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Contents

Chapter 1: Course Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Section 1.1: Exam 1 Fall 2003 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1



Section 1.4: Final Fall 03 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Section 1.5: Exam 1 Spring 2004 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12



Section 1.8: Final Spring 2004 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20


Section 1.10: Exam 2 Fall 2004 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28


Section 1.12: Final Fall 2004 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Section 1.13: Exam 1 Spring 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37



Section 1.16: Final Spring 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42




Section 1.20: Final Exam Fall 2006 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54


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Section 1.25: Quizzes Fall 2016 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74








1

Chapter 1: Course Documents

Section 1.1: Exam 1 Fall 2003

Exam 1 Math 3715 Fall 2003

1. Which of the following are propositions? For each statement that is a proposition,determine the truth value.

(2) a. x+ 4 = 8

This is not a proposition since the statement is sometimes true and sometimes false.

(2) b. ∃x(x+ 4 = 8)

This is a proposition since its truth value can be determined. The statement is true.

(2) c. ∀x(x+ 4 = 8)

This is a proposition since its truth value can be determined. The statement is false.

(3) 2. Give the truth table for p → (p ∧ q).

p q p ∧ q p → (p ∧ q)

T T T T

T F F F

F T F T

F F F T

(4) 3. Prove that (p ∧ q) → (p ∨ q) is a tautology.

Proof : Consider the truth table.

p q p ∧ q p ∨ q (p ∧ q) → (p ∨ q)

T T T T T

T F F T T

F T F T T

F F F F T

Proof : Consider

(p ∧ q) → (p ∨ q) ≡ [¬(p ∧ q)] ∨ (p ∨ q)

2

≡ (¬p∨¬q)∨ (p∨ q) (De Morgan’s law)

≡ ¬p ∨ ¬q ∨ p ∨ q (associative law)

≡ ¬p ∨ p ∨ ¬q ∨ q (commutative law)

≡ (¬p ∨ p) ∨ (¬q ∨ q) (associative law)

≡ T ∨ T (negation law)

≡ T (domination law).

4. Let p and q be the statements

p : John goes camping.

q : John gets poison ivy.

a. Write each of the following using quantifiers and connectives.

(2) i. John goes camping or John gets poison ivy.

p ∨ q

(2) ii. In order for John to go camping, it is necessary for John to get poison ivy.

p → q

(2) iii. If John goes camping, he will not get poison ivy.

p → ¬q

b. Write each of the following in words.

(3) i. The converse of p → q.

q → p

If John get poison ivy, then John goes camping.

(3) ii. The contrapositive of p → q.

¬q → ¬p

If John does not get poison ivy, then John does not go camping.

5. Give the negation of each proposition. Write you answer so that the negation symbolimmediately precedes the predicate.

(2) a. ∀x∀yP (x, y) ∃x∃y¬P (x, y)

3

(2) b. ∃x∃yP (x, y)

∀x∀y¬P (x, y)

(2) c. ∀x∃yP (x, y)

∃x∀y¬P (x, y)

(2) d. ∃x∀yP (x, y)

∀x∃y¬P (x, y)

(4) 6. Prove or disprove the following. If x and y are irrational, then xy is irrational.

See the text.

7. Let A = {1, 2, 7, a, b, α, β} and B = {2, 3, b, c, β, γ}. Also, suppose that the universal setis U = {1, 2, 3, 4, 5, 6, 7, a, b, c, α, β, γ, δ} List the elements of each of the following.

(1) a. A ∩ B = {2, b, β}

(1) b. A \B = {1, 7, a, α}

(2) c. A ∩B = {1, 7, a, α}

8. Let A = {0, 1} and B = {1, 2}.

(1) a. |A| = 2

(2) b. A×B = {(0, 1), (0, 2), (1, 1), (1, 2)}

(2) c. P(B) = {∅, {1}, {2}, {1, 2}}

(2) 9. State De Morgan’s laws for sets.

See the text or your class notes.

(4) 10. Prove that if A and B are sets, then A \B = A ∩ B.

Proof :

Suppose that x ∈ A \B. Then x ∈ A and x /∈ B. Since x /∈ B, x ∈ B. Also, since x ∈ Aand x ∈ B, x ∈ A ∩B and so A \B ⊆ A ∩B.

Now suppose that x ∈ A ∩ B. Then x ∈ A and x ∈ B. Since x ∈ B, x /∈ B. Also, sincex ∈ A and x /∈ B, x ∈ A \B and so A ∩B ⊆ A \B. Therefore, A \B = A ∩B.



(3) 11. Define f : R \ {0} → R by f(x) = 1x. Find f(N).

f(N) ={1, 1

2, 13, 14, . . .

}

4

(3) 12. Define g : R → R by g(x) = x2. Find g-1([-9, 9]).

f-1([-9, 9]) = [-3, 3]

13. Define f : N× N → N by f(m,n) = m · n.

(3) a. Is f one-to-one?

No. Note that f(2, 3) = f(1, 6).

(3) b. Is f onto?

Yes. For each n ∈ N, f(n, 1) = n.

(5) 14. Let f : X → Y and let A ⊆ X . Prove or disprove the following statement. Ifx /∈ A, then f(x) /∈ f(A).

This is not true.

Example 1: Let X = Y = {0, 1}, A = {0}, and define f : X → Y by f(x) = 1. Thenf(A) = f(X) = {1}. Note that 1 /∈ A and f(1) = 1 ∈ f(A).

(5) 15. Prove one of the following statements.

a. If f : X → Y and g : Y → Z are one-to-one, then g ◦ f : X → Z is one-to-one.

Proof : Let x1, x

2∈ X with x

16= x

2. Since f is 1–1, f(x

1) 6= f(x

2) and since g is 1–1,

g(f(x1)) 6= g(f(x2)). Therefore, g ◦ f is 1–1.

b. If f : X → Y and g : Y → Z are onto, then g ◦ f : X → Z is onto.

Proof : Let z ∈ Z. Since g is onto there exists y ∈ Y such that g(y) = z. Also, since f isonto, there exists x ∈ X such that f(x) = y. Therefore, (g ◦ f)(x) = g(f(x)) = g(y) = zand so g ◦ f is onto.

(3) 16. State the Fundamental Theorem of Arithmetic.

(3) 17. State the Division Algorithm.

(3) 18. Calculate [gcd(1000, 678)] · [lcm(1000, 678)].

5

19. For each of the following, determine whether the given integer is prime or composite.If it is prime, explain why it is prime. If it is composite, give its prime factorization.

(2) a. 240 = 24 · 3 · 5

(3) b. 167

This number is prime. The prime numbers less than√167 are 2, 3, 5, 7, and 11. None of

these primes divide 167.

20. (2) a. Calculate 47 div 7.

47 div 7 = 6

(2) b. Calculate 47 mod 7.

47 mod 7 = 5

21. (3) a. Find gcd(240, 140). (3) b. Find lcm(240, 140).

Since 240 = 24 · 3 · 5 and 140 = 22 · 5 · 7, gcd(240, 140) = 22 · 5 = 20 and lcm(240, 140) =24 · 3 · 5 · 7 = 1680.

(4) 22. Let a, b, c ∈ Z such that a|b and a|c. Prove that a|(b+ c).

Proof : Since a|b there exists j ∈ Z such that aj = b. Also, since a|c there exists k ∈ Zsuch that ak = c. Then (j+ k) ∈ Z and a(j+ k) = aj+ ak = b+ c. Therefore, a|(b+ c).



(3) 23. Find the base 10 expansion of (4C2)16 .

4 · 256 + 12 · 16 + 2 = 1218

(3) 24. Find the binary expansion of 81.

(1010001)2

25. Find each of the following.

(3) a. gcd(927, 412) = 103

927 = 412 · 2 + 103

412 = 103 · 4

gcd(927, 412) = 103

(3) b. gcd(341, 93) = 31

341 = 93 · 3 + 62

93 = 62 · 1 + 31

62 = 31 · 2

gcd(341, 93) = 31

6

(3) 26. Find the inverse of 7 modulo 12.

(3) 27. Find the inverse of 12 modulo 7.

(5) 28. Solve the following system of congruences.

x ≡ 1 (mod 2)x ≡ 1 (mod 3)x ≡ 3 (mod 5)

Mimic the proof of the Chinese Remainder Theorem.

Let a1 = 1, a2 = 1, and a3 = 3.

Also, let M1 = 15, M2 = 10, and M3 = 6.

Finally, let k1 = 1, k2 = 1, and k3 = 1 and x = 1 · 15 · 1 + 1 · 10 · 1 + 3 · 6 · 1 = 43.

Note that 43 ≡ 13 (mod 30).

29. (3) a. State Fermat’s Little Theorem.

(4) b. Calculate 72000mod3.

Since 3 ∤ 71000, Fermat’s Little Theorem implies that 72000 = (71000)2 ≡ 1 (mod 3).

7

(4) 30. Prove that there are no integers x and y such that x4 + y4 = 15.

Proof : Since 24 = 16, x, y ∈ {-1, 0, 1} which implies that x4 + y4 ≤ 2.

(3) 31. Define what it means for a set A to be countable.

(5) 32. Prove that if A is countable and B ⊆ A, then B is countable.

Proof : Since A is countable, |A| ≤ |N| and so there is a 1–1 function f : A → N. Defineg : B → N by g(x) = f(x) for all x ∈ B. To see that g is 1–1 let x1 , x2 ∈ B such thatx1 6= x2 . Since f is 1–1, g(x1) = f(x1) 6= f(x2) = g(x2). Therefore, f is 1–1 and so|B| ≤ |N| which implies that B is countable.

33. Find the 10th term of each of the following sequences.

(3) a. {1, 1, 2, 3, 5, 8, 13, 21, 34, 55}

(3) b. {3, 5, 9, 17, 33, 65, 129, 257, 513, 1025}

(4) 34. Use induction to prove that 2n ≤ n! for any natural number n ≥ 4.

See page 246.

Section 1.4: Final Fall 03

Final Exam Math 3715 Fall 2003

Name:

E-mail (optional):

Directions: Show all of your work and justify all of your answers. An answer withoutjustification will receive a zero.

35. Consider the statement “when there is snow on the ground, the outside temperature iscold.”

(2) a. Write the converse of the statement.

(2) b. Write the contrapositive of the statement.

(2) 36. Use a truth table to show that p → q ≡ ¬p ∨ q.

(3) 37. Show that [p ∧ (p → q)] → q (modus ponens) is a tautology.

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38. Let P (x, y) be the statement “x can fool y,” where the universe of discourse is the setof all people. Use quantifiers to express each of the following statements.

(1) a. Everybody can fool somebody.

(1) b. There is no one who can fool everyone.

(1) c. Everyone can be fooled by somebody.

(3) 39. Prove that if A, B, and C are sets, then (A ∩ B)× C ⊆ (A× C) ∩ (B × C).

40. Define f : Z → Z by f(x) = 2x and g : Z → Z by g(x) = x2.

(3) a. Is g ◦ f 1–1?

(3) b. Find the range of g ◦ f.

9

41. (1) a. Calculate 166 div 7.

(1) b. Calculate 166 mod 7.

42. (1) a. Find the prime factorization of 550.

(1) b. Find the prime factorization of 1176.

(1) c. Find the gcd(1176, 550).

(1) d. Find the lcm(1176, 550).


x ≡ 2 (mod 3)x ≡ 1 (mod 4)x ≡ 3 (mod 5)

10

(3) 44. Prove that if every even natural number can be written as the sum of two primenumbers, then every odd natural number greater than 5 can be written as the sum of threeprime numbers.

45. Let A and B be sets.

(2) a. Define what it means for |A| ≤ |B|.

(2) b. Prove that if A ⊆ B, then |A| ≤ |B|.

(3) 46. Use induction to prove that for all n ∈ N, the sum of the first n even integers is

n(n + 1)

(n∑

i=1

2i = n(n+ 1)

)

.

11

(2) 47. Give an example of a relation from R to Q.

48. Define a relation R on P(N) by A R B if and only if A ∩ B 6= ∅.

(1) a. Is the relation R reflexive.

(1) b. Is the relation R symmetric.

(2) c. Is the relation R antisymmetric.

(2) d. Is the relation R transitive.

12

49. Let S = {3, 5, 9, 15, 24, 45} and define a relation 4 on S by a 4 b if and only if a|b.Answer the following as true or false. Do not forget to justify your answer.

(1) a. 45 is a maximal element.

(1) b. 45 is the greatest element.

(1) c. 5 is minimal element.

(1) d. There is no least element.

(2) 50. Define what it means for a poset (S,4) to be dense.

Section 1.5: Exam 1 Spring 2004

Exam 1 Math 3715 Spring 2004

51. Which of the following are propositions.

(1) a. It is 3:00 A.M.

Yes.

(1) b. 4 + 3 = 7

Yes.

(1) c. x+ 3 = 7

No.

(1) d. Cleveland is the capital of Ohio.

Yes.

(3) 52. You are in a prison with two guards. One guard always tells the truth and theother always lies. You must choose one of two doors. One door leads to freedom and theother does not. You are allowed to ask one of the guards one question. What questionwould you ask and why?

