754 755 756 757 758 759 760 761 76 763 764 765 766 767 768 769 770 771 77 773 774 775 776 777 778 779 780 781 78 783 784 785 786 787 788 789 790 791 79 793 6 3 (upper right corner), we re really doing 6 30 = 180, which is 1 hundreds and 8 tens. So the 8 goes in the tens diagonal (with the 8 from the previous step), and the 1 goes up into the hundreds diagonal. Note that the diagonal lines are doing the same thing that we do when we add that extra zero in our algorithm to make sure that all the places line up. Let s look one more example in base 10 before we move to base 60. Example 10 Use the gelosia method to multiply 39 5. We begin by building a grid of the correct size 3 9 We now fill in the six multiplication blocks. Note that we can place a zero in the upper part of a block if the product of two digits is less than ten: Finally, we add up the diagonal place values, carrying where appropriate: 5 3 9 1 4 1 5 5 0 5 0 1 0 6 8 4 3 9 1 4 1 5 5 0 5 0 1 0 0 6 8 4 3 8 4 MAT107 Chapter 3, Lawrence Morales, 001; Page 6

794 795 796 797 798 799 800 801 80 803 804 805 806 807 808 809 810 811 81 813 814 815 816 817 818 819 80 81 8 83 84 85 86 87 88 89 830 831 83 833 834 835 836 837 838 839 We have an answer of 0,384, reading around the edge of the grid. It is not necessary to draw three separate grids to do this problem. One will suffice. But three are given here to show you the progression of steps. There are various gelosia grids in the Appendix at the end of the chapter that you can photocopy or trace so that you don t have to draw them by hand. Check Point I Example 11 Multiply 475 879 using the gelosia grid. Work in base 10. See the endnotes to check your answer. 3 Use the gelosia grid method to multiply 78.5 3.6 What do we do about decimals? It s simple. Simply do the calculations as before and then place the decimal point at the appropriate place in the final answer. 7 8. 5 1 1 4 5 3 4 4 3. 8 8 0 6 6 0 Think About It Suppose you had a child who had already learned their multiplication tables and you had to teach them either the modern multiplication algorithm or the gelosia grid system for multiplication. Which one do you think would be more effective? Why? Think About It The first diagonal (at the lower right corner) in the grid above corresponds to the hundredths place. Why? MAT107 Chapter 3, Lawrence Morales, 001; Page 7

840 841 84 843 844 845 846 847 848 849 850 851 85 853 854 855 856 857 858 859 860 861 86 863 864 865 866 867 868 869 870 871 87 873 874 875 876 877 878 879 880 881 88 883 884 885 We get a result of 860, but we still need to place our decimal point. Since there were a total of two digits to the right of the decimals (the 5 in 78.5 and the 6 in 3.6), we move the decimal place in a total of two places from the right. So our result is 8.60, which is, of course, correct. Check Point J Use the table method to multiply 615 48. Check the endnotes to check if your answer is correct. You can use the following grid to do your work. 4 The Babylonian Multiplication Table We are going to extend the table method of multiplication to the sexagesimal system shortly. But we have one more thing to take care of first. We said earlier that the Babylonians used multiplication tables to do complicated multiplication problems. So I have created a modern version of a sexagesimal multiplication table. It is titled Babylonian Multiplication Table in Base 60. You should have it in front of you for the following explanations and examples. As you look at the table, you will notice that there is a series of rows and columns labeled on the top and side of the table. To multiply two sexagesimal numbers together we move along the row containing the first number until intersects the column with the second number. For example, if we want to multiply the sexagesimal numbers 13 30, we do the following: 1. On the left side of the column, find the row labeled 13. (It has a 0 6) to its right.. Move along this row until you are in the column that is labeled 30 at the very top. (It will probably help for you to take a piece of paper and place in under row 13 to help you follow where you re at the table is also alternately shaded to help in this process.) 3. Where row 13 and column 30 intersect is the cell that gives the result. You should see a 6 30 in that cell. MAT107 Chapter 3, Lawrence Morales, 001; Page 8

886 887 888 889 890 891 89 893 894 895 896 897 898 899 900 901 90 903 904 905 906 907 908 909 910 911 91 913 914 915 916 917 918 919 90 91 9 93 94 Thus, in the sexagesimal system, 13 30 = 6,30 Let s check this by converting the result to base 10. 6,30 = 6 60+30 1 = 360+30 = 390 It s not hard to check that 13 30, in base 10, is 390. Hence, all is well. Example 1 Check Point K Example 13 Multiply 4 13 in sexagesimal. Row 4 intersects with column 13 at the cell 5,1. So 4 13 = 5,1 (Check: 4 13=31. But 5,1=5 60+1=300+1=31) Multiply 0 40 in sexagesimal. See the endnotes to check your answer. 5 Multiply 8 3 in sexagesimal. There is a row 8, but there is no column 3. So we think of the problem differently. 8 3 = 8 0 + 8 3 There is both a column 0 and a column 3, so we do it in two steps: 8 0=,40 8 3= 0,4 Total,64 = 3,4 Why did we convert,64 to 3,4? This is because,64 is not a valid base 60 number.,64 =,(60+4) = (+1),4 = 3,4 We carry 60 ones up to the next place and get 3,4. MAT107 Chapter 3, Lawrence Morales, 001; Page 9

95 96 97 98 99 930 931 93 933 934 935 936 937 938 939 940 941 94 943 944 945 946 947 948 949 950 951 95 953 954 955 956 957 958 959 960 961 96 963 964 965 966 967 968 969 970 Check Point L Multiply 0 4 in sexagesimal. See the endnotes to check your answer. 6 Babylonian Multiplication with Tables Let s now look at more complicated multiplication problems in the sexagesimal system. Example 14 Multiply (5,15) (3,5) in sexagesimal. This problem is slightly different than those we just finished. The major difference is that the table we have only allows us to multiply single pairs of numbers at a time. So, we resort to the gelosia grid method that we practiced earlier. We must notice, however, that we are in a base 60 system, so instead of carrying when we have 10, we will carry when we have 60! Also, the entries that we put into our gelosia grids will come from the Base 60 multiplication table. We start by creating the appropriate gelosia grid: 5 15 3 5 The entries into our multiplication blocks come from the following four products: 5 3 = 0,15 15 3 = 0,45 5 5 =,5 15 5 = 6,15 We can now put these into our grid, being certain to place numbers in the correct places. MAT107 Chapter 3, Lawrence Morales, 001; Page 30

