1. Questions for Discrete
Chapter 1

2. Problem 1.1-9 Let p  Swimming at the New Jersey Shore is allowed
Let q  Sharks have been spotted near the shore.
(c) Express  p  q in English
(d) Express p   q in English

3. Problem
Let p and q be the propositions given below. Express each of the following propositions using p, q, and logical connectives
p: It is below freezing
q: It is snowing
(b) It is below freezing but not snowing

4. Problem 1.1-15 Let p, q, and r be the propositions
p: Grizzly bears have been seen in this area
q: Hiking is safe on this trail
r: Berries are ripe along the trail
(b) Grizzly bears have not been seen in the area and hiking on the trail is safe, but berries are ripe along the trail.
(f) Hiking is not safe on the trail whenever grizzly bears have been seen in the area and berries are ripe along the trail.

5. Problem d
Determine if the following conditional statements are true.
If monkeys can fly, then =3

6. Problem (d)
Construct a truth table for the following compound propositon:
(d) (p  q)  (p  q)

7. p  q q  p p   q  p  q None of the above
1.1 Let p denote proposition “It is above freezing” and q denote “It is snowing”. Which below compound proposition represents “If it is above freezing, it is not snowing”
p  q
q  p
p   q
 p  q
None of the above

8. If it does not snow today, I will not ski tomorrow
1.2 State the contrapositive of the following: If it snows today, I will ski tomorrow
If it does not snow today, I will not ski tomorrow
If I do not ski tomorrow, it will not snow today.
If I do not ski tomorrow, it will snow today
If I do not ski tomorrow, it will not snow today
If I ski tomorrow, it will snow today
None of the above

9. Experience with Java or C++ is required Lunch includes soup or salad
1.3 Which of the following sentences involve the use of the inclusive OR (as opposed to the exclusive OR)
Experience with Java or C++ is required
Lunch includes soup or salad
To enter the country, you need a passport or a voter registration card
Publish or perish
None of the above involve the use of the exclusive OR.

10. Problem
Use a truth table to verify the following absorption law: p  (p  q)  p

11. 1.4 The statement “p is logically equivalent to q” means which of the following:
p is a model of q
pq is satisfiable
p  q is satisfiable
p  q is a tautology

12. 1.5 The statement “a  b  b  a” is called the
Associative law of 
Commutative law of 
The tautology of 
Satisfiability property of 

13. 1.6 The set {p, p} of wffs is
A model
A tautology
Unsatisfiable
De Morgan’s Law

14. 1.7 The statement “(ab)  a  b” is
Invalid
Absorption Law
DeMorgan’s law
Contraposition rule

15. Sect1. 3: May need paper & pen on following question
Sect1.3: May need paper & pen on following question. Let R denote “Jan is rich” and H denote “Jan is happy”. Which below is the negation of “Jan is rich and happy”?
R  H
 R  H
(R  H)
None of the above

16. Sect1. 3: Determine which of following is true for p(pq)  p
Sect1.3: Determine which of following is true for p(pq)  p. Use paper to calculate, if needed
p(pq)  p
Not (1), but p(pq)  p is satisfiable
p (pq)  p is not satisfiable.
None of the above

17. Let P(x) be “x has a cellular phone” and “C(x) be “x is in your class”
Let P(x) be “x has a cellular phone” and “C(x) be “x is in your class”. Which of the following is equivalent to “Everyone in your class has a cellular phone”?
x [P(x)  C(x)]
x[C(x)  P(x)]
x[C(x)  P(x)]
x[C(x)  P(x)]
Either “none” or “more than one” of the above”.

18. Let C(x) be “x is a comedian” and F(x) be “x is funny”
Let C(x) be “x is a comedian” and F(x) be “x is funny”. Which of below is a translation of “x[C(x)  F(x)]”
Some comedians are funny
Everyone is a comedian and is funny”
All comedians are funny
There is a comedian who is funny
Either “none” or “more than one” of above

19. Which of the following is a correct translation of “At least one of your friends are perfect”?
x[F(x)  P(x)]
x[F(x)  P(x)]
x[F(x)  P(x)]
x[F(x)  P(x)]
Either “none” or “more than one” of the above

20. Which of the following is the negation of “There is no dog that can talk”.
All dogs can talk
Some dogs can talk
Some dogs can’t talk
Some dogs can talk and some can’t
Either none or more then one of the above is true.

