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8 Number Theory
Dividing Without Division
Some problems of the form a+x = b had no solutions (values of x) among the
whole numbers so we created the integers. Some equations of the form a·x = b have
no solutions even among the set of integers, so we created rational numbers. This
chapter concentrates on equations of the form a·x = b that do have integer solutions.
If a = 2 and b = 6, there is an integer solution to a·x = b. If we look at the 2-row
of the multiplication table and scan across until we find a 6, we find it in the 3column. 3 satisfies the equation 2·x = 6. There are lots of other values of a and b
where the equation can be satisfied by an integer value of x.
Where there is an integer solution to the equation a·x = b, we say that a divides
b, and symbolize this relation as a|b.1 We say a is a divisor of b.
Given two integers a and b, a divides b (symbolized a|b) if and only if there is
some integer x such that a·x = b. In symbols, a|b ≡ ∃x(a·x = b). If a|b, then we
say that a is a divisor of b.
Some things about the | operator are easily proved. For example,2 to prove that
∀x (x|x) (every number divides itself) we just have to show that, no matter what
integer we take (call it a), there is an integer x (= 1) such that a·x = a. So a|a, no
matter what integer a is. So ∀x (x|x).„3
Another proof that is quite simple is that 1 divides every number or, in symbols,
∀x (1|x). Since, no matter what a we take, 1·a = a, then 1|a for all values of a.„
It is equally easy to prove that every number divides 0. Do it.
How many divisors does a number n have? What is the relationship between a
number and the cardinality of its divisors? We're looking for patterns in the numbers –
doing mathematics. To get data that might lead to a conjecture, we grab a bunch of
numbers. We look at the numbers from one to fifty, their divisors, and the number of
divisors of each number. Mathematicians use τ(n) (pronounced "tau of n") for the
number of divisors of some number n.
1
2
3
Following S.M. Dowdy, Mathematics: Art and Science (New York: John Wiley & Sons, Inc., 1971), p. 25.
Here and in what follows, I am implicitly limiting the domain (the universe of discourse, or the universal set)
to integers, so I don't have to keep saying things like x ∈ .
A black square is the convention used to indicate the end of a proof. In the old days, we used to write Q.E.D.
(meaning quod erat demonstrandum or 'which was to be proved') for the same purpose.
DIVIDING WITHOUT DIVISION
n
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
τ(n)
4
4
6
2
8
2
6
4
4
4
9
2
4
4
8
2
8
2
6
6
4
2
10
3
6
Consider one simple conjecture or hypothesis about the relationship between n
and τ(n). Looking at n = 3, 5, or 7, you might propose:
Divisors of n
1
1, 2
1, 3
1, 2, 4
1, 5
1, 2, 3, 6
1, 7
1, 2, 4, 8
1, 3, 9
1, 2, 5, 10
1, 11
1, 2, 3, 4, 6, 12
1, 13
1, 2, 7, 14
1, 3, 5, 15
1, 2, 4, 8, 16
1, 17
1, 2, 3, 6, 9, 18
1, 19
1, 2, 4, 5, 10, 20
1, 3, 7, 21
1, 2, 11, 22
1, 23
1, 2, 3, 4, 6, 8, 12, 24
1, 5, 25
τ(n)
1
2
2
3
2
4
2
4
3
4
2
6
2
4
4
5
2
6
2
6
4
4
2
8
3
111
n
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
Divisors of n
1, 2, 13, 26
1, 3, 9, 27
1, 2, 4, 7, 14, 28
1, 29
1, 2, 3, 5, 6, 10, 15, 30
1, 31
1, 2, 4, 8, 16, 32
1, 3, 11, 33
1, 2, 17, 34
1, 5, 7, 35
1, 2, 3, 4, 6, 9, 12, 18, 36
1, 37
1, 2, 19, 38
1, 3, 13, 39
1, 2, 4, 5, 8, 10, 20, 40
1, 41
1, 2, 3, 6, 7, 14, 21, 42
1, 43
1, 2, 4, 11, 22, 44
1, 3, 5, 9, 15, 45
1, 2, 23, 46
1, 47
1, 2, 3, 4, 6, 8, 12, 16, 24, 48
1, 7, 49
1, 2, 5, 10, 25, 50
CONJECTURE: If n is odd and n > 1, then τ(n) = 2.
