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Chapter 3
Teaching Notes, Chapter 3: The preparation of middle-grades mathematics teachers sometimes
does not include experience with reading and writing careful mathematical arguments. Since we must
provide middle-grades teachers with not only the specific knowledge they will teach but also the
mathematical context for that knowledge, experience with reading and writing proofs is an essential
part of their preparation.
There are at least three paths that can be taken through Chapter 3. For classes involving students
from a variety of backgrounds, you might choose to spend a day or two on Section 3.1 with the dual
goals of helping the students understand the pitfalls to inductive thinking that make mathematical
proofs necessary and how simple deductive arguments related to even and odd numbers and
divisibility can overcome those difficulties. In this case you would want to give careful attention to
examples where a pattern seems to exist, but closer examination produces a counterexample.
A second path would involve study of Sections 3.1, 3.2, and 3.3. This will introduce all the proof
techniques used later in the text and provide students the opportunity to practice those techniques in a
variety of circumstances.
The third path would involve the study of all five sections in the chapter. Because Sections 3.4
and 3.5, together with 3.1, provide students with a brief introduction to number theory, this allows
you to broaden the set of topics covered in the course. Further, the task of proving some of the simple
theorems of number theory provides an immediate opportunity for students to practice their newly
acquired skills in theorem proving.
Teaching Notes, Section 3.1: A primary objective of this section involves helping students
understand why mathematical proof is necessary; why it is not sufficient just to observe a pattern in a
few examples. Careful attention to the discussion of the first two pages of Section 3.1 and
Exploratory Exercises 3 and 4 will assist in this objective.
In this section we begin a theme we will continue throughout the remaining sections and chapters:
Before attempting a proof one must spend some time in preparation. If you intend to join the authors
in emphasizing this theme, it is important to model that preparation as you discuss the simple
theorems of this section with the students. Trying examples and testing conjectures is a first step;
reviewing related definitions and theorems, a second; and a moment or so of reflection in which you
devise a plan to connect known information with the theorem to be proved, a third.
We recommend having students work in groups on Exploratory Exercises 1 and 2 and Exercise 7
before sending them home to work on problems alone.
Exploratory Exercise Set 3.1
1. (a) An integer n is odd provided that there is an integer k with n = 2k + 1. 27 is odd because
13 is an integer with 27 = 2(13) + 1.
(b) An integer n is even provided that there is an integer k with n = 2k. 202 is even because 101
is an integer with 202= 2(101).
(c) 8
(d) Answers will vary. For example, 5 + 7 = 12.
(e) Therefore x + y = (2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1). Since m + n + 1 is an
integer, x + y is even.
(f) If we had used the same letter to represent both x and y as an odd integer (for example, x = 2k
+ 1 and y = 2k + 1), the notation would have suggested that x and y are the same number.
(g) Theorem 3.
2. Examples would include sums such as 4 + 3 = 7.
Conjecture: If x is an even integer and y is an odd integer, then x + y is an odd integer.
35
36
Chapter 3
Proof: Since x is an even integer, there is an integer m with x = 2m. Since y is an odd integer,
there is an integer n with y = 2n + 1. Therefore x + y = 2m + (2n + 1) = 2m + 2n + 1 = 2(m + n) +
1. Since m + n is an integer, x + y is odd.
This is a proof of Theorem 4.
3. (a)
n
1
2
3
4
5
6
7
8
9
f(n)
41
43
47
53
61
71
83
97
113
Factors
1, 41 1, 43 1, 47 1, 53 1, 61 1, 71 1, 83 1, 97 1, 113
Is f(1) prime?
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
(b) f(n) is a prime number for all positive integers n.
(c) f(41) = 41 · 41 – 41 + 41 = 41 · 41. Hence f(41) is not a prime number.
(d) The conjecture is not true.
(e) Even though we may find many examples that suggest the pattern represented in a conjecture,
we cannot know that the conjecture is always true until we have designed a proof of the
conjecture.
4. (a) 2, 3, 5, 7, 11, 13, 17, 19, 23, 29
(b)
p
p
Positive divisors
Is it a prime?
2 1
p
of 2  1
2
22 – 1 = 3
1, 3
Yes
3
7
1, 7
Yes
5
31
1, 31
Yes
7
127
1, 127
Yes
p
(c) If n is of the form 2  1 where p is a prime, then n is a prime number.
(d) 211  1  2047. Since 2047 = 23 · 89, 2047 is not a prime.
5. Answers will vary.
Exercise Set 3.1
1. (a) 9 is an integer with 27 = 3 · 9.
(b) 0 is an integer with 0 = 12 · 0.
(c) 4 is an integer with 16 = 4 · 4.
(d) 6 is an integer with 12 = 2 · 6.
(e) 6y is an integer with 6xy = x(6y).
(f) st is an integer with rst = r(st).
2. (a) 1, 2, 4, 8, 16
(b) 1, 2, 3, 4, 6, 8, 12, 24 (c) 1, 2, 3, 6, 7, 14, 21, 42
3. (a) There is no integer k with 7 = 0 · k.
(b) 7 is an integer with 42 = 6 · 7.
(c) The prime numbers are numbers greater than 1.
(d) The only positive divisors of 17 are 1 and 17.
(e) The only divisors of 2 are 1 and 2. 2 is a divisor of all other even numbers.
4. (a) 7 is a factor of 28.
(b) 5 is not a factor of 36.
5. (a) 28 is divisible by 7.
(b) 36 is not divisible by 5.
6. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71
7. (a) Yes, 3 | (12 + 24) since 12 + 24 = 3 · 4 + 3 · 8 = 3 · (4 + 8) = 3 · 12.
(b) Yes, 5 | (25+ 15) since 25 + 15 = 5 · 5 + 5 · 3 = 5 · (5 + 3) = 5 · 8.
(c) Answers vary; for example 7 | 14 and 7 | 35 and 7| (14 + 35).
(d) b = ak; c = am; k + m; integer (e) Theorem 7
8. Since a | b, there is an integer k with b = ak. Since a | c, there is an integer m with c = am. Since a
| d, there is an integer n with d = an. Therefore, (b + c + d) = ak + am + an = a(k + m + n). Since
k + m + n is an integer, a | (b + c + d).
9. (a) Yes, since 12 · 5 = (4 · 3) · 5 = 4 · (3 · 5) = 4 · 15.
(b) Yes, since 6 · 11 = (3 · 2) · 11 = 3 · (2 · 11) = 3 · 22.
(c) Answers vary; for example 4 | 8 and 4 | 8 · 5.
(d) b = ak; kn; a | bn
(e) Theorem 9
An Introduction to Mathematical Proof
37
10. (a) 3 | 6 and 6 | 18 and certainly 3 | 18 since 18 = 6(3).
(b) Answers will vary. For example, 11 | 22 and 22 | 66. Note that 11 | 66.
(c) Since a | b, there is an integer k with b = ak. Since b | c, there is an integer n with c = bn.
Replacing b in this last equality by ak yields c = (ak)n = a(kn). Since kn is an integer,
a | c.
11. (a) Answers vary; for example 3 · 11 = 33 which is odd.
(b) If x and y are odd integers, then xy is an odd integer.
(c) Since x is odd, there is an integer k with x = 2k + 1. Since y is odd, there is an integer j with y
= 2j + 1. Then xy = (2k + 1)(2j + 1). Using the properties of addition and multiplication of
integers, we can rewrite this as:
xy = 4kj + 2k + 2j + 1 = 2(2kj + k + j) + 1
Since 2kj + k + j is an integer, xy is an odd integer.
(d) Theorem 5
12. (a) Answers will vary. For example, 4 · 6 = 24.
(b) If x is an even integer and y is an integer, then xy is an even integer.
