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Contemporary Mathematics

Chapter 1 Sets

Solutions 1.1 Basic Set Concepts YOUR TURN 1.1 1. Answers may vary. One possible solution: T = {wrench, screwdriver, hammer, plyers} 1.3 1.

∅ or { }. No numbers are divisible by 0.

1.5 1. M = {1,3,5,…} 1.7 1. I = i i is a m𝖚sical instr𝖚ment} Since listing out every musical instrument is a tedious, and perhaps difficult task, the solution should have this form (set builder). 1.9 1. We can count up the elements of B. Thus, B is finite. 2. We cannot count up the number of elements in the real numbers and ever finish. Thus, ℝ is infinite.

CHECK YOUR UNDERSTANDING 1. set 3. This is not a well-defined set. Whether a restaurant is in the top five is a matter of opinion, not a fact. 5. n(A) = 12 and n(B) = 12. However, apples and donuts are different. Thus, A is equivalent to B but they are not equal. We can write this as: A ~ B, but A ≠ B. 7. Roster method: {A,B,C,…,Z} Set builder notation: {x|x is an 𝖚pper − case letter of the English alphabet}

EXERCISES 1. Let P represent the set. Then, P = {red, yellow, bl𝖚e} 3. Let A represent the set. Then, A = {50,51,52,…,100}. 5. Let C represent the set. Then, C = {king, q𝖚een, rook, knight, bishop, pawn} 7. Let L represent the set. Then, L = {l|l is a lizard} 9. Let M represent the set. Then, M = {3n|n is a member of ℕ} 11. Let P represent the set. Then, P = {p|p is an edible plant} 13. ∅ , since no squares are circles. 15. Let C represent the set. Then, C = {Greg, Peter, Bobby, Marsha, Jan, Cindy} For more free, peer-reviewed, openly licensed resources visit OpenStax.org. 3/20/23

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Contemporary Mathematics 17. ∅ , since no polar bears live in Antarctica. 19. Let B represent the set. B = {b|b is a children's book written and ill𝖚strated by Mo Willems} (You could write out all the book titles and use the roster method, but this would be a very long list. Try to choose the shortest method for writing a set that still makes the set well-defined.) 21. Because we can list out all of the character names in the book, this is a well-defined set. 23. The size of the group, the definition of “old”, and the definition of “new tricks” is not given and are all subject to opinion. This set is not well-defined. 25. Since zebras cannot fly an airplane, this is the empty set. Thus, this is a well-defined set. 27. n(P) = 6 29. n( ∅ ) = 0 31. n(F) = 9 33. n(C) = n(ℕ) = ℵ0 35. n(L) = 14 37. n(A) = n( right, ac𝖚te, obt𝖚se ) = 3; n(B) = n( eq𝖚ilateral, scalene, isoceles ) = 3 Since both sets contain 3 elements, the two sets are equivalent. But the elements of the sets are not the same, so the two sets are not equal. Thus, A ~ B 39. n(A) = n red, orange, yellow = 3; n(B) = n( green, bl𝖚e, indigo, violet ) = 4. Since they do not contain the same number of elements, they are neither equivalent nor equal. 41. n(A) = n({ − 2, − 1,0,…}) = ℵ0; n(B) = n({2,3,5,…}) = ℵ0. The two sets do not contain the same elements since A contains 0 (for example) and B does not. However, they both have the same number of elements. Thus, A ~ B. 43. A = ∅ ; B = { }. Since these are both the empty set, both sets contain 0 elements. Thus A = B. 45. The set of natural numbers is an infinite set. 47. We could count or list out all the jazz venues in New Orleans, Louisiana, and we would finish. Thus, this set is finite. 49. We could count or list out all of the different types of cheeses and we would finish. Thus, this set is finite.

1.2 Subsets YOUR TURN 1.11 1. {heads, tails}; {heads}, {tails}; and ∅ 1.13 1. E is a subset of ℕ, so E ⊂ ℕ. 1.15 1. Multiples of 5 can be written as 5n. So, {5, 10, 15, …} = {m|m = 5n where n ∈ ℕ}.

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Contemporary Mathematics 1.17 1. There are multiple possible solutions. Each set must contain two players, but both players cannot be the same, otherwise the two sets would be equal, not equivalent. For example, {Maria, Shantelle} and {Angie, Maria}.

CHECK YOUR UNDERSTANDING 9. To be a subset of a set every member of the subset must also be a member of the set. To be a proper subset there must be at least one member of the set that is not also in the subset. 11. True. A ⊆ A means A = A or A ⊂ A. 13. equivalent

EXERCISES 1.

chocolate, vanilla chocolate, strawberry , vanilla, strawberry , chocolate , vanilla , strawberry , ∅

3. {mother, father, da𝖚ghter}, {mother, father, son}, {mother, da𝖚ghter, son}, {father, da𝖚ghter, son}, {mother, father}, {mother, da𝖚ghter}, {mother, son}, {father, da𝖚ghter}, {father, son}, {da𝖚ghter, son}, {mother}, {father}, {da𝖚ghter}, {son}, ∅ 5. A is a proper subset of D, A ⊂ D. 7. C is a proper subset of D, C ⊂ D. 9. The empty set is a proper subset of Z, ∅ ⊂ Z. 11. A and C contain the exact same elements and hence, A is equal to C. So, there are three ways to describe this symbolically: A = C, A ⊆ C, and C ⊆ A. 13. B and C have the same number of elements, but the elements are different. B is equivalent to set C, B ∼ C. 15. n({Adele, Beyonce, Cher, Madonna, Shakira}) = 5. The number of subsets is 25 = 32. 17. n {Peter, Pa𝖚l, Mary} = 3. The number of subsets is 23 = 8. 19. n({3}) = 1. The number of subsets is 21 = 2. 21. n({ }) = 0. The number of subsets is 20 = 1. 23. If n A = 12, then the number of subsets is 212 = 4096. 25. Answers will vary, but it must contain 4 letters and must leave out at least one of the letters l, a, s, t. For example: a,r,t,s . 27. Answers will vary, but only the letters a, r, t will be in the set. For example: Answer: t,a,r . 29. Answers will vary, but it must contain 5 letters and must leave out at least one of the letters r, a, t, e, s. For example: l,a,r,g,e . 31. Answers will vary, however, each subset will have three letters in it and they cannot contain all the same letters. For example: s,e,t and r,a,g .

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Contemporary Mathematics 33. Answers will vary, however, each subset will have five letters in it and they must contain all the same letters. For example: g,l,a,r,e and l,a,r,g,e . 35. Answers will vary. For example: { − 7, − 6, − 5, − 4, − 3, − 2, − 1} and {1,2,3,4,5,6,7}. 37. Answers will vary. For example: {0, − 3, − 6, − 9,…}. 39. True, even though the set of natural numbers is a subset of set U they both have the same cardinality, ℵ0, because both sets are countably infinite.

1.3 Understanding Venn Diagrams YOUR TURN 1.19

1. The set of eagles and the set of canaries are two disjoint subsets of the universal set of all birds. No eagle is a canary, and no canary is an eagle.

1.21

CHECK YOUR UNDERSTANDING 15. relationship 17. disjoint or non-overlapping 19. disjoint

EXERCISES 1. Symbols: T ⊂ U Words: The set of team sports is a subset of the set of all sports. 3. Let P represent the set of pencils and let U represent the set of all writing utensils. Symbols: P ⊂ U Words: The set of pencils is a subset of the set of all writing utensils. 5. Symbols: C ⊂ U and P ⊂ U Words: The set of crayons and the set of pencils are disjoint subsets of the set of writing utensils. 7. Symbols: Card Games ⊂ Games and Video Games ⊂ Games Words: The set of card games and the set of video games are disjoint subsets of the set of all games. 9. All birds have wings. This means that the universal set could be “Things with wings”, then “birds” is a subset.