To disprove a conjecture, we look for a counterexample. As in logic, a
counterexample is a particular instance where the conjecture is false. We note that the
conjecture is true in many cases (e.g., when n ∈ {3, 5, 7, 11, 13, …}). You should be
aware that even numbers that have τ(n) = 2 are not counterexamples – the conjecture
doesn't say "if τ(n) = 2 then n is odd." It is not a biconditional, so it doesn't say "n is
odd, if and only if τ(n) = 2." The conjecture is a conditional: "n is odd and n > 1" is
the antecedent; "τ(n) = 2" is the consequent. To show that it is false, you must find a
case where the antecedent is true (find an odd number greater than 1) and the
consequent is false. We find counterexamples in our table, where n ∈ {9, 15, 21, 25,
27, 33, 35, 39, …}. There is at least one counterexample. One is all it takes to refute a
conjecture. The conjecture is false.
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8 NUMBER THEORY
Integer Weights and Measures
Imagine weighing potatoes that come in whole
numbers of grams.4 We have a two-pan balance scale like
the one at right. Assume an unlimited supply of 5- and 7gram weights. To weigh a potato, we put the potato in one
pan of the balance scale and add weights to one or both
pans until the scale balances.
The question is: What sizes of potatoes can we weigh
with our apparatus and weights?
Using only the 5-gram weights, we can weigh potatoes
of 5, 10, 15, … grams. Using just 7-gram weights, we can weigh 7, 14, 21, … gram
spuds. We could put a 5- and a 7-gram weight in one pan to weigh a 12-gram potato.
If we put a 5-gram weight in one pan with the potato and a 7-gram weight in the other,
we can weigh a 2-gram potato, because the weight of the potato plus the 5-gram
weight balances the 7-gram weight.
To weigh a 3-gram spud, put a 7 in the pan with the potato and two 5's in the
other. Expressed as an equation, we have 3+7 = 2·5. A 4-gram? Put two 5's with the
potato and two 7's will balance. This corresponds to the equation 4+2·5 = 2·7. Or put
three 7's with the potato and five 5's in the other pan, for 4+3·7 = 5·5.
We can weigh a 1-gram potato in several ways. One way is with two 7's with the
potato balanced by three 5's, for 1+2·7 = 3·5. Figure out some of the other ways.
Since we can weigh a 1-gram potato, there's an algorithm for measuring any
weight. For example, a 6-gram potato weighs as much as six 1-gram potatoes. Multiply all the numbers by 6. Using the solution 1+2·7 = 3·5, we could put the potato
and 12 (= 6·2) 7-gram weights in one pan and 18 (= 6·3) 5-gram weights in the other,
corresponding to the equation 6+12·7 = 18·5. This algorithm does not give the most
efficient method. 6+3·5 = 3·7 uses fewer weights. But at least it shows that, if we
discover how to weigh a 1-gram potato, we can weigh any whole-number weight.
What if we use other than 5-gram and 7-gram weights? Suppose we have
unlimited supplies of 8-gram and 21-gram weights. If we can weigh a 1-gram potato,
we know we can weigh any potato. 1+3·21 = 8·8, so weigh a 1-gram potato by
putting three 21's into the pan with the potato and balancing it with eight 8's.
Would any combination of two weights work? Consider unlimited numbers of 6and 8-gram weights. There is no way to weigh a 1-gram potato or any odd number of
grams. Satisfy yourself that this is true. Then figure out why.
4
This whole thing is based on Chapter 3 of Sherman K. Stein, Mathematics: the Man-Made Universe, 2nd Ed.
(San Francisco: W.H. Freeman and Company, 1969).
INTEGER WEIGHTS AND MEASURES
113
More generally, suppose we have an unlimited number of weights of two types.
