# Solution Manual for Mathematical Reasoning for Elementary Teachers, 7th Edition

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Chapter 2 Sets and Whole Numbers
COOPERATIVE INVESTIGATION
Counting Cars and Trains
1. 8 trains: P, (LG)W, W(LG), RR, RWW,
WRW, WWR, WWWW
2. 16 trains: Y, PW, WP, (LG)R, R(LG),
(LG)WW, W(LG)W, WW(LG), RRW, RWR,
WRR, RWWW, WRWW, WWRW, WWWR,
WWWWW
3. The number of trains of lengths 3, 4, and 5 are
respectively 4, 8, and 16, revealing a doubling
pattern. Continuation of the pattern shows
there will be 32, 64, and 128 trains of
respective lengths 6, 7, and 8.
4. 2nโ1 of length n can be formed. It must be
shown that there are twice as many trains of
length n as there are of length n โ 1. There are
two ways to lengthen a train of length
n โ 1 by one unit: (1) add a white caboose,
giving us 2nโ 2 trains of length n, all of which
have white cabooses; and (2) lengthen the
caboose by one unit, giving us another 2nโ 2
trains of length n, all of which have nonwhite
cabooses. Altogether, the two methods give us
2n โ 2 + 2n โ 2 = 2(2n โ 2 ) = 2nโ1 different trains
of length n. This counts all of the trains of
length n, since any train of length n
corresponds to a unique train of length n โ 1
by either removing a white caboose or
shortening a nonwhite caboose by one unit.
Since the number of trains of length one is 20,
there are n โ 1 doublings to give us 2nโ1
trains of length n. Alternatively, think of
starting with a train of n white cars. Those
could be glued together in all possible ways to
form all possible trains of length n. To do this,
at each of the n โ 1 junctions between two
white cars one must make one of two
choicesโto glue or not to glue. Hence, there
are 2nโ1 possibilities in all, as above.
5. (i) 5 trains: WWWW, WWR, WRW, RWW,
RR
(ii) 8 trains: WWWWW, WWWR, WWRW,
WRWW, RWWW, WRR, RWR, RRW
(iii) The number of RW-trains of length eight
is F9 = 34 ; 10 of these trains have five
cars (WWRRR and its rearrangements).
(iv) The number of RW-trains of length 1, 2,
3, 4, 5, … is 1, 2, 3, 5, 8, …, the Fibonacci
numbers beginning with F2. Recall that
F1 = F2 = 1 and Fn + 2 = Fn +1 + Fn . Note
that adding a white caboose to the trains
of length n + 1 and adding a red caboose
to trains of length n yields all of the RWtrains of length n + 2. Thus, the number
of trains of length n is Fn +1 .
6. (i) 2 trains: P, RR
(ii) 3 trains: Y, (LG)R, R(LG)
(iii) 13 trains: Br, (DG)R, R(DG), Y(LG),
(LG)Y, PP, PRR, RPR, RRP,
(LG)(LG)R, (LG)R(LG), R(LG)(LG),
RRRR. None of these trains contain five
cars.
(iv) The number of W-trains of length 1, 2, 3,
4, 5, … is 0, 1, 1, 2, 3, …, again the
Fibonacci numbers where we agree to set
F0 = 0. Note that adding cabooses of
length n โ 1, n โ 2, n โ 3, …, 2 to the
W-trains of length 2, 3, 4, .., n โ 1 gives
all the W-trains of length n + 1, except
for the one additional train that consists of
a single rod. This corresponds to the
formula 1 + F1 + F2 + + Fn โ 2 = Fn .
Thus the number of W-trains of length n
is Fn โ1 .
7. Answers will vary.
Section 2.1
Sets and Operations on Sets
Problem Set 2.1
1. (a) {Arizona, California, Idaho, Oregon,
Utah}
(b) {Maine, Maryland, Massachusetts,
Michigan, Minnesota, Mississippi,
Missouri, Montana}
(c) {Arizona}
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(c) True. The sets contain the same elements.
Order doesnโt matter.
2. (a) {l, i, s, t, h, e, m, n, a, o, y, c}
(b) {a, e, m, t}
8. (a) True. 7 is in {6, 7, 23}, but the two sets
are not the same.
(c) โ
3. (a) {7, 8, 9, 10, 11, 12,13}
(b) False. The sets are equal.
(b) {9, 11, 13}
(c) False. 6 is not a member of {7}.
(c) Since 2 = 2 ยท 1, 4 = 2 ยท 2, 6 = 2 ยท 3,
8 = 2 ยท 4, 10 = 2 ยท 5, etc., the set is
{2, 4, 6, 8, 10, 12, 14, 16, 18, 20}.
9.
4. (a) {4, 8, 12, 16, 20}
(b) Since 3 = 2 ยท 1 + 1, 5 = 2 ยท 2 + 1,
7 = 2 ยท 3 + 1, 9 = 2 ยท 4 + 1, 11 = 2 ยท 5 + 1,
13 = 2 ยท 6 + 1, 15 = 2 ยท 7 + 1,
17 = 2 ยท 8 + 1, and 19 = 2 ยท 9 + 1, the set
is {3, 5, 7, 9, 11, 13, 15, 17, 19}.
(c) Since 1 = 12, 4 = 22, 9 = 32, and 16 = 42,
the set is {1, 4, 9, 16}.
5. Answers will vary.
(a)
{ x โU 11 โค x โค 14} or
{ x โU 10 < x 12} or
{ x โ N x = 2n for n โ N and n > 6}
(c)
B โฉ C = {a, b}
(d) A โช B = {a, b, c, d, e}
(e)
A = {f, g, h}
(f)
A โฉ C = {a, b}
(g)
A โช ( B โฉ C ) = {a, b, c, d, e}
10. (a) M = {45, 90, 135, 180, 225, 270, 315, …}
(b) L โฉ M = {90, 180, 270, …}
This can be described as the set of natural
numbers that are divisible by both 6 and
45, or as the set of natural numbers that
are divisible by 90.
{x โ N x = n for n โ N with n odd
11. (a) D = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36,
48, 72, 144}
(b) G โฉ D = {1, 2, 3, 6, 9, 18}
2
and n โฅ 5}
(c)
(b) A โฉ B = {a, b, c}
(c) 90
6. Answers will vary.
(a)
(a) B โช C = {a, b, c, h}
{ x โ N x is divisible by 3}
(c) 18
12. (a) A โฉ B โฉ C is the set of elements common
to all three sets.
7. (a) True. The sets contain the same elements.
(b) True. Every element in {6}โnamely, 6โ
is in {6, 7, 8}.
