Download Math 3000 Section 003 Intro to Abstract Math Homework 2

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)
(b)
(c)
(d)
(e)
declarative and a statement;
declarative and open;
imperative;
interrogative;
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)
(b)
(c)
(d)
P ∨Q
P ∧Q
(∼ P ) ∨ Q
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
Math 3000-003 Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions)
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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)
(b)
(c)
(d)
P (x) : x − 3 = 4; Q(x) : x ≥ 8; S = R
P (x) : x2 ≥ 1; Q(x) : x ≥ 1; S = R
P (x) : x2 ≥ 1; Q(x) : x ≥ 1; S = N
P (x) : x ∈ [−1, 2]; Q(x) : x2 ≤ 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 6= 4 or x ≥ 8 is true for all x 6= 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 x2 < 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
Math 3000-003 Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions)
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logical conditional directly and observe that x2 ≤ 1 ≤ 2 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
/ [−1, 2] or x2 ≤ 2, which is true
√ the equivalent disjunction x ∈
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).
P
T
T
F
F
Q
T
F
T
F
P ⇒Q
T
F
T
T
P ∧ (P ⇒ Q)
T
F
F
F
(P ∧ (P ⇒ Q)) ⇒ Q
T
T
T
T
• Section 2.8: Logical Equivalence
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
T
T
F
F
Q
T
F
T
F
∼P
F
F
T
T
∼Q
F
T
F
T
P ⇒Q
T
F
T
T
(∼ P ) ⇒ (∼ Q)
T
T
F
T
Q⇒P
T
F
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.)
Math 3000-003 Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions)
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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
P
T
T
F
F
Q
T
F
T
F
P ∨Q
T
T
T
F
Q∨P
T
T
T
F
P
T
T
F
F
Q
T
F
T
F
P ∧Q
T
F
F
F
Q∧P
T
F
F
F
(b) Associative Laws
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
T
T
F
T
T
T
F
(P ∨ Q) ∨ R
T
T
T
T
T
T
T
F
P ∨ (Q ∨ R)
T
T
T
T
T
T
T
F
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
F
F
F
F
F
F
Q∧R
T
F
F
F
T
F
F
F
(P ∧ Q) ∧ R
T
F
F
F
F
F
F
F
P ∧ (Q ∧ R)
T
F
F
F
F
F
F
F
(c) Distributive Laws
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
P ∨R
T
T
T
T
T
F
T
F
Q∧R
T
F
F
F
T
F
F
F
P ∨ (Q ∧ R)
T
T
T
T
T
F
F
F
(P ∨ Q) ∧ (P ∨ R)
T
T
T
T
T
F
F
F
Math 3000-003 Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions)
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
F
F
F
F
F
F
P ∧R
T
F
T
F
F
F
F
F
Q∨R
T
T
T
F
T
T
T
F
P ∧ (Q ∨ R)
T
T
T
F
F
F
F
F
5
(P ∧ Q) ∨ (P ∧ R)
T
T
T
F
F
F
F
F
(d) De Morgan’s Laws
P
T
T
F
F
Q
T
F
T
F
∼P
F
F
T
T
∼Q
F
T
F
T
P ∨Q
T
T
T
F
∼ (P ∨ Q)
F
F
F
T
(∼ P ) ∧ (∼ Q)
F
F
F
T
P
T
T
F
F
Q
T
F
T
F
∼P
F
F
T
T
∼Q
F
T
F
T
P ∧Q
T
F
F
F
∼ (P ∧ Q)
F
T
T
T
(∼ P ) ∨ (∼ Q)
F
T
T
T
• Section 2.10: Quantified Statements
10. Exercise 2.48: Determine the truth value of each of the following statements.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
∃
∀
∀
∃
∃
∀
∃
∀
∀
∃
x ∈ R : x2 − x = 0
n∈N:n+1≥2
√
x ∈ R : x2 = x
x ∈ Q : 3x2 − 27 = 0
x ∈ R, ∃ y ∈ R : x + y + 3 = 8
x, y ∈ R : x + y + 3 = 8
x, y ∈ R : x2 + y 2 = 9
x ∈ R, ∀ y ∈ R : x2 + y 2 = 9
x ∈ R : ∃ y ∈ R : x = y (new!)
x ∈ R : ∀ y ∈ R : x = y (new!)
Solution: (a) True (Examples: x ∈ {0, 1}p⊆ 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.
Math 3000-003 Intro to Abstract Math Homework 2, UC Denver, Spring 2012 (Solutions)
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(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 L1 : y = ax+p and L2 : y = bx+q be two lines in the plane. Then L1 and L2 are
perpendicular, denoted by L1 ⊥ L2 , 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 L1 : ax + by = p and L2 : cx + dy = q be two
lines in the plane. Then L1 ⊥ L2 if and only if ac + bd = 0. (v) Let p, q, r, s ∈ R2 be real
two-dimensional vectors and L1 : (x, y) = p + tr and L2 : (x, y) = q + ts be two lines in the
plane parametrized by the real scalar t ∈ (−∞, ∞). Then L1 ⊥ L2 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 2
12. 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 x2 = 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.