# Math 3000 Section 003 Intro to Abstract Math Homework 2

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1 Math 3000 Section 003 Intro to Abstract Math Homework 2 Department of Mathematical and Statistical Sciences University of Colorado Denver, Spring 2012 Solutions (February 13, 2012) Please note that these solutions are only suggestions; different answers or proofs are also possible. Section 2.1: Statements 1. Give one sentence each about abstract mathematics (or the super bowl) that is (a) declarative and a statement; (b) declarative and open; (c) imperative; (d) interrogative; (e) exclamatory. Solution: Answers may vary, but here are some examples: (a) The super bowl is the annual championship game of the National Basketball Association. (False, but a declarative statement.) (b) This day is the biggest day for U.S. food consumption. (Note that this day is not specified but left open - however, you may guess that it s Thanksgiving Day, but be surprised that according to Wikipedia Super Bowl Sunday beats Christmas and takes second place.) (c) Wiggle wiggle wiggle wiggle wiggle! (d) Who wants chicken wings? (e) Touchdown! Section 2.2: The Negation of a Statement 2. Exercise 2.8: State the negation of each of the following statements. (Avoid the awkwardness of using double negation.) (a) 2 is a rational number. (b) 0 is not a negative integer. (c) 111 is a prime number. Solution: (a) 2 is not a rational (better: an irrational) number. (b) 0 is a negative integer. (Careful here: positive is not the opposite of negative because 0 is neither positive nor negative; to include zero with the positive numbers you have to say nonnegative; and similarly, you have to say nonpositive to include 0 with the negative numbers.) (c) 111 is not a prime number. (Careful again: the opposite of prime is not composite, and vice versa, the opposite of composite is not prime; for example, both 0 and 1 are neither composite nor prime.) Section 2.3: The Disjunction and Conjunction of Statements 3. Exercise 2.10: Let P : 15 is odd and Q: 21 is prime. State each of the following in words, and determine whether they are true or false. (a) P Q (b) P Q (c) ( P ) Q (d) P ( Q). Solution: (a) 15 is odd or 21 is prime (True). (b) 15 is odd and 21 is prime (False). (c) 15 is even or 21 is prime (False). (d) 15 is odd and 21 is not prime (True). Section 2.4: The Implication

2 Math Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions) 2 4. Formulate four conditional sentences about abstract mathematics (or the super bowl) that correspond to the four possible truth assignments (T T, T F, F T, F F ) for protasis (condition) and apodasis (consequence). Explain why only one of your implications is logically false. Solution: Consider the following mathematical statement that be composed of two open sentences over the domain of positive integers: If P (x): x is an even prime number, then Q(x): x has exactly two divisors, one and itself. You will likely agree that this statement is generally true: although condition and consequence are both satisfied only for x = 2 (T T ), the general statement remains true also when the condition is violated for all odd primes (F T ) or for all even (or odd) composite numbers (F F ): the fact that the condition is false for x = 3 does not mean the implication is false; similarly, if x = 4 were a prime, then by definition it would have exactly two divisors, 1 and itself. Now imagine that your friend believes that 0 is a prime number, so that the condition P (0) would be true, but agrees with you that 0 has infinitely many divisors (all positive integers are divisors of 0), so that the consequence Q(0) is false (T F ). In this case, the implication is clearly false, and you should have no difficulties convincing your friend that 0 cannot be a prime number. You may have noticed that a logical explanation using non-mathematical sentences is quite difficult: for example, sentences like If Justin Bieber sang during the half-time show of Super Bowl XLVI, then the New England Patriots won or If LMFAO wiggled, then Tom Brady and Eli Manning were sexy and they knew it (if this sentences seems weird, just ignore it!) simply do not make much sense: they are not related, act on different domains, and include a linguistic shade that mathematical logic does not know: the distinction of conditional sentences as factual (in past or present), predictive (in future), and speculative (in past, present, of future expressed in subjunctive mood using a modal verb could, might, would,... ). To avoid this confusion, we sometimes distinguish the logical conditional P Q (read if P, then Q or P implies Q ) from the material conditional P Q (read not P or Q ): Justin Bieber did not sing during the half-time show of Super Bowl XLVI, or the New England Patriots won (true and sensible) or LMFAO did not wiggle, or Tom and Eli were sexy and they knew it (not sure about that one). Although logical and material conditionals are equivalent in mathematical logic, their meaning and interpretation may seem different based on our past experience, intuition, and prepossession with natural language. Section 2.5: More on Implications 5. Exercise 2.20: In each of the following, two open sentences P (x) and Q(x) over a domain S are given. Determine all x S for which P (x) Q(x) is a true statement. (Hint: Use the logical equivalence between the two statements P (x) Q(x) ( P (x)) Q(x).) (a) P (x) : x 3 = 4; Q(x) : x 8; S = R (b) P (x) : x 2 1; Q(x) : x 1; S = R (c) P (x) : x 2 1; Q(x) : x 1; S = N (d) P (x) : x [ 1, 2]; Q(x) : x 2 2; S = [ 1, 1] Solution: This exercise (quite impressively) shows how the equivalent material conditional P (x) Q(x) ( P (x)) Q(x) can facilitate our understanding of a logical implication P (x) Q(x): (a) The equivalent disjunction x 3 4 or x 8 is true for all x 7. Therefore, over the domain of real numbers, the implication P (x) Q(x) is true for all real numbers but 7. (b) The equivalent disjunction x 2 < 1 or x 1 is true for all x > 1. Therefore, over the domain of real numbers, the implication P (x) Q(x) is true for all real numbers greater than 1. (c) An immediate consequence from (b), over the domain of positive integers, the implication P (x) Q(x) is always true. (d) In this case, it is easier to use the