How can we decide whether we can weigh a 1-gram potato with them? Can we weigh
a 1-gram potato with 539-gram and 1619-gram weights? How can we find out? Make
a general conjecture about what kinds of sets of weights can not weigh a 1-gram
potato. Prove it.
Denote the two kinds of weights as a and b grams. In our first combination, a was
5 and b was 7. Represent the weight of the potato as w grams. Then there are only
three ways to weigh a potato.
(1) We can put the potato in one pan and weights (of either or both types) in the
other. This solution corresponds to the equation w = xa+yb, where x and y are whole
numbers (either of which could be 0).
(2) We can put zero or more a-gram weights in the pan with the potato and one or
more b-gram weights in the other. The equation for this is w+xa = yb.
(3) We can put some or none of the b-gram weights with the spud and a-gram
weights in the other pan, corresponding to the equation w+xb = ya.
There is no other way. There is no point in putting weights of one kind in both
pans; removing equal numbers of them from both pans won't change the balance. We
can just take equal numbers from both pans until there are none of them in one pan.
There are only three equations to consider: (1) w = xa+yb, (2) w+xa = yb, and
(3) w+xb = ya. In all of these, x and y are whole numbers (possibly zero).
Now we can rephrase our question about weights and potatoes as a question about
whole numbers (a number-theoretical question), as: Let a and b be natural numbers.
For which natural numbers w can we find whole numbers x and y such that at least
one of these three equations holds?
If we use integer numbers, we can simplify the problem by reducing the three
equations to just one. We observe that if we add (−4)·5 to both sides of the equation
1+4·5 = 3·7 (of type (2), above) we get 1+4·5+(−4)·5 = (−4)·5+3·7. Since
4·5+(−4)·5 = 0, the equation becomes 1 = (−4)·5+3·7, an equation of type (1),
above. In general, any equation of type (2) (i.e., of the form w+xa = yb) can be rewritten as w = (−x)a+yb. Similarly, the relation described by equations of type (3)
w+xb = ya can be expressed by an equation of type (1) by writing it as w =
ya+(−x)b.
Our question now becomes: If a and b are natural numbers, which natural
numbers w can be expressed in the form w = xa+yb for integers x and y? The
question about weighing a 1-gram potato becomes: For which natural numbers a and b
are there integers x and y such that 1 = xa+yb? We see that this means "Are there
multiples of a and b whose difference is 1?" Other problems can be expressed in
similar ways. For example, if we have two measuring sticks, one a inches long and the
other b inches long, can we measure a length of 1 inch?
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8 NUMBER THEORY
Exercise on Weights and Measures
1.
2.
3.
4.
5.
6.
7.
8.
Find three different ways of weighing a 1-gram potato with 4- and 7-gram
weights. Find three different ways of weighing a 1-gram potato with 8- and 13gram weights.
What do you think is the smallest potato that can be weighed with:
(a) 8- and 12-gram weights;
(b) 8- and 11-gram weights;
(c) 9- and 12-gram weights?
Why do you think so?
Are there integers x and y such that
(a) 1 = x·7+y·10?
(b) 1 = x·8+y·10?
(c) 1 = x·13+y·22?
(d) 1 = x·6+y·21?
If your answer is "yes," give values of x and y. If "no" explain why not.
Can a 1-gram potato be weighed with 5- and 8-gram weights in such a way that
(a) the potato is with the 5's and the 8's are in the other pan;
(b) the potato is with the 8's and the 5's are in the other pan?
Consider 5- and 12-gram weights. Show that it is possible without using more
than four 12-gram weights (you can use as many 5's as you need) to weigh
(a) a 1-gram potato,
(b) a 2-gram potato,
(c) a 3-gram potato,
(d) a 4-gram potato,
(e) a 5-gram potato,
(f) a 6-gram potato, and
(g) a 7-gram potato.
Can you weigh heavier potatoes within the same restrictions (no more than four
12s, as many 5s as you like)?
You have two unmarked jugs. One holds 5 litres and the other holds 7. How can
you get precisely one litre of water from a lake?