Ask one guard which door the other guard will claim is the door to freedom. Then choosethe opposite door. If the guard you ask is is the one that always lies, then he will lie aboutthe other guards answer which would be the truth. If the guard you ask is the one thatalways tells the truth, then he will truthfully report the other guards answer which wouldbe a lie.

53. Prove each of the following.

(3) a. ¬(p → q) ≡ p ∧ ¬q

Proof : Consider

¬(p → q)

≡ ¬(¬p ∨ q)

≡ p ∧ ¬q (De Morgans’s Laws).

13

(3) b. p → (q ∨ r) ≡ (p → q) ∨ r

Proof : Consider

p → (q ∨ r)

≡ ¬p ∨ (q ∨ r)

≡ (¬p ∨ q) ∨ r (associativity)

≡ (p → q) ∨ r.

54. Let the universe of discourse be the set of all animals. Consider the propositionalfunctions listed below.

D(x) : “x is a duck”

B(x) : “x is a beaver”

F (x) : “x can fly”

S(x) : “x can swim”

Express each of the following using the propositional functions above, universal quantifiers,and logical connectives.

(2) a. All animals are ducks.

∀xD(x)

(2) b. There is a beaver that can swim.

∃x[B(x) ∧ S(x)]

14

(2) c. All ducks can fly.

∀x[D(x) → F (x)]

(2) d. Ducks are not beavers.

∀x[D(x) → ¬B(x)]

55. Let P (x, y) be the statement “x · y = x” where the universe of discourse is the set ofall real numbers. Express each of the following in English and give its truth value.

(3) a. ∀x∃yP (x, y)

For all x ∈ R there exists y ∈ R such that x · y = x. This is true since x · 1 = x for allx ∈ R.

(3) b. ∃x∀yP (x, y)

There exists x ∈ R such that for all y ∈ R x ·y = x. This is true since 0 ·y = 0 for all y ∈ R.

(3) c. ∀x∀yP (x, y)

For all x ∈ R and for all y ∈ R, x · y = x. This is false since 2 · 3 6= 2.

(4) 56. Choose one of the following.

a. Prove or disprove that if a, b ∈ Q, then ab ∈ Q.

See your class notes.

b. Prove or disprove that if a, b ∈ R \Q, then ab ∈ R \Q.

See the text.

57. Let A = {1, 3, a, b, 9} and B = {1, 2, a, c, 7}.

(2) a. A ∪ B = {1, 2, 3, 7, 9, a, b, c} (2) b. A ∩ B = {1, a, }

58. Let A = {a, b} and B = {1, 2, 3}.

(2) a. A× B = {(a, 1), (a, 2), (a, 3), (b, 1), (b, 2), (b, 3)}

(2) b. B × A = {(1, a), (1, b), (2, a), (2, b), (3, a), (3, b)} B × A =

(2) 59. Let S = {1, 2, 3, 3, 4, 5, 5, 5}. Find |S|.

|S| = 5

60. Define f : R → R by f(x) = x2.

(3) a. f(N) = {1, 4, 9, 16, 25, 36, 49, 64, 81, . . .}

15

(3) b. f-1(N) = {1,

√2,√3, 2,

√5,√6,√7,√8, 3, . . .}∪{-1, -

√2, -

√3, -2, -

√5, -

√6, -

√7, -

√8, -3, . . .}



(2) 61. Find the prime factorization of858.

858 = 2 · 3 · 11 · 13

Find the prime factorization of 462.

462 = 2 · 3 · 7 · 11

(2) 62. Is 113 a prime number?

Yes. Note that 113 is not divisible by 2, 3, 5, 7, or 11 and 112 = 121 > 113.

63. (3) a. Let p, q ∈ Z. Define what it means for p to divide q.

(3) b. Use the definition above to show that if a, b, c ∈ Z such that a|b and b|c, then a|c.

Proof : Since a|b, there exists r ∈ Z such that ar = b. Also, since b|c, there exists s ∈ Zsuch that bs = c. Note that since r, s ∈ Z, rs ∈ Z. Now consider a(rs) = (ar)s = bs = c.Therefore, a|c.

64. Compute each of the following.

(2) a. 104 div 5

Since 104 = 20 · 5 + 4, 104 div 5 = 20.

(2) b. 104 mod 5

Since 104 = 20 · 5 + 4, 104 mod 5 = 4.


(3) a. gcd(600, 84)

600 = 23 · 3 · 52

84 = 22 · 3 · 7

gcd(600, 84) = 22 · 3 = 12

(3) b. lcm(600, 84)

600 = 23 · 3 · 52

84 = 22 · 3 · 7

lcm(600, 84) = 23 · 3 · 52 · 7 = 4200

(3) 66. Use the Euclidean algorithm to compute gcd(414, 662).

662 = 1 · 414 + 248

414 = 1 · 248 + 166

248 = 1 · 166 + 82

166 = 2 · 82 + 2

82 = 2 · 41 + 0

gcd(414, 662) = 2

16

(3) 67. Let a, b ∈ N such that gcd(a, b) = 3 and lcm(a, b) = 18. Find a · b.

a · b = gcd(a, b) · lcm(a, b) = 18 · 3 = 51

(2) 68. Does 2 have an inverse modulo5?

Yes. Note that 2 · 3 ≡ 1 (mod 5).

(2) 69. Does 6 have an inverse modulo3?

No. Note that 6 · 1 ≡ 0 (mod 3) and6 · 2 ≡ 0 (mod 3).


x ≡ 1 (mod 2)x ≡ 2 (mod 3)x ≡ 2 (mod 5)


Let a1 = 1, a2 = 2, and a3 = 2.

Also, let M1 = 15, M2 = 10, and M3 = 6.


Note that 47 ≡ 17 (mod 30).

71. (2) a. Find the prime factorization of 4096.

4096 = 212

(3) b. State Fermat’s Little Theorem.

(3) c. Compute 4096 mod 13.

Since 4096 = 212 = 213−1, 4096 ≡ 1 (mod 13) by Fermat’s Little Theorem.

(4) 72. Let f : X → Y and suppose that A,B ⊆ Y . Prove that f-1(A ∩ B) =

f-1(A) ∩ f

-1(B).

Proof : First suppose that x ∈ f-1(A ∩ B). Then f(x) ∈ A ∩ B ⊆ A and f(x) ∈ A ∩ B ⊆

B. Since f(x) ∈ A, x ∈ f-1(A). Likewise, since f(x) ∈ B, x ∈ f

-1(B). So we have

that x ∈ f-1(A) and x ∈ f

-1(B) which implies that x ∈ (f

-1(A) ∩ f

-1(B)). Therefore,

f-1(A ∩B) ⊆ f

-1(A) ∩ f

-1(B).

17

Now suppose that x ∈ f-1(A) ∩ f

-1(B). Since x ∈ f

-1(A), f(x) ∈ A and since x ∈ f

-1(B),

f(x) ∈ B. Hence, f(x) ∈ A ∩ B. Therefore, x ∈ f-1(A ∩ B) and so f

-1(A) ∩ f

-1(B) ⊆

f-1(A ∩B).

Since f-1(A ∩ B) ⊆ f

-1(A) ∩ f

-1(B) and f

-1(A) ∩ f

-1(B) ⊆ f

-1(A ∩ B), f

-1(A ∩ B) =

f-1(A) ∩ f

-1(B).

(4) 73. Prove that the composition of two one-to-one functions is one-to-one.

Proof : Suppose that f : X → Y and g : Y → Z are 1–1. Let x1, x

2∈ X with x

16= x

2.

Since f is 1–1, f(x1) 6= f(x2) and since g is 1–1, g(f(x1)) 6= g(f(x2)). Therefore, g ◦ f is1–1.


74. For each of the following, give the next term of the sequence.

(3) a. {1, 9, 25, 49, . . .}

81

(3) b. {1, 2, 2, 4, 8, 32, . . .}

256

75. For each of the following, determine whether or not the given equation has a solutionin the set of integers. Justify your answer.

(3) a. x2 + y3 = 3

x = 2, y = -1

(3) b. x2 + y2 = 7

If there exist x, y ∈ Z such that x2+y2 = 7, then x and y can be assumed to be nonnegative.Also, x, y ∈ {0, 1, 2}. So we check that 22 +22 > 7 and 22+12 < 7. Therefore, there are nointegers x, y such that x2 + y2 = 7.

(5) 76. Use induction to prove that for all n ∈ N, the sum of the first n even integers is

n(n + 1)

(n∑

i=1

2i = n(n+ 1)

)

.

Proof : First, note that since 2 = 1(1 + 1) the statement is true for n = 1. Now suppose

that k ∈ N such that

k∑

i=1

2i = k(k + 1) and consider

k+1∑

i=1

2i

18

=

k∑

i=1

2i+ 2(k + 1)

= k(k + 1) + 2(k + 1) (inductive hypothesis)

= (k + 1)(k + 2)

as desired.

(6) 77. Find all n ∈ N for which 2n+ 1 ≤ n2? Prove your answer.

For n ∈ N, n ≥ 3, 2n+ 1 ≤ n2.

Proof : Note that for n = 3, 2n + 1 = 7 and n2 = 9. Now suppose that k ∈ N such thatsuch that k ≥ 3 and 2k + 1 ≤ k2. Then

2k + 1 ≤ k2

2k + 3 ≤ k2 + 2

2(k + 1) + 1 ≤ k2 + k

2(k + 1) + 1 ≤ k(k + 1)

2(k + 1) + 1 ≤ (k + 1)(k + 1)

2(k + 1) + 1 ≤ (k + 1)2

as desired.

(2) 78. Explain the difference between induction and strong induction.

79. For each of the following relations, determine whether or not the relation is reflexive,symmetric, antisymmetric, or transitive.

(4) a. Let R be the relation on N defined by (a, b) ∈ R if and only if gcd(a, b) = 2.

Reflexive: No.

gcd(1, 1) = 1

Symmetric: Yes.

gcd(a, b) = gcd(b, a)

Antisymmetric: No.

gcd(4, 2) = gcd(2, 4)

2 6= 4

Transitive: No.

gcd(4, 6) = 2, gcd(6, 8) = 2, gcd(4, 8) 6= 2

19

(2) 80. Let S be a set and R a relation on S. Define what it means for R to be anequivalence relation.

(2) 81. Let S be a set and R a relation on S. Define what it means for R to be a partialorder relation.

(4) 82. Prove or disprove the following statement.

A relation cannot be both symmetric and antisymmetric.

The statement is false. Consider (R,=)

83. Let S = {2, 3, 4, 5, 6, . . .} = N \ {1} and consider (S, |).

(3) a. Verify that (S, |) is a poset.

Reflexive: Note that a|a for all a ∈ S.

Antisymmetric: Suppose that a|b and b|a. Then a ≤ b and b ≤ a which implies that a = b.

Transitive: Suppose that a|b and b|c. Then a|c by theorem 1 of section 2.4.

(2) b. Are any elements from S minimal? If so, which ones.

Yes. The prime numbers and only the prime numbers are minimal.

(2) c. Are any elements from S maximal? If so, which ones.

No. If a ∈ S, then a|2a and so a is not maximal.

(2) d. Is there a minimum element? If so, which one?

No. Let a ∈ S and suppose that p is a prime number such that p 6= a. Then a ∤ p.

(2) e. Is there a maximum element? If so, which one?

No. Let a ∈ S then there is a prime number p such that p ∤ a.

f. Let A = {3, 6, 8, 12}.

(2) i. Does A have any lower bounds? If so, find them.

No. Note that 2 is a lower bound of {6, 8, 12} and 3 is a lower bound of {3, 6, 12}. Since2 ∤ 3 and 3 ∤ 8, there is no lower bound for A.

(2) ii. Does A have any upper bounds? If so, find them.

20

Yes. Note that 3, 6, 8, and 12 all divide 24. So any multiple of 24 is an upper bound of A.

Section 1.8: Final Spring 2004

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84. Consider the statement “when there is snow on the ground, the outside temperature iscold.”

(2) a. Write the converse of the statement.

(2) b. Write the contrapositive of the statement.

(2) c. Write the inverse of the statement.


(6) 86. Show that [¬q ∧ (p → q)] → ¬p (modus tollens) is a tautology.

21

87. Let T (x, y) be the statement “x is taller than y,” where the universe of discourse is theset of all people. Use quantifiers to express each of the following statements.

(2) a. Everybody is taller than somebody.

(2) b. There is no one who is taller than everyone.

(2) c. Everyone is shorter than somebody.

88. (2) a. Calculate 147 div 7. (2) b. Calculate 147 mod 7.

89. (4) a. Find the prime factorizations of 600 and 300.

(2) b. Find the gcd(630, 300). (2) c. Find the lcm(630, 300).

22


x ≡ 2 (mod 3)x ≡ 1 (mod 4)x ≡ 3 (mod 5)

(4) 91. Find the base 4 expansion of 1023.

