All of Andy's Assignment's shuffled

Question 1

Identify the correct steps involved while proving (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) is true when p, q, and r have the same truth value and it is false otherwise

Answer 1

1. Suppose p, q, and r are all true. Then, (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) is true as each clause has an unnegated variable.
2. Suppose p, q, and r are all true. Then, (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) is false as each clause has an unnegated variable.
3. Now, suppose one of the variables p, q, or r is true and the other two are false. Then, the clause that contains the negation of the true variable will be true.
4. We get a similar conclusion when one of the variables is false and the other two are true.
5. Hence, (p ∨ ¬q) ∧ (q ∨ ¬r) ∧ (r ∨ ¬p) is true when p, q, and r have the same truth value and it is false otherwise.

Question 2

Is the conditional statement ¬(p → q)→ ¬q tautology?

p q p → q ¬(p → q) ¬q ¬(p → q) → ¬q
T F F
T F F T
F T F
F F F

Answer 2

p q p → q ¬(p → q) ¬q ¬(p → q) → ¬q
T T T F F T
T F F T T T
F T T F F T
F F T F T T

Question 3

Let p, q, and r be the propositions

p: You have the flu.
q: You miss the final examination.
r: You pass the course.

Identify the English translation of the compound proposition (p → ¬r) ∨ (q → ¬r).

Answer 3

If you have the flu then you will not pass the course, or if you miss the final exam then you will not pass the course.

Question 4

Four friends have been identified as suspects for unauthorized access into a computer system. They have made statements to the investigating authorities. Alice said, “Carlos did it.” John said, “I did not do it.” Carlos said, “Diana did it.” Diana said, “Carlos lied when he said that I did it.”

If the authorities know that exactly one of the four suspects is telling the truth, who did it & why?

Answer 4

John

John did it. There are four cases to consider:

If Alice is the sole truth-teller, then Carlos did it, but this means that John is telling the truth, a contradiction.

If John is the sole truth-teller, then Diana must be lying, so she did it, but then Carlos is telling the truth, a contradiction.

If Carlos is the sole truth-teller, then Diana did it, but that makes John truthful, again a contradiction.

So, the only possibility is that Diana is the sole truth-teller. This means that John is lying when he denied it, so he did it.

Question 5

Four friends have been identified as suspects for unauthorized access into a computer system. They have made statements to the investigating authorities. Alice said, “Carlos did it.” John said, “I did not do it.” Carlos said, “Diana did it.” Diana said, “Carlos lied when he said that I did it.”

If the authorities know that exactly one is lying, who did it?

Answer 5

Carlos

Since Carlos and Diana are making contradictory statements, the liar must be one of them. Therefore, Alice is telling the truth, so Carlos did it. John and Diana are telling the truth as well here, and it is Carlos who is lying.

Question 6

Use De Morgan’s laws to find the negation of the given statement.

Rita will not move to Oregon and will not move to Washington.

Answer 6

The De Morgan’s Law used is ¬(p ∨ q) ≡ ¬p Λ ¬q.

Question 7

Determine whether the given compound propositions is satisfiable.
(p ∨ q ∨ ¬r) ∧ (p ∨ ¬q ∨ ¬s) ∧ (p ∨ ¬r ∨ ¬s) ∧ (¬p ∨ ¬q ∨ ¬s) ∧ (p ∨ q ∨ ¬s)

The compound proposition (p ∨ q ∨ ¬r) ∧ (p ∨ ¬q ∨ ¬s) ∧ (p ∨ ¬r ∨ ¬s) ∧ (¬p ∨ ¬q ∨ ¬s) ∧ (p ∨ q ∨ ¬s) is ______________

Answer 7

not satisfiable

Since p occurs in four of the five disjunctions, we can make p true, and then make q false. Thus, the given proposition is satisfiable

Question 8

Consider the following propositions:

L: The file system is locked.
Q: New messages will be queued.
N: The system is functioning normally.
B: New messages will be sent to the message buffer.

The goal of this exercise is to determine whether the following system specifications are consistent:

• If the file system is not locked, then new messages will be queued.
• If the file system is not locked, then the system is functioning normally, and conversely.
• If new messages are not queued, then they will be sent to the message buffer.
• If the file system is not locked, then new messages will be sent to the message buffer.
• New messages will not be sent to the message buffer.

What are the symbolic representations of the given system specifications?

Answer 8

¬L → Q, ¬L ↔ N, ¬Q → B, ¬L → B, and ¬B, respectively.

The first specification is “If the file system is not locked, then new messages will be queued.” This is represented symbolically as ¬L → Q.

The second specification is “If the file system is not locked, then the system is functioning normally, and conversely.” This is represented symbolically as ¬L ↔ N.

The third specification is “If new messages are not queued, then they will be sent to the message buffer.” This is represented symbolically as ¬Q → B.

The fourth specification is “If the file system is not locked, then new messages will be sent to the message buffer.” This is represented symbolically as ¬L → B.

The fifth specification is “New messages will not be sent to the message buffer.” This is represented symbolically as ¬B.

Question 9

Identify the propositions in the form “p if and only if q” in English.
If you watch television your mind will decay, and conversely. (Check all that apply.)

Answer 9

• Your mind will decay if and only if you watch television.
• Your mind will not decay if and only if you watch television.
• Your mind will decay if and only if you do not watch television.
• Your mind will not decay if and only if you do not watch television.

The statement “If you watch television your mind will decay, and conversely” is translated into the form “if p, then q” as “Your mind will decay if and only if you watch television,” and “Your mind will not decay if and only if you do not watch television.”

Question 10

Complete the truth table given below for the given proposition.
p ⊕ (p ∨ q)

Answer 10

p	 q	(p ∨ q)	p ⊕ (p ∨ q)
T	 T	    T	                F
T	 F	    T	                F
F	 T	    T	                T
F	 F	    F	                F

Question 11

Show that each of these conditional statements is a tautology by using truth tables.
Complete the truth table given below for the conditional statement [(p ∨ q) ∧ (p → r) ∧ (q → r)] → r.

Is the given conditional statement is a tautology?

