Solutions

Module 1 Set Theory

Exercise Set 1 Solutions

Exercise 4

a. 5P, 6P, 7P, 8P

b. 4D, 5D, 6D, 7D, 8D, 9D

c. 5H, 6H, 7H, 9H

d. There are 6 groups. You should have listed 2 of these groups: 1964: 4N, 4D, 4Q 1965: 5P, 5N, 5D, 5Q, 5H 1966: 6P, 6N, 6D, 6Q, 6H 1967: 7P, 7N, 7D, 7Q, 7H 1968: 8P, 8N, 8D, 8Q 1969: 9D, 9Q, 9H

Exercise 5

5. a collection of objects

Exercise 6

6. the objects in a set

Exercise 7

7. The null set is a set that contains no elements that is, it is empty.

Exercise 9

9. Yes.

N = {4N, 5N, 6N, 7N, 8N}

Q = {4Q, 5Q, 6Q, 7Q, 8Q, 9Q}

S = {4N, 4D, 4Q}

W = {6P, 6N, 6D, 6Q, 6H}

Y = {8P, 8N, 8D, 8Q}

T = { }

Exercise 10

10.

a. yes; any element of N will suffice 4N, 5N, 6N, 7N or 8N

b. no

c. no

d. yes

Exercise 11

11. Answers may vary. Any element in P will suffice 5P, 6P, 7P or 8P

Exercise 12

12.

a. F	b. T	c. T	d. T	e. F
f. T	g. T	h. T	i. F

Exercise 14

a. 4	b. 5	c. 6	d. 6	e. 3	f. 4
g. 4	h. 0	i. 3	j. 5	k. 5	l. 5
m. 25	n. 0	o. Z ~ S; P~H, H ~ Y; N ~ W;X ~ V; Q ~ D (it's not necessary to include any null subsets)

Exercise 15

15.

a. Ø	b. {9H}	c. {4H}
d. {6Q}	e. Ø

Exercise 17

17. Two sets, A and B, are disjoint if \(A \cap B = Ø\) that is to say that their intersection is empty.

Exercise 18

18. 5D, 6P, 7P, 8P, 5P, 5N, 5Q, 5H. yes, there are 8 coins.

Exercise 19

19.

a. {4N, 5N, 6N, 7N, 8N, 4D, 4Q}

b. {4Q, 5Q, 6Q, 7Q, 8Q, 9Q, 6P, 6N, 6D, 6H}

c. {5P, 6P, 7P, 8P, 4D, 5D, 6D, 7D, 8D, 9D}

d. {7P, 7N, 7D, 7Q, 7H, 8P, 8N, 8D, 8Q}

Exercise 23

23.

a. {2, 4}	b. {3, 5}
c. { } or Ø	d. {1, 2, 3, 4, 5, 6, 8}
e. {1, 2, 3, 4, 5, 7}	f. {2, 3, 4, 5, 6, 7, 8}
g. {6, 7, 8, 9}	h. {1, 3, 5, 7, 9}
i. {1, 2, 4, 6, 8, 9}	p. {1, 3, 5, 7}
j. {1, 3, 5}	q. {3, 5, 6, 7, 8, 9}
k. {6, 8}	r. {2,3,4,5,7}
l. {1, 2, 4}	s. {2, 4}
m. {7}	t. {6, 8, 9}
n. B or {2,4,6,8}	u. {7, 9}
o. C or {3,5,7}	v. {1, 3, 5, 6, 7, 8, 9}
w. {7, 9}	x. {3, 5, 6, 7, 8, 9}
y. T	z. T
aa. F	bb. F
cc. T	dd. F
ee. T	ff. F
gg. F	hh. T
ii. T	jj. F

Exercise 25

25.

a. {a, b, c, ..., z}

b. {124, 126, 128, ..., 698}

c. {101, 102, 103, ..., 999}

d. {x|x is a past president of the U.S. up until 1995}

e. {Reagan, Bush, Clinton}

Exercise 30

30.

a. 3; 8; answers may vary

b. 4; 6; answers may vary

c. 2; 12; answers may vary

d. 12; 2; answers may vary

e. 8; 3; answers may vary

f. 6; 4; answers may vary

Exercise 31

31. e. union

Exercise 32

32.

c. no

d. yes; LYC, LYQ, LYT

e. intersection;{LYC, LYQ, LYT}

Exercise 34

34.

a. {}

b. {}, {P}

c. { }, {G},{F}, {F,G}

d. {}, {X}, {Y}, {Z}, {X,Y}, {X,Z}, {Y,Z}, {X,Y,Z}

e. { }, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}

Exercise 35

35.

a. 1	b. 2	c. 4	d. 8
e. 16	f. 32	g. \(2^{n}\)

Exercise 36

36. 35

Exercise 37

37. 35

Exercise 38

38. yes

Exercise 39

39.

a. bc

b. \(b^{2}\)

c. \(c^{2}\)

Exercise 40

40. One possibility is let A = {m} and B = {n}. Then \(A \times B = {(m,n)}\) but \(B \times A = {(n,m)}\). Therefore, \(A \times B \nleq B \times A\) (be sure to write your own solution)

Exercise 41

41.

a. {(3,2), (3,6), (4,2), (4,6)}

b. {(6,5), (7,5), (8,5), (9,5)}

c. { }

d. {(a,a)}

e. {(x,x), (x,y), (y,x), (y,y)}

f. {(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)}

g. {((9,4),D), ((9,4), {a,b,c}), (C,D), (C, {a,b,c})}

h. {({5,6,7,8,9}, g), ({5,6,7,8,9},{4,3})}

Exercise Set 2 Solutions

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

18-56: Answers to these exercises are the Venn diagrams at the end of Exercise Set 2.

Exercise Set 3 Solutions

Exercise 1

1. Answers may vary. The overlap is the intersection. There are 8 regions.

Exercise 2

2.

a.	b.
c.	d.

Exercise 4

4.

a.

b.

c.

Exercise 5

5. Some possibilities are: triangle, circle, square; large, small and any other value.

Exercise 8

8.

a.

b. SRT, SYT, SGT, LGT, SYC, LYT, SRC, SGC, LRT, LGC, LRC, LYC

c. SRT, SYT, SGT, LGT, SYC, LYT, SRC, SGC, LRT, LGC, LRC, LYC

d. The elements are exactly the same

e. \(B \cup Q)^{c} = B^{c} \cap Q^{c}\)

Exercise 9

9.

a.

b. SRT, LRQ, LRT, SRQ

c. SRT, LRQ, LRT, SRQ

d. The elements are exactly the same.

e. \((R^{c} \cup C)^{c} = R \cap C^{c}\)

Exercise 10

10. For part a and b, the final shadings are the same as shown on the Venn diagram below. The equation you can write is: \((B \cup Q)^{c} = B^{c} \cap Q^{c}\)

Exercise 11

11. For part a and b, the final shadings are the same as shown on the Venn diagram below. The equation you can write is: \((A \cap B)^{c} = A^{c} \cup B^{c}\)

Exercise 12

12.

a. \(R \cup S\)

b. (\(M^{c} \cup N^{c})^{c}\)

c. \(F^{c} \cap G\)

d. \((H \cap I^{c})^{c}\)

e. \(P \cup Q^{c}\)

f. \((S^{c} \cup T)^{c}\)

Exercise 14

14. Only a proof to 14.c. is provided

\((B^{c} \cup C)^{c} = (({2, 4, 6, 8}}}^{c} \cup {3, 5, 7}^{c}\) = \({1, 3, 5, 7, 9} \cup {3, 5, 6})^{c}\) = \(({1, 3, 5, 7, 9})^{c}\) = {2, 4, 6, 8}.

\(B \cap C^{c} = {2, 4, 6, 8} \cap ({1, 3, 5, 7, 9})^{c}\) = \({2, 4, 6, 8} \cap {2, 4, 6, 8}\) = {2, 4, 6, 8}.

Since \((B^{c} \cup C)^{c}\) and \(B \cap C^{C}\) have the exact same elements, they are equal. Therefore, \((B^{c} \cup C)^{c} = B \cap C^{c}\).

Exercise 15

15.

a, b: final shadings are the same as shown on the Venn diagram below.

c. {2,3,4,5}

d. {2,3,4,5}

e. They are the same equal.

Exercise 16

16.

a, b: final shadings are the same as shown on the Venn diagram below.

c. {1,2,3,4,5}

d. {1,2,3,4,5}

e. They are the same.

Exercise 17

17.

a. (\(X \cap Y) \cup (X \cap Z\))

b. \((P \cup Q^{c}) \cap (P \cup R\))

c. \(K^{c} \cap (L \cup M)\)

d. \(D \cup (E^{c} \cap F\))

Exercise Set 4 Solutions

Exercise 1

1.

Exercise 2

2.

Exercise 3

3.

a. \((A \cup B \cup C)^{c}\)	b. A - (\(B \cup C)\)
c. (A \cap B) - C	d. B - (\(A \cup C)\)
e. \((A \cap B \cap C\))	f. \((A \cap C) - B\)
g. \((B \cap C) - A\)	h. \(C - \(A \cup B)\)

Exercise 4

4.

a. 43	b. 51	c. 46	d. 38
e. 18	f. 25	g. 13	h. 89

Exercise 5

5.

a. n(\(A - (B \cup C)\)) = 11

b. n(A) = 69

c. n(\(B \cap C\)) = 51

d. n(\((A \cup B) - C\)) = 71

e. n(\(C^{c}\)) = 73

f. n(\(B \cup C\)) = 160

g. n(B - C) = 60

h. n(\((A \cap B) \cup (B \cap C) \cup (A \cap C)\)) = 66

i. n(\((A \cap B) \cup (B \cap C) \cup (A \cap C) - (A \cap B \cap C)\)) = 23

Exercise 6

6.

a. 34	b. 5	c. 69	d. 37	e. 13
f. 3	g. 21	h. 9	i. 23

Exercise 7

7.

a. 12

b. 2

c. 6

Exercise 8

8.

a. 9	b. 5	c. 10	d. 43
e. 37	f. 18	g. 3

Exercise 9

9.

a. 86	b. 49	c. 31
d. 25	e. 4	f. 27

Exercise 10

10. This is one way to show it.

Exercise 11

11. This is one way to show it.

Exercise 12

12. This is one way to show it.

Homework Solutions

Exercise 1

1.

a. {a, c}

c. {e, v, w, z}

e. {r, u, x}

g. {a, c, r, u}

i. {x}

k. Ø or { }

m. 10

n. 2

Exercise 3

3. a. \((N^{c} \cap P)^{c}\)

Exercise 4

4. a. \(A^{c} \cup (E \cap F\))

Exercise 5

5.

a.

c. 15

e. 5

g. 9

Exercise 6

6.

a. 8

b. 3

Exercise 7

7.

a. T

c. F

e. T

Exercise 8

8.

a.

c.

e.

Exercise 9

9. a. \(A \cup (B \cap C)\)

Exercise 10

10.

a. { }

c. { }, {a}, {b}, {a, b}

Exercise 11

11.

a. {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (4, 1), (4, 2), (4, 4)}

c. {({a, c}, {a, c}), ({a, c}, 5), (5, {a, c}), (5, 5)}

e. {(x, x)}

Exercise 12

12.

a. {LBC, LYC, LGC}

c. {LRC, LBC, LGC, LYC, SRC}

Module 2 Counting and Numerals

Exercise Set 1 Solutions

Exercise 8

8. The sets have the same number of elements. Cardinality is equal.

Exercise 10

10.

a) 1) \(1 \leftrightarrow A\)

b. 1) \(2 \leftrightarrow B, 3 \leftrightarrow C\) 2) \(2 \leftrightarrow C, 3 \leftrightarrow B\)

c. 1) \(4 leftrightarrow D, 5 leftrightarrow E, 6 leftrightarrow F\) 2) \(4 leftrightarrow D, 5 leftrightarrow F, 6 leftrightarrow E\) 3) \(4 leftrightarrow E, 5 leftrightarrow D, 6 leftrightarrow F\) 4) \(4 leftrightarrow E, 5 leftrightarrow F, 6 leftrightarrow D\) 5) \(4 leftrightarrow F, 5 leftrightarrow D, 6 leftrightarrow E\) 6) \(4 leftrightarrow F, 5 leftrightarrow E, 6 leftrightarrow D\)

Exercise 11

11.

a) 1) \(M \leftrightarrow M\)

b. 1) \(x \leftrightarrow x, y \leftrightarrow z\) 2) \(x \leftrightarrow z, y \leftrightarrow x\)

c. 1) \(1 \leftrightarrow 1, 2 \leftrightarrow 2\) 2) \(1 \leftrightarrow 2, 2 \leftrightarrow 1\)

d. 1) \(1 \leftrightarrow 1, 2 \leftrightarrow 2, 3 \leftrightarrow 3\) 2) \(1 \leftrightarrow 1, 2 \leftrightarrow 3, 3 \leftrightarrow 2\) 3) \(1 \leftrightarrow 2, 2 \leftrightarrow 1, 3 \leftrightarrow 3\) 4) \(1 \leftrightarrow 2, 2 \leftrightarrow 3, 3 \leftrightarrow 1\) 5) \(1 \leftrightarrow 3, 2 \leftrightarrow 1, 3 \leftrightarrow 2\) 6) \(1 \leftrightarrow 3, 2 \leftrightarrow 2, 3 \leftrightarrow 1\)

e. 1) \(1 \leftrightarrow 3, 2 \leftrightarrow 4, 3 \leftrightarrow 5\) 2) \(1 \leftrightarrow 3, 2 \leftrightarrow 5, 3 \leftrightarrow 4\) 3) \(1 \leftrightarrow 4, 2 \leftrightarrow 3, 3 \leftrightarrow 5\) 4) 1 \leftrightarrow 4, 2 \leftrightarrow 5, 3 \leftrightarrow 3 5) 1 \leftrightarrow 5, 2 \leftrightarrow 3, 3 \leftrightarrow 4 6) 1 \leftrightarrow 5, 2 \leftrightarrow 4, 3 \leftrightarrow 3

Exercise 12

12.

a. 1

b. 2

c. 6

Exercise 14

14.

a. | | | | | | | | | | |

b. Make 512 stroke marks; make 2,000,000 stroke marks.

Exercise 16

16.

a. | | | | | | | | | | | | | | | | |

b.

Exercise 17

17. 36

Exercise 18

18. Answers may vary. Some possibilities are to make 123 strokes or to make 8 I's and 53 strokes.

Exercise 19

19.

a. 301,020	b. 4,010,507
c. 35,000	d. 110,023

Exercise 20

20.

a.

b.

c.

d.

e.

Exercise 21

21. Base Ten. They only have symbols up to a million in Egyptian. This makes writing very large numerals too cumbersome and possibly virtually impossible.

Exercise 24

24.

a. XXXII	b. DLXI
c. DCCVIII	d. MMLIII

Exercise 25

25.

a. 2,687

b. 1,232

Exercise 26

26.

a. \(\bar{\text{CCCXXX}}\)DCCCII

b. \(\bar{\bar{\text{LXX}}}\)MDCLI

Exercise 27

27.

a. CCCCCCCCCXXXXIIII or DCCCCXXXXIIII

b. CCCCXXXXXXXXXIIIIIIIII or CCCCLXXXXVIIII

Exercise 29

29.

a. DCCXLVIII

b. \(\bar{\text{XIV}}\)CDLXXX

c. CMLXIX

d. CDXLII

Exercise 30

30.

a. \(\bar{\text{XIX}}\)CDLIII

b. MMDCCCXLIX

c. MCMXCVI

Exercise 32

32. 3, 1, 3, 1, 3, 1

Exercise 33

33.

a. 10	b. 1	c. 2
d. 7	e. 7

Exercise 34

34.

a. 3

b. 2

c. 9

d. 900 strokes

e. 180 5 stroke tallies

Exercise 35

35.

Hindu-Arabic: 3, 3, 7, 5, 4

Stroke: 143, 400, 1000000, 30009, 2124

Tally: 31, 80, 200000, 6005, 428

Egyptian: 8, 4, 1, 12, 9

Roman: 6, 2, 1, 5, 7

Exercise Set 2 Solutions

Exercise 1

1.

a.

b.

c.

d.

Exercise 2

2.

a. 109

b. 6063

c. 40

d. 2815

e.7800

Exercise 3

3.

1.	5.	9.	13.	17.
2.	6.	10.	14.	18.
3.	7.	11.	15.
4.	8.	12.	16.	19.