(4) 92. Let a, b ∈ Z such that gcd(a, b) = 6 and lcm(a, b) = 36. Find a · b.

(6) 93. Give an example of a set and a relation on the set. Is your relation reflexive,symmetric, antisymmetric, or transitive?

23

94. Consider the poset ({{1}, {2}, {4}, {1, 2}, {1, 4}, {2, 4}, {3, 4}, {1, 3, 4}, {2, 3, 4}},⊆) .

(2) a. Find the maximal elements.

(2) b. Find the minimal elements.

(2) c. Is there a greatest element?

(2) d. Is there a least element?

(2) e. Find all upper bounds of {{2}, {4}}.

(2) f. Find the least upper bound of {{2}, {4}}, if it exists.

(2) g. Find all lower bounds of {{1, 3, 4}, {2, 3, 4}}.

(2) h. Find the greatest lower bound of {{1, 3, 4}, {2, 3, 4}} if it exists.

24

(6) 95. Find the sum-of-products expansion for the Boolean function f(x, y, z) = (x+y)z.

(6) 96. Use a K-map to simplify the Boolean expression

w x y z + w x y z + w x y z + w x y z + w x y z + w x y z + w x y z.

(6) 97. Prove that√3 is irrational.

25

(4) 98. Use induction to prove that the sum of the first n natural numbers isn(n+ 1)

2.



(6) b. Prove that if A ⊆ B, then |P(A)| ≤ |P(B)|.

(6) 100. Use the pigeonhole principle to show that if six natural numbers are selectedfrom the first ten, then there must be at least one pair of these numbers whose sum is 11.



101. Determine whether or not each of the following is a proposition.

(1) a. 2 + 4 = 9

Yes.

(1) b. Is it true that 2 + 4 = 9?

No.

102. For each of the following, identify the bound and free variables.

(2) a. ∃xP (x, y)

Bound: x

Free: y

(2) b. ∃x∀yS(x, y, z)

Bound: x, y

Free: z

(3) 103. Use a truth table to prove ¬(p ↔ q) ≡ (p ↔ ¬q).

p q p ↔ q ¬(p ↔ q) p ↔ ¬qT T T F F

T F F T T

F T F T T

F F T F F

(4) 104. Prove that (p ∨ q) ∧ (¬p ∨ r) → (q ∨ r) is a tautology.

Proof : Consider

(p ∨ q) ∧ (¬p ∨ r) → (q ∨ r)

≡ ¬[(p ∨ q) ∧ (¬p ∨ r)] ∨ (q ∨ r)

≡ [¬(p ∨ q) ∨ ¬(¬p ∨ r)] ∨ (q ∨ r)

≡ [(¬p ∧ ¬q) ∨ (p ∧ ¬r)] ∨ (q ∨ r)

26

≡ (¬p ∧ ¬q) ∨ (p ∧ ¬r) ∨ q ∨ r

≡ (¬p ∧ ¬q) ∨ q ∨ (p ∧ ¬r) ∨ r

≡ [(¬p∨q)∧ (¬q∨q)]∨ [(p∨r)∧ (¬r∨r)]

≡ [(¬p ∨ q) ∧ T] ∨ [(p ∨ r) ∧ T]

≡ (¬p ∨ q) ∨ (p ∨ r)

≡ (¬p ∨ p) ∨ (q ∨ r)

≡ T ∨ (q ∨ r)

≡ T.

27

105. Let A(x, y) be the statement “x annoys y.” Express each of the following using thepropositional function A(x, y) and logical operators.

(2) a. There is someone who annoys everyone.

∃x∀yA(x, y)

(2) b. There is someone whom everyone annoys.

∃y∀xA(x, y)

(2) c. Murray annoys everyone except Maggie.

[∀x(x 6= Maggie → A(Murray, x)] ∧ ¬A(Murray,Maggie)

106. Let p, q, r, and s be the following statements.

p : Green creatures have five legs.

q : Creatures with five legs have wings.

r : Frogs are green creatures.

s : Creatures with wings can fly.

a. Use words to state the following propositions.

(2) i. p → q

If green creatures have five legs, then creatures with five legs have wings.

(2) ii. ¬r ∧ s

Frogs are not green creatures and creatures with wings can fly.

b. Use p, q, r, and s and logical operators to express the following propositions.

(2) i. Green creatures have five legs and creatures with wings can fly.

p ∧ s

(2) ii. If frogs are green creatures, then green creatures do not have five legs.

r → ¬p

c. (3) i. Use p, q, r, and s to give a valid argument for the statement “frogs can fly.”

We are given that frogs are green creatures (r) and green creatures have five legs (p).Therefore, frogs have five legs. Since frogs have five legs and creatures with five legs havewings (q), frogs have wings. Since frogs have wings and creatures with wings can fly (s),frogs can fly.

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(2) ii. Why did the valid argument yield a false conclusion?

The propositions are false.

107. For each of the following, determine whether or not the number is rational or irrational.Justify your answer.

(2) a. .543

Rational.

.543 = 5431000

(4) b.√5

Irrational.

The proof is similar to example 21 on page 66.

(3) 108. Show that there are irrational numbers x and y such that xy is rational.

See example 26 on page 69.

109. Let A = {1, 2, 3}, B = {α, β}, and C = {1, 2, a, 1, a}.

(2) a. A× B = {(1, α), (1, β), (2, α), (2, β), (3, α), (3, β)}

(2) b. P(B) = {∅, {α}, {β}, B = {α, β}}

(2) c. |B| = 2

(2) d. |C| = 3



110. For each n ∈ N, let An= {1, 2, 3, . . . , n}.

(3) a.100⋃

n=1

An= {1, 2, 3, . . . , 100} (3) b.

100⋂

n=1

An= {1}

(6) 111. Suppose that A and B are subsets of a universal set U . Prove that A \ B =A ∩B.

29

Proof : To see that A \B ⊆ A∩B, suppose that x ∈ A \B. Then x ∈ A and x /∈ B. Sincex /∈ B, x ∈ B. So we have that x ∈ A and x ∈ B which implies that x ∈ A∩B. Therefore,A \B ⊆ A ∩B. To see that A∩B ⊆ A \B, suppose that x ∈ A ∩B. Since x ∈ B, x /∈ B.So we have that x ∈ A and x /∈ B which implies that x ∈ A \ B. Hence, A ∩ B ⊆ A \ B.Since A \B ⊆ A ∩ B and A ∩B ⊆ A \B, A \B = A ∩ B.

112. Define f : Z → Z by f(x) = |x| for all x ∈ Z. Let A = {0, -1, 2} and B ={0, -1, -2, -3, . . .}.

(4) a. f(A) = {0, 1, 2}

(4) b. f-1(A) = {0, 2, -2}

(3) c. f-1(f(B)) = Z

(3) d. f(f-1(B)) = {0}

(6) 113. Let f : X → Y, A ⊆ X, and B ⊆ X . Prove that f(A ∪ B) = f(A) ∪ f(B).

Proof :

First, suppose that y ∈ f(A ∪ B). Then there exists x ∈ A ∪ B such that f(x) = y.Since x ∈ A ∪ B x ∈ A or x ∈ B. Since x ∈ A or x ∈ B, y = f(x) ∈ f(A) or y =f(x) ∈ f(B). Hence, y = f(x) ∈ f(A) ∪ f(B) and so f(A ∪ B) ⊆ f(A) ∪ f(B). Nowsuppose that y ∈ f(A)∪ f(B). Since y ∈ f(A)∪ f(B), y ∈ f(A) or y ∈ f(B). Without lossof generality, suppose that y ∈ f(A). Then there exists x ∈ A such that f(x) = y. Sincex ∈ A ⊆ A ∪ B, y = f(x) ∈ f(A ∪ B). Therefore, f(A) ∪ f(B) ⊆ f(A ∪ B) and the proofis complete.

114. (4) a. gcd(630, 300) = (4) b. lcm(630, 300) =

630 = 2 · 32 · 5 · 7

300 = 22 · 3 · 52

gcd(630, 300) = 2 · 3 · 5 = 30

lcm(630, 300) = 22 · 32 · 52 · 7 = 6300

(2) 115. State the division algorithm.

(5) 116. Let a, b,m ∈ Z with m ≥ 1. Also, suppose that qa= a div m, r

a= a mod m,

qb= b div m, and r

b= b mod m. Prove that ab ≡ r

arb(mod m).

Proof : Note that

Therefore, ab ≡ rarb(mod m).

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(6) 117. Solve the following system of congruencies.

x ≡ 1 (mod 3)x ≡ 1 (mod 4)x ≡ 2 (mod 5)


Let a1 = 1, a2 = 1, and a3 = 2.

Also, let M1 = 20, M2 = 15, and M3 = 12.


Note that 157 ≡ 37 (mod 60).




(4) 118. Given that e6 > 6!, prove that en > n! for all n ≥ 6.

(4) 119. Show that for each n ∈ N,

n∑

i=1

i =n(n + 1)

2.



(4) b. Prove that if A ⊆ B, then |A| ≤ |B|.

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121. For each of the following, determine whether or not there are integer solutions to thegiven equation.

(4) a. x2 + 3y = 7 (4) b. 2x2 + y2 = 10

122. For each of the following, find the next term of the sequence. Do not forget to justifyyour answer.

(4) a.{1, 1

2, 4, 1

8, 16, 1

32, . . .

}(4) b. {1, 3, 6, 10, 15, 21, 28, . . .}

(4) c. {1, 1, 2, 5, 8, 13, 21, 34, . . .}

(4) 123. Let S be the set of all real-valued sequences that are eventually 0. Show that Sis countable. Hint: Define a 1–1 function f : S → Q.

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124. For each of the following relations, determine whether the relation is reflexive, sym-metric, antisymmetric, or transitive.

(4) a. Define R on the set of integers Z by a R b if a divides b.

(4) b. Define R on the set of real numbers R by a R b if a < b.

(4) 125. Let R be a relation on a set S. Show that if R2 ⊆ R, then R is transitive.

126. (2) a. State Goldbach’s conjecture.

(2) b. Do you think Goldbach’s conjecture is true?

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(5) 127. Prove that (p → q) ∧ (q → r) and (p ∨ q) → r are logically equivalent.

(5) 128. Prove that [¬p ∧ (p ∨ q)] → q is a tautology.

129. Express each of the following using predicates and quantifiers.

(5) a. Every integer is odd or even. (5) b. No integer is both odd and even.

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(5) 130. Murray is talking to three sisters named Abby, Lucy, and Maggie. He has ratherimpolitely asked the sisters their relative ages (i.e., who is the oldest, youngest, etc.). In anattempt to confuse Murray, Abby and Lucy make the statements listed below. Assumingthey are telling the truth, can Murray now determine their relative ages.

Abby: “If Maggie is not the oldest, then I am.”

Lucy: “If Maggie is not the youngest, then I am the oldest.”

(5) 131. Let A, B, and C be sets. Show that (A ∩B)× C = (A× C) ∩ (B × C).

(5) 132. Let A be a set with 10 elements. What is the cardinality of P(P(A))?

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133. Let f(x) = ⌊x⌋. Find each of the following.

(5) a. f-1(1) (5) b. f

-1(Z) (5) c. f

-1 ((12, 3))

(5) 134. Prove that there is an infinite number of primes.

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(5) 135. Find the binary expansion of (105)10.

(5) 136. Prove that (a mod m)(b mod m) ≡ ab (mod m).

36


b. Calculate each of the following.

(5) i. 7728 mod 29 (5) ii. 7729 mod 29

(5) 138. Prove that n2 ≤ 2n for all n ∈ N.

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(5) 139. Define a relation R on N by a R b if b = 3a. Determine whether or not R isreflexive, symmetric, antisymmetric, or transitive.

(5) 140. What value of Boolean variables satisfy xy = x+ y?

(5) 141. Find the sum-of-products expansion of f(x, y, z) = xy + xz + yz.



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(4) 142. Show that p → (p → q) ≡ p → q.

Proof : Consider the following truth ta-ble.

p q p → q p → (p → q)

T T T T

T F F F

F T T T

F F T T

Proof : Consider

p → (p → q)

≡ p → (¬p ∨ q)

≡ ¬p ∨ (¬p ∨ q)

≡ (¬p ∨ ¬p) ∨ q

≡ ¬p ∨ q

≡ p → q.

(6) 143. State the converse, contrapositive, and inverse of the following implication.

If it is cold outside, it is snowing.

Converse: If it is snowing, it is cold outside.

Contrapositive: If it is not snowing, it is not cold outside.

Inverse: If it is not cold outside, it is not snowing.

(6) 144. Consider the following statements made by three sisters.

Abby: I am the oldest.

Sophie: I am older than Lucy.

Lucy: I am the oldest.

Determine the relative ages of Abby, Sophie, and Lucy given that exactly one of them islying.

Solution: Abby is the oldest, Sophie is the middle child, and Lucy is the youngest.

Since the statements made by Abby and Lucy are contradictory, one of them is lying.Therefore, Sophie is telling the truth when she says that she is older than Lucy. So Lucy isnot the oldest and is lying. Therefore, Abby is telling the truth when she claims to be theoldest.

