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1 2 3 4 History of Math for the Liberal Arts 5 6 CHAPTER 4 7 8 The Pythagoreans 9 10 11 Lawrence Morales 12 13 14 Seattle Central Community College MAT107 Chapter 4, Lawrence Morales, 2001; Page 1 15 Table of Contents 16 Table of Contents .......................................................................................................................... 2 17 Part 1: Figurative and Friendly Numbers .................................................................................. 4 18 Introduction................................................................................................................................. 4 19 Meet Pythagoras ......................................................................................................................... 5 20 Pythagorean Arithmetica ............................................................................................................ 6 21 Figurative Numbers .................................................................................................................... 7 22 Friendly Numbers ..................................................................................................................... 17 23 Part 2: The Pythagorean Problem............................................................................................. 20 24 Proofs in Other Civilizations .................................................................................................... 22 25 Pythagorean Triples.................................................................................................................. 24 26 The Pythagorean “Formula” for Triples ................................................................................. 24 27 The Triples of Proclus............................................................................................................... 25 28 The Triples of Plato .................................................................................................................. 27 29 The Triples of Euclid................................................................................................................. 29 30 Part 3: Irrational Numbers (Incommensurables) and the Pythagoreans.............................. 31 31 Part 4: Modern Facts on Rational and Irrational Numbers................................................... 33 32 Starting Simple.......................................................................................................................... 33 33 Rational Numbers in More Detail............................................................................................. 34 34 Rational Numbers: Terminating and Repeating Decimals....................................................... 34 35 Non−Terminating, Repeating Decimals.................................................................................... 37 36 What About the Other Direction? ............................................................................................. 38 37 Irrational Numbers ................................................................................................................... 41 38 Proof by Contradiction ............................................................................................................. 42 39 Generating Hoards of Irrational Numbers ............................................................................... 44 40 Polynomial Equations............................................................................................................... 45 41 The Rational Roots Theorem .................................................................................................... 48 42 Using Rational Roots Theorem to Show Irrationality .............................................................. 50 43 Part 5: Pythagorean Geometry, A Brief Discussion ................................................................ 54 44 Conclusion ................................................................................................................................ 54 MAT107 Chapter 4, Lawrence Morales, 2001; Page 2 45 Part 6: Homework Problems ..................................................................................................... 55 46 Triangular Numbers.................................................................................................................. 55 47 Pentagonal Numbers................................................................................................................. 57 48 Oblong Numbers ....................................................................................................................... 57 49 Non-Standard Figurative Numbers........................................................................................... 58 50 Other Proofs of the Pythagorean Theorem............................................................................... 58 51 Pythagorean Triples via the Pythagoreans............................................................................... 64 52 Pythagorean Triples via Proclus .............................................................................................. 65 53 Pythagorean Triples via Plato .................................................................................................. 65 54 Pythagorean Triples via Euclid ................................................................................................ 66 55 Generalizing Pythagorean Facts .............................................................................................. 66 56 Rational and Irrational Numbers.............................................................................................. 67 57 Writing ...................................................................................................................................... 68 58 Part 6: Chapter Endnotes .......................................................................................................... 71 59 MAT107 Chapter 4, Lawrence Morales, 2001; Page 3 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 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 Part 1: Figurative and Friendly Numbers Introduction In the last few centuries of the second millennium B.C.E., we see many changes take place on political and economic fronts. Great civilizations such as Egypt and Babylon, powerful in the past, began to lose their power. New groups such as the Israelites (Hebrews), Assyrians, and Greeks began to emerge. As the Iron Age began, all sorts of new tools emerged that changed the way people lived their lives. Around this time, things like coins and an alphabet were being introduced, and trade became more and more a part of life. Along the coast of Asia Minor, in Greece, and in Italy, major trading towns emerged that took advantage of this new era. Along with this change came a shift in the kinds of questions people are believed to have begun to ask. For what appears to be the first time, people began to explore mathematical questions like “Why is it true that the diameter of a circle precisely cuts the circle into two equal pieces?” The question of Why? became much more prominent, whereas before people were content with asking How? (As in, “How do I compute what is owed to me for these goods?”) At this time, it became more important for people to demonstrate that certain things were true. To this day demonstrative mathematics rules the landscape of the field…modern mathematicians do no fully believe a mathematical statement until it is proven with one hundred percent certainty. (The “beyond−reasonable−doubt” threshold of proof is soundly rejected by almost all modern mathematicians.) The idea of demonstrating that something is true appears to have started with Thales of Miletus (c. 624−547 B.C.E.)1. Some believe that he lived in Egypt for a while2, gaining a knowledge of the work of the Egyptians. In Miletus, he was well known as a mathematician (among other things) and is the first known person in mathematics that is given credit for proving mathematical statements. This does not mean that others before him did prove such statements, only that they did not get credit for their efforts. He is thought to have “proved” the following, for example: a. A circle is bisected (cut into two equal pieces) by its diameter. b. The base angles of an isosceles triangle are equal. (Recall that an isosceles triangle is one in which two of the sides are the same length.) Figure 1 c. The vertical (opposite) angles formed by two intersecting straight lines are equal. In the figure shown, angles a and b are vertical and thus are equal, having the same number of degrees. These and other statements he proved are now considered basic facts of geometry and we accept them as true almost at face value. Thales is recognized for trying to find some logical reasoing that would demonstrate they were true instead of relying on intuiton to establish such facts. MAT107 Chapter 4, Lawrence Morales, 2001; Page 4 a b ∠a =∠ b Figure 2 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 Miletus begins a long line of mathematicians in search of “proof.” The history of the few hundred years of Greek mathematics is difficult to detail because so few primary sources of information are available (unlike the Egyptian and Babylonian records which we have in hand). Much of the information that we have comes from manuscripts and sources that were written well after this time period, even hundreds of years later. This means that what we have is a hypothetical account of Greek mathematics. But, from what we do have, we can try to piece together a little bit of history. One question often asked is how the Greeks were influenced by the mathematics of other civilizations like the Babylonians and Egyptans. In the past, these peoples were not given much credit for influencing Greek mathematics (widely considered an important development in the history of mathematics overall). However, recent evidence suggests that their influence may have been greater than first thought. Primarily, the Greeks themselves appear to express respect for the work of the East in their own writings3. Other internal connections related to arithmetic and astronomy also seem to exist.4 When studying Greek mathematics, one thing we see immediately is that early Greek mathematics is overshadowed by the achievement of Euclid (c 325 B.C.E.−265 B.C.E.). Euclid’s Elements, which we will look at more closely later, is a compliation of all the important mathematics known at the time, and, although not the first of its kind, is really the only one that survives due to its superiority. Finding informatin on other mathematicians’ work is not an easy task. Meet Pythagoras Pythagoras, pictured here5, is one of the best known early Greek mathematicians. He was born around 570 B.C.E. in Samos. This was about 50 years after Thales was born, so some have speculated that he studied with Thales. Others believe it was more likely that he studied under Anaximander.6 We do not know for sure if this is true. From what we know, we think that after studying under some teacher, he went to Egypt and then eventually to Babylon for seven years. (Note the connection between Pythagoras and the mathematics we’ve already studied in this course.) In Babylon, he probably learned Mesopotamian and Egyptian mystical rites, numbers, and music.7 Afterward he went back to Greece and eventually settled in Croton where he established his philosophical/religious society. He was a “mystic, philosopher, prophet, geometer, and sophist.”8 Figure 3 Pythagoras was the leader of a group of followers, now known as the Pythagoreans, which was a secret society based on following his teachings. Because of their secrecy as well as the Near Eastern practice of oral teaching, no firsthand records of their teachings, beliefs, or work exist. MAT107 Chapter 4, Lawrence Morales, 2001; Page 5 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 The Pythagorean Society was an elite, well−known group of people, numbering in the several hundreds. They were vegetarians, shared their possessions, believed in the transmigration and reincarnation of souls, as well as having other common beliefs and practices. This group was very unique due to their belief that one achieves union with the divine through numbers or mathematics. To the Pythagoreans, you could understand all that pertained to the universe by studying numbers. God created the universe by using numbers, said the Pythagoreans, and the universe still depends on them for its continued existence. The phrase “Everything is number” is an accurate description of their worldview. Members of the society would generally take up the study of four subjects. They were: 1. Arithmetica − the study of numbers and number theory (as opposed to basic calculations) 2. Harmonia − the study of music 3. Geometria − the study of geometry 4. Astrologia − the study of astronomy Later, in the Middle Ages, this became know as the “Quadrivium” and was adopted as the standard set of subjects to be studied while receiving a liberal arts education. (How would you like it if three out of four classes required for an A.A. degree were geometry, number theory, and astronomy?) Pythagoras divided his followers into two groups. One studied mystical−religious matters while the other studied scientific matters. For the first three years of their involvement, members were “listeners” who were silent as they heard their teacher speak. After that time, they were allowed into the mathematikoi group, where they would receive more complete teachings as well as be allowed the opportunity to express their own opinions and even elaborate on the teachings of Pythagoras. Pythagorean Arithmetica Prior to the Pythagoreans, it is generally believed that mathematicians in other cultures were primarily interested in mathematics for its applications to such fields as surveying, commerce, architecture, etc. The Pythagoreans, however, were more focused on the mystical aspects of numbers and so seem more conscious of more abstract principles associated to geometric figures and numbers. They believed that the universe could be interpreted by the study of numbers, which existed in a “self−contained realm.”9 As an outgrowth of their studies and their attitudes towards mathematics and numbers, the Pythagoreans helped to create what we might call “pure mathematics.” Question to think about: why would it be called “pure” mathematics? They also helped to establish the idea of “formal proof” as they placed an emphasis on finding results by following a chain of logical reasoning. One belief that they held was that geometry lies beneath physical objects, and that numbers lie beneath geometry. To them, numbers are what we would call the positive integers: {1,2,3,4,5,….}. They took these numbers and linked them to shape with what are called “figurative numbers.” MAT107 Chapter 4, Lawrence Morales, 2001; Page 6 198 199 200 201 202 203 Figurative Numbers Figurative numbers are numbers that can be represented as geometric figures. The easiest figurative number for us to recognize is the “square number.” These are numbers, which can be represented with dots in the shapes of squares. Here are the first few square numbers, which you probably recognize as “perfect squares.” 