Using the two jugs in question 7, what are all the different amounts of water you
could get in single trips to the lake? Explain how.
A banana is worth 17 cents and an orange 13 cents. What is the smallest number
of bananas that could be traded for oranges where equal values change hands?
Suppose someone gives some bananas plus 1 cent for some oranges and the
traders give and get equal value. What is the smallest number of bananas involved
in the trade?
PRIME NUMBERS
115
Prime Numbers
A prime number is any natural number that has exactly two distinct divisors
which are 1 and the number itself. In symbols, where Pn symbolizes "n is
prime," this is:
Pn ≡ ((n≠1) ∧ ∀x(x|n ⊃ ((x=1) ∨ (x=n)))).
From the definition, we can infer that, for any prime number n, τ(n)=2.
We see why the simple conjecture, above, did not hold. All primes except 2 are
odd, and primes have τ(n) = 2. The reason the first few odd numbers have τ(n) = 2 is
because they are primes. The conjecture failed for odd composite numbers.
A composite number is any natural number that has more than two distinct
divisors.
The proof of the theorem that there are infinitely many primes appears in Book IX
of Euclid's Elements. It uses a couple of premises that also have to be proved. One of
these premises is called a lemma (a proven result that is not important enough to be
called a theorem, but which is useful in the proof of a theorem).
LEMMA: If a natural number d divides each of two natural numbers a and b, then
d divides their difference a−b and their sum a+b. In symbols
((d | a) ∧ (d | b)) ⊃ ((d | a−b) ∧ (d | a+b))
PROOF: By the definition of "divides" we know that there are integers q1 and q2
such that a = q1d and b = q2d. So a−b = q1d−q2d = (q1−q2)d.
Since q1−q2 is an integer and a−b = (q1−q2)d, we conclude (again, by the
definition of "divides") that d divides a−b.„
A similar argument shows that d divides a+b. Write the argument/proof.
THEOREM 1: Any prime that divides the natural number that is 1 larger than the
product of several primes is not equal to any of those primes.
PROOF: Take any collection of n primes p1, p2, p3, …, pn and multiply them
together to form the product p1p2p3…pn. Add 1 to that product and call the result m
(so m = p1p2p3…pn+1). First we have to prove that none of the primes p1, p2, p3, …,
pn divides m. Each of p1, p2, p3, …, pn clearly divides the product p1p2p3…pn. If one
of the p's also divided the natural number m, then it would divide m and m−1. The
lemma shows that it would then have to divide their difference, which is 1. But the
only number that divides 1 is 1, and 1 is not a prime.
Now think of numbers larger than 1 that do divide m. If there is a prime divisor of
m then it cannot be one of the original collection of primes, as we just showed. If the
116
8 NUMBER THEORY
only divisor larger than 1 is m itself, then m is prime and was not in the original
collection of primes. If m is composite, take the smallest divisor of m other than 1.
Call it d. If d is composite, then it has smaller divisors. Those smaller numbers are
also divisors of m. But we said d was the smallest, so d must be prime. It cannot have
been in the original collection of primes (as argued above). So this smallest divisor is
prime and was not in the original collection of primes.„
THEOREM 2: THE INFINITUDE OF PRIMES. There is no end to the primes.
PROOF: (By reductio ad absurdum, or Indirect Proof)
Assume that there is an end to the primes, so that one could
make a complete list of all the primes. Say that there are n
primes in that list. Then the number obtained by
multiplying all the n primes together and adding 1 to the
product will have at least one divisor which is prime and is
different from all of the primes used in the product (by
Theorem 1). But the product was obtained by multiplying
all the primes there are. So our assumption must have been
mistaken, as it led to a contradiction – that there are only n
primes and there are more than n primes. Therefore our
assumption was false, and there is no end to the primes.„
Former students have suggested
that this and other proofs would
be more easily understood if
they were "based on actual
numbers." Some students have
offered a few examples of
numbers as a "proof" that some
conjecture about all numbers
was true. You should know by
now that instances or examples
are not proof. Consider, for
example, the fact that 31, 331,
3331, 33331, 333331, 3333331
and 33333331 are all prime.