(4) 145. Express the negation of the following so that all negation symbols immediatelyprecede predicates.

39

∀x∃y∀zP (x, y, z)

¬∀x∃y∀zP (x, y, z) ≡ ∃x∀y∃z¬P (x, y, z)

146. Consider the following statements where the universe of discourse for both x and y isthe set of all creatures in the animal kingdom.

D(x) : x is a dog

F (y) : y is a flea

B(x, y) : x has y on its back.

Express each of the following, usingD(x), F (y), B(x, y), quantifiers, and logical connectives.

(3) a. There is a dog with a flea on its back.

∃x∃y[D(x) ∧ F (y) ∧ B(x, y)]

(3) b. There is a dog with no fleas on its back.

∃x∀y[D(x) ∧ (F (y) → ¬B(x, y)]

(3) c. There is a creature that is both a dog and a flea.

∃x[D(x) ∧ F (x)]

(6) 147. Let N(x) be a propositional function and the universe of discourse be all realnumbers. Consider the following two propositions.

1. N(1).

2. ∀x[N(x) → N(x+ 1)].

Use propositions 1 and 2 to prove that N(x) is true for infinitely many real numbers x.

Proof : Suppose that A = {x ∈ R : N(x) is true} is finite. By proposition 1, A 6= ∅.Attempting a contradiction, suppose that A is finite. Then let y be the largest element ofA. By proposition 2, (y+1) ∈ A which contradicts the fact that y is the largest element ofA. Therefore, A is infinite.

148. For each of the following, A = {a, b, c}, B = {a, 1}, and the universal set is U ={a, b, c, 1, 2, α, β}. Find each of the following.

(3) a. A ∩ B = {a}

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(3) b. A ∪ B = {a, b, c, 1}

(3) c. A×B = {(a, a), (a, 1), (b, a), (b, 1), (c, a), (c, 1)}

(3) d. A = {1, 2, α, β}.

(3) e. P(B) = {∅, {a, 1}, {a}, {1}}



149. Define f : R → R by f(x) = x2 for all x ∈ R. Let A = {0, -1, 4} and B ={0, -1, -2, -3, . . .}.

(3) a. f(A) = {0, 1, 16}

(3) b. f-1(A) = {0, 2, -2}

(3) c. f-1(f(B)) = Z

(3) d. f(f-1(B)) = {0}

150. Define f : N → N by f(n) = n2 for all n ∈ N.

(3) a. Is f one-to-one?

Yes. Since√n2 = n for all n ∈ N, f(n1) = f(n2) if and only if n1 = n2 .

(3) b. Is f onto?

No. Note that f(N) = {1, 4, 9, 16, 25, . . .} 6= N.

(8) 151. Let f : X → Y, A,B ⊆ X, and C,D ⊆ Y . Prove or disprove two of thefollowing.

a. f(A ∪B) = f(A) ∪ f(B)

b. f(A ∩B) = f(A) ∩ f(B)

c. f(A) = f(A)

d. f-1(C ∪D) = f

-1(C) ∪ f

-1(D)

e. f-1(C ∩D) = f

-1(C) ∩ f

-1(D)

f. f-1(C) = f -1(C)

152. (3) a. State the Division Algorithm.

(4) b. Suppose that a, b,m ∈ Z with m ≥ 2 and that a ≡ b (mod m). Show that if d ∈ Z,then d|a if and only if d|b.

153. Given that 7700 = 22 · 52 · 7 · 11 and 5880 = 23 · 3 · 5 · 72 calculate the following.

(3) a. gcd(7700, 5880) = 22 · 5 · 7 (3) b. lcm(7700, 5880) = 23 ·3 ·52 ·72 ·11

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154. Find the prime factorization of each of the following.

(3) a. 3125 = 55 (3) b. 10!

(3) 155. Write the first 8 terms of the Fibonacci sequence.

1, 1, 2, 3, 5, 8, 13, 21

(3) 156. Give the sixth term of the following sequence. How did you find the sixth term?

{1, 4, 27, 256, 3125, . . .}

{11, 22, 33, 44, 55, 66, . . .}




x ≡ 2 (mod 3)x ≡ 3 (mod 4)x ≡ 2 (mod 5)x ≡ 5 (mod 7)


Then a1 = 2, a2 = 3, a3 = 2, and a4 = 5,

m1 = 3, m2 = 4, m3 = 5, and m4 = 7,

M1 = 140, M2 = 105, M3 = 84, and M4 = 60,

k1 = 2, k2 = 1, k3 = 4, and k4 = 2.

Let x = 2 · 140 · 2 + 3 · 105 · 1 + 2 · 84 · 4 + 5 · 60 · 2 = 560 + 315 + 672 + 600 = 2147.

Note that 2147 ≡ 47 (mod 420).



Since 17 ∤ 18, 1816 ≡ 1 (mod 17) by Fermat’s Little Theorem. Therefore, 1816mod17 = 1.

(4) 159. Are there integers x and y such that x2 + 2y2 = 7.

42

No. Note that if x, y ∈ Z such that x2 + 2y2 = 7, then x, y ∈ {0, 1, 2}. By checking allpossibilities, we see that there are no integers x and y such that x2 + 2y2 = 7.

(4) 160. Show that a subset of a countable set is countable.

See class notes.

(Page 253: 12) (4) 161. Prove that 3n < n! whenever n is a positive integer greater than6.

(4) 162. Prove that for any n ∈ N,

n∑

i=1

i3 = 13 + 23 + 33 + · · ·+ n3 =

(

n(n + 1)

2

)2

.

163. Let R be the relation on Z (the integers) defined by (a, b) ∈ R if and only if gcd(a, b) =1.

(3) a. Is R reflexive?

(3) b. Is R symmetric?

(3) c. Is R antisymmetric?

(3) d. Is R transitive?

(4) 164. Let S be a set and R a relation on S. Prove that if R is transitive, then R2 ⊆ R.

(4) 165. Give an example of an equivalence relation on Z. Prove that the relation is infact an equivalence relation.

166. Let A = {a, b}.

(3) a. How many relations are there on A?

24 = 16

(3) b. How many equivalence relations are there on A?

Two.

The equivalence relations A are {(a, a), (b, b)} and {(a, a), (b, b), (a, b), (b, a)}.

Section 1.16: Final Spring 2006

43

Final Exam Math 3715 Spring 2006

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167. All parts of this problem relate to inhabitants of the island of knights and knavescreated by Smullyan referenced in the text on page 20. Knights always tell the truth andknaves always lie. You encounter three people, A, B, and C. Determine if possible whatA, B, and C are if they address you in the ways described. If you cannot determine whatthese three people are, can you draw any conclusions?

a.

A : “Two of us are knights and one is a knave.”

B : “We are all knaves.”

C : “Two of us are knaves and one is a knight.”

b.

A : “We are all knights.”

B : “We are all knights.”

C : “We are all knights.”

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168. Show that (p ∧ q) → (p ∨ q) is a tautology.

(Page 40: 8) 169. Translate the following statements into English where

R(x) is “x is a rabbit,”

H(x) is “x hops,”

and the universe of discourse is the set of all animals.

a. ∀ x[R(x) → H(x)] b. ∃x[R(x) ∧H(x)]

(Page 254: 28) 170. For which nonnegative integers n is n2 < n!? Prove your answer usingmathematical induction.

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171. a. State De Morgan’s laws for sets.

b. Show that if A and B are sets, then A ∩B = A ∪B.

(Page 95: 10)172. Find sets A and B if A \B = {1, 5, 7, 8}, B \ A = {2, 10}, and A ∩ B = {3, 6, 9}.

173. Define f : Z → Z by f(z) = |z|. Let A = {0, 1, 2, 3, 4, . . .} and B = {0, -1, -2, -3, -4, . . .}

a. f(A ∩B) = b. f(A) ∩ f(B) =

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174. If the product of two numbers is 283553710 and their greatest common divisor is25345377, what is their least common multiple.

(Page 223: 5) 175. Prove that there are no solutions in positive integers x and y to theequation x4 + y4 = 625.

176. Solve the congruence x ≡ 5 (mod 7).

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177. Consider the relation R defined on the set of all people by (a, b) ∈ R if and only ifa is not taller than b (in other words, a’s height is less than or equal to b’s height). Is Rreflexive, symmetric, antisymmetric, and/or transitive?

178. Consider the poset (R,≤).

a. Does the set (0, 1) have a minimal element?

b. Does the set (0, 1) have a lower bound?

c. Does the set (0, 1) have a greatest lower bound?

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179. Find the sum-of-products expansion of F (x, y, z) = xy + z.

180. Simplify the following using K-maps or the Quine-McCluskey method.

w x y z + w x y z + w x y z + w x y z + w x y z + w x y z + w x y z



49

(1) 181. Show that p ∨ (p → q) is a tautology.

Proof : Note that p ∨ (p → q) ≡ p ∨ (¬p ∨ q) ≡ (p ∨ ¬p) ∨ q ≡ T ∨ q ≡ T.

(1) 182. Show that (p ∧ ¬q) ∧ (p → q) is a contradiction.

Proof : Note that (p∧¬q)∧ (p → q) ≡ (p∧¬q)∧ (¬p ∨ q) ≡ (p∧¬q)∧¬(p∧¬q) ≡ F.

183. Let R(x, y) be the statement “x is registered for y” where the universe of discoursefor x is the set of all students enrolled in YSU and the universe of discourse for y is the setof all courses offered at YSU.

a. Write each of the following in English.

(1) i. ∀x∃yR(x, y)

Every student at YSU is registered for a course.

(1) ii. ∃y∀xR(x, y)

There is a course at YSU in which all YSU students are registered.

b. Express each of the following in terms of R(x, y), quantifiers, and logical connectives sothat all negation symbols immediately precede predicates.

(1) i. There is a student at YSU that is registered for every course in which you areregistered.

∃x∀y(R(Your Name, y) → R(x, y)

(1) ii. No student at YSU is registered for every course in which you are registered.

∀x∃y(R(Your Name, y) ∧ ¬R(x, y)

184. All parts of this problem relate to inhabitants of the island of knights and knavescreated by Smullyan referenced in the text on page 20. Knights always tell the truth andknaves always lie. You encounter three people, A, B, and C. Determine if possible whatA, B, and C are if they address you in the ways described. If you cannot determine whatthese three people are, can you draw any conclusions?

(1) a.

A : “Two of us are knights and one is a knave.”

B : “We are all knaves.”

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C : “Two of us are knaves and one is a knight.”

Both A and B are knaves and C is a knight. Since a knight would not make the statement“we are all knaves,” B is a knave. Also, note that at least one of A and C is lying. If bothare lying, then B’s statement is true and a knave never tells the truth. Therefore, A is lyingand C is telling the truth.

(1) b.

A : “We are all knights.”

B : “We are all knights.”

C : “We are all knights.”

They are either all knights or all knaves. The statement “we are all knights” is either trueor false. If the statement is true, they are all knights. If on the other hand, the statementis false, they are all knaves.

185. Consider the following argument. If it is cold outside, I will wear my coat. It is notcold outside. Therefore, I will not wear my coat.

(1) a. Let p and q be the following statements.

p : It is cold outside.

q : I will wear my coat.

Give the logical form of the argument above.

Proof :

1. ¬p

2. p → q

3. ¬q

(1) b. Is the argument valid?

No. Note that [(p → q) ∧ ¬q] 9 ¬q.

(1) 186. Prove that if x is an irrational number then√x is irrational.

51

Proof : Attempting a contradiction, suppose that√x is rational. Then there exist p, q ∈ Q

such that√x = p

q. This implies that x = p2

q2which is rational. This contradicts the fact

that x is irrational.

(1) 187. Let A be a set. Show that A = A.

Proof : Note that A = {x : x /∈ A} = {x : x ∈ A} = A.

188. Let A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}.

(1) a. A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}

(1) b. A ∩ B = {4, 5}

(1) c. A \B = {1, 2, 3}

(1) d. B \ A = {6, 7, 8}

(1) 189. Let A = {2, 4, 6, 8, 10, 12, . . .} and B = {3, 6, 9, 12, 15, 18, . . .}. Find A ∩ B.

A ∩B = {6, 12, 18, . . . , }



190. Define f : R → R by f(x) = x2 + 2.

(4) a. f(R) = [2,∞) (4) b. f-1([0,∞)) = R

(4) 191. Show that if f and g are 1–1 functions, g ◦ f is 1–1.

Proof : Suppose that x1 , x2 ∈ dom(f) such that (g ◦ f)(x1) = (g ◦ f)(x2). Since g is 1–1,f(x1) = f(x2). Also, Since f is 1–1, x1 = x2 . Therefore, g ◦ f is 1–1.

(4) 192. Let A and B be sets such that A is finite. Suppose that f : A → B is a function.If |f(A)| = |A|, what can be concluded about f?

The function f is 1–1.

Proof : If f is not 1–1, then there are x, y ∈ A such that f(x) = f(y). Then |f(A)| < |A|which is a contradiction since |f(A)| = |A|.

