Download Problem Set 2 Solutions

Problem Set 2 Solutions
Section 2.2
1. Write the converse and contrapositive of each of the following conditional statements.
(a) If a = 5, then a2 = 25.
Converse: If a2 = 25 then a = 5.
Contrapositive: If a2 6= 25 then a 6= 5.
(b) If it is not raining, then Laura is playing golf.
Converse: If Laura is playing golf then it is not raining.
Contrapositive: If Laura is not playing golf then it is raining.
(c) If a 6= b, then a4 6= b4 .
Converse: If a4 6= b4 then a 6= b.
Contrapositive: If a4 = b4 then a = b.
(d) If a is an odd integer, then 3a is an odd integer.
Converse: If 3a is an odd integer then a is an odd integer.
Contrapositive: If 3a is an even integer then a is an even integer.
3. Write a useful negation of each of the following statements. Do not leave a negation as a prefix
of a statement. For example, we would write the negation of “I will play golf and I will mow
the lawn” as “I will not play golf or I will not mow the lawn.”
(a) We will win the first game and we will win the second game.
We will lose the first game or we will lose the second game.
(b) They will lose the first game or they will lose the second game.
They will win the first game and they will win the second game.
(c) If you mow the lawn, then I will pay you $20.
If you mow the lawn then I will not pay you $20.
(d) If we do not win the first game, then we will not play a second game.
If we do not win the first game then we will play a second game
(e) I will wash the car or I will mow the lawn.
I will not wash the car and I will not mow the lawn,
(f) If you graduate from college, then you will get a job or you will go to graduate school.
If you graduate from college, you will not get a job and will not go to graduate school.
(g) If I play tennis, then I will wash the car or I will do the dishes.
If I play tennis, then I will not wash the car and I will not do the dishes.
(h) If you clean your room or do the dishes, then you can go to see a movie.
If you clean the room or do your dishes, then you cannot go to see a movie.
(i) It is warm outside and if it does not rain, then I will play golf.
It is not warm outside, or it does not rain and I will not play golf.
4. Use truth tables to establish each of the following logical equivalencies dealing with biconditional
statements:
(a) (P ↔ Q) ≡ (P → Q) ∧ (Q → P )
P
T
T
F
F
Q
T
F
T
F
P ↔Q
T
F
F
T
P →Q
T
F
T
T
Q→P
T
T
F
T
(P → Q) ∧ Q → P
T
F
F
T
Noting that columns 3 and 6 are the same gives that (P ↔ Q) ≡ (P → Q) ∧ (Q → P )
(b) (P ↔ Q) ≡ (Q ↔ P )
P
T
T
F
F
Q
T
F
T
F
P ↔Q
T
F
F
T
Q↔P
T
F
F
T
Notice columns 3 and 4 are the same, verifying that (P ↔ Q) ≡ (Q ↔ P )
(c) (P ↔ Q) ≡ (¬P ↔ ¬Q)
P
T
T
F
F
Q
T
F
T
F
P ↔Q
T
F
F
T
¬P
F
F
T
T
¬Q
F
T
F
T
¬P ↔ ¬Q
T
F
F
T
Noting that columns 3 and 6 are the same gives that (P ↔ Q) ≡ (¬P ↔ ¬Q).
