Download Babylonian Mathematics - Seattle Central College

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