Does that prove that every
number that begins with a
bunch of threes followed by a
one is prime?
In the following table of natural numbers up to 420,
the primes are emphasized. Prime numbers seem to be
scattered almost randomly throughout the list. Some hints
of patterns tantalize us (look at sequences like 7, 23, 37, 53,
67, 83, 97, 113, 127), but patterns are less obvious (or simply do not exist) as we get
into higher and higher numbers. Is there a pattern in the sizes of the gaps between
successive primes? There is only one pair of primes (2 and 3) that have a gap of 1
between them, because 2 is the only even prime. Some pairs of primes have a gap of
only 2, even near the end of our list (347 and 349). Pairs of primes that differ by 2 are
called "twin primes." Will there continue to be twin primes even among the very
large numbers? Look at the distribution of primes. There are 4 primes among the first
10 numbers (40% of the first 10 numbers are primes). There are 8 among the first 20
numbers (still 40%). There are only 10 primes < 30 (33%); 12 < 40 (30%); 24 < 100
(24%); 44 < 200 (22%); and so on. Among the first 100 numbers, 24% are prime; 20%
of the next 100; 16% of the next 100; 15% of the next 100. The primes thin out as they
get larger.
CONJECTURE:5 there are infinitely many twin primes.
5
The Twin Prime Conjecture was made in 1923 by Hardy and Littlewood.
PRIME NUMBERS
117
This conjecture has never been proved or disproved. We don't know. We also
cannot prove that the conjecture is unprovable. It's an open question. Other conjectures
to which we have found no exceptions but which nobody has yet proved are:
1
16
31
46
61
76
91
106
121
136
151
166
181
196
211
226
241
256
271
286
301
316
331
346
361
376
391
406
2
17
32
47
62
77
92
107
122
137
152
167
182
197
212
227
242
257
272
287
302
317
332
347
362
377
392
407
3
18
33
48
63
78
93
108
123
138
153
168
183
198
213
228
243
258
273
288
303
318
333
348
363
378
393
408
4
19
34
49
64
79
94
109
124
139
154
169
184
199
214
229
244
259
274
289
304
319
334
349
364
379
394
409
5
20
35
50
65
80
95
110
125
140
155
170
185
200
215
230
245
260
275
290
305
320
335
350
365
380
395
410
6
21
36
51
66
81
96
111
126
141
156
171
186
201
216
231
246
261
276
291
306
321
336
351
366
381
396
411
7
22
37
52
67
82
97
112
127
142
157
172
187
202
217
232
247
262
277
292
307
322
337
352
367
382
397
412
8
23
38
53
68
83
98
113
128
143
158
173
188
203
218
233
248
263
278
293
308
323
338
353
368
383
398
413
9
24
39
54
69
84
99
114
129
144
159
174
189
204
219
234
249
264
279
294
309
324
339
354
369
384
399
414
10
25
40
55
70
85
100
115
130
145
160
175
190
205
220
235
250
265
280
295
310
325
340
355
370
385
400
415
11
26
41
56
71
86
101
116
131
146
161
176
191
206
221
236
251
266
281
296
311
326
341
356
371
386
401
416
12
27
42
57
72
87
102
117
132
147
162
177
192
207
222
237
252
267
282
297
312
327
342
357
372
387
402
417
13
28
43
58
73
88
103
118
133
148
163
178
193
208
223
238
253
268
283
298
313
328
343
358
373
388
403
418
14
29
44
59
74
89
104
119
134
149
164
179
194
209
224
239
254
269
284
299
314
329
344
359
374
389
404
419
15
30
45
60
75
90
105
120
135
150
165
180
195
210
225
240
255
270
285
300
315
330
345
360
375
390
405
420
CONJECTURE: Every odd natural number larger than 2 can be written as the
sum of a prime plus a power of 2 (e.g., p+20 or p+21 or p+23 or whatever).
CONJECTURE: Every odd natural number larger than 3 can be written as the
sum of a prime plus twice a square (i.e., as p+2·n2 where p is a prime and n is a
whole number).