Answer 11

p q r p ∨ q p → r (p ∨ q) ∧ (p → r) q → r
T T T T T T T
T T F T F F F
T F T T T T T
T F F T F F T
F T T T T T T
F T F T T T F
F F T F T F T
F F F F T F T

(p ∨ q) ∧ (p → r) ∧ (q → r) [(p ∨ q) ∧ (p → r) ∧ (q → r)] → r
T T
F T
T T
F T
T T
F T
F T
F T

Question 12

Let p, q, r be the propositions:

p: The user enters a valid password.
q: Access is granted.
r: The user has paid the subscription fee.

Express the following system specifications using the propositions p, q, and r and logical connectives (including negations).

Answer 12

Access is denied if the user has not paid the subscription fee.
¬r → ¬q

“The user has not paid the subscription fee” is the negation of r; this is the hypothesis. “Access is denied” is the negation of q; this is the conclusion. Therefore, the given system specification can be expressed as ¬r → ¬q.

Question 13

Let p, q, r be the propositions:

p: The user enters a valid password.
q: Access is granted.
r: The user has paid the subscription fee.

Express the following system specifications using the propositions p, q, and r and logical connectives (including negations).

Answer 13

If the user has not entered a valid password but has paid the subscription fee, then access is granted.
(¬p ∧ r) → q

The hypothesis is a conjunction of “The user has not entered a valid password” (¬p) and “The user has paid the subscription fee” (r). The conclusion is “Access is granted” (q). Therefore, the given system specification can be expressed as (¬p ∧ r) → q.

Question 14

Steps in the correct order to show that ¬p → (q → r) and q → (p ∨ r) are logically equivalent.

(Note: While proving, prove the equivalence from ¬p → (q → r) to q → (p ∨ r).)

Answer 14

First, consider ¬p → (q → r) ≡ p ∨ (q → r)

                                           ≡ p ∨ ¬q ∨ r

                                           ≡ ¬q ∨ p ∨ r

                                           ≡ q → (p ∨ r)

So, ¬p → (q → r) and q → (p ∨ r) are equivalent.

Question 15

An explorer is captured by a group of cannibals. There are two types of cannibals—those who always tell the truth and those who always lie. The cannibals will barbecue the explorer unless he can determine whether a particular cannibal always lies or always tells the truth. He is allowed to ask the cannibal exactly one question.

Why does the question “Are you a liar?” not work?

Answer 15

Both types of cannibals will answer “no.”

If the explorer encounters a truth-teller, then he will honestly answer “no” to her question. If she encounters a liar, then the honest answer to her question is “yes,” so he will lie and answer “no.” Thus everybody will answer “no” to the question, and the explorer will have no way to determine which type of cannibal she is speaking to.

Question 16

An explorer is captured by a group of cannibals. There are two types of cannibals—those who always tell the truth and those who always lie. The cannibals will barbecue the explorer unless he can determine whether a particular cannibal always lies or always tells the truth. He is allowed to ask the cannibal exactly one question.

Which of the following questions helps in determining whether the cannibal she is speaking to is a truth-teller or a liar?

Answer 16

• Is the colour of the sky blue?
• If I were to ask you if you always told the truth, would you say that you did?
• If I say that you are a liar, would I be correct?
• Do you always tell the truth?

There are several possible correct answers. One is the following question: “If I were to ask you if you always told the truth, would you say that you did?” Then if the cannibal is a truth teller, he will answer yes (truthfully), while if he is a liar, then, since in fact he would have said that he did tell the truth if questioned, he will now lie and answer no.

Question 17

Complete the truth table given below for the statement (¬p ∧ (p → q)) → ¬q.

The given statement is a ___________________

Answer 17

contingency

p q ¬p p → q ¬p ∧ (p → q) ¬q
T T F T F F
T F F F F T
F T T T T F
F F T T T T

(¬p ∧ (p → q)) → ¬q
                T
                T
                F
                T

Question 18

As a reward for saving his daughter from pirates, the King has given you the opportunity to win treasures hidden inside two of three trunks. The trunk that does not hold any treasure is empty. To win, you must select the correct trunks. The inscriptions on Trunks 1, 2, and 3 are “This trunk is empty,” “There is a treasure in Trunk 1,” and “There is a treasure in Trunk 2.” For each of the following statements, determine whether the Queen who never lies could state this, and if so, which two trunks contain the treasures.

The Queen can state, “Exactly two of the inscriptions are true.” T or F

If the Queen can state this, in which of the trunks are the treasures hidden?

Answer 18

True
Trunk 1 and Trunk 2

The Queen can say this, but one cannot determine the location of the treasures. If the inscription on Trunk 1 is false, then the treasures are in Trunks 1 and 2. If the inscription on Trunk 2 is false, then the treasures are in Trunks 2 and 3. The inscription on Trunk 3 cannot be false, since the inscriptions on Trunks 1 and 2 are contradictory.

Question 19

Identify the correct steps involved in proving p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) are logically equivalent

Answer 19

• If both p and q are false, then (p ∧ q) is false and (¬p ∧ ¬q) is true. This again implies that the second statement (p ∧ q) ∨ (¬p ∧ ¬q) is false.
• If p is true and q is false, then (p ∧ q) is false and (¬p ∧ ¬q) is true. This again implies that the second statement (p ∧ q) ∨ (¬p ∧ ¬q) is false.
• Thus, p ↔ q and (p ∧ q) ∨ (¬p ∧ ¬q) have same truth value; hence, they are logically equivalent.

Question 20

Write the given sentence in the “if p, then q” form.

You will reach the summit unless you begin your climb too late.

Answer 20

• If you do not begin your climb too late, then you will reach the summit.
• “q unless ¬p” is equivalent to “If p, then q.”

Question 21

Let R(x) is “x is a rabbit” and H(x) is “x hops,” and the domain consists of all animals.

Translate the statement ∃x(R(x) → H(x)) into English

Answer 21

• There exists an animal such that if it is a rabbit then it hops.
• Every animal that is a rabbit also hops.
• Every rabbit hops.

The existential quantifier states that the conditional statement applies to at least one animal. Thus, “There exists an animal such that if it is a rabbit, then it hops.”

Question 22

Identify the mistake, if there is one, in the following argument.

Let S(x, y) be “x is shorter than y.” Given the premise ∃sS(s, Max), it follows that S(Max, Max). Then, by existential generalization, it follows that ∃xS(x, x), so that someone is shorter than himself.

Answer 22

The argument is incorrect. We know that some s exists that makes S(s, Max) true, but we cannot conclude that Max is one such s.