204 Figure 4 Sn = 1 • 205 206 207 208 209 210 211 212 213 214 215 S2 = 4 S3 = 9 S4 = 16 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • You can visually see why they are called square numbers. The formula for the nth square number is given by: Sn = n2 Another example of figurative numbers, that are not as familiar to most of us, is that of the triangular numbers. The “triangular” numbers are numbers that can be represented as triangles. Each number can be represented with a series of dots. The triangular numbers are 1,3,6,10,15,21,28,… and are pictured below. Figure 5 T1 =1 T2 = 3 T3 = 6 T4 = 10 • • • • 216 217 • • • • • • • • • • • • • • • • The first figure (on the left) has 1 dot, the second figure has 3 dots, the third has 6 dots, etc. MAT107 Chapter 4, Lawrence Morales, 2001; Page 7 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 To distinguish the triangular numbers from each other, we’ll use the following notation to name them: T1 = 1, the first triangular number. T2 = 3, the second triangular number. T3 = 6, the third triangular number. Etc. Note that T3 has three dots on each side of the triangle, so we say its side has length of three. In general, the nth triangular number, which we denote by Tn, has n dots per side on the triangle. If you look closely, you can see how each succeeding triangular number is built by adding a diagonal row of dots to the previous triangular number. (We’ll need this for a proof later.) Figure 6 T1 =1 T2 = 3 T3 = 6 T4 = 10 • • • • 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 • • • • • • • • • • • • • • • • There are some interesting facts that emerge when we study triangular numbers in more detail. For example: FACT1: The sum of two consecutive triangular numbers always equals the square whose “side” is the same as the side of the larger of the two triangles. Figure 7 Huh? Read that again and then consider the following illustration of FACT 1. We are looking at the triangular numbers T4 and T5. These are two consecutive triangular numbers, as FACT1 requires. What is the sum of these two consecutive triangular numbers? Numerically, we have the sum to be: MAT107 Chapter 4, Lawrence Morales, 2001; Page 8 T4=10 T5=15 T4 + T5 = 10 + 15 = 25 250 = 52 251 252 253 254 255 256 257 258 Note that the sum is actually a perfect square, which we can interpret as a square with area 25 and with each side of length 5. FACT1 states this square is the same as the one whose side is the same as the larger of the two triangles. If you look at the larger of the two triangular numbers above, each side has 5 dots, which is what FACT1 is stating. Geometrically, you can see what is happening in the following diagram: Figure 8 Rotate me around + = T4 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 T5 This picture shows that when you add up these two triangular numbers, you do get a square and its side has length 5…this is the length of the side of the larger triangle that corresponds to T5. If you take any two consecutive triangular numbers then, and then geometrically fit them together, you will get a square.♦ FACT2: n(n + 1) . (This is 2 a modern representation…the Pythagoreans certainly did not have such notation.) The nth triangular number, Tn, is given by the formula Tn = Let’s first see if this is true for a triangular number whose value we already know. We know T5 = 15. The formula gives: T5 = 274 5 ( 5 + 1) 2 5× 6 = 2 30 = 2 = 15 MAT107 Chapter 4, Lawrence Morales, 2001; Page 9 275 276 277 278 279 280 281 282 283 284 285 286 The formula works for n = 5, as it should. You can try a few more if you want to check for yourself. Example 1 What is the 50th triangular number? Solution: We certainly don’t want to try to draw 50 triangles, or even just one with 50 dots on each side. The value of the formula becomes apparent as we can now simply compute T50 by using the formula carefully. T50 = 2 50 × 51 = 2 2550 = 2 = 1275 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 50 ( 50 + 1) Hence, the 50th triangular number has 1,275 dots that compose it. ♦ Check Point A Solution: Find the 100th triangular number, T100. See endnotes to check your answer.10 ♦ Think About It Why is T100 more than twice the size of T50 if 100 is only twice the size of 50? We can also ask the opposite question… Example 2 Is 105 a triangular number? Solution: To answer this question, we need to determine if there is a positive integer n that will make the following equation true: n(n + 1) 105 = 2 Think About It Why is it enough to solve this equation? To solve this equation, we will eventually need the quadratic formula. Recall that the quadratic formula states that if ax 2 + bx + c = 0 , then the solutions of MAT107 Chapter 4, Lawrence Morales, 2001; Page 10 313 this equation are x = 314 315 triangular number. − b ± b 2 − 4ac . We now proceed to check if 105 is a 2a Step Comments Starting equation n(n + 1) 2 210 = n(n + 1) 105 = 210 = n + n 0 = n 2 + n − 210 2 n= − 1 ± 12 − 4(1)(−210) 2(1) Carefully simplify the inside of the radical − 1 ± 841 2 − 1 ± 29 n= 2 n = 14 or n = −15 n= 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 Multiply both sides by 2. Distribute the n on the right side Set the equation equal to 0 so that it is in standard form and we can use the quadratic formula. Note that a = 1, b = 1, and c = −210 Apply the quadratic equation. Take the square root Simplify Since n must be a positive integer, we toss out the negative solution and are left to conclude that 105 must be the 14th triangular number. 9 Think About It If we don’t start with a triangular number like we did in this example, what kind of result should we expect from the quadratic formula? (See Check Point C for an example.) Check Point B Is 276 a triangular number? Solution: See endnotes to check your answer.11 MAT107 Chapter 4, Lawrence Morales, 2001; Page 11 336 337 338 339 340 341 342 343 344 345 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 Check Point C Is 333 a triangular number? Solution: See endnotes to check your answer.12 So we’ve used this formula for working with triangular numbers and even checked to see that it is true for n = 5. In mathematics, however, we would not call this a proof. It is simply a verification that the formula holds true for one (or maybe even several) value of n. If we want to prove that it’s true for every value of n, then we’ll need something more than this. (Although there are more formal ways of proof than the one we give here, this one at least provides a more general demonstration than simply testing it for one number.) Proof: We’ll start by recalling each succeeding triangular number is built by adding a diagonal row of dots to the previous triangular number (see above). In particular, to create the nth triangular number, we add n dots to the previous triangular number. As in illustration of this fact, look at the picture below. To get T2 we add 2 dots to T1 To get T3 we add 3 dots to T2. To get T4 we add 4 dots to T3. T1 T2 T3 T4 T5 Figure 9 This pattern continues so that we can state that to get Tn, we add n dots to Tn−1. The notation for Tn−1 simply means the triangular number before the nth one, Tn. If we were to write out a list of the first n triangular numbers, it would look something like this… T1, T2, T3, T4, T4, T5, T6, T7, …., Tn−2, Tn−1, Tn It may seem like odd notation because it is likely unfamiliar to you, but we need it for our proof. With this notation we can now use it to describe what we mean when we say that to create the nth triangular number, we add n dots to the previous triangular number. In this notation we would write Tn = Tn −1 + n MAT107 Chapter 4, Lawrence Morales, 2001; Page 12 382 383 384 385 386 387 388 389 390 The nth triangular number is gotten by taking n dots & adding them to the previous triangular number How do we get Tn−1? The same way as we do all the others: by adding (n−1) dots to the previous triangular number, which is Tn−2 (see list above). This means that Tn-1 = Tn-1 + n. So, we can rewrite our equation as: Tn = Tn −1 + n 391 392 393 394 395 = Tn − 2 + (n − 1) + n We can continue this process until we get all the way down to T1 = 1, the first triangular number: Tn = Tn −1 + n = Tn − 2 + (n − 1) + n 396 = Tn −3 + (n − 2) + (n − 1) + n = Tn − 4 + (n − 3) + (n − 2) + (n − 1) + n = ..... = 1 + 2 + .... + Tn − 4 + (n − 3) + (n − 2) + (n − 1) + n 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 This odd-looking string of symbols tells us how many dots there are in Tn. Now we do something very clever. We take two triangular numbers of the same size (their sides have lengths of n dots) and we add them together. That is, we physically join them. Here’s a picture of what that looks like: Tn = n dots per side + Tn = n dots per side Figure 10 When we put these two together, you can see below that we get a rectangle, where one side has n dots and the other side has n + 1 dots. Figure 11 MAT107 Chapter 4, Lawrence Morales, 2001; Page 13 n dots 419 420 421 422 423 424 425 426 (n +1) dots From the picture, we can see we have n rows and (n + 1) columns so the total number of dots is n(n + 1). But, this is two Tn’s so we can express this fact as follows: 2Tn = n(n + 1) ⇓ n(n + 1) Tn = 2 427 428 429 430 431 432 433 434 435 436 437 438 439 440 There it is! This is the formula we’ve been using. This proof works because it does not depend on particular values of n for it to be true. Because it is done in general terms using an arbitrary number, n, it always holds. ♦ This is the first formal “proof” we’ve seen this quarter and it comes at an appropriate place. We’ve said the Pythagoreans were among the first to develop the idea of formal proof on the basis of logical reasoning and this proof uses that kind of logical reasoning to establish itself. (The Pythagoreans went about their proofs in different ways since they did not have variables and modern notation to help them out, but we can still share the spirit of proof with them.) FACT3: The nth triangular number is the sum of the first positive n integers. For example, T10 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = ? 10(10 + 1) 10 ×11 110 = = = 55 . We know from the previous fact that T10 = 2 2 2 If you add up 1 + 2 + 3 + … + 10, you’ll see you get exactly 55. Proof: This fact basically says that Tn = 1 + 2 + 3 + 4 + .... + (n − 2) + (n − 1) + n . We’ve already proved it! Go back into the previous proof and see where we did it.♦ Proof #2: Here’s a “visual proof” of this fact. 441 442 443 444 445 446 447 448 449 450 451 Divide both sides by 2 MAT107 Chapter 4, Lawrence Morales, 2001; Page 14 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 Figure 12 Think About It Why does this picture “prove” this fact? This often happens. Mathematicians will be working hard at proving some statement that they think is important and in order to get to their desired destination, they either forced to or “accidentally” prove some other interesting fact along the way. It creates an interwoven web of mathematics that connects facts together in beautiful ways. Example 3 Find the sum of the first 100 positive integers. Solution: We want to determine 1+2+3+4+5+….+98+99+100. But, the sum of the first n positive integers is the nth triangular number, so this sum is equal to n(n + 1) Tn = . Hence the first 100 integers add up as follows: 2 100(100 + 1) 2 100 × 101 = 2 = 5050 T100 = 478 479 480 481 482 483 484 485 486 487 488 489 490 491 Where have we seen this before? ♦ There’s a fun story about one of the most famous and talented mathematicians of all time that’s related to this sum. Carl F. Gauss13 (1777−1855) is reported to have been assigned the task of adding up 1 through 100 by his teacher. At the age of seven, Gauss apparently could not be bothered by the tedious task before him so instead devised a scheme close to this: Take the sum 1+2+3+4+….+97+98+99+100 and rearrange them in pairs, as shown below. MAT107 Chapter 4, Lawrence Morales, 2001; Page 15 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 1 2 3 4 … 50 Figure 13 Now add up the pairs: 1 2 3 4 … 50 + 100 = 101 + 99 = 101 + 98 = 101 + 97 = 101 … + 51 = 101 50 pairs of 101 You can see that here are precisely fifty pairs and they each add up to 101. Therefore the sum is 50×101=5050, just as the formula predicted! We can see the connection between his method and the formula for Tn by observing that T100 = 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 + 100 + 99 + 98 + 97 … + 51 FACT4: 100(100 + 1) 100 × 101 100 = = × 101 = 50 × 101 = 5050 2 2 2 The sum of any consecutive odd numbers, starting with 1, is a square number. Here are some examples of what is meant by this statement: 1+3+5+7= 16 = 42 1+3+5+7+9= 25 = 52 1+3+5+7+9+11= 36 = 62 Does this always work? Consider the picture to the right. How does the picture verify this fact? The “proof” is left to the reader. (This is a famous line that you see throughout mathematics textbooks at the higher level. Students often joke that this is inserted whenever the author is too lazy to write out the proof him or herself.) MAT107 Chapter 4, Lawrence Morales, 2001; Page 16 Figure 14 • • • • • • • • • • • • • • • • • • • • • • • • • 532 533 There are even more patterns imbedded in the triangular numbers. For example, what pattern do you see below? 