People have run computer programs to test these conjectures on billions of
billions of billions … of numbers, and the conjectures have always been confirmed.
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8 NUMBER THEORY
But no matter how many confirming instances we find, instances are not a
deductive proof, and deductive proof is what counts in mathematics.
The biggest gap between successive primes in our sample is 14 (the first such gap
is from 113 to 127). Is there a limit to how large the gap can be between successive
primes? In this case, we can prove the answer:
THEOREM 3: There is no limit to the size of the gap between successive primes.
PROOF: The proof is a demonstration that we can construct a sequence of as
many consecutive composite numbers as we like. For example, suppose we can
construct a sequence of 5 consecutive composite numbers (e.g., 24, 25, 26, 27, 28).
That shows that there is a gap of at least 6 between successive primes (29−23=6).
Any prime would have to be smaller than the smallest of the five numbers (24) or
larger than the biggest number (28).
The construction6 involves the concept of a factorial. A factorial is a function of
a number. The factorial of 3 is written as 3! which means 3·2·1. To find the factorial
of any natural number, multiply the number by every natural number less than or equal
to itself. 27! is 27·26·25· … ·3·2·1.7
To find a gap of at least 5, we find a sequence of 4 consecutive composite
numbers. Calculate (4+1)! = 5! = 5·4·3·2·1 = 120. Then 120+2, 120+3, 120+4,
and 120+5 must all be composite numbers. 120+2 must be divisible by 2, because
120 has 2 as one of its factors, and 2 divides 2. By the lemma, above, since 2 divides
both 120 and 2, it divides their sum. By similar reasoning, 120+3 must be divisible by
3. And so on. The largest prime smaller than our first number (122) is 113. The next
prime after our last number (125) is 127. The gap between them is 14, which is larger
than 5.
To find 999 consecutive composite numbers (which separate two successive
primes by a gap of at least 1000), we calculate 1000! and add 2 to the result. We can't
start with n!+1, because n!+1 may be prime – look at 3!+1. 1000!+2 is our first
composite number. That and the next 998 numbers are all composite by the same
argument we used for 122, 123, etc., above.
This procedure works for any number, so we can construct a gap of any size,
without limit, between successive primes.„
If you try calculating n! for bigger and bigger values of n, you'll see that n! gets
very large very fast. You might be curious about how big a number 1000! is. We
calculated and printed the exact value in the previous edition of this book. It was a full
page (2,566 digits) long.
6
7
This is a constructive proof. We show that something exists by specifying the procedure to construct it.
We'll redefine "factorial" for whole numbers (to define 0!) later in the course.
PRIME NUMBERS
119
The approximate value of 1000! (in scientific notation) is 4.02387×102565. It has
no use outside of number theory. Such huge numbers are utterly useless even in
science. Carl Sagan8 says, "… the total number of elementary particles – protons and
neutrons and electrons – in the observable universe is about 1080. If the universe were
packed solid with neutrons, say, so there were no empty space anywhere, there would
still be only about 10128 particles in it." 1000! is a number that is about 102485 times as
big as the total number of elementary particles in the universe, and about 102440 times
as big as the number of neutrons that would fill the universe.
We can calculate it and add 2 to that huge number. The result, and the next 998
numbers after it, will all be composite. There will be a gap of at least 1000 between
successive primes.
You may have figured out that we could use a smaller number than 1000! to get a
gap of at least 1000 between successive primes. We could multiply together all the
primes less than 1000, add 2 to the result, and that and the next 998 numbers will all
be composite. That is, we don't have to divide by 4 or 6 or 9 to check if a number is
composite. If 4 or 6 divides it, then 2 will divide it. If 9 divides it, then 3 will divide it,
and so on.
Exercise on Primes
1.
2.
3.
4.
5.
6.
7.
8
Some primes can be written in the form n2+1 for some natural number n. For
example, 22+1 = 5, and 42+1 = 17. Find three more primes of the form n2+1.
