CHAPTER TEN
MATHEMATICAL SYSTEMS
Exercise Set 10.1
1. A binary operation is an operation that is performed on two elements, and the result is a single element.
2. A set of elements and at least one binary operation.
3. Each of these operations can be performed on only two elements at a time and the result is always a single
element.
a) 2 + 3 = 5
b) 5 – 3 = 2
c) 2 × 3 = 6
d) 6 ÷ 3 = 2
4. Closure, identity, each element must have a unique inverse, associative property.
5. Closure, identity, each element must have a unique inverse, associative property, commutative property.
6. Abelian group
7. If a binary operation is performed on any two elements of a set and the result is an element of the set, then
that set is closed under the given binary operation. For all integers a and b, a + b is an integer. Therefore,
the set of integers is closed under the operation of addition.
8. An identity element is an element in a set such that when a binary operation is performed on it and any
given element in the set, the result is the given element. The additive identity element is 0, and the
multiplicative identity element is 1.
Examples: 5 + 0 = 5, 5 × 1 = 5
9. When a binary operation is performed on two elements in a set and the result is the identity element for the
binary operation, then each element is said to be the inverse of the other. The additive inverse of 2 is (– 2)
since 2 + (– 2) = 0, and the multiplicative inverse of 2 is (1/2) since 2 × 1/2 = 1.
10. A specific example illustrating that a specific property is not true is called a counterexample.
11. No. Every commutative group is also a group.
12. Yes. For a group, the Commutative property need not apply.
13. d The Commutative property need not apply.
14. Squaring, finding square roots, finding the reciprocal, finding the absolute value
15. The associative property of addition states that (a + b) + c = a + (b + c), for any elements a, b, and c.
Example: (3 + 4) + 5 = 3 + (4 + 5)
16. The associative property of multiplication states that (a × b) × c = a × (b × c), for any real numbers
a, b, and c.
Example: (3 × 4) × 5 = 3 × (4 × 5)
17. The commutative property of multiplication stated that a × b = b × a, for any real numbers a, b, and c.
Example: 3 × 4 = 4 × 3
18. The commutative property of addition stated that a + b = b + a, for any elements a, b, and c.
Example: 3 + 4 = 4 + 3
20. 7 – 3 = 4, BUT 3 – 7 = – 4
19. 8 ÷ 4 = 2, but 4 ÷ 8 = ½
21. (6 – 3) – 2 = 3 – 2 = 1, but 6 – (3 – 2) = 6 – 1 = 5
22. (16 ÷ 4) ÷ 2 = 4 ÷ 2, = 2 but 16 ÷ (4 ÷ 2) = 16 ÷ 2 = 8
23. No. No inverse element
24. No. No inverse element
25. Yes. Satisfies 5 properties needed
26. Yes. Satisfies 4 properties needed
27. No. Not closed
28. No. Not closed
327
328 CHAPTER 10
29.
31.
33.
35.
37.
39.
Mathematical Systems
No. No identity or inverse elements
No. Not closed
Yes. Satisfies 4 properties needed
No. Not closed
ie.: 1/0 is undefined
No. Does not satisfy Associative property
No; the system is not closed, π + (– π ) = 0
which is not an irrational number.
30.
32.
34.
36.
38.
40.
No. Not all elements have inverses
No. Not all elements have inverses
No. Not all elements have inverses
No. Does not satisfy Associative property
No. Not closed
No; π
(1/ π ) = 1 which is not an irrational
number.
41. Yes. Closure: The sum of any two real
numbers is a real number. The identity element
is zero. Example: 5 + 0 = 0 + 5 = 5
Each element has a unique inverse.
Example: 6 + (– 6) = 0
The associative property holds:
Example: (2 + 3) + 4 = 2 + (3 + 4)
42. No. Closure: The product of any two real
numbers is a real number. The identity element
is one. Example: 5 y 0 = 0 y 5 = 5
Not every element has an inverse.
Example: 2 y ? = 1
The associative property holds:
Example: (2 y 3) y 4 = 2 y (3 y 4)
43. Answers will vary.
45. 9/19/29/39/49/59/69/79/89
90/91/92/93/94/95/96/97/98/99
44. 999
9 } 20
11 }
Exercise Set 10.2
1. The clock addition table is formed by adding all pairs of integers between 1 and 12 using the 12 hour clock
to determine the result. Example: If the clock is at 7 and we add 8, then the clock will read 3.
Thus, 7 + 8 = 3 in clock arithmetic.
2. 12 + 12 = 12. Start at 12 move clockwise 12 hours, the result is 12.
3. a) First add (6 + 9) on the clock, then add that result to 5 on the clock to obtain the final answer.
b) (4 + 10) + 3 = 2
(2) + 3 = 5
4. a) Start at the first number on the face of the clock, then count counterclockwise the number being
subtracted. The number you end at is the difference.
b) 4 – 7 = 9
5. a) 5 – 9
5 + 12 = 17
17 – 9
b) 17 – 9 = 8
6. The system is commutative if the elements in the table are symmetric about the main diagonal.
7. If a binary operation is performed on any two elements of a set and the result is an element of the set, then
that set is closed under the given binary operation. For all integers a and b, a + b is an integer. Therefore,
the set of integers is closed under the operation of addition.
8. Yes. 12
9. Yes. One and 11 are inverses, 2 and 10 are inverses, 3 and 9 are inverses, 4 and 8 are inverses, 5 and 7 are
inverses, 6 is its own inverse, and 12 is its own inverse.