9. Use previously proven logical equivalencies to prove each of the following logical equivalencies:
(a) [¬P → (Q ∧ ¬Q)] ≡ P
¬P → (Q ∧ ¬Q)
≡ P ∨ (Q ∧ ¬Q)
≡P
Conditional Statement
Impossibility of Q ∧ ¬Q
(b) (P ↔ Q) ≡ (¬P ∨ Q) ∧ (¬Q ∨ P )
P ↔Q
≡ (P → Q) ∧ (Q → P )
≡ (¬P ∧ Q) ∧ (¬Q ∧ P )
Biconditional Statement
Conditional Statement
(c) ¬(P ↔ Q) ≡ (P ∧ ¬Q) ∨ (Q ∧ ¬P )
¬(P ↔ Q)
≡ ¬([P → Q) ∧ (Q → P )]
≡ ¬(P → Q) ∨ ¬(Q → P )
≡ (P ∧ ¬Q) ∨ (Q ∧ ¬P )
(d) (P → Q) → R ≡ (P ∧ ¬Q) ∨ R
Biconditional Statement
DeMorgan’s Law
Conditional Statement
(P → Q) → R
≡ ¬(P → Q) ∨ R
≡ (P ∧ ¬Q) ∨ R
Conditional Statement
Conditional Statement
(e) (P → Q) → R ≡ (¬P → R) ∧ (Q → R)
(P → Q) → R
≡ (¬P ∨ Q) → R
≡ (¬P → R) ∧ (Q → R)
Conditional Statement
Conditional with Disjunction
(f) [(P ∧ Q) → (R ∨ S)] ≡ [(¬R ∧ ¬S) → (¬P ∨ ¬Q)]
(P ∧ Q) → (R ∨ S)
≡ ¬(R ∨ S) → ¬(P ∧ Q)
≡ (¬R ∧ ¬S) → (¬P ∨ ¬Q)
Contrapositive
DeMorgan’s Law
(g) [(P ∧ Q) → (R ∨ S)] ≡ [(P ∧ Q ∧ ¬R) → S]
(P ∧ Q) → (R ∨ S)
≡ (P ∧ Q ∧ ¬R) → S
Conditional with Disjunction
(h) [(P ∧ Q) → (R ∨ S)] ≡ (¬P ∨ ¬Q ∨ R ∨ S)
(P ∧ Q) → (R ∨ S)
≡ (P ∧ Q ∧ ¬R) → S
≡ ¬(P ∧ Q ∧ ¬R) ∨ S
≡ ¬P ∨ ¬Q ∨ R ∨ S
Conditional with Disjunction
Conditional Statement
DeMorgan’s Law
(i) ¬[(P ∧ Q) → (R ∨ S)] ≡ (P ∧ Q ∧ ¬R ∧ ¬S)
¬[(P ∧ Q) → (R ∨ S)]
≡ (P ∧ Q) ∧ ¬(R ∨ S)
≡ P ∧ Q ∧ ¬R ∧ ¬S
Conditional Statement
DeMorgan’s Law
12. Let x be a real number, Consider the following conditional statement:
If x3 − x = 2x2 + 6, then x = −2 or x = 3.
Which of the following statements have the same meaning as the conditional statement and
which ones are negations of the conditional statement? Explain each conclusion.
The original statement is of the form P → (Q ∨ R).
(a) If x 6= −2 and x 6= 3, then x3 − x 6= 2x2 + 6.
(¬Q ∧ ¬R) → ¬P
This is the contrapositive, so it is logically equivalent.
(b) If x = −2 or x = 3, then x3 − x = 2x2 + 6.
(Q ∨ R) → P
This is the converse of the statement, so it is not equivalent nor is it the negation of the
original statement.
(c) If x 6= −2 or x 6= 3, then x3 − x 6= 2x2 + 6.
(¬Q ∨ ¬R) → ¬P
This is neither the negation of the statement nor an equivalent statement.
(d) If x3 − x = 2x2 + 6 and x 6= −2, then x = 3.
(P ∧ ¬Q) → R
This is equivalent to the original statement.
(e) If x3 − x = 2x2 + 6 or x 6= −2, then x 6= 3.
(P ∨ ¬Q) → ¬R
This is neither the negation nor an equivalent statement.
(f) x3 − x = 2x2 + 6x, x 6= −2, and x 6= 3.
P ∧ (¬Q ∧ ¬R)
This is the negation of the original statement.
(g) x3 − x 6= 2x2 + 6x or x = −2 or x = 3.