(The conjecture that there are infinitely many such primes has never been proved
or disproved.)
Some primes are one less than a square (they can be written in the form n2−1 for
some natural number n). For example, 22−1 = 3. Can you find other primes of the
form n2−1? Make a conjecture and prove it.
If n is a natural number larger than 1, is there always at least one prime between n
and 2n? Experiment for n less than 200 (use the table of primes). Make a
conjecture.
In Theorem 1, above, we examined a way of making new primes by multiplying
primes together and adding 1 to the result, and then looking at the numbers that
divide that result. Use that procedure starting with the primes 11 and 13. What
new primes does it produce? Start with the primes 2, 11, and 13. What primes
does this produce?
Compute 6!. Check that 6!, 6!+2, 6!+3, 6!+4, 6!+5, 6!+6, are not prime. Is 6!+1
prime?
If d divides a and d divides b, prove that d divides a+b. Prove that d divides a·b.
If a prime divides a+b, must it divide at least one of a and b? (This is the
converse of the lemma, above.)
Carl Sagan, Cosmos (New York: Ballantine Books, 1985), p. 181.
120
8 NUMBER THEORY
Mathematical (Perfect) Induction
In Chapter 1 we said, "There are 'laws of right mathematical reasoning' that only
or mostly apply within mathematics. The principle of mathematical induction (often
called "perfect induction") is uniquely mathematical.
We usually contrast induction with deduction as two distinct methods of
reasoning. Ambiguity plagues us, because mathematical induction is a special mode of
deductive reasoning.
Here is how proof by (mathematical) induction works. A mathematician dreams
up and states a conjecture that she supposes may be true for all natural numbers. She
then checks and proves that the statement of the conjecture is true for the number 1.
She also proves that if the statement is true for some natural number n, it will have to
be true for the next number n+1. The principle of mathematical or perfect induction
permits her to infer that the statement is true for every natural number.
The principle of mathematical or perfect induction is: If a statement about
natural numbers can be proved to be true for the number 1, and if it can be
proved that, whenever the statement is true for an arbitrarily-chosen natural
number n it must be true for n+1, then it must be true for all n ∈ .
To state this principle symbolically as a rule of inference, we can represent "the
statement S is true for the number x" as S(x). We can then say:
S(1), S(n) ⊃ S(n+1) ⇒ ∀x(S(x)).
The rule argues from the particular (a proof about the particular number 1 and
about an arbitrary other particular number and its successor) to the general (a statement that applies to all natural numbers). But unlike empirical induction, the reasoning
is deductive. That is, if the premises (the statement about 1 and the conditional statement about n and n+1) are true, the conclusion cannot be false. The truth of the
conclusion is guaranteed by the truth of the premises.
As an example, consider the sequence of natural numbers 1, 2, 3, 4, …. A clever
mathematician (like you) might notice that the triangular numbers are the sums of the
numbers in this sequence. That is, the first triangular number is 1, which is the sum of
the first 1 term(s). The second triangular number (3) is the sum of the first 2 terms.
The third (6) is the sum of the first 3, and so on. The fourth triangular number is the
sum of the first four natural numbers.
We can re-order the first four numbers as (1+4)+(2+3), which is 2·5. If we look
at the sum of the first five natural numbers (the fifth triangular number), the sum is
1+2+3+4+5. There is an odd number of summands, so we can't order them in pairs
to replace the long addition with a simple multiplication, as we did with the first four.
Creative thinking suggests that we figure out what double the sum is, and then take
half the result. The sum of the first five natural numbers is half of
MATHEMATICAL (PERFECT) INDUCTION
121
(1+2+3+4+5)+(1+2+3+4+5). Now we can re-order these numbers in pairs, as
(1+5)+(2+4)+(3+3)+(4+2)+(5+1), giving 5·6 = 30. Half of 30 (15) is the sum of
the first five natural numbers. Returning to the sum of the first four numbers, we get
(1+2+3+4)+(1+2+3+4) = (1+4)+(2+3)+(3+2)+(4+1) = 4·5 = 20. Half of 20 is
10, the sum of the first four natural numbers.
n
We introduce a new symbolism for the sum of n numbers.