10. (2 + 3) + 8 = 2 + (3 + 8)
11. Yes. 6 + 9 = 3 and 9 + 6 = 3
5 + 8 = 2 + 11
1=1
SECTION 10.2
12. Yes, the five properties are met.
1) The system is closed. All results are from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
2) The identity element is 12.
3) Each element has an inverse.
4) The associative property holds true.
5) The system is commutative.
13. a) Identity element = 5
b) Add inverse of 2, which is 3.
2+3=5
15. Yes. Commutative, symmetrical around main
diagonal
17. Identity element = C, Row 3 is identical to top
row and column 3 is identical to left column
19. The inverse of A is B, because A operate B = C
and B operate A = C.
21.
24.
27.
30.
33.
36.
39.
42.
45.
46.
49.
52.
55.
58.
61.
64.
4 + 7 = 11
10 + 4 = 2
3 + (8 + 9) = 3 + 5 = 8
(6 + 10) + 12 = 4 + 12 = 4
7–4=3
3–9=6
1 – 12 = 1
8 – 8 = 12
+
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4+5=3
5–2=3
3–4=5
See above.
4+4=1
3–6=4
3 – (2 – 6) = 3 – 3 = 7
22.
25.
28.
31.
34.
37.
40.
43.
4
5
6
1
2
3
4
5
6
1
2
3
4
5
45.
1+6=1
4–5=5
4–6=4
5+4=2
7+6=6
2–4=5
3+5=8
16. No. Not commutative, Non-symmetrical
around main diagonal
18. There is no identity. While the top row = 3rd row,
the left column ≠ any other column.
20. The inverse of A is A, because A operate A = A.
8+7=3
4 + 12 = 4
(8 + 7) + 6 = 3 + 6 = 9
(7 + 8) + (9 + 6) = 3 + 3 = 6
11 – 8 = 3
5 – 10 = 7
6 – 10 = 8
12 – 12 = 12
6
1
2
3
4
5
6
47.
50.
53.
56.
59.
62.
14. a) Identity element = 8
b) Add inverse of 3, which is 5.
+
1
2
3
4
5
6
7
23.
26.
29.
32.
35.
38.
41.
44.
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1
2
48.
51.
54.
57.
60.
63.
9+8=5
12 + 12 = 12
(6 + 4 ) + 8 = 10 + 8 = 6
(7 + 11) + (9 + 5) = 6 + 2 = 8
4 – 12 = 4
3 – 10 = 5
5 – 5 = 12
5–8=9
3
4
5
6
7
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1
2
3
4
5
6
7
1
2
3
4
5
6
7
6+4=4
2–6=2
2 + (1 – 3) = 2 + 4 = 6
6+5=4
2–3=6
(4 – 5) - 6 = 6 – 6 = 7
329
330 CHAPTER 10
Mathematical Systems
65. Yes. Satisfies 5 required properties
66. No, not necessarily. It may not have an inverse,
identity element, or satisfy the Commutative or
Associative properties.
67. a) {0, 1, 2, 3}
airplane
b)
c) Yes. All solutions are members of the
original set.
d) Identity element is 0.
0 = 0, 1
3 = 0,
e) Yes; 0
2
2 = 0, 3
1=0
2)
3=3
3=2
f) (1
and 1
(2
3) = 1
1=2
2=1=2
3
g) Yes; 3
h) Yes, system satisfies five properties needed.
68. a) {*, 5, L}
69. a) {r, s, t, u} b)
c) Yes. All solutions are members of the
original set.
d) Yes, the identity element is t.
e) Yes; r
r = t, s
u = t,
t
t = t, u
s=t
s)
u =u
u=r
f) (r
and r
(s
u) = r
t=r
g) Yes; s
r = u and r
s=u
h) Yes, system satisfies five properties needed.
b)
b)
c) Yes. All solutions are members of the
original set.
d) Identity element is L.
5 = L, 5
e) Yes; *
f) (*
5)
and *
5=L
(5
* = L, L
L=L
5=5
5) = *
*=5
* = * and *
L=*
g) Yes; L
h) Yes, system satisfies five properties needed.
70. a) {3, 5, 8, 4} b)
c) Yes. All solutions are members of the
original set.
d) Identity element is 4.
e) Yes. 3
4
f) (5
8 = 4, 5
5 = 4, 8
3 = 4,
4=4
8)
and 5
4=3
(8
4=3
4) = 5
8=3
5=3=5
8
g) Yes. 8
h) Yes, system satisfies five properties needed.
71.a) {f, r, o, m}
b)
c) The system is closed. All elements in the table
are elements of the set.
d) (r o) f = m f = m
e) (f r) m) = r m = f
f) Identity element is f.
g) Inverse of r is m since m r = f.
h) Inverse of m is r since r m = f.
72. a) No, there is no identity element.
b) (1 w 3) w 4 ≠ 1 w (3 w 4)
4w4 ≠ 1w3
73. a) Is closed; all solutions are members of the
original set.
b) Identity = ‰
c) Inverse: of ‰ is ‰ ; of M is M; of is d) (M8 ) 8 M = 8 M = M
M 8 ( 8 M) = M 8 M = ‰
Not associative since M
‰
e) 8 M = M
M8=
Not commutative since M ≠ SECTION 10.2
74. Not closed: y ^ x = a and a is not a member of
the set {w, x, y}
No identity element, and therefore no inverses.