2 13 = 1 = T1 13 + 2 3 = 9 = T2 534 2 13 + 2 3 + 3 3 = 36 = T3 2 13 + 2 3 + 3 3 + 4 3 = 100 = T4 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 2 Can express your pattern in words?14 (The proof is a bit tough, so we’ll omit it here.) All of these facts deal with specific properties of the triangular numbers and are nice examples of what we would call (simple) number theory, the study of numbers. This is basically what the Pythagoreans would call arithmetica in the quadrivium. This, you can see, is much different than simply adding numbers together to keep track of animals, people, or possessions. In this way, the Pythagoreans are different from the Egyptians and Babylonians. The Pythagoreans also studied other figurative numbers such as squares, rectangles, pentagons, and higher numbers, using these shapes to classify numbers. Friendly Numbers Another area of study for the Pythagoreans related to numbers was “friendly” (or amicable) numbers. Two numbers are amicable if each is the sum of the proper divisors of the other. For example, 284 and 220 are friendly numbers. The proper divisors of 220 are 1,2,4,5,10,11,20,22,44,55,110. The sum of these is 284. On the other hand, the proper divisors of 284 are 1,2,4,71,142. The sum of these is 220. Because of their relationship to each other, this pair of numbers achieved a “mystical aura”15 about them and later superstition said that two talismans with these two numbers on them would symbolize a perfect friendship between the two people who wore them. These numbers also played a role in sorcery, astrology, magic, horoscopes, etc. It was not until Pierre de Fermat (in 1646) that another pair of friendly numbers was found. They are 17,296 and 18,416. The divisors of 17,296 are: {1, 2, 4, 8, 16, 23, 46, 47, 92, 94, 1081, 184, 368, 188, 376, 752, 2162, 4324, 8648} The divisors of 18,416 are: {1, 2, 4, 8, 16, 1151, 2302, 4604, 9208} Oh, but wait. It was found that Arab al−Banna16 (1256−1321) discovered this exact pair in 1300, well before Fermat was born. You will not always hear his name associated with this pair of friendly numbers, which is a common occurrence in mathematics history − someone who Challenge Find the proper divisors of 1184 and 1210 and verify that these are friendly numbers…warning, do not undertake this unless you have lots of time to blow. MAT107 Chapter 4, Lawrence Morales, 2001; Page 17 573 574 575 576 577 578 579 580 581 discovered something does not get credit for it while someone else who lived later does. Other great names in the history of mathematics have searched for friendly pairs including Descartes and Euler. But the search for friendly numbers is not confined to the “great minds” of mathematics. In 1866, 16−year old Nicolo Paganini found that 1184 and 1210 are friendly numbers. Fermat17 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 Descartes18 Euler19 Figure 15 Figure 16 Euler himself gave 30 pairs of them in 1747 and later extended his list to 59 pairs. He was a calculating monster! Just for fun, here’s a pair of friendly numbers that was discovered recently (2000) by Pedersen20: Friendly Number1: 101912069888090336515444350802308297159793341076299379130248147409037917819405 12685965874751421209161744634973401968229045017461807880895272226747149 Friendly Number2: 101923245069831450648512550193810277344799075309653872181279759957437998660440 91895019397550855288689851283216607038140496702852401487947300696132851 Yes, those are both individual numbers, so long that they overflow onto two lines of the page. MAT107 Chapter 4, Lawrence Morales, 2001; Page 18 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 Any suggestions as to how they were found? As of May, 2000, there have been at least 657,000 pairs of friendly numbers found. Here is a list of the top 10 discovery leaders. As you can see, there has been some busy mathematics at work. Of course, there is no way they can check these large numbers by hand. Instead, number theorists look for patterns and rules (“theorems”) that allow them to search for such numbers with some level of efficiency. In this day and age, the computer doesn’t hurt either! Discoverer # of pairs Te Riele/Pedersen 399076 Pedersen 168608 Borho&Battiato 37785 Einstein 16935 Wiethaus 10401 Te Riele 7285 Einstein&Moews 4247 Borho&Hoffmann 3471 Moews&Moews 2614 Ball 1938 Garcia 1247 Keep in mind that Pythagoras (and/or his followers) found ONE pair of friendly numbers that we know of. But that contribution helped to get the question out there for people to think about as well as helped to establish number theory as its own branch of mathematics. The Pythagoreans also appeared to have studied perfect, deficient, and abundant numbers. These are topics that we’ll leave for the reader to explore alone if interest is generated. MAT107 Chapter 4, Lawrence Morales, 2001; Page 19 619 620 621 622 623 624 625 641 642 643 644 645 646 647 648 Part 2: The Pythagorean Problem The most famous result that Pythagoras is famous for is, of course, the Pythagorean Theorem. We’ve all seen it before. In a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse (longest side). In modern notation, we write: a2 + b2 = c2 626 627 Almost everyone gives credit to Pythagoras for Figure 17 c 628 first discovering this fact. By “discovering” we 629 mean he (or his followers?) was the first to prove 630 that this relationship holds in all right triangles. 631 We already know that the Babylonians at least 632 knew of the relationship (Plimpton 322) even c 633 though they may not have proved it. The 634 Egyptians were also aware of the relationship a 635 and used ropes to measure out proper lengths of b 636 right triangles. a 637 b 638 639 The actual Think About It 640 proof that they gave is unknown to us. It is likely that it was very geometrical What does the figure and involved taking a geometric object, “dissecting” it into shown have to do with the pieces, and then rearranging the pieces into a new figure that Pythagorean Theorem? demonstrated the Pythagorean relationship. Here’s one such proof for you to consider. For many of you, this may very well be the very first time you’ve ever seen this famous theorem proven. Proof: We consider the two following pictures c a a a b a b b c 649 650 651 c b c a b a c b b b b c a b Figure 18 MAT107 Chapter 4, Lawrence Morales, 2001; Page 20 a 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 Both squares are to be considered to be the same size. The length of a side is (a+b). The values for a, b and c in each picture are to be considered consistent in their lengths across both pictures. The square on the left has an area of (a + b) 2 since the area of any square is its side squared. But the figure on the left is also made up of six pieces…two small squares and four triangles. If we add up their areas we should get the area of the whole square. The area of the squares are a 2 and b 2 . There are also four triangles to consider. Thus the area of the triangle on the life, AL, is: 1 AL = a 2 + b 2 + 4 a × b 2 1 The 4 a × b comes from the fact that there are 4 equally−sized right 2 triangles in the left picture and the area of any right triangle is one half its base times its height. Now if we take another square of the same size and “dissect” it differently as shown in the picture to the right, we still get the same area, namely (a + b) 2 . However, there are only 5 pieces in this dissection (one square with area c 2 and four triangles), so we’ll add them up to get the area on the right, AR: 1 AR = 4 a × b + c 2 2 However the two areas are the same, since AL = AR = (a + b) 2 . They were, after all, both squares of the same size so we equate the two and then simplify from there. AL = AR 1 1 a 2 + b2 + 4 a × b = 4 a × b + c2 2 2 ⇓ a 2 + b2 = c 2 This finishes the proof as it established the relationship that we all know and love.♦ There are literally thousands of proofs of the Pythagorean Theorem, many of them published in books devoted just to this subject. Even a former president of the United States is credited with MAT107 Chapter 4, Lawrence Morales, 2001; Page 21 686 687 688 689 690 691 692 693 694 695 696 697 698 699 700 701 702 703 704 presenting a unique proof of this theorem. See the endnotes for some websites with some interesting proofs to examine.21 Proofs in Other Civilizations Many civilizations had proofs of the Pythagorean Theorem, including the Chinese. The picture22 is the Chinese hsuanthu, which is believed to show one of the oldest demonstrations of the Pythagorean Theorem in history. It is dated by some people to as far back as 1100 B.C.E. Another related proof comes from Bhaskara (1114-1185), a Hindu mathematician. His proof is demonstrated interactively on the internet.23 His proof, like many of his time and culture, consisted solely of a picture with the word “Behold” written nearby24. Figure 19 below is the basic picture that goes with Bhaskara’s proof. Figure 19 Figure 20 705 706 707 708 709 710 711 712 713 714 715 716 717 Other proofs of the Pythagorean Theorem revolve around a common theme: dissect a square into pieces that somehow demonstrate the Pythagorean Theorem. (This is why they are often called “dissection proofs.”) In most cases, the figure is dissected into pieces to show that it is equivalent to having three new squares, one with sides of length a, one with sides of length b, and one with sides of lengths c. Furthermore, these three squares must be dissected and/or arranged to show that the area of the squares with sides of lengths a and b, when added together, are exactly equal to the area of the square with sides of length c. In the figure below, the square with side of length a and the square with side of length b would each be cut up into several pieces. Those pieces would then be arranged so that they fit exactly into the square with sides of length c. The hard part, of course, is determining how to cut up the first two squares so that all the pieces exactly fit into the third! MAT107 Chapter 4, Lawrence Morales, 2001; Page 22 718 719 720 721 Figure 21 Area of a square with sides Of length a Area of a square with sides Of length b c a2 + b2 = c2 c a Area of a square with sides Of length c a 722 b b MAT107 Chapter 4, Lawrence Morales, 2001; Page 23 723 724 725 726 727 728 729 730 731 732 733 734 735 Pythagorean Triples Closely related to this theorem is the problem of finding integers, a, b, and c that satisfy the Pythagorean Theorem. We’ve already seen that the Babylonians had a tablet (Plimpton 322) that listed several of these. We called them Pythagorean Triples. The set of three numbers (3,4,5) is commonly known as such a Pythagorean triple. The Pythagorean “Formula” for Triples The Pythagoreans are given credit for finding a formula for generating these trios of numbers. In our modern notation it would look like the following: 2 736 737 738 739 740 741 This formula is good for any odd value of m ≥ 3 and generates a series of triples. This formula essentially assigns the following values to a, b, and c: a=m m 2 −1 b= 2 2 m +1 c= 2 742 743 744 745 746 747 2 m2 + 1 m2 −1 = + m 2 2 2 Example 4 Find the Pythagorean triple that corresponds to m = 3. Solution: 2 2 32 + 1 32 − 1 2 = +3 2 2 2 748 749 750 751 2 9 +1 9 −1 2 = +3 2 2 2 2 2 5 = 4 +3 Well, look at that! m = 3 produces the triple (3,4,5) ♦ MAT107 Chapter 4, Lawrence Morales, 2001; Page 24 752 753 754 755 756 757 758 759 760 761 762 763 764 765 Check Point D Find the Pythagorean triple that corresponds to m = 5. Solution: See the endnotes to check you answer.25 One interesting thing to note is that if (3,4,5) is a triple, so is (6,8,10) and (9,12,15), etc. In fact, any “multiple” of the triple (3,4,5) is also a triple. We can represent this generic triple as (3k,4k,5k), where k is a positive integer. Proof: (3k,4k,5k) is a Pythagorean Triple. We need to show that these satisfy a2 + b2 = c2 a 2 + b 2 = (3k ) 2 + ( 4k ) 2 = 9k 2 + 16k 2 = 25k 2 = 5k × 5 k 766 = (5k ) 2 = c 2 ⇓ (3k ) 2 + (4k ) 2 = (5k ) 2 767 768 769 770 771 772 773 774 775 776 777 This shows that (3k,4k,5k), which is any multiple of (3,4,5), is also a Pythagorean Triple. ♦ The Triples of Proclus The writer Proclus (411-485) claims that Pythagoras devised an algorithm for obtaining triples.26 In our own notation, it would amount to finding a, b, and c with the following equations: a = 2n + 1 b = 2n 2 + 2n c = 2n 2 + 2 n + 1 778 779 780 781 782 where n is an integer and n ≥ 1 . (n does not have to be odd for these equations to work, as we show below.) MAT107 Chapter 4, Lawrence Morales, 2001; Page 25 783 784 785 786 787 Example 5 Find the Pythagorean Triple for n = 2. Solution: We simply use the formulas as given: a = 2(2) + 1 = 5 b = 2(2) 2 + 2(2) = 2(4) + 4 = 12 788 789 790 791 792 793 794 795 796 797 798 799 800 801 802 803 804 805 806 c = 2(2) 2 + 2(2) + 1 = 8 + 4 + 1 = 13 This gives the triple (5,12,13) which can easily be checked.♦ Check Point E Find the Pythagorean Triple for n = 10 Solution: See the endnotes to check your answer.27 To convince ourselves that these three expressions for a, b, and c do indeed give us Pythagorean Triples, we propose the following proof: Proof: a +b 2 We want to show that the Proclus equations for a, b, and c satisfy the Pythagorean Theorem. To verify this, we need to show that they satisfy the equation a 2 + b 2 = c 2 . We will substitute the corresponding formulas for a, b, and c into this equation to see what we get. Read carefully and follow along as we go back and review lots of polynomial algebra. We’ll do both sides of the equation separately so can compare what’s happening. Step 1 2 (2n + 1)2 + (2n 2 + 2n )2 (2n + 1)(2n + 1) + (2n 2 + 2n)(2n 2 + 2n) 4 n 2 + 2 n + 2n + 1 + 4n 4 + 4n 3 + 4n 3 + 4 n 2 4n 4 + 8n 3 + 8n 2 + 4n + 1 4n 4 + 8n3 + 8n 2 + 4n + 1 807 808 809 810 811 812 813 814 815 816 c2 2 (2n 3 4 5 = (2n 2 + 2n + 1)(2n 2 + 2n + 1) You fill in what goes here when multiplying 4n 4 + 8n 3 + 8n 2 + 4n + 1 4n 4 + 8n3 + 8n 2 + 4n + 1 2 ) + 2n + 1 2 Step 1: Starting point. Step 2: Substitute Proclus’ equations into a, b, and c. Step 3: What does it mean to square something? Answer: To multiply it by itself. Step 4: Multiply the polynomials by each other…very, very carefully. Step 5: Combine like terms. Think About It Step 6: Note that we have the same thing on each side, so we can conclude that If these equations did not satisfy the the quantities in Step 1 were also Pythagorean Theorem, what would equal, concluding this proof. ♦ you get in Step 5? MAT107 Chapter 4, Lawrence Morales, 2001; Page 26 817 818 819 820 821 822 These two methods of generating Pythagorean triples (from the Pythagoreans and as described by Proclus) appear to be equivalent since they seem to generate the same list of triples. Here’s a table that illustrates this fact: Table 1 Triples From Original Equation m a b c 1 − − − 2 − − − 3 3 4 5 4 − − − 5 5 12 13 6 − − − 7 7 24 25 8 − − − 9 9 40 41 10 − − − 11 11 60 61 12 − − − 13 13 84 85 823 824 825 826 827 828 829 830 831 832 833 834 835 836 837 838 Triples from Proclus’ Equations n a b c 1 3 4 5 2 5 12 13 3 7 24 25 4 9 40 41 5 11 60 61 6 13 84 85 7 15 112 113 8 17 144 145 9 19 180 181 10 21 220 221 11 23 264 265 12 25 312 313 13 27 364 365 Think About It What makes these two methods essentially equivalent? Unfortunately, neither of these methods generates all possible Pythagorean triples. The method of Plato28 will demonstrate that clearly by finding a triple that neither of these can generate. The Triples of Plato Plato’s method says that you can generate a triple (a,b,c) with the following equations: a = 2p b = p 2 −1 839 c = p2 +1 840 841 842 843 Example 6 What Pythagorean Triple is generated by p = 4? MAT107 Chapter 4, Lawrence Morales, 2001; Page 27 Figure 22 844 845 Solution: Using the equation we get: a = 2(4) = 8 b = 4 2 − 1 = 15 846 c = 4 2 + 1 = 17 847 848 849 850 851 852 853 854 855 856 857 858 859 860 861 862 863 864 865 866 867 868 869 870 871 872 873 874 875 876 877 878 879 880 881 882 883 The triple generated is (8,15,17). This is not on either of the two lists before which shows there are more triples out there than the Pythagoreans (apparently) knew about. ♦ Check Point F What Pythagorean Triple is generated by p = 10? Solution: Table 2 See the table to check your answer. Here is a table that lists the first few triples generated with Plato’s equations. You can see that it has very little in common with the other lists. Most striking is that all of the values for a in Plato’s table are even numbers, while all the values for a in the Pythagorean tables are odd. Think About It Why does one table given even values for a while the other gives odd values of a? Proof: Triples From Plato's Equations p a b c 1 − − − 2 4 3 5 3 6 8 10 4 8 15 17 5 10 24 26 6 12 35 37 7 14 48 50 8 16 63 65 9 18 80 82 10 20 99 101 11 22 120 122 12 24 143 145 13 26 168 170 The equations of Plato satisfy the Pythagorean Theorem. To prove this, we proceed as before. We need to substitute Plato’s equations for a,b, and c and then simplify what we get to make sure that we get equal expressions on both sides of a equal sign. MAT107 Chapter 4, Lawrence Morales, 2001; Page 28 a +b 2 Step 1 2 (2 p )2 + ( p 2 − 1)2 2 (p (2 p)(2 p) + ( p 2 − 1)( p 2 − 1) 3 ( p 2 + 1)( p 2 + 1) 4 p 2 + p 4 − p 2 − p 2 +1 4 What do you get here? p + 2 p +1 5 p4 + 2 p2 +1 p4 + 2 p2 + 1 = p4 + 2 p2 + 1 ( 4 884 885 886 887 888 889 890 891 892 893 894 895 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 911 c2 ) 2 2 ) +1 2 Step 1: Starting point. Step 2: Substitute Plato’s equations into a, b, and c. Step 3: What does it mean to square something? Answer: To multiply it by itself. Step 4: Multiply the polynomials by each other…very, very carefully Step 5: Combine like terms. Step 6: Note that we have the same thing on each side, so we can conclude that the quantities in Step 1 were also equal, concluding this proof.