¬P ∨ (Q ∨ R)
This is equivalent to the original statement.
Section 2.3
1. Use the roster method to specify the elements in each of the following sets and then write a
sentence in English describing the set.
2
(a) {x
1 ∈ R| 2x + 3x − 2 = 0}
2 , −2
(b) {x ∈ Z | 2x2 + 3x − 2 = 0}
{−2}
(c) {x ∈ Z | x2 < 25}
{0, ±1, ±2, ±3, ±4}
(d) {x ∈ N | x2 < 25}
{1, 2, 3, 4}
(e) {y
∈1 Q
| |y − 2| = 2.5}
− 2 , 29
(f) {y ∈ Z | |y − 2| ≤ 2.5}
{0, 1, 2, 3, 4}
5. Use set builder notation to specify the following sets:
(a) The set of all integers greater than or equal to 5.
{x ∈ Z| x ≥ 5}
(b) The set of all even integers.
{x ∈ Z | x is even}
(c) The set of all positive rational numbers.
{x ∈ Q | x > 0}
(d) The set of all real numbers greater than 1 and less than 7.
{x ∈ R | 1 < x < 7}
(e) The set of all real numbers whose square is greater than 10.
{x ∈ R | x2 > 10}
6. For each of the following sets, use English to describe the set and when appropriate, use the
roster method to specify all of the elements of the set.
(a) {x ∈ R | − 3 ≤ x ≤ 5}
The set of all reals at least as large as -3 but no larger than 5.
(b) {x ∈ Z | − 3 ≤ x ≤ 5}
The set of all integers at least as large as -3 but no larger than 5.
{−3, −2, −1, 0, 1, 2, 3, 4, 5}
(c) {x ∈ R | x2 = 16}
The set of all reals such that x2 = 16
{−4, 4}
(d) {x ∈ R | x2 + 16 = 0}
The set of all reals such that x2 = −16
{} or ∅
(e) {x ∈ Z | x is odd}
The set of all odd integers.
(f) {x ∈ R | 3x − 4 ≥ 17}
The set of all reals such that x ≥ 7
Section 2.4
1. For each of the following, write the statement as an English sentence and then explain why the
statement is false.
(a) (∃x ∈ Q)(x2 − 3x − 7 = 0).
There exists a rational number x such that x2 − 3x − 7 = 0.
The statement is false since the roots to this quadratic equation are x =
are irrational.
√
3± 37
,
2
which
(b) (∃x ∈ R)(x2 + 1 = 0).
There exists a real number x such that x2 = −1.
√
This is false because the roots of this quadratic equation are x = ± −1 = ±i, which
are imaginary.
(c) (∃m ∈ N)(m2 < 1).
There exists a natural number m such that m2 < 1.
This is false because there are no natural numbers smaller than 1 and 12 6< 1.
2. For each of the following, use a counterexample to show that the statement is false. Then write
the negation of the statement in English, without using symbols for quantifiers.
(a) (∀m ∈ Z)(m2 ) is even.
Counterexample: any odd integer, since the square of an odd integer is odd.
Negation: There exists an integer m such that m2 is odd.
(b) (∀x ∈ R)(x2 > 0).
Counterexample: The only counterexample is x = 0.
Negation: There exists a real number x such that x2 ≤ 0.
√
(c) For each real number x, x ∈ R.
√
Counterexample: x = 2, giving 2 6∈ R.
√
Negation: There exists a real number x such that x is not in the set of real numbers.
(d) (∀ m ∈ Z) m
3 ∈Z .
Counterexample: m = 2 gives 32 6∈ Z.
√
(e) (∀a ∈ Z)( a2 = a).
√
Counterexample: a = −1 There exists an integer a such that a2 6= a is not in the set
of integers.
(f) (∀x ∈ R)(tan2 x + 1 = sec2 x).
Counterexample: x = π2 since tangent is undefined there because cos
There exists a real number x such that tan2 x + 1 6= sec2 x
π
2
= 0.