971 97 973 974 975 976 977 978 979 980 981 98 983 984 985 986 987 988 989 990 991 99 993 994 995 996 997 998 999 1000 1001 100 1003 1004 1005 1006 1007 1008 1009 1010 1011 101 1013 1014 1015 1016 Example 15 The first diagonal is the ones, the second is 60 s, the third is 60 s, and the last is 60 3 s. By adding (and carrying only if we have groups of 60), we get the following: From what our table tells us, (5,15) (3,5) = 0,17,56,15. We should definitely check this by converting to base 10. 5,15 = 5 60+15 = 315 3,5 = 3 60+5 = 05 17,56,15 = 17 60 + 56 60 + 15 = 64,575 To finish the check we simply check that 315 05=64,575 in base 10, which it is. Multiply (,40,50) (35,18) 40 50 5 15 0 0 15 45 3 6 5 15 5 5 15 0 0 0 15 45 3 6 17 5 15 5 56 15 35 18 MAT107 Chapter 3, Lawrence Morales, 001; Page 31

1017 1018 1019 100 101 10 103 104 105 106 107 108 109 1030 1031 103 1033 1034 1035 1036 1037 1038 1039 1040 1041 104 1043 1044 1045 1046 1047 1048 1049 1050 1051 105 1053 1054 1055 1056 1057 1058 1059 1060 1061 106 Check Point M Entries into the blocks come from the following products in the Multiplication Table: 35 = 1,10 and 18 = 0,36 40 35 = 3,0 and 40 18 = 1,0 50 35 = 9,10 and 50 18 = 15,0 We therefore get the following. Keep in mind that we are carrying ONE up to the next diagonal when we have 60 in a diagonal: In the third diagonal (60 s), we have 36+1+0+9 = 97 = 60+37 =1,37. That s why you see the 37 in the third diagonal s total box. In the next diagonal up, we carry one so that we have 0+10+3+1=34. We can check this pretty easily, again by converting to base 10. 35,18 = (35 60 1 ) + (18 60 0 ) =,118,40,50 = ( 60 ) + (40 60 1 ) + (50 60 0 ) = 9650 118 9650 = 0,438,700 40 50 1 3 9 1 10 0 10 35 0 1 15 34 36 0 018 37 5 0 Finally we check our total: 1,34,37,5,0 = (1 60 4 )+ (34 60)+ (39 60 ) + (5 60 1 )+(0 60 0 ) = 0,438,700 Multiply (5,50) (10,15) in sexagesimal. See the endnotes to check your answer. 7 Use the grid below to do your work. MAT107 Chapter 3, Lawrence Morales, 001; Page 3

1063 1064 1065 1066 1067 1068 1069 1070 1071 107 1073 1074 1075 1076 1077 1078 1079 1080 1081 108 1083 1084 1085 1086 1087 1088 1089 1090 1091 109 1093 1094 1095 1096 1097 1098 1099 1100 1101 110 1103 1104 The Babylonians also had numbers with fractional parts. When multiplying these numbers together, the process will be the same. Example 16 Example 17 Multiply (4;16) (40;5) 16 5 We recall that 4;16 means 4 + while 40;5 means 40 + 60 60 The method we use is essentially the same as what we did when we multiplied numbers in base 10 that had fractional parts. Since we need to move two sexagesimal places, we get 16,10;46,40. You should check this in base 10 to make sure it works. Multiply (30,10;50) (3,0;30) in sexagesimal. Here is the appropriate table. 4; 16 16 10 16 0 40 40 10 6 ; 0 0 40 5 46 40 30 10; 50 1 0 1 30 30 30 3 10 3 16 40 0 0 40 0 15 5 5 ; 51 0 0 030 1 5 0 The result here is 1,40,51,1;5,0 MAT107 Chapter 3, Lawrence Morales, 001; Page 33

1105 1106 1107 1108 1109 1110 1111 111 1113 1114 1115 1116 1117 1118 1119 110 111 11 113 114 115 116 117 118 119 1130 1131 113 1133 1134 1135 1136 1137 1138 1139 1140 1141 Check Point N Multiply (18,40;15) (30,10;35) in sexagesimal. See the endnotes to check your answer. 8 Babylonian Multiplication on Tablets 35 The Babylonian tablets that we know about usually show multiplication by placing the numbers in particular places relative to each other on the tablet. For example, let s look at the (fake) tablet shown here. (This might be a round bun tablet.) The two numbers to the left of the vertical line are 35 and 13. These are the two numbers being multiplied. The scribe or student would use the vertical line (usually) to separate these from the answer, which is on the right of the vertical line. 13 Here, we see that 35 13=7,35, which you can easily confirm by looking up 35 13 on the base 60 multiplication table. The tables we ve been constructing have been primarily for our own use and do not necessarily reflect the way the Babylonians would have done multiplication. Remember, our goal was to learn how to do multiplication in another base. Babylonian Division The Babylonians used the fact that division is the opposite of multiplication. That is, they A 1 recognized that = A. That is, dividing A by B is the same as multiplying A by the B B reciprocal of B. To suit their division needs, the Babylonian had tables of reciprocals so that they could do these calculations quickly and efficiently (even though they probably memorized most of the table entries after a while.) Below you can see a picture of a Babylonian reciprocal table: 9 MAT107 Chapter 3, Lawrence Morales, 001; Page 34

114 1143 1144 1145 1146 1147 1148 1149 1150 1151 115 1153 1154 1155 1156 1157 1158 1159 1160 1161 116 1163 1164 1165 1166 1167 1168 1169 1170 This is probably going to be next to impossible for you to read, so I ve placed some of the more common reciprocals at the bottom of the Base 60 Multiplication Table. Here s an example: If you want to know the reciprocal of 3 (which is 1/3) in sexagesimal notation, you simply go to the column labeled 3 and read the sexagesimal number that is underneath it. From this table you will see that 1/3 = 0;1,5,30. (This particular reciprocal is highlighted in the boxes on the table above. Can you read them off the picture? If not, it s okay that s why I put them on the multiplication table.) Example 18 Divide (,40) 8 in sexagesimal. We only need to convert this to a multiplication to proceed as before. Since the reciprocal of 8 is 0;7,30 (from the table) we can rewrite this as,40 0;7,30. Here s the appropriate table of computations. 40 0 4 0 14 40 ;7 1 0 0 0 0 30 0 0 Our final answer, after adjusting two sexagesimal places to the left, is 0;0 MAT107 Chapter 3, Lawrence Morales, 001; Page 35