♦ You would think that between these two methods (we’ll count the two Pythagoras methods as one since they produce equivalent lists) would generate all of them. After all, we get both evens and odds for value of a. Unfortunately, they don’t do the job. One of the first (perhaps the very first) mathematician to give a method to generate all Pythagorean triples was Euclid. The Triples of Euclid Euclid is one of the great names of Greek mathematics. His work, The Elements, is one of the classics of mathematics and is the basis of modern high−school geometry (yes, you have him, in part, to thank for your high−school geometry class). Euclid’s method is slightly different from the others we’ve seen so far. For this method, we let g and h be positive integers with g > h. Euclid shows that a Pythagorean triple (a,b,c) is generated with the following equations: a = 2 gh b = g 2 − h2 c = g 2 + h2 912 MAT107 Chapter 4, Lawrence Morales, 2001; Page 29 Figure 23 913 914 915 916 917 Example 7 Generate a Pythagorean Triple with g = 4 and h = 3. Solution: Using Euclid’s equations we get the following: a = 2(4)(3) = 24 b = 4 2 − 32 = 7 918 c = 4 2 + 32 = 25 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 This generates the triple (7,24,25), which does not appear to be in any of the tables we have seen so far. ♦ Check Point G Generate a Pythagorean Triple with g = 10 and h = 7. Solution: See the endnotes for the answer.29 Unfortunately, trying to prove that this method generates all Pythagorean Triples is a bit tough in a course like this, so we won’t be able to explore it here. But we can show that the equations Euclid gives satisfy the Pythagorean Theorem, a 2 + b 2 = c 2 . That proof is left for the reader in the exercises. MAT107 Chapter 4, Lawrence Morales, 2001; Page 30 937 938 939 940 941 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 960 961 962 963 964 965 966 967 968 969 970 971 972 Part 3: Irrational Numbers (Incommensurables) and the Pythagoreans a , where a and b b are integers and b is not 0. They are essentially the “well−behaved” fractions: no decimals, roots, or other weirdness…just nice integers. (While studying the Pythagoreans, we will restrict ourselves to the positive integers since they did not recognize negative numbers.) An irrational number is a number that is not rational. That is, it cannot be written as the ratio of two integers. Irrational numbers are also called incommensurables. As you may know, a rational number is a number that can be written in the form Some believe that the discovery that incommensurable numbers exist is the most important discovery of the Pythagoreans.30 It was important not only to mathematics but also to the Pythagoreans and their beliefs. They assumed that a ratio was limited to containing only integers, which is simply an extension of their belief that “everything is [whole] number.” If there exist numbers that are not whole numbers, or ratios of whole numbers, then their fundamental philosophy, on some level, was being challenged. It is not unreasonable to speculate that such a discovery would have been disturbing to them. To understand why the Pythagoreans would have made such an assumption, it is helpful to explore what they meant when they talked about rational numbers. We need to keep in mind that the Pythagoreans did not have modern symbols for numbers. To them, a number represented the length of a line segment. It was therefore associated with a geometric object. Given any two line segments (numbers), they believed that it is always possible to find a third line segment that would evenly divide into both of the original line segments. For example, let’s say the following figures represent enlarged line segments of different lengths: The Pythagoreans believed that if you searched long and hard enough, you could find a smaller line segment that divided each of these two line segments up. Here’s what that might look like: You can see here that this smaller (darker) line segment will evenly divide the large segment into 12 pieces while it will divide the shorter segment into 8 pieces. The Pythagoreans believed that it was only a matter of time before they could find such a “dividing segment,” even if it had to be very, very small. For example, the following two line segments have a very small common measure: MAT107 Chapter 4, Lawrence Morales, 2001; Page 31 973 974 975 976 977 978 979 980 981 982 983 984 985 986 987 988 989 990 991 992 993 994 995 996 997 998 999 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 When we look for a common measure, we find it to be much smaller, but still available… With this common measure, two numbers are said to be “commensurable” (rational). Ratios that cannot be expressed by whole numbers with a common measure like this (which they did not originally believe existed) are now said to be “incommensurable.” The word “incommensurable” is alogos in Greek (which means “without ratio”) or arrhetos (inexpressible). The Greek word logos (speech or word) represented the ratio of two numbers. The discovery of alogos, usually translated as “irrational,” was a challenge to their theory of proportions and may have led them to reconsider it completely.31 (By they way, key terms in our mathematics vocabulary come from very geometric interpretations. Why do we read x2 as “x squared?” It is because a square with a length of x has an area of x2. Thus, the measure for area gets associated with a square and the term sticks. The same goes for x3. The volume of a cube with length x is x3 so we say “x cubed.” This is how the Greeks viewed numbers…as line segments. Their “squares” of numbers were thought of as geometrical square objects. Finally, a number multiplied by itself three times was viewed as a geometric object – a cube. Because of their views and interpretations of numbers, the Greeks had a tremendous impact not only on how we do mathematics but also on how 1 we talk about mathematics. Their influence lasts until today.) It is uncertain how the Pythagoreans discovered the incommensurables. Most believe it was due to their investigation of a square with unit length. (Unit length means the side has a length of 1.) 1 c They might have asked the question, “What is the length of the diagonal of this square?” A quick application of the Pythagorean Theorem shows Figure 24 that c = 2 . (You should do this on scratch paper to convince yourself that it is true.) However, the Pythagoreans would have been interested in finding two commensurable lengths (whole numbers) that could be placed into a ratio that would represent the exact value of c. (Remember that the Babylonians had a method of finding a fractional estimate of a square root. The third estimate of 2 using this method is 17 / 12 ≈ 1.416666... ., while the actual value of 2 is closer to 1.4142…) As they tried to find this pair of commensurable numbers, they could not do so and eventually proved that the pair did not exist at all. A proof of the incommensurability of 2 is given later in the chapter. MAT107 Chapter 4, Lawrence Morales, 2001; Page 32 1010 1011 1012 1013 1014 1015 1016 1017 1018 1019 1020 1021 1022 1023 1024 1025 1026 1027 1028 1029 1030 1031 1032 1033 1034 1035 1036 1037 1038 1039 1040 1041 1042 1043 1044 1045 1046 1047 1048 1049 1050 1051 1052 1053 1054 Part 4: Modern Facts on Rational and Irrational Numbers The discovery of irrational numbers by the Pythagoreans is certainly one of their most important contributions in the history of mathematics. These numbers have since been studied and dissected by thousands (if not millions) of mathematicians through the ages as they have strived to understand the basic structure of number systems. In this section, we will look at some of the interesting facts about rational and irrational numbers, concentrating mainly on more modern approaches to these numbers. So while we are not necessarily doing “historical” mathematics, we can appreciate the fact that groups such as the Pythagoreans and their work have survived and been extended for thousands of years. Starting Simple The positive whole numbers are numbers that belong to the set: {1,2,3,4,….}. These are used for basic counting and are sometimes called the natural numbers. If we take any two natural numbers and add them together, we get another natural number. 3 + 4 = 7 , and 7 is a natural number. Likewise, multiply two whole numbers together and we get another whole number. These two properties are actually pretty important in number theory and we say that the natural numbers are closed under addition and the natural numbers are closed under multiplication. This simply means that when you take any two natural numbers and either add them or multiply them, the result will be a member of the natural numbers. To illustrate this property of closure from another point of view, consider the set {1,2,3}, having only three numbers in it. If you take any two of these numbers and add them together, do your get another number in that set? Well, 1 + 2 = 3 , and 3 is certainly in the set. However, 1 + 3 = 4 , but 4 is not in the set. So, this set is not closed under addition. The natural numbers are certainly not closed under all basic operations in arithmetic. For example, they are not closed under subtraction. An example of this is 1 − 8 = −7 . While 1 and 8 are certainly in the natural number set, when you subtract them you get a negative number, which is not a member of the natural number set, so we don’t have closure under subtraction. In fact, in order to get closure, we have to add all of the negative numbers as well as 0 to cover our backs. So in one sense, we could say that a quest to obtain closure under subtraction leads to the “creation” of the set of integers. The last basic mathematical operation, division, also provides problems in the context of the natural numbers. Take any two natural numbers and divide them and rarely will you get another natural number. For example, while 6 ÷ 3 = 2 does the job, 3 ÷ 4 certainly does not. Therefore, the natural numbers are not closed under division. In fact, if you want closure under division, one way to do it is to expand the natural numbers to include fractions, or what we shall call the a rational numbers. These are numbers that can be written in the form , where a and b are b integers and b is not zero. We see here another case where a desire to extend the property of closure leads to the “creation” of a new set of numbers, the rational numbers. MAT107 Chapter 4, Lawrence Morales, 2001; Page 33 1055 1056 1057 1058 1059 1060 1061 1062 1063 1064 1065 1066 1067 1068 1069 1070 1071 1072 1073 1074 1075 1076 1077 1078 1079 1080 1081 1082 1083 1084 1085 1086 1087 1088 1089 1090 Rational Numbers in More Detail The definition that we gave above for rational numbers is one that is deserving of some extra commentary. First of all, notice that the definition states that a rational number is one “that can be put in the a a form . It isn’t quite enough to simply say that rational numbers are those “of the form ” b b 3 can be because there are infinitely many ways to express a given fraction. (For example, 4 6 3 2 written as or or in infinitely many other ways.) The reason that this is not enough is 8 4 2 that we don’t want our definition of rational numbers to depend on the particular way that we decide to express that number. 32 Second, observe that we require that b is non−zero. This requirement is in place to avoid division 0 by 0, which has no mathematical meaning. (While is generally 0, except when n is 0, the n n value of is not defined in our basic mathematical system.) Think About It 0 Rational Numbers: Terminating and Repeating Decimals One of the more interesting characteristics of rational numbers is that when you divide two integers and obtain a decimal representation for the number you get one of two results: n undefined in 0 our basic mathematical system? Why is 1) The decimal part either terminates. 2) The decimal part does not terminate but repeats at some point. 3 which is exactly equal to 0.75. The decimal part of this 4 terminates at the 5, once and for all. To see an example of the second kind, one only look at the 1 familiar fraction . The decimal value of this numbers is 0.33333333…., where the 3’s never 3 stop or terminate but repeat indefinitely. We sometimes write this number as 0.3 , where the bar over the 3 indicates what part of the number repeats. An example of the first kind is If you’re at all curious like I am, you wonder which rational numbers have decimal representations that terminate and which ones have decimal representations that repeat instead of terminate. MAT107 Chapter 4, Lawrence Morales, 2001; Page 34 1091 1092 1093 1094 1095 1096 6275 . Any terminating 10000 decimal can be written as a fraction with a denominator that is a power of 10, simply because we 327 221151 and 0.221151 = . are operating in a base−10 system. For example, 0.327 = 1000 1000000 To consider those that terminate, let’s take the example of 0.6275 = In the case of 0.6275, we can see that the denominator, 10000 can be rewritten as follows: 6275 6275 = 10000 10 × 10× 10 × 10 6275 = 2×5× 2×5× 2×5× 2×5 1097 1098 1099 1100 1101 1102 1103 1104 What is noteworthy here is that the denominator factors into several 2’s and 5’s, both prime numbers. Any terminating decimal, since it can be written as a fraction with a power of 10 as a denominator, has the same property. Namely, the prime factors of the denominator are 2 and 5 only. Even if we write the fraction in lowest terms, we still see the same property emerge. For example: 6275 269 = 10000 400 269 = 4 × 100 269 = 2 × 2 × 100 1105 1106 1107 1108 1109 1110 1111 1112 1113 1114 1115 1116 1117 1118 1119 1120 1121 1122 Once again, we notice that the denominator of the reduced fraction has only 2 and 5 as its prime factors. We are now prepared to state which rational numbers have terminating decimals: FACT 1 A rational fraction a/b (in lowest terms) has a terminating decimal representation if and only if the integer b has no prime factors except for 2 and 5. (Please note that in order for this fact to hold true, the fraction must be in lowest terms. Otherwise, all bets are off.) 1 1 has a terminating decimal representation, does not as its denominator has neither 2 3 a 2 nor 5 as a factor of its denominator. So, while MAT107 Chapter 4, Lawrence Morales, 2001; Page 35 1123 Example 8 1124 1125 1126 1127 1128 1129 Use Fact 1 above to determine if 37 has a terminating decimal. 