3 Math Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions) 3 logical conditional directly and observe that x for all x S = [ 1, 1]. Therefore, over the domain S = [ 1, 1], the implication P (x) Q(x) is always true. Alternatively, the same conclusion follows from the equivalent disjunction x / [ 1, 2] or x 2 2, which is true for all real numbers x / [ 2, 2] and thus for all real numbers in the domain S = [ 1, 1]. Section 2.6: The Biconditional 6. Exercise 2.22: Let P : 18 is odd and Q: 25 is even. State P Q in words. Is P Q true or false? Solution: 18 is odd if and only if 25 is even. This biconditional is logically true. If you are not convinced, state the two implications 18 is odd if 25 is even and 18 is odd only if 25 is even separately (rewrite as if..., then... if you prefer) and formulate them as material conditionals: 18 is odd or 25 is odd and 18 is even or 25 is even both of which are correct. Section 2.7: Tautologies and Contradictions 7. Exercise 2.32: For statements P and Q, show that (P (P Q)) Q is a tautology. Then state (P (P Q)) Q in words. (This is an important logical argument form, called modus ponens.) Solution: If P is true, and if P implies Q, then Q is true. (Using natural language, this means that a correct inference from a correct condition always yields a correct consequence). Section 2.8: Logical Equivalence P Q P Q P (P Q) (P (P Q)) Q T T T T T T F F F T F T T F T F F T F T 8. Exercise 2.34: For statements P and Q, the implication ( P ) ( Q) is called the inverse of the implication P Q. (a) Use a truth table to show that these statement are not (!) logically equivalent. (b) Find another implication that is logically equivalent to ( P ) ( Q) and verify your answer. Solution: (a) The truth table below shows that the truth values of P Q and its inverse ( P ) ( Q) are different when exactly one of the two statements P and Q is true and the other one is false. (b) A logically equivalent implication to ( P ) ( Q) is its contrapositive Q P which is itself equivalent to the material conditional ( Q) P. P Q P Q P Q ( P ) ( Q) Q P T T F F T T T T F F T F T F F T T F T F T F F T T T T T Section 2.9: Some Fundamental Properties of Logical Equivalence 9. Verify (mathematically) or explain (logically) correctness of the laws in Theorem 2.18 on page 49 in your text book. (These laws are very important and we will use them a lot, so please make sure that you understand their meaning.)

4 Math Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions) 4 Solution: Use the truth tables below to verify these laws, and your common sense to convince yourself of their correctness. Commutative and associative laws should be clear, but do think a little (and maybe formulate a few examples) about distributive and De Morgen s laws. (a) Commutative Laws (b) Associative Laws (c) Distributive Laws P Q P Q Q P T T T T T F T T F T T T F F F F P Q P Q Q P T T T T T F F F F T F F F F F F P Q R P Q Q R (P Q) R P (Q R) T T T T T T T T T F T T T T T F T T T T T T F F T F T T F T T T T T T F T F T T T T F F T F T T T F F F F F F F P Q R P Q Q R (P Q) R P (Q R) T T T T T T T T T F T F F F T F T F F F F T F F F F F F F T T F T F F F T F F F F F F F T F F F F F F F F F F F P Q R P Q P R Q R P (Q R) (P Q) (P R) T T T T T T T T T T F T T F T T T F T T T F T T T F F T T F T T F T T T T T T T F T F T F F F F F F T F T F F F F F F F F F F F