(c) Since x is even, there is an integer k with x = 2k. Then xy = (2k)(y). Using the properties of
addition and multiplication of integers, we can rewrite this as:
xy = 2(ky)
Since ky is an integer, xy is an even integer.
(d) Theorem 6
13. (a) 7(100) + 4(10) + 4
(b) According to the Rule of 2, since 2 | 4, 2 | 744.
(c) 744 = 2 · 372
(d) 10 = 2 · 5; Theorem 9; Theorem 8
14. (a) 6(100) + 5(10) + 4
(b) According to the Rule of 3, since 3 | (6 + 5 + 4) , 3 | 654.
(c) 654= 3 · 218
(d) a(99 + 1) +b(9 + 1) + c = a(99) + a(1) + b(9) + b(1) + c = a(99) + b(9) + (a + b + c).
99 = 3 · 33; Theorem 9; 9 = 3 · 3; Theorem 9
By Theorem 8, 3 | a(99) + b(9) + (a + b + c) and hence 3 | z.
15. z can be written in expanded notation as z = a(100) + b(10) + c.
4 | 100 since 100 = 4(25). Therefore 4 | [a(100)] by Theorem 9.
By hypothesis, 4 divides the number formed by the rightmost two digits of z, that is,
4 | [b(10) + c]. Since 4 | [a(100)] and 4 | [b(10) + c], then by Theorem 7, 4 | z.
16. z can be written in expanded notation as z = a(100) + b(10) + c.
5 | 100 since 100 = 5(20). Therefore 5 | [a(100)] by Theorem 9.
5 | 10 since 10 = 5(2). Therefore 5 | [b(10)] by Theorem 9.
If c is 0 or 5, 5 | c. Since 5 | [a(100)] and 5 | b(10) and 5 | c, then by Theorem 8, 5 | c.
17. z can be written in expanded notation as z = a(100) + b(10) + c.
100 = 99 + 1; 10 = 9 + 1.
Therefore z = a(99 + 1) +b(9 + 1) + c = a(99) + b(9) + a·1 + b·1 + c = a(99) + b(9) + (a + b + c).
9 | 99 since 99 = 9·11. Therefore 9 | [a(99)] by Theorem 9.
9 | 9 since 9 = 9 · 1. Therefore 9 | [b(9)] by Theorem 9.
9 | (a + b + c) is the hypothesis of the “Rule for 9.”
It follows by Theorem 8 that 9 divides a(99) + b(9) + (a + b + c) and hence 9 | z.
18. z can be written in expanded notation as z = a(100) + b(10) + c
= a(99 + 1) + b(11 – 1) + c
= a(99) + a + b(11) – b + c
= a(99) + b(11) + (a – b + c)
11 | 99 since 99 = 11·9. Therefore 11 | [a(99)] by Theorem 9.
11 | 11 since 11 = 11·1. Therefore 11 | [b(11)] by Theorem 9.
11 | (a – b + c) is the hypothesis of the “Rule for 11.”
It follows by Theorem 8 that 11 divides a(99) + b(11) + (a – b + c) and hence 11| z.
38
Chapter 3
19. (a) We would write the integers in expanded form as a(1000) + b(100) + c(10) + d and
then make the arguments looking at the divisibility properties of this decomposition.
(b) If a | b, a | c, a | d, and a | e, then a | (b + c + d + e).
(c) Since a | b, a | c, a | d, and a | e, there are integers j, k, l, and m with b = aj, c = ak, d = al,
and e = am. Thus b + c + d + e = aj + ak + al + am = a (j + k + l + m). Since j + k + l + m
is an integer, a | (b + c + d + e).
(d) z can be written in expanded notation as z = a(1000) + b(100) + c(10) + d.
2 | 1000 since 1000 = 2(500). Therefore 2 | [a(1000)] by Theorem 9. 2 | 100 since
100 = 2(50). Therefore 2 | [b(100)] by Theorem 9. 2 | 10 since 10 = 2(5).
Therefore 2 | [c(10)] by Theorem 9. “2 | d” is the hypothesis of the “Rule for 2.”
Since 2 | [a(1000)] and 2 | [b(100)] and 2 | [c(10)] and 2 | d, then by (c) of this
exercise, 2 | z.
20. (a) a | a because a = a · 1. Thus | is reflexive.
(b) In Exercise 10 we proved that if a | b and b | c, then a | c. This is the statement of the
transitive property for “divides.”
(c) 6 divides 12 but 12 does not divide 6.
(d) Since a | b there is an integer k with a = bk. Since b | a there is an integer n with b = an.
Replacing a by bk in the second equation yields b = bkn. Thus kn = 1 meaning k = n = 1 or k
= n = – 1. Since a and b are positive, the desired result follows.
Teaching Notes, Section 3.2: Depending on the level of preparation of your students, you may
want to do much of the work of this section in groups. At the very least, have students work on
Exploratory Exercises 1 and 2 before sending them home to work on problems alone.
The collection of exercises, Exercises 11 through 15, provide the basis of a good group or
individual project to summarize the material learned in Sections 3.1 and 3.2.
Exploratory Exercise Set 3.2
1. (a) A  (B  C) = {s, t} and (A  B)  (A  C) = {s, t}
(b) Find in text.
(c) (A  B)  (A  C)  A  (B  C); Let a  A  (B  C); Let a  (A  B)  (A  C).
(d) Definition of intersection; definition of union; Looking back at the two previous
statements we see that a is definitely an element of A and it is also in either B or C; a is an
element of either (A  B) or (A  C) and hence by definition is in the union of these two sets.
Each element of A  (B  C) is an element of (A  B)  (A  C).
(e) Let a  (A  B)  (A  C). Then by the definition of union, a  (A  B) or a  (A  C). If a
 (A  B), then a  A and a  B. Similarly, if a  (A  C), then a  A and a  C. In either
case, a  A. Further, a  B or a  C. By the definitions of intersection and union,
a  A  (B  C).
(f) Since A  (B  C)  (A  B)  (A  C) and (A  B)  (A  C)  A  (B  C), the two sets
are equal.
2. (a) 2 + ; Answers will vary. For example 1 + 2
(b) Find in text.
(c) Suppose x is a rational number and y is an irrational number and x + y is a rational number.
Since y is an irrational number, it by definition cannot be written as quotient of two integers.
(d)
m k m
m
; ,
; Subtracting
from both sides of the equation yields
n l n
n
An Introduction to Mathematical Proof
y=
39
k
m nk  ml

=
. Since nk – ml is an integer and nl is a non-zero integer, this is a
l
nl
n
contradiction of the hypothesis that y is irrational.
(e) We assumed that the negation of this statement is true and this led to a contradiction.
Therefore the statement itself must be true.
3. (a) (1) 1. 3 | (1 – 4) and 3 | (4 – 1)
2. 3 | (5 – 2) and 3 | (2 – 5)
3. 3 | (0 – 6) and 3 | (6 – 6)
(2) Find in text.
(3) Reflexive, symmetric, transitive
(b) (1) 3 | (a – a)
(2) b  a; (a – b) = 3k; Therefore – (a – b) = – 3k or (b – a) = 3(–k). Thus 3 | (b – a) or
b  a.
(3) a  c; (a – b) = 3k; 3 | (b – c); (b – c) = 3l; Thus (a – b) + (b – c) = 3k + 3l or
(a – c) = 3k + 3l = 3(k + l). Thus 3 | (a – c) or a  c.
Exercise Set 3.2
1. (a) {v, w}
(b) Find in this section.
(c) v and w are the two elements in A  B, and each of these elements is also in A.
2. (a) Let x  A  B. We will show that this means x  A.
(b) (1) A, B
(2) A
3. (a) A  B = {1, 2, 3, 5}. Since every element of A is an element of A  B, A  A  B.