We know that some s exists that makes S(s, Max) true, but we cannot conclude that Max is one such s. Therefore, this first step is invalid.

Question 23

Use the rules of inference to show that if ∀ x (P(x) ∨ Q(x)) and ∀ x ((¬P(x) ∧ Q(x)) → R(x)) are true, then ∀ x(¬R(x) → P(x)) is also true, where the domains of all quantifiers are the same.

1. We will show that if the premises are true, then (¬R(a) → P(a)) for every a.
2. Suppose ¬R(a) is true for some a.
3. For such an a, universal modus tollens applied to the second premise gives us ¬(¬P(a)˄Q(a)).

Answer 23

We want to show that the conditional statement ¬R(a) → P(a) is true for all a in the domain; the desired conclusion then follows by universal generalization. Thus we want to show that if ¬R(a) is true for a particular a, then P(a) is also true. For such an a, universal modus tollens applied to the second premise gives us ¬(¬P(a) ∧ Q(a)). By rules from propositional logic, this gives us P(a) ∨ ¬Q(a). By universal instantiation from the first premise, we have P(a) ∨ Q(a). Now by resolution, we can conclude P(a) ∨ P(a), which is logically equivalent to P(a), as desired.

Question 24

Let P(x, y) be the statement “Student x has taken class y,” where the domain for x consists of all students in your class and for y consists of all computer science courses at your school.

Match the following quantifications (in the left column) with their corresponding expressions (in the right column).

∃y∀xP(x, y)      
∀x∀yP(x, y)
∃x∃yP(x, y)
∃x∀yP(x, y)
∀y∃xP(x, y) 
∀x∃yP(x, y)

There is a student in your class who has taken every computer science course.

Some student in your class has taken some computer science course.

Every student in your class has taken at least one computer science course.

Every student in your class has taken every computer science course.

Every computer science course has been taken by at least one student in your class.

There is a computer science course that every student in your class has taken.

Answer 24

The expression for the quantification ∃x∃yP(x, y) is “Some student in your class has taken some computer science course.”

The expression for the quantification ∃x∀yP(x, y) is “There is a student in your class who has taken every computer science course.”

The expression for the quantification ∀x∃yP(x, y) is “Every student in your class has taken at least one computer science course.”

The expression for the quantification ∃y∀xP(x, y) is “There is a computer science course that every student in your class has taken.”

The expression for the quantification ∀y∃xP(x, y) is “Every computer science course has been taken by at least one student in your class.”

The expression for the quantification ∀x∀yP(x, y) is “Every student in your class has taken every computer science course.”

Question 25

Translate each of these nested quantifications into an English statement that expresses a mathematical fact. The domain in each case consists of all real numbers.

∀x∀y(((x ≥ 0) ∧ (y < 0)) → (x – y > 0))

Answer 25

For every nonnegative number, one can find a negative number so that the first number minus the second is positive.

One can find a nonnegative number so that for any positive number chosen, the first number minus the second is positive.

An English statement for the given quantifier is “A nonnegative number minus a negative number is positive.”

Question 26

Express each of these statements using quantifiers. Then form the negation of the statement so that no negation is to the left of a quantifier. Next, express the negation in simple English.

F(x) be the expression “x can swim,” and G(x) be “x can catch fish,” and the domain of discourse is pigs. The statement is “There exists a pig that can swim and catch fish.”

Answer 26

The expression for the statement is ∃ x (F(x) ∧ G(x)), and its negation is ∀ x¬(F(x) ∧ G(x)) and the sentence is “No pig can both swim and catch fish.”

Let F(x) be “x can swim” and let G(x) be “x can catch fish,” where the domain of discourse is pigs. Then our original statement is ∃x (F(x) ∧ G(x)). Its negation is ∀x¬(F(x) ∧ G(x)), which could also be written ∀x (¬F(x) ∨ ¬G(x)) by De Morgan’s law. In English this is “No pig can both swim and catch fish,” or “Every pig either is unable to swim or is unable to catch fish.”

Question 27

Let F(x, y) be the statement “x can fool y,” where the domain consists of all people in the world. Use quantifiers to express each of these statements.

Nancy can fool exactly two people.

Answer 27

∃ y1 ∃ y2(F(Nancy, y1) ∧ F(Nancy, y2) ∧ y1 ≠ y2 ∧ ∀ y(F(Nancy, y) → (y = y1 ∨ y = y2)))

The statement “Nancy can fool exactly two people.” can be expressed as ∃y1∃y2(F(Nancy, y1) ∧ F(Nancy, y2) ∧ y1 ≠ y2 ∧ ∀y(F(Nancy, y) → (y = y1 ∨ y = y2))).

Question 28

Let the domain of discourse be the set of all students in your class.
If L(x, y) means x has learned programming language y, express the statement "All students in this class have learned at least one programming language" using quantifiers.

Answer 28

∀x∃y L(x, y)
The statement "All students in this class have learned at least one programming language" can be expressed using quantifiers as ∀x∃y L(x, y)

Question 29

Let the statements “If I play hockey, then I am sore the next day,” “I use the whirlpool if I am sore,” and “I did not use the whirlpool” be the given premises.

Identify the rule of inference that is used to arrive at the conclusion that I am not sore.

Answer 29

modus tollens

If modus tollens is used from the back, the given conclusion is arrived.

Question 30

Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expression involving logical connectives).

¬ ∀ x ∃ yP(x, y)

Answer 30

∃ x ∀ y ¬P(x, y)

We need to use the transformations of De Morgan’s Laws for Quantifiers. In other words, we push all the negation symbols inside the quantifiers, changing the sense of the quantifiers as we do so, because of the equivalences of De Morgan’s Laws for Quantifiers. In addition, we need to use De Morgan’s laws to change the negation of a conjunction to the disjunction of the negations and to change the negation of a disjunction to the conjunction of the negations. We also use the fact that ¬(¬p) ≡ p. Using these method, we can rewrite the given expression as ∃x∀y ¬P(x, y).

Question 31

Rewrite each of these statements so that negations appear only within predicates (that is, so that no negation is outside a quantifier or an expression involving logical connectives).