Exercise 4

4. A dot represents the number one and a line segment represents the number five. Any combination of 1-3 line segments and/or 1-4 dots forms any numeral up to 19.

Exercise 5

5.

a. 90

b. 320

c. 162

Exercise 6

6.

a. 3974	b. 1946	c. 3300
d. 32454	e. 7319

Exercise 7

7.

a.

b.

c.

d.

e.

Exercise 8

8.

a. 27 rem. 156

b. 9 rem. 6709

c. 16 rem. 13

d. 2 rem. 142040

Exercise 9

9. A 19 at the second level up is 380. The Mayan numeral for 380 is shown below.

Exercise 10

10. You can't have higher than the 17 at the second level up because 18 at that level represents 360, which should be represented on the third level up. For all other levels, the highest numeral can be 19.

Exercise 11

11.

a.

b.

c.

d.

e.

Exercise 12

12.

a.

b.

c.

Exercise 13

13.

a. 10

b. 12

c. 3

Exercise 14

14.

	Chinese	Mayan
a. 15
b. 100
c. 1000
d. 9999

Exercise Set 3 Solutions

Exercise 1

1.

a. 2

b. 2

c. 2

d. 2

e. 2

Exercise 2

2.

a. 27	b. 13, 1, 14
c. 6, 1, 1, 8	d. 3, 0, 1, 1, 5
e. 1, 1, 0, 1, 1, 4	f. 1, 1, 0, 1, 1
g. 27, 4

Exercise 3

3.

a. 18	b. 9, 0, 9
c. 4, 1, 0, 5	d. 2, 0, 1, 0, 3
e. 1, 0, 0, 1, 0, 2	f. 1, 0, 0, 1, 0

Exercise 4

4.

a. 45	b. 22, 1, 23
c. 11, 0, 1, 12	d. 5, 1, 0, 1, 7
e. 2, 1, 1, 0, 1, 5
f. 1, 0, 1, 1, 0, 1, 4
g. 1, 0, 1, 1, 0, 1

Exercise 5

5.

a. three, zero, two, base six

b. one, zero, one, one, base two

c. four, three, five, base seven

Exercise 6

6.

a. \(5016_{\text{eight}}\)

b. \(101001_{\text{two}}\)

Exercise 7

7.

2048, 1024, 512, 256, ...

Exercise 8

8.

a. 19

b. 65

c. 63

Exercise 9

9. 3125, 625, 125, 25, 5, 1

Exercise 10

10. 8

Exercise 11

11. 2187, 729, 243, 81, 27, 9, 3, 1

Exercise 12

12.

a. 500

b. 3711

c. 1093

Exercise 13

13.

a. Six: 7776, 1296, 216, 36, 6, 1

b. Seven: 16807, 2401, 343, 49, 7, 1

c. Nine: 59049, 6561, 729, 81, 9, 1

d. Eight: 32768, 4096, 512, 64, 8, 1

e. Four: 1024, 256, 64, 16, 4, 1

f. Ten: 100000,10000,1000,100,10,1

g. Twelve: 20736, 1728, 144, 12, 1

h. Eleven: 14641, 1331, 121, 11, 1

Exercise 14

14. \(n^{10}, n^{9}, n^{8}, n^{6}, n^{5}, n^{4}, n^{2}, n^{1}\)

Exercise 15

15.

a. \(3 \times 815 + 6 \times 811 + 2 \times 87\)

b. \(3 \times 515 + 4 \times 510 + 2 \times 57\)

c. \(4 \times 1114 + 3 \times 1110 + 2 \times 110\)

d. \(1 \times 214 + 1 \times 211 + 1 \times 24\)

Exercise 16

16.

a. \(9^{6}\)

b. \(9^{0}\)

c. \(9^{12}\)

d. \(9^{10}\)

Exercise 17

17.

a. \(2401000300_{\text{five}}\)

b. \(30040001600_{\text{eight}}\)

c. \(7000804000_{\text{twelve}}\)

d. \(201010200_{\text{three}}\)

Exercise 18

18.

a. \(6^{17}\)

b. \(9^{19}\)

Exercise 19

19. 3: 0, 1, 2

Exercise 20

20. 4: 0, 1, 2, 3

Exercise 21

21. 5: 0, 1, 2, 3, 4

Exercise 22

22. 6: 0, 1, 2, 3, 4, 5

Exercise 23

23. 8: 0, 1, 2, 3, 4, 5, 6, 7

Exercise 24

24. No, because the only digits that can occur in a Base Seven numeral are 0, 1, 2, 3, 4, 5 and 6.

Exercise 25

25.

a. Seven

b. There is no highest base, it can occur in any base higher than Six.

Exercise 26

26. 11

Exercise 27

27. 12

Exercise 28

28. 13

Exercise 29

29.

a. 671	b. 1291	c. 1830
d. 3496	e. 1358	f. 2231

Exercise 30

30.

a. 6	b. 12	c. 20	d. 30
e. 42	f. 56	g. 72	h. 90
i. 132	j. 156	k. 182	l. 420

Exercise 31

31. \(13201154320050146_{\text{eleven}}\) is larger. Exercise 30 illustrates that if two numerals have the same exact digits, the one with the higher base has a higher value since each place value is worth more. Explanations may vary.

Exercise 32

32. \(4 \times 1213 + 10 \times 129 + 11 \times 123\)

Exercise 33

33. \(600T000E0000W00_{\text{thirteen}}\)

Exercise Set 4 Solutions

Exercise 1

1. Ask a friend how your art work is.

Exercise 2

2.

a. 4, 4, 4, 4, 4, 4

b. 4,16,64,256

c. 5

d. 36

Exercise 3

3. 2, 2, 2, 2, 2

Exercise 4

4. 3, 3, 3

Exercise 5

5. 4, 4, 4

Exercise 6

6.a.

i. 42 ii. 21, 0 iii. 10, 1, 0 iv. 5, 0, 1, 0 v. 2, 1, 0, 1, 0 vi. 1, 0, 1, 0, 1, 0 vii. See first row of part d.

b. See second row of part d.

c. See third row of part d.

d. 101010 Base Two 001120 Base Three 000222 Base Four

Exercise 7

7.

a. \(110100_{\text{two}}\)

b. \(1221_{\text{three}}\)

c. \(310_{\text{four}}\)

Exercise 8

8.

a. \(3020_{\text{four}}\)

b. \(1212_{\text{seven}}\)

c. \(1011001_{\text{two}}\)

d. \(100210_{\text{three}}\)

e. \(1E8_{\text{thirteen}}\)

Exercise 9

9.

a. \(12202_{\text{three}}\)

b. \(10E_{\text{twelve}}\)

c. \(10011011_{\text{two}}\)

d. \(1110_{\text{five}}\)

Exercise 10

10.

a. \(T23T_{\text{eleven}}\)

b. \(625T_{\text{thirteen}}\)

Exercise 11

11.

a. \(59T_{\text{twelve}}\)

b. \(1506_{\text{eight}}\)

c. \(2305_{\text{seven}}\)

d. \(1131_{\text{nine}}\)

Exercise 13

13. 10

Exercise 14

14. The last digit of the previous numeral must have been a 9 and the last digit of the new numeral will be 0.

Exercise 15

15. 99,999,999

Exercise 16

16. 1,000,000,000

Exercise 17

17. 3

Exercise 18

18. The last digit of the previous numeral must have been a 2 and the last digit of the new numeral will be 0.

Exercise 19

19. \(1000000_{\text{three}}\)

Exercise 20

20. \(2222222_{\text{three}}\)

Exercise 21

21.

a. \(1202011_{\text{three}}\)

b. \(2220012_{\text{three}}\)

c. \(1010110_{\text{three}}\)

d. \(2100220_{\text{three}}\)

e. \(2121000_{\text{three}}\)

Exercise 22

22.

a. \(1200101_{\text{three}}\)

b. \(1202220_{\text{three}}\)

c. \(2110012_{\text{three}}\)

d. \(2110022_{\text{three}}\)

Exercise 23

23. The digit "3" can't be in the numeral.

Exercise 24

24. \(0_{\text{four}}, 1_{\text{four}}, 2_{\text{four}}, 3_{\text{four}}, 10_{\text{four}}, 11_{\text{four}}, 12_{\text{four}}, 13_{\text{four}}, 20_{\text{four}}, 21_{\text{four}}, 22_{\text{four}}, 23_{\text{four}}, 30_{\text{four}}, 31_{\text{four}}, 32_{\text{four}}, 33_{\text{four}}, 100_{\text{four}}, 101_{\text{four}}, 102_{\text{four}}, 103_{\text{four}}, 110_{\text{four}}, 111_{\text{four}}, 112_{\text{four}}, 113_{\text{four}}\)

Exercise 25

25. 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 10000, 10001, 10010, 10011, 10100

Exercise 26

26. 75, 76, 77, 100, 101, 102, 103, 104

Exercise 27

27.

a. \(3026_{\text{eight}}, 3030_{\text{eight}}\)

b. \(1234_{\text{five}}, 1241_{\text{five}}\)

c. \(101011_{\text{two}}, 101101_{\text{two}}\)

Exercise 28

28.

a. \(12221_{\text{three}}\), 160

b. \(2405_{\text{eight}}\), 1285

c. \(1310_{\text{three}}\), 205

Exercise Set 5 Solutions

Exercise 1

1.

a. \(\frac{1}{4}\)	b. \(\frac{1}{16}\)	c. \(\frac{1}{64}\)
d. \(\frac{1}{3}\)	e. \(\frac{1}{25}\)	f. \(\frac{1}{8}\)

Exercise 2

2.

a. 35\(\frac{1}{4}\)	b. 168\(\frac{7}{11}\)
c. 15\(\frac{1}{2}\)	d. 27\(\frac{1}{2}\)

Exercise 3

3.

a. 27\(\frac{13}{36}\)	b. 45\(\frac{27}{64}\)
c. 90\(\frac{9}{49}\)	d. 7\(\frac{5}{8}\)
e. 26\(\frac{26}{27}\)	f. 1475\(\frac{5}{72}\)

Exercise 4

4.

a. binary: 00110101; decimal: 53

b. binary: 10011010; decimal: 154

c. binary: 01010100; decimal: 84

d. binary: 00000000; decimal: 0

e. binary: 11111111; decimal: 255

Exercise 5

5.

a. binary: 00110101; hex: 35

b. binary: 10011010; hex: 9A

c. binary: 01010100; hex: 54

d. binary: 00000000; hex: 00

e. binary: 11111111; hex: FF

Exercise 6

6.

a. binary: 01011110; decimal: 94

b. binary: 11100101; decimal: 229

c. binary: 00111001; decimal: 57

d. binary: 00011111; decimal: 31

e. binary: 10011000; decimal: 152

f. binary: 00101010; decimal: 42

g. binary: 00000111; decimal: 7

h. binary: 01000000; decimal: 64

Exercise 7

7.

a. I LOVE MATH!

b. TEACHING IS A CHALLENGING, BUT REWARDING CAREER!

Exercise 8

8.

a. Hex: 48, 45, 4C, 50, 21 Binary: 01001000,01000101, 01001100, 01010000, 00100001

b. Hex: 2,45,20,48,41,50,50,59,2E Binary: 01000010, 01000101,00100000,01001000,01000001,01010000, 01010000, 01011001, 00101110

c. This will depend on your name!

Exercise 9

9.

a. binary: \(1001001_{\text{two}}\) hex: \(49_{\text{sixteen}}\)

b. binary: \(1111010_{\text{two}}\) hex: \(7A_{\text{sixteen}}\)

c. binary: \(110010_{\text{two}}\) hex: \(32_{\text{sixteen}}\)

d. binary: \(11111010_{\text{two}}\) hex: \(FA_{\text{sixteen}}\)

e. binary: \(1111101000_{\text{two}}\) hex: \(3E8_{\text{sixteen}}\)

Homework Solutions

Exercise 1

1. b. There is no matching possible because the cardinality of the two sets is different

Exercise 2

2. One possibility is: \(1 \leftrightarrow 5, 2 \leftrightarrow 10, 3 \leftrightarrow 15, 4 \leftrightarrow 20\), etc. Give your own answer

Exercise 3

3.

(1) \(SBC \leftrightarrow SRC, SBT \leftrightarrow SRT, SBQ \leftrightarrow SRQ\)

(2) \(SBC\leftrightarrow SRC, SBT \leftrightarrow SRQ, SBQ \leftrightarrow SRT\)

(3) \(SBC \leftrightarrow SRT, SBT \leftrightarrow SRC, SBQ \leftrightarrow SRQ\)

(4) \(SBC \leftrightarrow SRT, SBT \leftrightarrow SRQ, SBQ \leftrightarrow SRC\)

(5) \(SBC \leftrightarrow SRQ, SBT \leftrightarrow SRT, SBQ \leftrightarrow SRC\)

(6) \(SBC \leftrightarrow SRQ, SBT \leftrightarrow SRC, SBQ \leftrightarrow SRT\)

Exercise 4

4.

a. 16

c. 1962

d. 744

f. 3,031,020

h. 50,703

i. 1395

k. 971

Exercise 5

5.

a. CCCXLII

c.

e.

Exercise 6

6.

a. \(59T_{\text{twelve}}\)

d.

Exercise 7

7. \(7T4E_{\text{twelve}}\)

Exercise 10

10. a. \(17342575T_{\text{eleven}}\)

Exercise 11

11. a. \(539100E0_{\text{twelve}}\)

Exercise 13

13.

a. \(10122_{\text{three}}\)

b. 98

Exercise 14

14.

a. \(300640000_{\text{seven}}\)

c. \(40T00E00_{\text{thirteen}}\)

Exercise 15

15.

a. \(2 \times 911 + 5 \times 97 + 3 \times 94\)

c. \(8 \times 1210 + 11 \times 125 + 10 \times 123\)

Exercise 16

16.

a. \(39\frac{1}{3}\)

b. \(16\frac{7}{9}\)

Exercise 18

18.

a. \(302.301_{\text{four}}\)

c. \(101.1001_{\text{two}}\)

Module 3 Addition and Subtraction

Exercise Set 1 Solutions

Exercise 1

1.

b. 3

d. 4

f. 7

Exercise 2

2.

a. {t, u, v, w, x, y, z}

b. 7

c. no

d. no

e. A & B must have no elements in common.

Exercise 4

4.

a. {LBT, LBC, LBQ, LRT, LRC, LRQ,LGT, LGQ, LGC, LYT, LYC, LYQ, SBT, SRT, SGT, SYT}

b. 16

c. no

d. L and T have elements in common.

Exercise 5

5.

a. Let A = {x, y} and B = {a, b, c, d} 2+4 = n(A) + n(B) = \(n(A \cup B)\) = n({x, y, a, b, c, d}) = 6 Therefore, 2 + 6 = 6	Since n(A) = 2, n(B) =4, and \(A \cap B =Ø\), then by substituting n(A) for 2 and n(B) for 4 by substituting n(A) for 2 and n(B) for 4 by computing \(A \cup B\) by counting the elements in \(A \cup B\)
b. Let A = {x, y} and B = {a, b} 2+2 = n(A)+n(B) = \(n(A \cup B)\) = n({x, y, a, b}) = 4 Therefore, 2 + 2 = 4	Since n(A) = 2, n(B) =2, and \(A \cap B = Ø\), then by substituting n(A) for 2 and n(B) for 2 by the set theory definition of addition by computing \(A \cup B\) by counting the elements in \(A \cup B\)
c. Let A = {x, y, z} and B = { } 3 + 0 = n(A) + n(B) = \(n(A \cup B)\) = n({x, y, z}) = 3 Therefore, 3 + 0 = 3	Since n(A) = 3, n(B) =0, and \(A \cap B = Ø\), then by substituting n(A) for 3 and n(B) for 0 by the set theory definition of addition by computing \(A \cup B\) by counting the elements in \(A \cup B\)

Exercise 6

6.

a. K	b. H	c. B	d. O	e. S
f. P	g. B	h. N	i. H	j. B
k. N	l. H	m. each pair is equal.