(b) See Preparation for Theorems 10 and 11.
(c) The first sentence in our proof will be, “Let a be an arbitrary element of A.” We will then try
to show that a must be an element of A  B.
(d) Let a be an arbitrary element of A. By the definition of A  B, a is also an element of
A  B. Thus A  A  B.
4. First do preparation similar to that done prior to the proof of Theorem 11. Choose sets A and B
and work through an example, review the definitions of intersections, unions and complements of
sets, and reflect on a strategy for showing two sets are equal.
Show A  B  A  B . Let x A  B . Then x A  B . Said another way, “it is not true that
xA and xB.” We can rewrite this statement using DeMorgan’s Laws for the negation of
conjunction as “x  A or x B.” Thus x A or x  B . By definition of union, x  A  B .
Show A  B  A  B . Let x  A  B . Then x  A or x  B . Said in words, “x is not an
element of A or x is not an element of B.” By DeMorgan’s Law for negation of conjunction, this
statement can be rewritten as “It is not true that x is an element of A and x is an element of B.”
x  A  B. x A  B .
Since we have now shown set containment in both directions, we have shown that the two sets are
equal.
5. First do preparation similar to that done in Exploratory Exercise 1. Choose sets A, B, and C and
work through an example, review the definitions of intersection and union of sets and reflect on a
strategy for showing two sets are equal.
40
Chapter 3
Show that A  (B  C)  (A  B)  (A  C).
Let a  A  (B  C). Then, by definition of union, a  A or a  B  C. If a  A then the
definition of union ensures that a is an element of both (A  B) and (A  C). If a  B  C,
then by definition of intersection, a  B and a  C. It follows that in either case a is an
element of both (A  B) and (A  C). Thus a  (A  B)  (A  C) and we have proved
A  (B  C)  (A  B)  (A  C).
The proof can then be finished by showing that (A  B)  (A  C)  A  (B  C) using a similar
argument.
6. First do preparation similar to that done in Exploratory Exercise 2. Look at several products of
the form (rational number)(irrational number), review the definitions of rational and irrational
numbers and review the way one formulates a proof by contradiction.
Suppose x is a non-zero rational number, y is an irrational number and xy is a rational number.
m
. Further, since x is non-zero, m
n
k
 0. Since we are supposing xy is rational, there are integers k and l (l  0) with xy = . By
l
k
m
m
nk
replacing x with
, we see that
y=
or y =
. Since nk is an integer and ml is a non-zero
l
n
n
ml
Since x is rational, there are integers m and n (n  0) with x =
integer, this result contradicts our assumption that y is irrational.
j
m
and let y =
where j, k, m, and n are integers and k and n are
n
k
nj  km
not equal to 0. Then x + y =
. Since nj  km and kn are integers and
kn
7. (a) Let x =
kn is not equal to 0, then x + y is a rational number.
j
m
and let y =
where j, k, m, and n are integers and k and n are
n
k
jm
not equal to 0. Then xy =
. Since jm and kn are integers and kn is not
kn
(b) Let x =
equal to 0, then xy is a rational number.
(c) False. Let x =  and let y = . Both  and  are irrational numbers, but
x + y = 0 which is a rational number.
(d) False. Let x = 2 and let y =  2 . Both 2 and  2 are irrational
numbers, but xy = 2 which is a rational number.
8. Since d | b then by Theorem 9, d | bq. Since d | bq and d | r, then by Theorem 7, d | bq + r.
9. (a) x – y = x + (–1)y. Since d | y, then, by Theorem 9, d | (–1)y. It follows from
Theorem 7 that if d | x and d | (–1)y, then d | (x + (–1) y) or d | x – y.
(b) Since a = bq + r, then r = a – bq. Since d | a and d | b, d | bq by Theorem 9 and
d | r by the result in (a).
10. (a) If n is not even, then n2 is not even. Alternatively, If n is odd, then n2 is odd.
(b) Since n is odd, then by Theorem 5, n · n = n2 is odd.
11. (a) 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 36; 3 · 6 = 18, 3 · 9 = 27, 3 · 12 = 36, 9 · 12 =
108. In each case the product of two smooth integers is smooth.
(b) If x and y are smooth, then xy is smooth.
(c) Since x is smooth, there is an integer k such that x = 3k. Since y is also smooth,
An Introduction to Mathematical Proof
41
there is an integer j with y = 3j. Then xy = (3k)(3j) = 9kj = 3(3kj). Since 3kj is an integer, xy
is three times an integer, and hence xy is smooth.
12. (a) Answers will vary. For example, 7 = 2(3) + 1 and 13 = 4(3) + 1 are rough integers and since
(7)(13) = 91 = 3(30) + 1, their product is rough.
(b) If x and y are rough integers, xy is rough.
(c) Since x is rough, there is an integer k such that x = 3k + 1. Since y is also
rough, there is an integer j with y = 3j+ 1. Then xy = (3k + 1)(3j + 1) = 9kj + 3k + 3j + 1 =
3 (3kj + k + j) + 1. Since 3kj + k + j is an integer, xy is rough.
13. (a) 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35; 5·8 = 40 which is rough since 40 = 13(3)
+ 1. 8·11 = 88 which is rough since 88 = 3(29) + 1.
(b) If x and y are abrasive, then xy is rough.
(c) Since x is abrasive, there is an integer k such that x = 3k + 2. Since y is also abrasive, there is
an integer j with y = 3j+ 2. Then xy = (3k + 2)(3j + 2) = 9kj + 6k + 6j + 4 = 9kj + 6k + 6j + 3
+ 1 = 3(3kj + 2k + 2j + 1) + 1. Since 3kj + 2k + 2j + 1 is an integer, xy is rough.
14. By way of contradiction, suppose that a2 is smooth and a is not smooth. Since a is not smooth, a
must be either rough or abrasive.
Case 1: Suppose that a is rough. By Exercise 12, a · a = a2 is rough.
Case 2: Suppose that a is abrasive. By Exercise 13, a · a = a2 is rough.
Therefore it is not possible for a2 to be smooth and a not be smooth. We have a contradiction.
3 and x is rational. Then there are integers m and n (n  0) with
m
x= 3 =
n
m
such that m and n have no common factors other than 1. Since 3 =
, we can square both
n
15. Suppose x =
sides of the equation to learn that
3=
m2
or 3n2 = m2
n2
Since m2 is the product of 3 and the integer n2, m2 is smooth. By Exercise 14 this means that m is
smooth so there is an integer k with m = 3k. Thus m2 = (3k)2 = 9k2. Hence 3n2 = 9k2 and
dividing by 3 reveals that n2 = 3k2. Using Exercise 14 once again indicates that since n2 is
smooth, n is smooth. Now we have shown that both m and n are smooth, that is, 3 is a divisor or
factor of both m and n. This contradicts our assumption that m and n share no common positive
divisors other than 1.
16. (1) For all integers a, a  a(mod 5) since 5 | (a – a). Thus the relation is reflexive.
(2) If a  b(mod 5), then 5 | (a – b). Thus there is an integer k with (a – b) = 5k. Therefore
– (a – b) = – 5k or (b – a) = 5(–k). Thus 5 | (b – a) or b  a (mod 5). Thus the relation is
symmetric.
(3) If a  b(mod 5) then 5 | (a – b). Thus there is an integer k with (a – b) = 5k. Similarly, if 5 |
(b – c) there is an integer l with (b – c) = 5l; Thus (a – b) + (b – c) = 5k + 5l or (a – c) = 5k +
5l = 5(k + l). It follows that 5 | (a – c) and that a  c(mod 5). Thus the relation is transitive.
17. Since n2 is even, by Theorem 12, n is even. Thus there is an integer k with n = 2k. It
follows that n2 = (2k)( 2k) = 4k2, and n2 is divisible by 4.