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Contemporary Mathematics

11. “All almonds are nuts, and all pecans are nuts”, so the universal set will be “Nuts”. Both of these are subsets, but the statement “but no almonds are pecans” means that the two circles do not touch each other.

13. Lizards ⊂ Reptiles. This means that “Reptiles” is the universal set and “Lizards” is a subset.

15. Ladybugs ⊂ Insects and Ants ⊂ Insects, so the universal set can logically be “Insects”. Both are subsets, but the statement “no Ants are Ladybugs” means that the two circles do not touch each other.

17. Since A and B are disjoint subsets of U, the circles for A and B cannot touch.

19. T will be a circle inside the rectangle labeled U.

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Contemporary Mathematics

21. “J ⊂ M” means that M will be the universal set and J will be a subset inside it.

23. M is the universal set with J and M represented by circles inside it. “R is disjoint from J” means that the two circles will not touch.

25. The complement is the set that contains all the rest of the elements in U that are not in A. Thus, A' = 0,1,2,3,4,5,9 27. Since A is the empty set, all of the elements in U that are not in A means that A' = U = {0,1,2,3,4,5,6,7,8,9} 29. Since A contains all of the elements of U, there are no elements left to list. Thus, A' = ∅ . 31. The complement is the set that contains all the rest of the elements in U that are not in A. Thus, A' = {0} 33. The complement is the set that contains all the rest of the elements in U that are not in A. Thus, A' = {Doc, Dopey, Sleepy, Sneezy} 35. The complement is the set that contains all the rest of the elements in U that are not in A. Thus, A' = {Bashful, Dopey, Grumpy, Happy, Sleepy, Sneezy} 37. Since A is the empty set, all of the elements in U that are not in A means that A' = U = {Bashful, Doc, Dopey, Grumpy, Happy, Sleepy, Sneezy} 39. The complement is the set that contains all the rest of the elements in U that are not in A. Thus, A' = {6,7,8,…} 41. The complement is the set that contains all the rest of the elements in U that are not in A. Thus, A' = {2,3,4,…} 43. Since A = {2, 3, 4, 9}, then A' = {0,1,5,6,7,8}

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Contemporary Mathematics 45. Since A={n,e,t}, then A' = {l,i,s}

1.4 Set Operations with Two Sets YOUR TURN 1.23

1. The only element in common is a. Thus, A ∩ B = {a}.

1.25

1. The intersection is the set of vowels. Thus, A ∩ B = B = {a, e, i, o, 𝖚}.

1.27

1. A ∪ B = {red, yellow, bl𝖚e, orange, green, p𝖚rple}.

1.29

1. n A ∪ B = n A + n B − n A ∪ B = 23 + 17 − 7 = 33

1.31

1. A or B = A ∪ B = {h, a, p, y, w, e, s, o, m}.

2. A and C = A ∩ C = {a, h}. 3. B or C = B ∪ C = {a, w, e, s, o, m, t, h}. 4. (A and C) and B = (A ∩ C) ∩ B = {a, h} ∩ {a, w, e, s, o, m} = {a}. 1.33.1. A ∩ B = {3,5,7}. 2. A ∪ B = {1,2,3,5,7,9}. 3. A ∩ B' = {1,9}. 4. n(A ∩ B') = 2.

CHECK YOUR UNDERSTANDING 21. union 23. A ∩ B 25. B (This is because the union is the set that contains all elements that lie in either of the two sets or both sets.) 27. n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

EXERCISES 1. B ∪ C= {4,8,12,16,20,24,32,40} 3. D ∩ C = {40} 5. A ∩ (C ∪ D) = A ∩ {8,10,16,20,24,30,32,40,50} = {8,10} 7. D ∪ A ∩ C = D ∪ {8} = {8,10,20,30,40,50} 9. B ∩ A ∩ D = B ∩ {10} =∅ For more free, peer-reviewed, openly licensed resources visit OpenStax.org. 3/20/23

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Contemporary Mathematics 11. B ∪ A ∪ D = B ∪ {2,4,6,8,10,12,20,30,40,50} = {2,4,6,8,10,12,16,20,30,40,50} 13. S and P = S ∩ P = {p,l} 15. P or M = P ∪ M = {p,l,o,t,m,a}. 17. L and M = L ∩ M = M = {m,a,p} 19. D or M or P = D ∪ M ∪ P = {a,d,g,l,m,o,p,t} 21. (S or D) and P = (S ∪ D) ∩ P = {s,a,m,p,l,e,d,o,g} ∩ P = {l,o,p} 23. U or P and S = U ∪ (P ∩ S) =U = {a,b,c,…,z} 25. A ∪ B = b,l,a,c,k,r,e,d 27. (A ∩ B)' = ∅ ' =U = a,b,c,…,z 29. B' = {a,b,c,f,g,h,i,j,k,l,m,n,o,p,q,s,t,𝖚,v,w,x,y,z} A ∩ B' = A = {b,l,a,c,k} 31. A ∩ B is the set of elements in the overlapping region for A and B: A ∩ B = {a,c,e} 33. (A ∪ B)'is the set that contains all elements inside the rectangle, but not in A ∪ B = {p,l,a,c,e,b,r,s}. So, (A ∪ B)' = {d,f,g,h,i,j,k,m,n,o,q,t,𝖚,v,w,x,y,z} 35. A' = {b,d,f,g,h,i,j,k,m,n,o,q,...,z} B ∩ A' = B = {b,r,s} 37. A ∪ B = B = {0,1,2,3,...} 39. A ∪ B = B (A ∪ B)' = B' = {…, –3, –2, –1} 41. A' = {..., − 2, − 1,0} B ∩ A' = {0} 43. n(A) = 3, n(B) = 3, n(A ∩ B) = 2 n(A ∪ B) = n(A) + n(B) − n(A ∩ B) =3+3−2 =4

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Contemporary Mathematics 45. n(A) = 4, n(B) = 3, n(A ∩ B) = 0 n(A ∪ B) = n(A) + n(B) − n(A ∩ B) =4+3−0 =7 47. Using the formula: n(A) = 9, n(B) = 15, n(A ∩ B) = 7, then n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 9 + 15 − 7 = 17 Or, we can add up the numbers inside the regions of A ∪ B, and get n(A ∪ B) = 2 + 7 + 8 = 17 49. Since A ⊂ B, A ∪ B = B. Thus, n(A ∪ B) = n(B) = 52 + 22 = 74.

1.5 Set Operations with Three Sets YOUR TURN 1.35 1. n(B) = n(AB + ) + n(AB − ) + n(B + ) + n(B − ) = 3 + 1 + 8 + 2 = 14 2. n(B') = n(U) − n(B) = 100 − 14 = 86 3. n(B ∪ Rh + ) = 14 + 36 + 37 = 87 1.37 1. A ∩ (B ∩ C) = {0,1,2,3,4,5,6} ∩ {0,6,12} = {0,6}. 2. (A ∩ B) ∪ (A ∩ C) = {0,2,4,6} ∪ {0,3,6} = {0,2,3,4,6}. 3. (A ∪ C') ∩ (B ∪ C') = {0,1,2,3,4,5,6,7,8,10,11} ∩ {0,1,2,4,5,6,7,8,10,11,12} = {0,1,2,4,5,6,7,8,10,11}.

CHECK YOUR UNDERSTANDING 29. central 31. parentheses, complement

EXERCISES 1. There are 10 gamers club members.

3. Javier plays video and card games.

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Contemporary Mathematics

5. 1 + 2 + 0 + 4 + 10 + 3 = 20. There are 20 gamers who are in the set Board ∪ Card.