3. For each of the following statements
• Write the statement as an English sentence that does not use the symbols for quantifiers.
• Write the negation of the statement in symbolic form in which the negation symbol is
not used.
• Write a useful negation of the statement in an English sentence that does not use the
symbols for quantifiers.
√ (a) (∃x ∈ Q) x > 2 .
√
There exists a √
rational number x such that x > 2.
(∀x ∈ Q)(x ≤ 2)
√
For each rational number x, x ≤ 2.
(b) (∀x ∈ Q)(x2 − 2 6= 0).
For all rational numbers x, x2 − 2 6= 0.
(∃x ∈ Q)(x2 − 2 = 0)
There exists a rational number x such that x2 − 2 = 0.
(c) (∀x ∈ Z)(x is even or x is odd).
For each integer x, x is even or x is odd.
(∃x ∈ Z)(x is odd and x is even)
There exists an integer x such that x is odd and x is even.
√
√ (d) (∃x ∈ Q) 2 < x < 3 .
√
√
There exists a √
rational number
x such that 2 < x < 3.
√
(∀x ∈ Q)(x ≤ 2 or x ≥ 3) √
√
For all rational numbers x, x ≤ 2 or x ≥ 3.
(e) (∀x ∈ Z)(If x2 is odd, then x is odd).
For all integers x, if x2 is odd then x is odd.
(∃x ∈ Z)(x2 is odd and x is even)
There exists an integer x such that x2 is odd and x is even.
(f) (∀n ∈ N)[If n is a perfect square, then (2n − 1) is not a prime number].
For all natural numbers n, if n is a perfect square, then (2n − 1) is not a prime number.
(∃n ∈ N)(n is a perfect square and (2n − 1) is a prime number)
There exists a natural number n such that n is a perfect square and (2n − 1) is a prime
number.
(g) (∀n ∈ N)(n2 − n + 41 is a prime number).
For all natural numbers n, n2 − n + 41 is a prime number.
(∃n ∈ N)[(n2 − n + 41) is not a prime number] There exists a natural number n such
that n2 − n + 41 is not a prime number.
(h) (∃x ∈ R)(cos(2x) = 2cos x).
There exists a real number x such that cos(2x) = 2cos x.
(∀x ∈ R)(cos(2x) 6= 2cos x)
For each real number x, cos(2x) 6= 2cos x.
4. Write each of the following statements as an English sentence that does not use the symbols
for quantifiers.
(a) (∃ m ∈ Z)(∃ n ∈ Z)(m > n)
There exists integers m and n such that m > n.
(b) (∃ m ∈ Z)(∀ n ∈ Z)(m > n)
There exists an integer m such that for all integers n, m > n.
(c) (∀ m ∈ Z)(∃ n ∈ Z)(m > n)
For each integer m, there exists an integer n such that m > n.
(d) (∀ m ∈ Z)(∀ n ∈ Z)(m > n)
For each integer m and for each integer n, m > n.
(e) (∃ n ∈ Z)(∀ m ∈ Z)(m2 > n)
There exists an integer n such that for each integer m, m2 > n.
(f) (∀ n ∈ Z)(∃ m ∈ Z)(m2 > n)
For each integer n, there exists an integer m such that m2 > n.
9. An integer m is said to have the divides property provided that for all integers a and b, if m
divides ab, then m divides a or m divides b.
(a) Using the symbols for quantifiers, write what it means to say that the integer m has the
divides property.
(∃m ∈ Z)(m|ab → (m|a ∨ m|b))
(b) Using the symbols for quantifiers, write what it means to say that the integer m does
not have the divides property.
(∃m ∈ Z)(m 6 |ab → (m 6 |a ∧ m 6 |b))
(c) Write an English sentence stating what it means to say that the integer m does not have
the divides property.
For an integer m, m does not divide ab if m does not divide a and m does not divide b.