Teaching Notes, Section 3.3: Again, we encourage group work on Exploratory Exercises 1
through 3 before asking students to attempt proofs on their own.
After having done one or two induction proofs, students have a tendency to address those proofs
mechanically without much thought about what is actually being accomplished. Careful attention to
Exploratory Exercise 5 and Exercises 18 and 19 are a useful antidote to this tendency.
42
Chapter 3
Exploratory Exercise Set 3.3
1. (a) 16; 25; 36
(b) n2
2. (a) 1 + 3 + 5 +…+ (2n 1) = n2
(b) P(1): 1 = 12 or 1 = 1 which is true.
P(2): 1 + 3 = 22 or 1 + 3 = 4 which is true.
P(3): 1 + 3 + 5 = 32 or 1 + 3 + 5 = 9 which is true.
(c) P(k): 1 + 3 + 5+…+ (2k 1) = k2
P(k + 1): 1 + 3 + 5+…+ (2k 1) + (2(k + 1) – 1) = (k + 1)2 or
1 + 3 + 5 + … + (2k  1) + (2k + 1) = (k + 1)2
(d) Basis Step: P(1) is the statement that 1 = 12 or 1 = 1 which is true.
(e) (k + 1)2
(f) We have shown that P(n) is true for n = 1, and we have shown that for each positive integer k,
when P(k) is true, P(k + 1) is true. By the Principle of Induction, P(n) is true for all positive
integers n.
n
n
3. (a) 4 divides 6  2 .
(b) Find in text.
(c) P(1): 4 | (6 – 2) or 4 | 4 which is true.
P(2): 4 | (62 – 22) or 4 | 32 which is true.
P(3): 4 | (63 – 23) or 4 | 208 which is true.
(d) P(k): 4 divides 6 k  2 k . P(k + 1): 4 divides 6 k 1  2 k 1 .
6 k 1  2 k 1 = 6 · 6k – 2k+1 = (4 + 2) 6k – 2k+1 = 4 · 6k + 2 · 6k – 2 · 2k = 4 · 6k + 2 (6k – 2k).
(e) Basis Step: P(1) is the statement 4 | (6 – 2) or 4 | 4 which is true.
(f) In (d) we see that 6 k 1  2 k 1 = 4 · 6k + 2(6k – 2k). By definition of “divides,” 4 divides 4 · 6k
and by induction hypothesis, 4 | (6k – 2k). Theorem 9 then asserts that 4 | 2(6k – 2k) and
Theorem 7 then allows us to conclude that 4 | 6 k 1  2 k 1 .
4. (a)
10(11)
= 55
2
(b) S = 1 + 2 + 3 + … + 98 + 99 + 100
S = 100 + 99 + 98 + … + 3 + 2 + 1
2S = 101 + 101 + 101 + …. + 101+ 101 +101
Since the right side of the equation is the sum of 100 copies of 101,
2S = 100(101) or S = 5050.
(c) S = 1 +
2 +
3 + … + (n – 2) + (n – 1) + n
S = n + (n – 1) + (n – 1) + … +
3 +
2 +
1
2S = (n + 1) + (n + 1) + (n + 1) + …. + (n + 1) + (n + 1) + (n + 1)
Since the right side of the equation is the sum of n copies of (n + 1),
2S = n(n + 1) or S =
n(n  1)
.
2
5. Answers will vary.
Exercise Set 3.3
1.
Basis Step: 3 =
3(1)(1  1)
.
2
Inductive Step: We wish to prove the statement “If P(k) and k  1, then P(k + 1).”
An Introduction to Mathematical Proof
43
Our inductive hypothesis, then, is the statement P(k): 3 + 6 + 9 + …+ 3k =
3k ( k  1)
.
2
Add 3(k + 1) to both sides of the equation given in the inductive hypothesis and then perform
algebraic manipulation on the right hand side of the resulting equation with these results.
3k ( k  1)
+ 3(k + 1)
2
3k ( k  1) 6( k  1)
3 + 6 + 9 + …+ 3k + 3(k + 1) =
+
2
2
3 + 6 + 9 + …+ 3k + 3(k + 1) =
3 + 6 + 9 + …+ 3k + 3(k + 1) =
(k  1)(3k  6) 3(k  1)( k  2)
=
2
2
Since this last statement is P(k + 1), we have shown that when P(k) is true, P(k + 1) is true.
1(1  1)(2  1  1) 6
2. Basis Step: 12 =
= =1
6
6
Inductive Step: We wish to prove the statement “If P(k) and k  1, then P(k + 1)”
Our hypothesis, then, is the statement P(k): 12  2 2  3 2    k 2 
k (k  1)( 2k  1)
.
6
Add (k + 1)2 to both sides of the equation given in the inductive hypothesis and then perform
algebraic manipulation on the right hand side of the resulting equation with these results.
12  2 2  3 2    k 2  (k  1) 2 
k (k  1)( 2k  1)
 (k  1) 2
6
(k  1)[ k (2k  1)  6(k  1)]
6
(k  1)[( k  2)( 2k  3)]
12  2 2  3 2    k 2  (k  1) 2 
6
(k  1)[(( k  1)  1)( 2(k  1)  1)]
12  2 2  3 2    k 2  (k  1) 2 
6
12  2 2  3 2    k 2  (k  1) 2 
Since this last statement is P(k + 1), we have shown that when P(k) is true, P(k + 1) is true.
1(1  1) 
Basis Step: 1 = 
.
 2 
2
3.
3
Inductive Step: We wish to prove the statement “If P(k) and k  1, then P(k + 1)”
 k (k  1) 
Our inductive hypothesis, then, is the statement P(k): 1  2  3    k  
 2 
3
3
3
2
3
Add (k + 1)3 to both sides of the equation given in the inductive hypothesis and then perform
algebraic manipulation on the right hand side of the resulting equation with these results.
 k (k  1) 
3
1  2  3    k  (k  1)  
  (k  1)
2


2
2
k (k  1)
4(k  1) 3
13  23  33    k 3  (k  1) 3 

4
4
2
3
3
3
3
3
44
Chapter 3
(k  1) 2 (k 2  4(k  1))
4
2
2
(k  1) (k  4k  4)
(k  1) 2 (k  2) 2
13  2 3  33    k 3  (k  1) 3 

4
4
13  2 3  33    k 3  (k  1) 3 
 (k  1)( k  2) 
1  2  3    k  (k  1)  

2


3
3
3
3
2
3
Since this last statement is P(k + 1), we have shown that when P(k) is true, P(k + 1) is true.
4. Basis Step: 1 =
1(3(1)  1)
= 1.
2
Inductive Step: We wish to prove the statement “If P(k) and k  1, then P(k + 1).”
P(k) is the statement, 1 + 4 + 7 + …+ (3k – 2) =
k (3k  1)
. Add (3(k +1) – 2) = 3k + 1 to
2
both sides of this equation and manipulate the results algebraically.
1 + 4 + 7 + …+ (3k – 2) + (3k + 1) =
k (3k  1)
+ (3k + 1)
2
1 + 4 + 7 + …+ (3k – 2) + (3k + 1) =
k (3k  1)  2(3k  1)
2
3k 2  5k  2
2
(k  1)(3k  2) (k  1)(3(k  1)  1)

1 + 4 + 7 + …+ (3k – 2) + (3k + 1) =
2
2
1 + 4 + 7 + …+ (3k – 2) + (3k + 1) =
Since this last statement is P(k + 1), we have shown that when P(k) is true, P(k + 1) is true.
5. Basis Step: 1 < 21 = 2.
Inductive Step: We wish to prove the statement “If P(k) and k  1, then P(k + 1).”
