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A University-Level Understanding of Number with an eye towards elementary education

Shannon Ezzat

Section 9.1 Chapter 1 Questions

Subsection 9.1.1

Exercises Exercises

1.
Solution.
(a) Statement. (b) Statement. (c) Not a statement (question). (d) Statement (truth may be disputed, but it’s truth-apt). (e) Not a (classical) statement (liar paradox; no consistent truth value). (f) Not a statement (command).
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2.
Solution.
True. “Dolphins are mammals” is true and “dolphins live primarily underwater” is true; a disjunction is true if at least one disjunct is true.
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3.
Solution.
  1. \(P \land Q\) (true only when both \(P\) and \(Q\) are true).
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  2. \(P \lor Q\) (false only when both are false).
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  3. \(\lnot(P \lor Q)\) (NOR; true only when both are false).
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  4. A tautology (e.g., \(P \lor \lnot P\)).
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4.
Solution.
Open sentences and domains: (a) Open; domain typically \(\mathbb{R}\text{.}\) (b) Open; domain: the set of dogs. (c) Statement. (d) Open; domain: propositions/statements. (e) Statement. (f) Statement.
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5.
Solution.
(a) \(x=1\) works. (b) Choose \(D\) to be any mastiff. (d) No value makes “\(P\) is both true and false” true in classical logic.
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6.
Solution.
Its truth value is exactly \(\lnot P\text{:}\) it is true when \(P\) is false and false when \(P\) is true.
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7.
Solution.
  1. P (cow awake) Q (pig asleep) P∧Q
    T T T
    F T F
    T F F
    F F F
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  2. P (accepts cash) Q (accepts credit) P∨Q
    T T T
    F T T
    T F T
    F F F
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  3. P (prepares career) Q (poor use of time) P∨Q
    T T T
    F T T
    T F T
    F F F
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  4. A (Anna happy) J (Jordan happy) ¬A ∧ ¬J
    T T F
    F T F
    T F F
    F F T
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8.
Solution.
  1. “Gravity is a force.”
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  2. “City \(C\) has a population less than 10 million.”
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  3. “Harold did not eat a burger or did not eat fries for lunch.”
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  4. “Every penguin has seen Antarctica.”
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  5. “Someone on campus is neither a student nor a staff-member.”
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  6. “There exists a beverage that is hot and sweet.”
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9.
Solution.
  1. “Peter is happy.”
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  2. “There exists a book that is neither informative nor interesting.”
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  3. Equivalent reading: “No sculptures are made of marble.”
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  4. “All films are silent.”
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  5. \(n=11\text{.}\)
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  6. “There exists a bus with four tires.”
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10.
Solution.
  1. Parrots are a subset of birds: the “parrots outside birds” region is empty.
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  2. Place an \(X\) in the intersection of widgets and gizmos (nonempty intersection).
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  3. Shade the intersection of mountains and volcanoes to indicate emptiness.
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  4. Place an \(X\) in the part of “songs” outside “classical” (∃ non-classical song).
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  5. Place an \(X\) in “articles” outside “has sources”.
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  6. Use three sets (athletes, strict diet, exercises daily): put an \(X\) in athletes outside the intersection of the last two.
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  7. With sets vegans, honey-eaters, chicken-eaters: put an \(X\) in vegan∩honey; shade vegan∩chicken empty.
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  8. Mammals ⊂ oxygen-breathers; also place an \(X\) in oxygen-breathers outside mammals.
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11.
Solution.
  1. \(2^4=16\) rows (four propositional variables: \(P,Q,R,S\)).
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  2. P Q R P∨Q (P∨Q)∧R
    T T T T T
    F T T T T
    T F T T T
    F F T F F
    T T F T F
    F T F T F
    T F F T F
    F F F F F
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  3. Not equivalent. For example, with \((P,Q,R)=(T,F,T)\) we have \((P\lor R)\land Q = F\) while \((P\lor Q)\land R = T\text{.}\)
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Subsection 9.1.2 Solutions to “Conditional Questions”

Exercises Exercises

1.
Solution.
Conditionals (truth-apt): (a), (c), (d). Items (b), (e), (f) are questions/commands, hence not statements in classical logic.
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2.
Solution.
Invalid. A direct proof of “for all even \(n\text{,}\) \(n^3\) is even” must begin with an arbitrary even \(n=2k\) and show \(n^3=8k^3\) is even. A single example (\(n=2\)) does not prove the universal statement.
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3.
Solution.
  1. If I do not eat my hat, then the Blue Jays did not win the World Series.
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  2. If a shark does not die, then it did not stop moving.
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  3. If the roads are not slippery, then the weather is not cold.
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  4. If \(n\) cannot be written as the sum of an even and an odd number, then \(n\) is not odd.
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4.
Solution.
  1. False (counterexample: \(p=2\text{,}\) then \(p^2=4\) is even). True if restricted to odd primes.
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  2. False (triangle: \(180^\circ\)). True if restricted to polygons with at least four sides.
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  3. True (by definition of “bilingual”).
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  4. (No statement provided.)
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5.
Solution.
Write \(a=2r+1,\, b=2s+1,\, c=2t+1,\, d=2u+1\text{.}\) Then
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\((a b c d)=(2r+1)(2s+1)(2t+1)(2u+1)=2k+1\) for some integer \(k\text{,}\) hence the product is odd. (The sum of four odd integers is even.) Thus “their sum or their product is odd” is true (indeed, the product is always odd).
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6.
Solution.
Biconditionals: (b) and (e). Item (c) is not a sound biconditional as stated unless a domain is specified; with \(n\in\mathbb{Z}\) it becomes trivially true but uninformative. Items (a) and (d) are not biconditionals (and (d) is false: \(x=\pm1\) both solve \(x^4=1\) over the integers).
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7.
Solution.
No specific statements were provided, so no counterexamples can be evaluated here.
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Subsection 9.1.3 Solutions to “Set Theory Questions”