119 1171 We can check this by noting that,40 = 60 + 40 = 160. So, in base 10 we 117 are just dividing 160 by 8, which is 0, of course. 1173 1174 1175 Example 19 1176 Divide (13,40;30) 50. 1177 1178 1179 The reciprocal of 50 is 0;1,1, so we want (13,40;30) (0;1,1) 1180 1181 13 40; 30 118 0 0 0 1183 0 13 40 30 0;1 1184 1185 8 6 1186 16 36 0 0 1 1187 1188 1189 1190 4 36 0 1191 Let s check to see if this works: 13,40;30 = (14 60 1 ) + (40 60 0 ) + 30 60 = 80.5. 1193 So we are computing 80.5 50 = 16.41 1194 1195 Now checking our Total: 16;4,36 = 16 + 4/60 + 36/60 = 16.41, a match. MAT107 Chapter 3, Lawrence Morales, 001; Page 36

1196 1197 1198 1199 100 101 10 103 104 105 106 107 108 109 110 111 11 113 114 115 116 117 118 119 10 Check Point O Divide (36,1;17) 7 in sexagesimal. See the endnotes to check your answer. 30 Division on Babylonian Tablets To show division on clay tablet, the Babylonians would have done something like what is shown for (1,5) 15. You can probably see the 1,5 on the top to the left of the vertical line. The 15 (divisor) is on the left side of the second row it s reciprocal 0;4 (from reciprocal table) is immediately next to it. Hence, we have to know (from the context) that these are actually two different numbers. The answer to the right of the vertical line is obtained by multiplying 1,5 0;4 to get 5;40. (Check that this is correct.). Note that there is not a sexagesimal place on the tablet to tell us that 40 is a fractional part of the number. We would have to be able to discern that from the rest of what s on the table. MAT107 Chapter 3, Lawrence Morales, 001; Page 37

11 1 13 14 15 16 17 18 19 130 131 13 133 134 135 136 137 138 139 140 141 14 143 144 145 146 147 Part 4: Babylonian Root Approximations Babylonians and Square Roots The Babylonians were able to estimate the value of square roots to a reasonable degree of accuracy with an iterative process using basic arithmetic. The best way to see how they did this is with an example where we try to explain each of the steps carefully. Example 0 Let s find an estimate for 11 as the Babylonians might have. The first thing to notice is that 11 is between 3 and 4 (since 3 = 9 and 4 = 16). Since 11 is closer to 3 than it is to 4, this is the Babylonian s first estimate for 11. Note: the first estimate will always come from determining between what two whole numbers the desired square root lies and then choosing the whole number that s closest to the radicand (number inside the sign) as the first estimate. We need to keep in mind that the square root of N is found by identifying some number a such that a a = N. In other words, two numbers multiplied together need to give N. In this example, we re looking for two numbers (presumably the same) that multiply to give 11. Since our first guess is 3, we need to find a number x such that 3 x = 11. Solving this equation for a gives 11 us x =. 3 11 11 When we square x we see that = 13.444. This is too big (since 3 9 we want 11 when we square). But 3 = 9 is too small. Hence, our actual 148 square root is somewhere between 3 and 11. Since we don t know exactly 3 where in between these two numbers it lies, we will assume that it is half way 149 150 151 15 153 between them. To find the point that is half way between two numbers, we only need to take the average of the two numbers. In this case, we get: 11 9 11 3 + + 3 = 3 3 0 3 = 0 10 = = 6 3 MAT107 Chapter 3, Lawrence Morales, 001; Page 38

155 154 Therefore, 10 3 is our second estimate for 11. 10 3.333 which, when 3 156 squared, gives 11.11, which is okay if you don t mind that much error. 157 Note: Every estimate after the first is always found by computing the average 158 of the previous estimate and the number that multiplies by the previous 159 estimate to give the number under the radical sign. 160 161 To compute the third estimate, we need to find the number that, when 16 multiplied by 10, gives 11. Thus we need to find x such that: 3 163 164 165 166 167 168 169 170 171 17 173 174 175 10 x = 11 3 3 33 x = 11 = 10 10 33 Note that 10.89, so our actual answer has to between 33 10 10 and 10 3 We now average this with our previous estimate: 10 33 100 99 + + 3 10 = 30 30 199 30 = 199 = 3.3166666 60 This is third estimate for 11. As stated before, it is the average of two 199 numbers. When we square this, we find that 11.000777. This is 60 accuracy to three decimal places. Not bad. The graphic below attempts to map out the process for you visually. The first, second and third estimates are in dotted circles. MAT107 Chapter 3, Lawrence Morales, 001; Page 39

176 177 178 179 180 181 18 183 184 185 186 187 188 189 190 191 19 193 194 195 196 197 198 199 1300 1301 130 1303 1304 1305 1306 1307 1308 Example 1 Estimate 34 3< 11< 4 11 3 = 11 3 11 3 + 3 10 = 3 10 33 = 11 3 10 10 33 + 3 10 199 = 60 First estimate: 34 is between 5 and 6, but is closer to 6 (since 34 is closer to 36 that it is to 5). So our first estimate is 6. Note that this time, the first guess is closer to the higher of the two integers between which 34 lies. Second estimate: First find what number times 6 gives 34. 34 17 6x = 34 x = =. (Always reduce fractions when possible.) Now take 6 3 the average of this with the first estimate. 17 18 17 6 + + 3 = 6 3 35 3 = 35 = 5.8333... 6 Thus 35 6 is our second estimate. Check: 35 34.0777..., only accurate to 6 one decimal place. MAT107 Chapter 3, Lawrence Morales, 001; Page 40