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Chapter 2 Sets and Whole Numbers
(b) A โช ( B โฉ C ) is the set of elements in B
but not in C, together with the elements
of A.
(c) ( A โฉ B ) โช C is the set of elements in C or
in both A and B.
(d) A โช ( B โฉ C ) is the set of all elements that
are not in A, together with additional
elements that are in both B and C.
(c) Answers may vary.
14. No. It is possible that there are elements of A
that are also elements of B but not C, or C but
not B. For example, let A = {1, 2}, B = {2, 3},
C = {3}. A โช B = A โช C = {1, 2, 3}, but
B โ C.
15. No, it is not necessarily true that F = G. For
example, if D is any subset of the whole
numbers, then F = {0, 1} and G = {7, 9} is a
counterexample. In other words, suppose that
D = {2, 4, 6, 8} and F and G are defined as
before. D โฉ F = โ
= D โฉ G , but F โ G.
16. (a) Since
A โฉ B = {6, 12, 18} ,
(e)
A โช B โช C is the set of all elements in A,
B, and/or C.
A โฉ B = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11,
13, 14, 15, 16, 17, 19, 20}
Since
A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} and
B = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16,
17,19, 20},
(f)
A โฉ B โฉ C is the set of elements that are
in both B and C, but not in A.
A โช B = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14,
15, 16, 17, 19, 20}
Since
A โช B = {2, 3, 4, 6, 8, 9, 10, 12, 14,
15, 16, 18, 20}
A โช B = {1, 5, 7, 11, 13, 17, 19} .
Since
A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} and
13. (a) A and B must be disjoint sets contained in
C.
B = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14,
16, 17, 19, 20},
A โฉ B = {1, 5, 7, 11, 13, 17, 19} .
(b) Answers may vary.
(b) The results of part (a) suggest that
A โฉ B = A โช B and A โช B = A โฉ B
17. (a) R โฉ C is the set of red circles. Draw two
red circlesโone large and one small.
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Solutions to Problem Set 2.1
(b) L โฉ H is the set of large hexagons. Draw
two large hexagonsโone red and one
blue.
(c) T โช H is the set of shapes that are either
triangles or hexagons. Draw two large
triangles, two large hexagons, two small
triangles, and two small hexagonsโa red
and a blue of each.
(d) L โฉ T is the set of large triangles. Draw
two large trianglesโone red and one
blue.
(e) B โฉ C is the set of blue shapes that are
not circles. That is, the set of four
elements consisting of the large and small
blue hexagons and the large and small
blue triangles.
(f)
H โฉ S โฉ R is the set of hexagons that are
both small and red. Draw one small red
hexagon.
18. (a) L โฉ T
(b) B โฉ (T โช H ) or B โฉ C
(c)
S โชT
(d) ( B โฉ C ) โช R
19. (a) Answers will vary. One possibility is
B = set of students taking piano lessons,
C = set of students learning a musical
instrument.
of shapes used. The number of cards is
probably 3 ร 3 ร 2 ร 3 = 54 . We cannot
be sure of this result without knowing for
sure that the deck does not contain any
different shapes, colors, etc.
(b) Answers will vary.
(c) Answers will vary.
21. Answers will vary.
22. Children often interpret โorโ as the โexclusive
orโ in which either one of two possibilities
holds, but not both. Have the child consider
the statement โTomorrow, we will have music
class or we will go to the gymโ, and point out
that this remains true even if both music class
and gym class take place.
23. Answers will vary. Sample answer:
The natural numbers (also called the counting
numbers) are a subset of the whole numbers.
Zero is a whole number but not a natural
number, but all natural numbers are a subset of
the whole numbers.
24. Joe always seems to find two sets which are
disjoint. His teacher can use any pair of sets
such as the set of all boys in his class and the
other set of all students in the class of those
who have brown eyes. If the two sets have an
overlap, Joe won’t get the addition that he
expected.
25. (a) 8 regions
(b) Answers will vary.
(c) Answers will vary.
20. (a) There appear to be 3 choices for shape, 3
choices for color, 2 choices for shading
(shaded or not), and 3 choices for number
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(b) By counting, there are 14 regions.
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Chapter 2 Sets and Whole Numbers
(c) Verify that A โฉ B โฉ C โฉ D has no region
by observing that the region for A โฉ D is
entirely contained in B โช C . Likewise,
the region for B โฉ C is entirely contained
in A โช D. The other missing region is
Aโฉ B โฉC โฉ D .
(d) Yes. Each loop contains 8 different
regions, and there are 16 regions all
together.
30. Note that in this diagram, O refers to the
region outside both circles.
31. Answers will vary. (The loop shown on the
bottom could have represented negative
instead of positive, reversing all signs.)
Note that in this diagram, Oโ refers to the
region outside all circles.
26. These Venn diagrams show that
Aโฉ B = Aโช B :
Aโฉ B
32. B. 4 students like all three foods, so there must
be a region common to all three loops in the
Venn diagram.
Aโช B
27. These Venn diagrams show that
Aโช B = Aโฉ B :
Aโช B
33.
A โฉ B = {6, 12, 18}
These are the elements that are divisible by
both 2 and 3.
Aโฉ B
28. (a) There are 6 choices for shape,
2 choices for size, and 2 choices for color,
so the number of pieces is 6 ร 2 ร 2 = 24 .
(b) 6 ร 2 ร 2 ร 2 = 48
29. (a) There are eight subsets: โ
, {P}, {N},
{D}, {P, N}, {P, D}, {N, D},
{P, N, D}.
(b) There are 16 subsets: โ
, {P}, {N}, {D},
{P, N}, {P, D}, {N, D},
{P, N, D}, {Q}, {P, Q}, {N, Q},
{D, Q}, {P, N, Q}, {P, D, Q},
{N, D, Q}, {P, N, D, Q}.
(c) Half of the subsets of {P, N, D, Q}
contain Q.
(d) The number of subsets doubles with each
additional element, so a set with n
elements has 2n subsets.
34. Choice H (42) only. It is both divisible by 7 (as
in set V) and divisible by 3 (as in set W).
35. The number must be a multiple of both 10 and
12. The question asks for the least number
with that property, so the answer is
LCM(10, 12) = 3 โ
2 โ
2 โ
5 = 60, choice D.
36.
There are 28 cello students, of whom 12 also
play violin Therefore, there are 28 โ 12 = 16
who only play cello. There are 23 violin
students, of whom 12 also play cello.
Therefore, there are 23 โ 12 = 11 who only
play violin. Thus, there are 16 + 12 + 11 = 39
students who play either the cello or violin.
There are 54 students, so there are
54 โ 39 = 15 students who do not play either
cello or violin.