160 Solution: To settle this issue, we really only need to find the prime factorization of 160. Any method you’ve learned in the past will do. Here’s one possible way to find such a factorization. 160 = 8 × 20 = 23 × 4 × 5 1130 = 23 × 2 2 × 5 = 25 × 5 1131 1132 1133 1134 1135 Since the only prime factors in the denominator are 2 and 5, then we know that this fraction does have a decimal representation that terminates. ♦ Example 9 1136 1137 1138 1139 Use Fact 1 above to determine if Solution: To settle this issue, we really only need to find the prime factorization of 600. 600 = 6 × 100 1140 1141 1142 1143 1144 1145 1146 = 2 × 3 × 100 At this point we can stop since we see that 3 is part of the prime factorization. Since 2 and 5 are not the only primes present, we can conclude that this fraction does not have a decimal representation that terminates. Check Point H Use FACT 1 above to determine if 1147 1148 1149 1150 1151 1152 1153 1154 1155 1156 455 has a terminating decimal. 600 Solution: 3022 has a terminating decimal. 3125 Check the endnote for the answer.33 As humans faced questions like how to divide a loaf of bread among 5 people, the natural idea of fractions arose. MAT107 Chapter 4, Lawrence Morales, 2001; Page 36 1157 1158 1159 1160 1161 1162 1163 1164 1165 1166 1167 1168 1169 1170 1171 1172 1173 1174 1175 Non−Terminating, Repeating Decimals We now know what kinds of rational numbers have decimal representations that terminate. That takes care of one of the stated possibilities for rational numbers. Recall that one of the following is true: 1) The decimal part either terminates. 2) The decimal part does not terminate but repeats at some point. Certainly, if the decimal part of a fraction does not terminate then the first half of the second possibility holds. But what about the second, more noteworthy part? Namely, that it repeats? Why should we expect that it would repeat, rather than simply keep going in some chaotic, unpredictable manner? In this section, we’ll try to look at why this is true. The following fractions all have decimal representations that not only do not terminate, but also repeat at some point: 1 = 0.16 6 1 = 0.3 3 3 = 0.27 11 733 = 0.814 900 To see why the decimal portions repeat, we will look at a particular example. Consider 3 ÷ 7 . To find the decimal representation of this number, we can divide as normal: 0.4285714 7 3.000000 The 3 in the first step reappears and so the pattern will repeat after this point. 1176 − 28 20 −14 60 − 56 40 − 35 50 − 49 10 − 07 30 − 28 MAT107 Chapter 4, Lawrence Morales, 2001; Page 37 1177 1178 1179 1180 1181 1182 1183 1184 1185 1186 1187 1188 1189 1190 1191 1192 1193 1194 1195 1196 1197 1198 1199 1200 1201 1202 1203 1204 1205 1206 1207 1208 1209 1210 1211 1212 1213 1214 1215 1216 1217 1218 While things seem to go along pleasantly for a while, we notice that several steps into the division, the 3 reappears and “resets the clock,” so to speak. After this point, you will see the pattern repeat itself all over again. It need not be that the first step is the one that reappears and hence causes things to repeat. It could take place anywhere. Try to divide 209/700 and you’ll see it’s a 6 that repeats at some point. In our extended example above, since we are dividing by 7, the possible remainders that could arise in any particular step are 1, 2, 3, 4, 5, and 6. Note that the remainders in this case are 2,6,4,5,1, and 3 and we’ve cycled through all the possible remainders before we get back to 3 and notice that we have not managed to get the decimal to terminate. We are then plunged back into the process anew. We can approach the general case of a/b in a similar way. That is, when b is divided into a, the only possible remainders that we could get are 1, 2, 3, 4,…,b-2, and b-1. So, “a recurrence of the division process is certain. When the division process recurs a cycle is started and the result is a periodic decimal.”34 a can be expressed as a b terminating decimal or as a non-terminating, repeating decimal. That is, if we have a rational number, then we know that it has to behave in one of these two ways. We have, at this point, “demonstrated” that any rational number What About the Other Direction? We still have one more question to consider. That is, if we have a decimal number that terminates or repeats, do we necessarily know that we have a rational number? This is very different that saying what we already have; namely, that if you have a rational number then it must terminate or repeat. We are considering the opposite direction of this whole mess and mathematics treats them as very separate questions. (The force behind this distinction is basic logic, the tool which mathematics uses to establish and prove facts and theorems in this field. We will ignore the subtleties of this matter here, but it should be pointed out that there are delicate issues that need to be paid attention to in cases such as these.) The first case, when it terminates, is easy to dispose of. For example, 0.321 terminates and we know that we can read this as “three hundred twenty one thousandths.” This translation into 321 words, and our existence in the base-ten system, allows us to write this number as . 1000 The second case is a little more interesting. For example, if you have the number 0.345 , do you a know for sure that you could write this number in the form of , where a and b are integers and b b is non-zero? To answer this question, we consider a formal example. MAT107 Chapter 4, Lawrence Morales, 2001; Page 38 1219 1220 1221 1222 1223 1224 1225 1226 1227 1228 1229 1230 1231 Example 10 If possible, write 0.345 as a rational number. Solution: To do this, we start by letting x = 0.345 . That is, x = 0.34545454545... Now consider that 10 x = 3.4545454545... (Check it if you are not convinced.) Also, consider that 1000 x = 345.45454545... (Again, you should check this.) Now we subtract these two results from each other: 1000 x − 10 x = 345.45454545.... − 3.4545454545... 1232 990 x = 342 1233 1234 1235 1236 1237 This is the key step because it completely eliminates the repeating part of the number, leaving us with only whole numbers with which to work. This equation can be solved for x by dividing both sides by 990 and we get: 342 990 1238 x= 1239 1240 1241 However, we said at the beginning that x = 0.345 . So, we have shown that: 1242 1243 1244 1245 1246 1247 1248 1249 1250 1251 1252 1253 1254 1255 1256 1257 1258 1259 1260 0.345 = 342 990 This means that 0.345 is indeed rational. ♦ The central idea in this last example is that if we are given a repeating decimal, we should be able to multiply the number by two different powers of 10, subtract the results to eliminate the repeating decimal, and thens solve for the actual fraction that is being sought. The hardest part is to figure out what powers of 10 to multiply by, how to describe the general case, and how it should be handled. Let’s try another example first. Example 11 Write 0.21378378… as a rational fraction. Solution: We start by letting x = 0.21378378... Our first power of 10 that we will multiply by is designed to isolate the repeating part of the decimal to the right of the decimal point. In this case, the 21 is not part of the repeating part and so those two digits need to “move” to MAT107 Chapter 4, Lawrence Morales, 2001; Page 39 1261 1262 1263 1264 1265 1266 1267 1268 1269 1270 1271 1272 1273 1274 1275 1276 1277 the left of the decimal point. Or, another way to state it is that the decimal point must move two places to the right. We do this by multiplying by 100, so we consider: 100 x = 21.378378378... The other power of 10 that we will multiply by is designed to allow us to subtract the two powers so that the result will eliminate the repeated part. To do this, consider: 100000 x = 21378.378378... Notice that this has three more zeros than our previous multiple (100) precisely because we need to move the decimal point three more places to the right since three digits repeat. It might be more obvious why this works to see them subtracted: 100000 x − 100 x = 21378.378378... − 21.378378... = 21357 99900 x = 21357 1278 1279 If we divide, we find that x = 1280 1281 1282 1283 1284 1285 1286 1287 1288 1289 1290 1291 1292 1293 1294 1295 1296 1297 1298 1299 1300 1301 1302 21357 99900 A quick check will show that this is, indeed, equal to 0.21378378….♦ This last example gives us a strategy for how to pick our powers of ten that we will ultimately use to get our desired result. The first power of ten should isolate the repeating part of the decimal on the right side of the decimal point. The second power of ten should move the decimal point an additional number of spaces that correspond to how many digits are repeating in the given number. We’ll do one more example before moving on: Example 12 Write 0.387219821982198… as a rational fraction. Solution: We start by letting x = 0.387219821982198... Note that four digits repeat here. We isolate 219821982198 on the right by letting: 1000 x = 387.219821982198... To move the decimal an additional four places to the right we consider: MAT107 Chapter 4, Lawrence Morales, 2001; Page 40 1303 1304 1305 1306 1307 10000000 x = 3872198.21982198.... Subtracting these two equations gives: 10000000 x − 1000 x = 3872198.2198... − 387.2198... 9999000 x = 3871811 3871811 x= 9999000 A check on a calculator shows this to be accurate. ♦ 1308 1309 1310 1311 1312 1313 1314 1315 1316 1317 1318 1319 1320 1321 1322 1323 1324 1325 1326 1327 1328 1329 1330 1331 1332 1333 1334 1335 1336 1337 1338 1339 1340 1341 1342 1343 Check Point I Write 0.46283462834…as a rational number. Solution: Check the endnote for answer.35 So, this concludes this portion of our examination of rational numbers. There are actually many holes that we’ve left unfilled, but you could take a whole course on the topic, so we have to stop somewhere. Right? Irrational Numbers At this point, we might ask the question: “If the decimal does not terminate or does not repeat, then what are you left with?” In this case, we have what are called irrational numbers. That is, if a real number is not rational, then it is irrational. Irrational numbers thus have the property that they never terminate and they never repeat. This leaves us with the reality that irrational numbers have decimal expansions that go on forever and never establish any pattern. While that may seem a bit sad or scary, it’s actually pretty convenient because it allows us to work with and recognize numbers such as 2 and π, both of which are irrational. Furthermore, if you lay all the rational numbers (which include all the integers) out on the real number line, you will find that even though there are infinitely many of them, they do not “fill up” the number line. The irrational numbers fill up all those holes so that we have complete “density” on the familiar number line. Another way to say this is that the real number line (which is just a picture of the set of real numbers) is completely made up of both rational and irrational numbers. Because of the definition of irrational numbers and its dependence on what rational numbers are, irrational numbers are a little harder to work with. Thus, if you have a number and you want to show that it’s irrational, then you essentially have to show that it never terminates and it never repeats. This is much more subtle and difficult than showing you can produce a terminating or repeating decimal expansion. In order to see just how tricky this can get, let’s take the classic MAT107 Chapter 4, Lawrence Morales, 2001; Page 41 1344 1345 1346 1347 1348 1349 1350 1351 1352 1353 1354 1355 1356 1357 1358 1359 1360 1361 1362 1363 1364 1365 1366 1367 1368 1369 1370 1371 1372 1373 1374 1375 1376 1377 1378 1379 1380 1381 1382 1383 1384 1385 example of 2 and show that it is irrational. That is, let’s show that it is NOT rational (i.e. does not terminate or repeat). Proof by Contradiction To show that 2 is not rational, we will employ a method of proof that is very common in mathematics. It is called proof by contradiction, and its strategy is exposed in its name. The idea behind such a proof is to assume the opposite of what you want to actually prove. With that assumption, you then explore what such a statement would imply logically, moving along step by step. Each step must be a logical deduction of the previous one(s). You continue to do this looking for a statement or result that is either blatantly/obviously false or which contradicts something you definitely know to be true or have assumed to be true (for good reasons) in your problem. When you get such a false, contradictory statement, then the only conclusion you can reasonably reach is that one of your steps or your original opposite assumption is false. But since all the intermediate steps follow strict rules of logic, only your opposite assumption can be the culprit and is therefore false. This method is not natural or intuitive to follow. Budding mathematicians generally need lots of time and practice before the art of writing proofs with this method becomes a comfortable tool with which to work. But we are going to use the method here to prove that 2 is not rational (i.e. irrational) so you can see how the process works. First however, here are some facts to think about and recognize before we get started: •Any even number can be written as 2k , where k is an integer. •Any odd number can be written as 2k + 1 , again where k is an integer. •The product of two even numbers is even. (Can you prove it?) •The product of two odd numbers is odd. (Can your prove it?) •The product of an even number and an odd number is even. (What about this one!) Proof: Show that 2 is irrational. We start by assuming that the opposite is true. That is, we will assume that a 2 is rational and can therefore can be written in the form , where this b fraction is in lowest terms , a and b are integers, and b is not zero. The assumption that the fraction is in lowest terms is key to the argument. If it’s not in lowest terms, then reduce the fraction so that it is. a . b This is simply a result of assuming that Logical Step 1: 2= 2 is rational. MAT107 Chapter 4, Lawrence Morales, 2001; Page 42 1386 1387 1388 1389 1390 1391 1392 1393 1394 1395 1396 1397 1398 1399 1400 1401 1402 1403 1404 1405 1406 1407 1408 1409 1410 1411 1412 a2 Logical Step 2: 2 = 2 b This logically follows from squaring both sides of the previous step. Logical Step 3: 2b 2 = a 2 This follows from multiplying both sides of the previous step by b 2 . Logical Step 4: a 2 must be an even number. This statement follows from the previous step because any number that can be written in the form 2k , where k is an integer, is an even number. Since b 2 is an integer, then 2b 2 is of the form 2k , and so 2b 2 is even. Therefore, a 2 , being equal to 2b 2 , is also even. Logical Step 5: a must be an even number. This follows from the only way that two (non-zero) numbers can multiply by each other to get an even number is if both of them are even. (Odd times odd is odd. But even times even is even.) So, if a 2 is even then, since it’s just the product of a times a, then a must be even. Think About It Prove that even times even equals even and odd times odd equals odd. What about odd times even? Logical Step 6: a can be written as a = 2k This comes from the basic definition of what it means to be even. Logical Step 7: 2k 2 = b 2 This statement comes from substituting our previous result, a = 2k , into the result from Logical Step 3. a 2 = 2b 2 ( 2k ) 2 = 2b 2 4k 2 = 2b 2 2k 2 = b 2 1413 1414 1415 1416 1417 1418 1419 1420 1421 1422 Logical Step 8: b 2 is an even number. Since b 2 = 2k 2 , it is written in the form 2c , so it is even. Logical Step 9: b is an even number. Since b 2 is even, b is even. See the logic of Logical Step 5. At this point, notice that we have concluded that both a (Step 5) and b (Step 9) are even. MAT107 Chapter 4, Lawrence Morales, 2001; Page 43 a can be reduced. An even number divided by an even b number, since they both have a factor of 2 in them, can be reduced. 