Exercise 8

8.

a. Y, B

b. D, B

c. same

d. B, S

e. D, S

f. same

Exercise 11

11.

a. Commutative property of addition

b. Commutative property of addition

c. Associative property of addition

d. Associative property of addition

Exercise 13

13. W + W + W, W + R, R + W, L

Exercise 14

14. 1 + 1 + 1 = 3, 1 + 2 = 3, 2 + 1 = 3, 3 = 3

Exercise 15

15. P,W + W + W + W, W + W + R, W + R + W, R + W + W, W + L, L + W, R + R

Exercise 16

16. 4 = 4, 1 + 1 + 1 + 1 = 4, 1 + 3 = 4, 3 + 1 = 4, 2 + 2 = 4, 1 + 1 + 2 = 4, 1 + 2 + 1 = 4, 2 + 1 + 1 = 4

Exercise 17

17. There are eleven trains: R + P, L+ L, R + R + R, W + W + W + L, D, W + W + W + W + W + W, W + Y, W + W + W + W + R, W + L + R, W + W + P, W + W + R + R

Exercise 18

18. 1 + 1 + 1 + 1 + 1 + 1 = 6, 1 + 5 = 6, 1 + 1 + 4 = 6, 1 + 1 + 1 + 1 + 2 = 6, 1 + 1 + 2 + 2 = 6, 1 + 1 + 1 + 3 = 6, 6 = 6, 2 + 2 + 2 = 6, 1 + 2 + 3 = 6, 3 + 3 = 6, 2 + 4 = 6

Exercise 19

19.

a. R	b. Y	c. N
d. R	e. P	f. W

Exercise 20

20.

b. because 30 = 22 + 8

c. because 156 = 96 + 60

d. because 80 = 0 + 80

e. because 231 = 195 + 36

f. because 987 = 967 + 20

Exercise 21

21.

b. because 30 + 69 = 99

c. because 19 + 51 = 70

d. because 0 + 32 = 32

e. because 489 + 11 = 500

f. because 65 + 136 = 201

Exercise 22

22.

a. left

b. right

Exercise 23

23. yes

Exercise 24

24. yes

Exercise 25

25.

a. <

b. >

Exercise 26

26. {0, 1, 2, 3, . . .}

Exercise 27

27. yes

Exercise 28

28. no

Exercise 29

29.

a. closed (since 0 + 0 = 0)

b. not closed; 1 + 1 = 2 is a counterexample

c. closed; mult of 2: proof is left to you!

d. not closed; 1 + 3 = 4 is a counterexample

Exercise Set 2 Solutions

Exercise 1

1.

a.

b.

c.

d.

Exercise 2

2.

a. B

b. D

c. G

Exercise 3

3.

a. DDDCCCCCCAAAAA

b. EEEDCCCCCBBAAAAA

c. FEB

d. E

Exercise 4

4.

a.

b.

c.

Exercise 5

5.

a.	b.	c.
d.	e.

Exercise 6

6. \(33_{\text{four}} + 31_{\text{four}} = 130_{\text{four}}\)

Exercise 7

7. \(21_{\text{seven}} + 16_{\text{seven}} = 40_{\text{seven}}\)

Exercise 8

8. \(1111_{\text{two} + 1101_{\text{two} = 11100_{\text{two}}\)

Exercise 9

9. \(16_{\text{nine}} + 14_{\text{nine}} = 31_{\text{nine}}\)

Exercise 9

10. \(120_{\text{three}} + 111_{\text{three}} = 1001_{\text{three}}\)

Exercise 9

11. \(17_{\text{eight}} + 15_{\text{eight}} = 34_{\text{eight}}\)

Exercise 13

13.

a. \(2_{\text{nine}}\)	b. \(2_{\text{seven}}\)	c. \(T_{\text{twelve}}\)
d. \(8_{\text{eleven}}\)	e. \(4_{\text{six}}\)	f. \(1_{\text{two}}\)
g. \(1_{\text{three}}\)	h. \(3_{\text{four}}\)	i. \(2_{\text{five}}\)

Exercise 14

14.

a. \(14_{\text{eleven}}\)	b. \(12_{\text{five}}\)
c. \(11_{\text{eight}}\)	d. \(11_{\text{thirteen}}\)
e. \(12_{\text{four}}\)	f. \(13_{\text{seven}}\)
g. \(11_{\text{three}}\)	h. \(10_{\text{two}}\)
i. \(12_{\text{six}}\)	j. \(13_{\text{nine}}\)

Exercise 15

15.

a. Base Three Addition Table

b. Base Six Addition Table

c. Base Two Addition Table

Exercise Set 3 Solutions

Exercise 1

1.

a. 50

b. 51

c. 47

d. 52

Exercise 2

2.

a. 69

b. 70

c. 74

Exercise 3

3.

a. 47

b. 43

c. 53

d. 53

Exercise 4

4.

a. \(1065_{\text{thirteen}}\)	b. \(10110_{\text{two}}\)
c. \(12222_{\text{five}}\)	d. \(1168_{\text{nine}}\)
e. \(110100_{\text{two}}\)	f. \(10211_{\text{three}}\)
g. \(T450_{\text{twelve}}\)	h. \(11521_{\text{seven}}\)

Exercise 6

6.

a. 43 + 47 = (40 + 3) + (40 + 7) = (40 + 40) + (3 + 7) = 80 + 10 = 90

b. 88 + 54 = (80 + 8) + (50 + 4) = (80 + 50) + (8 + 4) = 130 + 12 = 100 + 30 + 10 + 2 = 100 + 40 + 2 = 142

Exercise 7

7. The first student adds the ones (9 + 7 = 16), then the tens (50 + 60 = 110) and thirdly, the hundreds (800 + 400 = 1200). Next those three intermediate sums are added together (16 + 110 + 1200 = 1326). The second student is doing basically the same thing except the hundreds are added first, then the tens and ones are added. The first student is adding right to left whereas the second student is adding left to right.

Exercise 9

9.

a. 12,109	b. 111,463
c. 16,668	d. 128,334

Exercise 10

10.

a. 8,805	b. 10,463
c. 13,029	d. 106,003

Exercise 11

11.

a. \(1300_{\text{six}}\)	b. \(1392_{\text{eleven}}\)
c. \(1166_{\text{eight}}\)	d. \(11001_{\text{two}}\)
e. \(3201_{\text{five}}\)	f. \(1065_{\text{thirteen}}\)
g. \(10110_{\text{two}}\)	h. \(12222_{\text{five}}\)
i. \(1168_{\text{nine}}\)	j. \(110100_{\text{two}}\)
k. \(10211_{\text{three}}\)	l. \(T450_{\text{twelve}}\)
m. \(11521_{\text{seven}}\)

Exercise 12

12.

a. seven

b. thirteen

c. twelve

d. three - thirteen

e. four - thirteen

Exercise 13

13.

a. Base Eight; \(132_{\text{eight}}\)

b. Base Nine: \(82_{\text{nine}}\)

c. Base Three; \(110_{\text{three}}\)

d. Base Ten; 123

Exercise 14

14. She thought of 68 as 70 - 2. So, 47 + (70 - 2) = (47 + 70) - 2 = 117 - 2 = 15

Exercise 15

15. First, he added the tens together, (40 + 60 = 100), then he added the ones together (8 + 7 = 15) and finally he added the two preliminary sums together (100 + 15 = 115).

Exercise 18

18. 1115; LATTICE SHOWN BELOW

Exercise 19

19.

a. \(41_{\text{six}}\)

b. \(WOE_{\text{thirteen}}\)

Exercise 20

20. Estimate: $46; Actual Sum: $45.18

Exercise 22

22. You'll underestimate if you happen to round down on almost every item. You'll overestimate if you happen to round up on almost every item.

Exercise Set 4 Solutions

Exercise 2

2.

a. 7

b. 3

c. 4

Exercise 4

4.

a. Let A = {t, u, v, w, x, y, z} and B = {w, y, z}. Since n(A) = 7, n(B) = 3 and \(B \subseteq A\), 7 - 3 = n(A) - n(B) by substituting n(A) for 7 and n(B) for 3 = n(A \ B) by the set theory definition of substitution = n({t, u, v, x}) by computing A \ B = 4 by counting the elements in A \ B Therefore, 7 - 3 = 4.

a. Let A = {t, u, v, w, x, y, z} and B = {w, y, z}. Since n(A) = 7, n(B) = 3 and \(B \subseteq A\), 7 - 3 = n(A) - n(B) by substituting n(A) for 7 and n(B) for 3 = n(A \ B) by the set theory definition of substitution = n({t, u, v, x}) by computing A \ B = 4 by counting the elements in A \ B Therefore, 7 - 3 = 4.

a. Let A = {t, u, v, w, x, y, z} and B = {w, y, z}. Since n(A) = 7, n(B) = 3 and \(B \subseteq A\), 7 - 3 = n(A) - n(B) by substituting n(A) for 7 and n(B) for 3 = n(A \ B) by the set theory definition of substitution = n({t, u, v, x}) by computing A \ B = 4 by counting the elements in A \ B Therefore, 7 - 3 = 4.

Exercise 5

5.

a.

b.

Exercise 6

6.

Exercise 7

7.

a.	b.
c.	d.

Exercise 8

8.

\(32_{\text{five}}; 14_{\text{five}}; 13_{\text{five}}\) \(32_{\text{five}} – 14_{\text{five}} = 13_{\text{five}}\)

Exercise 9

9.

\(21_{\text{eight}}; 7_{\text{eight}}; 12_{\text{eight}}\) \(21_{\text{eight}} – 7_{\text{eight}} = 12_{\text{eight}}\)

Exercise 10

10.

\(210_{\text{three}}; 21_{\text{three}}; 112_{\text{three}}\) \(210_{\text{three}} – 21_{\text{three}} = 112_{\text{three}}\)

Exercise 12

12. L

Exercise 13

13.

a. B

b. Y

c. N

d. N

e. S

Exercise 14

14. This is done with the C-Strips.

Exercise 15

15. No; 7 –10 is not a whole number

Exercise 16

16.

a. closed

b. not closed; 2-4 is a counterexample

Exercise 17

17.

a. 12 – 4 = 8 ; 12 – 8 = 4

b. 170 – 130 = 40 ; 170 – 40 = 130

c. 80 – 62 = 18 ; 80 – 18 = 62

Exercise 18

18.

a. 4 = 7 –3

b. 8 = 9 –1

Exercise 19

19. No. 4 < 7 but it isn't true that 7 < 4

Exercise 21

21.

a. 2; 9 = 7 + 2	b. 7; 13 = 6 + 7
c. 80; 88 = 8 + 80	d. 56; 70 = 14 + 56

Exercise 22

22. \(41_{\text{five}}\)

Exercise 23

23. \(1_{\text{two}}\)

Exercise 24

24. \(4E_{\text{twelve}}\)

Exercise 25

25.

a. -5

b. 3

Exercise 26

26.

a. 6

b. 7

c. 5

d. 2

e. -7

f. -3

g. -12

h. 0

Exercise 27

27. vector a: 6; vector c: -5; vector d: 3

Exercise Set 5 Solutions

Exercise 1

1.

a. \(314_{\text{thirteen}}\)	b. \(151_{\text{nine}}\)
c. \(1225_{\text{six}}\)	d. \(134_{\text{five}}\)
e. \(12_{\text{three}}\)

Exercise 2

2.

a. 1618	b. \(31_{\text{four}}\)
c. \(363_{\text{twelve}}\)	d. \(45_{\text{six}}\)
e. \(111_{\text{two}}\)	f. \(254_{\text{nine}}\)
g. \(33_{\text{five}}\)	h. \(234_{\text{seven}}\)
i. \(2_{\text{three}}\)	j. \(236_{\text{eight}}\)

Exercise 3

3. Show the check by adding the answer to the subtrahend and making sure the sum is the minuend.

Exercise 4

4.

a. 6717	b. \(1214_{\text{five}}\)
c. \(1424_{\text{eight}}\)	d. \(49E_{\text{thirteen}}\)
e. \(1010_{\text{two}}\)

Exercise 5

5. Show the check by adding the difference (answer) to the subtrahend and making sure the sum is the minuend.

Exercise 6

6.

a. 6697	b. \(1144_{\text{five}}\)
c. \(1774_{\text{eight}}\)	d. \(4WE_{\text{thirteen}}\)
e. \(111_{\text{two}}\)

Exercise 7

7. Show the check by adding the difference (answer) to the subtrahend and making sure the sum is the minuend.

Exercise 10

10.

a. 9462

b. \(122_{\text{four}}\)

c. \(101_{\text{two}}\)

Exercise 11

11. \(1_{\text{two}}\)

Exercise 12

12.

a. 4677

b. \(134_{\text{five}}\)

Exercise 13

13.

a. Since the minuend is \(120_{\text{three}}\), the next place value block is a base three block. Since the subtrahend is \(12_{\text{three}}\), we have to figure out what to add to \(12_{\text{three}}\) to make a block. \(12_{\text{three}}\) is a long and two units, so I have to add 1 unit, 1 long and 2 flats to make a block. Therefore, ass 2 flats, a long a unit to the minuend, \(120_{\text{three}}\), which gives 1 block, 1 flat and 1 unit for the new minuend. If I remove the largest block from this new minuend, I should have the answer of 1 flat and 1 unit.So the answer is \(101_{\text{three}}\). I've left the drawing up to you. See #13 below to see how to start the problem with an abbreviated drawing.

b. Since the minuend is \(213_{\text{four}}\), the next place value block is a base four block. Since the subtrahend is \(133_{\text{four}}\), we have to figure out what to add to \(133_{\text{four}}\) to make a block. \(133_{\text{four}}\) is a flat, 3 longs and three units, so I have to add 1 unit and 2 flats to make a block. Therefore, add 2 flats and a unit to the minuend, \(213_{\text{four}}\), which gives 1 block and 2 longs for the new minuend. If I remove the largest block from this new minuend, I should have the answer of 2 longs. So the answer is \(20_{\text{four}}\). You can use abbreviations for the drawing and start by representing each original number as shown below. I've left the part to add on and final subtraction for you

Exercise 14

14.

a. She is breaking up the subtrahend into three parts (100 + 50 + 2), subtracting one place value at a time. 634 – 100 = 534; 534 – 50 = 484; 484 – 2 =482.

b. He is using a complementary method. If you add 48 to 152, you get 200. So he adds 48 to the minuend (634 + 48 = 682) and then subtracts 200.

Homework Solutions

Exercise 1

1. Provide your own sets. Here is one solution.

a. First, we must define two sets, A and B, that are disjoint with each other, such that one set has 5 elements and the other has 3 elements.

Let A = {1, 2, 3, 4, 5} and B = {x, y, z}. Since n(A) = 5, n(B) = 3 and \(A \cap B = Ø\), we can find the sum of 5 and 3 by first forming the union of A and B and then counting the number of elements in their union. \(A \cup B\) = {1, 2, 3, 4, 5, x, y, z}. Since n(\(A \cup B\)) = 8, then 5 + 3 = 8.

Exercise 2

2. a. since 79 = 66 + 13

Exercise 3

3. a. closed under both operations

Exercise 4

4.

a.

c.

Exercise 5

5.

a.

c.

Exercise 6

6.

a. \(12325_{\text{six}}\)

b. \(2535_{\text{five}}\)

Exercise 13

13. a. 6

Exercise 14

14.

a. Base 9

c. Bases Three through Thirteen

Module 4 Multipication

Exercise Set 1 Solutions

Exercise 1

1.

b. Step 1:

Step 2:

c. Step 3:

d. both trains have the same length

Exercise 2

2.

a. 2 dark green

b. 6 red

c. 12, 12

d. same length

e. \(R \times D = D \times R ; 2 \times 6 = 6 \times 2\)

Exercise 3

3.

a. 7 light green

b. 3 black

c. 21, 21

d. same length

e. \(K \times L = L \times K ; 7 \times 3 = 3 \times 7\)

Exercise 4

4.

a. 12 white

b. 1 hot pink

c. 12, 12

d. same length

e. \(H \times W = W \times H ; 12 \times 1 = 1 \times 12\)

Exercise 5

5.

a. D

b. W

c. R

d. P

e. L

f. K

g. L

h. R

i. R

Exercise 6

6.

a. These are the four trains, but they may have been listed them in a different order.

b. Using the trains as listed above, Train 1: \(R \times D = H\), or \(2 \times 6 = 12\) Train 2: \(P \times L = H\), or \(4 \times 3 = 12\) Train 3: \(D \times R = H\), or \(6 \times 2 = 12\) Train 4: \(H \times W = H\), or \(12 \times 1 = 12\)

Exercise 7

7. Red, R

Exercise 8

8.

a. 3 + 3 + 3 + 3 + 3 + 3 = 18

b. 9 + 9 = 18

c. 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 = 18

Exercise 9

9.

a. \(4 \times 3 = 3 + 3 + 3 + 3 = 6 + 3 + 3 = 9 + 3 = 12 ; 3 \times 4 = 4 + 4 + 4 = 8 + 4 = 12\)

b. \(2_{\text{five}} \times 3_{\text{five}} = 3_{\text{five}} + 3_{\text{five}} = 11_{\text{five}} ; 3_{\text{five}} \times 2_{\text{five}} = 2_{\text{five}} + 2_{\text{five}} + 2_{\text{five}} = 11_{\text{five}}\)

Exercise 10

10. Only the answer, and not the individual steps, are shown

a.

b.

c.