(x ^ w) ^ x = y ^ x = a
x ^ (w ^ x) = x ^ y = w
Not associative since a
75. No inverses for  and *
(* 8 *) 8 T =  8 T = *
* 8 (* 8 T) = * 8 T = 
Not associative since * ≠ 
w
y ^ x = a and x ^ y = w
Not commutative since a ≠ w
76. (a
a)
∆= ∆
∆=a
a
(a
∆)=a
0=≺
Not associative since a
≺
∆
U=a
≺=≺
≺
77. No identity element and therefore no inverses.
(d ⇔ e) ⇔ d = d ⇔ d = e
d ⇔ (e ⇔ d) = d ⇔ e = d
Not associative since e
e ⇔d=e
d ⇔ e=d
Not commutative since e ≠ d
Not commutative since ≺ ≠ a
78. No inverses for 0, 2, 3, and 4
79. a)
+ E O
E E O
O O E
b) The system is closed, the identity element is E,
each element is its own inverse, and the system
is commutative since the table is symmetric
about the main diagonal. Since the system has
fewer than 6 elements satisfying the above
properties, it is a commutative group.
83. a) All elements in the table are in the set
{1, 2, 3, 4, 5, 6} so the system is closed.
The identity is 6. 5 and 1 are inverses of each
other, and 2, 3, 4, and 6 are their own inverses.
Thus, if the associative property is assumed,
the system is a group.
b) 4 ∞ 5 = 2, but 5 ∞ 4 = 3
d
80. a)
E O
E E E
O E O
b) The identity is 0, but since E has no inverse,
the system is not a group.
81. Student activity - Answers will vary.
82. Student activity - Answers will vary.
83. Examples of associativity
(2 ∞ 3) ∞ 4 = 5 ∞ 4 = 3 and
2 ∞ (3 ∞ 4) = 2 ∞ 5 = 3
(1 ∞ 3) ∞ 5 = 4 ∞ 5= 2 and
1 ∞ (3 ∞ 5) = 1 ∞ 4 = 2
331
332 CHAPTER 10
84. a) Is closed
(C
Mathematical Systems
85. a)
Identity = F
D)
A=E
* R S T U V
R V T U S I
S U I V R T
T S R I V U
U T V R I S
V I U S T R
I R S T U V
R * (T * V) = R * U = S
(R * T ) * V = U * V = S
Is Associative since S = S
b) Is closed
c) R * S = T
S*R=U
Not Commutative since T ≠ U
R  (S  V) = R  T = U
A=F
C (D A) = C C = F
Is Associative since F = F
Inverses of:
A
E = F, B
B = F,
C
C = F, D
D = F,
E
A = F, F
F=F
C B=A
B C=E
Not Commutative since E ≠ A
86. 43 = 64 ways
87.
+
0
1
2
3
4
0
0
1
2
3
4
1
1
2
3
4
0
2
2
3
4
0
1
3
3
4
0
1
2
89.
+
0
1
2
3
0
0
1
2
3
1
1
2
3
4
2
2
3
4
0
3
3
4
0
1
4
4
0
1
2
3
88.
+
0
1
2
3
4
5
0
0
1
2
3
4
5
1
1
2
3
4
5
0
2
2
3
4
5
0
1
3
3
4
5
0
1
2
4
4
5
0
1
2
3
I
R
S
T
U
V
I
5
5
0
1
2
3
4
89. 1) Add # in left column to # in top row
2) Divide by 4
3) Replace remainder in table
Exercise Set 10.3
1. A modulo m system consists of m elements, 0 through m – 1, and a binary operation.
2. a) a is congruent to b modulo m, written a ≅ b (mod m), means a and b have the same remainder when
divided by m.
b) 13 and 3 have the same remainder, 3, when divided by 5.
3. In a modulo 5 system there will be 5 modulo classes. When a number is divided by 5 the remainder
will be a number from 0 – 4.
0 1 2 3 4
0 1 2 3 4
5 6 7 8 9
10 11 12 13 14
……………………………………
SECTION 10.3
333
4. In any modulo system, modulo classes are developed by placing all numbers with the same remainder
in the same modulo class.
5. In a modulo 12 system there will be 12 modulo classes. When a number is divided by 12 the remainder
will be a number 0 – 11.
6. In a modulo n system there will be n modulo classes. When a number is divided by n the remainder
will be a number from 0 – (n–1).
7. 27 ≅ ? (mod 5) c or d 27, 12, and 107 have
the same remainder, 2, when divided by 5.
9. Thursday = Day 4
30 ≅ 2 (mod 7)
Saturday
8. 167 ≅ ? (mod 7) b or d 106, 71, and 22
have the same remainder, 1, when divided by 7.