Exercises Exercises

1.
Solution.
  1. Universe: the Latin alphabet letters.
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  2. Universe: \(\mathbb{N}\) (natural numbers).
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  3. Universe: common household pets.
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  4. Universe: \(\mathbb{Z}\text{.}\)
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  5. Universe: English sentences.
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2.
Solution.
  1. \(\{\text{platypuses},\ \text{echidnas}\}\) (the monotremes).
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  2. \(\{\text{AB},\ \text{BC},\ \text{MB},\ \text{NB},\ \text{NL},\ \text{NS},\ \text{ON},\ \text{PE},\ \text{QC},\ \text{SK}\}\text{.}\)
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  3. \(\{\}\text{.}\)
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  4. \(\{11,13,17,19,23,29\}\text{.}\)
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  5. \(\{q\in\mathbb{Q}\mid |q| < 4\}\) is infinite; a roster with ellipses: \(\{\ldots,-\dfrac{7}{2},-2,-\dfrac{3}{2},-1,0,1,\dfrac{3}{2},2,\dfrac{7}{2},\ldots\}\text{.}\)
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  6. Any three, e.g., \(\{\text{Canadiens},\ \text{Maple Leafs},\ \text{Canucks}\}\text{.}\)
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3.
Solution.
\(C=\{1,4,5,10\}\text{.}\)
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  1. \(A\cup B=\{0,1,2,3,4,7,8,9\}\text{.}\)
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  2. \(B\cap C=\{1,4\}\text{.}\)
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  3. \(A\cap B\cap C=\{1\}\text{.}\)
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  4. \(B-A=\{0,4,9\}\text{.}\)
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4.
Solution.
For example, \(\{0,1,2,\dots,19\}\) (subset of \(\mathbb{N}_0\)) has cardinality \(20\text{.}\)
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5.
Solution.
Within \(\mathbb{Z}\text{,}\) \(\sim\mathbb{N}=\mathbb{Z}\setminus\mathbb{N}=\{\ldots,-3,-2,-1,0\}\) (assuming \(\mathbb{N}=\{1,2,3,\ldots\}\)).
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6.
Solution.
If \(x\in A\cap B\text{,}\) then \(x\in A\) and \(x\in B\text{,}\) so \(A\cap B\subseteq A\) and \(A\cap B\subseteq B\text{.}\) For unions, generally \(A\cup B\) is not a subset of both. Counterexample: \(A=\{1\}\text{,}\) \(B=\{2\}\text{;}\) then \(A\cup B=\{1,2\}\) is not a subset of \(A\) (nor of \(B\)). (In fact, \(A\subseteq A\cup B\) and \(B\subseteq A\cup B\text{.}\))
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7.
Solution.
Let \(E_0=\{2n\mid n\in\mathbb{N}_0\}\text{.}\)
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  1. \(B=\mathbb{Z}\setminus E_0\text{.}\)
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  2. \(B=\mathbb{N}_0\setminus E_0\) (the odd nonnegative integers).
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  3. \(B=\mathbb{N}\setminus (E_0\setminus\{0\})\) (the positive odd integers).
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  4. \(B=\mathbb{Q}\setminus E_0\text{.}\)
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8.
Solution.
  1. \(6\text{.}\)
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  2. \(10\) (squares of \(1\) through \(10\)).
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  3. \(72\) (there are \(90\) two-digit numbers; \(18\) contain a \(7\)).
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  4. \(70\) (arithmetic progression step \(5\text{:}\) \((385-40)/5+1=70\)).
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Subsection 9.1.4 Solutions to “Explanatory and Critical Thinking Questions”

Exercises Exercises

1.
Solution.
Disagree. “This statement is false” is self-referential and yields a contradiction under classical bivalence: if it’s true, it says it’s false; if it’s false, it must be true. Because it cannot be consistently assigned a truth value, it is not treated as a (well-formed) statement in classical logic.
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2.
Solution.
The flaw is assuming questions/commands can be given truth values without rephrasing. Truth values apply to propositions (declarative sentences). A question or command must first be converted into a proposition (e.g., “You will give me the snacks”) before truth evaluation is meaningful.
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3.
Solution.
Not correct. The student stated the inverse. The correct contrapositive of “If square then four sides” is: “If a shape does not have four sides, then it is not a square.”
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4.
Solution.
They used a single example (\(n=2\)). A universal conditional must be proved for arbitrary even \(n\text{:}\) write \(n=2k\text{,}\) then \(n^2=4k^2\text{,}\) which is divisible by 4.
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5.
Solution.
Sets ignore multiplicity. \(\{1,2,2,3\}=\{1,2,3\}\text{,}\) which has three elements, not four.
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6.
Solution.
Disagree. The empty set is defined to have no elements; its existence is postulated (and unique) in standard set theory. Knowing it exists does not require it to contain anything.
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7.
Solution.
The union collects elements that are in either set, so it generally cannot be a subset of both. Example: \(A=\{1\}\text{,}\) \(B=\{2\}\text{;}\) then \(A\cup B=\{1,2\}\) is not a subset of \(A\) or of \(B\text{.}\) By contrast, \(A\cap B\) is always a subset of each.
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