∑ i is read as "the sum
i=1
of all values of i as the value of i goes from 1 to n," or just "the sum of all i from 1 to
n." The symbol is the Greek capital-letter sigma, and stands for "summation."
So our conjecture is that the sum of the first n natural numbers is equal to
n·(n+1)÷2. We test to see if this statement holds for n = 1. We calculate 1·(1+1)÷2
= 1·(2)÷2 = 2÷2 = 1. The conjecture is true. Now we try to prove the conditional
that says "If the conjecture is true for n then the conjecture will be true for n+1."
Since we are trying to prove a conditional, we use conditional proof (CP). We assume
the antecedent, and try to prove the consequent.
The antecedent says that the conjecture is true for n – that is, that the sum of the
n
first n natural numbers is given by the formula ∑ i = n·(n+1)÷2. The sum of the first
i=1
n+1 natural numbers will be Sn+(n+1).9 A little simple algebra gives:
n
∑ i+(n+1) =
i=1
n(n+1)
n(n+1) 2(n+1) n(n+1)+2(n+1) (n+1)(n+2)
+ (n+1) =
+
=
=
2
2
2
2
2
(n+1)((n+1)+1)
. The last expression is exactly what you get when you substitute
2
(n+1) for n in n·(n+1)÷2. We have proved that the conjecture is true for n+1, based
on our assumption that it was true for n. By rule CP, we infer that if it is true for n,
then it is true for n+1.
=
Having proved the conjecture true for 1 and having proved the conditional, the
principle of mathematical induction ensures that it is true for all values of n.„
In the sequence of square numbers 1, 4, 9, 16, 25, … the successive elements are
the sums of odd numbers, as 1, 1+3, 1+3+5, 1+3+5+7, 1+3+5+7+9,…. The fifth
square number is the sum of the first five odd numbers, and so on. That is, the nth
square number is the sum of the first n odd numbers. The odd natural numbers form a
sequence 1, 3, 5, 7, 9, …. The nth odd number is equal to 2·n−1. Thus, the first odd
number is 2·1−1 or just 1. The second is 2·2−1 or 3. And so on. So the sum of the
9
That is, it will be the sum of the first n natural numbers plus n+1.
122
8 NUMBER THEORY
n
first n terms of the sequence of odd numbers is ∑ (2i−1). Our previous exploration of
i=1
th
square numbers also led us to believe that the n square number was n2. We'll prove
this conjecture using the principle of mathematical induction.
It is clear that when n is 1, the sum of the first n odd numbers is n2; that sum is
just 1, and 1 = 12.
We try to prove the conditional by rule CP. The "Assumed Premise" is that the
conjecture is true for some arbitrarily chosen n, so that the sum of the first n odd
n
numbers is ∑ (2i−1) = n2. To get the sum of the first n+1 odd numbers we take the
i=1
sum of the first n odd numbers and add the (n+1)th odd number. The (n+1)th odd
n+1
number is 2·(n+1)−1. So we have ∑ (2i−1) = n2+(2·(n+1)−1) = n2+2n+2−1 =
i=1
n
n+1
i=1
i=1
n2+2n+1 = (n+1)2. Therefore, if ∑ (2i−1) = n2, then ∑ (2i−1) = (n+1)2. Thus,
whenever the conjecture holds for n, it must also hold for n+1. So, by the principle of
induction, the conjecture is true for all n.„
Exercise on Perfect Induction
1.
Look at the sequence of squares. We can calculate the sums of the first n squares.
The sequence of sums is 12=1, 12+22 = 5, 12+22+32 = 14, 12+22+32+42 = 30,
and so on. Use mathematical induction to prove the conjecture that
n
∑ i2 =
i=1
2.
3.
n(n+1)(2n+1)
.
6
Use mathematical induction to prove that the largest binary (base 2) number that
can be represented by n binary digits (bits) is 2n−1.
Use mathematical induction to prove that the largest n-digit decimal number is
10n−1.