7. Only 1 member only plays board games.

9. 10 + 3 = 13 Thirteen members play video and card games.

11. Four donors have type AB + blood.

13. 8 + 50 + 2 + 4 = 64. There are 64 donors with type A blood.

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Contemporary Mathematics

15. There are 52 donors with O + blood.

17. 8 + 50 + 52 + 10 = 120. She can receive blood from 120 of the donors.

19. n(A ∪ Rh + ) = 8 + 2+ 50 + 4 + 11 + 52 = 127

21.

23.

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Contemporary Mathematics 25.

27.

29.

31. A ∪ B ∩ C = red,yellow,bl𝖚e,orange,green,violet ∩ red,green,indigo = {red, green} 33. U ∩ B ∪ C = U ∩ orange,green,violet,red,indigo = {red, orange, green, indigo, violet} 35. Find A ∩ (B ∩ C)' = {red,yellow,bl𝖚e} ∩ ({green})' = {red,yellow,bl𝖚e} ∩ {red,orange,yellow,bl𝖚e,indigo,violet} = {red, yellow, bl𝖚e} 37. A and B and C' = A ∩ B ∩ C' = {24} ∩ {20,22,24,26,28,29} = 24 39. Find A or B and C' = (A ∪ B) ∩ C' = {20,21,22,24,27,28} ∩ {20,22,24,26,28,29} = {20,22,24,28} 41. (A and C) and B' = (A ∩ C) ∩ B' = {21,27} ∩ {21,23,25,26,27,29} = {21,27}

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Contemporary Mathematics 43.

47.

CHAPTER REVIEW

1.1 Basic Set Concepts 1. set 3. {v,w,x,y,z}

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Contemporary Mathematics 5.

∅ or

7. {z ∈ ℤ|z is divisible by 2} or {2z|z ∈ ℤ} 9. The set is well-defined because the 5 types of apples are listed out and is not ambiguous. 11. n(A) = n({Alabama, Alaska, Arkansas, Arizona}) = 4 13. n(A) = 3 and n(B) = 4 so the two sets are neither equivalent nor equal. 15. n(A) = 3 and n(B) = 3 so the two sets are equivalent. However, since A and B each contain different elements, they are not equal. Thus A ~ B.

1.2 Subsets 17. A is a proper subset of B since B contains one more element than A, A ⊂ B. 19. A is a subset of set B since they both contain the same elements, A ⊆ B. 21. {0}, ∅ 23. n({top hat, thimble, iron, shoe, battleship, cannon}) = 6. 26 = 64 The number of subsets is 64. 25. Answers will vary. Any subset that is equal to t, e, a must contain those exact letters. The possible subsets are {e, a, t}, {a, t, e}, {t, a, e}, {e, t, a}, {a, e, t}. 27. Answers will vary. Any pair of sets with the exact same 3 natural numbers as members, for example: {1,2,3} and {3,2,1}.

1.3 Understanding Venn Diagrams 29. Symbolically: Planes ⊂ Modes of Transportation, Trains ⊂ Modes of Transportation, and Planes ∩ Trains = ∅ Words: The sets of planes and trains are two disjoint subsets of the set of all modes of transportation. 31.

33. V' = {1,2,3,…,17}. 35. A' = {i,r}

1.4 Set Operations with Two Sets 37. S ∪ B = {s,c,r,a,b,l,e} For more free, peer-reviewed, openly licensed resources visit OpenStax.org. 3/20/23

Contemporary Mathematics 39. B and Q = B ∩ Q ={} 41. C ∪ R = {c, r, a, b, i, s, k} 43. n(S 𝖚nion R) = n(S ∪ R) = n {s,c,r,a,b,l,e,i,k} =9 45. n(A ∪ B) = 5 + 7 + 2 = 14

1.5 Set Operations with Three Sets 47. n(B ∩ C) = 2 + 8 = 10 49. (S ∪ R) ∩ L' = {s,t,a,i,r,e} ∩ {r,t,a} = {a,r,t}

CHAPTER TEST 1. The set is not well-defined because “small” is an opinion. 3. finite since there are a finite number of cats 5. finite since there are 5 letters in the set 7. A or B = A ∪ B = {32,35,36,38,40,41,44,47,48,50} 9. Set A is equivalent to set C because n(A) = 6 = n(C), but not all their elements are the same so they are not equal. 11. B ∩ C = {32,48} A ∩ (B ∩ C) = ∅ 13. B' = {31,33,34,35,37,38,39,41,42,43,45,46,47,49,50} A ∩ B' = {35,38,41,47,50} (A ∩ B') ∪ C = {31,32,35,38,41,42,47,48,50} 15. A ∪ B = {l, o, d, g, e} 17.

19. A + or B + or AB + = A + ∪ B + ∪ AB + = 40 + 3 + 12 = 55

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Chapter 2 Logic

Solutions 2.1 Statements and Quantifiers YOUR TURN 2.1 1. This is a logical statement since it is a complete sentence and it can be determined to be true or false. It is false. 2. This is a logical statement since it is a complete sentence and it can be determined to be true or false. It is true. 3. This is not a logical statement since it is a question, and it cannot be determined as to whether it is true or false. 2.3 1. The word not is removed to form the negation. Ted Cruz was born in Texas. 2. The word not is added to form the negation. This requires changing “has not” to “does not have” in order to make the sentence read properly.: Adele does not have a beautiful voice. 3. The word not is removed to form the negation. This requires changing “not convert” to “do not convert” in order to make the sentence read properly.: Leaves do not convert carbon dioxide to oxygen during the process of photosynthesis. 2.5 1. Because this is the negation of the given statement, the symbolic form of this statement is: ~(~p) = p. 2. ~r is the negation of r. In words, this reads: Woody and Buzz Lightyear are not best friends. 2.7 1. Some triangles are squares. (The negation of “No A are B” is “Some A are B”.) 2. No vegetables are green. (The negation of “Some A are B” is “No A are B”.) 3. All apples are sweet. (The negation of “Some A are not B” is “All A are B”.)

CHECK YOUR UNDERSTANDING 1. logical statement 3. “Marigolds are not yellow flowers.” is the negation of, thus, it is written: ~p. 5. premises 7. Quantifiers

EXERCISES 1. This is a complete sentence that makes a claim and can be either true or false. Thus, it is a logical statement and it is true.

3. This is a request and cannot be found to be true or false. Thus, it is not a logical statement. 5. This is a complete sentence that makes a claim and can be either true or false. Thus, it is a logical statement and it is true. 7. This is a complete sentence, even though it is an equation, that makes a claim and can be either true or false. Thus, it is a logical statement and it is false. 9. p: Grammy award winning singer, Lady Gaga was not born in Houston, Texas. 11. r: Coco Chanel said, “The most courageous act is still to think for yourself aloud.” 13. Bozo is a clown. 15. Vanilla is not the most popular flavor of ice cream. 17. Elephant and Piggy are best friends. 19. All donuts are round. 21. Some circles are squares. 23. The ancient Greek philosopher Plato did not say, “The greatest wealth is to live content with little.” 25. ~p: Shana’s hair is not red. 27. ~s: Some wolves do not howl at the moon. 29. u: I love New York. 31. q: Some squares are not parallelograms. 33. Kermit is not a green frog. 35. Some dogs do not go to heaven. 37. ~(~p) = p. So, we have: Some pizza comes with pineapple on it. 39. ~(~t) = t, so we have: Thelonious Monk is a famous jazz pianist. 41. Some car companies build cars in Michigan. 43. All even numbers are divisible by two. 45. Some odd numbers are not divisible by three. 47. All living creatures need water to survive.