Exercise 11

11.

b.	c. 6
e.	f. 6
g. same length	h. (\(R \times L ) \times P) = R \times (L \times P)\)
i. \((2 \cdots 3) \cdots 4 = 2 \cdots (3 \cdots 4)\)	j. \((2 \cdots 3) \cdots 4 = 6 \cdots 4 = 24; 2 \cdots (3 \cdots 4) = 2 \cdots 12 = 24\)
k. yes

Exercise 12

12.

b.	c. 10
e.	f. 10
g. same length	h. \((Y \times R ) \times L) = Y \times (R \times L)\)
i. \((5 \cdots 2) \cdots 3 = 5 \cdots (2 \cdots 3)\)	j. \((5 \cdots 2) \cdots 3 = 10 \cdots 3 = 30; 5 \cdots (2 \cdots 3) = 5 \cdots 6 = 30\)
k. yes

Exercise 13

13. Do the order of operations to simplify each side of each equation in a - d.

Exercise 14

14. There are various combinations you could use for part a.

Exercise 15

15.

a. 3 purple strips and 3 yellow strips.	b. same as part a
c. \(L \times (P + Y) = (L \times P) + (L \times Y)\)	d. \(4 \times (2 + 3) = (4 \times 2) + (4 \times 3)\)
e. \(4 \times (2 + 3) = 4 \times 5 = 20\) and \((4 \times 2) + (4 \times 3) = 8 + 12 = 20\)

Exercise 16

16.

a. 2 white strips and 2 dark green strips	b. same as part a
c. \(R \times (W + D) = (R \times W) + (R \times D)\)	d. \(2 \times (1 + 6) = (2 \times 1) + (2 \times 6)\)
e. \(2 \times (1 + 6) = 2 \times 7 = 14\) and \((2 \times 1) + (2 \times 6) = 2 + 12 = 14\)

Exercise 17

17. Do the order of operations to simplify each side of each equation in a - d.

Exercise 18

18. There are various combinations you could use for part a and b.

Exercise 19

19.

a. R; \(1 \cdots 2 = 2\)	b. B; \(1 \cdots 9 = 9\)	c. D; \(1 \cdots 6 = 6\)
d. S; \(1 \cdots 11 = 11\)	e. N; \(8 \cdots 1 = 8\)	f. H; \(12 \cdots 1 = 12\)

Exercise 20

20. W, W

Exercise 21

21.

a. \(764 \cdots (1000 - 1) = 764 \cdots 1000 - 764 \cdots 1 = 764000 - 764 = 763236\)

b. \(324 \cdots (100 + 2) = 324 \cdots 100 + 324 \cdots 2 = 32400 + 648 = 33048\)

c. \(83 \cdots (74 + 26) = 83 \cdots 100 = 8300\)

Exercise 22

22. Make sure you know all the properties and can provide examples.

Exercise 23

23.

a. {(3,2), (3,6), (4,2), (4,6)}

b. {(6,5), (7,5),(8,5),(9,5)}

c. { }

d. {(a,a)}

e. {(x,x), (x,y), (y,x), (y,y)}

f. {(1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5)}

Exercise 24

24. a. Any two sets, one having 5 elements and one having 2 elements can be used. This is just one possible solution. Let M = {a,b,c,d,e} and let N = {x,y}. Since n(M) = 5 and n(N) = 2, then \(5 \times 2 = n(M \times N) = n({(a,x), (a,y), (b,x), (b,y), (c,x), (c,y), (d,x), (d,y), (e,x), (e,y)}) = 10\).

Exercise 25

25.

a. \(8 \times 6\)

b. \(3 \times 10\)

c. \(8 \times 6\)

d. \(5 \times 5\)

e. \(4 \times 14\)

f. \(3 \times 7\)

Exercise 28

28. c. \(6 \times 4\)

Exercise 30

30.

a.

b.

c.

d.

e. yes

Exercise 31

31. The distributive property of multiplication over addition.

Exercise 32

32. This should be done like exercise 30. Explain the steps as well as show the pictures.

Exercise 33

33.

a. 8

b. 8

c. same length

Exercise Set 2 Solutions

Exercise 2

2.

\(13 \times 29\) 1 -----> 29 2 -----> 58 4 ----->116 8 ----->232 ---------------------------- 232 + 116 + 29 = 377

\(29 \times 13\) 1 ----->13 2 ----->26 4 ----->52 8 ----->104 16 ----->208 ---------------------------- 208 + 104 + 52 + 13 = 377

Exercise 3

3.

\(27 \times 14\) 1 -----> 14 2 -----> 28 4 -----> 56 8 -----> 112 16 -----> 224 ---------------------------- 224 + 112 + 28 + 14 = 378

\(14 \times 27\) 1 -----> 27 2 ----->54 4 ----->108 8 ----->216 ---------------------------- 216 + 108 + 54 = 378

Exercise 4

4.

Exercise 5

5.

a. Answer: 3,591

b. Answer: 29,070

e. Answer: 3,886

f.

Exercise 6

6.

a. \(221_{\text{four}}\)	b.
c. \(1102_{\text{four}}\)	d. \(2_{\text{four}} \times 221_{\text{four}} = 1102_{\text{four}}\)

Exercise 7

7.

a. \(221_{\text{four}}\)	b.
c. \(1102_{\text{four}}\)	d. \(2_{\text{four}} \times 221_{\text{four}} = 1102_{\text{four}}\)

Exercise 8

8. Make two piles of base three blocks, each having 2 flats, 2 longs and a unit. The two equal piles are shown below.

Then, combine the piles together and make appropriate exchanges. After making exchanges, this is what it looks like:

Write the base three numeral represented after making all exchanges. The answer is \(1212_{\text{three}}\).

Exercise 9

9. Make three piles of base three blocks, each having 2 flats, 2 longs and a unit. The three equal piles are shown below.

Then, combine the piles together and make appropriate exchanges. After making exchanges, this is what it looks like:

Write the base three numeral represented after making all exchanges. The answer is \(2210_{\text{three}}\).

Exercise 10

10.

a. a flat

b. a flat

c. a flat

d. \(L \times L\) = F

Exercise 11

11.

a. a block

a. a block

a. a block

d. \(F \times L\) = B

Exercise 12

12.

a. a long block

a. a long block

a. a long block

d. \(B \times L\) = LB

Exercise 13

13.

a. F

b. B

c. LB

d. B

e. LB

Exercise 15

15.

a. 3, 1 and 2

b. 3 blocks, 1 flat and 2 longs

Exercise 16

16. Only the answers are given here. Make sure you show it with blocks and EXPLAIN.

a. \(1032_{\text{four}}\)

b. \(10122_{\text{three}}\)

Exercise 17

17.

a. a long block

b. a long block

c. a long block

d. \(F \times F\) = LB

Exercise 18

18.

a. U	b. L	c. F	d. B
e. L	f. F	g. B	h. F
i. B	j. LB

Exercise 20

20. The answer is \(20010_{\text{three}}\). Make sure you show it with blocks and EXPLAIN.

Exercise 21

21. The entire table is shown:

Exercise 22

22.

a. \(23_{\text{five}} \times 32_{\text{five}}\) = (2L + 3U) \(\times\) (3L + 2U) = \(2L \times 3L + 2L \times 2U + 3U \times 3L + 3U \times 2U\) = 6F + 4L + 9L + 6U = 6F + 13L + 6U = 1B + 1F + 2F + 3L + 1L + 1U = 1B + 3F + 4L + 1U = \(1341_{\text{five}}\)

\(42_{text{eight}} \times 53_{\text{eight}}\) = (4L + 2U) \(\times\) (5L + 3U) = \(4L \times 5L + 4L \times 3U + 2U \times 5L + 2U \times 3U\) = 20F + 12L + 10L + 6U = 20F + 22L + 6U = 2B + 4F + 2F + 6L + 6U = 2B + 6F + 6L + 6U = \(2666_{\text{eight}}\)

Exercise 23

23.

\(212_{\text{four}} \times 102_{\text{four}}\)

= (2F + 1L + 2U) \(\times\) (1F + 2U)

= \(2F \times 1F + 2F \times 2U + 1L \times 1F + 1L \times 2U + 2U \times 1F + 2U \times 2U\)

= 2LB + 4F + 1B + 2L + 2F + 4U

= 2LB + 1B + 6F + 2L + 4U

= 2LB + 1B + 1B + 2F + 2L + 1 L

= 2LB + 2B + 2F + 3L

= \(22230_{\text{four}}\)

Exercise 24

24.

a. \(361_{\text{nine}} \times 15_{\text{nine}}\) = (3F + 6L + 1U) \(\times\) (1L + 5U) = \(3F \times 1L + 3F \times 5U + 6L \times 1L + 6L \times 5U + 1U \times 1L + 1U \times 5U\) = 3B + 15F + 6F + 30L + 1L + 5U = 3B + 21F + 31L + 5U = 3B + 2B + 3F + 3F + 4L + 5U = 5B + 6F + 4L + 5U = \(5645_{\text{nine}}\)

b. \(111_{\text{two}} \times 11_{\text{two}}\) = (1F + 1L + 1U) \(\times\) (1L + 1U) = \(1F \times 1L + 1F \times 1U + 1L \times 1L + 1L \times 1U + 1U \times 1L + 1U \times 1U\) = 1B + 1F + 1F + 1L + 1L + 1U = 1B + 2F + 2L + 1U = 1B + 1B + 1F + 1U = 1LB + 1F + 1U = \(10101_{\text{two}}\)

Exercise 25

25.

a. Answer: \(1422_{\text{six}}\)	b. Answer: \(1469_{\text{eleven}}\)	c. \(100011_{\text{two}}\)
d. \(T09_{\text{twelve}}\)	e. \(20211_{\text{three}}\)	f. \(2122_{\text{four}}\)

Exercise 26

26. Only a, b and c are shown here. Make sure you do d and e.

Exercise 27

27.

a. \(1103_{\text{five}}\)	b. \(1T58_{\text{thirteen}}\)	c. \(1022_{\text{three}}\)
d. \(16015_{\text{seven}}\)	e. \(16015_{\text{seven}}\)	f. \(101310_{\text{four}}\)

Exercise 29

29. \(2144_{\text{five}}\)

Exercise 30

30. \(31653_{\text{seven}}\)

Exercise 31

31. \(10001111_{\text{two}}\)

Exercise 32

32. \(1220212_{\text{three}}\)

Exercise 33

33. \(15417_{\text{twelve}}\)

Exercise 34

34. \(115503_{\text{six}}\)

Exercise 35

35. \(2012220_{\text{three}}\)

Exercise 36

36. \(424204_{\text{six}}\)

Exercise 37

37. \(101001010_{\text{two}}\)

Exercise 38

38. \(T73E8_{\text{twelve}}\)

Exercise 39

39. \(615032_{\text{eight}}\)

Exercise 40

40. \(1033230_{\text{four}}\)

Exercise 41

41. \(2321411_{\text{five}}\)

Exercise 42

42. \(1563526_{\text{seven}}\)

Exercise 43

43. \(414025_{\text{nine}}\)

Exercise 44

44. \(8T6697_{\text{eleven}}\)

Homework Solutions

Exercise 1

1. a. 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 32

Exercise 3

3. a. P; \(4 \cdots 1 = 4\)

Exercise 4

4. a. {(3,0), (3,1), (3,6), (x,0), (x,1), (x,6)}

Exercise 6

6.

a. B

c. LB

e. FB

Exercise 10

10.

a. Associative Property of Multiplication

Illustrate this equation is true: \((2_{\text{five}} \cdots 3_{\text{five}}) \cdots 4_{\text{five}} = 2_{\text{five}} \cdots (3_{\text{five}} \cdots 4_{\text{five}}\))

Left side: \((2_{\text{five}} \cdots 3_{\text{five}}) \cdots 4_{\text{five}} = 11_{\text{five}} \cdots 4_{\text{five}} = 44_{\text{five}}\)

Right side: \(2_{\text{five}} \cdots (3_{\text{five}} \cdots 4_{\text{five}}) = 2_{\text{five}} \cdots 22_{\text{five}}) = 44_{\text{five}}\)

Both expressions equal \(44_{\text{five}}\), so \((2_{\text{five}} \cdots 3_{\text{five}}) \cdots 4_{\text{five}} = 2_{\text{five}} \cdots (3_{\text{five}} \cdots 4_{\text{five}}\))

Distributive Property of Multiplication of Addition:

Illustrate this equation is true: \(2_{\text{five}} \cdots (3_{\text{five}} + 4_{\text{five}}) = (2_{\text{five}} \cdots 3_{\text{five}}) + (2_{\text{five}} \cdots 4_{\text{five)}}\)

Left side: \(2_{\text{five}} \cdots (3_{\text{five}} + 4_{\text{five}}) = 2_{\text{five}} \cdots (12_{\text{five}}) = 24_{\text{five}}\)

Right side: \((2_{\text{five}} \cdots 3_{\text{five}}) + (2_{\text{five}} \cdots 4_{\text{five}}) = 11_{\text{five}} + 13_{\trxt{five}} = 24_{\text{five}}\)

Both expressions equal \(24_{\text{five}}\), so \(2_{\text{five}} \cdots (3_{\text{five}} + 4_{\text{five}}) = (2_{\text{five}} \cdots 3_{\text{five}}) + (2_{\text{five}} \cdots 4_{\text{five}}\))

Module 5 Binary Operations

Exercise Set 1 Solutions

Exercise 2

2.

a. 16	b. 65	c. 2	d. 42	e. 4
f. 18	g. 14	h. 41	i. 20

Exercise 3

3.

a. m ! n = n ! m; Yes; Proof: If m ! n =n ! m, then ! is commutative. m ! n = 2 and n ! m = 2. Since both expressions (m ! n and n ! m) equal 2, then ! is commutative.

b. \(m \bigoplus n = n \bigoplus m\); Yes; Proof: If \(m \bogoplus n = n \bigoplus m\), then \(\bigoplus\) is commutative. \(m \bigoplus n = 3mn\) and \(n \bigoplus m = 3nm\). Since 3mn = 3nm, then \(m \bigoplus n = n \bigoplus m.\) Therefore, is commutative.

c. m # n = n # m; Yes; Provide your own an example using numbers.

d. m ʘ n = n ʘ m; Yes; Proof: If m ʘ n = n ʘ m, then ʘ is commutative. m ʘ n = 2m + 2n, and n ʘ m = 2n + 2m. Since 2m + 2n = 2n + 2m, then m ʘ n = n ʘ m. Therefore, ʘ is commutative.

e. \(m \XBox n = n \XBox m\); No; Provide your own counterexample using numbers.

f. m , n = n , m; Yes; Proof: If m , n = n , m, then , is commutative. \(m , n = m^{2} + n^{2}\) and \(n , m = n^{2} + m^{2}\). Since \(m^{2} + n^{2} = n^{2} + m^{2}\), then m , n = n , m. Therefore ,, is commutative.

g. m * n = n * m; No; Provide your own counterexample using numbers.

h. m )( n = n )( m; No; Provide your own counterexample using numbers.