1309 Third estimate: First find what number times 35 6 35 04 gives 34. 1310 1311 131 1313 1314 x = 34 x =. Now we take the average of this with the previous 6 35 estimate. 35 04 15 14 + + 6 35 = 10 10 449 10 = 449 = 5.83095381 40 1315 Thus our third estimate is 449. When we square this we see 40 449 1316 34.00000567. This is accurate to about five decimal places. 40 1317 1318 1319 Check Point P 130 131 Find the first three estimates for 9. 13 133 See the endnotes to check your answer. 31 134 135 Of course, we have only described their method here and we ve used modern notation to do the 136 work. You can imagine how hard it would be to try to all of this in base 60. We ll have to leave 137 that for a more advanced class. 138 139 An Alternate Method of Estimating Roots 1330 1331 Other mathematics historians believe that the Babylonians used yet another method to obtain an 133 estimate for the square root of a given number: 1333 1334 h a + h a +, a 1335 1336 where a is a square number that is close to (but not bigger than) the number whose square root 1337 is being estimated. Essentially, this method requires that you rewrite the number you want to 1338 take the square root of in the form a + h, where a and h are integers. If you can do that, then 1339 you can use the formula to obtain an estimate. MAT107 Chapter 3, Lawrence Morales, 001; Page 41

1340 1341 134 1343 1344 1345 1346 1347 1348 1349 1350 1351 135 1353 1354 1355 1356 1357 1358 1359 1360 1361 136 1363 1364 1365 1366 1367 1368 1369 1370 1371 Example Example 3 Check Point Q Estimate 13. Note that 13 is between the two perfect squares of 9 and 16. With that in mind we pick a to be 3 since 3 is close to 13 but not bigger than 13. (We would not want to pick a to be 4 since that would make a bigger than 13.) With the choice of a = 3, then h would be 4 since a + h = 3 + 4 = 13. With the choices of a and h made, we can use the formula to estimate 13 : 13 = 3 3+ = 3 3 + 4 4 (3) This gives a rough approximation for the square root of 13. Estimate 47 Note that 47 is between the perfect squares 36 and 49, so we choose a to be 6, since 6 = 36 is close to 47 but not greater than 47. With a = 6, then h has to be equal to 11. So we get: 47 = 6 6 + This completes the estimate. = 11 6 1 Use the formula to estimate 76. See endnotes for the answer. 3 + 11 11 (6) MAT107 Chapter 3, Lawrence Morales, 001; Page 4

137 1373 1374 1375 1376 1377 1378 1379 1380 1381 138 1383 1384 1385 1386 1387 1388 1389 1390 1391 139 1393 1394 1395 1396 1397 1398 1399 1400 1401 140 Plimpton 3 One on the most famous Babylonian tablets in existence is called Plimpton 3 (circa 1700 B.C.E.). Among the square root problems that the Babylonians undertook was the question of the relation between the side of a square and its diagonal. This is just a special case of the Pythagorean theorem, which gives the relation between the lengths of the legs of a right triangle and its hypotenuse. a + b = c This is a result that we ve all used numerous times. Even though this theorem is named after Pythagoras, who lived in the sixthcentury B.C.E. (well after the Babylonians), the result of the Pythagorean theorem was known by civilizations well before Pythagoras arrived on the scene, including the Babylonians and Egyptians. c The tablet is made up of four columns of numbers. Upon close examination, it appears that this table is actually a list of what are called Pythagorean Triples. These are sets of three numbers that satisfy the Pythagorean Theorem. For example (3,4,5) is a Pythagorean triple because 3 + 4 = 5. However, the triples it lists are very obscure. For example, one triple it lists is (119,10,169). Others are even more awkward. The a why how s and why s of this table are hard to unearth. Katz 33 provides an interesting explanation of how the columns might have been generated, but the details are probably more than we want to venture into here. b MAT107 Chapter 3, Lawrence Morales, 001; Page 43

1403 1404 1405 1406 1407 1408 1409 1410 1411 141 1413 1414 1415 1416 1417 1418 1419 140 141 14 143 144 145 146 147 148 149 1430 1431 143 1433 1434 1435 Part 5: Babylonian Algebra The Babylonians were able to solve systems of equations and this, in turn, allowed them to solve a variety of quadratic equations. They did not have the quadratic formula, as we do, and they did not have variables, as we do. Instead, it appears that they had proscribed steps that they would follow in order to solve these types of problems. We will explore how they solved systems of equations first and then look at how that allowed them to solve quadratic equations. Systems of Equations We start with a simple example 34 of how the Babylonians would solve a simple system of linear equations. Example 4 Solve the system 1 l + w = 7 4 l+ w= 10 The first thing to note is that the Babylonians did not have variables or equations like those shown above. Instead, their problems were given and solved in words. For example, the problem above would most likely be given in terms of a rectangle. The first equation might have come from a statement such as The length plus one fourth of the width is seven. The second equation might have come from Length plus width is ten. The scribe would then turn around and ask that the dimensions of the length and width be found. To do this, the scribe might have written something like what you see in the left column below. The center column is a more modern algebraic representation of what is happening. Finally, the right column is one that gives commentary on what is happening. Babylonian Modern Commentary 7 4= 8 4l+ w= 8 Multiply the first equation by 4 to clear l + w = 10 the fractions. The 7 4 is computed to get the new right side of the equation. 8 10 = 18 3l = 18 Subtract the two equations from each other and what is left is 3l = 18. 1 l = 6 Multiply both sides of the resulting 18 = 6 (the 3 equation by 1/3. length) 10 6 = 4 (the width) w = 10 6 = 4 Use the value of l to get w If you were to just look at the left column, it might be difficult to see what they are doing. But when you place that information alongside a modern version of the problem, you see that they are essentially doing exactly what we do without the use of variables. This is typical of MAT107 Chapter 3, Lawrence Morales, 001; Page 44