10. 4 + 161 = 165 and 165 ÷ 7 = 23, remainder 4
Day 4 = Thursday
11. 4 + 366 = 370 and 370 ÷ 7 = 52, remainder 6
Day 6 = Saturday
12. 5 years = (5 y 365) days = 1825 days
4 + 1825 = 1829 and 1829 ÷ 7 = 261, remainder 2
Day 2 = Tuesday
13. 3 years, 34 days = (3)(365 + 34) days = 1129 days
4 + 1129 = 1133 and 1133 ÷ 7 = 161, remainder 6
Day 6 = Saturday
14. 4 + 463 = 467 and 467 ÷ 7 = 66, remainder 5
Day 5 = Friday
15. 728 days / 7 = 104 remainder 0
16. 3 yrs. 27 days = 1122 days
1122 / 7 = 160 remainder 2
17. Answers will vary.
21. Answers will vary.
Thursday
18. Answers will vary.
22. Answers will vary.
14 ≅ 4 (mod 5)
25. 8 + 6 = 14
27. 1 + 9 + 12 = 22
19. Answers will vary.
23. Answers will vary.
28. 9 - 3 = 6
20. Answers will vary.
24. Answers will vary.
15 ≅ 0 (mod 5)
26. 5 + 10 = 15
22 ≅ 2 (mod 5)
6 ≅ 1 (mod 5)
29. 5 - 12 = 3
3 ≅ 3 (mod 5)
30. 7 y 4 = 28
28 ≅ 3 (mod 5)
31. 8 y 9 = 72
72 = 2 (mod 5)
32. 10 - 15 = 0
0 ≅ 0 (mod 5)
33. 4 - 8 = 1
1 ≅ 1 (mod 5)
35. (15 y 4) - 8 = 60 – 8 = 52
37. 15 (mod 5) ≅ 0
41. 60 (mod 9) ≅ 6
45.
49.
53.
57.
-5 (mod 7) ≅ 2
135 (mod 10) ≅ 5
2 + 2 ≅ 4 (mod 5)
5 y 5 ≅ 7 (mod 9)
34. 3 - 7 = 1
52 ≅ 2 (mod 5)
38. 23 (mod 7) ≅ 2
42. 75 (mod 8) ≅ 3
46.
50.
54.
58.
-7 (mod 4) ≅ 1
-12 (mod 4) ≅ 0
4 + 5 ≅ 3 (mod 6)
3 y { } ≅ 5 (mod 6)
No solution
1 ≅ 1 (mod 5)
36. (4 – 9)7 = (-5)7 = 5(7) = 35
39. 84 (mod 12) ≅ 0
43. 30 (mod 7) ≅ 2
47.
51.
55.
59.
Saturday
-13 (mod 11) ≅ 9
3 + 4 = 7 ≅ 1 (mod 6)
4 - 5 ≅ 5 (mod 6)
3 y { } ≅ 1 (mod 6)
No solution
35 ≅ 0 (mod 5)
40. 43 (mod 6) ≅ 1
44. 53 (mod 4) ≅ 1
48.
52.
56.
60.
-11 (mod 13) ≅ 2
6 + 5 ≅ 3 (mod 8)
4 y 5 ≅ 6 (mod 7)
3 y { } ≅ 3 (mod 12)
{1, 5, 9}
334 CHAPTER 10
Mathematical Systems
61. 4 y { } ≅ 4 (mod 10)
{1, 6}
65. 3 y 0 ≅ 05 (mod 10)
68. a) flying 7 R 4
c) resting 30 R 0
e) flying
70. a)
b)
c)
d)
62. 2 – 6 ≅ 4 (mod 8)
63. 4 – 7 ≅ 9 (mod 12)
66. 4 y { } ≅ 5 (mod 8)
No solution
67. a) 2016, 2020, 2024,
67. c) 2552, 2556, 2560,
20,28, 2032
2564, 2568, 2572
b) 3004
69. a) 28/8 = 3 R 4 resting 2nd day
b) 60/8 = 7 R 4 resting 2nd day
c) 127/8 = 15 R 7 am/pm practice
d) no am practice
71. The manager’s schedule is repeated every seven
weeks. If this is week two of her schedule, then
this is her second weekend that she works, or week
1 in a mod 7 system. Her schedule in mod 7 on any
given weekend is shown in the following table:
Weekend (mod 7):
Work/off 0 1 2 3 4 5 6
w w w w w w o
a) If this is weekend 1, then in 5 more weeks
(1 + 5 = 6) she will have the weekend off.
b) 25 ≅ 7 = 3, remainder 4. Thus 25 ≅ (mod 7)
and 4 weeks from weekend 1 will be weekend 5.
She will not have off.
c) 50 ≅ 7 = 7, remainder 1. One week from
weekend 1 will be weekend 2. It will be 4 more
weeks before she has off. Thus, in 54 weeks she
will have the weekend off.
b) flying 11 R 2
d) flying 7 – 6 = 1
f) 7 – 20 = 3
20/10 = 2 R 0 twice a day
49/10 = 4 R 9 twice a day
103/10 = 10 R 3 twice a day
78/10 = 7 R 8 yes, rest
72. a) 6 ≅ 1 (mod 5)
If this is week 3, then 3 + 1 ≅ 4 (mod 5) indicates
the 3 P.M. - 11 P.M. shift.
b) 7 ≅ 2 (mod 5)
If this is week 4, then 4 + 2 ≅ 1 (mod 5) indicates
the 7 A.M. - 3 P.M. shift.
c) 11 ≅ 1 (mod 5)
If this is week 1, then 1 + 1 ≅ 2 (mod 5) indicates
the 7 A.M. - 3 P.M. shift.
73. The waiter’s schedule in a mod 14 system is given
in the following table:
Day: 0 1 2 3 4 5 6 7 8 9 10 11 12 13
shift: d d d d d e e e d d d d e e
Note: This is his second day shift which is day 1
in the mod 14 system.
a) 20 ≅ 14 = 1, remainder 6. Six days from day 1 is
day 7 which is the evening shift.
b) 52 ≅ 14 = 3, remainder 10. Ten days from day 1
is day 11, which is the day shift.
c) 365 ≅ 14 = 26, remainder 1. One day from day 1
is day 2, which is the day shift.