¬ ∃ y(Q(y) ∧ ∀ x¬R(x, y))

Answer 31

∀ y(¬Q(y) ∨ ∃ xR(x, y))

We need to use the transformations of De Morgan’s Laws for Quantifiers. In other words, we push all the negation symbols inside the quantifiers, changing the sense of the quantifiers as we do so, because of the equivalences of De Morgan’s Laws for Quantifiers. In addition, we need to use De Morgan’s laws to change the negation of a conjunction to the disjunction of the negations and to change the negation of a disjunction to the conjunction of the negations. We also use the fact that ¬(¬p) ≡ p. Using these method, we can rewrite the given expression as ∀y(¬Q(y) ∨ ∃xR(x, y)).

Question 32

Translate these system specifications into English, where the predicate S(x, y) is “x is in state y” and where the domain for x and y consists of all systems and all possible states, respectively.

∀x(S(x, malfunctioning) ∨ S(x, diagnostic))

Answer 32

Every system is either malfunctioning or in a diagnostic state.

The expression ∀x(S(x, malfunctioning) ∨ S(x, diagnostic)) denotes that all systems are malfunctioning or all systems are in a diagnostic state, which can be simply stated as “Every system is either malfunctioning or in a diagnostic state.”

Question 33

Translate these system specifications into English, where the predicate S(x, y) is “x is in state y” and where the domain for x and y consists of all systems and all possible states, respectively.

∀x¬S(x, working)

Answer 33

Every system is working.
Every system is not working.
Some system is not working.
Some system is working.

• The expression ∀x¬S(x, working) denotes that all systems are not working, which can be stated in two ways:
• No system is working.
• Every system is not working.

Question 34

Suppose that the domain of the propositional function P(x) consists of −5, −3, −1, 1, 3, and 5.
Express ∃x((x ≥ 0) ∧ P(x)) without using quantifiers, instead using only negations, disjunctions, or conjunctions.

Answer 34

P(1) ∨ P(3) ∨ P(5)

The formal translation is as follows: ((−5 ≥ 0) ∧ P(−5)) ∨ ((−3 ≥ 0) ∧ P(−3)) ∨ ((−1 ≥ 0) ∧ P(−1)) ∨ ((1 ≥ 0) ∧ P(1)) ∨ ((3 ≥ 0) ∧ P(3)) ∨ ((5 ≥ 0) ∧ P(5)). Since only three of the x’s in the domain meet the condition, the answer is equivalent to P(1) ∨ P(3) ∨ P(5).

Question 35

Let N(x) be the statement “x has visited North Dakota,” where the domain consists of the students in your school. Click and drag any of the given English statements and place them next to the quantifications provided. A quantification may be matched to more than one English statement provided.

∃xN(x)
∀xN(x)
¬∃xN(x)
∃x¬N(x) and ¬∀xN(x)
∀x¬N(x)

• There is a student at your school who has visited North Dakota.
• At least one student at your school has visited North Dakota.
• All students at your school have not visited North Dakota.
• Each student at your school has not visited North Dakota.
• Every student at your school has visited North Dakota.
• There is a student at your school who has not visited North Dakota.
• At least one student at your school has not visited North Dakota.
• Some students at your school have not visited North Dakota.
• It is not the case that each student at your school has visited North Dakota.
• All students at your school have visited North Dakota.
• It is not the case that a student at your school has visited North Dakota.
• No student at your school has visited North Dakota.
• Some students at your school have visited North Dakota.

Answer 35

There is a student at your school who has visited North Dakota.
At least one student at your school has visited North Dakota.
Some students at your school have visited North Dakota.

Every student at your school has visited North Dakota.
All students at your school have visited North Dakota.

It is not the case that a student at your school has visited North Dakota.
No student at your school has visited North Dakota.

There is a student at your school who has not visited North Dakota.
At least one student at your school has not visited North Dakota.
Some students at your school have not visited North Dakota.
It is not the case that each student at your school has visited North Dakota.

All students at your school have not visited North Dakota.
Each student at your school has not visited North Dakota.

Question 36

Let C(x, y) mean that student x is enrolled in class y, where the domain for x consists of all students in your school and the domain for y consists of all classes being given at your school. Express each of these statements by a simple English sentence.

∃x1∃x2∀z((x1 ≠ x2) ∧ (C(x1, z) → C(x2, z)))

Answer 36

There exist two distinct people, the second of whom is enrolled in every class that the first is enrolled in.

The statement ∃x1∃x2∀z((x1 ≠ x2) ∧ (C(x1, z) → C(x2, z))) can be written as “There exist two distinct people, the second of whom is enrolled in every class that the first is enrolled in.”

Question 37

Consider the following propositions.

Let S(x) denote “x can keep a secret,” where the domain is the set of all people. Express the negation of the proposition “No one can keep a secret” using quantifiers, and then express the negation in English.

Answer 37

∃xS(x); Some people can keep a secret.

The original statement is ¬∃xS(x), and the negation is ∃xS(x); “Some people can keep a secret.”

Question 38

Identify the error or errors in this argument that supposedly shows that if ∀x(P(x)∨Q(x)) is true then ∀x P(x) ∨ ∀x Q(x)) is true.

Step Reason

1. ∀x(P(x) ∨ Q(x)) Premise
2. P(c) ∨ Q(c) for arbitrary c Universal instantiation from (1)
3. P(c) for arbitrary c Simplification from (2)
4. ∀xP(x) Universal generalization from (3)
5. Q(c) for arbitrary c Simplification from (2)
6. ∀xQ(x) Universal generalization from (5)
7. (∀xP(x)) ∨ (∀xQ(x)) Disjunction from (4) and (6)

Answer 38

The errors are in step 3 and step 5, because the simplification rule is (P(x) ∧ Q(x)) → P(x).

Question 39

Identify the error or errors in this argument that supposedly shows that if ∀x(P(x)∨Q(x)) is true then ∀x P(x) ∨ ∀x Q(x)) is true.
Identify the error in steps 3 and 5.

Step Reason

1. ∀x(P(x) ∨ Q(x)) Premise
2. P(c)∨Q(c) for arbitrary c Universal instantiation from (1)
3. P(c) for arbitrary c Simplification from (2)
4. ∀xP(x) Universal generalization from (3)
5. Q(c) for arbitrary c Simplification from (2)
6. ∀xQ(x) Universal generalization from (5)
7. (∀xP(x)) ∨ (∀xQ(x)) Disjunction from (4) and (6)

Answer 39

Simplification applies to conjunctions, not disjunctions.

The error is that simplification applies to conjunctions, not disjunctions.