1436 1437 1438 1439 1440 1441 144 1443 1444 1445 1446 1447 1448 1449 1450 1451 145 1453 1454 Babylonian algebra problems. We call them algebra problems because we can take what they do and represent their steps in familiar algebraic notation to see that they are doing more than simple arithmetic. Indeed, they are employing algebraic reasoning to solve simple geometric problems. One common type of algebra problem that appears on tablets is the following: Find two numbers if their sum and product are given. Many of these problems were given in the context of the dimensions and measurements of a rectangle. Example 5 Length plus width is 14. Length times width is 45. What is the length and what is the width? To solve this, we will show the calculations that may have appeared on a tablet. The Babylonians did not provide a formula or explicit process. They often just wrote down their calculations, leaving us to figure out what they were doing, and why they were doing it. Steps Tablet Translation to Words Computations Step 1 14 = 7 Take half the sum of the length and the width, which is 7. Step 7 7= 49 Square 7 to get 49. Step 3 49 45 = 4 Subtract the area to get 4. Step 4 4 = Take the square root to get. Step 5 7+ = 9 The length is half the sum plus the square root. Length is 9. 1455 1456 1457 1458 1459 1460 1461 146 1463 1464 1465 Step 6 7 = 5 The width is half the sum minus the square root. Width is 5. Woah! What is going on here? First of all, notice that the two dimensions found satisfy both conditions. 9 + 5 = 14, and 9 5 = 45. To see what is happening, let s translate what they are doing into modern algebraic notation. In Step1, the scribe takes half the sum of the length and width. This sum is 14, so half of it is 7. That s easy enough. But why does he take half? The answer to that question comes from realizing that if you know that two numbers add up to 14, then the first guess to take for each of the numbers is MAT107 Chapter 3, Lawrence Morales, 001; Page 45

1466 1467 1468 1469 1470 1471 147 1473 1474 1475 1476 1477 1478 1479 1480 1481 148 1483 1484 1485 1486 1487 1488 1489 1490 1491 149 1493 1494 1495 1496 1497 1498 1499 1500 1501 150 1503 1504 1505 1506 1507 exactly half the sum. 7 + 7 is certainly 14. However, it is rare that the length and width will be the same. However, if one number is a more than 7, then the other number must be 7 minus a. That is, we can write our two numbers as: 7 + a and 7 a Let s say that the length is 7 + a and the width is 7 a. Note that if you add them, you get (7 + a) + (7 a) = 14 + a a = 14. So, all we need to do is find the value of a that satisfies the second requirement; namely, the product must be equal to 45. So, we take the product of these two numbers to get: ( a)( a) 7+ 7 = 45 When we multiply the left side out, we get: a 49 = 45 In this process, notice that we had to square 7 (to get 49) that s Step above. Step3 says to subtract the area (from 49) to get 4. Note that if you subtract 45 from both sides of this last equation you get which we will rewrite as 4 a = 0 4 = a. To solve this for a, we would take the square root of both sides, which is what Step4 says to do basically. When we do this, we get two answers: a =±. However, the Babylonians did not recognize negative answers, so the only answer they would have given was. Now that we know what a is, we can find the length and width, since we designated the length to be 7 + a and the width to be 7 a. Step5 says the length is the half sum (7) plus the square root (a), so length is 7 + = 9. Likewise, Step6 tells us to get the width by taking 7 = 5. While this explanation is a bit long, it does show that the Babylonians are clearly using algebraic reasoning to find the answer to their question. MAT107 Chapter 3, Lawrence Morales, 001; Page 46

1508 1509 1510 1511 151 1513 1514 1515 1516 1517 1518 1519 150 151 15 153 154 155 156 157 158 159 1530 1531 Example 6 Use this example as a guide for your homework problems on this topic. Check Point R Solve the following problem using the Babylonian technique. Show an algebraic representation of what is going on. Length plus width is 18. Length times width is 7. Find each. Tablet Translation Into Words Computations 18 = 9 Half the length plus width is 9 9 9 81 Modern Algebra 1 (Length + Width) = 9 ( 9+ a) + ( 9 a) = 9 9+ a 9 a = 7 81 a = 7 81 7 = a 9 = a 9 = a 3 = a Length = 9+ a = 9+ 3= 1 = Square 9 to get 81 ( )( ) 81 7 = 9 Subtract the area to get 9 9 = 3 Take the square root to get 3 9+ 3= 1 (the length) 9 3= 6 (the width) Half the sum plus the root is length. Length is 1. Half the sum minus the root is the width. Width is 6. Width = 9 a = 9 3= 6 Solve the following problem using the Babylonian technique. Show an algebraic representation of what is going on. Length plus width is 30. Length times width is 00. Find each. Check the endnote for the solution. 35 Another common type of algebra problem that appears on tablets is the following: Find two numbers if their difference and product are given. This is very similar to those we ve just examined. We will see that the same process, with only a slight modification, will work fine. MAT107 Chapter 3, Lawrence Morales, 001; Page 47

153 1533 1534 1535 1536 1537 1538 1539 1540 1541 154 1543 1544 1545 1546 1547 1548 1549 1550 1551 155 1553 1554 1555 1556 1557 1558 1559 1560 1561 156 Example 7 The length of a rectangle minus its width is 8. The length times the width is 84. What are the dimensions? We resort back to Babylonian methods. Steps Tablet Computations Translation to Words Step1 8 = 4 Take half the difference of the length and the width, which is 4. Step 4 4= 16 Square 4 to get 16. Step3 16 + 84 = 100 Add the area to get 100. Step4 100 = 10 Take the square root to get 10. Step5 10 + 4 = 14. Length is 14 The length is the square root plus half the difference. Length is 14. Step6 10 4 = 6. Width is 6 The width is square root minus the half the difference. Width is 6. If you compare this to our previous problems and examples, you see an almost identical pattern, with only small modifications. As before, let s see what is happening algebraically. In Step 1, we note that we are once again taking half of the given difference (as opposed to half the given sum.) In this case, half of 8 is 4. We reintroduce the variable a as before and observe that we can let the length be a + 4 and the width be a 4. (The length is larger since when we take length and subtract width, we get a positive number.) We can now check that our condition of l w= 8. ( 4) ( 4) l w= a+ a = a a+ 4+ 4 = 8 With expressions for l and w, we can multiply them to satisfy the second condition. lw = 84 ( a )( a ) + 4 4 = 84 When we multiply this out on the left side, we get: a 16 = 84 MAT107 Chapter 3, Lawrence Morales, 001; Page 48