1423 This means that 1424 1425 1426 1427 BUT… a was b already in lowest terms and therefore reduced. This result contradicts that, so something is wrong. Since each of the logical steps that we walked through are perfectly legitimate, the ONLY thing that could be “wrong” is our assumption that 2 is rational. If it’s not rational, then it’s irrational. This concludes the proof.♦ Way back in at the beginning of this process we said that the fraction 1428 1429 1430 1431 1432 1433 1434 1435 1436 1437 1438 1439 1440 1441 1442 1443 1444 1445 1446 1447 1448 1449 1450 1451 1452 1453 1454 1455 1456 1457 1458 1459 1460 1461 This result, as we’ve seen, was a major discovery by the Pythagoreans, and created a major problem in their philosophical outlook of the world. No one knows exactly how the Pythagoreans proved this fact, and there are many possibilities, but their discovery (and other interests) played a crucial role in establishing the field of number theory. Other numbers can be shown to be irrational with similar arguments that depend on this method of proof, but we will not explore them in great detail here. But it is important to recognize that there are an infinite number of irrational numbers and some of them are very familiar to us. For example, 3 is irrational. The same goes for 6 . Unfortunately, these proofs by contradiction can get increasingly complex so we want to be able to find and identify irrational numbers with other methods. (If you’re interested in seeing a proof that 3 is irrational, you can read a proof in the endnotes.36 ) Generating Hoards of Irrational Numbers FACT 2 If a is any irrational number and r is any rational number except for zero, then all of the following are irrational: a r 1 a + r , a − r , r − a, ar , , , − a, r a a This latest fact (which we will not prove here) says that if you combine an irrational number and a rational number (except maybe for zero) together with addition, subtraction, multiplication, or division, you will end up with an irrational number. This allows us to generate a whole bunch of irrational numbers without even trying. For example, since we know that 2 is irrational, then we automatically know that all of the following are irrational: 2 +3 , 2 −5, 2+ 1 2 , 4 3 MAT107 Chapter 4, Lawrence Morales, 2001; Page 44 1462 1463 1464 1465 1466 1467 1468 1469 1470 1471 1472 1473 1474 1475 1476 1477 1478 1479 1480 1481 1482 1483 1484 1485 1486 1487 1488 1489 1490 1491 1492 1493 1494 1495 1496 1497 1498 1499 1500 1501 1502 1503 1504 Unfortunately, this does not tell us anything about other numbers like 2 + 5 since these two numbers do not satisfy the condition that one be rational and the other irrational. They’re both irrational so Fact 2 will not say anything about 2 + 5 . So, we see that there are many irrational numbers and many ways to generate even more. An endless supply exists for anyone in need of such numbers. We still have to consider the question of how to identify a number as being irrational without having to do a lengthy proof. To do this for certain kinds of numbers, we turn to a brief study of polynomial equations. Polynomial Equations By using polynomial equations we can develop a common method for establishing the irrationality of a large class of numbers. To do this, we focus not on the numbers themselves but on relatively simple algebraic equations that have these numbers as solutions. Example 13 Find an equation that has 3 as a solution. Solution To do this, we proceed as follows: Let x = 3 Square both sides to get: x2 = 3 Subtract 3 from both sides so that the equation is equal to 0: x2 − 3 = 0 Thus, x 2 − 3 = 0 , has as a solution x = 3 if we travel backwards in the logic of this problem. ♦ Similarly, other numbers that look like they could be irrational satisfy certain equations: 3 5 5 7 9 93 satisfies satisfies satisfies satisfies x2 −5 = 0 x2 −7 = 0 x3 − 9 = 0 x 5 − 93 = 0 MAT107 Chapter 4, Lawrence Morales, 2001; Page 45 1505 1506 1507 1508 1509 1510 1511 1512 1513 1514 1515 1516 1517 1518 1519 1520 1521 1522 1523 1524 1525 1526 1527 1528 1529 1530 1531 1532 If you’re unsure of any of these, you’re encouraged to solve the equations to verify that the stated solutions are indeed valid. All of these equations are called polynomials and are the focus of this section. We start by looking at one of the simplest of the polynomials, the quadratic polynomial. This is an expression of the form ax 2 + bx + c , where a, b, and c are called the coefficients. The quadratic should be familiar to you by now as you have used the quadratic formula to solve them many times before and have even developed methods for graphing the parabolic curves that are associated with these equations. A cubic polynomial, or a polynomial of degree three, looks like ax 3 + bx 2 + cx + d . Cubic polynomials may not be as familiar to you as they are generally not seen much until you study precalculus. We can extend this idea by defining a polynomial of degree n (where n is a positive integer) by stating that it has the form: c n x n + c n − a x n −1 + ... + c1 x + x 0 , with c n not zero. The numbers c n , c n −1 ,..., c 2 , c1 , c 0 are called the coefficients of the polynomial. Example 14 Identify the degree, n, and the coefficients, c n , of the polynomial: 4 x 5 + 2 x 4 − 5x 3 + 7 x − 3 = 0 Solution The degree of the polynomial is n = 5 since 5 is the highest power of x that is present. The coefficients are as follows: c5 = 4 c4 = 2 c3 = −5 1533 c2 = 0 c1 = 7 c 0 = −3 1534 1535 1536 1537 1538 1539 1540 Note that c 2 = 0 because there is no x 2 term, and so we can pretend that the term 0x 2 is hiding between the − 5x 3 and the + 7 x terms. ♦ Check Point J Identify the degree, n, and the coefficients, c n , of the polynomial: 8x 6 − 5x 4 + 6 x 2 − 8x + 3 MAT107 Chapter 4, Lawrence Morales, 2001; Page 46 1541 1542 1543 1544 1545 1546 1547 1548 1549 1550 1551 1552 Solution Eventually, we will use polynomial equations to determine whether or not certain numbers are irrational or not, but before we do that, we first need to make sure we know how to check if a given number is a solution to a polynomial. This is relatively easy, since all it entails is substituting the number into the polynomial and checking to see if we get a true statement. The only hard part is making sure that we don't make any errors in the process. An example is probably helpful. Example 15 Is 1553 1554 1555 Check the endnotes for the answer.37 2 a solution of the polynomial 10 x 3 + 6 x 2 + x − 2 = 0 ?38 5 Solution 2 everywhere we see an x. 5 To check this, we substitute 1556 1557 3 2 2 2 2 10 x 3 + 6 x 2 + x − 2 = 10 + 6 + − 2 5 5 5 8 4 2 = 10 + 6 + − 2 125 25 5 80 24 2 = + + −2 125 25 5 80 120 50 250 = + + − 125 125 125 125 250 250 = − 125 125 =0 1558 1559 1560 Since we get a true statement, we can say that 1561 1562 1563 polynomial.♦ 1564 1565 1566 1567 2 is a solution to the 5 In the previous example, if we had ended with a false statement, then our conclusion would have 2 been that was not a solution to the given equation. 5 Check Point K Is 1 a solution to the polynomial x 4 − 3x 2 − 5 x + 9 = 0 ? 2 1568 MAT107 Chapter 4, Lawrence Morales, 2001; Page 47 1569 1570 1571 1572 1573 Solution Check Point L Is 1574 1575 1576 1577 1578 1579 1580 1581 1582 1583 1584 1585 1586 1587 1588 1589 1590 1591 1592 1593 Solution 1607 1608 1609 1610 1 a solution to the polynomial 15 x 3 − 23x 2 + 9 x − 1 = 0 ? 3 See endnote for answer.40 We are now ready to link polynomials to the question of whether or not a number is rational or irrational. The Rational Roots Theorem A root of a polynomial equation is the same thing as a solution of the polynomial set equal to 0. The two words are used interchangeably. The following theorem/fact gives us what we need to make the connection between polynomials and rational or irrational numbers. Theorem Given any polynomial with integer coefficients, such as: c n x n + c n − a x n −1 + ... + c 2 x 2 + c1 x 1 + c 0 a , where this fraction is in b lowest terms, then a is a divisor of c 0 and b is a divisor of c n . 1594 1595 1596 1597 1598 1599 1600 1601 1602 1603 1604 1605 1606 See endnote for answer.39 If this equation does have a rational root, say This theorem basically tells us what the candidates are for possible rational solutions to a polynomial equation. If we can identify what they are, then we can use that information to determine whether or not a number is irrational or not. Let's first see how this theorem works. Example 16 What are the possible rational roots of the following equation?41 2 x 3 − 9 x 2 + 10 x − 3 = 0 Solution a . By the Rational b Roots Theorem, a must be a divisor of −3. The divisors of −3 are +1, −1, +3, and −3. Also, b must be a divisor of +2. The divisors of +2 are +1, −1, +2, and −2. Here is where it gets interesting. Any rational root of this equation is If there is a rational root to this equation, we will call it MAT107 Chapter 4, Lawrence Morales, 2001; Page 48 a , so we now have to combine all of these possibilities b with each other to get all possible rational roots. Taking a = +1 with all possible values for b gives: 1 1 1 1 , , , 1 −1 2 − 2 1611 going to look like 1612 1613 1614 1615 1616 Taking a = −1 with all possible values for b gives: −1 −1 −1 −1 , , , 1 −1 2 − 2 1617 1618 1619 Taking a = 3 and a = −3 with all possible combinations for b gives: 3 3 3 3 −3 −3 −3 −3 , , , , , , , 1 −1 2 − 2 1 −1 2 − 2 1620 1621 1622 1623 1624 A careful examination of this list shows that many of them are repeats, so if we list only the unique possibilities from these lists, we have the following as all possible rational roots of the given polynomial: 1 1 3 3 − 1, + 1, − , + , − 3, + 3, − , + 2 2 2 2 1625 1626 1627 1628 If we want to know which of these work, we have to painstakingly substitute each of them in to determine if each one works or not. I'll let the reader verify 1 that 1, , and 3 all work. (If you have and know how to use a graphing 2 calculator, this process is a little bit easier.) 1629 1630 1631 1632 1633 1634 1635 1636 1637 1638 1639 1640 1641 1642 1643 1644 1645 1646 FACT 3 A polynomial of degree n has at most n roots, whether they are rational or not. This fact is useful because it gives you some idea of when to stop looking for rational roots. In Example 16 above, the polynomial was degree 3 so that means it has at most 3 roots. Once we 1 had determined that 1, , and 3 worked, we could have stopped because we know that there 2 could be no more. By this Fact, a degree-4 polynomial has at most 4 roots. It is possible that it could have fewer than that, but it will certainly not have any more than 4. Example 17 What are the rational roots of the equation: 3x 3 − 10 x 2 − 9 x + 4 MAT107 Chapter 4, Lawrence Morales, 2001; Page 49 1647 1648 1649 Solution By the Rational Roots Theorem, we know that a must divide 4, so a could be ± 1 , ± 2 , or ± 4 . Also, b must divide 3, so b could be ± 1 or ± 3 . Hence the a combinations for are as follows: b 1 1 2 2 4 4 ± , ± ,± , ± , ± , ± 1 3 1 3 1 3 1650 1651 1652 1653 1654 1655 Generally, it is easy to check if integers are solutions first, so we by checking 1, −1, 2, −2, 4, and −4, we find that −1 and 4 are solutions. (The details are left out here.) This means there is at most one more solution to this polynomial 1 and good old-fashioned brute force will reveal that is the remaining root. ♦ 3 1656 1657 1658 1659 1660 1661 1662 1663 1664 1665 1666 1667 1668 1669 1670 1671 1672 1673 1674 1675 1676 1677 1678 1679 1680 1681 1682 1683 1684 1685 1686 Check Point M Find the rational roots of the polynomial 2 x 2 + 9 x − 5 = 0 by using the Rational Roots Theorem. Solution See the endnote for the answer.42 Using Rational Roots Theorem to Show Irrationality We now move to the punch line for all of this work. We want to be able to determine if a given number is irrational or not by using the Rational Roots Theorem. We saw earlier that 2 is irrational, but we used a long, intricate proof by contradiction to do it. We now have the tools to give a much more compact proof. Example 18 Use the Rational Roots Theorem to show that 2 is irrational. Solution We first find a polynomial that has bill:43 2 as a solution. The following fits the x2 − 2 = 0 a , for this b equation will have to meet the requirements that a is a divisor of -2, so a could be 1, -1, 2, or -2; and b is a divisor of 1 and is thus either -1 or 1. So, the list of possible roots for this equation is as follows: From the Rational Roots Theorem, we know that any rational root, MAT107 Chapter 4, Lawrence Morales, 2001; Page 50 1687 1688 1689 1690 1691 1692 1693 1694 1695 1696 1697 1698 1699 1700 1701 1702 1703 1704 1705 1706 1707 1708 1709 1710 1711 1712 1713 1714 1715 1716 1717 1718 1719 1720 1721 1722 1723 1724 1725 1726 1727 1728 1729 1, −1, 2, −2 There are no other possible candidates for solutions that are rational. Since 2 is not on this list, then 2 could not be rational and is therefore irrational! ♦ Note that it is not necessary to do any substitution here. We are only interested in generating a list of all of the possible rational roots for this polynomial and then comparing that to the number we are examining. Example 19 8 is irrational. Use the Rational Roots Theorem to show that Solution The polynomial that has 8 as a solution is x 2 − 8 = 0 . By the Rational Roots Theorem, the list of possible rational roots for this equation is: +1, −1, +2, −2, +4, −4, +8, -8 8 is not rational and is therefore irrational. ♦ 8 is not on this list, Since Example 20 Use the Rational Roots Theorem to show that 3 6 is irrational. Solution We start by letting x = 3 6 . Cubing both sides gives: x3 = 6 Setting this equal to 0 gives: x3 − 6 = 0 This polynomial (by Rational Roots Theorem) has the following list of possible rational roots: +1, −1, +2, −2, +3, −3, +6, −6 Since 3 6 is not on this list, 3 6 is not rational and is therefore irrational. ♦ An alternative proof by contradiction for this last example would be much more lengthy and much more difficult to follow. So, even though the Rational Roots Theorem takes a little time to understand and master, once we do that we have a very handy way of proving that a given number is irrational. As we mentioned earlier, this method will not work for all irrational numbers. In fact, it's only suited for numbers n x . MAT107 Chapter 4, Lawrence Morales, 2001; Page 51 Think About It Why can't the Rational Roots Theorem be used to prove that π is irrational? 