2.2 Compound Statements YOUR TURN 2.9 1. Today we will go skiing, but last night it did not snow. q

and

not p

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q˄~p 2. Today we will go skiing if and only if it snowed last night. q

if and only if

q↔p 3. Last night it snowed or today we will not go skiing. p

not q

p𝗏 ~q 4. If it snowed last night, then today we will go skiing. If

then

p→q 2.11 Negation is the least dominant, so it occurs first. Then, since conjunction and disjunction have the same dominance, they occur in order from left to right. The conditional comes next and the biconditional is last. 1.

p 𝗏 q 𝖠 ~r

~p → q 𝗏 r

~p 𝗏 ~q ↔ ~ p ˄ q

. This is another one of DeMorgan’s laws and is always true.

CHECK YOUR UNDERSTANDING 9. compound statement 11. biconditional, ↔ 13. Conjunction, ˄; disjunction, 𝗏 (in any order)

EXERCISES 1. Translating the statements to letters first: If p, then r. and then to symbols: p→r 3. Translating the statements to letters first: If (not q), then (not s). and then to symbols: ~q→~s 5. Translating the statements to letters first: If not p and q, then s. and then to symbols: ~p ˄ q →s 7. Translating the statements to letters first: q and p and then to symbols: q ˄ p 9. Translating the statements to letters first: If not t, then not s. and then to symbols: ~t→~s

11. Translating the statements to letters first: It is not the case that not p and not q. Here, the “It is not the case” negates the entire clause after “that”. and then to symbols: ~ ~p˄ ~q 13. Tracy Chapman plays guitar, or all rock bands include guitarists. 15. Tracy Chapman plays guitar and Joan Jett plays guitar if and only if all rock bands include guitarists. 17. It is not the case that Tracy Chapman plays the guitar and Joan Jett does not play the guitar. 19. If the data set is symmetric then the median, mean, and mode are equal. 21. If the data set is not symmetric then the median, mean, and mode are not equal. 23. An “if-then” statement can also be written as “p implies q”. This helps to clarify the statement when a connective like “and” or “or” are used between two “if-then” statements. So, we have: The data set is symmetric implies the median is not the middle number or the mean is the average implies the median, mean, and mode are equal. 25. p→(q 𝗏 r) 27. p 𝗏 r 𝗏 ~q 29.

p˄r 𝗏s ˄t

31. p→ r 𝗏 s ↔ ~t

2.3 Constructing Truth Tables YOUR TURN 2.13 1. p is true and q is true. Therefore, the conjunction is true and true, which is true. 2. q is true and r is false. Therefore, the conjunction is true and false, which is false. 3. r is false, so ~r is true. p is true. Therefore, the conjunction is true and true, which is true. 2.15 1. Applying the dominance of connectives, ~q ˄ p ˅ r can be written more clearly as (~q ˄ p) ˅ r . Set up a table with each statement and its truth value in individual columns. Then, make columns for each of the pieces of the compound statement and determine their truth values. In order for a conjunction to be true, both component statements must be true. In order for a disjunction to be true, at least one of the two component statements must be true. p

~q ˄ p

(~q ˄ p) ˅ r

The statement ~q ˄ p ˅ r is false. 2. Applying the dominance of connectives, p ˅ q ˄ ~r can be written more clearly as (p ˅ q) ˄ (~r). Set up a table with each statement and its truth value in individual columns. Then, make columns for each of the pieces of the compound statement and determine their truth values. In order for a conjunction to be true, both component statements must be true. In For more free, peer-reviewed, openly licensed resources visit OpenStax.org. 4

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order for a disjunction to be true, at least one of the two component statements must be true. p

p˅q

(p ˅ q) ˄ (~r)

The statement p ˅ q ˄ ~r is true. 3. Set up a table with each statement and its truth value in individual columns. Then, make columns for each of the pieces of the compound statement and determine their truth values. In order for a conjunction to be true, both component statements must be true. In order for a disjunction to be true, at least one of the two component statements must be true. p

p˄r

~(p ˄ r)

~(p ˄ r) ˄ q

The statement ~(p ˄ r) ˄ q is true. 2.17 In order for a statement to be valid, it must be true always. 1. p

p ˅ ~p

The statement is valid. 2. p

~p ˅ ~q

The statement is not valid.

CHECK YOUR UNDERSTANDING 15. True 17. 4

EXERCISES 1. p: 7 × 3 = 21. Since p is true, the statement ~p is false.

3. ~r: The acceleration of gravity is 9.81m/sec2. Since ~r is true, the statement r is false. 5. t: Broccoli is a vegetable. Since t is true and since ~(~t) = t, the statement ~(~t) is true. 7. In order for a conjunction to be true, both of the component statements must be true. Since p is true and q is false, the statement p ˄ q is false. 9. In order for a disjunction to be true, at least one of the component statements must be true. Since p is true and q is false, ~p is false and ~q is true. Thus, the statement ~p ˅ ~q is true. 11. In order for a conjunction to be true, both of the component statements must be true. Since p is true and r is true, the statement p ˄ r is true. 13. In order for a conjunction to be true, both of the component statements must be true. Since r is true, then ~r is false. Thus, the statement q ˄ ~r is false. 15. In order for a conjunction to be true, both of the component statements must be true. p

p˄r

~(p ˄ r)

The statement ~(p ˄ r) is false. 17. In order for a conjunction to be true, both of the component statements must be true. In order for a disjunction to be true, at least one of the component statements must be true. p

q˄r

~p ˅ (q ˄ r)

The statement ~p ˅ (q ˄ r) is false. 19. In order for a disjunction to be true, at least one of the component statements must be true. Note that q ˅ r ˅ p is the same as (q ˅ r) ˅ p. p

q˅r

(q ˅ r) ˅ p

The statement q ˅ r ˅ p is true. 21. In order for a conjunction to be true, both of the component statements must be true. In order for a disjunction to be true, at least one of the component statements must be true. p

~p ˅ q

(~p ˅ q) ˄ r

23. In order for a conjunction to be true, both of the component statements must be true. In order for a disjunction to be true, at least one of the component statements must be true. p

~p ˄ q

(~p ˄ q) ˅~r

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25. ~r ˄ q ˄ p p

~r ˄ q

~r ˄ q ˄ p

27. ~p 𝗏 q ˄ r p q

~p ˅ q

~p ˅ q ˄ r

True

False 29. q

~q ˄ q

p T T F F

q T F T F

~p F F T T

~q ˅ q

p T T F F

q T F T F

~p F F T T

31. ~q F T F T

~p ˄ ~q F F F T

p˄q T F F F

p ˄ q˅ ~p T F T T

33.

Valid 35.

Not Valid

2.4 Truth Tables for the Conditional and Biconditional YOUR TURN

2.19 1. Valid p

~p ˅ q

q → (~p ˅ q)

q˄p

~p → (q ˄ p)

p˄q

~(p ˄ q)

~p ˅ ~q

~( p ˄ q) ↔ ~p ˅~q

q˄p

~p ↔ (q ˄ p)

p→q

~p ˅ q

(p → q) ↔ (~p ˅ q)

2. Not valid

2.21 1. Valid

2. Not valid

3. Valid

4. Valid p

~p ~q p ˄ q

(p ˄ q)→r

~p ˅ ~q

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(p ˄ q →r)↔ (~p ˅ ~q) ˅ r

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CHECK YOUR UNDERSTANDING 19. one-way contract 21. hypothesis 23. biconditional

EXERCISES 1.

p q ~p T

~p → q

3. p q ~p F

~p ↔ q

~p ˄ q

(~p ˄ q) → r

~p ˅ q

(~p ˅ q) ↔ ~r

9. p

~p ˅ r

p → ~r

(~p ˅ q) ↔ (p → ~r)

11. If Giacomo works with Faheem, then Faheem is not a software engineer.

, then →

r p

T T

not p. ~p

~p r → ~p F

The statement is false. 13. The software application was completed on time if and only if Giacomo worked with Faheem. s

if and only if

↔

s s

s↔r

T T

The statement is true. 15. If the software application was completed on time, then Ann is a project manager, but Faheem is not a software engineer. If

then

→

q ˄ ~p

s → (q ˄ ~p)

T T

p q T

and ˄

not p

~p)

The statement is false. 17. The software application was not completed on time if and only if Faheem is a software engineer or Giacomo did not work with Faheem.