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Solutions to Problem Set 2.2
Section 2.2
Sets, Counting, and the Whole Numbers
Problem Set 2.2
8. (a) The solutions for ( x โ 1)( x โ 9) = 0 are
x = 1 and x = 9, so C = {1, 9} and
n(C) = 2.
(b) n(D) = 2 because D = {40, 80}.
1. (a) 13(th): ordinal
first: ordinal
(b) fourth: ordinal; second: ordinal
93: cardinal. In common usage, it
represents the number, 93, correct out of
100.
9. (a) The correspondence 0 โ 1, 1 โ 2,
2 โ 3,…, w โ w + 1,… shows W โฌ N.
(b) The correspondence 1 โ 2, 3 โ 4,โฆ ,
n โ n + 1,โฆ shows D โฌ E.
(c) The correspondence
2. (a) fourth: ordinal
13: cardinal
(b) 2213762789: nominal; 23rd: ordinal
18: cardinal.
3. (a) Equivalent, since there are five letters in
the set {A, B, M, N, P}
(b) Not equivalent since the sets have
different numbers of elements.
(c) Equivalent, say by the correspondence
o โ t,n โ w,e โ o
(d) Not equivalent, since the set {0} has one
element and โ
has no elements.
1 โ 10, 2 โ 100 = 102 ,โฆ , n โ 10n ,โฆ
shows that the sets are equivalent.
10. (a) Finite. The number of grains of sand is
large, but finite.
(b) Infinite
(c) Infinite
11. (a) Answers will vary. For example,
Q1 โ Q2 , Q3 โ Q4 , and so on. (Note
that P need not be the centerโit can be
any fixed point inside the small circle.)
4. (a) Equivalent, since there are four elements
in each set.
(b) Equivalent, since there are 24 elements in
each set.
(c) Not equivalent, since {0, 1} has two
elements and the other set is infinite.
5. (a) n(A) is 0, 1, 2, 3, or 4, since it has strictly
fewer elements than of set B.
(b) Answers will vary. For example,
Q1 โ Q2 , Q3 โ Q4 , and so on.
(b) n(C) is 5, 6, 7, …, any whole number
greater than or equal to 5.
6. Yes, A = {1, 8, 27, 64}, B = {California,
Arizona, New Mexico, Texas}.
Since both sets have the same number of
elements, A โฌ B. We could have the
correspondence 1 โ California,
8 โ Arizona, 27 โ New Mexico,
64 โ Texas.
12. (a) Answers will vary. For example,
Q1 โ Q2 , etc.
7. (a) n(A) = 7 because
A = {21, 22, 23, 24, 25, 26, 27}.
(b) n(B) = 0 because B = โ
. There is no
natural number x such that x + 1 = x.
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29
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Chapter 2 Sets and Whole Numbers
that A โฉ B = A โช B . Thus A โช B has no
additional elements besides those in
A โฉ B, and so neither A nor B has any
additional elements. Since A = A โฉ B
and B = A โฉ B , we conclude that A = B.
(Caution: This reasoning would not be
valid if infinite sets were allowed.)
(b) Answers will vary. For example,
Q1 โ Q2 , etc.
13. (a) True. A set B cannot have fewer elements
than its subset A.
(b) False. Possible counterexample:
A = {1}, B = {2, 3}. Then n(A) = 1 and
n(B) = 2, so n(A) < n(B), but A โ B .
(c) True. The union A โช B does not include
any more elements than just A, so the
elements of B must already be elements
of A.
(d) True. If an element of A was not also an
element of B, then this element would be
missing in A โฉ B but included in A.
14. (a) True.
16. (a) 1000 รท 6 = 166.666โฆ, so the largest
element of S is 166 ยท 6 = 996. Therefore,
n(S) = 166.
(b) The elements in F โฉ S are the multiples
of 30. Since 1000 รท 30 = 33.33โฆ, the
largest element of F โฉ S is 33 ยท 30 =
990. Therefore n ( F โฉ S ) = 33.
(c)
17. Three of the four regions in the 2-loop diagram
account for 500 of the households. Thus there
are 200 households in the overlapped region
representing the number of households having
both a TV and a computer.
(b) True.
(c) False. For example, if
B = { x | x is any positive integer except 3} ,
then B = {3} .
(d) False. This is true only if A and B are
disjoint.
15. (a) n( A โฉ B) โค n( A)
The set A โฉ B contains only the elements
of A that are also elements of B. That is,
A โฉ B โ A . Thus, A โฉ B cannot have
more elements than A.
(b) n( A) โค n( A โช B )
18. Use the fact that the eight regions pictured
below are mutually disjoint sets and apply the
strategy of working backwards, that is, start
with n( A โฉ B โฉ C ) = 7 . Next use n( A โฉ B) ,
n( B โฉ C ) , and n( A โฉ C ) to find the values
10, 5, and 8, respectively. Then use n(A), n(B),
and n(C) to find the values 15, 28, and 10,
respectively, and finally n(U) to find the value
17.
The set A โช B contains all of the
elements of the set A and any additional
elements of B that are not already
included. Then A โ A โช B . Therefore,
A โช B must have at least as many
elements as A.
(c) Since n( A โฉ B) = n( A โช B) and
A โฉ B โ A โ A โช B , we can conclude
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Solutions to Problem Set 2.2
19. The 3-loop Venn diagram can be filled in the
manner of problem 18. We now see there are
20 + 5 + 25 = 50 percent of the students who
like just one sport and 5 percent do not like
any of the three sports.
20. Let A, H, and P be the sets of students taking
anthropology, history, and psychology,
respectively. Then n(A) = 40, n(H) = 11,
n(P) = 12, n( A โฉ H โฉ P) = 3, n( A โฉ H ) = 6,
and n( A โฉ P) = 6. Note: If one chooses to
solve this problem by using a Venn diagram,
there are several possibilities, since the
number of students taking both history and
psychology cannot be determined. However,
the answers to parts (a), (b), and (c) are
uniquely determined.
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23. Answers will vary. One possible answer uses a
cup with 5 marbles. Ask โhow many marbles
are in the cup?โ to get the response โfive.โ
Now remove a marble, and again ask how
many marbles are in the cup, eliciting the
response โfour.โ Continue to remove one
marble at a time, until no marbles remain in
the cup. Explain that the number of marbles in
the empty cup is โzero.โ
24. Answers will vary, but should include the
concept that if one strip is shorter than a
second strip, and the second strip is shorter
than the third, then the first must be shorter
than the third.
25. 400 is a square number since 400 is the
product of two squares, 4 and 100. Also,
400 = 202 , as illustrated below.