193. For each of the following, find the next term of the sequence. How did you determinethe next term?

(4) a. {2, 3, 6, 18, 108, . . .}

18 · 108 = 1954

For each n ∈ N, an+2 = a

n· a

n+1 .

(4) b. {1, 4, 27, 256, . . .}

52

55 = 3125

For each n ∈ N, an= nn.

(4) 194. Let a, b ∈ N such that a · b = 168 and gcd(a, b) = 2. Find lcm(a, b).

lcm(a, b) = 84


115 = 2 · 57 + 1

57 = 3 · 38 + 1

28 = 2 · 14 + 0

14 = 2 · 7 + 0

7 = 2 · 3 + 1

3 = 2 · 1 + 1

1 = 0 · 2 + 1

(115)10

= (1110011)2

196. (4) a. State the division algorithm.

(4) b. Show that if n is an odd integer, then n2 ≡ 1 (mod 4).

Proof : Since n is odd, there exists k ∈ N such that n = 2k + 1. Then n2 = 4k2 + 4k + 1.So n2 mod 4 = 1 = 1 mod 4 which implies that n2 ≡ 1 (mod 4).

(4) c. Let m ∈ N and a, b ∈ Z. Show that if a mod m = b mod m, then a ≡ b (mod m).

Proof : See class notes.

197. (4) a. Let a, b ∈ Z and m ∈ N. Define what it means for a to be congruent to bmodulo m.

(4) b. Let a, b ∈ Z and m ∈ N. Prove that if a ≡ b (mod 2m), then a ≡ b (mod m).

Proof : Since a ≡ b (mod 2m), 2m|(a− b) which implies that m|(a− b) by theorem 1(iii)on page 202. Therefore, a ≡ b (mod m).




x ≡ 1 (mod 2)x ≡ 1 (mod 3)x ≡ 2 (mod 7)

53


Then a1= 1, a

2= 1, and a

3= 2,

m1 = 2, m2 = 3, and m3 = 7,

M1= 21, M

2= 14, and M

3= 6,

k1 = 1, k2 = 2, and k3 = 6.

Let x = 1 · 21 · 1 + 1 · 14 · 2 + 2 · 6 · 6 = 21 + 28 + 72 = 121.

Note that 121 ≡ 37 (mod 42).



Since 17 ∤ 11, 1116 ≡ 1 (mod 17) by Fermat’s Little Theorem. Therefore, 1116mod17 = 1.

(3) c. Calculate 3417mod17.

By Fermat’s Little Theorem, 3417 ≡ 34 (mod 17) Therefore, 3417mod17 = 0.

200. (4) a. Let X be a set. Give an example of a 1–1 function f : X → P(X).

Define f : X → P(X) by f(x) = {x}. If y 6= z, then f(y) = {y} 6= {z} = f(z). Therefore,f is 1–1.

(4) b. Is your function onto?

No. See class notes.

(4) 201. Show that for any natural number n greater than 3, n+ 1 ≤ n!.

Proof : For each n ∈ N, let P (n) be the statement “n+1 ≤ n!.” Since 3+ 1 = 4 ≤ 6 = 3!,P (3) is true. Now suppose that P (k) is true and consider

(k + 1) + 1

≤ k! + 1

≤ k! + k!

= 2k!

54

≤ (k + 1)k!

= (k + 1)!.

Hence, P (k + 1) is true and so P (n) is true for all n ∈ N by the principle of mathematicalinduction.

(4) 202. Show that every natural number greater than 5 can be expressed as 3x + 4ywhere x and y are nonnegative integers. Hint: Verify this statement for the integers 6, 7,8, and 9. Use strong induction.

Proof : For each n ∈ N, let P (n) be the statement “n = 3x + 4y for some nonnegativeintegers x and y.” Since 6 = 3 · 2+4 · 0, 7 = 3 · 1+4 · 1, 8 = 3 · 0+4 · 2, and 9 = 3 · 3+4 · 0,P (6), P (7), P (8), and P (9) are all true. Now suppose that k ∈ N such that k ≥ 9 and P (j)is true for all j such that 6 ≤ j ≤ k. Then P (k − 3) is true and so there exist integers xand y such that 3x + 4y = k − 3. Then 3x + 4(y + 1) = k + 1 and so P (k + 1) is true.Therefore, P (n) is true for all n ≥ 6 by the principle of strong induction.

(4) 203. Let R be a relation on a set A that is symmetric and antisymmetric. Show thatif (a, b) ∈ R, then a = b.

Proof : Suppose that a, b ∈ A such that (a, b) ∈ R. Since R is symmetric, (b, a) ∈ R whichimplies that a = b since R is antisymmetric.

204. Consider the relation R on the set of integers defined by (a, b) ∈ R if and only if a|b.

(3) a. Is R reflexive?

(3) b. Is R symmetric?

(3) c. Is R antisymmetric?

(3) d. Is R transitive?

205. Consider the poset (P(N),⊆).

(3) a. Is there a maximum element.

N

(4) b. Find all lower bounds of {{1, 2}, {1, 3}}.

∅, {1}

Section 1.20: Final Exam Fall 2006


Name:

55

E-mail (optional):



207. Let P (m,n) be the statement “m has at least n days” where m is a variable repre-senting a month and n is a variable representing a number.

a. Express each of the following in words.

(4) i. ∀mP (m, 26)

(4) ii. ∀n∃mP (m,n)

b. Express each of the following with quantifiers and P (m,n).

(4) i. There is a month that has no more than 28 days.

(4) ii. There are two months that have at least 31 days.

56

208. Determine whether the following arguments are valid.

(4) a. Anyone who gets an A on the final, passes the class. John did not pass the class.Therefore, John did not get an A on the final.

(4) b. Anyone who gets an A on the final, passes the class. John passed the class. Therefore,John got an A on the final.

(4) c. Anyone who gets an A on the final, passes the class. John did not get an A on thefinal. Therefore, John did not pass the class.

(5) 209. Let A and B be sets. Show that A = B if and only if P(A) = P(B).

57

(5) 210. Let A1 , A2, . . . , Anbe sets. Use induction or strong induction to show that

A1 ∪A2 ∪ · · · ∪ An= A1 ∩ A2 ∩ · · · ∩A

n.

(5) 211. Let A be a nonempty set and f : A → P(A). Show that f is not onto.

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(5) 212. Let a and b be integers and m a natural number. Show that ifa mod m = b mod m, then a ≡ b (mod m).

(Page 218: 20) 213.

(4) a. gcd(22 · 33 · 55, 25 · 33 · 52) (4) b. lcm(22 · 33 · 55, 25 · 33 · 52)

(Page 230: 23) (4) 214. Find gcd(111, 201).

59


x ≡ 3 (mod 4)x ≡ 2 (mod 5)x ≡ 4 (mod 7)

216. Define a relation R on P(N) by A R B if and only if A ∩ B = ∅.

(3) a. Is the relation R reflexive.

(3) b. Is the relation R symmetric.

(3) c. Is the relation R antisymmetric.

(3) d. Is the relation R transitive.

(3) e. Is there an element of P(N) that is related to every element of P(N)?

60

(5) 217. Let R be an equivalence relation on a set A and suppose that a, b ∈ A. Provethat a R b if and only if [a] ⊆ [b].

(4) 218. Prove the Boolean identity x+ y = x · y.

(Page 760: 3) (4) 219. Find the sum-of-products expansion for the Boolean functionF (x, y, z) = (x+ z)y.

(5) 220. In negotiating a business deal, a salesman states “you will not pay more thanone-half the price of any customer.” If you are the only customer that does not pay $20 forthe item, how much will you pay? Hint: After calculating your answer, carefully rereadthe salesman’s statement.


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221. Consider the statements below.

p: It is raining.

q: It is cloudy.

Express each of the following as an English sentence.

(1) a. p ∧ q

It is raining and cloudy.

(1) b. p ∨ (¬p ∧ q)

It is either raining or it is not raining and cloudy.

(1) c. p → q

If it is raining, then it is cloudy.

(1) 222.

Show that (p ∧ ¬q) ∨ ¬p ∨ q is a tautology.

(p ∧ ¬q) ∨ ¬p ∨ q

≡ ¬¬(p ∧ ¬q) ∨ ¬p ∨ q

≡ ¬(¬p ∨ q) ∨ ¬p ∨ q

≡ ¬(¬p ∨ q) ∨ (¬p ∨ q)

≡ ¬(p → q) ∨ (p → q)

≡ T

223. This question relates to the island of knights and knaves created by Smulyan, whereknights always tell the truth and knaves always lie. You encounter two people, A andB. Can you determine what A and B are if they address you as below? If not, can youeliminate some of the possibilities?

(1) a.

A : I am a knave and B is a knight.

B : Says nothing.

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They are both knaves. Since a knight will never claim to be a knave, A is not a knight.Since A is a knave, we know his statement is false. Therefore, B must also be a knave.

(1) b.

A : We are both knights.

B : We are both knights.

They are either both knights or both knaves. If a knave and a knight were walking together,the knight would not claim that they were both knights.

(1) c. Explain why the following will not happen.

A : We are both knaves.

B : We are both knaves.

Since knights do not lie, neither A nor B can be a knight. On the other hand, since knavesdo not tell the truth, they cannot both be knaves.

224. Let the universe of discourse be the set of all natural numbers. Express each of thefollowing as an English sentence (using mathematical symbols if you like) and give the truthvalue of each.

(1) a. ∃ n ∀ m(n ≤ m)

There is a natural number that is less than or equal to every other natural number. Thisis true.

(1) b. ∀ j ∃ k(j < k)

Every natural number is less than some other natural number. This is true.

(1) c. ∃ s ∀ t(t < s)

There is a natural number that is greater than every other natural number. This is false.

(1) 225. Determine whether or not the following is a valid argument.

If cats can fly, so can pigs. Cats can fly. Therefore, pigs can fly.

This is valid argument using modus ponens.

(1) 226. Let n be a positive integer. Prove that n is odd if and only if n2 is odd.

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Proof : First, suppose that n is odd. Then there exists an integer k such that n = 2k + 1.Then n2 = (2k + 1)2 = 4k2 + 4k + 1 which is odd.

Now suppose that n2 is odd. To see that n is odd, assume the contrary. So there is aneven integer k such that n = 2k. Then n2 = (2k)2 = 4k2 which is even, contradicting thehypothesis.

(1) 227. Prove that the square root of an irrational number is irrational.

Proof : We prove the contrapositive. Suppose that x ∈ R such that√x is rational. Then√

x2= x is rational.

(1) 228. Prove that if two sets A and B have the same power set, then A = B.

Proof : Since A and B have the same power set, A ∈ P(B) which implies that A ⊆ B.Likewise, B ∈ P(A) which implies that B ⊆ A. Therefore, A = B.


(10) 229. State DeMorgan’s Law’s for sets and prove one of them.

230. For each prime number p, let Ap= {p, 2p, 3p, 4p, . . .}.

(10) a. Show that⋃

p∈P

Ap= N\{1}. Hint: Recall the Fundamental Theorem of Arithmetic.

Proof : Suppose that n ∈ N\{1}. Then by the Fundamental Theorem of Arithmetic, thereexists a prime number q such that q|n. Then n ∈ A

q⊆ ⋃

p∈P

Ap.

(10) b. Show that⋂

p∈P

Ap= ∅.

Proof : Suppose that n ∈ N. Then there exists a prime number q such that n < q. Thenn /∈ A

q⊇ ⋂

p∈P

Ap. Therefore,

⋂

p∈P

Ap= ∅.

(10) 231. Show that the composition of two 1–1 functions is 1–1.

Proof : Suppose that f : A → B and g : B → C are 1–1. Let x, y ∈ A such that (g ◦ f)(x)= (g ◦ f)(y). Then

(g ◦ f)(x) = (g ◦ f)(y)

g[f(x)] = g[f(y)]

f(x) = f(y) (g is 1–1)

64

x = y (f is 1–1).

(10) 232. Give an example of a function f : Z → N that is both 1–1 and onto.

Define f : Z → N by f(z) =

{

-2z + 2, if z ≤ 0

2z − 1, if z > 0

233. Suppose f : X → Y and A ⊆ X .

(10) a. Show that A ⊆ f-1(f (A)) .

Proof : Suppose that x ∈ A. Then f(x) ∈ f(A) which implies that x ∈ f-1(f (A)) .

(10) b. Show that A = f-1(f (A)) if f is 1–1.

Proof : Suppose that x ∈ f-1(f (A)). Then f(x) ∈ f(A). So there exists y ∈ A such that

f(x) = f(y). Since f is 1–1, x = y and so x ∈ A.

(10) 234. For each of the following, determine a simple rule or formula for the sequence.Use your answer to write the next term.

a. {5, 8, 11, 14, 17, 20, . . .}

xn= 2 + 3n

b. {1, 2, 2, 4, 8, 32, 256, . . .}

xn+2 = x

n· x

n+1

(10) 235. Find the quotient and remainder from the Division Algorithm when

a. -37 is divided by 5

q = -8

r = 3

b. -9 is divided by 10

q = -1

r = 1

(10) 236. Suppose that a, b, c ∈ Z and m ∈ N. Show that if a ≡ b (mod m) and b ≡ c(mod m), then a ≡ c (mod m).