Question 40

Let Q(x) be the statement “x + 1 > 2x,” where the domain consists all integers. Identify the quantifications that have the truth value “true.”

Answer 40

Q(0)
Q(−1)
∃ x¬Q(x)
∃ xQ(x)

a) If x = 0, 0 + 1 > 2 · 0, we know that Q(0) is true.
b) If x = –1, (−1) + 1 > 2 · (−1), we know that Q(−1) is true.
c) If x = 1, 1 + 1 ≯ 2.1, we know that Q(1) is false.

From part (a), we know that there is at least one x that makes Q(x) true, so ∃xQ(x) is true.
From part (c) we know that there is at least one x that makes Q(x) false, so ∃x¬Q(x) is true.

Question 41

Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contraposition.

Answer 41

In order to prove the contrapositive, assume that n is odd. Then, we can write n = 2k + 1 for some integer k.

Then, 3n + 2 = 3(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1. This number is again of the form 2p + 1 for the integer p = 3k + 2; hence, it is odd.

Thus, if n is odd, then 3n + 2 is odd.

Question 42

Prove that if n is an integer and 3n + 2 is even, then n is even using a proof by contradiction.

Answer 42

Suppose that 3n + 2 is even and that n is odd.

Since 3n + 2 is even, so is 3n.

We know that if we subtract an odd number from an even number, we get an odd number.

As 3n is even and n is odd, 3n – n should be odd, but 3n – n = 2n is even. This is a contradiction.

Therefore, our supposition was wrong; hence n is even.

Question 43

The steps in the correct order to show that if n is a positive integer, then n is even if and only if 7n + 4 is even. Suppose that n is even.

Answer 43

We need to prove two things, since this is an “if and only if” statement. First, let us prove directly that if n is even then 7n + 4 is even. Since n is even, it can be written as 2k for some integer k. Then, 7n + 4 = 14k + 4 = 2(7k + 2). This is 2 times an integer, so it is even as desired. Next, we give a proof by contraposition that if 7n + 4 is even then n is even. So, suppose that n is not even, i.e., that n is odd. Then, n can be written as 2k +1 for some integer k. Thus, 7n + 4 = 14k +11 = 2(7k + 5) + 1. This is 1 more than 2 times an integer; so it is odd. That completes the proof by contraposition.

Question 44

Let A = {a, b, c}, B = {x, y}, and C = {0, 1}.

Identify C × A × B.

Answer 44

{(0, a, x), (0, a, y), (0, b, x), (0, b, y), (0, c, x), (0, c, y), (1, a, x), (1, a, y), (1, b, x), (1, b, y), (1, c, x), (1, c, y)}

The ordered n-tuple (a1, a2, …, an) is the ordered collection that has a1 as its first element, a2 as its second element, …, and an as its nth element.
The Cartesian product of the sets C, A, and B, denoted by C × A × B, is the set of all ordered 3-tuples (a3, a1, a2), where a3 ∈ C, a1 ∈ A, and a2 ∈ B. Hence, C × A × B = {(a3, a1, a2) ∣ a3 ∈ C ∧ a1 ∈ A ∧ a2 ∈ B}.

Thus, the Cartesian product C × A × B can be written as {(0, a, x), (0, a, y), (0, b, x), (0, b, y), (0, c, x), (0, c, y), (1, a, x), (1, a, y), (1, b, x), (1, b, y), (1, c, x), (1, c, y)}.

Question 45

Use the following building blocks in the right column to assemble a direct proof that the product of two odd numbers is odd.

Not all blocks belong in the proof.

• Suppose that the product of two odd numbers is odd.
• By definition of an odd integer, that means that nm is odd.
• Then n = 2k+1 and m = 2k+1 for some integer k.
• Distributing, we get nm = 2(2kq + k +q) + 1
• Suppose that nm is odd.
• Suppose that n and m are odd numbers.
• Then n = 2k+1 for some integer k and m = 2q+1 for some integer q.
• By multiplying these equations, we obtain nm = (2k+1)(2q+1).

Answer 45

• Suppose that n and m are odd numbers.
• Then n = 2k+1 for some integer k and m = 2q+1 for some integer q.
• By multiplying these equations, we obtain nm = (2k+1)(2q+1).
• Distributing, we get nm = 2(2kq + k +q) + 1
• By definition of an odd integer, that means that nm is odd.

Question 46

Determine whether each of these sets is the power set of a set, where a and b are distinct elements.
{Ø, {a}, {Ø, a} }

Answer 46

This cannot be the power set of any set.

Ø, {a}, and {Ø, a} are the subsets. These are not valid subsets of a set.

Question 47

Determine whether each of these sets is the power set of a set, where a and b are distinct elements.
{Ø, {a}, {b}, {a, b} }

Answer 47

This is the power set of {a, b}.

Ø, {a}, {b}, and {a, b} are the subsets.

Question 48

Find the cardinality of the set A.

A = {ø, {ø}, {ø, {ø} } }

Answer 48

3, as this set contains exactly three different elements, each of which is a set

Cardinality is the number of elements in a set. The set {Ø, {Ø}, {Ø, {Ø} } } contains three elements Ø, {Ø}, and {Ø, {Ø} }.

Question 49

Arrange these steps in the correct order to prove that if you have an 8-gallon jug of water and two empty jugs with capacities of 5 gallons and 3 gallons, respectively, then you can measure 4 gallons by successively pouring some of or all of the water in a jug into another jug in the correct order.

• Top off the 3-gallon jug from the 5-gallon jug, and there will be (1, 4, 3), with four gallons in the 5-gallon jug.
• Pour the contents of the 3-gallon jug back into the 8-gallon jug, leaving (6, 2, 0).
• Fill the 5-gallon jug from the 8-gallon jug, leaving the contents (3, 5, 0), where the ordered triple is used to record the amount of water in the 8-gallon jug, the 5-gallon jug, and the 3-gallon jug, respectively.
• Empty the 5-gallon jug’s contents into the 3-gallon jug, leaving (6, 0, 2), and then fill the 5-gallon jug from the 8-gallon jug, producing (1, 5, 2).
• Fill the 3-gallon jug from the 5-gallon jug, leaving (3, 2, 3).