Not s

if and only if

not r

↔

~r)

p r s ~s T T T F

~r F

p ˅ ~r T

~s ↔ ( p ˅ ~r) F

The statement is false. 19. Ann is a project manager implies Faheem is a software engineer if and only if the software If

then

if and only if

→

↔

application was completed on time implies Giacomo worked with Faheem.

p q

then

→

q→p s→r

(q → p) ↔ (s → r)

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T T T

The statement is true. 21. p

p ˅ ~q

(p ˅ ~q) → q

Some outcomes are not true. The statement is not valid. 23. p

p→q

(p → q) ↔ q

One outcome is not true. The statement is not valid. 25. p

p˅q

~(p ˅ q)

~p ˄ ~q

~(p ˅ q) ↔ ~p ˄ ~q

All outcomes are true. The statement is valid. 27. p

p→q

(p → q) → r

Some outcomes are not true. The statement is not valid. 29. p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

p˅q T T T T T T F F

q˄r T F F F T F F F

p˅r T T T T T F T F

p ˅ (q ˄ r) T T T T T F F F

(p ˅ q) ˄ (p ˅ r) T T T T T F F F

p ˅ (q ˄ r) ↔ (p ˅ q) ˄ (p ˅ r) T T T T T T T T

All outcomes are true. The statement is valid.

2.5 Equivalent Statements YOUR TURN 2.23 1. If Elvis Presley wore capes, then some superheroes wear capes. 2.

If some superheroes wear capes, then Elvis Presley wore capes.

If Elvis Presley did not wear capes, then no superheroes wear capes.

If no superheroes wear capes, then Elvis Presley did not wear capes.

2.25 Since the statement p → q is false, this means that p is true and q is false because that is the only time the conditional is false. 1. We want q → p: If my friend does not live in California, then my friend lives in San Francisco. Since F → T is T, this is true. 2. We want ~p → ~q: If my friend does not live in San Francisco, then my friend lives in California. Since F → T is T, this is true. 3. We want ~q → ~p: If my friend lives in California, then my friend does not live in San Francisco. Since T → F is F, this is false.

CHECK YOUR UNDERSTANDING 25. Always true OR valid OR a tautology. 27. logically equivalent 29. converse; inverse

EXERCISES 1. Build the truth table for each statement and the biconditional:

p q q→p ~p ~q ~p→~q

q→p ↔ ~p→~q

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Since the biconditional is always true, the two statements are logically equivalent. 3. Build the truth table for each statement and the biconditional:

p q ~p ~q ~p→~q ~q→~p T T F F

T F T F

F F T T

F T F T

T T F T

T F T T

~p→~q ↔ ~q→~p T F F T

Since the biconditional is not always true, the two statements are not logically equivalent. 5. Build the truth table for each statement and the biconditional:

p q ~p ~q ~p→q p 𝗏 ~q T T F F

T F T F

F F T T

F T F T

T T T F

~p→q ↔ p 𝗏 ~q

T T F T

T T F F

Since the biconditional is not always true, the two statements are not logically equivalent. 7. Build the truth table for each statement and the biconditional:

p q ~p ~q p ˄ q ~ p ˄ q

~p ˄~q ~ p ˄ q ↔ ~p ˄ ~q

Since the biconditional is not always true, the two statements are not logically equivalent. 9. Build the truth table for each statement and the biconditional:

p q r q 𝗏 r p ˄ q p ˄ r p ˄ (q 𝗏 r)

p˄q 𝗏 p˄r

p ˄ (q 𝗏 r)↔(p ˄ q) ˅ (p ˄ r)

T T T

T T F

T F T

T F F

F T T

F T F

F F T

F F F

Since the biconditional is always true, the two statements are logically equivalent. 11. a) If Six is afraid of Seven, then Seven ate Nine. b) If Seven ate Nine, then Six is afraid of Seven. c) If Six is not afraid of Seven, then Seven did not eat Nine. d) If Seven did not eat Nine, then Six is not afraid of Seven. 13. a) If Tom Brady is a quarterback, then he does not play soccer. b) If Tom Brady does not play soccer, then he is a quarterback. c) If Tom Brady is not a quarterback, then he plays soccer. d) If Tom Brady plays soccer, then he is not a quarterback. 15. a) If the shape does not have three sides, then it is not a triangle. b) If the shape is not a triangle, then it does not have three sides. c) If the shape has three sides, then it is a triangle. d) If the shape is a triangle, then it has three sides. 17. a) If penguins cannot fly, then some birds can fly. b) If some birds can fly, then penguins cannot fly. c) If penguins can fly, then no birds can fly. d) If no birds can fly, then penguins can fly. 19. a) If no Pokémon are little ponies, then Bulbasar is a Pokémon. b) If Bulbasar is a Pokémon, then no Pokémon are little ponies. c) If some Pokémon are little ponies, then Bulbasar is not a Pokémon. d) If Bulbasar is not a Pokémon, then some Pokémon are little ponies. 21. The hypothesis of the conditional statement is the “if” statement: p: Clark Kent is Superman. Knowing the story of Superman, we know that Clark Kent is actually Superman. Thus, this statement is true. 23. “If Clark Kent is not Superman, then Lois Lane is a reporter.” In symbols, this statement is ~p → ~q, thus it is the inverse. This is true since F→ T is T. 25. The inverse is logically equivalent to the converse.

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27. The conclusion of the conditional statement is the “then” statement: q: Donnie Wahlberg was a member of New Kids on the Block. The New Kids on the Block is a music group that includes Donnie Wahlberg as one of its members. Thus, statement q is true. 29. “If the Masked Singer is a music competition, then Donnie Wahlberg was not a member of New Kids on the Block.” In symbols, this statement is ~p → ~q, thus it is the inverse. This is false since T→ F is F. 31. The hypothesis of the conditional statement is the “if” statement: p: All whales are mammals. This is true. 33. “If some fish are whales, then some whales are not mammals.” In symbols, this statement is ~q → ~p, thus it is the contrapositive. This is true since F→ F is T. 35. The converse is q → p. In words: If no fish are whales, then all whales are mammals. Since T→ T is T, the converse is true. 37. The conclusion of the conditional statement is the “then” statement: q: Some circles are not symmetrical. From geometry, we know that this is false. 39. The contrapositive is ~q → ~p. In words: If all circles are symmetrical, then no parallelograms are rectangles. Since T→ F is F, the contrapositive is false.

2.6 De Morgan’s Laws YOUR TURN 2.27 ~ p→q is logically equivalent to p˄ ~q. a. Edna Mode made a new superhero costume and it includes a cape. b. I had pancakes for breakfast, and I did not use maple syrup. 2.29 1. If we let p: Eric needs to replace the light bulb, q: Marcus left the light on all night, r: Dan broke the bulb, then the statement has the symbolic form: p → (q ˅ r). Thus ~(p → (q ˅ r)) is p ˄ ~(q ˅ r) which is equivalent to p ˄ ~q ˄ ~r. In words we have: Eric needs to replace the light bulb, and Marcus did not leave the light on all night, and Dan did not break the bulb. 2. If we let p: Trenton went to school, q: Regina went to work, r: Merika cleaned the house, then the statement has the symbolic form: (p ˄ q) →r. Since ~((p ˄ q) →r) is p ˄ q ˄ ~r, in words we have: Trenton went to school, and Regina went to work, and Merika did not clean the house.