64. 6 – 7 ≅ 8 (mod 9)
74. The truck driver's schedule is repeated every 17
days as indicated by the following table:
Days
Activity
0-2
N.Y. - Chicago
3
Rest in Chicago
4-6
Chicago - L.A.
7-8
Rest in L.A.
9 - 13
L.A. - N.Y.
14 – 16
a) 30 ≅ 13 (mod 17) indicates that he will be
driving from L.A. to N.Y.
b) 70 ≅ 2 (mod 17) indicates that he will be
driving from N.Y. to Chicago.
c) 2 years = 730 days ≅ 16 (mod 17)
SECTION 10.3
75. a)
b)
c)
d)
e)
f)
+ 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2
Yes. All the numbers in the table are from the
set {0, 1, 2, 3}.
The identity element is 0.
Yes. element + inverse = identity
0+0=0
1+3=0
2 + 2 =0
3+1=0
(1 + 3) + 2 0 + 2 = 2
1 + (3 + 2) = 1 + 1 = 2
Associative since 2 = 2.
Yes, the table is symmetric about the main
diagonal.
1+3=0=3+1
76. a)
b)
c)
d)
e)
f)
g)
h)
77. a)
b)
c)
d)
e)
f)
g)
0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2
Yes. All the elements in the table are from
the set {0, 1, 2, 3}.
Yes. The identity element is 1.
elem. inverse = identity
0 none = 1 1 1 = 1 2 none = 1
3 3 = 1 Elements 0 and 2 do not have
inverses.
(1 3) 0 = 3 0 = 0
1 (3 0) = 1 0 = 0 Yes, Associative
Yes. 2 3 = 2 = 3 2
No. Not all elements have inverses.
78. a)
b)
c)
d)
e)
f)
g)
+ 0 1 2 3 4 5 6 7
0 0 1 2 3 4 5 6 7
1 1 2 3 4 5 6 7 0
2 2 3 4 5 6 7 0 1
3 3 4 5 6 7 0 1 2
4 4 5 6 7 0 1 2 3
5 5 6 7 0 1 2 3 4
6 6 7 0 1 2 3 4 5
7 7 0 1 2 3 4 5 6
Yes. All the numbers in the table are from the
set {0, 1, 2, 3, 4, 5, 6, 7}.
The identity element is 0.
elem. + inverse = identity
0+0=0 1+7=0 2+6=0 3+5=0
4+4=0 5+3=0 6+2=0 7+1=0
(1 + 2) + 5 = 3 + 5 = 0
1 + (2 + 5) = 1 + 7 = 0 Yes, Associative
Yes. 2 + 4 = 6 = 4 + 2
Yes. All five properties are satisfied.
Same answer as problem 63 part h.
0 1 2 3 4 5 6
0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6
2 0 2 4 6 1 3 5
3 0 3 6 2 5 1 4
4 0 4 1 5 2 6 3
5 0 5 3 1 6 4 2
6 0 6 5 4 3 2 1
Yes. All the elements in the table are from
the set {0, 1, 2, 3, 4, 5, 6}.
Yes. The identity element is 1.
No. elem. Æ inverse
0 Æ none 1 Æ1
2Æ4 3Æ5 4Æ2 5Æ3 6Æ6
The element 0 does not have an inverse.
(1 2) 4 = 2 4 = 1
1 (2 4) = 1 1 = 1 Yes, Associative
Yes. 2 3 =6 = 3 2
No. 0 does not have an inverse.
For the operation of division in modular systems, we define n ÷ d = n y i, where i is the multiplicative inverse of d.
79. 5 ÷ 7 ≅ ? (mod 9)
80. ? ÷ 5 ≅ 5 (mod 9)
Since 7 y 4 = 28 ≅ 1 (mod 9), 4 is the inverse of 7.
Since 5
5 ≅ 5 (mod 9), 1 ≅ 5 (mod 5)
Thus, 5 ÷ 7 = 0 R 2
5
7 ≅ 2 (mod 9)
?=2
335
?=7
336 CHAPTER 10
81. ? ÷ ? ≅ 1 (mod 4)
1 ÷ 1 ≅ 1 )mod 4)
3 ÷ 3 ≅ 1 (mod 4)
Mathematical Systems
0 ÷ 0 is undefined.
2 ÷ 2 ≅ 1 (mod 4)
? = {1, 2, 3}
82. 1 ÷ 2 ≅ ? (mod 5)
2(1/2) ≅ 3?
1 ≅ 6 (mod 5)
?=3
1=1
83. 5k ≅ x (mod 5) 5(1) ≅ 0 (mod 5)
5(2) = 10 ≅ 0 (mod 5)
x=0
84. 5k + 4 ≅ x (mod 5) 5(1) + 4 = 9 ≅ 4 (mod 5)
5(2) + 4 = 14 ≅ 4 (mod 5)
x=4
85. 4k – 2 ≅ x (mod 4) 4(0) – 2 = -2 ≅ 2 (mod 4)
4(1) – 2 = 2 ≅ 2 (mod 4) 4(2) – 2 = 6 ≅ 2 (mod 4)
x=2
86. Check the numbers divisible by 5 until you find
one that is also congruent to 2 in modulo 6.
20 ≅ 2 (mod 6) and 20 is also divisible by 5.
87. (365 days)(24 hrs./day)(60 min./hr.) = 525,600 hrs.