Proof : Since a ≡ b (mod m), m|(a− b) and since b ≡ c (mod m), m|(b− c). So c divides[(a− b) + (b− c)] = a− c which implies that a ≡ c (mod m).


(10) 237. Give the prime factorization of each of the following.

a. 132 = 22 · 3 · 11 b. 450 = 2 · 32 · 52

65

(10) 238. Calculate each of the following.

a. gcd(450, 132) = 6 b. lcm(450, 132) = 22 · 32 · 52 · 11 = 9900

(10) 239. Convert the following to decimal notation.

(101001)2 = 32 + 8 + 1 = 41

(10) 240. Convert the following from decimal notation to binary notation.

123 = (1111011)2

241.

(10) a. Find the inverse of 7 modulo 11.

8

(10) b. Solve the congruence 7x ≡ 6 (mod 11).

7x ≡ 6 (mod 11)

8 · 7x ≡ 8 · 6 (mod 11)

x ≡ 4 (mod 11)

(10) 242. Find the set of all integers that are divisible by 6 and congruent to 1 modulo5.

x ≡ 0 (mod 6)x ≡ 1 (mod 5)


Then a1 = 0, a2 = 1,

m1 = 6, m2 = 5,

M1 = 5, M2 = 6,

k1 = 5, and k2 = 1.

Let x = 0 · 5 · 1 + 1 · 6 · 1 = 6.

{6 + 30z : z ∈ Z}


66

x ≡ 1 (mod 2)x ≡ 2 (mod 3)x ≡ 1 (mod 5)x ≡ 0 (mod 7)


Then a1 = 1, a2 = 2, a3 = 1, a4 = 0,

m1 = 2, m2 = 3, m3 = 5, m4 = 7,

M1 = 105 ≡ 1 (mod 2), M2 = 70 ≡ 1 (mod 3), M3 = 42 ≡ 2 (mod 5), M4 = 30 ≡ 2 (mod 7),

k1 = 1, k2 = 1, k3 = 3, and k4 = 4.

Let x = 1 · 105 · 1 + 2 · 70 · 1 + 1 · 42 · 3 + 0 · 30 · 4 = 371.

Note that 371 ≡ 161 (mod 210).

244. Let A ⊆ Z with |A| = 5.

(10) a. Prove or give a reasonable explanation of the fact that there exist x, y ∈ A suchthat x 6= y and x ≡ y (mod 4).

Proof : Every element of A is congruent modulo 4 to 0, 1, 2, or 3. Since A has five elements,there must be two distinct elements of A that are congruent to the same integer modulo 4.

(Bonus) (5) b. Generalize the previous problem as much as possible. Do not prove yourgeneralization.

If A ⊆ Z with n elements and k ∈ N with k < n, then there are x, y ∈ A with x 6= y suchthat x ≡ y (mod k).

(10) 245. Prove that 5782 ≡ 1 (mod 8).

Proof : Since 57 ≡ 1 (mod 8), 5782 ≡ 182 (mod 8).



Name:


67


247. Let P (m,n) be the statement “m has at least n days” where m is a variable repre-senting a month and n is a variable representing a number.

a. Express each of the following in words.

(1) i. ∀mP (m, 26)

(1) ii. ∀n∃mP (m,n)

b. Express each of the following with quantifiers and P (m,n).

(1) i. There is a month that has no more than 28 days.

(1) ii. There are two months that have at least 31 days.

68

248. Determine whether the following arguments are valid.

(1) a. Anyone who gets an A on the final, passes the class. John did not pass the class.Therefore, John did not get an A on the final.

(1) b. Anyone who gets an A on the final, passes the class. John passed the class. Therefore,John got an A on the final.

(1) 249. In negotiating a business deal, a salesman states “you will not pay more thanone-half the price of any customer.” If you are the only customer that does not pay $20 forthe item, how much will you pay? Hint: After calculating your answer, carefully rereadthe salesman’s statement.

69

(1) 250. Prove that there exist irrational numbers x and y such that xy is rational.

(2) 251. Let A and B be sets. Show that A = B if and only if P(A) = P(B).

70

(2) 252. Let A be a nonempty set and f : A → P(A). Show that f is not onto.

(2) 253. Let A1, A

2, . . . , A

nbe sets. Use induction or strong induction along with

De Morgan’s Laws to show that A1 ∪A2 ∪ · · · ∪ An= A1 ∩ A2 ∩ · · · ∩A

n.

(2) 254. Let a and b be integers and m a natural number. Show that a mod m = bmod mif and only if a ≡ b (mod m).

71

(Page 218: 20) 255.

(1) a. gcd(22 · 33 · 55, 25 · 33 · 52) (1) b. lcm(22 · 33 · 55, 25 · 33 · 52)

(1) 256. Solve the following.

3x ≡ 4 (mod 7)


x ≡ 3 (mod 4)x ≡ 2 (mod 5)x ≡ 4 (mod 7)

72

(2) 258. Define a relation R on the set of integers by x R y if and only if |x| = |y|. Provethat R is an equivalence relation. Give the equivalence classes of R.

(2) 259. Let R be an equivalence relation on a set A that is antisymmetric. Show thatfor all a, b ∈ A, a R b if and only if a = b.

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260. Define a relation 4 on the set of natural numbers by m 4 n if and only if m|n.

(2) a. Prove that (N,4) is a partially ordered set.

(1) b. Does (N,4) have a maximal element?

(1) c. Does (N,4) have a minimal element?

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Section 1.25: Quizzes Fall 2016

Quiz 1(09-02-16)

Name:

Directions: Show all of your work and justify all of your answers.

(2) 1. Complete the following truth table.

p q p → q q → p (p → q) ∨ (q → p)

T T T T T

T F F T T

F T T F T

F F T T T

(2) 2. All parts of this problem relate to the inhabitants of the island of knights andknaves created by Smullyan, where knights always tell the truth and knaves always lie. Anexplorer encounters two people, A and B, on the island. The explorer knows that one is aknight and the other is a knave. In an attempt to determine who is the knight and who isthe knave, the explorer begins asking questions. For each of the questions below, determinewhether or not the answer to the question will satisfy the explorer’s curiosity.

a. The explorer asks A “Are you a knight?”

Every inhabitant of the island will claim to be a knight. So the answer to this question willbe of no help to the explorer.

b. The explorer asks A “Are you both knights?”

If A’s answer is yes, then A is a knave and B is a knight. If A’s answer is no, then A is aknight and B is a knave.

c. The explorer asks A “Will B say that he is a knight?”

If A’s answer is yes, then A is a knight and B is a knave. If A’s answer is no, then A is aknave and B is a knight.

Quiz 2

Name:


(1) 1. Is the proposition p ∨ ¬q equivalent to the proposition q → ¬p?

No.

Note that when p and q are both true, p ∨ ¬q is true and q → ¬p is false.

Alternatively, a truth table can be used.

(1) 2. Is the proposition ¬p ∨ q equivalent to the proposition ¬q → ¬p?

Yes.

Note that ¬p ∨ q is equivalent to p → q which is the contrapositive of ¬q → ¬p.

Alternatively, a truth table can be used.

3. Let P (x, y, z) be the statement “xy = z” where the universe of discourse is {0, 1, 2}.Write each of the following statements in English (using math symbols where appropriate).Evaluate (carefully) the truth value of each statement.

(1) a. ∀z ∃x ∃y P (x, y, z)

For all z there exist x and y such that xy = z.

True.

For z = 0, 1, 2, there are x and y such that xy = z. Note that P (0, 0, 0), P (1, 1, 1), andP (1, 2, 2) are all true.

(1) b. ∀x ∀y ∃z P (x, y, z)

For all x and y there exists z such that xy = z.

False.

Note that ∀z P (2, 2, z) is false.

Quiz 3

Name:


(1) 1. Express the negations of each of the following in such a way that all negationsymbols immediately precede predicates.

a. ∀ x ∀ yP (x, y)

¬ ∀ x ∀ yP (x, y)

≡ ∃ x ∃ y ¬P (x, y)

b. ∃ x ∀ yP (x, y)

¬ ∃ x ∀ yP (x, y)

≡ ∀ x ∃ y ¬P (x, y)

(2) 2. For each of the following, determine whether or not the given argument is valid.

Consider the statements below.

D(x) : x is a door

F (x) : x eats food

P (x) : x is a pet

a. All pets eat food. Max is a pet. Therefore, Max eats food.

Valid.

Statement Reason

(i) ∀x[P (x) → F (x)] Premise

(ii) P (Max) Premise

(iii) F (Max) Universal modus ponens

b. All pets eat food. Max eats food. Therefore, Max is a pet.

Invalid.

This is an example of the fallacy of affirming the conclusion.

c. All pets eat food. All doors are pets. Therefore, all doors eat food.

Valid.

78

Statement Reason

(i) ∀x[P (x) → F (x)] Premise

(ii) ∀x[D(x) → P (x)] Premise

(iii) ∀x[D(x) → F (x)] Universal transitivity

Quiz 4

Name:


(2) 1. Let A = {1, 3, {3}, {1, 3}}. Answer the following as true or false (write the entireword).

a. 3 ∈ A

True.

b. 3 ⊆ A

False.

c. {3} ∈ A

True.

d. {3} ⊆ A

True.

e. {1} ∈ A

False.

f. {1} ⊆ A

True.

g. |A| = 2

False.

h. |P(A)| = 16

True.

(2) 2. Consider the sets 0 = ∅, 1 = {0}, 2 = {0, 1}, and 3 = {0, 1, 2}. Write the set 3using only the empty set symbol, set brackets, and commas.

3 = {0, 1, 2} = {∅, {∅}, {0, 1}} = {∅, {∅}, {∅, {∅}}}

Quiz 5

Name:


1. For each n ∈ N, let An=[0, 1

n

]. Find each of the following. Express your answer in the

most simplified form.

(1) a.∞⋃

n=1

An= [0, 1] (1) b.

∞⋂

n=1

An= {0}

(1) 2. Suppose that A and B are sets. Show that A \B = A ∩ B.

Proof :

Suppose that x ∈ A \ B. Then x ∈ A and x /∈ B. So x ∈ B. Since x ∈ A and x ∈ B,x ∈ A ∩ B. Hence, A \B ⊆ A ∩B.

Now suppose that x ∈ A ∩ B. Then x ∈ A and x ∈ B. So x ∈ A and x /∈ B which meansthat x ∈ A \B. Therefore, A \B ⊆ A ∩ B.

(1) 3. Suppose that A and B are sets such that A∪B = A ∩B. What can you concludeabout A and B?

The two sets are equal.

Proof : Note that A ⊆ A ∪ B = A ∩ B ⊆ B and B ⊆ A ∪ B = A ∩ B ⊆ A. Since A ⊆ Band B ⊆ A, A = B.

Quiz 6

Name:


(4) 1. Define f : R → R by f(x) = x2. Find each of the following.

a. f ([0, 2]) = [0, 4]

b. f-1([0, 2]) =

[-√2,√2]

c. f ((-∞, 0] ∩ [0,∞)) = f ({0}) = {0}

d. f ((-∞, 0]) ∩ f ([0,∞)) = [0,∞) ∩ [0,∞) = [0,∞)

Quiz 7(10-14-16)

Name:


(1) 1. Write the first ten terms of the following sequence.

{(-1)n

(12

)+ (-1)2n

(12

)}∞

n=1= {0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, . . .}

2. For each of the following, determine a simple rule or formula for the sequence. Use youranswer to write the next term.

(1) a.{

11, 12, 24, 68, 2416, 120

32, 720

64, . . .

}={

n!2n

}∞

n=0

(1) b. {1, 3, 6, 10, 15, 21, 28, 36, 45, . . .} =

{n∑

i=1

i

}∞

n=1

Quiz 8

Name:


(1) 1. Suppose that p1, p2 , p3, p4 , and p5 are distinct prime numbers, a = p41· p0

2· p7

3· p0

4· p2

5,

and b = p31· p2

2· p4

3· p5

4· p2

5. Find each of the following.

a. gcd(a, b) = p31· p0

2· p4

3· p0

4· p2

2b. lcm(a, b) = p4

1· p2

2· p7

3· p5

4· p2

5

(2) 2. Suppose that a and b are positive numbers. Prove or reasonably explain thatab = gcd(a, b) · lcm(a, b).

Proof :

Recall that there exists prime numbers p1 , p2, p3 , . . . , pnand nonnegative integers

a1 , a2 , . . . , an, b1 , b2 , . . . , bn such that

a = pa11pa22

· · · pann,

b = pb11pb22· · · pbn

n,

gcd(a, b) = pmin(a1,b

1)

1pmin(a

2,b

2)

2· · · pmin(a

n,b

n)

n, and

lcm(a, b) = pmax(a1,b

1)

1pmax(a

2,b

2)

2· · · pmax(a

n,b

n)

n.