21. None or more than 1 of the above
Which of the below is a counterexample to the following in the domain of real numbers: x[(x<0)  (x2 ≠ 2)] 2 -2
-2 2
None or more than 1 of the above

22. This statement is only true if x ≥ 1. This statement is always true
(Problem 35 in 1.4): Which of the below choices are true for the following universally quantified statement: x (x2 ≥ x)
This statement is only true if x ≥ 1.
This statement is always true
Zero is a counterexample to this statement
The value x=-1 is a counterexample
Either none or more than 1 of the above.

23. (#37 in 1.4) Let E(x) denote “x is an elite flyer”, M(x) denote “x flies more than 25,000 miles each year” F(x) denote “x takes more than 25 flights each year”. Which of below statements correctly represents the following: “A passenger on an airline Is an elite flyer if they fly more than 25,000 miles or take more than 25 flights during the year
x [M(x)  F(x)  E(x)]
x [E(x)  M(x)  F(x) ]
x [M(x)  F(x)  E(x)]
x [E(x)  M(x)  F(x)]
None or more than one of above.

24. (Problem 61 in 1.4) Given the below premises, does (d) follow from (a-c)?
Babies are illogical
x[B(x)  L(x)]
Nobody is despised who can manage a crocodile
If a person can manage a crocodile are not despised
x[ M(x)   D(x)]
Illogical persons are despised
x[  L(x)  D(x)]
Babies cannot manage crocodiles
x[B(x)  M(x)]
Conclusions:
If x is a baby then by (a), x is illogical by the first premise
Then by the third premise, x is despised
By the second premise, if x could mange a crocodile,then x would not be depised.
By 2 and 3 above, x cannot manage a crocodile, which establishes (d) is true.

25. Prob Which of below is the correct translation for the following wff for the domain of real numbers: x y (x<y)
There is a largest real number
There is no largest real number
There is a smallest real number
There is no smallest real number
None of the above

26. xyP(x,y) where P(x,y) is “x has taken y”
Prob : Which of the below is a correct translation of the following sentence: Every student in this class has taken at least one computer science class. ASSUME: The domain of students are those in this class and the domain of classes are the computer science courses.
xyP(x,y) where P(x,y) is “x has taken y”
xyP(x,y) where P(x,y) is “x has taken y”.
xyP(x,y) where P(x,y) is “x has taken y”.
xyP(x,y) where P(x,y) is “x has taken y”
None or more than one of the above

27. None or more than one of the above
Prob Which of the following formulas express the statement: A negative number does not have a square root that is a real number. The domain is all real numbers
x((x<0)  y(x2=y)
x((x<0)  y(x=y2)
x((x<0)  y(x=y2)
x((x<0)  y(x2=y)
None or more than one of the above

28. 1.5-31 What is the negation of the following formula (after simplifying): xy[P(x,y) Q(x,y)]
None of the above

29. If the domain is all integers, find a counterexample for the following expression: xy(y2 = x)
None of the above or or than 1 of above is a counterexample.

30. 1.6-3 What rule of inference is used in the following” If it snows today, the university will close. The university is not closed today. Therefore it has not closed today.
Modus ponens
Disjunctive syllogism
Resolution
Modus tollens
None or else more than one of the above.

31. 1.6.37(a) Express the following statement using quantifiers: Every student in this class has taken exactly two mathematics courses at this school. (b) Next, form the negation of this statement so that no negation is to the left of a quantifier and then simplify it into a simple English statement.
Let T(x,y) be the statement that x has taken y. Here, x ranges over students in this class and y ranges over mathematics classes at this school.
One formal representation of original sentence is xyz[y≠z T(x,y)  T(y,z) w{T(x,w)  (w=y  w=z)}]
Negation of formal representation is xyz[y≠z (T(x,y) T(y,z) w{T(x,w)  w≠y  w≠z)}]
There is someone in this class for whom no matter which two distinct math courses you consider, these are not [the two and only two math courses this person has taken].