1730 1731 1732 1733 1734 1735 1736 1737 1738 1739 1740 1741 1742 1743 1744 1745 1746 1747 1748 1749 1750 1751 1752 1753 1754 1755 1756 1757 1758 1759 1760 1761 1762 1763 1764 1765 However, you will not be able to show that π is irrational using the Rational Roots Theorem. (The proof that π is irrational was developed relatively late in the history of mathematics and requires mathematics that goes well beyond what this chapter or course provides.) Check Point N Check Point O Use the Rational Roots Theorem to show that 4 Example 21 What is 2 added to − 2 , rational or irrational? Solution We know that each of these two numbers are irrational. But what about their sum? It's easy to see that they add up to 0, which is rational. Therefore, it's possible for the sum of two irrational numbers to be rational. Example 22 What is 2 + 3 , rational or irrational?46 Solution Certainly, this one is not as straightforward. To determine this number's personality, we start by letting x = 2 + 3 . Now subtract 2 from both sides to get: x− 2 = 3 Now square both sides carefully to get the following: (x − 2 ) = ( 3 ) 2 )(x − 2 ) = 3 (x − 2 x 2 − 2 2x + 2 = 3 1767 1768 1769 1770 5 is irrational.45 Earlier in this chapter, we talked about how we could generate a long list of irrational numbers by adding, subtracting, multiplying, or dividing a rational number (except maybe 0) to/with/by an irrational number. (See FACT 2 on page 44.) What about adding two irrational numbers? What does that give you? 2 1766 20 is irrational.44 Use the Rational Roots Theorem to show that Rearranging this we have: x 2 −1 = 2 2x MAT107 Chapter 4, Lawrence Morales, 2001; Page 52 1771 1772 1773 Squaring both sides again we get: (x 2 ) ( 2 −1 = 2 2x ) 2 x 4 − 2 x 2 + 1 = 8x 2 1774 x 4 − 10 x 2 + 1 = 0 1775 1776 1777 1778 1779 1780 1781 1782 1783 1784 1785 1786 1787 1788 1789 1790 1791 1792 1793 1794 1795 This is the equation to which we want to apply the Rational Roots Theorem. By that Theorem, the only possible rational roots would be +1 or −1. Since 2 + 3 is clearly larger than either of these two possibilities, 2 + 3 could not be rational and is therefore irrational. ♦ Check Point P Show that Solution 2 − 5 is irrational. Check the endnote for results.47 This concludes our study of modern facts on irrational numbers. Although the Pythagoreans did not have these sophisticated tools and theorems, the ideas they discovered and explored left a lasting impact on the future of mathematics, especially with respect to the field we now call number theory. As time has passed, mathematicians have sought to learn more and more about the qualities, properties, and behavior of both rational and irrational numbers. Much of what they have learned has evolved into increasingly complex theories and facts about numbers and their natures that are now catalogued in the history of the subject. MAT107 Chapter 4, Lawrence Morales, 2001; Page 53 1796 1797 1798 1799 1800 1801 1802 1803 1804 1805 1806 1807 1808 1809 1810 1811 1812 1813 1814 1815 1816 1817 1818 1819 1820 1821 1822 1823 1824 1825 Part 5: Pythagorean Geometry, A Brief Discussion The Pythagoreans are also known for some their studies of geometry, but their achievements here are more questionable.48 Of course, the famous Pythagorean theorem can be considered a study in geometry, but beyond that, it appears that they also investigated the five regular polyhedra, which we will study more in the next chapter and are shown here49: There is conflicting opinion on whether or not the Pythagoreans actually knew how to construct all five of these. Calinger conjectures that because the iron pyrite crystals found in southern Italy have faces that are composed of regular pentagons, the Pythagoreans became curious about figures such as these.50 There is also some speculation that the Pythagoreans did “geometric algebra,” showing such facts as (a + b) 2 = a 2 + 2ab = b 2 by drawing appropriate pictures. We’ll also save this topic for the next chapter when we look at geometric algebra. Conclusion The Pythagoreans, while known for their famous Theorem, also investigated many other ideas in mathematics. Their basic belief that numbers were the substance of the universe drove them to study mathematics in ways that allowed them to interpret the universe around them through a lens of mathematics. This is a unique approach and sets the stage for the development of number theory as well as the general attempt to interpret the world around us by investigating related mathematics. Whether we see it or not, mathematics is used all around us to fuel the technological change that is going on and to find new and better ways to view the world around us. As we have seen, the Pythagoreans and their studies branch off into many areas that they are not generally known for but for which they should be given due credit. MAT107 Chapter 4, Lawrence Morales, 2001; Page 54 1826 1827 1828 1829 1830 1831 1832 1833 1834 1835 1836 1837 1838 1839 1840 1841 1842 1843 1844 1845 1846 1847 1848 1849 1850 1851 1852 1853 1854 1855 1856 1857 1858 1859 1860 1861 1862 1863 1864 1865 1866 1867 Part 6: Homework Problems Triangular Numbers Find each of the following triangular numbers: 1) T7 3) T234 _______ 2) T18 4) T1200 For each of the following numbers, determine whether or not they are triangular numbers. If they are, indicate which one they are. If not, show why not. Use the quadratic formula in all calculations, and show all work neatly. 5) 595 7) 3,488 _______ 6) 41,616 8) 8,515 In around 100 C.E., Plutarch noted that if a triangular number is multiplied by 8 and then 1 is added to the result, then the resulting number would be a square number.51 9) a. b. 10) 11) Use Plutarch’s statement to do the following: Show that it is true for T2. Draw a picture that clearly shows this fact for T2 geometrically. Show that Plutarch’s statement is true for T5. Show that Plutarch’s statement is true for T20. 12) Show that Plutarch’s statement is true for any triangular number, Tn. (You will need to work algebraically with variables rather than particular numbers. The algebraic expression you obtain after multiplying by 8 and adding 1 should be factorable.) _____ Write each of the following numbers as the sum of three or fewer triangular numbers.52 13) 56 14) 69 15) 185 16) 287 _____ MAT107 Chapter 4, Lawrence Morales, 2001; Page 55 1868 1869 1870 1871 1872 1873 1874 1875 1876 1877 1878 1879 1880 1881 1882 1883 17) In 1872, the mathematician who is best know for discovering calculus, Lebesgue, showed two things related to triangular numbers.53 • Every positive integer is the sum of a square number (possibly 02) and two triangular numbers. • Every positive integer is the sum of two square numbers and triangular number Show that Lebesgue’s findings are true for the following numbers: a. 9 _____ d. 100 12 + 2 2 + 3 2 + 4 2 + .... + n 2 = a. b. c. n(n + 1)(2n + 1) 6 To verify this formula works in particular cases, you would physically add up the first n square numbers on the left and then plug the appropriate value of n into the formula on the right to see that it matches what you got on the left. Verify that the formula works for: The sum of the first three square numbers. The sum of the first four square numbers. The sum of the first five square numbers. 19) A famous Hindu mathematician, Aryabhata, showed (in around 500 C.E.) that you could add up the first n triangular numbers with the following formula: T1 + T2 + T3 + T4 + ... + Tn = 1896 1897 1898 1899 1900 1901 1902 1903 1904 1905 1906 1907 1908 1909 1910 1911 c. 81 18) The greatest mathematician in all of antiquity, Archimedes, found that you could add up the first n square numbers with a relatively simple formula: 1884 1885 1886 1887 1888 1889 1890 1891 1892 1893 1894 1895 b. 44 a. b. c. n(n + 1)(n + 2) 6 To verify this formula works in particular cases, you would physically add up the first n triangular numbers on the left and then plug the appropriate value of n into the formula on the right to see that it matches what you got on the left. For example, to add up the first three triangular numbers, you would let n = 3 and substitute into the formula. This would give you the sum of T1 + T2 + T3 = 1 + 3 + 6 = 10 Verify that the formula works for the sum of the first 10 triangular numbers by using the formula to compute the sum AND by adding up the first 10 triangular numbers. We certainly don’t want to have to add up the first 100 triangular numbers, but use the formula to easily find the sum. What is the sum of the first 1000 triangular numbers? MAT107 Chapter 4, Lawrence Morales, 2001; Page 56 1912 1913 1914 1915 1916 1917 1918 1919 1920 1921 1922 1923 1924 1925 1926 1927 1928 1929 1930 1931 1932 1933 1934 1935 1936 1937 1938 1939 1940 1941 1942 1943 1944 Pentagonal Numbers A pentagonal number is one that can be arranged in the shape of a pentagon (five sides). The first three pentagonal numbers are shown and the number of dots in a figure corresponds to the value of the pentagonal number. (The first one is “trivial.”). The nth Pentagonal 3n(n − 1) number is given by Pn = n + . 2 n=2 P2=5 20) Use the formula to compute P4 and then draw it. (Use a ruler or straightedge, please.) 21) Use the formula to compute P5 and then draw it. (Use a ruler or straightedge, please.) 22) What is the value of P15? 23) What is the value of P25? 24) What is the value of P100? For each of the following numbers, determine whether or not they are pentagonal numbers. If they are, indicate which one they are. If not, show why not. Use the quadratic formula in all calculations, and show all work neatly. 25) 205 26) 590 27) 5735 28) 2,150 Oblong Numbers An “oblong” number54 gives the number of dots in a rectangular grid having one more row than it has columns. The first few oblong numbers are shown below: O1=2 • • 1945 1946 1947 1948 1949 n=1 P1=1 O2=6 • • • O3=12 • • • • • • • • • • • • • • • • • • • • O4=20 • • • • • • • • • • • • • • • 29) Explain why the nth oblong number is given by O n = n(n + 1) . 30) Use the formula given in the previous problem to find the 100th oblong number 31) Show algebraically and geometrically that any oblong number is the sum of two equal triangular numbers. MAT107 Chapter 4, Lawrence Morales, 2001; Page 57 n=3 P3=12 1950 32) Show algebraically and geometrically that O n − n 2 = n . 1951 1952 1953 1954 1955 1956 1957 1958 1959 33) Show algebraically that O n + n 2 = T2 n . (To find an expression for T2 n , you will need to plug 2n into the general formula for a triangular number.) ___ 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 Non-Standard Figurative Numbers 34) Consider the following sequence of figures. Dn is the total number of dots in the figure. n=1 n=2 n=3 n=4 Dn = 8 Dn = 13 Dn = 19 Dn = 26 a. Find a general formula for Dn which can be used to find the number of dots in the nth figure. Explain in words or show how you got your formula. (Hint: Combine the formulas for two known types of figurative numbers.). Show that your formula works for n = 3 and 4. b. Use your formula to find the number of dots in the 50th figure. c. One of these figures has 818 dots. Which one is it? Use the formula from part (a) to and the quadratic formula to answer this. Show all algebraic steps. Other Proofs of the Pythagorean Theorem 35) Many civilizations had proofs of the Pythagorean theorem, including the Chinese. The attached picture is the Chinese hsuan-thu. The square that is in a diamond orientation is what we’re most interested in. Notice that it is made up of four triangles surrounding a small center square in the middle of the diamond. (The triangles can be a bit hard to see, but look hard enough and you should see four of them arranged in such a way that they form and surround a small square in the middle.) Label the shorter legs of the triangles with lengths a, and the longer legs with length b. Label the hypotenuse of each with length c. a. Carefully cut the inner square/diamond into four triangles and one small square and then rearrange them to prove visually that a 2 + b 2 = c 2 . When you are convinced you have a correct diagram, glue them together on a sheet and add any labels you think help MAT107 Chapter 4, Lawrence Morales, 2001; Page 58 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019 2020 2021 2022 2023 2024 2025 b. c. demonstrate the proof. Hints: You don’t need the four outer triangles; the shape you get may not be a square; make copies of the diagram before cutting in case you need another copy. There’s also copies on the web site. HINT: You will find this same drawing and a “related” proof inside this chapter that should get you going with dissecting the pieces properly. Find the following WITHOUT assigning particular values to a, b, and c. i. The area of the diamond, in terms of c. ii. The area of all four inner triangles combined, in terms of a and b only. iii. The area of the small square, in terms of a and b only. Use your answers from part (a) and part (b) to algebraically “prove” the Pythagorean Theorem. Your proof should only use the variables a, b, and c. Briefly explain how your rearranged drawing finishes the proof. 36) In 1876, James Garfield, who would eventually be the 20th president of the United States, published a proof of the Pythagorean Theorem. Research what it was and present it. To receive credit you must include a neat drawing similar to his (use a ruler) as well as a proof that goes along with the drawing. The proof should be in your own words and reflect your own understanding of the problem. Your write-up, logic, and organization will be the main basis of your score. Specifically state where you got your information from or no credit will be given. (No printed web sites will be accepted…it must be your written summary that you present.) 37) Research and present, in your own style, one proof of the Pythagorean Theorem that we have not talked about in this chapter. Your write-up, logic, and organization will be the main basis of your score. Specifically state where you got your information from or no credit will be given. (No printed web sites will be accepted…it must be your written summary that you present.) 