5 Math Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions) 5 (d) De Morgan s Laws P Q R P Q P R Q R P (Q R) (P Q) (P R) T T T T T T T T T T F T F T T T T F T F T T T T T F F F F F F F F T T F F T F F F T F F F T F F F F T F F T F F F F F F F F F F Section 2.10: Quantified Statements P Q P Q P Q (P Q) ( P ) ( Q) T T F F T F F T F F T T F F F T T F T F F F F T T F T T P Q P Q P Q (P Q) ( P ) ( Q) T T F F T F F T F F T F T T F T T F F T T F F T T F T T 10. Exercise 2.48: Determine the truth value of each of the following statements. (a) x R : x 2 x = 0 (b) n N : n (c) x R : x 2 = x (d) x Q : 3x 2 27 = 0 (e) x R, y R : x + y + 3 = 8 (f) x, y R : x + y + 3 = 8 (g) x, y R : x 2 + y 2 = 9 (h) x R, y R : x 2 + y 2 = 9 (i) x R : y R : x = y (new!) (j) x R : y R : x = y (new!) Solution: (a) True (Examples: x {0, 1} R). (b) True (Proof: n 1 for all n N). (c) False (Counterexample: x = 1 R but ( 1) 2 = 1). (d) True (Examples: x { 3, 3} Q). (e) True (Example: (x, y) = (3, 2) R R). (f) False (Counterexample: (x, y) = (3, 1) R R). (g) True (Example: (x, y) = (3, 0) R R). (h) False (Counterexample: (x, y) = (3, 1) R R). (i) True (Proof: We need to show that given any real number x, there exists a real number y so that x = y. This is (almost) trivial: Let x be the real number that we are given, and choose y = x.). (j) False: This statement says that there exists a real number x such that x = y for all real numbers y, or worded slightly differently, that there exists a real number that is equal to all real numbers. This statement is clearly false. Section 2.11: Characterizations of Statements 11. Exercise 2.52: Give a definition of each of the following, and then state a characterization of each.

6 Math Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions) 6 (a) two lines in the planes are perpendicular (b) a rational number Solution: (a) Possible definition: Two lines in the plane are said to be perpendicular if they form congruent adjacent angles (a T-shape). Possible characterizations: (i) Two lines in the plane are perpendicular if and only if they intersect at an angle of 90 degrees (you can say a right angle if you define (or assume the reader knows) that a right angle is an angle that measures 90 degrees, or π/2 radians). (ii) Two lines in the plane are perpendicular if and only if they have opposite reciprocal slopes (the product of their slopes is 1) or if one line is horizontal and the other line is vertical (because the slope of a vertical line is usually described as undefined or infinity, you need to treat vertical and horizontal lines as a special case). (iii) Two lines in the plane are said to be perpendicular if the dot product between the two direction vectors that describe these lines equals zero. (iv) Let a, b, p, q R be real numbers and L 1 : y = ax+p and L 2 : y = bx+q be two lines in the plane. Then L 1 and L 2 are perpendicular, denoted by L 1 L 2, if and only if ab = 1. (Note that this characterization does not say that vertical and horizontal lines are not perpendicular, because vertical lines can not be represented as shown here and thus do not fall into the domain of this result.) (iv) Let a, b, c, d, p, q R be real numbers and L 1 : ax + by = p and L 2 : cx + dy = q be two lines in the plane. Then L 1 L 2 if and only if ac + bd = 0. (v) Let p, q, r, s R 2 be real two-dimensional vectors and L 1 : (x, y) = p + tr and L 2 : (x, y) = q + ts be two lines in the plane parametrized by the real scalar t (, ). Then L 1 L 2 if and only if r s = 0. (b) Possible definition: A number that can be expressed as the simple fraction of an integer and a positive integer is called a rational number. The same definition using more symbols: A number r is rational, denoted by r Q, if and only if there exists an integer p Z and a positive integer q N such that r = p/q (recall that the letter Q is derived from the word quotient ). The same definition using only symbols: Let the set of rational numbers be defined by Q := {r : p Z, q N : r = p/q} = {p/q : (p, q) Z N}. Possible characterizations: (i) A real number is rational if and only if it is not irrational (of course, this definition only makes sense if we already know or have defined what real and irrational numbers are). (ii) A real number is rational if and only if it has a finite or repeating decimal expansion. (iii?) The following characterization is wrong: Let p, q R be two real numbers. Then the number r = p/q is rational if and only p Z and q N. (The if direction is true but it is not difficult to find counterexample for the only if direction: r = p/q is also rational if p = q I because then r = 1 Q, among others. In other words, given the representation r = p/q, the condition (p, q) Z N is sufficient but not necessary for r Q.) Additional Exercises for Chapter Exercise 2.60: Rewrite each of the implications below using (1) only if and (2) sufficient. (a) If a function f is differentiable, then f is continuous. (b) If x = 5, then x 2 = 25. Solution: (a) A function is differentiable only if it is continuous. The differentiability of a function is sufficient for its continuity. Differentiability of a function is a sufficient condition for its continuity. (In other words, it is not possible that a function is differentiable but not continuous. This condition is not necessary, however: it is also possible that a function is continuous but not differentiable). (b) A number equals 5 only if its square equals 25 (note that the inverse is not correct: the square of a number equals 25 not only if that number is 5, but also if that number is (positive) 5.) A value of 5 is sufficient for that number s square being 25 (but it is not necessary: another possibility would be a value of (positive) 5). Please let me know if you have any questions, comments, corrections, or remarks.

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