CHECK YOUR UNDERSTANDING 31. ~p ˄ ~q

33. De Morgan’s Law

EXERCISES 1. p˄ ~q 3. ~p 𝗏 q 5. ~(p ˄ q) is equivalent to ~p ˅ ~q: Sergei does not play right wing or Patrick does not play goalie. 7. ~(p ˅ q) is equivalent to ~p ˄ ~q: Luigi is not a plumber and he is a video game character. 9. ~(p ˄ q) is equivalent to ~p ˅ ~q: No people like broccoli or my siblings liked broccoli. 11. p˄ ~q 13. ~p𝗏 ~q 15. p ˄ q ˄ r ≡ p ˄ q ˄ r 17. p 𝗏 q ˄ r ≡ p ˄ r 𝗏 q ˄ r 19. ~(p → q) ≡ p ˄ ~q: The student scored an 85 on the final exam and they did not receive an A in the class. 21. ~(p → q) ≡ p ˄ ~q: The student did not do their homework and they played their video games. 23. ~(p → q) ≡ p ˄ ~q: A racecar driver was pulled over for speeding and they made it to the track on time for the race. 25. ~((p ˄ q) → r) ≡ (p ˄ q) ˄ ~r: George Boole invented Boolean algebra, Thomas Edison invented the light bulb, and Pacman is the best video game ever. 27. ~((p ˅ q) → r) ≡ (p ˅ q) ˄ ~r: Billie Holiday sang the blues or Cindy Lauper sang about true colors, and John Lennon was a Beatle. 29. ~(p → q) ≡ p ˄ ~q: All rock stars are men and Pat Benatar is a rock star. 31. ~(p → q) ≡ p ˄ ~q: Yellow and blue make green, but some colors are not beautiful. 33. p q p → q ~p ~p ˅ q

(p → q) ↔(~p ˅ q)

Because the biconditional is always true, the logical property is valid. 35. p q ~p T

~p → q p ˅ q T

(~p → q) ↔ (p ˅ q) T

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Because the biconditional is always true, the logical property is valid.

2.7 Logical Arguments YOUR TURN 2.31 1. The premises are p → q and p. Thus, the conclusion is q: Some people like history. 2. The premises are p → q and p. Thus, the conclusion is q: Some people do not like reading. 3. The premises are p → q and p. Thus, the conclusion is q: The polygon is not an octagon. 2.33 1. The premises are p → q and q → r. Thus, the conclusion is p → r: If my roommate does not go to work, then my roommate will not be able to pay their bills. 2. The premises are p → q and q → r. Thus, the conclusion is p → r: If penguins cannot fly, then we will watch the news. 3. The premises are p → q and q → r. Thus, the conclusion is p → r: If Marcy goes to the movies, then she will buy water.

CHECK YOUR UNDERSTANDING 35. valid 37. deductive 39. sound

EXERCISES 1. The premises are p → q and ~q. Thus, the conclusion is ~p. This is denying the consequent. 3. This the chain rule for conditional arguments. 5. This fits none of the forms. Hence, it is none of these. 7. The premises are p → q and p. Thus, the conclusion is q. This is the law of detachment. 9.

∴ ~q.

11. Therefore, David Radcliffe played harry Potter. 13. Therefore, we will make an apple pie. 15. Therefore, Greg and Ralph are not friends. 17. ∴~p 19. Therefore, some dragons do not breathe fire.

21. ~p → ~r 23. p → r 25. If Mr. Spock is a science officer, then James T. Kirk is the captain. 27. The premises are p → q and ~q. Thus, the conclusion is ~p. This is denying the consequent. Therefore, my siblings did not drink milk out of the carton. 29. The premises are p → q and q → r. Thus, the conclusion is p → r. This is the chain rule for conditional arguments. Therefore, if mathematics is fun, then students will score a 100 on their final exam. 31. The premises are p → q, q → r, and r → s. Applying the chain rule for conditional statements twice, the conclusion is p → s. Therefore, if the toddler is not tall, then they will not eat any cookies. 33. Using a truth table: p q P → q ~p

(p → q) ˄ ~q

((p → q) ˄ ~q) →~q

Because this is not a tautology, the statement is not valid. Using a Venn Diagram:

The statement is not valid. 35. Using a truth table: p q ~p

~p ˅ q

(~p ˅ q) ˄ p

((~p ˅ q) ˄ p) → q

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Using a Venn Diagram:

The statement is valid.

CHAPTER REVIEW

2.1 Statements and Quantifiers 1. negation 3. This is a question. It is not a logical statement and it is neither true nor false. 5. This is a logical statement since it can be determined to be true or false. It is false. 7. ~q: No dogs are Labrador retrievers. 9. Some noodles are made with wheat. (For example: Gluten-free noodles are not made with wheat, so the best we can conclude is “some”.)

2.2 Compound Statements 11. Connectives 13. If Tweety Bird is a bird, then Sylvester will not chase him. If

then

not t

p → ~t 15. If Bugs is not a Bunny or Tweety Bird is a bird, then Sylvester is not a cat. 17. Everything to the left of → and to the right of → must be done prior to applying →. p 𝗏 q ˄ r → ~s ˄ t

2.3 Constructing Truth Tables 19. disjunction 21. p ~r p ˄ ~r T

The compound statement is false. 23.

~(~r)

~p ˅ q

(~p ˅ q) ˄ (~(~r))

The compound statement is true. 25. p q ~p ~q

~p ˅ q (~p ˅ q) ˅ ~q

The statement is valid since it is always true.

2.4 Truth Tables for the Conditional and Biconditional 27. true; false 29. p

p→q

~(p → q)

p ˄ ~q

~(p → q) ↔ (p ˄ ~q)

31. Cosmos and Wanda will not grant Timmy’s wishes if and only if Timmy Turner is ten years old r

if and only if

or he does not have fairly odd parents. or not q In symbols: ~r ↔( t ˅ ~q) q

t ˅ ~q

~r ↔( t ˅ ~q)

Thus, the statement is false.

2.5 Equivalent Statements 33. converse (or q → p) 35. q: Novak’s team will win the game. 37. This is ~p → ~q, so it is the inverse.

2.6 De Morgan’s Laws 39. ~p ˄ ~q

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41. ~(~p ˄ q) ≡ p ˅ ~q 43. Using symbols first, the statement is: (p → q) ˅ r. So, in symbols, the negation is: ~((p → q) ˅ r ) ≡ ~(p → q) ˄ ~r ≡ p ˄ ~q ˄ ~r. In words: Thomas Edison invented the phonograph, and albums are not made of vinyl, and the transistor radio was not the first portable music device.

2.7 Logical Arguments 45. law of detachment 47. By the Law of Denying the Consequent, the conclusion is: Therefore, the Tampa Bay Buccaneers won Super Bowl LV. 49. By the Law of Detachment, the conclusion is: Therefore, Kamala Harris is the president of the US Senate.

CHAPTER TEST 1. This is a logical statement because it can be determined to be true or false. It is true. Answer: true logical statement 3. This is a logical statement since it can be determined to be true or false. It is false. 5. ~(p ˄ ~q) ≡ ~p ˅ q 7. Frodo is a hobbit or Gandalf is a wizard if and only if Frodo and Samwise will take the ring to Mordor. 9. r ˅ ~q ˄ p 11. ((~p) → q) ↔ r 13. p q p ˄ q ~q p ˅ ~q F

(p ˄ q ) ↔ (p ˅ ~q )

The proposition is true. 15. In symbolic form: (q ˅ ~p) ↔ ~s s

q ˅ ~p

(q ˅ ~p) ↔ ~s

T T

p q T

The statement is false. 17. If the hypothesis of a conditional statement is true, then the conditional can be true or false, depending upon whether the conclusion of the conditional statement is true or false respectively. In this case, if the hypothesis is true and the conditional statement is true, then the conclusion must be true. Therefore, Phil Mickelson won the Player’s Championship.