Answer 49

• Fill the 5-gallon jug from the 8-gallon jug, leaving the contents (3, 5, 0), where the ordered triple is used to record the amount of water in the 8-gallon jug, the 5-gallon jug, and the 3-gallon jug, respectively.
• Fill the 5-gallon jug from the 8-gallon jug, leaving the contents (3, 5, 0), where the ordered triple is used to record the amount of water in the 8-gallon jug, the 5-gallon jug, and the 3-gallon jug, respectively.
• Pour the contents of the 3-gallon jug back into the 8-gallon jug, leaving (6, 2, 0).
• Empty the 5-gallon jug’s contents into the 3-gallon jug, leaving (6, 0, 2), and then fill the 5-gallon jug from the 8-gallon jug, producing (1, 5, 2).
• Top off the 3-gallon jug from the 5-gallon jug, and there will be (1, 4, 3), with four gallons in the 5-gallon jug.

Fill the 5-gallon jug from the 8-gallon jug, leaving the contents (3, 5, 0), where we are using the ordered triple to record the amount of water in the 8-gallon jug, the 5-gallon jug, and the 3-gallon jug, respectively. Next fill the 3-gallon jug from the 5-gallon jug, leaving (3, 2, 3). Pour the contents of the 3-gallon jug back into the 8-gallon jug, leaving (6, 2, 0). Empty the 5-gallon jug’s contents into the 3-gallon jug, leaving (6, 0, 2), and then fill the 5-gallon jug from the 8-gallon jug, producing (1, 5, 2). Finally, top off the 3-gallon jug from the 5-gallon jug, and we’ll have (1, 4, 3), with four gallons in the 5-gallon jug.

Question 50

Use the following building blocks to assemble a proof that if m and n are integers, and mn is even, then m is even or n is even. Not all blocks belong in the proof. Before you start, you might want to write out such a proof on paper, and carefully consider whether a direct proof, a proof by contraposition or a proof by contradiction is the most appropriate.

Therefore, nm = (2k + 1) (2k + 1) = 4k² + 4k + 1 = 2 (2k² + 2k) + 1.

Suppose that m is odd and n is odd.

We will give a proof by contradiction.

Therefore, we have finished our direct proof that one of the numbers n, m must be even.

By definition of odd number, m = 2p+1 for some integer p, and n = 2q+1 for some integer q.

This contradicts our assumption that nm is even. Thus, we have proved by contradiction that m must be even, or n must be even.

We will give a proof by contraposition.

We will give a direct proof. Suppose that nm is even.

By definition of odd number, m = 2k + 1 and n = 2k + 1 for some integer k.

By definition of odd number, this means that nm is odd.

Suppose that nm is odd.

By definition of even number, nm = 2k for some integer k. Therefore, n=2 and m = k or n=k and m=2.

Therefore, nm = (2q+1)(2p+1) = 4pq + 2q + 2p + 1 = 2(2pq + q + p) + 1.

Therefore, we have proved the contraposition of the desired statement and thus completed the proof by contraposition.

Answer 50

• We will give a proof by contradiction.
• Suppose that m is odd and n is odd.
• By definition of odd number, m = 2p+1 for some integer p, and n = 2q+1 for some integer q.
• Therefore, nm = (2q+1)(2p+1) = 4pq + 2q + 2p + 1 = 2(2pq + q + p) + 1.
• By definition of odd number, this means that nm is odd.
• Therefore, we have proved the contraposition of the desired statement and thus completed the proof by contraposition.

Question 51

Use set builder notation to give a description of each of these sets.

{0, 3, 6, 9, 12}

Answer 51

{3n | n = 0, 1, 2, 3, 4}

In set builder notation, we characterize all the elements in a set by stating the property or properties they must have to be members.

Question 52

Arrange these steps in the correct order to prove that there are no solutions in integers x and y to the equation 2x2 + 5y2 = 14.

Answer 52

If |y| ≥ 2, then 2x^2 + 5y^2 ≥ 2x^2 + 20 ≥ 20,

so, the only possible values of y to try are 0 and ±1.

In the former case, we would be looking for solutions to 2x2 = 14.

In the latter case to 2x2 = 9.

Clearly, there are no integer solutions to these equations,

so, there are no solutions to the original equation.

If |y| ≥ 2, then 2x2 + 5y2 ≥ 2x2 + 20 ≥ 20, so the only possible values of y to try are 0 and ±1. In the former case we would be looking for solutions to 2x2 = 14 and in the latter case to 2x2 = 9. Clearly, there are no integer solutions to these equations, so there are no solutions to the original equation.

Question 53

Suppose that A = {2, 4, 6}, B = {2, 6}, C = {4, 6}, and D = {4, 6, 8}. Identify the pairs of sets in which one is a subset of the other (in any order). (

Answer 53

B, D

B is a subset of A. A = {2, 4, 6} and B = {2, 6}.

C is a subset of A. A = {2, 4, 6} and C = {4, 6}.

C is a subset of D. D = {4, 6, 8} and C = {4, 6}.

Question 54

Prove or disprove that if a and b are rational numbers, then ab is also rational.

Answer 54

• a = 2 and b = 1/2, then ab = 21/2 is a counterexample that disproves the statement.
• a = 2 and b = 1/2, then ab = 21/2 is a counterexample that proves the statement.
• a = 3 and b = 1/2, then ab = 31/2 is a counterexample that disproves the statement.
• If a = p / q and b = r / s are rational numbers, then ab can be written as )r / s= (pr / qs) , where , q, r, and s are integers. Thus, ab is rational.

An assertion like this one is implicitly universally quantified—it means that for all rational numbers a and b, ab is rational. To disprove such a statement it suffices to provide a counterexample. There are two counter examples for the given statement.

1) Take a = 2 and b = 1/2. then ab = 21/2 = 2 [Math Processing Error] .
2) Take a = 3 and b = 1/2, then a^b = 3^(1/2) = √3 .

Question 55

Steps to show that at least three of any 25 days chosen must fall in the same month of the year.

Answer 55

• Assume there were at most two chosen days falling into any one month.
• Then, there will be at most 2.12 = 24 chosen days in 12 months.
• This contradicts the assumption that we have 25 days.
• Hence, three chosen days must fall in the same month.

Question 56

Prove that if n is a perfect square, then n + 2 is not a perfect square.

Answer 56

• Assume n = m2, for some nonnegative integer m.
• If m = 0, then n + 2 = 2, which is not a perfect square. The smallest perfect square greater than n is (m + 1)2.
• Lets assume m ≥ 1.
• Expand (m + 1)2 to obtain (m + 1)2 = m2 + 2m + 1 = n + 2m + 1 > n + 2 + 1 > n + 2.