Exercise 4

4.

a. 17

b. 4

c. 2

d. 216

e. 36

f. 125

Exercise 5

5.

a. 270	b. 34	c. 27	d. 6	e. 6
f. 629	g. 6	h. 4	i. 12

Exercise 6

6.

a. (a ! b)q c = a ! (b! c); yes; Proof: If (a ! b) ! c = a ! (b! c), then ! is associative. (a ! b) !c = 2! c = 2, and a ! (b! c) = a ! 2 = 2. Since both (a ! b) ! c and a ! (b! c) are equal to 2, then ! is associative.

b. \((a \bigoplus b) \bigoplus c = a \bigoplus (b \bioplus c)\); yes; Proof: If \((a \bigoplus b)r c = a \bigoplus (b \bigoplus c)\), then \(\bigoplus\) is associative. \((a \bigoplus b) \bigoplus c = 3ab \bigoplus c = 3 \cdots 3ab \cdots c = 9abc,\) and \(a \bigoplus (b \bigoplus c) = a \bigoplus 3bc = 3 \cdots a \cdots 3bc = 9abc\). Since both expressions \([(a \bigoplus b) \bigoplus c\) and \(a \bigoplus (b \bigoplus c)]\) are equal to 9abc, then \(\bigoplus\) is associative.

c. (a # b)# c = a # (b # c); yes; Provide your own example using numbers.

d. (a ʘ b) ʘ c = a ʘ (b ʘ c); no; Provide your own counterexample using numbers.

e. \((a \XBox b) \XBox c = a \XBox (b \XBox c)\); no; Provide your own counterexample using numbers.

f. (a ,b) , c = a , (b, c); no; Provide your own counterexample using numbers.

g. (a * b)* c = a * (b * c); no; Provide your own counterexample using numbers.

Exercise 7

7. a @ (b + c) = (a @ b) + (a @ c)

Exercise 8

8. a + (b @ c) = (a + b) @ (a + c)

Exercise 9

9.

a. a & (b $ c) = (a & b) $ (a & c)

b. a $ (b & c) = (a $ b) & (a $ c)

Exercise 10

10.

a. a ! (b r c) = (a ! b) r (a ! c)

b. no

c. Provide your own counterexample using numbers.

d. \(a \bigoplus (b ! c) = (a \bigoplus b) ! (a \bigoplus c)\)

e. no

f. Provide your own counterexample using numbers.

Exercise 11

11.

a. \(a + (b \cdots c) = (a + b) \cdots (a + c)\)

b. no

c. Provide your own counterexample using numbers.

Exercise 12

12.

a. \(a \cdots (b \cdots c) = (a \cdots b) \cdots (a \cdots c)\)

b. yes

c. Provide your own example using numbers.

Exercise 13

13.

a. a , (b @ c) = (a , b) @ (a , c)

b. no

c. Provide your own counterexample using numbers.

d. a @ (b , c) = (a @ b) , (a @ c)

e. no

f. Provide your own counterexample using numbers.

Exercise 14

14. You are on your own here. Be creative.

Exercise 15

15.

a. m n = n m; no; provide your own counterexample using numbers

b. m n = n m; yes; provide your own proof

c. (m n) x = m (n x); no; provide your own counterexample using numbers

d. (m n) x = m y (n x); no; provide your own counterexample

e. a (b + c) = (a b) + (a c); no; provide your own counterexample

f. a (b + c) = (a b) + (a c); no; provide your own counterexample

g. a (b y c) = (a b) y (a c); no; provide your own counterexample

h. a (b c) = (a b) (a c); no; provide your own counterexample

Exercise 16

16. \((a + b) \cdots c = (a \cdots c) + (b \cdots c)\); yes; provide your own example

Exercise 17

17. \((a \cdots b) + c = (a + c) \cdots (b + c)\); no; provide your own counterexample

Exercise 18

18. \((a + b) \div c = (a \div c) + (b \div c)\); yes; provide your own example

Exercise 18

19. \(a \div (b + c) = (a \div b) + (a \div c)\); no; provide your own counterexample

Exercise 20

20.

a. (b + c) a = (b a) + (c a); no; provide your own counterexample

b. (b + c) a = (b a) + (c a); no; provide your own counterexample

c. (b c) a = (b a) (c a); no; provide your own counterexample

d. (b c) a = (b a) (c a); no; provide your own counterexample

Homework Solutions

Exercise 1

1.

a. 23	b. 20	c. 2a + b	d. 26	g. 2
j. 90	m. 1	p. 10	s. 26	v. 17
y. 45

Exercise 2

2. a. 19

Exercise 3

3.

a. 20

c. 36

e. 21

Module 6 Integers

Exercise Set 1 Solutions

Exercise 1

1. {..., -3, -2, -1}

Exercise 2

2. {..., -3, -2, -1, 0, 1, 2, 3, ...}

Exercise 3

3.

a. 4

b. 8

c. 2

d. 1

e. 0

Exercise 4

4.

a. 7

b. 13

c. 4

d. \(\frac{3}{7}\)

e. \(\frac{3}{7}\)

f. 6

Exercise 5

5. a. 6, -6

Exercise 6

6. a. 19, -19

Exercise 7

7. 0

Exercise 8

8. none

Exercise 9

9.

a. -5

b. +3

Exercise 10

10.

a. -3 since the terminal point of the last vector landed on -3.

b. -9 since the terminal point of the last vector landed on -9.

c. 4 since the terminal point of the last vector landed on 4.

d. 1 since the terminal point of the last vector landed on 1.

e. -2 since the terminal point of the last vector landed on -2.

f. -60 since the terminal point of the last vector landed on -60.

g. -1 since the terminal point of the last vector landed on -1.

h. -3 since the terminal point of the last vector landed on -3.

Exercise 11

11. thermometer

Exercise 12

12. walk two blocks east

Exercise 13

13. The answers for these are the same as for those in exercise 10.

Exercise 14

14.

a. 5

b. -9

c. -3

d. 6

Exercise 15

15.

a. RRRRR

b. GG

c. RRRR

Exercise 16

16. 0; reasons may vary

Exercise 17

17. Answers will vary, but there will always be the same number of negative counters as positive counters in each representation.

Exercise 18

18.

a. -2

b. -2

c. -2

d. -2

e. -3

f. +3

g. +3

h. +4

Exercise 19

19. Answers will vary, but there will be 4 more positive counters than negative counters in each representation.

Exercise 20

20. Answers will vary, but there will be 4 more negative counters than positive counters in each representation.

Exercise 21

21.

a. (1) Combine 5 reds and 3 reds: RRRRR + RRR (2) There are 8 reds, RRRRRRRR, which represents -8. (3) Therefore, -5 + -3 = -8.

b. (1) Combine 6 reds and 9 greens: RRRRRR + GGGGGGGGG = R R R R R R G G G G G G G G G (2) Remove 6 red-green pairs (zero), leaving GGG, which represents -3. (3) Therefore, -6 + 9 = 3.

c. (1) Combine 8 reds and 6 greens: RRRRRRRR + GGGGGG = R R R R R R R R G G G G G G (2) Remove 6 red-green pairs (zero), leaving RR, which represents -2. (3) Therefore, -8 + 6 = -2.

d. (1) Combine 5 reds, 3 reds and 6 greens: RRRRR + RRR + GGGGGG = R R R R R R R R G G G G G G (2) Remove 6 red-green pairs (zero), leaving RR, which represents -2. (3) Therefore, -5 + -3 + 6 = -2.

e. (1) Combine 6 reds, 9 greens and 1 red: RRRRRR + GGGGGGGGG + R = R R R R R R R G G G G G G G G G (2) Remove 7 red-green pairs (zero), leaving GG which represents +2. (3) Therefore, -6 + 9 + -1 = +2.

f. (1) Combine 4 greens, 6 greens and 5 reds: GGGG + GGGGGG + RRRRR = G G G G G G G G G G R R R R R (2) Remove 5 red-green pairs (zero), leaving GGGGG, which represents +5. (3) Therefore, +4 + +6 + -5 = +5.

Exercise 22

22.

a. (1) 6 – 3 means remove 3 greens from a collection of counters representing 6. (2) Represent 6 with 6 greens: GGGGGG (3) Remove 3 greens, leaving GGG, which represents +3. (4) Therefore, 6 – 3 = 3

b. (1) 4 – 6 means remove 6 greens from a collection of counters representing 4. (2) Represent 4 with 4 greens: GGGG (3) Add 2 red-green pairs (zero) to the collection: G G G G G G R R (4) Remove 6 greens, leaving RR, which represents -2. (5) Therefore, 4 – 6 = -2

c. (1) -7 – -6 means remove 6 reds from a collection of counters representing -7. (2) Represent -7 with 7 reds: RRRRRRR (3) Remove 6 reds, leaving R, which represents -1. (4) Therefore, -7 – (-6) = -1

d. (1) -3 – -7 means remove 7 reds from a collection of counters representing -3. (2) Represent -3 with 3 reds: RRR (3) Add 4 red-green pairs (zero) to the collection: R R R R R R R G G G G (4) Remove 7 reds, leaving GGGG, which represents +4. (5) Therefore, -3 – (-7) = +4

e. (1) 4 – -3 means remove 3 reds from a collection of counters representing +4. (2) Represent +4 with 4 greens: GGGG (3) Add 3 red-green pairs (zero) to the collection: G G G G G G G R R R (4) Remove 3 reds, leaving GGGGGGG, which represents +7. (5) Therefore, 4 – (-3) = +7

f. (1) -2 – 5 means remove 5 greens from a collection of counters representing -2. (2) Represent -2 with 2 reds: RR (3) Add 5 red-green pairs (zero) to the collection: R R R R R R R G G G G G (4) Remove 5 greens, leaving RRRRRRR, which represents -7. (5) Therefore, -2 – 5 = -7

g. (1) 5 – 5 means remove 5 greens from a collection of counters representing 5. (2) Represent 5 with 5 greens: GGGGG (3) Remove 5 greens, leaving nothing, which represents 0. (5) Therefore,5 – 5 = 0

h. (1) 5 – (-5) means remove 5 reds from a collection of counters representing 5. (2) Represent 5 with 5 greens: GGGGG (3) Add 5 red-green pairs (zero) to the collection: G G G G G G G G G G. R R R R R (4) Remove 5 reds, leaving GGGGGGGGGG, which represents 10. (5) Therefore, 5 – (-5) = +10

i. (1) -4 – (- 4) means remove 4 reds from a collection of counters representing -4. (2) Represent -4 with 4 reds: RRRR (3) Remove 4 reds, leaving nothing, which represents 0.. (5) Therefore, -4 – -4 = 0.

j. (1) -6 – 6 means remove 6 greens from a collection of counters representing -6. (2) Represent -6 with 6 reds: RRRRRR (3) Add 6 red-green pairs (zero) to the collection: R R R R R R R R R R R R G G G G G G (4) Remove 6 greens, leaving RRRRRRRRRRRR, which represents -12. (5) Therefore, -6 – 6 = -12

Exercise 23

23.

a. (1) 5 – 8 means remove 8 greens from a collection of counters representing 5. (2) Represent 5 with 5 greens: GGGGG (3) Add 3 red-green pairs (zero) to the collection: G G G G G G G G R R R (4) Remove 8 greens, leaving RRR, which represents -3. (5) Therefore, 5 – 8 = -3

b. (1) 5 + -8 means combine 5 greens and 8 reds: GGGGG + RRRRRRRR = G G G G G R R R R R R R R (2) Remove 5 red-green pairs (zero), leaving RRR, which represents -3. (3) Therefore, 5 + -8 = -3.

c. The answers are the same.

Exercise 24

24. There are various ways to do this problem. But in any case, the final collection for each should give the same answer.

Exercise 25

25.

a. 1st way: Remove 3 greens from a collection representing -5. 2nd way: Combine a collection of 5 reds and 3 reds together.

b. 1st way: Remove 8 greens from a collection representing +4. 2nd way: Combine a collection of 4 greens and 8 reds together.

c. 1st way: Remove 7 reds from a collection representing -9. 2nd way: Combine a collection of 9 reds and 7 greens together.

d. 1st way: Remove 10 reds from a collection representing -10. 2nd way: Combine a collection of 10 reds and 10 greens together.

Exercise 26

26.

a. 4 + (-5)

b. -9 + (-7)

c. -3 + (+4)

Exercise 27

27.

a. 4 + -5 + -7 + -6 + +9

b. -43 + (+75) + 12 + 63 + (-9)

Exercise 28

28.

a. -6

b. +8

c. negative, negative

d. positive, positive

e. opposite

Exercise 29

29.

a. When the minuend and subtrahend are the same number

b. 0

Exercise 30

30.

a. 10 – 4 = +6	e. 3 – 10 = -7
b. -7 – 5 = -12	f. -5 – (-8) = +3
c. 8 – (-3) = +11	g. 6 – (-6) = +12
d. -9 – -1 = -8	h. 9 – 9 = 0

Exercise 31

31.

a. 1 – (-5) = +6

b. 13 – 10 = +3

c. 3 – 8 = -5

Exercise 32

32.

a. -92 – -71 = -21	d. -122 – 76 = -198
b. 92 – -81 = +173	e. 208 – 389 = -181
c. -110 – -200 = +90	a. 46 + -76 + 92 = 138 + -76 = 62

Exercise 33

33.

a. 46 + -76 + 92 = -30 + 92 = 62

b. -63 + 94 + 45 + -71 = 31 + 45 + -71 = 76 + -71 = 5

Exercise 34

34.

a. 46 + -76 + 92 = 138 + -76 = 62

b. -63 + 94 + 45 + -71 = 139 + -134 = 5

Exercise 35

35.

a. Closed

a. Closed

a. Closed

d. Not Closed; one counterexample is: 7 – 10 = -3

e. Closed

f. Not Closed; one counterexample is: -11 – -20 = +9

g. Not closed; one counterexample is: 1 + 1 = 2

h. Not closed; one counterexample is: 1 – -1 = 2

Exercise Set 2 Solutions

Exercise 1

1.

a. R R R R R G

b. R R R R R R R R R R R R R R R G G G

c. RRRRRRRRRRRR = -12

Exercise 2

2.

a. R R R R R R G G

b. R R R R R R R R R R R R R R R R R R G G G G G G

c. RRRRRRRRRRRR = -12

Exercise 3

3.

a. R R R R

b. R R R R R R R R R R R R

c. -12

Exercise 4

4.

a. \(4 \times -2\) means to combine 4 sets of 2 reds: RR + RR + RR + RR = RRRRRRRR = -8

b. \(3 \times 5\) means to combine 3 sets of 5 greens: GGGGG + GGGGG + GGGGG = GGGGGGGGGGGGGGG = +15

c. \(5 \times -3\) means to combine 5 sets of 3 reds: RRR + RRR + RRR + RRR + RRR = RRRRRRRRRRRRRRR = -15

d. \(7 \times 2\) means to combine 7 sets of 2 greens: GG + GG + GG + GG + GG + GG + GG = GGGGGGGGGGGGGG = +14

e. \(0 \times -3\) means to combine 0 sets of 3 reds = 0

Exercise 5

5.

a. R R R R R R R R R R R R R R G G G G G G G G G G G G G G

b. R R R R R R R R R R R R R R G G

c. R R R R R R R R R R R R = -12

Exercise 6

6.

a. R R R R R R R R R R R R G G G G G G G G G G G G

b. R R R R R R R R R R R R

c. -12

Exercise 7

7.

a. \(-5 \times 3\) means to remove 5 sets of 3 greens from a collection representing zero. (1) Let zero be represented by 15 reds and 15 greens: R R R R R R R R R R R R R R R G G G G G G G G G G G G G G G (2) Remove 5 sets of 3 greens from the above collection, which leaves RRRRRRRRRRRRRRR, which represents -15. Therefore, \(-5 \times 3 = -15\).

b. \(-3 \times 2\) means to remove 3 sets of 2 greens from a collection representing zero. (1) Let zero be represented by 6 reds and 6 greens: R R R R R R G G G G G G (2) Remove 3 sets of 2 greens from the above collection, which leaves RRRRRR, which represents -6. Therefore, \(-3 \times 2 = -6\)

c. \(2 \times -3\) means to combine 2 sets of 3 reds: RRR + RRR = RRRRRR = -6. Therefore \(2 \times -3 = -6\)

d. \(-2 \times 3\) means to remove 2 sets of 3 greens from a collection representing zero. (1) Let zero be represented by 6 reds and 15 greens: R R R R R R G G G G G G (2) Remove 2 sets of 3 greens from the above collection, which leaves RRRRRR, which represents -6. Therefore, \(-2 \times 3 = -6\).

e. \(3 \times 2\) means to combine 3 sets of 2 greens: GG + GG + GG = GGGGGG = 6. Therefore \(3 \times 2 = 6\)

f. \(0 \times -4\) means to combine 0 sets of 4 reds = 0

g. \(-4 \times 0\) means to remove 4 sets of 0 counters from a collection representing zero. (1) Choose any representation of zero. One possibility is to let zero be represented by 6 reds and 6 greens. R R R R R R G G G G G G (2) Remove 0 sets of 0 counters from the above collection, which leaves the original collection, which represents 0. Therefore, \(-4 \times 0 = 0\).

Exercise 8

8.