19. p q ~p

~p ˅ q q → p (~p ˅ q) ↔ (q → p)

Since the argument is not always true, then the argument is not valid.

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Chapter 3 Real Number Systems and Number Theory

Solutions 3.1 Prime and Composite Numbers YOUR TURN 3.1 1. 54 is divisible by 9. When 54 is divided by 9, the result is 6 with no remainder. Also, 54 can be written as the product of 9 and 6. 3.3 1. Since 3 + 4 + 2 + 8 + 8 + 7 = 32 and 32 is not divisible by 9, the number 342887 is not divisible by 9. 3.5 1. Since the last digit of 87762 is not 0, it is not divisible by 10. 3.7 1. The square root of 1429 is approximately 37.8. None of the primes less than 37 (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31) divide into 1429 evenly, thus 1429 is prime. 3.9 1. Since 5,067,322 is even (the last digit is 2), the number is divisible by 2. Hence, it is a composite number. 3.11 1. Start with factors of 90 that we know, say 2, then continue factoring each subsequent factor until we only have primes raised to powers, as necessary: 90 = 2 × 45 = 2 × 5 × 9 = 2 × 5 × 32. So, the prime factorization of 90 is 2 × 32 × 5. 3.13 1. We notice that 280 is divisible by 4 since 80 is divisible by 4, then we continue to factor until all the leaves are primes:

The prime factorization of 280 is 23 × 5 × 7. 3.15 1. The divisors of 270: 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270 The divisors of 99: 1, 3, 9, 11, 99. Thus, the GCD is 9. 3.17 1. The prime factorization of 120 is 23 × 3 × 5 The prime factorization of 200 is 23 × 52 Therefore, the GCD of 120 and 200 is 23 × 5 = 40. 3.19 1. The largest team size is represented by the GCD of 21, 35, and 28. 21 = 3 × 7

35 = 5 × 7 28 = 22 × 7 Since the GCD is 7, the largest team size that can be formed is 7 students. 3.21 1. The prime factorization of 20: 20 = 22 × 52 The prime factorization of 28: 28 = 22 × 7 The LCM of 20 and 28 is: 22 × 52 × 7 = 140. 3.23 1. Using lists: A list of multiples of 18: 18, 36, 54, 72, 90, 108, 126, 144, 162, 180, 198, 216, 234, 252, 270, 288, 306, 324, 342, 360, 378, … A list of multiples of 24: 24, 48, 72, 96, 120, 144, 168, 192, 216, 240, 264, 288, 312, 336, 360, 384, … A list of multiples of 40: 40, 80, 120, 160, 240, 280, 320, 360, 400, … The LCM is: 360. Using prime factorization: The prime factorization of 18: 2 × 32 The prime factorization of 24: 24 = 23 × 3 The prime factorization of 40: 40 = 23 × 5 The LCM is: 23 × 32 × 5 = 360 3.25 1. We need to find the LCM of 130 and 900. 130 = 2 × 5 × 13 900 = 22 × 32 × 52 The LCM is 22 × 32 × 52 × 13 = 11700. The first person to receive both giveaways would be the person who submits the 11,700th submission.

CHECK YOUR UNDERSTANDING 1. primes: 13, 701; composite: 56 (even), 213(divisible by 3), 48 (divisible by 2 and 3) 3.

410 = 2 × 5 × 41 144 = 24 × 32 GCD is 2 5. The GCD of 30, 20, and 70 is 10. Therefore, the maximum number of bags that can be filled with the same number of each item is 10.

EXERCISES 1. The last digit is even, so 24 is divisible by 2. For more free, peer-reviewed, openly licensed resources visit OpenStax.org. 2

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3. The last digit is not even, so 1,345,321 is not divisible by 2. 5. The sum of the digits is 2 + 1 + 0 = 3, which is divisible by 3. Therefore 210 is divisible by 3. 7. The last digit is 0, therefore 130 is divisible by 5. 9. The last digit is not 0 or 5, therefore 1,345,321 is not divisible by 5. 11. The sum of the digits is 2 + 1 + 0 = 3, which is not divisible by 9. Therefore 210 is not divisible by 9. 13. The sum of the digits is 4 + 8 = 12, which is divisible by 3. Therefore 48 is divisible by 3. The number 48 is also divisible by 4. Since 48 is divisible by both 3 and 4, it is divisible by 12. 15. The sum of the digits is 5 + 3 + 5 + 5 + 3 + 2 + 4 = 27, which is divisible by 3. Therefore, 5,355,324 is divisible by 3. The last two digits form the number 24 which is divisible by 4. Since 5,355,324 is divisible by both 3 and 4, it is divisible by 12. 17. 31 is prime since it is not divisible by 2, 3 or 5. 97 is prime since it is not divisible by 2, 3, 5, or 7. 188 is not prime since it is divisible by 2. The square root of 389 is approximately 19.7. It is not divisible 2 since it is not even. The sum of the digits is 20 which is not divisible by 3, so it is not divisible by 3. The last digit is neither 0 nor 5, so it is not divisible by 5. The primes 7, 11, 13, 19 do not divide into it evenly. Thus, is it prime. The primes are 31, 97, and 389. 19. The square root of 53 is approximately 7.3. It is not divisible by 2, 3, 5 or 7, so it is prime. Its prime factorization is 53, which can also be written as 53 × 1. 21.

345 = 3 × 5 × 23 23.

36068 = 22 × 71 × 127 8211679 = 7 × 592 × 337 25. 45 = 32 × 5 245 = 5 × 72 The GCD is 5. 27. 56 = 23 × 7 44 = 22 × 11 The GCD is 22 = 4.

29.

1746 = 2 × 32 × 97 28324 = 22 × 73 × 97 The GCD is 2 × 97 = 194. 31. 19 = 19 × 1 45 = 32 × 5 70 = 2 × 5 × 7 The GCD is 1. 33. 3927473 = 11 × 67 × 732 82709 = 11 × 73 × 103 1210121 = 112 × 73 × 137 The GCD is 11 × 73 = 803. 35. 20: 1, 2, 4, 5, 10, 20 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 GCD: 4 Common divisors of 20 and 84: 1, 2, 4. All divide the GCD. 37. All common divisors of two numbers divide the GCD of those two numbers. 39. The GCD of 14, 10, and 8 is 2. Yasmin and Cameron can build a maximum of 2 piles. 41. 420 = 22 × 3 × 5 × 7 300 = 22 × 3 × 52 1610 = 2 × 5 × 7 × 23 GCD = 2 × 5 = 10 She can have a maximum of 10 teams. 43. 11 = 11 × 1 24 = 23 × 3 LCM = 23 × 3 × 11 = 264 45. 200 = 23 × 52 450 = 2 × 32 × 52 LCM = 23 × 32 × 52 = 1800 47. 36 = 22 × 32 42 = 2 × 3 × 7 70 = 2 × 5 × 7 LCM = 22 × 32 × 5 × 7 = 1260

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49.

4450864 = 24 × 114 × 19 339889 = 112 × 532 157339 = 72 × 132 × 19 LCM = 24 × 72 × 114 × 132 × 19 × 532 = 103,533,011,838,256 (A calculator will only take this out to 9 places, then gives it in scientific notation. You will need to do some of this either by hand or with a computing program that permits more than 10 digits.) 51. Each week has 7 days. The LCM of 7 and 30 is 210. It will be 210 days until the lunar month begins on Saturday again. 53. 1680 = 24 × 3 × 5 × 7 2031 = 3 × 677 LCM = 24 × 3 × 5 × 7 × 377 = 1137360 It will take 1,137,360 days until they are both at their perihelion again.