Expand (m + 1)2 to obtain (m + 1)2 = m2 + 2m + 1 = n + 2m + 1 > n + 2 + 1 > n + 2.

• Hence, n + 2 is not a perfect square.

An integer a is a perfect square if there is an integer b such that a = b2.

Question 57

two of the numbers 65^1000 − 8^2001 + 3^177, 79^1212 − 9^2399 + 2^2001, and 24^4493 − 5^8192 + 7^1777 is nonnegative (for the purpose of this agreement, we will think of 0 as carrying a positive sign).

Identify the correct proof of the given statement.

Answer 57

Of these three numbers, at least two must have the same sign, since there are only two signs. The product of two with the same sign is nonnegative.

Of these three numbers, at least two must have the same sign (both positive or both negative), since there are only two signs. (It is conceivable that some of them are zero, but we view zero as positive for the purposes of this problem.) The product of two with the same sign is nonnegative.

Question 58

Consider the statement that min(a, min(b, c)) = min(min(a, b), c) whenever a, b, and c are real numbers.

Prove min(a, min(b, c)) = min(min(a, b), c) whenever a, b, and c are real numbers. Assume, c is the smallest real number.

(Note: In your proof, consider the left side of the equation first.)

Answer 58

• Therefore, on the right-hand equals c.
• It follows that <= min(b,c).
• On the right-hand side, c <= min(a,c), so, min(min(a,c), b) = c.
• On the right-hand side, c <= min(a,c), so, min(min(a,b), c) = c.
• So, the left-hand side equals c.
• Suppose c is the smallest of the three real numbers.
• It follows that c >= min(a,b).

Suppose c is the smallest of the three real numbers. It follows that c ≤ min(b, c). So, the left-hand side equals c. On the right-hand side, c ≤ min(a, b), so, min(min(a, b), c) = c. Therefore, right hand side equals c.

Question 59

Find the truth set of each of the given predicates if the domain is a set of integers.
P(x): x < x^2

Answer 59

{x ∈ Z ∣ x < x^2} = {x ∈ Z | x ≠ 0 ∧ x ≠ 1}

Negative integers certainly satisfy this inequality, as do all positive integers greater than 1. However, 0 ≮ 0^2 and 1 ≮ 1^2. Thus the truth set is {x ∈ Z | x ≠ 0 ∧ x ≠ 1} = {…, −3, −2, −1, 2, 3, …}.

Question 60

Consider these statements about the real number:

(i) x is rational.
(ii) x/2 is rational.
(iii) 3x − 1 is rational.

Put the steps in correct order to prove that (i) implies (ii).

• Suppose x is rational.
• Then, x/2 = p/(2q). Since q ≠ 0. 2q ≠ 0.
• So, x = p/q where p and q are integers with q ≠ 0.
• Hence, x/2 is rational.

Answer 60

• Suppose x is rational.
• So, x = p/q where p and q are integers with q ≠ 0.
• Then, x/2 = p/(2q). Since q ≠ 0. 2q ≠ 0.
• Hence, x/2 is rational.

Suppose that x = p/q where p and q are integers with q ≠ 0. Then x/2 = p/(2q), and this is rational, since p and 2q are integers.

Question 61

Determine whether each of these sets is finite, countably infinite, or uncountable. For those that are countably infinite, exhibit a one-to-one correspondence between the set of positive integers and that set.

The integers greater than 10.

Answer 61

• The set is countably infinite.
• The set is finite.
• The set is countably infinite with one-to-one correspondence 1 ↔ 11, 2 ↔ 12, 3 ↔ 13, and so on.
• The set is countably infinite with one-to-one correspondence 1 ↔ 10, 2 ↔ 12, 3 ↔ 17, and so on.
• The one-to-one correspondence is given by n ↔ (2n + 10).

The integers in the set are 11, 12, 13, 14, and so on.

Question 62

Determine whether each of these sets is finite, countably infinite, or uncountable. For those that are countably infinite, exhibit a one-to-one correspondence between the set of positive integers and that set.

The set A × Z^+ where, A = {2, 3}

Answer 62

• The set is countable.

- The set is countably infinite with one-to-one correspondence 1 ↔ (2,1), 2 ↔ (3,1), 3 ↔ (2,2), 4 ↔ (3,2),and so on.

Question 63

Find a formula for m∑k=0⌊^3√k⌋ , when m is a positive integer.

Answer 63

If we write down the first few terms of this sum we notice a pattern. It starts (1 + 1 + 1 + 1 + 1 + 1 + 1) + (2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2) + (3 + 3 + 3 + … + 3) + …. There are seven 1s, then 19 2s, then 37 3s, and so on; in general, the number of i’s is (i + 1)^3 – i^3 = 3i^2 + 3i + 1. So, we need to sum i(3i^2 + 3i + 1) for an appropriate range of values for i. We must find this range. It gets a little messy at the end if m is such that the sequence stops before a complete range of the last value is present. Let n=⌊^3√m⌋−1 . Then there are n + 1 blocks, and (n + 1)^3 – 1 is where the next-to-last block ends. The sum of those complete blocks is n∑i=1 i(3i^2+3i+1)=n∑i=1 3i^3+3i^2+i=n(3n+4)(n+1)^2/4 . The remaining terms in our summation all have the value n + 1 and the number of them present is m – ((n + 1)3 – 1). Our final answer is therefore m∑k=0 ⌊^3√k⌋=n(3n+4)(n+1)^2/4+(n+1)(m−(n+1)^3 +1) , where once again, n=⌊^3√m⌋−1.

Question 64

Find the value of each of the sums.

8∑j = 0 (2^(j + 1) − 2^j) = _____

Answer 64

511

We could evaluate this sum by breaking it into two summations. Both could be computed from the formula for the sum of a geometric progression. There is an easier solution though. All powers of 2 cancel except −20 when j = 0 and 29 when j = 8.

