Solutions for third week’s assignments

Also available as PDF.

According to the percentages, Primestar has 16% of 12 million or 1.92 million. C-Band then has 15% or 1.8 million. Primestar has 120 thousand more subscribers.

However, the slices appear of drastically different sizes. I suspect the satellite dish is tilted “upwards” like a real dish, distorting the slices’ areas.

- C
- G
- E
- A
- None of the above! They meant B, but 1 = {2}^{0} is a positive integer and a power of two. The authors meant “two raised to the power of each of the five least positive integers”. I hadn’t realized this at first, or else I would not have given this one.
- D
- H
- F

11. \{0,1,2,3,4\}

17. \{2,4,8,16,32,64,128,256\}

30. \{x|x\text{ is an even natural number}\} is a direct translation,
but \{2x|x {J}^{+}\} is shorter.
Another possibility is \{x|x > 0,x\text{ is an even integer}\}.

32. One form is \{35 + 5i|i J,0 ≤ i ≤ 12\}.

62. - 12\mathrel{∉}\{3,8,12,18\}.

63. 0 \{- 2,0,5,9\}.

66. \{6\}\mathrel{∉}\{3,4,5,6,7\}. But
note that \{6\} ⊂\{3,4,5,6,7\}.

68. false

71. true

74. true

78. true (assuming a typical meaning for
“\mathop{\mathop{…}}”)

Part a. Three chocolate bars are contain a total of 660 calories. The point of this exercise is to ensure you recognize that sets are unordered, so \{r,s\} = \{s,r\} and you only include it once. The list is as follows: \{r\}, \{r,s\}, \{r,c\}, \{r,g\}, \{r,v\}, \{s,c\}, and \{s,g\}.

Part b. Five bars is 1100 calories. The list is \{r,s,v\}, \{r,s,g\}, \{r,s,c\}, \{r,c,v\}, \{r,c,g\}, and \{r,g,v\}.

8. \{M,W,F\}⊄\{S,M,T,W,Th\}.

10. \{\text{a},\text{n},\text{d}\} ⊂\{\text{r},\text{a},\text{n},\text{d},\text{y}\}.

12. ∅⊂∅.

14. \{2,1∕3,5∕9\} ⊂ ℚ.

24. true

26. false

28. false

30. true

32. false

34. false

- B
- F
- A
- C
- E
- D

10. Y ∩ Z = \{b,c\}.

17. X ∪ (Y ∩ Z) = \{a,b,c,e,g\}.

18. Y ∩ (X ∪ Z) = \{a,b,c\} = Y
because Y ⊂ X ∪ Z.

23. X \ Y = \{e,g\}.

24. Y \ X = \{b\}.

The set consisting of all the elements of A along with those elements of C that are not in B.

The set consisting of elements in A but not in C as well as elements in B but not in C. If we consider the union of A, B, and C to be the universal set, then this is the set of all elements in A complement along with all elements in B complement.

61. X ∪∅ = X,
and the conjecture is that the union of any set with the empty set is the set
itself.

62. X ∩∅ = ∅,
and the conjecture is that the intersection of any set with the empty set is the empty
set.

72. A × B = \{(3,6),(3,8),(6,6),(6,8),(9,6),(9,8),(12,6),(12,8)\}

B × A = \{(6,3),(8,3),(6,6),(8,6),(6,9),(8,9),(6,12),(8,12)\}

73. A × B = \{(d,p),(d,i),(d,g),(o,p),(o,i),(o,g),(g,p),(g,i),(g,g)\}.

B × A = \{(p,d),(i,d),(g,d),(p,o),(i,o),(g,o),(p,g),(i,g),(g,g)\}, alas,
no pigdog in sight.

117. A \ B = A implies
that A ∩ B = ∅.

118. B \ A = A is true
only when B = A = ∅.

121. If A ∪∅ = ∅,
then A = ∅.

122. A ∩∅ = ∅ for
any set A.

123. If A ∩∅ = A,
then A = ∅.

124. A ∪∅ = A for
all sets A.