(a) n( A โ ( A โฉ H ) โช ( A โฉ P))
= n( A) โ n( A โฉ H ) โ n( A โฉ P )
+ n( A โฉ H โฉ P )
= 40 โ 6 โ 6 + 3 = 31
Note: One might argue that the answer is
zero, since all of the students in this
situation are also taking mathematics.
(b) n( A โช H ) = n( A) + n( H ) โ n( A โฉ H )
= 40 + 11 โ 6 = 45
(c) n(( H โฉ A) โ P )
= n( H โฉ A) โ n( H โฉ A โฉ P)
= 6โ3=3
26. Annabelle recognized that โthirdโ and โsixthโ
are nominal. There isnโt a cardinal number in
the conversation, so there is no addition in this
setting. Annabelle laughed with the math
professor, as did her parents.
21. (a)
27. (a) The numbers that have been inserted are
bold. Explanations to Zack will vary.
(b)
22. Answers will vary. Both โnumberโ and โcolorโ
are abstract concepts that can be taught by
giving concrete examples.
(b) Answers will vary
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Chapter 2 Sets and Whole Numbers
28. (a) Richard is multiplying by 2, rather than
squaring. He must think that the exponent
2 is just a multiplier.
(b) Drawing pictures of squares of size 1 ร 1,
2 ร 2, 3 ร 3, 4 ร 4, and 5 ร 5 would show
that areas donโt double, they grow much
faster than that.
31. (a)
Row 0
1
Row 1
1
1
Row 2
1
2
1
Row 3
1
3
3
1
Row 4
1
4
6
4
1
(b) The table is the same as Pascalโs triangle
each entry in the table is the sum of the
numbers in the row just above that are
directly above and one column to the left.
Using this pattern, we get
Row 5: 1 5 10 10 5 1 and
Row 6: 1 6 15 20 15 6 1.
29. (a) Jeff is multiplying the base by the
exponent, which is the wrong procedure.
In this case, he is multiplying by three,
not taking numbers to the third power.
32. Evelynโs assumptions give the following
diagram.
(b) Use cubes whose sides are of length 1 cm,
2 cm, 3 cm, 4 cm, and 5 cm.
30. (a) Six ways: yrg, rgy, gyr, ygr, gry, ryg
(b) Six. The numbers 1, 2, 3 can be viewed as
the bottom, middle, top positions for balls
in the can. Therefore, each of the six ways
to place the balls in the can are a distinct
one-to-one correspondence.
(c) There are 12 ways to place the first egg,
and 11 empty spaces in which to place the
second egg. Therefore, the first two eggs
can be put into the carton in 12 11
ways. This leaves 10 empty choices for
the third egg, so there are 12 11 10
ways to place the first three eggs.
Continuing in this way shows there are
12 11 10 9 8 7 6 5 4
3 2 1 = 479,001,600 ways to
place the dozen eggs.
(d) If {1, 2, 3, โฆ, 12} and {a, b, c, d, e, f, g,
h, i, j, k, l} represent the set of eggs and
the set of spaces in the carton from part
(c), the number of one-to-one
correspondences is the same as the
number of ways to put the eggs into the
carton, namely
12 11 10 9 8 7 6 5 4
3 2 1 = 479,001,600.
The zero values are obtained from Evelynโs
assumption that anyone with a VCR also has a
TV. Evelyn should question the survey
because only 94 of the 100 households are
represented. (Evelynโs assumption may be
wrongโfor example, there may be people who
use a monitor instead of a TV to view
videotapesโbut these are very unlikely to
account for 6% of a given sample.) Also, the
stereo and VCR category is missing
33. Consider the 35 students taking Arabic: 20
take only Arabic, 7 take both Arabic and
Bulgarian (and possibly Chinese), telling us
that 8 students are taking Arabic and Chinese
and not Bulgarian. Similarly, there are 5
students taking Bulgarian and Chinese but not
Arabic. The rest of the values in the Venn
diagram are now easy to fill in. In particular, 3
students take all three languages, and 26 are
not taking any of the three languages.
34. Since we are given that k < l and l < m, we can
choose sets K, L, and M satisfying
K โ L โ M and n(K) = k, n(L) = l, and
n(M) = m. By the transitive property of set
inclusion (see Section 2.1, or just look at a
Venn diagram) we know that K โ M and so
k 19)
(f) Closed (since the sum of two multiples of
3 is a multiple of 3: 3a + 3b = 3(a + b))
9. (a) Commutative property of addition
(b) Closure property
(c) Additive-identity property of zero
(b)
(d) Associative and commutative properties
(e) Associative and commutative properties
10. (a) (1 + 20) + (2 + 19) + (3 + 18) + + (10 + 11)
= 21 + 21 + 21 + + 21 (for 10 terms)
= (10)(21) = 210
(b) Associative and commutative properties
11. (a)
(b)
15. (a) comparison
12. (a)
(b) measurement (number-line)
16. (1) missing addend
(b)
(b) take-away
17. Answers will vary. For example,
13. (a) 5 + 7 = 12
7 + 5 = 12
12 โ 7 = 5
12 โ 5 = 7
(b) 4 + 8 = 12
8 + 4 = 12
12 โ 8 = 4
12 โ 4 = 8
(a) Take-away: Maritza bought a booklet of
20 tickets for the amusement park rides.
She used 6 tickets for the roller coaster.
How many tickets does she have left?
Copyright ยฉ 2015 Pearson Education, Inc.
36
Chapter 2 Sets and Whole Numbers
(b) Missing Addend: Nancyโs school has 23
proof-of-purchase coupons for graphing
calculators. Her school will be entitled to
a free overhead projection calculator with
40 proof-of-purchase coupons. How many
more coupons do they need to get the free
calculator?
(c) Comparison: Oak Ridge Elementary
School has an enrollment of 482 students
and Crest Hill School has an enrollment
of 393 students. How many more students
does Oak Ridge have than Crest Hill?
(d) Measurement: A fireman climbed up 11
rungs on a ladder, but the smoke was too
thick and he came down 3 rungs. How
many rungs up the ladder is the fireman?
18. Jeff has read through page 240. Therefore, the
number of pages is 257 โ 240 = 17 pages or
257 โ 241 + 1 = 17 pages.
19. Use the guess and check method. Some
answers will vary.
22. Write all possible combinations of 3 different
numbers whose sum is the indicated number in
the triangle. Then, place numbers which occur
in more than one sum at the vertices.