Note that for each i, 1 ≤ i ≤ n, one of aiand b

iis max (a

i, b

i) and the other is min (a

i, b

i).

So

ab

= pa11pa22

· · · pann

· pb11pb22· · ·pbn

n

= pa11pb21pa22pb22· · · pan

npbnn

= pmax(a1,b

1)

1pmin(a

1,b

1)

1pmax(a

2,b

2)

2pmin(a

2,b

2)

2· · ·pmax(a

n,b

n)

npmin(a

n,b

n)

n

= pmin(a1,b

1)

1pmin(a

2,b

2)

2· · · pmin(a

n,b

n)

n· pmax(a

1,b

1)

1pmax(a

2,b

2)

2· · ·pmax(a

n,b

n)

n

= gcd(a, b) · lcm(a, b).

Quiz 9

Name:


1. Find the hexadecimal expansion of each of the following.

(1) a. (952)10

952 = 59 · 16 + 8

59 = 3 · 16 + 11

3 = 0 · 16 + 3

(952)10 = (3B8)16

(1) b. (1001 1111 0101 1011)2= (9F5B)

16

(2) 2. Prove that for any n ∈ N, 102n − 1 is divisible by 99.

Proof : Since 102n =

2n+1 digits︷︸︸︷

10000 · · ·0, 102n − 1 =

2n digits︷︸︸︷

9999 · · ·9 which consists of an even number ofdecimal digits all of which are equal to 9. Therefore, 102n − 1 is divisible by 99.

Quiz 10

Name:


1. Determine whether each of the following sets is countable or uncountable. Prove orreasonably justify your answer.

(1) a. The set of real numbers between 0 and 1 that have only 1s in their decimal repre-sentation.

Countable.

Suppose that A is the set described above. Define f : N → A by f(1) = 0.11, f(2) = 0.1,f(3) = 0.11, f(4) = 0.111, etc.

(1) b. The set of real numbers between 0 and 1 that have only 0s and 1s in their decimalrepresentation.

Uncountable.

Mimic the solution to example 5 of page 173 using 0 and 1 instead of 4 and 5.

(1) c. The set of irrational numbers.

Uncountable.

If R \ Q is countable, then R = Q ∪ (R \ Q) is countable. Since R is uncountable, R \ Qmust be uncountable.

86



2. Let p, and q be the following statements.

p : “John likes oranges.”

q : “Jane likes apples.”

a. Use words to state the following propositions.

(10) i. p → q

If John likes oranges, then Jane likes apples.

(10) ii. p ∨ ¬q

Either John likes oranges or Jane does not like apples.

(10) b. Use p and q and logical operators to express the following proposition.

If John likes oranges, then Jane does not like apples and if Jane likes apples, then Johndoes not like oranges.

(p → ¬q) ∧ (q → ¬p)

(10) 3. Show that (p ∧ q) → p is a tautology.

(p ∧ q) → p

≡ ¬(p ∧ q) ∨ p

≡ (¬p ∨ ¬q) ∨ p

≡ ¬p ∨ ¬q ∨ p

≡ (¬p ∨ p) ∨ ¬q

≡ T ∨ ¬q

≡ T

(10) 4. Show that p⊕ q ≡ (p ∨ q) ∧ (¬p ∨ ¬q).

p⊕ q ≡ (p ∨ q) ∧ ¬(p ∧ q) ≡ (p ∨ q) ∧ (¬p ∨ ¬q)

5. Let the universe of discourse be the set of all animals. Consider the propositionalfunctions listed below.

F (x) : “x can fly”

S(x) : “x can swim”

87

Express each of the following using the propositional functions above, universal quantifiers,and logical connectives.

(10) a. Any animal that can swim, can also fly.

∀x[P (x) → S(x)]

(10) b. All animals can either swim or fly but no animal can do both.

∀x[P (x)⊕ S(x)]

(10) 6. Let P (x, y) be any propositional function. Give a proof or reasonable explanationof the following.

∃ y ∀ xP (x, y) → ∀ x ∃ yP (x, y)

Explanation: The first statement says that there exists y such that for all x, P (x, y) is true.The second statement says that for all x there exists y such that P (x, y) is true. Supposethe first statement is true. Then there is some element c such that P (x, c) is true for all x.Then the statement ∀ x ∃ yP (x, y) is true since ∀ xP (x, c) is true.

Proof :

Statement Reason

(i) ∃y∀xP (x, y)] Premise

(ii) ∀xP (x, c) for some c Existential instantiation

(iii) ∀x∃yP (x, y) Existential generalization

(10) 7. Is the following argument valid?

All birds can fly. Sam cannot fly. Therefore, Sam is not a bird.

Yes.


B(x) : x is a bird

F (x) : x can fly

Statement Reason

(i) ∀x[B(x) → F (x)] Premise

(ii) ¬F (Sam) Premise

(iii) ¬B(Sam) Universal modus tollens

88

(10) 8. Is the following argument valid?

All birds can fly. Sam can fly. Therefore, Sam is a bird.

No.


B(x) : x is a bird

F (x) : x can fly

Statement Reason

(i) ∀x[B(x) → F (x)] Premise

(ii) B(Sam) → F (Sam) Universal instantiation

(iii) F (Sam) Premise

(iv) B(Sam) Fallacy of affirming the conclusion

(10) 9. Prove that the sum of two odd integers is even.

Proof : Suppose that m and n are odd integers. Then there exist integers j and k suchthat m = 2j + 1 and n = 2k + 1. Then m + n = (2j + 1) + (2k + 1) = 2j + 2k + 2 =2(j + k + 1) which is even.

(10) 10. Prove that if x is an irrational number then√x is irrational. Hint: Use a proof

by contradiction or contraposition.

Proof : Suppose that√x is rational. Then there exist p, q ∈ Q such that

√x = p

q. This

implies that x = p2

q2which is rational.

(10) 11. Determine whether or not the given equation has a solution in the set of integers.

x4 + x2 = 18

No. If x ∈ Z such that x4 + x2 = 18, then x can be assumed to be nonnegative since theexponents are even and less than 2 since 24 + 22 = 20. So we need only check that 0 + 0 =0 and 1 + 1 = 2. Therefore, there are no integer solutions to the equation.

89



12. Let A = {1, 2, 3} and B = {3, 4}. Find each of the following.

(10) a. A ∪ B ={1, 2, 3, 4}

(10) b. A ∩ B = {3} (10) c. A \B = {1, 2}

(10) d. A× B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

(10) e. P(A) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, A}

(10) 13. Suppose that A, B, and C are sets such that A ⊆ B and B ⊆ C. Prove thatA ⊆ C.

Proof : Suppose that x ∈ A. Then x ∈ B since A ⊆ B. Since x ∈ B and B ⊆ C, x ∈ C.Therefore, A ⊆ C.

(10) 14. Give an example of a universal set U and two sets A ⊆ U and B ⊆ U such thatA ∪B 6= A ∪B.

Let U be any nonempty set, A ⊆ U, and B = U \ A. Then A ∪ B = U = ∅ and A ∪ B =B ∪A = U.

15. Define f : Z → ω by f(x) = |x|. Let A = {0, 1, 2, 3, 4, . . .} and B = {0, -1, -2, -3, -4, . . . }.Find each of the following.

(10) a. f(A ∩ B) = f ({0}) = {0} (10) b. f(A) ∩ f(B) = ω ∩ ω = ω

(10) c. f(A) = f ({-1, -2, -3, -4, . . . }) = N (10) d. f(A) = ω = ∅

(10) 16. Suppose that f : X → Y and g : Y → Z are 1–1. Prove that g ◦ f is 1–1.

Proof : Suppose that x, y ∈ X such that (g ◦ f)(x) = (g ◦ f)(y). Then

g [f(x)] = g [f(y)]

f(x) = g(y) (since g is 1–1)

x = y (since f is 1–1).

Therefore, g ◦ f is 1–1.

90

17. Suppose that an= a

n−1 · an−2 · · · a1 · a0 for all n ∈ N. List the first five terms of thesequence {a

n}∞n=0 for each of the following values of a

0.

(10) a. a0 = 1

a1 = a0 = 1

a2 = a1 · a0 = 1 · 1 = 1

a3 = a2 · a1 · a0 = 1

{an}∞n=0 = {1, 1, 1, 1, 1, · · · }

(10) b. a0 = 2

a1 = a0 = 2

a2 = a1 · a0 = 2 · 2 = 4

a3 = a2 · a1 · a0 = 4 · 2 · 2 = 16

{an}∞n=0 = {2, 2, 4, 16, 256, · · ·}

18. For each of the following, determine a simple rule or formula for the sequence. Useyour answer to write the next term.

(10) a. {1, 2, 2, 3, 3, 3, 4, 4, 4, 4, . . .}

Each natural number n is repeated n times.

The next term is 5.

(10) b. {2, 8, 18, 32, 50, 72, 98, 128, 162, . . .}

= {2 · 1, 2 · 4, 2 · 9, 2 · 16, 2 · 25, 2 · 36, 2 · 49, 2 · 64, 2 · 81, . . .}

= {2n2}∞n=1

The next term is 200.


(10) a. 71 div 6 = 11 (10) b. 71 mod 6 = 5

Since 71 = 11 · 6 + 5, 71 div 6 = 11 and 71 mod 6 = 5.

(10) 20. Give two positive solutions and two negative solutions to following.

x ≡ 4 (mod 7)

The positive solutions are 4, 11, 18, 25, 32, 39, . . . .

The negative solutions are -3, -10, -17, -24, . . . .

21. Give the prime factorization of each of the following.

(10) a. 132 = 22 · 3 · 11 (10) b. 7! = 7 ·6 ·5 ·4 ·3 ·2 ·1 = 24 ·32 ·5 ·7

91



(10) 22. Give the prime factorizations of 300 and 168.

300 = 22 · 3 · 52

168 = 23 · 3 · 7

(10) 23. Find the gcd(300, 168).

gcd(300, 168) = 22 · 3 = 12

(10) 24. Find the lcm(300, 168).

lcm(300, 168) = 23 · 3 · 52 · 7 = 4200

(10) 25. Express the gcd(300, 168) as a linear combination of 300 and 168.

300 = 168 + 132

168 = 132 + 36

132 = 3 · 36 + 24

36 = 24 + 12

24 = 2 · 12 + 0

12

= 36− 24

= 36− (132− 3 · 36)

= 4 · 36− 132

= 4(168− 132)− 132

= 4 · 168− 5 · 132

= 4 · 168− 5(300− 168)

= 9 · 168− 5 · 300

(10) 26. Find the hexadecimal expansion of (103)10 .

103 = 6 · 16 + 7

6 = 0 · 16 + 6

(103)10

= (67)16


(67)16

= (110 0111)2

92

(10) 28. Does 5 have an inverse modulo 6?

Yes.

5 · 5 ≡ 1 (mod 6)

So 5 is an inverse of 5 modulo 6.

(10) 29. Does 3 have an inverse modulo 6?

No.

Check that 3x ≡ 1 (mod 6) has no solution.


x ≡ 1 (mod 2)x ≡ 3 (mod 5)x ≡ 2 (mod 7)


a1 = 1 a2 = 3 a3 = 2

M1 = 35 M2 = 14 M3 = 10

k1 = 1 k2 = 4 k3 = 5

x = 1 · 35 · 1 + 3 · 14 · 4 + 2 · 10 · 5 = 303

303 ≡ 23 (mod 70)

(10) 31. State Fermat’s Little Theorem.

(10) 32. Calculate 1007 mod 7.

1007 ≡ 100 (mod 7)

100 mod 7 = 2

(10) 33. Calculate 100100 mod 11.

100100 = (10010)10

(10010)10 ≡ 1 (mod 11)

100100 mod 11 = 1



93

Name:


(10) 34. Are ¬(p → q) and (¬p ∨ ¬q) logically equivalent?

(10) 35. This problem relates to inhabitants of the island of knights and knaves createdby Smullyan referenced in the text. Knights always tell the truth and knaves always lie.You encounter three people, A, B, and C. Determine if possible what A, B, and C are ifthey address you in the ways described. If you cannot determine what these three peopleare, can you draw any conclusions?

A : “We are all knaves.”

B : “A is a knight.”

C : Says nothing.

94

36. Consider the following statements where the universe of discourse for both x and y isthe set of all creatures in the animal kingdom.

D(x) : x is a dog

C(y) : y is a cat

A(x, y) : x will chase y

Express each of the following using D(x), C(y), A(x, y), quantifiers, and logical connectives.

(10) a. There is a dog that will chase any cat.

(10) b. There is a cat that no dog will chase.

(10) c. There is a creature that is neither a dog nor a cat that will chase itself.

(10) 37. Give the negation of the following proposition. Write you answer so that thenegation symbol immediately precedes the predicate.

∀x ∃y ∃z P (x, y, z)

(10) 38. Prove that if A and B are sets such that P(A) = P(B), then A = B.