P(k) is the statement, k < 2k. Multiplying both sides of this statement by 2 yields
2k < 2·2k = 2k + 1
Since k  1, k + 1  k + k = 2k so k + 1  2k < 2k + 1. Since k + 1 < 2k + 1 is P(k + 1), we are
done.
6. Basis Step: P(1) is the statement 5| (9 – 4) or 5 | 5 which is true.
9 k 1  4 k 1  9  9 k  4  4 k  (5  4)9 k  4  4 k  5  9 k  4(9 k  4 k ) . By definition of
“divides,” 5 divides 5 · 9k and by induction hypothesis, 5 | (9k – 4k). Theorem 9 then asserts that
5 | 4(9k – 4k) and Theorem 7 then allows us to conclude that 5 | 9 k 1  4 k 1 .
7. Basis Step: Since 31  1 = 2, 2 | (31  1).
Inductive Step: Assume that k is a natural number and 2 divides 3 k  1 . Show this
ensures that 2 divides 3k 1  1 . We observe that
3k 1  1
= 3  3k  1
= (2  1)  3k  1
= 2  3k  3k  1
= 2  3k  (3k  1)
An Introduction to Mathematical Proof
45
Now we have the connection with P(k) that we wished. By P(k), the inductive hypothesis, 2
divides 3 k  1 . Theorem 7 ensures that 2 divides 3k 1  1 .
8. Basis Step: 3 | (13 – 1), since 3 | 0.
Inductive Step: Assume that k is a natural number and 3 divides k3 – k. Show this
ensures that 3 divides (k + 1)3 – (k + 1). Observe that
(k + 1)3 – (k + 1) = k3 + 3k + 3k + 1 – k – 1
= k3 + 5k
We have to somehow connect this result to the inductive hypothesis, 3 | k3 – k. Observe that k3 +
5k = k3 – k + 6k. Since 3 | 6 by Theorem 9, 3 | 6k. By Theorem 7 together with the inductive
hypothesis, 3 | (k + 1)3 – (k + 1).
1
1 1
1
2 1
1
1
3
 ;

 ;



1 2 2 1 2 2  3 3 1 2 2  3 3  4 4
1
1
1
1
n
Conjecture:




1 2 2  3 3  4
n(n  1) n  1
1
1
 .
(b) Basis Step:
1 2 2
9. (a)
Inductive Step: Assume that k is a natural number and
1
1
1
1
k
1
. Adding
to both sides of this




1 2 2  3 3  4
k (k  1) k  1
(k  1)( k  2)
equation yields
1
1
1
1
1
k
1






1 2 2  3 3  4
k (k  1) (k  1)( k  2) k  1 (k  1)( k  2)
k (k  2)
1
=

(k  1)( k  2) (k  1)( k  2)
k 2  2k  1
(k  1)( k  2)
(k  1) 2
(k  1)
=
=
(k  1)( k  2) (k  2)
=
Since this last statement is P(k + 1), we have shown that when P(k) is true, P(k + 1) is true.
10. n3 – n = n(n2 –1) = n(n –1)(n + 1). Since (n –1), n, (n + 1) are three consecutive integers, one of
them is divisible by 3. By Theorem 9, n(n –1)(n + 1) is divisible by 3 so n3 – n is divisible by 3.
11. Basis Step: If a | b1 then a certainly divides b1.
Inductive Step: Assume that k is a natural number and that if a | b1, a | b2, a | b3 ,…,
a | bk, then a | (b1 + b2 + … + bk).
If a | b1, a | b2, a | b3, …, a | bk, a | bk + 1, then the inductive hypothesis asserts that a |
(b1 + b2 + … + bk). Observe that b1 + b2 + … + bk + bk + 1 = (b1 + b2 + … + bk) + bk+1.
Using the fact that a | bk+1 and Theorem 7 we see that a | [(b1 + b2 + … + bk) + bk+1]
or a | (b1 + b2 + … + bk + bk + 1).
12. Basis Step: When n = 1, the number consists of a single digit a. The number
represented by the right digit of a single digit number is that number. If 2 | a, then certainly 2 | a.
Inductive Step: Suppose k is a positive integer and the “Rule for 2” is true for
numbers with k-digits. Consider any positive integer z with k + 1 digits, and suppose
that 2 divides the number formed by the rightmost digit of z. Observe that z can
be written in expanded form as z = a(10)k + x where x is an integer with k digits.
The rightmost digit of z is the same as the right-most digit of x. By the inductive hypothesis,
2 | x. Further, since k > 1, 2 | (10)k. Using Theorems 7 and 9, it follows that 2 | z.
46
Chapter 3
13. Basis Step: When n = 1, the number consists of a single digit a. The number
represented by the right-most two digits of a single digit number is that number. If
4 | a, then certainly 4 | a.
Inductive Step: Suppose k is a positive integer and the “Rule for 4” is true for
numbers with k-digits. Consider any positive integer z with k + 1 digits, and suppose
that 4 divides the number formed by the rightmost two digits of z. Observe that z can
be written in expanded form as z = a(10)k + x where x is an integer with k digits. If
k = 1, then z is a number with two digits so 4 | z. If k > 1, then the number formed by
the rightmost two digits of z is the same as the number formed by the rightmost two
digits of x. By the inductive hypothesis, 4 | x. Further, since k > 1, 4 | (10)k. Using
Theorems 7 and 9, it follows that 4 | z.
14. Basis Step: When n = 1, the number consists of a single digit a. If a = 0 or a = 5, then 5 | a.
Inductive Step: Suppose k is a positive integer and the “Rule for 5” is true for
numbers with k-digits. Consider any positive integer z with k + 1 digits, and suppose
that the right-most digit of z is 0 or 5. Observe that z can be written in expanded form as z =
a(10)k + x where x is an integer with k digits. If the right-most digit of z is 0 or 5 then the rightmost digit of x is 0 or 5. By the inductive hypothesis, 5 | x. Further, since k > 1, 5 | (10)k. Using
Theorems 7 and 9, it follows that 5 | z.
15. If in our arrangement of dominos we make sure that (a) the rth domino must fall and
(b) that for each k  r, when the kth domino falls, the (k + 1)st domino must fall, we
can be sure that all the dominos from the rth domino onward will fall.
16. Basis Step: 2(3) + 1 < 23 since 7 < 8.
Inductive Step: Suppose k is an integer  3 and that 2k + 1 < 2k. We need to show that
2(k+ 1) + 1 < 2k + 1 or that 2k + 3 < 2k + 1. Multiply both sides of the inductive hypothesis by 2 to
get 2(2k + 1) < 2(2k) = 2k + 1. Observe that 2(2k+1) can be written as 2k + 3+ (2k – 1) so we have
2k + 3+ (2k – 1) < 2k + 1
Since k  3, (2k – 1) > 0 so 2k + 3 < 2k + 1.
2
17. Basis Step: 5 = 25 < 32 = 25.
Inductive Step: Suppose k is an integer  5 and that k2 < 2k. We need to show that (k + 1)2 < 2k+1.
Let us focus on the left-hand side of this desired inequality. (k + 1)2 = k2 + 2k + 1. We know
something about k2 so let us focus attention on 2k + 1. Since k ≥ 5, 2k + 1  2k + k ≤ 3k < 5k.
Since k  5, k2 = k · k  5k. It follows that 2k + 1 < k2. Thus, putting all this together with the
inductive hypothesis, we have
(k + 1)2 = k2 + 2k + 1 < k2 + k2 = 2k2 < 2 · 2k = 2k + 1
18. (a) We can form postage of 8 cents with one 3-cent stamp and one 5-cent stamp.