(a)
1+3+6
2+3+5
4+5+1
(b)
1+4+6
2+3+6
5+4+2
(c)
1+5+6
2+4+6
3+4+5
(a) ((8 โ 5) โ 2) โ 1 = 0
23. Blake has one more marble than before, and
Andrea has one less than before, so now Blake
has two more marbles than Andrea. Use a
small number of marbles, say five each, to
demonstrate what happened. After giving
Blake one marble, he has 6 and Andrea has 4,
and thus clearly Blake has two more marbles
than Andrea.
(b) 8 + (5 โ 2) + 1 = 12
24. Answers will vary.
(c) (8 + 5) โ (2 + 1) = 10
25. Answers will vary.
(a) (8 โ 5) โ (2 โ 1) = 2
(b) 8 โ (5 โ 2) โ 1 = 4
20. Use the guess and check method. Some
answers will vary.
21. (a) First fill in the squares by noting that
3 โ 1 = 2, 4 โ 2 = 2, and 7 โ 2 = 5. Then
complete the circles.
26. (a) 0 is the cardinal number of the empty set,
n{โ
} = 0. If the set A has a elements in it
(i.e., n(A) = a), then
n( A) + n(โ
) = n( A โช โ
). But,
A โช โ
= A, since there are no elements in
โ
to add to the set A. Thus,
n( A โช โ
) = n( A) = a.
(b) Use the guess and check method.
(b) Answers will vary, but the major idea is
to give an example of starting with a
group of children in the room, playing the
role of A, and adding all others in the
room who are over 80 years old. As there
is no addition to A, the number of
elements in it doesnโt change, and so
a + 0 = a.
Copyright ยฉ 2015 Pearson Education, Inc.
Solutions to Problem Set 2.3
37
27. Answers will vary but should include that given two of the minuend, subtrahend, and answer, the third can be
found. Furthermore, students often struggle when problems are presented in โnon-conventionalโ ways. (This
is worth the struggle as it is the beginning of algebra.) Many elementary school students have difficulty when
there is a blank in the front of the number sentence.
28. Answers will vary.
29. First find the top row and the left column. For example, the fourth entry in the left column must be 6 โ 2 = 4,
and then the third entry in the top row is 5 โ 4 = 1, and so on. The completed table is shown below.
+
5
4
1
6
9
2
0
8
7
3
3
8
7
4
9
12
5
3
11
10
6
9
14
13
10
15
18
11
9
17
16
12
6
11
10
7
12
15
8
6
14
13
9
4
9
8
5
10
13
6
4
12
11
7
0
5
4
1
6
9
2
0
8
7
3
7
12
11
8
13
16
9
7
15
14
10
5
10
9
6
11
14
7
5
13
12
8
2
7
6
3
8
11
4
2
10
9
5
1
6
5
2
7
10
3
1
9
8
4
8
13
12
9
14
17
10
8
16
15
11
30. They must both love some of the same mice. In
fact, there are five mice that Angel and Jane
both love since 8 = 7 + 6 โ n (where there are
n mice they both love).
33. (a) Yes, Carmenโs method is correct using the
associativity property and a good
understanding of position value.
(b)
31. (a) This student is โsubtracting upโ, which is
a common mistake.
(b) Answers will vary. For example: After
reminding her about place value, ask her
to do a two-digit subtraction such as
73 โ 35 and reflect on the similarities
between how to subtract two-digit
numbers and how to subtract three-digit
numbers. That could be followed with
736 โ 327 = ? as a next level problem that
is easier than the one in the problem.
32. Phillip does two things incorrectly. First, he
sees that he canโt subtract the 9 from the 6, so
he goes to the next column to do an exchange.
He changes the 8 to a 7, but then, instead of
having 16 ones, he adds 1 to 6 to give 7. He
has misunderstood that the 1 he got from the
exchange is actually 10, not 1 unit. Phillip then
proceeds to subtract up anyway (he mistakenly
writes 7 โ 9 as 2).
60 + 11 = 71
โ (30 + 8) = โ38
30 + 3 = 33
34.
Note that elements of A โฉ B are counted
twice when calculating n(A) + n(B). We
compensate by subtracting n( A โฉ B) ,
giving n( A โช B).
35. Let
T = {n โ N n is a multiple of 12 and n โค 200}
and
F = {n โ N n is a multiple of 5 and n โค 200} .
The number we need to find is
n(T โช F ) = n(T ) + n( F ) โ n(T โฉ F )
= 16 + 40 โ 3 = 53
Copyright ยฉ 2015 Pearson Education, Inc.
38
Chapter 2 Sets and Whole Numbers
22 = 21 + 1
23 = 10 + 10 + 3
24 = 21 + 3
25 = 15 + 10
36.
(c) For example, choose 73 and 74.
73 = 45 + 28, and 74 = 45 + 28 + 1. Three
is the maximum number of triangular
numbers needed.
37.
(d) Any whole number may be written as the
sum of at most three triangular numbers.
38. {0}, since 0 โ 0 = 0.
39. (a) {2, 3, 4, 5, 6, 7, 8, 9, โฆ} since 2 + 2 = 4,
2 + 3 = 5, 2 + 4 = 6, etc.
(b) {0, 1}
(c) No, because the set of all whole numbers,
{0, 1, 2, โฆ}, fits the given description.
(d) Whole numbers that must be in C are all
even numbers โฅ 2. Zero and odd numbers
may or may not be in C.
40. (a) Use the formula tn =
n(n + 1)
.
2
n
1
2
3
4
5
6
tn
1
3
6
10
15
21
n
7
8
9
10
11
12
tn
28
36
45
55
66
78
n
13
14
15
tn
91
105
120
(b) 11 = 10 + 1
12 = 6 + 6
13 = 10 + 3
14 = 10 + 3 + 1
15 = 15
16 = 15 + 1
17 = 15 + 1 + 1
18 = 15 + 3
19 = 10 + 6 + 3
20 = 10 + 10
21 = 21
41. It can be shown that Sameer is correct. The
easiest way to find a sum is to use the largest
Fibonacci number possible as the next
summand, a procedure sometimes called a
โgreedy algorithm.โ For example, to express
100 as a Fibonacci sum, first write down 89
since it is the largest Fibonacci number less
than 100. This leaves 11, so next write down
8. The 3 that still remains to be accounted for
is a Fibonacci number, so 100 = 89 + 8 + 3.
42. (a) Corey is correct. The sum of two even
numbers can be thought of as the sum of
two rows having m columns with two
rows having n columns. The result is two
rows having m + n columns. Since the
result can always be represented using
two rows each having the same number of
columns, the result is always an even
number.
(b) Maya is mistaken. This can be shown by a
counterexample. For example, the sum of
the odd numbers 3 and 5 is the even
number 8.
43. Statement B does not correspond to the
number sentence 15 โ 8 = ไ. The other
problems illustrate take-away (A), comparison
(C), and missing addend (D).