39. For each n ∈ N, let An= {1, 2, 3, . . . , n}. Find each of the following.

(10) a.∞⋃

n=1

An

(10) b.∞⋂

n=1

An

95

(10) 40. Suppose that f : X → Y and g : Y → Z are onto. Prove that g ◦ f is onto.

41. (10) a. Find gcd(840, 198). (10) b. Find lcm(840, 198).


x ≡ 2 (mod 3)x ≡ 3 (mod 5)x ≡ 5 (mod 7)

96

(10) 43. Give an example of a function from N× N to N that is 1–1.

(10) 44. Prove that the set of all sequences on N is uncountable.

97

(10) 45. Prove that for all n ≥ 4, 2n ≤ n!.

(10) 46. Prove that every integer n ≥ 2 can be expressed as a linear combination of 3and 5. Hint: Show that the statement is true for 2, 3, 4, and 5. Use strong induction.

98

(10) 47. Let X be a set and define a relation 4 on P(X) by A 4 B if and only if A ⊆ B.

(10) a. Is the relation 4 reflexive.

(10) b. Is the relation 4 symmetric.

(10) c. Is the relation 4 antisymmetric.

(10) d. Is the relation 4 transitive.

(10) 48. Suppose that R is a relation on a set A that is symmetric and transitive. Also,suppose that for all x ∈ A there is a y ∈ A such that (x, y) ∈ R. Show that R is reflexive.

99



(10) 1. State De Morgan’s Laws.

2. All parts of this problem refer to the following statement.

If George is a goat, then Sam is a cow.

(10) a. State the converse.

If Sam is a cow. then George is a goat.

(10) b. State the contrapositive.

If Sam is not a cow, then George is not a goat.

(10) c. State the inverse.

If George is not a goat, then Sam is not a cow.

(10) 3. Show that (p → r) ∧ (q → r) ≡ (p ∨ q) → r. Hint: Rewrite each statementwithout using “→”.

Proof :

(p → r) ∧ (q → r)

≡ (¬p ∨ r) ∧ (¬q ∨ r)

≡ (¬p ∧ ¬q) ∨ r (disjunction distributes over conjunction)

≡ ¬(p ∨ q) ∨ r (De Morgan’s Laws)

≡ (p ∨ q) → r

Alternatively, use a truth table.

(10) 4. Consider the statements below.

c : Maggie is a cat.d : Maggie is a dog.

Express the following argument in words.

100

c ∨ d¬d∴ c

Maggie is either a cat or a dog. Maggie is not a dog. Therefore, Maggie is a cat.

5. Consider the following argument.

If 5 > 7, then 6 = 16. It is known that 5 > 7. Therefore, 6 = 16.

(10) a. Is the conclusion true?

No.

6 6= 16

(10) b. Is the argument valid?

Yes. The argument makes use of modus ponens.

6. Let P (x, y) be the statement “x < y” where the universe of discourse is R. Express eachof the following in words and give the truth value.

(10) a. ∀x ∃y P (x, y)

For all x ∈ R, there is y ∈ R such that x < y.

This is a true statement.

(10) b. ∃y ∀x P (x, y)

There exists y ∈ R such that x < y for all x ∈ R.

This statement is false. There is no largest real number. Also, note that if such a y existed,then we would have y < y which is a contradiction.

(10) 7. Consider the propositional functions below. The universe of discourse is the setof all students.

M(x): x is a mathematics major.D(x): x likes discrete mathematics.

Express the following argument using the above statements, quantifiers, and logical opera-tors.

101

All mathematics majors like discrete mathematics. John is a mathematics major. Therefore,John likes discrete mathematics.

∀x [M(x) → D(x)] (Given)M(John) (Given)

∴ D(John) (Universal modus ponens)

(10) 8. Express the negation of the following so that the negation symbol immediatelyprecedes the propositional function.

∀x ∀y P (x, y)

¬[∀x ∀y P (x, y)] ≡ ∃x ∃y ¬P (x, y)

9. For each of the following, determine whether or not the argument is valid.

(10) a.

p ∨ q¬q

p → r∴ r

Valid.

1. p ∨ q

2. ¬q

3. p

4. p → r

5. r

(hypothesis)

(hypothesis)

(elimination)

(hypothesis)

(modus ponens using 3 and 4)

(10) b.

p ∨ q¬p ∨ q

∴ q

Valid. Note that this is a special case of resolution.

Alternatively, note that p is either true or false. If p is true, then q is true by secondstatement. If p is false, then q is true by first statement.

Alternatively, use a truth table.

102



(10) 10. Give the prime factorization of 7!.

7!

= 7 · 6 · 5 · 4 · 3 · 2 · 1

= 7 · 3 · 2 · 5 · 2 · 2 · 3 · 2

= 24 · 32 · 5 · 7

(10) 11. Give a rule or formula for the following sequence.

{-14, 29, - 3

16, 425, - 5

36, 649, . . .} =

{

(-1)n

n

(n+1)2

}∞

n=1

(10) 12. Give the base 7 expansion of (312)10.

312 = 44 · 7 + 4

44 = 6 · 7 + 2

6 = 0 · 7 + 6

(624)7 = (312)10

(10) 13. Prove that the product of an irrational number and a nonzero rational numberis irrational. You may use without proof the fact that the product of two rational numbersis rational.

Proof : Suppose that r ∈ Q, x ∈ R \Q, and r 6= 0. Toward a contradiction, suppose thatxr ∈ Q. Since r ∈ Q, 1

r∈ Q. So xr · 1

r∈ Q. This is a contradiction since xr · 1

r= x /∈ Q.

(10) 14. Suppose that a and b are integers such that the remainder is 1 when each isdivided by 3. Prove that the remainder is also 1 when their product ab is divided by 3.

Proof : Since the remainder is 1 when each of a and b is divided by 3, there are j, k ∈ Zsuch that a = 3j + 1 and b = 3k + 1. Then ab = (3j + 1)(3k + 1) = 9jk + 3j + 3k + 1 =3(3jk + j + k) + 1. So the remainder when ab is divided by 3 is 1.

(10) 15. Prove that 22n − 1 is composite for all natural numbers n ≥ 2. Hint: Induction

is neither needed nor advised.

103

Proof : Note that 22n − 1 =

(2n

+ 1) (

2n − 1

)which is composite since neither of

(2n

+ 1)

and(2n − 1

)is 1. Furthermore, 2

2n − 1 is divisible by 3 for all n ∈ N. Recall that for any

three consecutive integers, one is divisible by 3. So one of(2n − 1

), 2

n

, and(2n

+ 1)is

divisible by 3. Since 2n

is not divisible by 3, one of(2n − 1

)and

(2n

+ 1)is divisible by 3.

(10) 16. Prove that√5 is irrational.

Proof : Toward a contradiction, suppose that√5 is rational. Then there are integers p and

q such that√5 = p

qand p and q are relatively prime. Then

p

q=

√5

p2

q2= 5

p2 = 5q2.

Since 5|p2 and 5 is prime, 5|p. So there exists a natural number k such that p = 5k. So wehave that

p2 = 5q2

(5k)2 = 5q2

25k2 = 5q2

5k2 = q2.

Then 5|q2 which implies that 5|q since 5 is prime. This is a contradiction since p and q arerelatively prime. Therefore,

√5 is irrational.

(10) 17. Prove that 2n+1 ≤ n! for all natural numbers n ≥ 5.

Proof : For each n ∈ N let P (n) be the statement “2n+1 ≤ n!”. Note that P (5) is true since

25+1

= 64 ≤ 120 = 5!. Now suppose that k ∈ N such that P (k) is true. Then we have

2k+1 ≤ k!

2 · 2k+1 ≤ 2 · k!

2 · 2k+1 ≤ (k + 1)k!

2k+2 ≤ (k + 1)!

104

2(k+1)+1 ≤ (k + 1)!.

Therefore, P (k + 1) is true.

(10) 18. Prove that 5n − 2

n

is divisible by 3 for all natural numbers n.

Proof : For each n ∈ N let P (n) be the statement “5n − 2

n

is divisible by 3”. Note that

P (1) is true since 51 −2

1= 3. Now suppose that k ∈ N such that P (k) is true. Then 5

k −2k

is divisible by 3. Then 5(

5k − 2

k)

is also divisible by 3. Note that

5(

5k − 2

k)

= 5 · 5k − 5 · 2k

= 5 · 5k − (3 + 2) · 2k

= 5k+1 − 3 · 2k − 2 · 2k

= 5k+1 − 3 · 2k − 2

k+1.

Since both 5k+1 − 3 · 2k − 2

k+1and 3 · 2k

are divisible by 3, 5k+1 − 3 · 2k − 2

k+1+ 3 · 2k

=5k+1 − 2

k+1is divisible by 3. Therefore, P (k + 1) is true.

(10) 19. Suppose that A ⊆ {1, 12, 13, 14, 15, . . .} and that A 6= ∅. Prove that A has a largest

element. Hint: Let B ={n ∈ N : 1

n∈ A

}and use the Well-Ordering Principle.

Proof : Let B ={n ∈ N : 1

n∈ A

}. Since A 6= ∅, B 6= ∅. By the Well-Ordering Principle,

B has a least element, l. Since l ≤ b for all b ∈ B, 1l≥ 1

bfor all b ∈ B. This means that 1

l

is the greatest element of{

1b: b ∈ B

}= A.

105



1. In all parts of this problem, the universal set U = {a, b, c, d, 1, 2, 3, 4}, A = {a, b, 1, 2},and B = {b, c, 2, 3}. List the elements in each of the following sets.

(10) a. A ∪B = {a, b, c, 1, 2, 3} (10) b. A ∩ B = {b, 2}

(10) c. A \B = {a, 1} (10) d. Ac

= {c, d, 3, 4}

2. Suppose that A and B are sets such that |A| = 3 and |B| = 2. Find each of the following.

(10) a. |A× B| = 6 (10) b. |P(A× B)| = 26= 64

(10) 3. Suppose that A, B, C, and D are sets such that A ⊆ C and B ⊆ D. Prove that(A ∩B) ⊆ (C ∩D).

Proof : Suppose that x ∈ A ∩B. Then x ∈ A and x ∈ B. Since x ∈ A and A ⊆ C, x ∈ C.Likewise, because x ∈ B and B ⊆ D, x ∈ D. Since x ∈ B and x ∈ D, x ∈ B ∩D.

(10) 4. Suppose that A and B are both subsets of a universal set U such that A ⊆ B.Prove that B

c ⊆ Ac

.

Proof : Suppose that x ∈ Bc

. Then x /∈ B. Since x /∈ B and A ⊆ B, x /∈ A. Therefore,x ∈ A

c

.

(10) 5. Does there exist a town in which there is a resident who transports all the residentsof the town who do not transport themselves?

No. Suppose that Jane (also called Joan) is a resident of the town who transports all of theresidents of the town who do not transport themselves. If Joan transports herself, then Janedoes not transport Joan. Therefore, Joan does not transport herself. On the other hand,if Joan does not transport herself, then Jane transports her which implies that Joan doestransport herself. So Jane transports herself if and only if she does not transport herself.Therefore, such a town does not exist.

(10) 6. Define f : Q → Q by f(

p

q

)

= 1q. Explain why f is not a function.

Since f(12

)= 1

2and f

(24

)= 1

4, f is not well-defined.

7. Define f : [0,∞) → [0,∞) by f(x) =√x.

(10) a. List the elements f(N).

f(N) = {√1,√2,√3,√4, . . .}

(10) b. List the elements f-1(N).

f-1(N) = {1, 4, 9, 16, 25, . . .}

106

(10) 8. Prove that in any set of 9 integers chosen from 2, 3, 4, . . . , 21, 22, there are atleast two with a common divisor greater than 1. Hint: How many prime numbers are therebetween 2 and 22?

Proof : For each integer z ∈ {2, 3, 4, . . . , 22}, let pzbe the largest prime that divides z. Let

P = {2, 3, 5, 7, 11, 13, 17, 19}. Now suppose that A ⊆ {2, 3, 4, . . . , 22} such that |A| = 9.Define f : A → P by f(z) = p

z. Since |A| = 9 and |P | = 8, f is not 1–1 by the Pigeonhole

Principle. Therefore, there are z1 , z2 ∈ {2, 3, . . . , 22} such that pz1

= pz2. So p

z1= p

z2is a

prime number that divides both z1 and z2 .

Note 2: In other words, 9 of 21 pigeons fly into 8 pigeonholes. So at least one pigeonholecontains more than one pigeon.

(10) 9. Prove that in any group of 37 people there are at least 4 born in the same monthof the year.

Proof : Let X be the set of 37 people and Y be the months of the year. Define f : X → Yby f(x) is the month of the year in which x was born. Note that |X| = 3|Y |+1. So by the

Generalized Pigeonhole Principle, there is y ∈ Y such that |f -1(y)| ≥ 3+ 1 = 4. So at least

4 people in the set X were born in the same month of the year.

Note 3: In other words, 3 · 12 + 1 = 37 pigeons fly into 12 pigeonholes. So at least onepigeonhole contains at least 3 + 1 = 4 pigeons.