1563 1564 1565 1566 1567 1568 1569 1570 1571 157 1573 Step above tells us to square 4, which is 16. The multiplication above does this. Step 3 says to add the area to the square. Note that if you add 16 to both sides of the last equation above, you get a = 100. Step 4 says to take the square root, which is 10. Finally, Step5 and Step6 tell us how to find length and width, just as before. When we put these side by side, we see this: Steps Tablet Computations Translation to Modern Algebra Step 1 8 = 4 1 ( ) 4 l w = Step 4 4= 16 ( l w) = ( a+ 4) ( a 4) = a a+ 4+ 4 = 8 lw = 84 a+ 4 a 4 = 84 ( )( ) a 16 = 86 Step 3 16 + 84 = 100 a = 100 Step 4 100 = 10 a = 10 Step 5 10 + 4 = 14. Length is Length = a + 4= 10+ 4= 14 14. Step 6 10 4 = 6. Width is 6. Width = a 4= 10 4= 6 MAT107 Chapter 3, Lawrence Morales, 001; Page 49

1574 1575 1576 1577 1578 1579 1580 1581 158 1583 1584 1585 1586 1587 1588 1589 1590 1591 159 1593 1594 1595 1596 1597 1598 Example 8 Use this example as a guide for your homework problems on this topic. Check Point S Babylonian Quadratics Solve the following problem using the Babylonian Technique. Show an algebraic representation of what is going on. The length of a rectangle minus its width is 10. The length times the width is 96. What are the dimensions? Tablet Computations Translation to Words Modern Algebra 10 = 5 Take half the difference of the 1 length and the width, which is ( ) 5 l w = 5. l w= a+ 5 a 5 = a a+ 5+ 5 = 10 5 5= 5 Square 5 to get 5. ( a )( a ) ( ) ( ) lw = 96 + 5 5 = 96 a = 5 96 5 + 96 = 11 Add the area to get 11. a = 11 11 = 11 Take the square root to get 11. a = 11 11+ 5 = 15. The length is the square root Length is 16. plus half the difference. Length is 16. 11 5 = 6. Width is 6. The width is square root minus the half the difference. Width is 6. Length = a + 5= 11+ 5= 16 Width = a 5= 11 5= 6 Solve the following problem using the Babylonian Technique. Show an algebraic representation of what is going on. The length of a rectangle minus its width is 8. The length times the width is 308. What are the dimensions? See endnote for answer. 36 The last few examples have been instances where a system of equations was to be solved. We have primarily shown what we believe to be the procedures that the Babylonians were likely to use. MAT107 Chapter 3, Lawrence Morales, 001; Page 50

1599 1600 1601 160 1603 1604 1605 1606 1607 1608 1609 1610 1611 161 1613 1614 1615 1616 1617 1618 1619 160 161 16 163 164 165 166 167 168 169 1630 1631 However, there are other ways to solve the last worked out example. In a more modern algebra class we might approach the problem a little differently. We will do so here as a transition into quadratic equations and how the Babylonians solved them. Example 9 The length plus the width of a rectangle is 18. The length times the width is 7. What are the dimensions of the rectangle? We start by letting the length be l and the width be w. We now have a system of equations: l+ w= 18 lw = 7 Take the first equation and solve for l to get l = 18 w. We can now substitute this into the second equation to get the following: lw = 7 (18 ww ) = 7 w w = 18 7 This gives a quadratic equation, which, when set equal to 0, can be solved either by factoring or with the quadratic formula. w 18w+ 7 = 0 ( w )( w ) 1 6 = 0 w= 1 or w= 6 If w = 1, then the length is 18 1 = 6, and if w = 6, then the length is 18 6 = 1. In either case, we get the two distinct dimensions, just as the Babylonians did. The last example gives us hints as to how the Babylonians would solve quadratic equations. Note that in the last example, solving the given system of equations reduced to solving a quadratic equation, so if we reverse directions, we see that solving a quadratic should be equivalent to solving a system of equations. Since the Babylonians definitely knew how to do the latter, they presumably could do the former. And indeed they could. The technique involves taking a given quadratic equation and creating from it a system of equations. MAT107 Chapter 3, Lawrence Morales, 001; Page 51

163 1633 1634 1635 1636 1637 1638 1639 1640 1641 164 1643 1644 1645 1646 1647 1648 1649 1650 1651 165 1653 1654 1655 1656 1657 1658 1659 1660 1661 166 1663 1664 1665 Example 30 The area of a square plus six times the side gives 16. What is the side? We will resort to using modern notation to help us, but the spirit of the solution will be consistent with what the Babylonians might have done. In this problem, if we let the length of the side by x then we can translate it into an equation. If the side of a square is x units long, then the area of the square is x. Six times the side would be translated as 6x. Therefore, the equation we have to solve is: Area + Six times the side give 16 x + 6x= 16 The Babylonian solution amounts to the following in modern notation: x( x+ 6) = 16 We ve simply factored an x out of the left side. If we let y = x+ 6, then we get x ( y) = 16. Also, if y = x + 6, then y x = 16, we find that we have created a system of equations in two variables. The system is: y x= 6 xy = 16 Here we have something very similar to length minus width is 6 and length times width is 16. But this is a typical Babylonian systems-of-equations problem. We can set this up as before. MAT107 Chapter 3, Lawrence Morales, 001; Page 5

1666 1667 1668 1669 1670 1671 167 1673 1674 1675 1676 1677 1678 1679 1680 1681 168 1683 1684 1685 1686 1687 1688 1689 Example 31 Tablet Computations Translation to Words Modern Algebra 6 = 3 Take half the difference of x and y, which is 3. 1 ( ) 3 y x y x= a+ 3 a 3 = a a+ 3+ 3 = 6 3 3= 9 Square 3 to get 9. xy = 16 a+ 3 a 3 = 16 ( )( ) a = ( ) ( ) 9 16 9+ 16= 5 Add the area to get 5. a = 5 5 = 5 Take the square root to get 5. a = 5 5+ 3= 8. y is the square root plus half y = y = 8 the difference. y = 8.. a + 5= 11+ 5= 16 5 3=. The width is square root x = x is. minus the half the difference. a 5= 11 5= 6 x is. We can check this easily. If a square has a side of length, then its area is 4. Six times the side is 1. Adding the area and six times the side gives 4 + 1 = 16, which does satisfy our conditions. Solve x 8x= 9 using the Babylonian technique. We start by factoring out on the left: x( x 8) = 9. We can rewrite this as x y = 8, the difference of two unknowns. This gives us a system of equations: We let the inside of the parentheses by y so that y = ( x 8) x y = 8 xy = 9 We introduce our variable a to help us along and build our familiar table. We can let x= a+ 4 and y = a 4 so that the difference is 8. MAT107 Chapter 3, Lawrence Morales, 001; Page 53