(525,600)/(4) = 131,400 rolls 131400 ≅ 0 (mod 4)
88. 1 yr. 21 days = 365 + 21 = 386 days
386/5 = 77 R 1
Halfway up the mountain
89. If 10 is subtracted from each number on the wheel,
23 11 3 18 10 19 2 10 16 4 24
becomes
13 1 20 8 0 9 19 0 6 21 14 which is equivalent to
M A T H
I S
F U N
Review Exercises
1. A set of elements and at least one binary operation.
2. A binary operation is an operation that can be performed on two and only two elements of a set. The result is
a single element.
3. Yes. The sum of any two integers is always an integer.
4. No. Example: 2 – 3 = – 1, but – 1 is not a natural number.
5. 9 + 10 = 19 ≅ 7 (mod 12)
6. 5 + 12 = 17 ≅ 5 (mod 12)
7. 8 – 10 = -2 ≅ 10 (mod 12)
8. 4 + 7 + 9 = 20 ≅ 8 (mod 12)
9. 7 – 4 +6 = 9 ≅ 9 (mod 12)
10. 2 – 8 – 7 = -13 ≅ 11 (mod 12)
11. a) The system is closed. If the binary operation is ‰ then for any elements a and b in the set, a  b is a member
of the set.
b) There exists an identity element in the set. For any element a in the set, if a ‰ i = i ‰ a = a, then i is
called the identity element.
c) Every element in the set has a unique inverse. For any element a in the set, there exists an element b
such that a ‰ b = b ‰ a = i. Then b is the inverse of a, and a is the inverse of b.
d) The set is associative under the operation For elements a, b, and c in the set, (a ‰ b) ‰ c = a ‰ (b ‰ c).
12. An Abelian group is a group in which the operation has the commutative property.
13. Yes. Closure: The sum of any two integers is an
integer. The identity element is zero.
Yes, Associative Example: (2 + 3) + 4 = 2 + (3 + 4)
Each element has a unique inverse.
14. The set of integers with the operation of multiplication does not form a group since not all elements have
an inverse. 4 y ? = 1
REVIEW EXERCISES
15. Yes. Closure: The sum of any two rational #s is a
rational number.
The identity element is zero. Ex.: 5 + 0 = 0 + 5 = 5
Yes, Associative Example: (2 + 3) + 4 = 2 + (3 + 4)
Each element has a unique inverse. Ex. ; 6 + (– 6) = 0
16. The set of rational numbers with the operation of multiplication does not form a group since zero does not
have an inverse. 0 y ? = 1
17. There is no identity element. Therefore the system does not form a group.
18. Not Associative
Example: (! … p) … ? = p … ? = !
19. Not Associative
Example: (p ? p) ? 4 = L ? 4 = #
20. a) {
, ?, ∆ }
,
b)
c) Yes. All the elements in the table are from
, ?, ∆ }.
,
the set {
! … (p … ?) = ! … ! = U
p ? (p ? 4) = p ? L = 4
!≠U
# ≠4
21. 21 ÷ 3 = 7, remainder 0
21 ≅ 0 (mod 3)
22. 31 ÷ 8 = 3, remainder 7
31 ≅ 7 (mod 8)
23. 31 ÷ 6 = 5, remainder 1
31 ≅ 1 (mod 6)
24. 59 ÷ 8 = 7, remainder 3
59 ≅ 3 (mod 8)
25. 82 ÷ 13 = 6, remainder 4
82 ≅ 4 (mod 13)
26. 54 ÷ 4 = 13, remainder 2
54 ≅ 2 (mod 4)
d) The identity element is )-- .
inverse = identity
e) Yes. elem.
?
=
?=
∆
∆ =
=
f) Yes, Associative
(
∆ =?
?)
(?
∆)=
∆ =
27. 52 ÷ 12 = 4, remainder 4
52 ≅ 4 (mod 12)
=
?=
=?
∆
g) Yes. ∆
h) Yes, all five properties are satisfied.
28. 54 ÷ 14 = 3, remainder 12
54 ≅ 12 (mod 14)
29. 97 ÷ 11 = 8, remainder 9
30. 42 ÷ 11 = 3, remainder 9
42 ≅ 9 (mod 11)
31. 5 + 8 = 13 ≅ 4 (mod 9)
Thus, replace ? with 4.
32. ? – 3 ≅ 0 (mod 5)
0 – 3 ≅ 2 (mod 5)
2 – 3 ≅ 4 (mod 5)
Replace ? with 3.
1 – 3 ≅ 3 (mod 5)
3 – 3 ≅ 0 (mod 5)
337
97 ≅ 9 (mod 11)
33. 4 y ? ≅ 3 (mod 6)
4 y 0 ≅ 0 (mod 6)
4 y 1 ≅ 4 (mod 6)
4 y 2 = 8 ≅ 2 (mod 6) 4 y 3 = 12 ≅ 0 (mod 6)
4 y 4 = 16 ≅ 4 (mod 6) 4 y 5 = 20 ≅ 2 (mod 6)
There is no solution. ? = { }
338 CHAPTER 10
Mathematical Systems
34. 6 – ? ≅ 5 (mod 7)
6 – 0 ≅ 6 (mod 7)
6 – 2 ≅ 4 (mod 7)
6 – 4 ≅ 2 (mod 7)
Replace ? with 1.