\(-3 \times 6\): Remove 3 sets of 6 greens from zero; 5G and 23R; -18

\(-3 \times 6\): Remove 3 sets of 6 greens from zero; 18R; -18

\(-4 \times 4\): Remove 4 sets of 4 greens from zero; 2G and 18R; -16

\(-4 \times -4\): Remove 4 sets of 4 reds from zero; 18G and 2R; 16

\(-5 \times 2\): Remove 5 sets of 2 greens from zero; 4G and 14R; -10

\(-2 \times 5\): Remove 2 sets of 5 greens from zero; 3G and 13R; -10

\(-5 \times -2\): Remove 5 sets of 2 reds from zero; 11G and 1R; 10

\(-2 \times -5\): Remove 2 sets of 5 reds from zero; answers will vary; 10

\(-3 \times 3\): Remove 3 sets of 3 greens from zero; 1G and 10R; -9

\(-3 \times -3\): Remove 3 sets of 3 reds from zero; 12G and 3R; 9

\(-7 \times 2\): Remove 7 sets of 2 greens from zero; answers will vary; -14

\(-2 \times 8\): Remove 2 sets of 8 greens from zero; 2G and 18R; -16

\(-2 \times -8\): Remove 2 sets of 8 reds from zero; 18G and 2R; 16

Exercise 9

9.

a. positive

b. negative

c. zero

d. positive

Exercise 10

10.

a. Closed

b. Closed

c. Not Closed: one counterexample is: \(-2 \times -8 = 16\)

d. Not Closed: one counterexample is: \(-1 \times -1 = 1\)

e. Closed

f. Closed

Exercise 11

11.

a. true

b. true

c. true

d. false; 2 < 3; Multiply both sides by 5: \(2 \times 5 > 3 \times 5\) is false.

e. true

f. false: 2 < 3; Multiply both sides by -5: \(2 \times -5 > 3 \times -5\) is false.

Exercise 12

12.

a. If a > b and b > c, then a > c

b. If a > b, then a + c > b + c

c. If a > b, then ap > bp

d. If a > b, then an < bn

Homework Solutions

Exercise 2

2.

a. 16

c. 2

e. 8

Exercise 3

3. a. 4 and -4

Exercise 4

4.

a. +7

c. -9

Exercise 5

5.

a. 2 – (-5)

c. 3 – 12

Exercise 6

6. -8 + 5 + (-3) = -6, since the terminal point of the last vector landed on -6.

Exercise 8

8.

a. 5 – 11 = -6

c. 5 – (-11)= 16

Exercise 9

9.

a. (1) -4 + (-3) means to combine 4 reds and 3 reds: RRRR + RRR = RRRRRRR, which represents -7

(2) Therefore, -4 + -3 = -7

Exercise 10

10.

a. (1) 6 – 8 means remove 8 greens from a collection of counters representing +6.

(2) Represent +6 with 6 greens: GGGGGG

(3) Add 2 red-green pairs (zero) to the collection:

G G G G G G G G

R R

(4) Remove 8 greens, leaving RR, which represents -2.

(5) Therefore, 6 – 8 = -2

Exercise 11

11.

a. Closed

b. Not closed; a counterexample is 3 – 10 = -7

c. Closed

Exercise 12

12.

a. \(-5 \times 3\) means to remove 5 sets of 3 greens from a collection representing zero.

(1) Let zero be represented by 15 reds and 15 greens:

R R R R R R R R R R R R R R R

G G G G G G G G G G G G G G G

(2) Remove 5 sets of 3 greens from the above collection, which leaves

RRRRRRRRRRRRRRR, which represents -15. Therefore, \(-5 \times 3 = -15\).

Homework Solutions

HW #2

2.

a. 16

c. 2

e. 8

HW #3

3. a. 4 and -4

HW #4

4. a. +7

HW #5

5.

a. 2 – (-5)

c. 3 – 12

HW #6

6. -8 + 5 + -3 = -6, since the terminal point of the last vector landed on -6.

HW #8

8.

a. 5 – 11 = -6

c. 5 – (-11) = 16

HW #9

9.

a. (1) -4 + -3 means to combine 4 reds and 3 reds: RRRR + RRR = RRRRRRR, which represents -7

(2) Therefore, -4 + -3 = -7

HW #10

10.

a. (1) 6 – 8 means remove 8 greens from a collection of counters representing +6.

(2) Represent +6 with 6 greens: GGGGGG

(3) Add 2 red-green pairs (zero) to the collection:

G G G G G G G G

R R

(4) Remove 8 greens, leaving RR, which represents -2.

(5) Therefore, 6 – 8 = -2

HW #11

11.

a. Closed

b. Not closed; a counterexample is 3 – 10 = -7

c. Closed

HW #12

12.

a. \(-5 \times 3\) means to remove 5 sets of 3 greens from a collection representing zero.

(1) Let zero be represented by 15 reds and 15 greens:

R R R R R R R R R R R R R R R

G G G G G G G G G G G G G G G

(2) Remove 5 sets of 3 greens from the above collection, which leaves RRRRRRRRRRRRRRR, which represents -15. Therefore, \(-5 \times 3 = -15\).

Module 7 Division

Exercise Set 1 Solutions

Exercise 1

1.

a. 2

b. R

c. R, R

Exercise 2

2.

a. P, W, P, W 9, 2, 4, 1, 9, 4, 2, 1

b. L, R, L, R, 11, 3, 3, 2, 11, 3, 3, 2

c. R, blank (or 0) , R, blank (or 0), 8, 4, 2, blank (or 0), 8, 2, 4, blank (or 0)

d. W, L, W, L, 7, 4, 1, 3, 7, 1, 4, 3

e. R, blank (or 0), R, blank (or 0), 6, 3, 2, blank (or 0), 6, 2, 3, blank (or 0)

Exercise 3

3. 2 equal piles of 18 units each

Exercise 4

4.

a. \(210_{\text{four}\)

b. Methods will vary.

c. \(102_{\text{four}}\)

d. \(210_{\text{four}}, 102_{\text{four}}\)

Exercise 5

5.

a. \(1100_{\text{three}}\)

b. Methods will vary.

c. \(200_{\text{three}}\)

d. \(1100_{\text{three}}, 200_{\text{three}}\)

Exercise 7

7.

a. \(210_{\text{four}}\)

b. Methods will vary.

c. \(30_{\text{four}}\)

d. \(210_{\text{four}}, 30_{\text{four}}\)

Exercise 9

9.

a. \(100 \div 4\)

b. Partitioning into subsets

c. Disburse 100 marbles into 4 equal subsets.

d.

e. \(4 \cdots 25\)

f. Count how many marbles were placed in one of the equal subsets.

g. 25

Exercise 10

10.

a. \(80 \div 8\)

b. Repeated Subtraction

c. Put 8 ounces in a subset at a time.

d.

e. \(10 \cdots 8\)

f. Count how many subsets, each having 8 ounces, were made.

g. 10

Exercise 11

11.

a. \(144 \div 16\)

b. Repeated Subtraction

c. Count out 16 pages at a time.

d.

e. \(9 \cdots 16\)

f. Count how many short stories, each having 16 pages, were made.

g. 9

Exercise 12

12.

a. \(500 \div 4\)

b. Partitioning into subsets

c. Disburse $500 into 4 equal subsets.

d.

e. \(4 \cdots 125\)

f. Count how much money was placed in one of the equal subsets.

g. $125

Exercise 13

13.

a. \(8 \div 1\frac{1}{3}\)

b. Repeated Subtraction

c. Put 1\(\frac{1}{3}\) cups into a subset at a time.

d.

e. \(6 \cdots 1\frac{1}{3}\)

f. Count how many subsets were formed.

g. 6

Exercise 14

14.

a. \(65 \div 13\)

b. Partitioning into subsets

c. Disburse 65 baseballs into 13 equal subsets.

d.

e. \(13 \cdots 5\)

f. Count how baseballs were placed in one of the equal subsets.

g. 5

Exercise 15

15.

d.

e. \(12 \cdots 4\)

g.

h. \(4 \cdots 12\)

Exercise 16

16.

d.

e. \(4 \cdots 50\)

g. Put 4 into a subset at a time.

h. \(50 \cdots 4\)

Exercise 17

17.

d. Draw 50 subsets, then disburse 150 amongst the 50 subsets.

e. \(50 \cdots 3\)

g.

h. \(3 \cdots 50\)

Exercise 18

18.

d. Draw 35 subsets, then disburse 140 amongst the 35 subsets.

e. \(35 \cdots 4\)

g.

h. \(4 \cdots 35\)

Exercise 19

19.

d.

e. \(5 \cdots 19\)

g. Put 5 into a subset at a time.

h. \(19 \cdots 5\)

Exercise 19

19.

d.

e. \(5 \cdots 19\)

g. Put 5 into a subset at a time.

h. \(19 \cdots 5\)

Exercise 19

19.

d.

e. \(5 \cdots 19\)

g. Put 5 into a subset at a time.

h. \(19 \cdots 5\)

Exercise 22

22.

a. \(56 \div 8\)

b. \(40 \div 5\)

Exercise 23

23.

a. \(18 \div 9\)

b. \(10 \div 2\)

Exercise 24

24. a represents how many subsets and b represents how many are in each subset.

Exercise 25

25.

a. Since 32 = 8 \(\cdots\) 4, then \(32 \div 8\) = 4

b. Since 56 = 8 \(\cdots\) 7, then 56 \(\div\) 8 = 7

c. Since 32 = 2 \(\cdots\) 16, then 32 \(\div\) 2 = 16

d. Since 0 = 13 \(\cdots\) 0, then 0 \(\div\) 13 = 0

e. Since 12 = 1 \(\cdots\) 12, then 12 \(\div\) 1 = 12

f. Since X = 1 \(\cdots\) X, then X \(\div\) 1 = X

g. Since 0 = Y \(\cdots\) 0, then 0 \(\div\) Y = 0

Exercise 26

26.

c. Since 48 = 6 \(\cdots\) 8, then 48 \(\div\) 6 = 8

d. Since there is no whole number solution to make the equation 35= 4 \(\cdots\) __ true, 35 \(\div\) 4 is not defined under whole numbers.

e. Since 48 = 1 \(\cdots\) 48, then 48 \(\div\) 1 = 48

f. Since there is no whole number solution to make the equation 55 = 7 \(\cdots\) __ true, 55 \(\div\) 7 is not defined under whole numbers.

g. Since 0 = 8 \(\cdots\) 0, then 0 \(\div\) 8 = 0

e. 203 r. 314

f. 248 r. 14

Exercise 27

27.

a. There is no number that will make the equation, 6 = 0 \(\cdots\) ____ true, since any number put in the blank will make the right hand side of the equation zero, which can never equal the left-hand side of the equation, which is 6. Therefore, \(6 \div 0\) is not defined.

b. There is no number that will make the equation, 18 = 0 \(\cdots\)____ true, since any number put in the blank will make the right hand side of the equation zero, which can never equal the left-hand side of the equation, which is 18. Therefore, \(18 \div 0\) is not defined.

c. There is no number that will make the equation, M = 0 \(\cdots\)____ true, since any number put in the blank will make the right handside of the equation zero, which can never equal the left-hand side of the equation, which is M (since it is assumed that M is not equal to zero). Therefore, \(M \div 0\) is not defined.

Exercise Set 2 Solutions

Exercise 1

1.

a. 6 r. 43

b. 7 r. 20

c. 8 r. 8

d. 3 r. 194

e. 5 r. 55

f. 5 r. 165

g. 9 r. 239

Exercise 2

2.

a. 45 r. 15

b. 12 r. 27

c. 32 r. 34

d. 81 r. 29

Exercise 3

3.

a. 45 r. 15

b. 12 r. 27

c. 32 r. 34

d. 81 r. 29

e. 203 r. 314

f. 248 r. 14

Exercise Set 3 Solutions

Exercise 1

1.

a. F F L L U U

b. i. LUU, LUU, LUU, LUU, LUU ii. U

c. UUUUU

d. quotient: LUU, remainder: U

e. \(12_{\text{three}}r. 1_{\text{three}}\)

f. \(12_{\text{three}} \cdots 12_{\text{three}} + 1_{\text{three}} = 221_{\text{three}} + 1_{\text{three}} = 222_{\text{three}}\)

Exercise 2

2.

a. \(13_{\text{five}}r. 3_{\text{five}}\)

b. \(32_{\text{six}}r. 21_{\text{six}}\)

c. \(11_{\text{four}}r. 10_{\text{four}}\)

d. \(110_{\text{two}}r. 1_{\text{two}}\)

Exercise 3

3.

a. \(23_{\text{seven}}r. 3_{\text{seven}}\)

b. \(102_{\text{three}}\)

Exercise Set 3 Solutions

Exercise 1

1.

a. F F L L U U

b. i. LUU, LUU, LUU, LUU, LUU ii. U

c. UUUUU

d. quotient: LUU, remainder: U

e. \(12_{\text{three}}r. 1_{\text{three}}\)

f. \(12_{\text{three}} \cdots 12_{\text{three}} + 1_{\text{three}} = 221_{\text{three}} + 1_{\text{three }}= 222_{\text{three}}\)

Exercise 2

2.

a. \(13_{\text{five}}r. 3_{\text{five}}\)

b. \(32_{\text{six}}r. 21_{\text{six}}\)

c. \(11_{\text{four}} r. 10_{\text{four}}\)

d. \(110_{\text{two}} r. 1_{\text{two}}\)

Exercise 3

3.

a. \(23_{\text{seven}}r. 3_{\text{seven}}\)

b. \(102_{\text{three}}\)

Exercise 4

4. \(12_{\text{three}}r. 1_{\text{three}}\)

Exercise 5

5. \(13_{\text{five}}r. 3_{\text{five}}\)

Exercise 6

6. \(11_{\text{four}}r. 10_{\text{four}}\)

Exercise 7

7. \(32_{\text{six}}r. 21_{\text{six}}\)

Exercise 8

8. \(110_{\text{two}}r. 1_{\text{two}}\)

Exercise 9

9. \(23_{\text{seven}}r. 3_{\text{seven}}\)

Exercise 10

10. \(102_{\text{three}}\)

Exercise 11

11. \(31_{\text{five}}r. 2_{\text{five}}\)

Exercise 12

12. \(111_{\text{two}}r. 101_{\text{two}}\)

Exercise 13

13. \(61_{\text{eight}}r. 6_{\text{eight}}\)

Exercise 14

14. \(T2_{\text{thirteen }}r. E_{\text{thirteen}}\)

Exercise 15

15. \(23_{\text{twelve}}r. 1E_{\text{twelve}}\)

Homework Solutions

Exercise 1

1.

a. \(72 \div 4\)

b. Repeated subtraction

c. Put 4 lollipops in a bag at a time by repeatedly subtracting 4 from 72 until you run out of lollipops.

d.

e. \(18 \cdots 4\)

f. Count how many subsets of 4 you made, how many bags of lollipops there are.

g. 18

Exercise 7

7.

a.

b.

c. Partitioning into subsets is easier, because you only need to make 2 equal subsets as opposed to subtracting 2 at a time.

Exercise 11

11.

a. the number of subsets

b. amount in each subset

Exercise 13

13. \(56_{\text{seven}}r. 22_{\text{seven}}\)

Exercise 15

15. \(285_{\text{eleven}}r. 167_{\text{eleven}}\)

Exercise 17

17. \(2111_{\text{three}}r. 210_{\text{three}}\)

Exercise 18

18.

a. Not closed; \(-1 \div -1 = 1\)

c. Not closed; \(1 \div 0\) is undefined

Exercise 19

19. No; \(10 \div 2 = 5\), but \(2 \div 10 = 1/2\). Therefore, \(10 \div 2 \nleq 2 \div 10\). Use a different counterexample in your answers.