3.2 The Integers YOUR TURN 3.27 1.

3.29 1.

–63 > –213 –213 < –63

3.31 1. |38| = 38 3.33 1. (–18) + 11 = –7. Since |–18| > |11|, the answer matches the sign of –18. 3.35 1. 45 – (–26) = 71. Since subtracting a negative number is the same as adding a positive number, this is the same as 45 + 26 = 71. 3.37 1. Christian’s net worth = amount due from Chanel + amount owed to Jeff = 180 + (–91) = 89 Christian’s net worth is $89. 3.39 1. (–18) x 13 = –234. Since the numbers have opposite signs, the product is negative. 3.41 1.

− 77 − 11 = 7. Since the signs of the numbers match, the division will result in a positive number.

CHECK YOUR UNDERSTANDING 7. 9. |–7| = 7 11. (–3) x (–12) = 36

EXERCISES 1. The integers are: 4, –17, 8, –300 3. 5. 7. Since –3 is to the right of –10 on a number line, this is a false statement. 9. Since 7 is to the right of –6 on a number line, this is a true statement. 11. Since 18 is smaller than 20, this is a true statement. 13. |76| = 76 and |–76| = 76. The two numbers are 76 and –76. 15. |–67| = 67 17. |61| = 61 19. This is the same as asking which two numbers have an absolute value of 23. The two numbers are 23 and –23. 21. 47 + 200 = 247 23. (–86) + 104 = 18 25. (–45) – (–26) = –45 + 26 = –19 27. 38 x 12 = 456 For more free, peer-reviewed, openly licensed resources visit OpenStax.org. 6

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29. 14 x (–5) = –70 31. (–412) x 504 = –207,648 33. (–130) ÷ (–65) = 2 35. 2000 ÷ (–40) = –50 37.

( − 9) + ( − 17) + ( − 18) + ( − 14) + ( − 17) + ( − 19) + ( − 11) = − 105 = − 15 7 7

The average low temperature in Fargo that week was –15 degrees centigrade. 39. 495 +150 + 250 – 50 – 110 – 300 = 435 Heath will have $435 in their account. 41. 127 – 154 + 350 – 83 + 306 – 158 = 388 The politician has a net gain of 388 voters.

3.3 Order of Operations YOUR TURN 3.43 1. 43 + 18 x 15 = 43 + 270 = 313 3.45 1. 3 × 63 − 18 = 3 × 216 − 18 = 648 − 18 = 630 3.47 1.

3.49

12 −4

+ 8 × 9 × 23 − 24 × 25 = 12 + 8 × 3 × 8 − 24 × 5 12

−4

= − 3 + 6 × 8 − 60 = − 3 + 48 − 60 = 45 – 60 = –15 2 1. 8 − 25 − 2 = 8 − 25 − 4 = 8 − 21 = 8 − 3 = 5 7

CHECK YOUR UNDERSTANDING 13. Exponents 15. Parentheses

EXERCISES 1. Addition and subtraction 3. addition and subtraction; multiplication and division 5. 4 − 5 × 6 = 4 − 30 = − 26 7. 18 + 8 × 13 = 18 + 104 = 122 9. 50 ÷ 2 + 8 = 25 + 8 = 33 11. 13 + 4 × 32 = 13 + 4 × 9 = 13 + 36 = 49 13. 62 × 5 − 13 × 92 = 36 × 5 − 13 × 81 = 180 − 1053 = − 873 15. 450 ÷ 32 − 56 ÷ 22 = 450 ÷ 9 − 56 ÷ 4 = 50 − 14 = 36

17. 38 × 6 − 4 + 5 × 18 ÷ 10 = 228 − 4 + 90 ÷ 10 = 228 − 4 + 9 = 224 + 9 = 233 19. 600 ÷ 12 × 5 + 40 − 62 = 600 ÷ 12 × 5 + 40 − 36 = 50 × 5 + 40 − 36 = 250 + 40 − 36 = 290 − 36 = 254 21. 10 × 63 ÷ 2 × 5 + 32 − 240 ÷ 8 × 9 = 10 × 216 ÷ 2 × 5 + 9 − 240 ÷ 8 × 9 = 2160 ÷ 2 × 5 + 9 − 30 × 9 = 1080 × 5 + 9 − 270 = 5400 + 9 − 270 = 5409 − 270 = 5139 23. (4 + 3) × 2 = 7 × 2 = 14 25. (14 − 25) × 2 = ( − 11) × 2 = − 22 27. (3 + 2)3 = 53 = 125 29. (45 − 60)2 = ( − 15)2 = 225 31. (3 × 4)3 ÷ 8 + 5 = (12)3 ÷ 8 + 5 = 1728 ÷ 8 + 5 = 216 + 5 = 221 33. 4 − (6 + 3) × 5 + 11 = 4 − 9 × 5 + 11 = 4 − 45 + 11 = − 41 + 11 = − 30 35. 15 × (4 − 1)2 + 5 × (17 + 2 × 3) + 4 = 15 × (3)2 + 5 × (17 + 6) + 4 = 15 × 9 + 5 × (23) + 4 = 135 + 115 + 4 = 250 + 4 = 254 37. 4 × (5 + 2 × (6 + 8)) + 10 × 32 = 4 × (5 + 2 × 14) + 10 × 9 = 4 × (5 + 28) + 90 For more free, peer-reviewed, openly licensed resources visit OpenStax.org. 8

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= 4 × 33 + 90 = 132 + 90 = 222 39. 21 − 3 × (5 × (2 + 6) − 3 × (18 − 11) + 25) ÷ 2 = 21 − 3 × (5 × 8 − 3 × 7 + 25) ÷ 2 = 21 − 3 × (40 − 21 + 25) ÷ 2 = 21 − 3 × (19 + 25) ÷ 2 = 21 − 3 × 44 ÷ 2 = 21 − 132 ÷ 2 = 21 − 66 = − 45

3.4 Rational Numbers YOUR TURN 3.51 1. Since 94 factors into the following pairs of factors: 1 x 94 and 2 x 47, we see that 94 is not a perfect square. 2. The prime factorization of 441 is 32 × 72. We note that both 3 and 7 are squared, so that leads us to test 212, which is 441. Using a calculator to find the square root of 441 also yields 21, which is an integer. Thus 441 is a perfect square. 3.53 1. a × d = 8 × 26 = 208 b × c = 14 × 12 = 168 Since these products are not equal, the two fractions are not equivalent. 3.55 1. 3.57 1. 3.59 1.

124 297 21 40

+ 3 By using Desmos, we get 0.4415084175. Using the fraction button, we get 16391. 125

37125

− 8 = 13 Because 13 is prime and isn’t a factor of 40, this is in lowest terms. 40

99 = 32 × 11,

300 = 22 × 3 × 52 The LCM of 99 and 300 is 22 × 32 × 52 × 11 = 9900 10 17 10 × 100 17 × 33 1000 561 439 − = − = − = 99 300 99 × 100 300 × 33 9900 9900 9900

5 + 5 131 3.61 1. 914 = 9 × 14 = 14 14

3.63 1. Since the digit to the right of the second decimal place is 9, which is greater than 5, we write the decimal to two places: 18.62, then add 1 to the 2. This yields the rounded number as 18.63. 5

3.65 1. 17.03347 = 17.033475× 10 = 1703347 100000 10

3.67 1. 45.63 ÷ 17.13 ≅ 2.664 (to three decimal places) 3.69 1.

2+1÷ 3 = 5 10

2+1÷ 3 5 10