32. Identify the rule of inference is used in the following argument: It is either hotter than 100 degrees today or the pollution is dangerous. It is less than less than 100 degrees outside today. Therefore the pollution is dangerous!
Hypothetical Syllogism
Conjunction
Disjunctive Syllogism
Modus tolens
resolution

33. Disjunctive Syllogism Resolution None of the above
1.6 What rule of inference is used in this argument? No man is an island. Manhattan is an island. Therefore, Manhattan is not a man.1
Modus Ponens
Modus Tollens
Disjunctive Syllogism
Resolution
None of the above

34. 1.6 What conclusions can be drawn from the following premises: What is good for corporations is good for the United States. What is good for the United States is good for you. What is good for corporations is for you to buy lots of stuff
It is good for the U.S. for you to buy lots of stuff
It is good for you to buy lots of stuff
It is bad for you to buy lots of stuff
It is bad for the U.S. for you to buy lots of stuff
Both (1) and (2)

35. All steps are justified
Assuming “x P(x)  x Q(x)” as a premise, list the first step in the following derivation which is not justified by a rule of inference. If all steps are correct, choose (7) below.
x P(x)
P(c)
x Q(x)
Q(c)
P(c)  Q(c)
x[P(x)  Q(x)]
All steps are justified

36. 1. 7. 5 Suppose m+n and n+p are even
1.7.5 Suppose m+n and n+p are even. Consider the first step incorrect step in the following proposed proof that m+p is even. If proposed proof is correct, choose (7) below.
Assume that m+n is even.
Then m+n =2s for some integer s.
Assume that n+p is even.s
Then m+p = 2t for some integer t.
(m+n)+(n+p) = 2s + 2t
m+p = 2s +2t -2n = 2(s+t-n) is even
All steps in proposed proof are correct

37. Let r be rational and s be irrational. Assume that r+s is rational
Problem : Using a proof by contradiction, that the sum of a rational number and an irrational number is irrational.
Let r be rational and s be irrational.
Assume that r+s is rational
Then we can express r+s as a fraction a/b, where a and b are both integers and b≠0.
Since r is rational, it also can be expressed as a fraction c/d with d≠0.
Since s = (r+s) – r = a/b + c/d = (ad +cb)/bd, it follows that s is rational .
This contradicts the fact that s is irrational
It follows that r+s is irrational.

38. 1.7.15 Use a proof by contraposition to show that if x+y ≥ 2, then x ≥ 1 or y ≥ 1.
To prove by contraposition, we must show that “if x<1 and y< 1, then x+y < 2”.
But if x<1 and y<1, then x + y < 1+1 =2.

39. A proof by contraposition
Show that if n is an integer and n3 + 5 is odd, then n is even using
A proof by contraposition
Must show that if n is an odd integer, then n3 + 5 is even
But if n is odd, then n = 2p +1 for some integer p.
Then n3 + 5 = (2p+1)3 + 5 =(8p3 + 12p2 + 6p +1) +5
= 8p3 + 12p2 + 6 = 2(4p3 + 6p2 + 3)
This shows n3 + 5 is even
A proof by contradiction
Assume that n is odd and show a contradiction occurs

40. 1.7.33 Show that the following statements are equivalent:
x is irrational
3x + is irrational
x/2 is irrational
METHOD: Show (1)  (2)  (3)  (1)

41. 1.7.35 Are the solution steps shown correct.
The steps shown are correct, with one exception.
Since both sides are squared at one step, this will method will produce two roots.
However, if you check these two roots in original equation, one of them satisfy original solution and and the other does not satisfy the original solution.

42. Problem 1.8.1: Prove that n2 +1 ≥ 2n when 1≤n≤4
What proof technique should be used here?
Does this proof technique provide a solution to this problem?

43. Problem 1.8.7: If x and y are real numbers, prove that the max(x,y) + min(x,y) = x + y.
Case 1: x ≥ y
y = max(x,y) and x = max(x,y) so x + y = max(x,y) + min(x,y).
Case 2: x < y
Similar

44. Problem1.8.7: Define |x| by |x| = x if x ≥ 0 and |x| = -x if x < 0. Prove the triangle inequality which states that if x and y are real numbers, then |x| + |y| ≥ |x + y|.
Comment – Solution takes a lot of time unless break up into special cases efficiently.

45. Problem : Prove that given a real number x, there exists unique numbers n and  such that x = n -  and 0 ≤  < 1.
If x is an integer, let n = x and  = 0. Clearly x = n -  and 0≤ <1. No other solution is possible in this case since if the integer n is greater than x, then n is at least x+1, which would make ≥1.
If x is not an integer, let n = x and  = n - x. Clearly 0 ≤  < 1 since  = n – x = x - x ≥ 0. This is the only  that will work with this n, and n cannot be larger since  is constrained to be less than 1.
COMMENTS:
The colored print above is a way of summarizing the required proof.
A more formal proof in each case could be given by assuming that n’ and ’ also satisfied the conditions (x = n’ - ’ and 0 ≤ ’ < 1) and arguing it follows that n = n’ and  = ’.
Clearly the summary statement in colored print provides a shorter argument.