Question 65

Identify the Boolean product of A and B, where  
A = [1 0 0 1]
      [0 1 0 1]
      [1  1 1  1]
B = [1 0]
      [1  1]
      [0 1]
      [1  1]

Answer 65

[1 1]
[1 1]
[1 1]

A matrix all of whose entries are either 0 or 1 is called a zero–one matrix. Let A = [aij] be an m × k zero–one matrix and B = [bij] be a k × n zero–one matrix. Then, the Boolean product of A and B, denoted by A ⊙ B, is the m × n matrix with (i, j)th entry cij, where cij = (ai1 ∧ b1j) ∨ (ai2 ∧ b2j) ∨ ⋯ ∨ (aik ∧ bkj).

Question 66

For the sequence an = an – 1 – an – 2, a0 = 2, and a1 = –1, the values of the first six terms are a0 = __, a1 = __, a2 = __, a3 = __, a4 = __, a5 = __.

Answer 66

a0 = –1

Plug n = 1, 2, 3, 4, 5 into the formula: an = an – 1 – an – 2.

Question 67

For the sequence an = na(n – 1) + a^2n – 2, a0 = −1 and a1 = 0, the values of the first six terms are a0 = __, a1 = __, a2 = __, a3 = __, a4 = __, a5 = __.

Answer 67

a0 = –1 and a1 = 0

Plug n = 2, 3, 4, 5 into the formula: an = na(n – 1) + a^2n – 2

Question 68

A = [1 1]
[0 1]

B = [0 1]
[0 1]

Identify A ∧ B

Answer 68

[0 1]
[0 1]

A matrix all of whose entries are either 0 or 1 is called a zero–one matrix. We can write a∧b={1 if a=b=1
{0 otherwise

Let A = [aij] and B = [bij] be m × n zero–one matrices. The meet of A and B is the zero–one matrix with (i, j)th entry aij ∧ bij. The meet of A and B is denoted by A ∧ B.

Question 69

A = [1 1]
[0 1]

B = [0 1]
[0 1]

Identify A ⊙ B

Answer 69

B = [0 1]

[0 1]

Question 70

Find these terms of the sequence {an}, where an = 2(–3)^n + 5^n.

a0

Answer 70

3

Plug n = 0 into the formula: an = 2 · (−3)^n + 5^n.

a0 = 2 · (−3)^0 + 5^0 = 2(1) + 1

Question 71

Find these terms of the sequence {an}, where an = 2(–3)^n + 5^n.

a5

Answer 71

2,639

Plug n = 5 into the formula: an = 2 · (−3)^n + 5^n.

a5 = 2 · (−3)^5 + 55 = –486 + 3,125

Question 72

A person deposits $1,000 in an account that yields 9% interest compounded annually.

Set up a recurrence relation for the amount in the account at the end of n years.

Answer 72

an = 1.09a(n – 1)

The amount after (n − 1) years is multiplied by 1.09 to give the amount after n years, since 9% of the value must be added to account for the interest.

Question 73

A person deposits $1,000 in an account that yields 9% interest compounded annually.

Find an explicit formula for the amount in the account at the end of n years.

Answer 73

an = 1000(1.09)^n

The amount after (n − 1) years is multiplied by 1.09 to give the amount after n years, since 9% of the value must be added to account for the interest.

Question 74

A person deposits $1,000 in an account that yields 9% interest compounded annually.

How much money will the account contain after 100 years?

Answer 74

a100 ≈ $5,529,041

The amount after (n − 1) years is multiplied by 1.09 to give the amount after n years, since 9% of the value must be added to account for the interest.

Question 75

Find the solution to the recurrence relation by using an iterative approach.

Identify the steps for the recurrence relation an = 2a(n – 1) – 3 with the initial condition a0 = –1.

Answer 75

an = −3 + 2(−3) + 4an − 2

an = −3 + 2(−3) + 4(−3) + 8an − 3

an = −3 + 2(−3) + 4(−3) + 8(−3) + 16an − 4

Continuing the same manner, an = −3(1 + 2 + 4 + · · · + 2n − 1) + 2nan − n = −3(2n − 1) + 2n(−1) = −2n + 2 + 3

Question 76

What are the values of these sums if S = {1, 3, 5, 7}?

∑ j∈S j^2 = _____

Answer 76

84

∑ j∈S j^2=1^2+3^2+5^2+7^2

Question 77

Find the value of A + B.
A = [-1 0 5 6]
[-4 -3 5 -2]

B = [-3 9 -3 4]
[0 -2 -1 2]

Answer 77

[-4 9 2 10]
[-4 -5 4 0 ]

The sum of two matrices of the same size is obtained by adding elements in the corresponding positions.

Question 78

Identify the correct steps involved in proving that the set Z^+ × Z^+ is countable

Answer 78

Define the sets Ai := {(i, n) | n ∈ Z^+}, where i belongs to Z^+. We can list the sets as A1 = {(1, n) | n ∈ Z^+}, A2 = {(2, n) | n ∈ Z^+}, and so on.

For each i, define f(n) = (i, n), ∀n ∈ Z^+. Clearly, it is a one-to-one correspondence from the set of positive integers to the set Ai.

From the definition of cardinality, each Ai is a countable set.

Thus, we can think of Z^+ × Z^+ as a union of countable number of countable sets A1, A2, …, An, ….

We know that the union of countable number of countable sets is countable. Thus, Z^+ × Z^+ is countable.

Let us define the sets Ai := {(i, n) | n ∈ Z^+}, where i belongs to Z^+. We can list the sets as A1 = {(1, n) | n ∈ Z^+}, A2 = {(2, n) | n ∈ Z^+}, and so on. For each i, define f(n) = (i, n), ∀n ∈ Z^+. Clearly, it is a one-to-one correspondence from the set of positive integers to the set Ai.
From the definition of cardinality, each Ai is a countable set. Thus, we can think of Z^+ × Z^+ as a union of countable number of countable sets A1, A2, …, An, …. We know that the union of countable number of countable sets is countable. Thus, Z^+ × Z^+ is countable.

Question 79

A = [-1 2]
[1 3]

Find (A^(–1))^3

Answer 79

[-37/125 18/125]
[9/125 -1/125]

Use the value of (A^(–1))^2 to arrive at (A^(–1))^3

Question 80

A = [0 -1]
[7 2]
[-4 -3]

A = [4 -1 2 3 0]
[-2 0 3 4 1]

Answer 80

[2 0 -3 -4 -1]
[24 -7 20 29 2]
[-10 4 -17 -24 -3]

To multiply matrices A and B, compute the (i, j)^th entry of the product AB by adding all the products of elements from the i^th row of A with the corresponding element in the j^th column of B.