(b) (1) Replace one of the 5-cent stamps used in forming postage for k cents by two 3-cent
stamps. This will produce postage for k + 1 cents.
(2) Suppose that in forming k cents of postage we use no 5-cent stamps. Since k
 8, we must have used at least three 3-cent stamps. Replace three 3-cent stamps by
two 5-cent stamps to form postage for k + 1 cents.
19. Basis Step: We can produce 12 cents in postage using three 4-cent stamps.
Inductive Step: Let k  12 be an amount of postage that we can form using 4-cent and
5-cent stamps.
Case 1: Suppose that in forming k cents of postage we use at least one 4-cent stamp.
Then if we replace this 4-cent stamp by a 5-cent stamp, we will form k + 1 cents in
postage.
Case 2: Suppose that in forming k cents of postage we use no 4-cent stamps. Since k
 12, we must have used at least three 5-cent stamps. Replace three 5-cent stamps by
four 4-cent stamps to form postage for k + 1 cents.
An Introduction to Mathematical Proof
47
Teaching Notes, Section 3.4: Because the notions of iteration and recursion are among the three
topics that the Principles and Standards for School Mathematics identify as being key topics in the
K – 12 curriculum, we focus on the iterative properties of each of the algorithms we examine in this
text. This leads to a somewhat different presentation of the Euclidean algorithm in this section and
the Sieve of Eratosthenes in Section 3.5. However, experience with this presentation will serve
students well when they look at some of the iterative algorithms of graph theory.
Exploratory Exercises 1 – 4 are important in helping students understand the significance of
Theorem 20 to the development of number theory.
Related Website: Visit this web site for links to a broad range of topics in number theory, both
elementary and sophisticated :http://archives.math.utk.edu/topics/numberTheory.html
Exploratory Exercise Set 3.4
1. (a) (1)
(2)
5
3
9
3
3
3
4
(3)
(4)
5
5
3
13
(b) (1) 3(3) + 0(5) = 9
(3) 1(3) + 2(5) = 13
(c) All potatoes
2. (a) (1)
7
3
3
5
5
3
(2) 3(3) + (–1) 5 = 4
(4) (–1)(3) + 2(5) = 7
(2)
9
12
6
6
21
6
6
48
Chapter 3
(3)
Not possible
(4)
6
6
(5)
9
33
6
9
9
9
9
6
(6)
Not Possible
(b) (1) 2(6) + 0(9)= 12
(2) 2(6) + 1(9) = 21
(3) Not possible
(4) 1(6) + 0(9) = 6
(5) (–2)(6) + 5(9) = 33 (6) Not possible
(c) Potatoes whose weights in ounces are a multiple of 3.
3. (a) (1)
(2)
5
17
5
7
7
3
5
5
(3)
1
7
7
5
5
5
(b) (1) Since 1 = 3(5) + (–2) 7, 2 = 6(5) + (–4)(7). Thus we can weigh a 2-ounce potato by
placing four 7-ounce weights on the pan with the potato and six 5-ounce weights on the
other pan.
(2) Since 1 = 3(5) + (–2) 7, 4 = 12(5) + (–8)(7). Thus we can weigh a 4-ounce potato by
placing eight 7-ounce weights on the pan with the potato and twelve 5-ounce weights on
the other pan.
(3) Since 1 = 3(5) + (–2) 7, 9 = 27(5) + (–18)(7). Thus we can weigh a 9-ounce potato by
placing eighteen 7-ounce weights on the pan with the potato and twenty-seven 5-ounce
weights on the other pan.
(4) Since 1 = 3(5) + (–2) 7, n = (3n)(5) + (–2n)(7). Thus we can weigh a n-ounce potato by
placing (2n) 7-ounce weights on the pan with the potato and (3n) 5-ounce weights on the
other pan.
An Introduction to Mathematical Proof
49
4. Answers will vary. Discussion should include the fact that the integer coefficients guaranteed by
Theorem 20 describe how to distribute weights on the scale to weigh a potato of d ounces.
Further, if n = kd is a multiple of d, we can multiply the equation from Theorem 20 by k to learn
how to weigh a potato weighing n ounces. Since every positive integer is a multiple of 1, the same
reasoning ensures that if gcd(a, b) = 1, we can weigh all potatoes that weigh an integer number of
ounces.
Exercise Set 3.4
1. (a) (1) 1, 2, 3, 6; 1, 3, 5, 15
(2) 1, 3
(3) 3
(b) (1) 1, 2, 3, 4, 6, 12; 1, 2, 3, 6, 9, 18, 27, 54 (2) 1, 2, 3, 6 (3) 6
(c) (1) 1, 3, 7, 21; 1, 3, 7, 9, 21, 63 (2) 1, 3, 7, 21
(3) 21
(d) (1) 1, 2, 4, 7, 14, 28; 1, 2, 3, 6, 9, 18, 27, 54 (2) 1, 2 (3) 2
(e) (1) 1, 17; 1, 2, 3, 4, 6, 9, 12, 18, 36 (2) 1 (3) 1
2. (a) If the groups are to have equal sizes, the number of students in the group must be a divisor of
the number of students in the class. We want the largest divisor that divides both 21 and 35.
Each group should contain 7 children.
(b) Since each basket must contain the same number of apples, the number of baskets must divide
18. Similarly, the number of baskets must divide 24. We want the largest integer that divides
18 and 24. There are 6 baskets.
(c) In each case we are looking for the greatest common divisor of two integers.
3. (a)
Equation
A
B
R
(1) 62 = 4(14) + 6
62
14
6
(2) 14 = 2(6) + 2
14
6
2
(3) 6 = 3(2) + 0
6
2
0
gcd(62, 14) = 2.
(b) gcd(62, 14) = gcd(14, 6) = gcd(6, 2) = gcd(2, 0) = 2
4. (a) gcd (82, 36) = 2
Equation
A
B
R
82 = 2(36) + 10
82
36
10
36 = 3(10) + 6
36
10
6
10 = 1(6) + 4
10
6
4
6 = 1(4) + 2
6
4
2
4 = 2(2) + 0
4
2
0
(b) gcd(82, 36) = gcd(36, 10) = gcd(10, 6) = gcd(6, 4) = gcd(4, 2) = 2.
5. (a) 15 = 2(6) + 3; 6 = 2(3) + 0, gcd(6, 15) = 3
(b) 54 = 4(12) + 6; 12 = 2(6) + 0, gcd(12, 54) = 6
(c) 63 = 3(21) + 0, gcd(21, 63) = 21
(d) 54 = 1(28) + 26; 28 = 1(26) + 2; 26 = 13(2) + 0; gcd(28, 54) = 2
(e) 36 = 2(17) + 2; 17 = 8(2) + 1; 2 = 2(1) + 0; gcd(17, 36) = 1
6. (a)
Equation
A
B
R
180 = 1(144) + 36
180
144
36
144 = 4(36) + 0
144
36
0
gcd(180, 144) = 36
50
Chapter 3
(b)
Equation
315 = 1(220) + 95
220 = 2(95) + 30
95 = 3(30) + 5
30 = 6(5) + 0
A
315
220
95
30
B
220
95
30
5
R
95
30
5
0
gcd(315, 220) = 5
(c)
Equation
A
B
R
4576 = 1(3360) + 1216
4576 3360 1216
3360 = 2(1216) + 928
3360 1216
928
1216 = 1(928) + 288
1216
928
288
928 = 3(288) + 64
928
288
64
288 = 4(64) +32
288
64
32
64 = 2(32) + 0
64
32
0
gcd(4576, 3360) = 32
7. 2 = 9(14) + (2)(62)
Equation
Solve for remainders
Substitute the remainder from
(1) into *
(1) 62 = 4(14) + 6
62 – 4(14) = 6
(2) 14 = 2(6) + 2
14 – 2(6) = 2*
14 – 2(62 – 4(14)) = 2
9(14) + (– 2)(62) = 2
8.
