Tuto Logic

MATH 220 - HOMEWORK 3 1

Problem (2.2). Consider the sets A, B, C, andD below. Which of the followingstatements are true? Give an explanation for each false statement.

A = {1, 4, 7, 10, 13, 16, . . .}B = {x ∈ Z : x is odd}

C = {x ∈ Z : x is prime and x 6= 2}D = {1, 2, 3, 5, 8, 13, 21, 34, 55, . . .}

(a) 25 ∈ A

(b) 33 ∈ D

(c) 22 /∈ A ∪D

(d) C ⊆ B

(e) ∅ ∈ B ∩D

(f) 53 /∈ C.

Solution.

(a) The statement 25 ∈ A is true. Note that A = {3n − 2 : n ∈ N}. Since3 · 9− 2 = 25 and 9 ∈ N, we must have 25 ∈ A.

(b) The statement 33 ∈ D is false. The set D is the set whose elements areFibonacci numbers. That is, if Fn is the n-th term, then

Fn = Fn−1 + Fn−2.

Since D is listed in increasing order, it is clear that 33 6∈ D.

(c) The statement 22 6∈ A ∪ D is false. Since A = {3n − 2 : n ∈ N}, 8 ∈ N,and 3 · 8− 2 = 22 we conclude that 22 ∈ A. Therefore 22 ∈ A ∪D.

(d) The statement C ⊆ B is true. Since all prime numbers larger than 2 areodd, C consists of the set of all odd prime integers and is hence a propersubset of B.

(e) The statement ∅ ∈ B ∩D is false. Both B and D contain only integers aselements. Since ∅ is a set, it is not a member of B, nor is it a member ofD. Therefore ∅ cannot be a member of the intersection B ∩D.

(f) The statement 53 6∈ C is false. To show that 53 is prime, it is enoughto simply check that all integers greater than 1 and less than 8 (since82 = 64 > 53) are not divisors of 53.

Remark. Many of you have some very creative ways for creating “display”style mathematics. Use \[ and \] as the beginning and ending markers ofmathematics which you wish to center (on a separate line). Alternately, youcan use $$ (as both beginning and ending markers) in place of \[ and \]. Fortext that you wish to center, use the center environment.


Problem (2.4). Consider the open sentence P (x) : x(x−1) = 6 over the domainR.

(a) For what values of x is P (x) a true statement?

(b) For what values of x is P (x) a false statement?

Solution. Over R, the values of x that make P (x) true are x = −2 and x = 3.These values can be obtained by solving x(x− 1) = 6 as shown below.

x(x − 1) = 6

x2 − x = 6

x2 − x− 6 = 0

(x− 3)(x+ 2) = 0

On the other hand, if x is an element of the set R−{−2, 3}, then P (x) is a falsestatement.

Remark. If you would like to create multiple rows of mathematics that alignspecifically at, say, an equal sign, then the align or align* environment willdo the job. The align environment will additionally number each row of math-ematics for future referencing, whereas align* will not.

\begin{align*}

x (x-1) & = 6 \\

x^2 - x & = 6 \\

x^2 - x - 6 & = 0 \\

(x-3)(x+2) & = 0

\end{align*}

All rows are horizontally aligned according to the placement of the ampersandsymbol (&), and the double backslash signifies the end of a row.


Problem (2.6). For the open sentence P (A) : A ⊆ {1, 2, 3} over the domainS = P({1, 2, 4}), determine:

(a) all A ∈ S for which P (A) is true.

(b) all A ∈ S for which P (A) is false.

(c) all A ∈ S for which A ∩ {1, 2, 3} = ∅.

Solution. Note that

S = P({1, 2, 4}) = {∅, {1}, {2}, {4}, {1, 2}, {2, 4}, {1, 4}, {1, 2, 4}}.

(a) If A ∈ {∅, {1}, {2}, {1, 2}} then A ⊂ {1, 2, 3} and P (A) would be true.

(b) If A ∈ {{4}, {2, 4}, {1, 4}, {1, 2, 4}}, then A 6⊆ {1, 2, 3} and P (A) would befalse.

(c) The only possible choices of A such that A ∩ {1, 2, 3} = ∅ are either theemptyset itself, or if A is a set whose elements are not contained in {1, 2, 3}.Hence A = ∅ or A = {4}.

Remark. The sentence

If A is {{4}, {2, 4}, {1, 4}, {1, 2, 4}}, then P (A) is true.

is an incorrect response to part (a). The set A must be a member of S, andthe members of S are sets that contain integers. So A must be a set whosemembers are integers. The members of the set in the sentence above are notintegers—the members are sets of integers.


Problem (2.14). State the negation of each of the following statements.

(a) At least two of my library books are overdue.

(b) One of my two friends misplaced his homework assignment.

(c) No one expected that to happen.

(d) It’s not often that my instructor teaches that course.

(e) It’s surprising that two students received the same exam score.