Module 8 Number Theory

Exercise Set 1 Solutions

Exercise 1

1.

a. 7	b. 3
c. 8	d. 1
e. 2	f. 8
g. 2	h. 8
i. 0	j. 2
k. 4	l. 0
m. 7	n. 8

Exercise 2

2. same answers as for exercise 1

Exercise 3

3. Yes.

Exercise 4

4.

a. Ck: 6 + 7 = 13 → 4 \(4 \nleq 2\); mistake

b. Ck: 5 + 1 = 6 6 = 6; correct

c. Ck: 2 + 3 = 5 \(5 \nleq 4\); mistake

d. Ck: 8 + 2 + 5 = 15 → 6 6 = 6; correct

e. Ck: 5 + 2 + 2 = 9 → 0 \(0 \nleq 3\); mistake

f. Ck: 1 + 7 + 0 = 8 8 = 8; correct

Exercise 5

5.

a. \(2 \cdots 8 = 16 → 7\) 7 = 7; correct

b. \(2 \cdots 0 = 0\) \(0 \nleq 8\); mistake

c. \(5 \cdots 1 = 5\) 5 = 5; correct

d. \(2 ·\cdots 7 = 14 → 5\) \(5 \nleq 4\); mistake

e. \(0 \cdots 1 = 0\) 0 = 0; correct

f. 3 · 7 = 21 → 3 \(3 \nleq 4\); mistake

g. \(2 \cdots 7 = 14 → 5\) 5 = 5; correct

h. \(5 \cdots 2 = 10 → 1\) \(1 \nleq 0\); mistake

i. \(2 \cdots 6 = 12 → 3\) 3 = 3; correct

Exercise 6

6. Do the problems, then check your answers using digital roots.

Exercise 7

7.

a. 0 + 5 = 5 5 = 5; correct

b. 8 + 0 = 8 8 = 8; correct

c. 5 + 3 = 8 \(8 \nleq 6\); mistake

d. 8 + 6 = 14 → 5 \(5 \nleq 4\); mistake

e. 5 + 4 = 9 → 0 0 = 0; correct

f. 1 + 4 = 5 5 = 5; correct

Exercise 8

8. Do the problems, then check your answers using digital roots.

Exercise 9

9.

a. \(1 \cdots 4 = 4; 4 + 7 = 11 → 2\) 2 = 2; correct

b. \(6 \cdots 0 = 0; 0 + 0 = 0\) 0 = 0; correct

c. \(8 \cdots 8 = 64 → 10 → 1; 1 + 1 = 2\) \(2 \nleq 4\); mistake

d. \(6 \cdots 3 = 18 → 0; 0 + 5 = 5\) 5 = 5; correct

Exercise 10

10. Do the problem, then check your answer using digital roots.

Exercise 11

11.

a. This is a division problem. The answer is 5.

b. This is a false statement since 35 is not a factor of 7.

c. This is a true statement since 7 is a factor of 35; (\(7 \cdots 5 = 35\))

d. This is a division problem. The answer is 5 r. 5.

e. This is a false statement since 56 is not a factor of 8.

f. This is a false statement since 7 is not a factor of 40.

g. This is a true statement since 12 is a factor of 60; (\(12 \cdots 5 = 60\))

h. This is a division problem. The answer is 2 r. 20.

i. This is a division problem. The answer is 14.

j. This is a false statement since 42 is not a factor of 3.

k. This is a true statement since 6 is a factor of 42; (\(6 \cdots 7 = 42\))

l. This is a division problem. The answer is 8.

m. This is a division problem. The answer is 50.

n. This is a true statement since 4 is a factor of 100; (\(4 \cdots 25 = 100\))

o. This is a false statement since 4 is not a factor of 90.

p. This is a false statement since 25 is not a factor of 5.

Exercise 13

13. The proof is written exactly like Example 1 shown above this exercise if a is replaced by x, b is replaced by y, and c is replaced by z.

Exercise 18

18. If a is a factor of b, then am = b for some whole number, m. If a is a factor of c, then an = c for some whole number n. Using substitution, bc = (am)(an) = a(amn), which shows a is a factor of bc.

Exercise 19

19. The proof is written exactly like the example shown above this exercise if a is replaced by c, b is replaced by a, and c is replaced by b.

Exercise 20

20.

a. 2|9,712 since the last digit is even.

b. 5,643 is odd, so it is not divisible by 2.

c. 5,690 is not divisible by 4 since 4 is not a factor of 90.

d. 63,868 is divisible by 4 since 4 is a factor of 68.

e. 854,100 is divisible by 4 since 4 is a factor of 0.

f. 8 is a factor of 12,345,248 since 8 is a factor of 248.

g. 54,094,422 is not divisible by 8 since 8 is not a factor of 422.

Exercise 21

21.

a. 5|9,750 since the last is 0.

b. 5|5,645 since the last is 5.

c. 5,696 is not divisible by 5 since the last digit is not 0 or 5.

d. 10|63,860 since the last digit is 0.

e. 854,105 is not divisible by 10 since the last digit is not 0.

Exercise 22

22.

a. 3|9,750 since 3|3, where 3 is the digital root of 9,750.

b. 3|5,645 is false since 3 is not a factor of 2, which is the d.r. of 5,645.

c. 3|5,696 is false since 3 is not a factor of 8, the d.r. of 5,696.

d. 3|63,860 is false since 3 is not a factor of 5, the d.r. of 63,860.

e. 3|854,115 since 3|6, where 6 is the d.r. of 854,115.

Exercise 23

23.

a. 9|9,753 is false since the digital root of 9,753 is 6, not 0.

b. 9|5,646 is false since the d.r. of 5,646 is 3, not 0.

c. 9|5,697 since the digital root of 5,697 is 0.

d. 9|63,576 since the digital root of 63,576 is 0.

e. 9|854,103 is false since the d.r. of 854,103 is 3, not 0.

Exercise 24

24.

a. 6|9,753 is false since 9,753 is not even, therefore not divisible by 2.

b. 6|5,645 is false since 5,645 is not even, therefore not divisible by 2.

c. 6|5,696 is false since 3 is not a factor of 8, which is the d.r. of 5,696.

d. 6|63,876 since it is even and is also divisible by 3, since the d.r. is 3.

e. 6|854,103 is false since 854,103 is not even, therefore not divisible by 2.

Exercise 25

25.

a. 15|9,753 is false since it is not divisible by 5.

b. 15|6,645 since both 3 and 5 are factors.

c. 15|1,690 is false since 3 is not a factor.

d. 15|63,872 is false since 5 is not a factor.

e. 15|654,105 since it is divisible by both 3 and 5.

Exercise 26

26. Include justification

a. true

b. false

c. true

d. true

Exercise 27

27. Include justification

a. true

b. false

c. false

d. true

e. false

Exercise Set 2 Solutions

Exercise 1

1. c. 1, 2, 3, 4, 6, 12

Exercise 2

2.

a. 1, 2

b. 1, 3

c. 1, 2, 4

d. 1, 5

e. 1, 2, 3, 6

g. 1, 2, 4, 8

h. 1, 3, 9

i. 1, 2, 5, 10

j. 1, 11

k. 1, 13

l. 1, 2, 7, 14

m. 1, 3, 5, 15

n. 1, 2, 4, 8, 16

Exercise 3

3.

a. 2, 3, 5, 7, 11, 13

b. The only factors are 1 and the number itself.

c. 4, 9, 16

d. They are perfect squares.

Exercise 4

4.

a. \(45 = 3 \cdots 3 \cdots 5\)

b. \(65 = 5 \cdots 13\)

c. \(200 = 2 \cdots 2 \cdots 2 \cdots 5 \cdots 5\)

d. \(91 = 7 \cdots 13\)

e. \(76 = 2 \cdots 2 \cdots 19\)

f. \(350 = 2 \cdots 5 \cdots 5 \cdots 7\)

g. \(189 = 3 \cdots 3 \cdots 3 \cdots 7\)

h. \(74 = 2 \cdots 37\)

i. \(512 = 2 \cdots 2 \cdots 2 \cdots 2 \cdots 2 \cdots 2 \cdots 2 \cdots 2 \cdots 2\)

j. \(147 = 3 \cdots 7 \cdots 7\)

Exercise 5

5. 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

Exercise 6

6. 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48

Exercise 7

7. 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100, 102, 104, 105, 106, 108, 110, 111, 112, 114, 115, 116, 117, 118, 119, 120

Exercise 8

8. \(11 \cdots 47\)

Exercise 9

9. 22.7

Exercise 10

10.

a. 11; 149

b. 13; 3 · 7 · 13

c. 19; 3 · 127

d. 19; 19 · 23

e. 19; 509

f. 23; 613

g. 23; 787

Exercise 11

11.

a. \(2 \cdots 2 \cdots 2 \cdots 5 \cdots 7\)

b. 281

c. \(2 \cdots 3 \cdots 47\)

d. 283

e. \(2 \cdots 2 \cdots 71\)

f. \(5 \cdots 3 \cdots 19\)

g. \(2 \cdots 11 \cdots 13\)

h. \(7 \cdots 41\)

i. \(25 \cdots 3 \cdots 3\)

j. \(17 \cdots 17\)

k. \(2 \cdots 5 \cdots 29\)

l. \(3 \cdots 97\)

m. \(2 \cdots 2 \cdots 73\)

n. 293

o. \(2 \cdots 3 \cdots 7 \cdots 7\)

p. \(5 \cdots 59\)

Exercise 12

12. 281 and 283

Exercise 13

13. No; at least one will be a multiple of 3.

Exercise 14

14.

a. 1, 2, 3, 6, 7, 14, 42

b. 1, 2, 5, 7, 10, 14, 35, 70

c. 1, 2, 7, 14

d. 14

e. 14

Exercise 15

15.

a. 1, 2, 4, 23, 46, 92

b. 1, 5, 23, 115

c. 1, 23

d. 23

e. 23

Exercise 16

16.

a. 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

b. 1, 2, 3, 6, 9, 18, 27, 54

c. 1, 3, 7, 9, 21, 63

d. 1, 3

e. 3

f. 3

Exercise 17

17.

a. 14	b. 23	c. 3
d. 34	e. 25	f. 42

Exercise 18

18. a. \(24 \cdots 3 \cdots 132\)

Exercise 19

19.

a. \(2^{4} \cdots 3^{2} \cdots 7^{6} \cdots 13^{2}\)

b. \(3^{4} \cdots 5^{2}\)

c. \(a^{4} \cdots c^{3} \cdots d^{3} \cdots e^{2}\)

d. \(a^{2} \cdots d^{3}\)

Exercise 22

22. a. 2, 7

Exercise 23

23. 1

Exercise 24

24.

a. m	b. 2m	c. m
d. 1	e. m

Exercise 25

25. a - g: 1, 2, 3, 6

h. same factors

Exercise 26

26. a - f: 6

Exercise 27

27.

a. 13

b. 18

c. 36

Exercise 26

28.

a. 38 r. 186

b. 9 r. 33

c. 7 r. 11

d. 8 r. 25

Exercise 29

29.

a. 13

b. 18

c. 36

Exercise 30

30. d. 22

Exercise 31

31. d. 31

Exercise 32

32. d. 1

Exercise 34

34. 14|n if 2|n and 7|n.

Exercise 35

35.

a. True since 2|742 and 7|742

b. False –7 is not a factor of 968

c. False –2 is not a factor of 483

Exercise 36

36. Think about why YOU think it doesn't work. This is a question only you can answer.

Exercise 37

37. Think about why YOU think it works. This is a question only you can answer.

Exercise 38

38.

a. 4|n and 3|n

b. 9|n and 2|n

Exercise 39

39.

a. 35|n if 5|n and 7|n

b. 28|n if 4|n and 7|n

c. 75|n if 3|n and 25|n

d. 56|n if 7|n and 8|n

e. c|n if 4|n, 27|n, 25|n and 11|n

f. d|n if 16|n, 3|n, 5|n and 11|n

Exercise 40

40.

a - e. 1

f. No; 1 is the smallest factor of every number.

Exercise Set 3 Solutions

Exercise 1

1.

a. 8, 16, 24, 32, 40, 48, 56, 64, 72, 80

b. 12, 24, 36, 48, 60, 72, 84, 96, 108, 120

c. 24, 48, 72

d. 24

e. no

Exercise 2

2.

a. 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60

b. 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 66, 72, 78, 84, 90

c. 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150

d. 60

e. 60

Exercise 5

5.

a. X, Y

b. Y, Z

c. X, Y

Exercise 7

7. \(2^{2} \cdots 3^{5} \cdots 5 \cdots 7^{3} \cdots 11^{2} \cdots 1^{3}\)

Exercise 8

8. \(2^{2} \cdots 3^{4} \cdots 7^{3} \cdots 11^{2} \cdots 13^{2} \cdots 1^{9}\)

Exercise 9

9. \(2^{4} \cdots 3^{6} \cdots 5^{4} \cdots 7^{6} \cdots 11^{2} \cdots 1^{9} \cdots 23^{2}\)

Exercise 10

10.

a. GCF: \(2^{2} \cdots 3^{2} \cdots 1^{3}\) LCM: \(2^{2} \cdots 3^{5} \cdots 5 \cdots 7^{3} \cdots 1^{3}\)

b. GCF: 2 LCM: \(2^{2} \cdots 3^{4} \cdots 7^{3} \cdots 11^{2} \cdots 13^{2} \cdots 1^{9}\)

c. GCF: \(c^{3} \cdots d\) LCM: \(a^{5} \cdots b^{4} \cdots c^{5} \cdots d^{3} \cdots e^{2}9\).

d. GCF: \(c \cdots d\) LCM: \(a^{6} \cdots b^{4} \cdots c^{4} \cdots d^{3} \cdots e^{7}\)

Exercise 11

11.

a. 120

b. 1400

c. 1274

d. 840

e. 1870

Exercise 12

12.

a. 3 · 2 · 2 · 7 · 3 · 5 = 1260

b. 2·3·3·11·7·5·2·2·2·17 = 942,480

c. 5·2·3·7·2·2·17·3·5 = 214,200

d. 3 · 5 · 2 · 3 · 2 · 5 · 7 = 6,300

Exercise 13

13. 14,112

Exercise 14

14. 2,835

Exercise 15

15. 3,705

Exercise 16

16. 2 and 20; 4 and 10

Exercise 17

17.

a. 21

b. 18,018

Exercise 18

18.

a. 73

b. 74,095

Exercise 19

19.

a. 37

b. 395,641

Exercise 20

20.

a. 6	b. 58,344,300

Exercise 25

25.

a. 2(4n + 10) is an even number.

b. 2(5k + 4) + 1 is an odd number.

c. 5x + 2 can be even or odd

Exercise 26

26. Let 2n = one even number, and let 2m = another even number. The sum is: 2n+2m = 2(n+m), which is in the form of an even number. Therefore, the sum of 2 even numbers is even.

Exercise 27

27. Let 2n + 1 = one odd number, and let 2m + 1 = another odd number. The sum is: 2n + 1 + 2m + 1 = 2n + 2m + 2 = 2(n + m + 1), which is in the form of an even number. Therefore, the sum of 2 odd numbers is even

Exercise 28

28. Let 2n = one even number, and let 2m + 1 = one odd number. The sum is: 2n + 2m + 1 = 2n + 2m + 1 = 2(n + m) + 1, which is in the form of an odd number. Therefore, the sum of an even number and an odd number is odd.

Exercise 29

29. Let 2n = one even number, and let 2m = another even number. The product is: 2n · 2m = 4mn = 2(2mn), which is in the form of an even number. Therefore, the product of two even numbers is even.

Exercise 30

30. Let 2n + 1 = one odd number, and let 2m + 1 = one odd number. The product is: (2n + 1)(2m + ) = 2nm + 2m + 2n + 1 = 2(nm + m + n) + 1, which is in the form of an odd number. Therefore, the product of two odd numbers is odd.

Exercise 31

31. Let 2n = one even number, and let 2m + 1 = one odd number. The product is: 2n(2m + 1) = 4nm + 2n = 2(2nm + n), which is in the form of an even number. Therefore, the product of an even number and an odd number is even.

Exercise 32

32. 3240

Exercise 33

33. 3240

Exercise 34

34. 31,375

Exercise 35

35.

a. 3,77

b.148,181

c.195,650

Exercise 36

36.

a. 3,775

b. 148,181

c. 195,650

d.[n –(k –1)]\((n + k) \div 2\) or (n –k + 1)\((n + k) \div 2\)

Exercise 37

37. multiples of 7

Exercise 38

38. 1 + 2 + 3 + . . . + 99 + 100

Exercise 39

39.

a. 5,050

b. 5,050

c. 35,350

Exercise 40

40.

a. 63,000

b. 41,151

c. 226,500

Exercise 41

41. multiples of 4

Exercise 42

42. 28 + 29 + 30 + . . . + 131 + 132

Exercise 43

43. 33,600

Exercise 44

44.

a. 54,450

b. 96,019

Exercise 45

45. 6

Exercise 46

46. 4, 8, 14

Exercise 47

47. 4, 6

Exercise 48

48. 4

Exercise 49

49. 8

Exercise 50

50. 3, 7

Exercise 51

51. 7

Exercise 52

52. 68, 178, 288, 398, 508, 618, 728, 838, 948

Exercise 53

53. 5

Exercise 54

54. 5

Exercise 55

55. 5 or 26

Exercise 56

56. mult. of 7

Exercise 57

57. 9 or 23

Exercise 58

58. 3, 5, or 9

Exercise 59

59. 9

Exercise 60

60. Yes; explanation not provided here

Homework Solutions

Exercise 1

1. GCF:14; LCM: 6,300

Exercise 3

3. a. GCF: 47; LCM: 49,350

Exercise 4

4. a. Since the number is even, we only need to make the resulting number divisible by 3. The digital root so far is 8. So, decide which single digits could be added to 8 to get a number that is still divisible by 3. Adding 1, 4 or 7 to 8 will work. So the possibilities are 4, 7 or 8.