1690 1691 169 1693 1694 1695 1696 1697 1698 1699 1700 1701 170 1703 1704 1705 1706 1707 1708 1709 1710 1711 171 1713 1714 1715 1716 1717 1718 Use this example as a guide for your homework problems on this topic. Check Point T Check Point U Tablet Computations Translation to Words Modern Algebra 8 = 4 Take half the difference of x and y, which is 4. 1 ( ) 4 x y x y = a+ a = 8 4 4= 16 Square 4 to get 16. xy = 9 a+ 4 a 4 = 9 ( )( ) ( 4) ( 4) a 16 = 9 16 + 9 = 5 Add the product to get 5. a = 5 5 = 5 Take the square root to get 5. a = 5 5 + 4 = 9. x is the square root plus half x = x = 9 the difference. x = 9. a + 4= 5+ 4= 9 5 4= 1. y is square root minus the half y = y is 1. the difference. y is 1 a 5= 5 4= 1 Note that we don t really need the last row and the value of y since our goal was to find x. Solve x + 4x= 1 using the Babylonian technique. See Examples above and plug in your result for a check. Solve x 4x= 1 using the Babylonian technique. See Examples above and plug in your result for a check. General Approaches to Quadratics To recap, solving Babylonian quadratics boils down to writing the quadratic as a system of equations. We need to be a little careful when we generalize, however. Note that the quadratic must be in the form x + bx = c to use this method. Not only that, but the value of c must be positive since the Babylonians did not recognize negative numbers. In this form, we can factor out an x on the left. For example, to solve x + 3x 4= 0, the problem would first have to be rewritten as x + 3x= 4 so that it could be rewritten as xx+ ( 3) = 4. MAT107 Chapter 3, Lawrence Morales, 001; Page 54

1719 170 171 17 173 174 175 176 177 178 179 1730 1731 173 1733 1734 1735 1736 1737 1738 1739 1740 If the quadratic took on a different form, such as x + c= bx, the Babylonians might have seen that as a completely different problem and had a different (but perhaps related) technique to tackle that kind of problem. We should immediately recognize that this is very different than how we solve quadratics today. In modern algebra, we can solve any quadratic of the form ax + bx + c = 0 by using the quadratic formula. Thousands of years of evolution, including the recognition and use of negative numbers helps us make the process much more general and compact. The Babylonians, on the other hand, did not have one general method for solving quadratics. They had to improvise given the nature of the problem given to them. Even though they did not have a formula to solve problems like these, that does not stop us from exploring what kind of formula their process would actually give. In general, suppose that the Babylonians had a system of equations such as: x + y = b xy = c If we follow the steps of the Babylonians, we should get a formula that will work for any system of equations that looks like this one. Tablet Translation Into Words Modern Algebra Computations b = b/ Half the length plus 1 b width is b /. ( l+ w ) = b b + a + a = b b b b b b b b = Square to get + a a = c 4 4 b a = c 4 b Subtract the product b c c= a 4 4 b Take the square root b c c = a 4 4 b c = a 4 b b Half the sum plus the b b + c root is x x = + c 4 4 If you look at the final result, you see the value of x is : MAT107 Chapter 3, Lawrence Morales, 001; Page 55

1741 174 1743 1744 1745 1746 1747 1748 1749 1750 1751 175 1753 1754 1755 1756 1757 1758 1759 1760 b b + c 4 You might call this the Babylonian Quadratic Formula for certain cases. (Of course, they did not have this available to them but their methods lead to this equation when you translate their steps into modern algebraic notation.) Think About It How can a more direct link be made between the modern quadratic equation and the Babylonian Quadratic Equation? If you look at this closely, it looks very much like what we have in the quadratic formula: b± b 4ac a With this latest development, we see that the methods of the Babylonians can at least be indirectly related to the modern quadratic formula. The question at this point, of course, is what that link really is. Much more can be said about Babylonian algebra and many more examples could be given, but we ll stop here. Hopefully, it is apparent that the Babylonians did engage in algebraic reasoning. Furthermore, even though their techniques may look different than the algebra that we are used to using and seeing, much of what the Babylonians did can be demystified by translating their work into modern notation and techniques. MAT107 Chapter 3, Lawrence Morales, 001; Page 56

1761 176 1763 1764 1765 1766 1767 1768 Part 6: Homework Problems Conversions Convert the following Babylonian numbers to modern sexagesimal notation (i.e. 45,1;30) and then determine their base-10 value. Please be sure your count the symbols carefully. 1769 1770 1771 177 1773 1774 1775 1776 1777 1778 1779 1780 1781 178 1783 1784 1785 1786 1787 1788 1789 1790 1791 179 1793 1794 1) ) Please note that #3 and 4 have dashed lines indicating the separation of whole and fractional parts. 3) 4) Convert the following decimal numbers to base 60. Write your final results in both modern sexagesimal notation (i.e. 45,1;40) and cuneiform notation. Please show your calculations and/or work. 5) 853 6) 10,000 7) 15 8) 350,000 9) 8.45 10) 53.68 11) 1,453.003 1) A Babylonian Translation Problem 1 5 8 Suppose a Babylonian teacher/student tablet (perhaps similar to the one shown) 37 contains a calculation that is trying to determine the area of a rectangular region. The scribe multiplies the length and the width and writes the following on a tablet: 1795 1796 1797 1798 1799 1800 13) Suppose you know that neither the width nor the length exceed 30 units in length. What can you say about the exact value of the number shown above? Explain or show how you got your exact value. MAT107 Chapter 3, Lawrence Morales, 001; Page 57