Equation
(1) 82 = 2(36) + 10
(2) 36 = 3(10) + 6
(3) 10 = 1(6) + 4
(4) 6 = 1(4) + 2
Solve for
remainders
82 – 2(36) = 10
36 – 3(10) = 6
10 – 1(6) = 4
6 – 1(4) = 2
Substitute the
remainder from
(3) into …
Substitute the
remainder from
(2) into …
6–1(10 –1(6)) = 2 2(36 – 3(10)) +
2(6) +(–1)(10) =2 (–1)(10) =2
2(36) + (– 7)(10)
=2
16(36) + (–7)(82) = 2
9. (a) 3 = (1)(15) + (2)(6)
Equation
15 = 2(6) + 3;
(b) 6 = (1)(54) + (4)(12)
Equation
54 = 4(12) + 6
(c) 21 = (0)(63) + (1)(21)
(d) 2 = (2)(28) + (1)(54)
Solve for remainders
1(15) + (2)(6) = 3
Solve for remainders
(1)(54) + (4)(12) = 6
Substitute the
remainder from
(1) into …
2(36) + (– 7)( 82
– 2(36)) = 2
16(36) + (–7)(82)
=2
An Introduction to Mathematical Proof
51
Equation
Solve for remainders
(1) 54 = 1(28) + 26
(2) 28 = 1(26) + 2
54  1(28) = 26
28  1(26) = 2*
(e) 1 = (17)(17) + (8)(36)
Equation
Solve for remainders
Substitute the remainder from
(1) into *
28  1(54  1(28)) = 2
2(28) + (1)(54) = 2
Substitute the remainder from
(1) into *
36  2(17) = 2
17  8(2) = 1*
(1) 36 = 2(17) + 2
(2) 17 = 8(2) + 1
17  8(36  2(17)) = 1
17(17) + (8)(36) = 1
10. gcd(2, 3) = 1 = 1(3) + (1)(2) = 3(3) + (4)(2) = 5(2) + (3)(3) [Answers will vary.]
11. (a) Multiples of 8: 8, 16, 24, 32, 40, …; Multiples of 12: 12, 24, 36, 48, …; lcm(8, 12) = 24
(b) Multiples of 6: 6, 12, 18, 24, 30, 36, …; Multiples of 15: 15, 30, 45, 60, …; lcm(6, 15) =
30
(c) Multiples of 18: 18, 36, 54, 72, …; Multiples of 27: 27, 54, 81, …; lcm(18, 27) = 54
(d) Multiples of 16: 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224, 240, 256, 272,
288, 304, 320, 336, 352, …; Multiples of 42: 42, 84, 126, 168, 210, 252, 294, 336, …;
lcm(16, 42) = 336.
12.
a
8
6
18
16
b
12
15
27
42
gcd(a, b)
4
3
9
2
lcm(a, b)
24
30
54
236
ab
96
90
486
472
Conjecture: gcd(a, b) · lcm(a, b) = a  b.
Since 6 · 15 = 90 and gcd(6, 15) = 3, lcm(6, 15) = 90/3 = 30.
Since 12 · 54 = 648 and gcd(12, 54) = 6, lcm(12, 54) = 648/6 = 108.
Since 21 · 63 = 1323 and gcd(21, 63) = 21, lcm(21, 63) = 1323/21 = 63.
Since 28 · 54 = 1512 and gcd(28, 54) = 2, lcm(28, 54) = 1512/2 = 756.
Since 17 · 36 = 612 and gcd(17, 36) = 1, lcm(17, 36) = 612/1 = 612.
The cyclist returns to the starting point on multiples of 4: 4, 8, 12, …. The walker returns to
the starting point on multiples of 10: 10, 20, 30, …. The first time they will be at the starting
point together will be the least common multiple of 4 and 10. lcm(4, 10) = 20.
(b) In dealing with cake 1, the cook has the option of dividing each piece into two smaller equalsized pieces, three smaller equal-sized pieces, and so on. In so doing the total number of
pieces into which Cake 1 is divided will be a multiple of 24 such as 48, 72, …. Similarly, the
total number of pieces into which Cake 2 is divided will be a multiple of 36 such as 72,
108,…. When both cakes are divided into 72 equal pieces, the caterer will be happy.
(c) In each case we are looking for the least common multiple of two integers.
15. Any integer k is a divisor of 0 since 0 = 0·k. A is the greatest divisor of A and is also a
divisor of 0. Thus gcd(A, 0) = A.
16. (a) gcd(A, B) = 1.
(b) Suppose d is a positive integer that divides both A and B. Then by Theorem 9, d | MA and d |
NB. By Theorem 7, d | MA + NB or d | 1. Since the only positive integer that divides 1 is 1, d
= 1.
17. (a) gcd(A, B) divides 9.
(b) Let d = gcd(A, B). Then d | A and d | B so d | MA, d | NB, and d | MA + NB by
Theorems 7 and 9. Thus d | 9.
13. (a)
(b)
(c)
(d)
(e)
14. (a)
52
Chapter 3
18. Basis Step: If p | a1, then p | a1. That is, if p divides a product consisting of a single factor, then p
divides that single factor.
Inductive Step: Assume that when p | a1 · a2 · a3 · … · ak for some positive integer k, it implies that
p divides one of the factors ai. Suppose that p | a1 · a2 · a3 · … · ak· ak + 1. The product a1 · a2 · a3 ·
… · ak · ak + 1 can be understood as a product of two factors. That is a1 · a2 · a3 · … · ak· ak + 1 =
(a1 · a2 · a3 · … · ak)( ak + 1). Since p | (a1 · a2 · a3 · … · ak)( ak + 1), by Theorem 21, p | a1 · a2 · a3 ·
… · ak or p | ak + 1. In the case that p | a1 · a2 · a3 · … · ak, our inductive hypothesis asserts that p
divides one of the factors ai for i = 1, 2, …, k. In either case, p divides one of the factors ai for i=
1, 2, …, k, k + 1.
19. (a) N = pk.
(b) q| pk so q | p or q | k.
(c) Since p and q are distinct primes, q does not divide p. Thus, q | k and hence there
is an integer m with k = qm. Thus N = pk = pqm. Therefore, pq | N.
20. Basis Step: If p1 is a prime that divides N, then certainly the product consisting of the single factor
p1 divides N.
Inductive Step: Our induction hypothesis is this: k is a positive integer with the property that
whenever k distinct primes, p1, p2, …, pk, divide an integer N then the product p1 p2 …  pk
divides N.
Suppose distinct primes p1, p2, …, pk, pk + 1 divide N. Since pk + 1 | N, there is an integer M with N
= (pk + 1)(M). None of p1, p2, …, pk divide pk + 1, so by Theorem 21, each divides M. By inductive
hypothesis, the product p1 p2 …  pk divides M or M = (p1 p2 …  pk)(s) for some integer s.
Thus N = (pk + 1)(M) = (pk + 1)( (p1 p2 …  pk)(s)). The product p1 p2 …  pk pk + 1 divides N.
21. Let d = gcd(A, B). By Theorem 20, there are integers M and N with d = MA + NB.
Since D | A and D | B, then D | MA + NB by Theorems 7 and 9. Thus D | d.
Teaching Notes, Section 3.5
Related Websites: Visit the Prime Pages at http://www.utm.edu/research/primes/ for encyclopedic
information about prime numbers including lists of largest known primes and a primality checker that
will check whether any “small” integer is prime (“small” defined as less than 253 – 1. Other websites
of interest include:
www.2357.a-tu.net/index.php?link=links
www.utm.edu/research/primes/largest.html
Exploratory Exercise Set 3.5
1. (a) (1) 110
(2) 111 (3) 3 · 37
(b) (1) 31
(2) 2, 137
(3) 2, 9, 13
(c) Conjecture: The set of new primes produced by the machine is disjoint from the set of primes
inserted in the machine.