44. C. Jordan started with his allowance, x, and
subtracted $4 from it because he spent that
amount. He is left with x โ 4 which is the $16
he has left. Thus x โ 4 = 16.
45. D. The number must be more than 500.
46. A. 45,749. Just subtracting the last two digits
in your head gives the only possible correct
answer.
47. C. 1975 + 16 + 4 = 1995
48. B. 54 + 17 = 71
Copyright ยฉ 2015 Pearson Education, Inc.
Solutions to Problem Set 2.4
39
Section 2.4
Multiplication and Division of Whole Numbers
Problem Set 2.4
1. (a) 3 5 = 15, set model (repeated addition)
(b) 6 3 = 18, number-line model
(c) 5 3 = 15, set model
(d) 3 6 = 18, number-line model
(e) 8 4 = 32, rectangular area model
(f) 3 2 = 6, multiplication tree model
2. (a) A set of dominoes, containing 11 stacks of 5 dominoes is modeled by a rectangular array.
11 ร 5 = 55
The set contains 55 dominoes.
(b) The outfits that can be made from 3 skirts and 6 blouses is modeled by a Cartesian product, shown
below.
3 ร 6 = 18, so she has 18 outfits she can wear.
b1
b2
b3
b4
b5
b6
s1
(s1 , b1 )
(s1 , b 2 )
(s1 , b3 )
(s1 , b 4 )
(s1 , b5 )
(s1 , b6 )
s2
(s 2 , b1 )
(s 2 , b2 )
(s 2 , b3 )
(s 2 , b4 )
(s 2 , b5 )
(s 2 , b6 )
s3
(s3 , b1 )
(s3 , b 2 )
(s3 , b3 )
(s3 , b 4 )
(s3 , b5 )
(s3 , b6 )
(c) A hike of 10 miles each day for 5 days
can be modeled by a number line.
5 ยท 10 = 50
He hiked 50 miles.
(d) The production of 35 widgits a day in a
5-day workweek can be modeled by a set.
5 ยท 35 = 175
The company makes 175 widgets in a
5-day workweek.
Copyright ยฉ 2015 Pearson Education, Inc.
40
Chapter 2 Sets and Whole Numbers
(e) The sunroom can be modeled by a
rectangular area.
(iv) Using the commutative property of
multiplication,
56,108 6 = 6 56,108
= 56,108 + 56,108 + 56,108
+ 56,108 + 56,108
+ 56,108 = 336,648
4. (a) Each of the a lines from set A intersects
each of the b lines from set B. Since the
lines for A are parallel and the lines for B
are parallel, the intersection points will all
be distinct.
(b)
9 ร 18 = 162
She needs 162 1-square-foot tiles.
(f) The outcomes that result from rolling a
die and flipping a coin can be modeled
with a multiplication tree.
5. (a) Not closed. For example, 2 ร 2 = 4 ,
which is not in the set.
(b) Closed. 0 ร 0 = 0, 0 ร 1 = 0, 1 ร 1 = 1,
1 ร 0 = 0 . All products are in the set.
(c) Not closed. For example, 2 ร 4 = 8 ,
which is not in the set.
(d) Closed. The product of any two even
whole numbers is always another even
whole number.
(e) Closed. The product of any two odd
whole numbers is always another odd
whole number.
(f) Not closed. For example, 2 ร 23 = 24 ,
which is not in the set.
6 ร 2 = 12
12 outcomes are possible.
3. (a) Answers will vary.
(b) (i)
4 ยท 9 = 9 + 9 + 9 + 9 = 36
(ii)
7 ร 536
= 536 + 536 + 536 + 536 + 536
+ 536 + 536 = 3752
(iii)
6 47,819
= 47,819 + 47,819 + 47,819
+ 47,819 + 47,819
+ 47,819 = 286,914
(g) Closed. 2m ร 2n = 2m + n for any whole
numbers m and n.
(h) Closed. 7 a ร 7b = 7 a + b for any whole
numbers a and b.
6. (a) Closed. The product of any two whole
numbers in the set W โ {5} is also in the
set, and 5 is not a product of any numbers
in the set (5 is prime).
(b) Not closed, since 2 ร 3 = 6 .
(c) Closed. The product of any two whole
numbers in the set W โ {2, 3} is also in
the set, and 2 and 3 are not products of
numbers in the set (2 and 3 are prime).
Copyright ยฉ 2015 Pearson Education, Inc.
Solutions to Problem Set 2.4
41
7. (a) Commutative property of multiplication
(b) Distributive property of multiplication over addition
(c) Multiplication-by-zero property
8. (a) Distributive property of multiplication over addition
(b) Associative property of multiplication
(c) Multiplicative identity property of 1
9. Commutative property: 5 ร 3 = 3 ร 5
10. Commutative property: 6 ร 2 = 2 ร 6
11. (a)
(b)
(c)
12. If we assume the numbers are in feet, the
swimming pool is 2 ft by (5 + 4) = 9 ft and
thus the pool is 18 ft2. The other side of the
equation means that for the two pools, one is
2 ft by 5 ft (10 ft2) and the other is 2 ft by 4 ft
(8 ft2). The larger pool and these two pools
next to each other show the Distributive
property.
13. The rectangle is a + b by c + d, so its area is
(a + b) (c + d). The rectangle labeled F is a
by c so its area is ac. Similarly, the areas of the
rectangles O, I, and L are given by the
respective products ad, bc, and bd. Summing
the areas of the four rectangles labeled F, O, I,
and L gives the area of the large rectangle, so
(a + b) (c + d) = ac + ad + bc + bd.
14.
a
+
b
+
c
aยทd
aยทe
aยทf
bยทd
bยทe
bยทf
cยทd
cยทe
cยทf
d + e + f
(a + b + c ) ยท (d + e + f)
=aยทd+aยทe+aยทf+bยทd+bยทe+bยทf
+cยทd+cยทe+cยทf
15. (a) Distributive property of multiplication
over addition:
7 ยท 19 + 3 ยท 19 = (7 + 3) ยท 19
= 10 ยท 19 = 190
(b) Distributive property of multiplication
over addition:
24 ยท 17 + 24 ยท 3 = 24 ยท (17 + 3) = 24 ยท 20
= 480
Copyright ยฉ 2015 Pearson Education, Inc.