1801 180 1803 1804 1805 1806 1807 1808 1809 1810 1811 181 1813 1814 1815 1816 1817 1818 1819 180 181 18 183 184 185 186 187 188 189 1830 1831 183 1833 1834 1835 1836 1837 1838 1839 1840 1841 184 1843 1844 1845 1846 14) Now assume that you flip the tablet over and discover that the scribe has indicated that the length of the rectangular region is the following: Based on your answer to Problem (13), what is the width of the rectangular region? Clearly explain your reasoning! Write your answer in cuneiform. (Hint: You may want to do computations in base 10 then convert to base 60.) Multiplication Use the Babylonian multiplication and reciprocal table (Table 1 on page 69) to do the following multiplication problems. You should use the gelosia grid method that was described in this chapter. (See the blank grids at the end of this chapter you may have to make copies before beginning your work, unless you create your own grids.) When you are done write the original multiplication problem and answer as a scribal student would do so on a tablet. 15) (4,0,15) (3,0) 16) (15,0,10) (8,7,4) 17) (0,33) (13,40) 18) (1;40,0) (15;18) 19) (1,5;35) (40,0) 0) (0,14,18) (7,16,15) Division Use the Babylonian multiplication and reciprocal table (Table 1 on page 69) to do the following division problems. When you are done write the original division problem and answer as a scribal student would do so on a tablet. 1) (11,8,;30) (15) ) (17,5,35) (50) 3) (40,30,0,10) (8) 4) (16,0;30) (48) Tablet Problems 5) The picture to the right represents a problem given to a student on a clay tablet. a. What problem is being described? Justify your answer. b. Do the problem using the table method and express your answer in Babylonian notation. One of the standard units of length in the Babylonian measurement system was the nindan (about 6 meters). One of the standard units of area in the Babylonian measurement system was the sar (about 36 square meters). MAT107 Chapter 3, Lawrence Morales, 001; Page 58

1847 1848 1849 1850 1851 185 1853 1854 1855 1856 1857 1858 1859 1860 1861 186 1863 1864 1865 1866 1867 1868 1869 1870 1871 187 1873 1874 1875 1876 1877 1878 1879 1880 1881 188 1883 1884 1885 1886 1887 1888 1889 1890 1891 6) The tablet to the right shows a rectangle with its sides labeled. Calculate the are area of the rectangle and write the result (in base 60 cuneiform) inside the rectable. 7) Suppose you are scribal student and your scribal teacher tells you to draw a picture of a rectangle that is 1,5 nindan long and 4,7;18 nindan wide. (These are given in base 60) You are to label the sides of the rectangle (in base 60 cuneiform notation of course), compute the area of the rectangle, and write that result in the interior of the rectangle. He has provided the student tablet here to the right in which you can carve your results. Root Approximations Use Babylonian methods to do find the first, second, and third approximations to the following square roots. Show all steps carefully. Leave all your work in terms of fractions.do not convert anything to decimals. 8) 7 9) 5 30) 3 31) 50 3) 90 33) 79 34) 66 35) 107 MAT107 Chapter 3, Lawrence Morales, 001; Page 59

189 1893 1894 1895 1896 1897 1898 1899 1900 1901 190 1903 1904 1905 1906 1907 1908 1909 1910 1911 191 1913 1914 1915 1916 1917 1918 1919 190 191 19 193 194 195 196 197 198 199 1930 1931 193 The Alternate Method of Estimating Roots Estimate the following square roots using the following equation (as described in this chapter): a + h a + 36) 10 37) 105 38) 30 39) 500 40) 85 41) 555 4) 000 43) 59 The Babylonians and Pythagorean Triples As we know, the Babylonians were well aware of Pythagorean triples and the relationship that holds between the legs of a right triangle and its hypotenuse ( a + b = c ). One natural geometric question that seems to have arisen in ancient civilizations is calculating the length of a diagonal of a square if you know the length of one side. For example, given a square with sides of length 3, c =? 3 what is the length of the diagonal of the square? (See picture.) Note that the diagonal of the square corresponds to the hypotenuse of the right triangle in the lower half corner of the square. 3 44) Use the Pythagorean theorem to find the diagonal of square with given side length. Keep your answer in radical notation do not convert to decimals.(for example, if you were to get 8 for an answer, you would want to completely simplify this to using rules of radicals.) a) Length = 3 b) Length = 5 c) Length = 7 d) Length = 10 e) Writing: Look at your answers to the previous four parts. What patterns do you see? Write about them and then use them to describe the length of a diagonal in a square where each side is x units long. f) Now take a square with a side of Length = x. What is the length of the diagonal, in terms x? Simplify completely and compare this to your observation(s) in part (e). They should be reconcilable. h a MAT107 Chapter 3, Lawrence Morales, 001; Page 60

1933 1934 1935 1936 1937 1938 1939 1940 1941 194 1943 1944 A Famous Babylonian Tablet 45) (Make sure you have done Problem (44) before you attempt this problem.) The tablet shown is a Babylonian clay tablet inscribed in about 1600 B.C.E. It is a part of the Yale University. It shows a square drawn on a tablet. To determine what this tablet is about, first note the number that labels the side of the square. This is obviously 30. There are two numbers in the middle of the square. The one on the bottom line is the following (with the sexagesimal line added for you): 1945 1946 1947 1948 1949 1950 1951 a. What is this number in base ten? Round to four decimal places. b. The number that is written along the diagonal of the square is the following (with the sexagesimal line added for you): 195 1953 1954 1955 1956 1957 1958 1959 1960 1961 196 1963 1964 1965 1966 1967 1968 1969 1970 1971 197 1973 1974 1975 1976 What is this number, rounded to 4 decimal places? c. Find the geometric relationship between these three numbers and then clearly explain what the scribe was trying to point out on this tablet. 46) Take your result from the previous problem and draw your own tablet similar to the one pictured above, except have the side of the square be 50 instead of 30. Your drawing should use cuneiform (not Hindu-Arabic) numbers. Below your drawing, clearly show/explain why your drawing is accurate (you may use Hindu-Arabic numbers and modern sexagesimal notation in your explanation). Babylonian Algebra Systems of Equations Solve the following problems using the systems of equations techniques discussed in this chapter. Show an algebraic representation of what is going on. See Example 6 for examples of how to write these problems up. 47) Length plus width is 14. Length times width is 33. Find the dimensions. 48) Length plus width is 34. Length times width is 80. Find the dimensions. 49) Length plus width is 3. Length times width is 31. Find the dimensions. 50) Length plus width is 3. Length times width is 10. Find the dimensions. MAT107 Chapter 3, Lawrence Morales, 001; Page 61