(d) (1) p is one of the three primes, hence p divides the product.
(2) Since p divides p1  p2  p3, p divides (1) p1  p2  p3 by Theorem 9. Since p
also divides p1  p2  p3 + 1, p divides (p1  p2  p3 + 1) + (1) p1  p2  p3 by Theorem 7.
Hence, p | 1.
(3) Since all primes are greater than 1, this is a contradiction.
2. (a) There are only a finite number of primes.
(b) (1) p is not one of the primes p1, p2, …, ps.
(2) The list p1, p2, …, ps is supposed to be a complete list of the primes.
3. (a) (1) Since 2 is one of the factors of N, 2 | N.
(2) Since 3 is one of the factors of N, 3 | N.
(3) Since 4 is one of the factors of N, 4 | N.
(4) Since 5 is one of the factors of N, 5 | N.
Since 2 | 2, Theorem 7 ensures that 2 | (N + 2).
Since 3 | 3, Theorem 7 ensures that 3 | (N + 3).
Since 4 | 4, Theorem 7 ensures that 4 | (N + 4).
Since 5 | 5, Theorem 7 ensures that 5 | (N + 5).
An Introduction to Mathematical Proof
53
(5)Each of these four numbers is greater than 1 and has a factor other than 1 and itself.
(b) For j = 2 to 10, j is a factor of K and j | j. Thus j | (K + j) and K + j is a composite number.
(c) Let N = 21! = 21  20  19 … 3  2  1. Then N + 2, N + 3, N + 4, …, N + 21 is a
sequence of 20 consecutive positive integers, none of which is prime.
(d) Let N = (R + 1)! = (R + 1)  R … 3  2  1. Then N + 2, N + 3, N + 4, …, N + (R + 1)
is a sequence of R consecutive positive integers, none of which is prime.
4. Answers will vary.
Exercise Set 3.5
1. (a)
1050
(b)
36
42
9
25
4
6
3
3
2
7
5
5
2
2
3
225
124
2. (a)
(b)
2
25
5
2
(c)
2
3
1184
10
11
3
(d)
440
4
5
31
44
2
9
62
2
8
5
148
4
2
2
2
4
2
37
2
3. 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113
127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199
4. (a) 15
(b) 25
(c) 35
(d) 46
5. (a) 34
(b) 74
(c) 114
(d) 153
6. 3 and 5, 5 and 7, 11 and 13, 17 and 19, 29 and 31, 41 and 43, 59 and 61, 71 and 73
7. (a) If a divisor is greater than 1, then the prime factors of the divisors are either 2’s and/or 3’s, and
there can be at most two 2’s and three 3’s as factors of any divisor.
(b) 1, 21, 22, 31, 32, 33, 2131, 2132, 2133, 2231, 2232, 2233
8. (a) If a divisor is greater than 1, the prime factors of the divisors are either 5’s and/or 7’s, and
there can be at most three 5’s and one 7 as factors of any divisor.
(b) 1,71, 51, 52, 53, 5171, 5271, 5371
9. (a) If a divisor is greater than 1, the prime factors of the divisors are either 3’s and/or 7’s, and
there can be at most four 3’s and two 7’s as factors of any divisor: 1, 31, 32, 33, 34, 71, 72, 3171,
3172, 3271, 3272, 3371, 3372, 3471, 3472.
(b) If a divisor is greater than 1, the prime factors of the divisors are either 2’s, 5’s and/or 13’s,
and there can be at most one 2, two 5’s, and two 13’s as factors of any divisor: 1, 21, 51, 52,
131, 132, 2151, 2152, 21131, 21132, 51131, 51132, 52131, 52132, 2151131, 2151132, 2152131, 2152132.
54
Chapter 3
10. 2233. Let d be the greatest common divisor. Since d divides 2233, the prime factors of d must be
2’s and 3’s. Further there are no more than two 2’s and three 3’s. Similarly, since d divides 2334,
the prime factors of d must be 2’s and 3’s and there can be no more than three 2’s and four 3’s.
Thus the greatest common divisor will have two factors of 2 and three factors of 3.
11. 5372. Let d be the greatest common divisor. Since d divides 5472, the prime factors of d must be
5’s and 7’s. Further there are no more than four 5’s and two 7’s. Similarly, since d divides
23 53 73, the factors must be 2’s, 5’s and 7’s, and there can be no more than three 2’s, three 5’s,
and three 7’s. Thus the greatest common divisor will have three factors of 5 and two factors of 7.
12. (a) 3372
(b) 237211
13. (a) Divisors of 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Divisors of 144: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
gcd(96, 144) = 48.
(b) 144 = 1(96) + 48; 96 = 2(48) + 0; gcd(96, 144) = the last non-zero remainder = 48.
(c) 96 = 253, 144 = 2432, gcd(96, 144) = 243 = 48.
14. (a) Divisors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48
Divisors of 220: 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220
gcd(48, 220) = 4.
(b) 220 = 4(48) + 28; 48 = 1(28) + 20; 28 = 1(20) + 8, 20 = 2(8) + 4, 8= 2(4) + 0 gcd(48, 220) =
the last non-zero remainder = 4.
(c) 48 = 243, 220 = 2251111, gcd(48, 220) = 22 = 4.
15. (a) 3 is the only prime that is one less than a square.
(b) n2 – 1 = (n  1)(n + 1). Thus 3 = 22  1 = (2  1) (2 + 1) is a prime. However, for larger
values of n, (n  1) and (n + 1) are factors of n2 – 1 other than 1 and the number n2 – 1.
16. No. Any six consecutive positive integers will contain three even numbers. The only even
number that is prime is 2.
17. Let d be a positive divisor of 350 that is bigger than 1. Let p1  p2 … pr be the prime
factorization of d. Since d divides 350, there is a positive integer k with 350= d  k or
350 = p1  p2  …  pr (prime factorization of k). Since the prime factorization of 350 is
unique by Theorem 23, each of the p’s must be a 3 and thus d is a power of 3.
18. (a) 4, 9, 16, 25
(b) Conjecture: n is a positive integer with exactly three positive divisors if and only if n is the
square of a prime integer.
(c) Show: If n has exactly three positive divisors, n is the square of a prime. n cannot be a prime
because primes have exactly two divisors. Further, both 1 and n are divisors of n, so the
prime factorization of n can involve only one prime divisor, call it p. If the prime
factorization of n is pr, then n has r + 1 distinct positive factors 1, p, p2, …, pr = n. Therefore
r must be 2 and n is p2.
Suppose n = p2. The divisors of n are 1, p, p2. n has exactly three divisors.
19. (a) 6, 10, 14, 15, 21, 22, 26
(b) n has exactly four positive factors if and only if n is a product of two distinct
primes.
(c) Show: If n has exactly four positive factors, then n is a product of two distinct
primes.
Since 1 and n are both factors of n, the prime factorization of n consists of at most two
factors, n = pq. If p = q, then n has only three positive factors (See Exercise 18.) Thus p and
q must be distinct.
Show: If n is a product of two distinct primes, n has exactly four positive factors.
Suppose n = pq where p and q are primes with p not equal to q. Clearly 1, p, q, and n are four
factors of n. Suppose a is some other factor of n. Then there is a positive integer k with n =
ak. Thus pq = ak = (prime factorization of a)(prime factorization of k). Theorem 23 asserts
An Introduction to Mathematical Proof
55
that both a and k must be primes and one must be p and one must be q. Thus there are exactly
four positive divisors of n: 1, p, q, n.