6 – 1 ≅ 5 (mod 7)
6 – 3 ≅ 3 (mod 7)
6 – 5 ≅ 1 (mod 7)
36. 10 y 7 ≅ ? (mod 12)
10 y 7 = 70; 70
12 ≅ 5, remainder 10
Thus, 10 y 7 ≅ 10 (mod 12).
Replace ? with 10.
38. ? y 7 ≅ 3 (mod 10)
0 y 7 ≅ 0 (mod 10)
2 y 7 = 14 ≅ 4 (mod 10)
4 y 7 = 28 ≅ 8 (mod 10)
6 y 7 = 42 ≅ 2 (mod 10)
8 y 7 = 56 ≅ 6 (mod 10)
10 y 7 = 70 ≅ 0 (mod 10)
Replace ? with 9.
40. 7 y ? 2 (mod 9)
7 y 0 ≅ 0 (mod 9)
7 y 2 = 14 ≅ 5 (mod 9)
7 y 4 = 28 ≅ 1 (mod 9)
7 y 6 = 42 ≅ 6 (mod 9)
7 y 8 = 56 ≅ 2 (mod 9)
Replace ? with 8.
1 y 7 ≅ 7 (mod 10)
3 y 7 = 21 ≅ 1 (mod 10)
5 y 7 = 35 ≅ 5 (mod 10)
7 y 7 = 49 ≅ 9 (mod 10)
9 y 7 = 63 ≅ 3 (mod 10)
35. ? y 4 ≅ 0 (mod 8)
0 y 4 ≅ 0 (mod 8)
1 y 4 ≅ 4 (mod 8)
2 y 4 = 8 ≅ 0 (mod 8)
3 y 4 = 12 ≅ 4 (mod 8)
4 y 4 = 16 ≅ 0 (mod 8)
5 y 4 = 20 ≅ 4 (mod 8)
6 y 4 = 24 ≅ 0 (mod 8)
7 y 4 = 28 ≅ 4 (mod 8)
Replace ? with {0, 2, 4, 6}.
37. 3 – 5 ≅ ? (mod 7)
3 – 5 = (3+7) – 5 = 5 ≅ 5 (mod 7)
Replace ? with 5.
39. 5 y ? ≅ 3 (mod 8)
5 y 0 ≅ 0 (mod 8)
5 y 2 = 10 ≅ 2 (mod 8)
5 y 4 = 20 ≅ 4 (mod 8)
5 y 6 = 30 ≅ 6 (mod 8)
Replace ? with 7.
41. a)
7 y 1 ≅ 7 (mod 9)
7 y 3 = 21 ≅ 3 (mod 9)
7 y 5 = 35 ≅ 7 (mod 9)
7 y 7 = 49 ≅ 4 (mod 9)
+
0
1
2
3
4
5
0
0
1
2
3
4
5
1
1
2
3
4
5
0
2
2
3
4
5
0
1
5 y 1 ≅ 5 (mod 8)
5 y 3 = 15 ≅ 7 (mod 8)
5 y 5 = 25 ≅ 1 (mod 8)
5 y 7 = 35 ≅ 3 (mod 8)
3
3
4
5
0
1
2
4
4
5
0
1
2
3
5
5
0
1
2
3
4
CHAPTER TEST
339
0 1 2 3
41. b) Since all the numbers in the table are elements of 42. a)
0 0 1 2 3
{0, 1, 2, 3, 4, 5}, the system has the closure
1 1 2 3 0
property.
c) The commutative property holds since the
2 2 3 0 1
3 3 0 1 2
elements are symmetric about the main
diagonal.
b) The identity element is 1, but because 0 and 2
d) The identity element is 0 and the inverses of
have no inverses, the system does not form a
each element are 0 – 0, 1 – 5, 2 – 4, 3 – 3,
group.
4 – 2, 5 – 1
e) If it is assumed the associative property holds as
illustrated by the example: (2 + 3) + 5 = 4 =
2 +(3 + 5), then the system is a commutative
group.
43. Day (mod 10): 0 1 2 3 4 5 6 7 8 9
Work/off : w w w o o w w o o o
a) If today is the first day of her work pattern, day 0, then 18 ≅ 8 (mod 10) indicates Toni will not be working
in 18 days.
b) 38 ≅ 8 (mod 10) indicates that Toni will have the evening off in 38 days.
Chapter Test
1. A mathematical system consists of a set of elements and at least one binary operation.
2. Closure, identity element, inverses, associative property, and commutative property.
3. No, the numbers greater than 0 do not have inverses.
4.
+ 1
1
2
3
4
5
2
2
3
4
5
1
3
3
4
5
1
2
4
4
5
1
2
3
6. 9 + 3 + 2 = 14 ≅ 4 mod 5
8. a)
b)
c)
d)
e)
5
5
1
2
3
4
1
2
3
4
5
5. Yes. It is closed since the only elements in the table
are from the set {1, 2, 3, 4, 5}. The identity element
is 5. The inverses are 1 – 4, 2 – 3, 3 – 2, 4 – 1, and
5 – 5. The system is associative. The system is
commutative since the table is symmetric about the
main diagonal. Thus, all five properties are satisfied.
7. 5 – 18 = (15 + 5) – 18 = 20 – 18 = 2 ≅ 2 mod 5
The binary operation is ∆ .
Yes. All elements in the table are from the set {W, S, T, R}.
The identity element is T, since T ∆ x = x = x ∆ T, where x is any member of the set {W, S, T, R}.