38) Consider the figure shown where triangle ABC is a right triangle (with right angle at C). Triangle BAD is also a right angle (with right angle at A).55 a. b. 2026 2027 c. 2028 2029 d. A a c Explain why triangles ABC and DBA B are similar to each other. D b C Use the fact that triangle ABC is similar to triangle DBA to show that ac AD = . b ac a2 . and other geometry as necessary to show that DC = Use the fact that AD = b b Prove that a 2 + b 2 = c 2 by relating the area of triangle ABD to the areas of triangles ABC and ACD. MAT107 Chapter 4, Lawrence Morales, 2001; Page 59 2030 Blank Page MAT107 Chapter 4, Lawrence Morales, 2001; Page 60 2031 2032 2033 2034 2035 2036 2037 2038 2039 2040 2041 2042 2043 2044 2045 C 39) Consider the two squares that are joined at B 1 a their sides as shown. a. What is the total area of both squares? b. Mark the point G between point A and point D so that AG is equal to the length b. a c. Draw the line segment BG. Label segment BG as c. Note the right triangle (∆BAG) that you have created. d. Draw the line segment GF. Why is GF = BG? Why is ∠BGF a right angle? e. Cut out the triangles, ∆ BAG and ∆GEF so that A D you have a total of three pieces…no more. f. Rearrange your three pieces to form a square. How does this prove the Pythagorean Theorem? Make sure you understand this before moving on. 1 b This proof is credited by some to be by Al-Sabi Thabit ibn Qurra al-Harrani (826−901). He is one of many talented mathematicians that have come out of the Arab world, and Baghdad in particular. To flee from religious persecution, he headed for Harran and was appointed as the court astronomer in Baghdad. It is there that Thabit received training in both mathematics and medicine. He did work in number theory and in particular worked on what are called amicable numbers. These are also known as friendly numbers. After making comments on Euclid’s work on these numbers, he found a way to show how certain numbers were amicable. Thabit’s other work focused on geometry, astronomy, mechanics (equilibrium of levers), philosophy, and other areas of mathematics. MAT107 Chapter 4, Lawrence Morales, 2001; Page 61 F b E 2046 Blank Page MAT107 Chapter 4, Lawrence Morales, 2001; Page 62 2047 2048 2049 2050 2051 2052 2053 2054 2055 2056 2057 2058 2059 2060 2061 2062 40) a. b. c. d. e. f. g. On the following figure, follow the given steps carefully to come up with a dissection proof of the Pythagorean Theorem. Use a ruler to draw all lines. All lines should be carefully drawn according to the instructions. Extend side RB downwards until it intersects side DE. Let the point of intersection on DE be F. Starting at point P, carefully draw a line segment that is parallel to side AB until it intersects side CB. Call this point G. Connect points F and G to form segment FG. FG should be parallel (or very close to it) to CD. From point F, draw a line that is perpendicular to BF until it intersects segment CD. Draw a line segment between side CD and GF so that a square is formed with side equal to CG. You should now be able to identify 7 distinct pieces that the squares on side a and side b have been cut into. Do not proceed until you have labeled them from 1 to 7. Cut these 7 pieces up very carefully and rearrange them inside the square on side c to prove the Pythagorean Theorem. R A P c b C a D MAT107 Chapter 4, Lawrence Morales, 2001; Page 63 B E 2063 Blank page MAT107 Chapter 4, Lawrence Morales, 2001; Page 64 2064 2065 2066 2067 Pythagorean Triples via the Pythagoreans 41) One author56 says, “the Pythagoreans were familiar with the formula 2 2068 2069 2070 2071 2072 2073 2074 2075 2076 2077 2078 2079 where m is an odd natural number.” This formula is related to Pythagorean triples. a. Show that this works for m = 7. Show all algebraic steps. b. Bunt also says "the equality may be checked readily." In part (a) you checked it for m = 7. Show that it’s true for all appropriate values of m. Show all steps and work. Hint: You do this by algebraically simplifying both sides of the equation until you get the two sides of the equation equal to each other, leaving the variable m in place. Do not put any specific values in for m. In this respect, it is similar to a problem you saw on the Egyptian homework.. (Be careful when multiplying these expressions out. A common error is to say that ( m 2 + 1) = m 4 + 1 , which is obviously false!) 2 2080 2081 2082 2083 _____ Find the Pythagorean Triple that corresponds to each of the following values of m. Use the 2 2084 2085 2086 2087 2088 2089 2090 2091 2092 2093 2094 2095 2096 2097 2098 2099 2100 2101 2102 2103 2104 2105 2 m2 + 1 m2 −1 = + m 2 2 2 2 m2 +1 m2 −1 + m 2 , for generating your triples. = Pythagorean Formula, 2 2 42) m = 15 43) m = 25 44) m = 99 45) m = 111 Pythagorean Triples via Proclus Find the Pythagorean Triple that corresponds to each of the following values of n in Proculs’ equations. Show a check that your generated triple satisfies the Pythagorean Theorem. 46) n = 15 47) n = 25 48) n = 50 49) n = 100 Pythagorean Triples via Plato Find the Pythagorean Triple that corresponds to each of the following values of p in Plato’s equations. Show a check that your generated triple satisfies the Pythagorean Theorem. MAT107 Chapter 4, Lawrence Morales, 2001; Page 65 2106 2107 2108 2109 2110 2111 2112 2113 2114 2115 2116 2117 2118 2119 2120 2121 2122 2123 2124 2125 2126 2127 50) p = 15 51) p = 25 52) p = 99 53) p = 100 Pythagorean Triples via Euclid Find the Pythagorean Triple that corresponds to each of the following values of g and h in Euclid’s equations. Show a check that your generated triple satisfies the Pythagorean Theorem. 54) g = 10, h = 11 55) g = 7, h = 2 56) g = 18, h = 10 57) g = 13, h = 5 58) Show that Euclid’s formulas for generating Pythagorean Triples satisfy the Pythagorean Theorem. Do not use numbers in your proof, only the variables g and h. Generalizing Pythagorean Facts 59) Algebraically show that the following is true: 2 2128 2129 2130 2 a+b a −b ab + = 2 2 An algebraic proof does NOT substitute values for a and b, but “proves” it by showing that the left and right sides of the equation are equivalent expressions. 2 2131 2132 2133 2134 2135 2 a −b a+b 60) Show that if you let a = 2 p and b = 2 in the formula, ab + , then you = 2 2 get the statement of Plato’s formula for Pythagorean Triples. Show/explain all work, algebra, and reasoning carefully.57 Do not use actual numbers to do this problem…work only with the variables given. 2 2 2136 2137 2138 2139 2140 2141 2142 2143 2144 2145 2 a −b a+b 61) Show that if you let a = (2n + 1) and b = 1 in the formula: ab + = , then 2 2 you get the statement of the Proclus’ formulas for Pythagorean Triples. Show/explain all work, algebra, and reasoning carefully.58 Do not use actual numbers to do this problem…work only with the variables given. 2 62) Show that (3,4,5) is the only Pythagorean triple involving consecutive positive integers. (Hint: Take some general triple (x, x + 1, x + 2) and use the Pythagorean triple to show that x must be equal to 3.) 59 _____ MAT107 Chapter 4, Lawrence Morales, 2001; Page 66 2146 2147 2148 2149 2150 2151 2152 2153 2154 2155 2156 Rational and Irrational Numbers Write each of the following as rational numbers in the from a/b, where a and b are integers. 63) 0.2459 64) 1.6523 65) 0.00023 66) 0.2984762 _____ Use FACT 1 from Page 35 to determine whether or not the following numbers have terminating decimal representations. Show and explain your reasoning clearly. 67) 17 80 68) 157 240 70) 3 2048 2157 2158 2159 2160 153 240 _____ 69) 2161 Write each of the following decimals as reduced rational numbers in the form 2162 2163 2164 2165 2166 2167 2168 2169 2170 2171 2172 2173 2174 2175 2176 2177 2178 2179 2180 2181 2182 2183 2184 2185 2186 are integers. See Example 11 on page 39 for more details. 71) a , where a and b b 72) 0.00341292929… 0.2878787... 73) 0.555125551255512… _____ 74) 0.3890909090… Write a polynomial equation that is set equal to 0 that has the given number as a solution. See Example 13 on Page 45 75) x = 13 76) x = 115 77) x=3 7 78) x = 5 20 79) x = 2 3 _____ ( ) 80) x = 5 3 2 Use the Rational Roots Theorem to list all the possible rational roots for the given polynomial. Then determine which numbers from that list are solutions to the polynomial. Show all work. 81) x 3 − 3x − 2 = 0 82) 2 x 3 + x 2 − 7 x − 6 = 0 83) 7 x 4 − 43x 3 + 55 x 2 + 7 x − 2 = 0 (Hint: there are only two rational roots.) 84) 3x 4 + 5 x 3 − 13x 2 − x + 6 = 0 (Hint: there are only three rational roots.) MAT107 Chapter 4, Lawrence Morales, 2001; Page 67 2187 2188 2189 2190 2191 2192 2193 2194 2195 2196 2197 2198 2199 2200 2201 2202 2203 2204 2205 2206 2207 2208 2209 2210 2211 2212 2213 2214 2215 2216 2217 2218 2219 2220 2221 2222 2223 2224 2225 2226 2227 85) 3x 5 + 2 x 4 − 22 x 3 − 2 x 2 + 3x = 0 (Hint: there are only three rational roots.) _____ Use the Rational Roots Theorem to prove that the given numbers are irrational. 86) 5 87) 90 88) 3 17 89) 5 2 90) 8 30 91) 7 35 3+ 5 93) 92) 5− 2 Writing Write a short essay on the given topic. It should not be more than one page and if you can type it (double−spaced), I would appreciate it. If you cannot type it, your writing must be legible. Attention to grammar is important, although it does not have to be perfect grammatically…I just want to be able to understand it. 94) Use the library or internet to research one of the major beliefs of the Pythagoreans, explaining possible reasons why such a belief may have existed and what possible impact it may have had on their studies. Cite all of your sources. 95) The Pythagoreans believed that all is number. Do you agree? Why or why not? 96) The Pythagoreans believed, according to some, that friendship is the highest virtue. Do you agree? Why or why not? If you disagree, which virtue do you believe is the “highest?” 97) The Pythagoreans believed that all of creation is built from rational numbers. When they discovered that there were numbers that were incommensurable, they encountered a conflict and contradiction. From your own experience or background, can you think of a religious system that has one central belief so important that, if proved wrong, it would create a similar conflict and contradiction? Explain what that is and how you think such a conflict should be handled if it were ever to arise. 98) The Pythagoreans ascribed virtues to many of the whole numbers. Use the Internet or library to research what some of these were and why they were made as they were by the Pythagoreans. Cite all of your sources. MAT107 Chapter 4, Lawrence Morales, 2001; Page 68 2228 2229 2230 The Chinese hsuan-thu 2231 MAT107 Chapter 4, Lawrence Morales, 2001; Page 69 2232 Blank page MAT107 Chapter 4, Lawrence Morales, 2001; Page 70 Part 6: Chapter Endnotes 2233 1 http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Thales.html Eves,Howard, History of Mathematics, page 72 3 Eves, page 74 4 Eves, page 74 5 http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Pythagoras.html 6 Calinger, Ronald, A Contextual History of Mathematics, page 69 7 Calinger, page 69 8 Calinger, page 68 9 Calinger, page 70 10 Solution to Check Point A: T100=5,050 11 Solution to Check Point B: Yes, it’s the 23rd triangular number 12 Solution to Check Point C: It’s not a triangular number: 2 n(n + 1) 2 2 666 = n + n 333 = n 2 + n − 666 = 0 n= −1 ± 2665 2 However, the square root here does not give an integer, so there is no way this could be a triangular number. (The triangular numbers will have positive integers as their labels, not decimals. For example, you can talk about the 44th triangular number, but you can’t talk about the 44.3425rd triangular number.) 13 http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Gauss.html The sum of the first n cubes is equal to the square of the nth triangular number. 15 Eves, page 76 16 http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Al-Banna.html 17 http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Fermat.html 18 http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Descartes.html 19 http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Euler.html 20 http://www.vejlehs.dk/staff/jmp/aliquot/knwnap.htm for a list of friendly numbers 21 http://www.cut-the-knot.com/pythagoras/; http://www.mcn.net/~jimloy/pythag.html; http://mathworld.wolfram.com/PythagoreanTheorem.html; http://jwilson.coe.uga.edu/emt669/Student.Folders/Morris.Stephanie/EMT.669/Essay.1/Pythagorean.html ; http://www.wol.pace.edu/schools/wp/lgravitz/lglesson.htm ; 22 http://www.roma.unisa.edu.au/07305/pythag.htm 23 http://www.mathcs.richmond.edu/~bpuhak/cmsc340/PythThm/PythThm.html 24 http://jwilson.coe.uga.edu/EMT668/EMT668.Student.Folders/HeadAngela/essay1/Pythagorean.html Other interesting proofs can be seen online at the following web addresses: Proof by Parallelograms: http://persweb.wabash.edu/facstaff/footer/Pythagoras.htm Proof by Moving Triangles (Java): http://www.utc.edu/~cpmawata/geom/geom7.htm Half Area Proof (Java): http://sunsite.ubc.ca/LivingMathematics/V001N01/UBCExamples/Pythagoras/pythagoras.html Proof by Rearrangement into Squares http://www.davis-inc.com/pythagor/proof2.html 14 MAT107 Chapter 4, Lawrence Morales, 2001; Page 71 25 Solution to Check Point D: This gives the triple (5,12,13), which is also well known. 26 Calinger, page 72 27 Solution to Check Point E: This yields the triple (21,220,221) 28 http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Plato.html 29 Solution to Check Point G: This yields the triple (140,51,149) 30 Calinger, page 74 31 Calinger, page 74 32 Niven, page 22. 33 Solution to Check Point H: The factorization of 3125 is 55, so it does terminate. There is no requirement that 2 has to be a prime factor, only that 2 and 5 are the only ones that could appear. 34 Niven, page 32. 35 Solution to Check Point I: 462830/999990 36 Proof that 3 is irrational (Niven, pages 43-44): First we note that an integer that is divisible by 3 will have the form 3n , and integers that are not divisible by 3 will have the form 3n + 1 or 3n + 2 . The following equations show that numbers of the from 3n + 1 or 3n + 2 are not divisible by 3. (3n )2 = 9n 2 = 3(3n 2 ) (3n + 1)2 = 9n 2 + 6n + 1 = 3(3n 2 + 2n ) + 1 (3n + 2)2 = 9n 2 + 12n + 4 = 3(3n 2 + 4n + 1) + 1 We will argue a proof by contradiction, so we assume that is rational. Then we can write and b are integers and b is not zero. As in the 2 case, we will assume that 3 as 3 = a , where a b a is in lowest terms. Squaring b a2 a 3 = gives 3 = 2 , which in turn gives 3b 2 = a 2 . That means that a 2 is divisible by 3 and therefore a is b b 2 2 divisible by 3. So, a = 3c from some integer c. Replace a by 3c in the equation a = 3b and we get (3c )2 = 3b 2 which means 9c 2 = 3b 2 , which implies 3c 2 = b 2 . This means that b 2 is divisible by 3 and so b is a divisible by 3. This leads us to the fact that both a and b are divisible by 3, so is reducible, which is a b a contradiction to our assumption that is in lowest terms. Therefore, 3 must not be rational and is therefore b irrational. 37 Solution to Check Point J 8x 6 − 5x 4 + 6 x 2 − 8x + 3 n=6 c6 = 8, c5 = 0, c 4 = −5, c3 = 0, c 2 = 6, c1 = −8, c0 = 3 38 Niven, page 56. 39 Solution to Check Point K 40 Yes Solution to Check Point K No MAT107 Chapter 4, Lawrence Morales, 2001; Page 72 41 42 Niven, page 58 Solution to Check Point M The possible rational roots are +5, -5, +5/2, -5/2, +1, -1, +1/2, -1/2 The actual roots are -5 and 1/2. 43 To see why this is true, we can start by letting: x= 2 We now square both sides to get: x2 = 2 Finally, set this equation equal to zero: 44 45 46 47 48 x2 − 2 = 0 Solution to Check Point N The equation you want to work with is x 2 − 20 = 0 The equation you want to work with is x4 −5 = 0 Solution to Check Point N Niven, page 62. Solution to Check Point P The equation you want to end up with is x 4 − 14 x 2 + 9 = 0 Calinger, page 76 http://www.scienceu.com/geometry/facts/formulas/form13a/ 50 Calinger, page 77 51 Burton, page 101. 52 Burton, page 101. 53 Burton, page 101 54 Burton, page 101 55 Burton, page 117 56 Bunt 57 Burton, page 117. 58 Burton, page 117. 59 Burton, page 117. 49 MAT107 Chapter 4, Lawrence Morales, 2001; Page 73 Blank page MAT107 Chapter 4, Lawrence Morales, 2001; Page 74