Exercise 5

5. The examples and counterexamples are not provided by these solutions. Make sure you provide them!

a. false

b. true

Exercise 6

6.

a. \(7 \cdots 53\)

d. \(7 \cdots 41\)

Exercise 7

7.

a. m

b. m

c. m

d. 1

Exercise 9

9.

a. 49,770

b. 35,800

Exercise 10

10.

a. 39

b. 80

Exercise 11

11.

a. 1, 2, 3; 6; perfect

b. 1; 1; deficient

c. 1, 2, 4; 7; deficient

Exercise 13

13. Examples and counterexamples are not provided here. Make sure you provide them.

a. False

b. True

c. False

d. True

Exercise 15

15. a. 84

Module 9 Rational Numbers

Warm Up Solutions

Exercise 1

1. R

Exercise 2

2. Y

Exercise 3

3. P

Exercise 4

4. D

Exercise 5

5. O

Exercise 6

6. D

Exercise 7

7. H

Exercise 8

8. B

Exercise 9

9. D

Exercise 10

10. O

Exercise 11

11. Y

Exercise 12

12. B

Exercise 13

13. B

Exercise 14

14. L

Exercise Set 1 Solutions

Exercise 1

1. \(\frac{1}{12} < \frac{1}{10} < \frac{1}{9} < \frac{1}{8} < \frac{1}{6} < \frac{1}{5} < \frac{1}{4} < \frac{1}{3} < \frac{1}{2}\)

Exercise 2

2.

a. \(\frac{1}{90}\)

b. \(\frac{1}{32}\)

Exercise 3

3. \(\frac{5}{12} < \frac{5}{10} < \frac{5}{9} < \frac{5}{8} < \frac{5}{6}\)

Exercise 4

4.

a. \(\frac{15}{37}\)

b. \(\frac{89}{100}\)

Exercise 5

5. \(\frac{4}{12} < \frac{4}{11} < \frac{4}{10} < \frac{4}{9} < \frac{4}{8} < \frac{4}{7} < \frac{4}{6} < \frac{4}{5}\)

Exercise 6

6. If you are comparing two fractions that have the same numerator, the one with the smaller denominator has the larger value.

Exercise 7, 8

7, 8. \(\frac{1}{2} < \frac{2}{3} < \frac{3}{4} < \frac{4}{5} < \frac{5}{6} < \frac{7}{8} < \frac{8}{9} < \frac{9}{10} < \frac{11}{12}\)

Exercise 9

9.

a. \(\frac{94}{95}\)

b. \(\frac{89}{100}\)

Exercise 10

10. \(\frac{13}{14} < \frac{25}{26} < \frac{34}{35} <\frac{45}{46} < \frac{51}{52} < \frac{71}{72} < \frac{99}{100}\)

Exercise 11

11. If you are comparing two fractions where in each fraction the numerator is one less than the denominator, the fraction with the bigger numerator and denominator has the larger value.

Exercise 12

12. Answers will vary. Try drawing a pie for each fraction, shading in 2 of 3 equal parts for 2/3 and 4 of 5 equal parts for 4/5. Which pie is more shaded?

Exercise 13

13. Answers will vary

Exercise 14

14. \(\frac{2}{4}\) and \(\frac{3}{6}\) and \(\frac{4}{8}\) and \(\frac{5}{10}\) and \(\frac{6}{12}\)

Exercise 15

15. \(\frac{4}{6}\) and \(\frac{6}{9}\) and \(\frac{8}{12}\)

Exercise 16-17

16 - 17: There are many possibilities using multiple strips.

Exercise 18

18. Multiply the numerator and denominator by the same number to get an equivalent fraction.

Exercise 19-27

19 - 27. Only the final answers are shown below. Make your own models for exercises 22 - 28 by following the examples making sure you show all the steps, define units, rows, columns, etc.

Exercise 19

19. \(\frac{3}{4} < \frac{4}{5}\)

Exercise 20

20. \(\frac{13}{20}\)

Exercise 21

21. \(\frac{5}{24}\)

Exercise 22

22.

a. \(\frac{8}{15}\)

b. \(\frac{15}{24}\)

Exercise 23

23.

a. \(\frac{2}{3} \cdots \frac{5}{6} = \frac{10}{18}\)

b. \(\frac{1}{2} \cdots \frac{7}{8} = \frac{7}{16}\)

Exercise 24

24.

a. \(\frac{4}{5} \cdots \frac{3}{5} = \frac{12}{25}\)

b. \(\frac{1}{7} \cdots \frac{3}{4} = \frac{3}{28}\)

Exercise 25

25. Answers may vary. One possibility: Let 1 = H; \(\frac{3}{4}\) of H = B; \(\frac{1}{3}\) of B = L; \(\frac{L}{H} = \frac{3}{12}\)

Exercise 26`

26.

a. H; B; L; L/H; 3/12

b. N; R; W; W/N; 1/8

c. N; R; L; L/N; 3/8

d. D; L; R; R/D; 2/6

Exercise 27-30

27 - 30. Make your own models. One possibility for each using the C-strip model is given, but there are other choices.

Exercise 27

27. Let 1 unit = O, then 1/10 = W and 1/5 = R. There are 2 W in R, so the answer is 2.

Exercise 28

28. Let 1 unit = B, then 1/9 = W and 1/3 = L. There are 3 W in L, so the answer is 3.

Exercise 29

29. Let 1 unit = H, then 1/6 = R and 2/3 = N. There are 4 R in N, so the answer is 4.

Exercise 30

30. Let 1 unit = P, then 1/4 = W and 3 = H. There are 12 W in H, so the answer is 12.

Exercise 31

31.

a. equivalent

b. equivalent

c. not equivalent

Exercise 32

32. There are infinitely many possibilities - you can use the multiple strips to find some.

Exercise 33

33. same as #31

Exercise 34

34.

a. \(\frac{14}{25}\)

b. \(\frac{13}{17}\)

c. \(\frac{18}{35}\)

d. \(\frac{15}{22}\)

Exercise 35

35. same as #31

Exercise 36

36.

a. \(\frac{4}{5} > \frac{5}{8}\)

b. \(\frac{12}{35} < \frac{11}{18}\)

c. \(\frac{13}{15} > \frac{14}{17}\)

Exercise 37

37. \(89 \cdots 95 < 90 \cdots 94\) since 8455 < 8460. Cross products check.

Exercise 38

38. It checks, show work

Exercise 39

39.

a. -4/70

b. 71/72

Exercise 40

40. Answers may vary. Some possibilities: 21/50, 22/50, 23/50, . . . , 29/50

Exercise 41, 42

41,42. 1/2

Exercise 43

43. Some possibilities: 51/150, 52/150, 53/150, . . . , 59/150

Exercise 44

44. Some possibilities: 21/60, 22/60, 23/60, . . . , 29/60

Exercise 45

45. 241/300, 242/300, 243/300, . . . , 249/300

Exercise 46

46. 83/112

Exercise 47

47. 75/112

Exercise 48

48. Draw 12 circles. 11 of the circles add up to 66, so put 6 in each circle. Answer: 6 students.

Exercise 49

49. Draw 7 circles, Since 3 circles add up to 36, put 12 in each circle. The other 4 circles represent students not buying lunch. Answer: 48 students

Exercise 50

50. Draw 8 circles. Since all 8 circles add up to 120, put 15 in each circle. 5 of the circles represent the females, the other three represent the males. Answer: 75 females and 45 males.

Exercise Set 2 Solutions

Exercise 1

1.

a. four tenths; \(\frac{4}{10} = \frac{2}{5}\)

b. twenty-six hundredths; \(\frac{26}{100} = \frac{13}{50}\)

c. three and eight hundredths; \(3\frac{8}{100} = 3\frac{2}{25}\); \(\frac{388}{100} = \frac{77}{25}\)

d. nine and eighty-five hundredths; \(9\frac{85}{100} = 9\frac{17}{20}\); \(\frac{985}{100} = \frac{197}{20}\)

e. seventeen and three hundred five thousandths; \(17\frac{305}{1000}= 17\frac{61}{200}\); \(\frac{17305}{1000} = \frac{3461}{200}\)

Exercise 2

2.

a. 0.14	b. 008
c. 4\(\frac{35}{100} = 4.35\)	d. 563\(\frac{8}{10} = 563.8\)
e. 3\(\frac{5}{100} = 3.05\)

Exercise 3

3.

a. 0.028	b. 0.32
c. 0.075	d. 0.66

Exercise 4

4.

a. 1.9; 1.90000

b. 4.0340

Exercise 5

5.

a. 3.5 > .9

b. 35.06 = 3.0600

c. 0.089 < 0.0908

Exercise 6

6.

a. 3.51 > 3.488

b. 35.061 < 35.35

c. 0.8933 < 0.0894

Exercise 7

7.

a. \(2 \cdots 5\)	b. \(2 \cdots 5 \cdots 2 \cdots 5\)
c. \(2 \cdots 5 \cdots 2 \cdots 5 \cdots 2 \cdots 5\)	d. \(2 \cdots 5 \cdots 2 \cdots 5 \cdots 2 \cdots 5 \cdots 2 \cdots 5\)
e. \(2 \cdots 5 \cdots 2 \cdots 5 \cdots 2 \cdots 5 \cdots 2 \cdots 5 \cdots 2 \cdots 5\)

Exercise 8

8. 2 and 5

Exercise 9

9. 3

Exercise 10

10. 2

Exercise 11

11. Well, what do you think?

Exercise 12

12.

a. 0.0875

b. 0.875

Exercise 13

13. Write a response in your own words.

Exercise 14

14.

a. \(\frac{3}{4} = \frac{3}{2 \cdots 2} \cdots \frac{5 \cdots 5}{5 \cdots 5} = \frac{75}{100} = 0.75\)

b. \(\frac{9}{20} = \frac{3 \cdots 3}{2 \cdots 2 \cdots 5} \cdots \frac{5}{5} = \frac{45}{100} = 0.45\)

c. \(\frac{9}{15} = \frac{3 \cdots 3}{3 \cdots 5} = \frac{3}{5} \cdots \frac{2}{2} = \frac{6}{10} = 0.6\)

d. \(\frac{18}{25} = \frac{2 \cdots 3 \cdots 3}{5 \cdots 5} \cdots \frac{2 \cdots 2}{2 \cdots 2} = \frac{72}{100} = 0.72}\)

e. \(\frac{5}{14} = \frac{5}{2 \cdots 7}\) This can't be written as a terminating decimal because the reduced factor has a prime factor of 7 (which is other than a 2 or 5) in the denominator.

Exercise 15

15. Do three of your own.

Exercise 16

16.

a. 0, 1, 2, 3, 4, 5

b. 0, 1, 2, 3, 4, 5, 6

c. 0, 1, 2, 3, 4, 5, 6, 7, 8

d. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

e. 0, 1, 2

Exercise 17

17. Do three of your own.

Exercise 18

18. a, b, d

Exercise 19

19.

a. There are 33 (including 0 as a remainder). A maximum of 32 digits could be in a sequence without repeating.

b. If you do the long division, you get .939393...

c. 2

d. 13 and 31

Exercise 20

20.

a. 0.\(\bar{5}\) or 0.555...

b. 0.\(\bar{714285}\) or 0.714285714285...

c. 0.1\(\bar{6}\) or 0.1666...

d. 0.\(\bar{6}\) or 0.666..

e. 0.\(\bar{63}\) or 0.636363...

f. 0.41\(\bar{6}\) or 0.41666..

g. 0.5\(\bar{3}\) or 0.5333...

h. 0.3\(\bar{5}\) or 0.3555...

i. 0.0\(\bar{75}\) or 0.0757575...

Exercise 21

21. x - 1

Exercise 22

22.

10x = 7.272727...;

100x = 72.727272...;

1000x = 727.272727...

Exercise 23

23.

a. 99x

b. 72

c. 99x = 72; x = \(\frac{72}{99}\)

d. x = \(\frac{8}{11}\)

Exercise 24

24.

a. \(\frac{4}{10} =\frac{2}{5}\)	b. \(\bar{4} = \frac{4}{9}\)
c. \(\frac{6}{100} = \frac{3}{50}\)	d. \(\bar{0.06} = \frac{6}{99} = \frac{2}{33}\)
e. \(\frac{9}{10}\)	f. \(\bar{0.9} = \frac{9}{9} = 1\) Surprise!
g. \(\frac{45}{100} = \frac{9}{20}\)	h. 0.\(\bar{45} = \frac{45}{99} = \frac{5}{11}\)
i. \(\frac{84}{1000} = \frac{21}{250}\)	j. 0.\(\bar{084} = \frac{84}{999} = \frac{28}{333}\)

Exercise 25

25.

a. \(\frac{2.8}{99} = \frac{28}{99 \cdots 10} = \frac{14}{495}\)

b. \(\frac{2.6}{9} = \frac{26}{9 \cdots 10} = \frac{13}{45}\)

c. \(\frac{0.06}{9} = \frac{6}{9 \cdots 100} = \frac{1}{150}\)

d. \(\frac{101}{999}\)

e. \(\frac{3.6}{9} = \frac{36}{9 \cdots 10} = \frac{2}{5}\)

Exercise 26

26. Be creative!

Exercise 27

27. No, since 9 is a perfect square. So, \(\sqrt{9} = 3\)

Exercise 28-30

28-30. Be creative!

Exercise 31

31.

a. rational (repeating decimal).

b. irrational (it has a pattern, but does not repeat!)

c. rational

d. irrational (since 80 is not a perfect square)

e. rational (since 100 is a perfect square; so \(\sqrt{100} = 10\)

f. irrational

Homework Solutions

Exercise 1

1.

a. K

c. D

e. D

g. H

Exercise 2

2.

a. H

c. P

e. R

Exercise 3

3. a. L

Exercise 7

7. \(\frac{1}{3} < \frac{3}{8}\)

Exercise 8

8. \(\frac{17}{24}\)

Exercise 9

9. \(\frac{7}{20}\)

Exercise 10

10. a. \(\frac{3}{20}\)

Exercise 11

11. a. \(\frac{5}{6} \cdots \frac{2}{3} = \frac{10}{18}\)

Exercise 12

12. a. \(\frac{2}{3} \cdots \frac{3}{8} = \frac{6}{24}\)

Exercise 13

13. a. B; D; R; R/B; 2/9

Exercise 14

14. a. 15

Exercise 16

16. a. \(\frac{18}{35}\)

Exercise 18

18. Some possibilities: 31/80, 32/80, 33/80, . . . , 49/80

Exercise 20

20. a. Draw 10 circles. Since 7 of the circles add up to 21, put 3 in each circle. All 10 circles add up to 30, so there are 30 students.

Exercise 21

21.

a. seven tenths

b. three and twenty-eight hundredths

Exercise 22

22.

a. \(\frac{3}{10} \cdots \frac{8}{10} = \frac{14}{100} = 0.14\)

b. \(\frac{122}{100} \cdots \frac{23}{10} = \frac{2086}{1000} = 2.806\)

Exercise 23

23.

a. \(\frac{11}{16} = \frac{11}{2 \cdots 2 \cdots 2 \cdots 2} \cdots \frac{5 \cdots 5 \cdots 5 \cdots 5}{5 \cdots 5 \cdots 5 \cdots 5} = \frac{6875}{10000} = 0.6875\)

c. \(\frac{1}{12} = \frac{1}{2 \cdots 2 \cdots 3}\)

This can't be written as a terminating decimal because the reduced factor has a prime factor of 3 (which is other than a 2 or 5. in the denominator).

Exercise 24

24.

a. \(\frac{7}{9}\)

c. \(\frac{235}{999}\)