42
Chapter 2 Sets and Whole Numbers
(c) Distributive property, associative
property, and/or multiplication property
of zero:
36 ยท 15 โ 12 ยท 45 = (12 ยท 3) ยท 15 โ 12 ยท (3 ยท
15) = 12 ยท (3 ยท 15) โ 12 ยท (3 ยท 15)
= (12 โ 12) ยท (3 ยท 15) = 0 ยท (3 ยท 15) = 0
16. (a) Commutative and Distributive properties:
7 โ
16 + 7 โ
24 = 7 โ
40 = 280
(b) Associative and multiplicative property of
zero:
23 โ
( 25 โ
35 โ
24 โ
0) = ( 23 โ
25 โ
35 โ
24) โ
0
=0
(c) Distributive property, commutative, and
associative properties:
36 โ
15 + 12 โ
55 = 3 โ
12 โ
15 + 12 โ
55
= 12 โ
3 โ
15 + 12 โ
55
= 12 โ
45 + 12 โ
55
= 12 โ
100 = 1200
17.
22. (a) 19 โ 5 = 14, which is greater than 5, so
we must subtract 5 again. 14 โ 5 = 9,
which is still greater than 5. Subtracting 5
again, 9 โ 5 = 4, which is less than 4, so
we are done. Because we subtracted 5
three times, the quotient is 3. The
remainder is 4. This is represented as
โ5
โ5
โ5
a = 19 โ 19 โ 5 = 14 โ 14 โ 5 = 9 โ
9 โ 5 = 4 (done)
(b) 18 โ 9 = 9. Since 9 โค 9, subtract 9 again
to get 9 โ 9 = 0. The remainder is 0. We
subtracted twice, so the quotient is 2.
โ9
โ9
a = 18 โ 18 โ 9 = 9 โ 9 โ 9 = 0 (done)
(c) 25 โ 8 = 17, which is greater than 8, so
we must subtract 8 again. 17 โ 8 = 9,
which is still greater than 8. Subtracting 8
again gives 9 โ 8 = 1, which is less than
8, so we are done. Because we subtracted
8 three times, the quotient is 3. The
remainder is 1.
โ8
โ8
โ8
a = 25 โ 25 โ 8 = 17 โ 17 โ 8 = 9 โ
9 โ 8 = 1 (done)
(d) Answered in each part separately.
23. (a) 14 โ 7 = 7. Since 7 โค 7, subtract 7 again
to get 7 โ 7 = 0. The remainder is 0. We
subtracted twice, so the quotient is 2.
โ7
โ7
a = 14 โ 14 โ 7 = 7 โ 7 โ 7 = 0 (done)
18. (a) 18 รท 6 = 3
(b) 18 รท 3 = 6
19. (a) 4 8 = 32, 8 4 = 32, 32 รท 8 = 4,
32 รท 4 = 8
(b) 6 5 = 30, 5 6 = 30, 30 รท 5 = 6,
30 รท 6 = 5
20. (a) 7 9 = 63, 9 7 = 63, 63 รท 7 = 9,
63 รท 9 = 7
(b) 11 5 = 55, 5 11 = 55, 55 รท 5 = 11,
55 รท 11 = 5
21. (a) Repeated subtraction
(b) Partition
(c) Missing factor or repeated subtraction
(b) 7 is already less than 14, so there is no
subtraction. The remainder is 7 and, since
there were no subtractions, the quotient
is 0.
(c) Answered in each part separately.
24. (a) Since 78 โ 13 = = = = = = gives 0,
78 รท 13 = 6.
(b) Since 832 โ 52 = = = = = = = = = = = = =
= = = gives 0, 832 รท 52 = 16.
(c) Since 96 โ 14 = = = = = = gives 12,
96 รท 14 = 6 R 12.
(Note: Stop pressing = when the result is
less than the divisor, 14.)
(d) Since 548,245 โ 45,687 = = = = = = = = =
= = = gives 1, 548,245 รท 45,687 = 12 R 1.
Copyright ยฉ 2015 Pearson Education, Inc.
Solutions to Problem Set 2.4
25. (a) y = (5 ยท 5) + 4 = 25 + 4 = 29
(b) 3x + 2 = 20, so x = 6.
26. (a) z = 3 โ
15 + 4 = 45 + 4 = 49
(b) 20 = 2 x (3) + 2 โ 20 = 6 x + 2 โ
18 = 6 x โ x = 3
(c) w + 2 = 6 ( 4) + 5 โ w + 2 = 24 + 5 โ
w + 2 = 29 โ w = 27
27. (a) 320 โ
315 = 320 +15 = 335
Next have a second group come forward,
and verify that 2 4 = 8. And so on.
(b) Have the children form rectangles of
various sizes, and then count off row by
row to obtain the total number of children
in the array.
(c) Choose how many teams to be formed
(the divisor) and see how large the teams
will be, with an equal number on each
team.
(d) For a given number of children on a team,
add teams until the total number of
children reaches a certain total. Then
count how many teams were needed.
(b) (32 )5 = 32โ
5 = 310
(c)