The inverse of R is S, since R ∆ S = T
(T ∆ R) ∆ W = R ∆ W = S
9. The system is not a group. It does not have the closure property since c  c = d, and d is not a member of {a, b, c}.
10. Since all the numbers in the table are elements of {1, 2, 3}, the system is closed. The commutative property
holds since the elements are symmetric about the main diagonal. The identity element is 2 and the inverses are
1 – 3, 2 – 2, 3 – 1. If it is assumed the associative property holds as illustrated by the example:
(1 ? 2) ? 1 = 2 = 1 ? (2 ? 3), then the system is a commutative group.
340 CHAPTER 10
Mathematical Systems
11. Since all the numbers in the table are elements of {@, $, &, %}, the system is closed. The commutative
property holds since the elements are symmetric about the main diagonal. The identity element is $ and the
inverses are @ – &, $ – $, & – @, % – %. It is assumed the associative property holds as illustrated by the
example: (@ O $) O % = & = @ O ($ O %), then the system is a commutative group.
64 ≅ 1 (mod 9)
12. 64 ÷ 9 = 7, remainder 1
13. 58 ÷ 11 = 5, remainder 3
3 ≅ 1 (mod 11)
14. 7 + 7 = 6 mod 8
15. 2 – 3 = (5 + 2) -3 = 4 ≅ 4 mod 5
16. 3 – 5 ≅ 7 (mod 9)
3 – 5 = (3 + 9) – 5 =12 – 5 ≅ 7 (mod 9)
12 – 5 ≅ 7 (mod 9)
Replace ? with 5.
17. 4 y 2 = 8 and 8 ÷ 6 = 1, remainder 2
4 y 2 ≅ 2 (mod 6)
Replace ? with 2.
18. 3 y ? y≅ 2 (mod 6)
3 y 0 ≅ 0 (mod 6)
3 y 1 ≅ 3 (mod 6)
3 y 2 ≅ 0 (mod 6)
3 y 3 ≅ 3 (mod 6)
3 y 4 ≅0 (mod 6)
3 y 5 ≅ 3 (mod 6)
There is no solution for ?
The answer is { }.
19. 103 ÷ 7 = 14, remainder 5
103 ≅ 5 (mod 7)
Replace ? with 5.
20. a)
0
1
2
3
4
0
0
0
0
0
0
1
0
1
2
3
4
2
0
2
4
1
3
3
0
3
1
4
2
4
0
4
3
2
1
b) The system is closed. The identity is 1.
However, 0 does not have an inverse, so
the system is not a group.
GROUP PROJECTS
Group Projects
1. a)
♣ A
A B
B C
C D
D A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
b) The system is closed. The identity is D.
c) (A ♣ B) ♣ C = C ♣ C = B
A ♣ (B ♣ C) = A ♣ A = B
Yes, Associative
d) A ♣ C = D
B♣B=D
C♣A=D
D♣D=D
All elements have inverses.
e) A ♣ B = C = B ♣ A
Yes, Commutative, symmetrical around the
main diagonal
Therefore, the system is a group.
3. a)
mod 3
y 0
0 0
1 0
2 0
1
0
1
2
2
0
2
1
3. a)
mod 5
y 0
0 0
1 0
2 0
3 0
4 0
1
0
1
2
3
4
2
0
2
4
1
3
3
0
3
1
4
2
4
0
4
3
2
1
2. a) Yes, see Group Project exercise 3. a).
b) Product = 0 when factors
0
mod 4, mod 6, mod 8, mod 9
c) Product = 0 when at least 1 factor = 0
mod 3, mod 5, mod 7
d) The systems in which the modulo is a composite
number system have factors
mod 4
y 0
0 0
1 0
2 0
3 0
1
0
1
2
3
2
0
2
0
2
3
0
3
2
1
mod 6
y 0
0 0
1 0
2 0
3 0
4 0
5 0
1
0
1
2
3
4
5
2
0
2
4
0
2
4
3
0
3
0
3
0
3
4
0
4
2
0
4
2
5
0
5
4
3
2
1
0.
341
342 CHAPTER 10
Mathematical Systems
3. a)
mod 7
y 0
0 0
1 0
2 0
3 0
4 0
5 0
6 0
1
0
1
2
3
4
5
6
2
0
2
4
6
1
3
5
3
0
3
6
2
5
1
4
4
0
4
1
5
2
6
3
5
0
5
3
1
6
4
2
6
0
6
5
4
3
2
1
mod 9
y 0
0 0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
8 0
1
0
1
2
3
4
5
6
7
8
2
0
2
4
6
8
1
3
5
7
3
0
3
6
0
3
6
0
3
6
4
0
4
8
3
7
2
6
1
5
5
0
5
1
6
2
7
3
8
4
6
0
6
3
0
6
3
0
6
3
mod 8
y 0
0 0
1 0
2 0
3 0
4 0
5 0
6 0
7 0
1
0
1
2
3
4
5
6
7
2
0
2
4
6
0
2
4
6
3
0
3
6
1
4
7
2
5
4
0
4
0
4
0
4
0
4
5
0
5
2
7
4
1
6
3
6
0
6
4
2
0
6
4
2
7
0
7
6
5
4
3
2
1
3. b) mod 3, mod 5, mod 7
7
0
7
5
3
1
8
6
4
2
8
0
8
7
6
5
4
3
2
1
c) mod 4, mod 6, mod 8, mod 9
d) Modulo systems that have composite numbers
have multiplicative inverses for all nonzero
numbers.