Solution. The following are negations of the statements given above.

(a) At most one of my library books is overdue.

(b) None of my friends misplaced their homework assignment.

(c) At least one person expected that to happen.

(d) It is often that my instructor teaches that course.

(e) It is not surprising that two students received the same exam score.

Remark. If the subject is singular, then the negation of the statement is simplythe negation of the predicate (i.e. the negation of the verb). For example, thenegation of

The sky is blue.

is the sentence

The sky is not blue.

However, the scheme does not work when the subject is quantified. Considerthe statement in part (a)

At least two of my library books are overdue.

and call it P . Suppose we have n books from the library.

Overdue Not Overdue P ∼ P0 n F T1 n− 1 F T2 n− 2 T F3 n− 3 T F...

......

...

The statement

At least two of my library books are not overdue.

cannot be considered as the negation of P . If n = 5 and we had exactly twooverdue books (and hence three books not overdue), then both P and the state-ment above would be true simultaneously. However, P and ∼ P can never bothbe true! They must always have opposite truth values.


Problem (2.16). For the sets A = {1, 2, . . . , 10} and B = {2, 4, 6, 9, 12, 25},consider the statements:

P : A ⊆ B. Q : |A−B| = 6.

Determine which of the following statements are true.

(a) P ∨Q

(b) P ∨ (∼ Q)

(c) P ∧Q

(d) (∼ P ) ∧Q

(e) (∼ P ) ∨ (∼ Q)

Solution. The set A is simply the set of integers between 1 and 10 inclusive.Clearly P is false. The set A − B consists of integers from A which are simul-taneously not in B. So A−B = {1, 3, 5, 7, 8, 10} and |A−B| = 6, which implyQ is true. The truth tables below show the truth value of each statement listedabove.

P ∼ P Q ∼ Q P ∨Q P ∨ (∼ Q) P ∧QF T T F T F F

P ∼ P Q ∼ Q (∼ P ) ∧Q (∼ P ) ∨ (∼ Q)F T T F T T

Remark. If a single truth table is too wide, consider using two or more truthtables. However, each truth table should be indepedent of the others. Noticehow the second truth table repeats the truth values P , ∼ P , Q, and ∼ Q.Without them, the other parts of the second table would not be as clear ormake as much sense if the second table were to be by itself.


Problem (2.22). Consider the statements:

P :√2 is rational. Q : 2

3is rational. R :

√3 is rational.

Write each of the following statements in words and indicate whether the state-ment is true or false.

(a) (P ∧Q) ⇒ R

(b) (P ∧Q) ⇒ (∼ R)

(c) ((∼ P ) ∧Q) ⇒ R

(d) (P ∨Q) ⇒ (∼ R)

Solution. The following are the above statements in words and whether theyare true or false. The truth values can be explained by the corresponding truthtables.

(a) If both√2 and 2

3are rational, then

√3 is rational. This statement is true.

P Q R P ∧Q (P ∧Q) ⇒ RF T F F T

(b) If both√2 and 2

3are rational, then

√3 is not rational. This statement is

true.

P Q R ∼ R P ∧Q (P ∧Q) ⇒ (∼ R)F T F T F T

(c) If√2 is not rational and 2

3is rational, then

√3 is rational. This statement

is false.

P ∼ P Q R (∼ P ) ∧Q ((∼ P ) ∧Q) ⇒ RF T T F T F

(d) If√2 is rational or 2

3is rational, then

√3 is not rational. This statement

is true.

P Q R ∼ R P ∨Q (P ∨Q) ⇒ (∼ R)F T F T F T


Problem (2.26). A college student makes the following statement:

If I don’t see my advisor today, then I’ll see her tomorrow.

For each of the following, determine whether this statement is true or false.

(a) The student doesn’t see his advisor either day.

(b) The student sees his advisor both days.

(c) The student sees his advisor on one of the two days.

(d) The student doesn’t see his advisor today and waits until next week to seeher.

Solution. Let P and Q be the statements

I don’t see my advisor today.

and

I’ll see her tomorrow.

respectively. The given statement, in symbols, is P ⇒ Q. The truth value ofthe implication P ⇒ Q can be determined using the truth table below.

P Q P ⇒ Q(a) T F F(b) F T T(c) T T Tor F F Tor F T T(d) T F T

For part (c), there are actually three possible cases:

1. The student does not see his advisor today (P is true) and sees his advisortomorrow (Q is true).

2. The student sees his advisor today (P is false) and not tomorrow (Q isfalse).

3. The student sees his advisor both days, and hence on one of the days (Pwould be false, whereas Q is true).

In all three cases, the implication P ⇒ Q is true.

Remark. Please note that the phrase “on one of those days” does not meanthat the student sees his advisor only once. If he sees his advisor on a dailybasis, then he certainly has seen his advisor “on one of those days.”

Tuto Logic

Documents

Transcript of Tuto Logic