43
y 3 โ
z 3 = ( y โ
z )3 or ( yz )3
28. (a) 48 โ
78 = (4 โ
7)8 = 288
(b) x 7 โ
x9 = x7 + 9 = x16
(c) (t 3 )4 = t 3โ
4 = t12
29. (a) 8 = 2 โ
2 โ
2 = 23
(b) 4 โ
8 = (2 โ
2) โ
(2 โ
2 โ
2) = 25
(c) 1024 = 210
(d) 84 = (23 )4 = 23โ
4 = 212
30. Use the guess and check method.
(a) m = 4; 3 โ
3 โ
3 โ
3 = 81
(b) n = 12
(c) p = 10
(d) q = 20
31. Answers will vary, but should include the fact
that, if we are dividing a by b, then if a

**b, the either a โ b = 0 or a โ b > 0. The remainder cannot be negative in either case because the process stops before that can happen. 33. Answers will vary. These are sample problems: Peter has a board 14 feet long, and each box he makes requires 3 feet of board. How many completed boxes can be made? Answer = 4. Tina has a collection of 14 antique dolls. A display box can hold at most three dolls. How many boxes does Tina require to display her entire collection? Answer = 5. Andrea has 14 one-by-one foot paving stones to place on her 3-foot-wide walkway. How many feet of walk can she pave, and how many stones will she have left over for a future project? Answer = 4 feet of walk paved, with 2 stones remaining. 34. Answers will vary, but should make the point that the division a b is approached by forming a train of length a, and then seeing how many cars of length b are needed to form an equally long train. 35. As listed in the theorem, starting with the second one, โข It doesnโt matter in which order you multiply two numbers. โข It doesnโt matter which way you group the terms when multiplying three numbers. โข Multiplying by 1 never changes the number (and 1 is the only number for which that happens.) โข Zero times any number is zero. 32. Answers will vary. The following are sample answers. (a) Have children hold hands in, say, groups of four. Have one group come to the front of the class, and verify that 1 4 = 4. Copyright ยฉ 2015 Pearson Education, Inc. 44 36. Chapter 2 Sets and Whole Numbers 0 = 2 ร (3 โ 3) 1 = 2(3โ 3) 2 = 2 + (3 โ 3) 3 = 3(3 โ 2) 4 = 3 + (3 โ 2) 5 = 23 โ 3 6 = 32 โ 3 7 = 3ร 3โ 2 8 = 3+ 3+ 2 9 = 3 ร 3 10 = ? 11 = 23 + 3 or (3 ร 3) + 2 12 = 32 + 3 13 = ? 14 = ? 15 = 3 ร (3 + 2) 16 = ? 17 = ? 18 = 2 ร 3 ร 3 37. This student struggles to see the relationship between multiplication and division exercises. In addition, students typically have more difficulty when the blank or box is at the beginning of the equation. 38. Answers will vary, but you could first ask for her solution. She may say that since the total cost of one nut and one bolt together is $1.00, and she is buying 18 of them, the total cost is $18.00. You could then ask her to justify her response: 18 โ 86 + 18 โ 14 = 18 โ (86 + 14) = 18 โ 100 = 1800 cents or $18.00. You could then ask her a question that would lead to her saying it is the distributive property. 39. The student subtracted the number of cupcakes Nelson baked in each pan instead of dividing to find the number of pans needed to bake all of the cupcakes. This is common because students understand they need to do something with the numbers, but they arenโt sure what. In this case, the student is not thinking of grouping the cupcakes into pans, which should be a signal that he should divide rather than subtract, which suggests removal. 40. Draw a rectangular pool of 30 + 2 feet in length and 20 + 3 feet in width. Then identify the partial products in the picture. 32 ร 23 = (30 + 2)( 20 + 3) = 30 ( 20) + 2 (20) + 3 (30) + 3 ( 2) = 736 41. The equation is x 2 โ 3 = 33 โ x 2 = 36 โ x = 6 or x = โ6. Since we are dealing only with whole numbers, Jing’s answer is x = 6. 42. Andrea is clever to realize that there can’t be ft2 added to ft; that is, area with a length measurement together in one equation can’t be correct. She needed to notice that 2(5 + 1) = 2ft ร 5 ft + 2ft ร 1ft so there really is a ft2 throughout the equation. 43. The operation is closed, commutative, and associative. The circle is the identity, since if it is either of the โfactorsโ the outcome of the operation is the other factor. Copyright ยฉ 2015 Pearson Education, Inc. Solutions to Chapter 2 Review Exercises 1, 0. Thus the number we seek is 59, which we note has a remainder of 1 when divided by 2. 44. (a) The magic multiplication constant is 4096 = 2 . 12 (b) The exponents form a magic addition square with the magic addition constant 12. It is clear how the multiplication square has been formed. For example, the product of the upper row is 23 ร 28 ร 21 = 23 +8 +1 = 212. That is, the 12 product is always 2 . (c) 23 28 21 22 24 26 27 20 25 3 8 1 2 4 6 7 0 5 45 48. D 49. B 50. B 51. A 52. C 53. C Chapter 2 Review Exercises 1. (a) S = {4, 9, 16, 25} P = {2, 3, 5, 7, 11, 13, 17, 19, 23} T = {2, 4, 8, 16} (b) P = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25} S โฉ T = {4, 16} S โช T = {2, 4, 8, 9, 16, 25} S โฉ T = {9, 25} 2. 27 = 33 6561 = 38 3 = 31 9 = 32 81 = 34 729 = 36 2187 = 37 1 = 30 243 = 35 3. (a) (b) If A โ B and A is not equal to B, then A โ B. 45. (a) Since 2 ร 185 = 370,500 โ 370 = 130, and 130 รท 2 = 65, the question is โHow many tickets must still be sold?โ (b) 67 รท 12 = 5 R 7. Thus 5 answers โHow many cartons will be filled?โ and 7 answers โHow many eggs are in the partially filled carton?โ 47. When divided by 5 the remainder is 4, so the number is in the list 4, 9, 14, 19, …, 94, 99. When divided by 4 the remainder is 3, the list of remainders are 0, 1, 2, 3, 0, 1, 2, 3, …, 0, 1, 2, 3, so the number we seek is one of 19, 39, 59, 79, or 99. When divided by 3 the remainders in this list of 5 numbers is 1, 0, 2, A โฉ (B โช C ) = ( A โฉ B) โช ( A โฉ C ) (c) (d) A โช โ = A 4. n( S ) = 3, n(T ) = 6, n( S โช T ) = n({s, e, t, h, o, r, y}) = 7, n( S โฉ T ) = n({e, t}) = 2 , 46. Note that 318 รท 14 = 22 R 10. (a) 22 is the answer to โHow many full rows of chairs are set up in the auditorium?โ (b) 10 is the answer to โHow many chairs are in the incomplete row at the rear of the auditorium?โ A โ Aโช B n( S โฉ T ) = n({s}) = 1, n(T โฉ S ) = n({h, o, r, y)} = 4 5. 1 โ a 36 โ f 4 โ b 49 โ g 9 โ c 64 โ h 16 โ d 81 โ i 25 โ e 100 โ j Copyright ยฉ 2015 Pearson Education, Inc. Chapter 2 Sets and Whole Numbers 46 6. There is a one-to-one correspondence between the set of cubes and a proper subset. For example, 1 โ 1 8 โ 2 27 โ 3 64 โ 4 125 โ 5 (c) (d) โ K 3 โ K (e) 7. 12. (a) A B = {(p, x), (p, y), (q, x), (q, y), (r, x), (r, y), (s, x), (s, y)} 8. (a) Suppose A = {a, b, c, d, e} and B = { , } . Then n(A) = 5, n(B) = 2, โ โ A โฉ B = โ and n( A โช B) = n({a, b, c, d, e, โ , โ }) = 7 . (b) Since n(A) = 4, n(B) = 2, and n(A B) = 8, the Cartesian product models 4 2 = 8. 13. Answers will vary. Since 36 = 3 ร 3 ร 4 , one possibility is 6โฒโฒ ร 6โฒโฒ ร 8โฒโฒ . (b) 14. Since 92 รท 12 = 7 R 8, there are eight rows (7 full rows, and a partial row of 8 soldiers in the back). 9. (a) Commutative property of addition: 7+3=3+7 15. (a) (b) Additive-identity property of zero: 7+0=7 10. (a) (b) (b) 11. (a) (c) (b) Copyright ยฉ 2015 Pearson Education, Inc.**

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