Introduction to Advanced Mathematics (7E) Solutions Manual.pdf

1 Logic and Proofs

1.1 Propositions and Connectives

1. (a) true (b) false (c) true (d) false(e) false (f) false (g) false (h) false

2. (a) Not a proposition

(b) False proposition

(c) Not a proposition. It would be a proposition if a value for x had beenassigned.

(d) Not a proposition. It would be a proposition if values for x and y had beenassigned.

(e) False proposition

(f) True proposition

(g) False proposition

(h) True proposition

(i) False proposition

(j) Not a proposition. It is neither true nor false.

3. (a)P ∼ P P∧ ∼ PT F TF T F

(b)P ∼ P P∨ ∼ PT F TF T T

(c)P Q ∼ Q P∧ ∼ QT T F FF T F FT F T TF F T F

(d)P Q ∼ Q Q∨ ∼ Q P ∧ (Q∨ ∼ Q)T T F T TF T F T FT F T T TF F T T F

(e)P Q ∼ Q P ∧ Q (P ∧ Q)∨ ∼ QT T F T TF T F F FT F T F TF F T F T

(f)P Q P ∧ Q ∼ (P ∧ Q)T T T FF T F TT F F TF F F T

1

1 LOGIC AND PROOFS 2

(g)P Q R ∼ Q P∨ ∼ Q (P∨ ∼ Q) ∧ RT T T F T TF T T F F FT F T T T TF F T T T TT T F F T FF T F F F FT F F T T FF F F T T F

(h)P Q ∼ P ∼ Q ∼ P∧ ∼ QT T F F FF T T F FT F F T FF F T T T

(i)P Q R Q ∨ R P ∧ (Q ∨ R)T T T T TF T T T FT F T T TF F T T FT T F T TF T F T FT F F F FF F F F F

(j)P Q R P ∧ Q P ∧ R (P ∧ Q) ∨ (P ∧ R)T T T T T TF T T F F FT F T F T TF F T F F FT T F T F TF T F F F FT F F F F FF F F F F F

4. (a) false (b) true (c) true (d) true(e) false (f) false (g) false (h) false(i) true (j) true (k) false (1) false

5. (a) No solution.

(b)P Q P ∨ Q Q ∨ PT T T TF T T TT F T TF F F F

Since the third and fourth columns are the same, the propositions areequivalent.


(c)P Q P ∧ Q Q ∧ PT T T TF T F FT F F FF F F F

Since the third and fourth columns are the same, the propositions areequivalent.

(d)P Q R Q ∨ R P ∨ (Q ∨ R) P ∨ Q (P ∨ Q) ∨ RT T T T T T TF T T T T T TT F T T T T TF F T T T F TT T F T T T TF T F T T T TT F F F T T TF F F F F F F

Since the fifth and seventh columns are the same, the propositions areequivalent.

(e)P Q R Q ∧ R P ∧ (Q ∧ R) P ∧ Q (P ∧ Q) ∧ RT T T T T T TF T T T F F FT F T F F F FF F T F F F FT T F F F T FF T F F F F FT F F F F F FF F F F F F F


(f)P Q R Q ∨ R P ∧ (Q ∨ R) P ∧ Q P ∧ R (P ∧ Q) ∨ (P ∧ R)T T T T T T T TF T T T F F F FT F T T T F T TF F T T F F F FT T F T T T F TF T F T F F F FT F F F F F F FF F F F F F F F

Since the fifth and eighth columns are the same, the propositions areequivalent.

(g)P Q R Q ∧ R P ∨ (Q ∧ R) P ∨ Q P ∨ R (P ∨ Q) ∧ (P ∨ R)T T T T T T T TF T T T T T T TT F T F T T T TF F T F F F T FT T F F T T T TF T F F F T F FT F F F T T T TF F F F F F F F



(h) No solution.

(i)P Q P ∨ Q ∼ (P ∨ Q) ∼ P ∼ Q ∼ P∧ ∼ QT T T F F F FF T T F T F FT F T F F T FF F F T T T T

Since the fourth and eighth columns are the same, the propositions areequivalent.

6. (a) equivalent (b) equivalent(c) equivalent (d) equivalent(e) equivalent (f) not equivalent(g) not equivalent (h) not equivalent

7. (a) ∼ P , true (b) P ∧ Q, true(c) P Q, true (d) P ∨ Q ∨ R, true

8. (a) Since P is equivalent to Q, P has the same truth table as Q. Therefore, Qhas the same truth table as P , so Q is equivalent to P .

(b) Since P is equivalent to Q, P and Q have the same truth table. Since Q isequivalent to R, Q and R have the same truth table. Thus, P and R havethe same truth table so P is equivalent to R.

(c) Since P is equivalent to Q, P and Q have the same truth table. That is, thetruth table for P has value true on exactly the same lines that the truthtable for Q has value true. Therefore the truth table for ∼ Q has value falseon exactly the same lines that the truth table for ∼ P has the value false.Thus ∼ Q and ∼ P have the same truth table.

9. (a) (P ∧ Q) ∨ (∼ P∧ ∼ Q) is neither.

P Q ∼ P ∼ Q P ∧ Q ∼ P∧ ∼ Q (P ∧ Q) ∨ (∼ P∧ ∼ Q)T T F F T F TF T T F F F FT F F T F F FF F T T F T T

(b) ∼ (P∧ ∼ P ) is a tautology.

P ∼ P P∧ ∼ P ∼ (P∧ ∼ P )T F F TF T F T

(c) (P ∧ Q) ∨ (∼ P∨ ∼ Q) is a tautology.

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

(d) (A ∧ B) ∨ (A∧ ∼ B) ∨ (∼ A ∧ B) ∨ (∼ A∧ ∼ B) is a tautology.


(A ∧ B) ∨ (A∧ ∼ B)∨A B ∼ A ∼ B A ∧ B A∧ ∼ B ∼ A ∧ B ∼ A∧ ∼ B (∼ A ∧ B) ∨ (∼ A∧ ∼ B)T T F F T F F F TF T T F F F T F TT F F T F T F F TF F T T F F F T T

(e) (Q∧ ∼ P )∧ ∼ (P ∧ R) is neither.

P Q R ∼ P Q∧ ∼ P P ∧ R ∼ (P ∧ R) (Q∧ ∼ P )∧ ∼ (P ∧ R)T T T F F T F FF T T T T F T TT F T F F T F FF F T T F F T FT T F F F F T FF T F T T F T TT F F F F F T FF F F T F F T F

(f) P ∨ [(∼ Q ∧ P ) ∧ (R ∨ Q)] is neither.

P Q R ∼ Q ∼ Q ∧ P R ∨ Q [(∼ Q ∧ P ) ∧ (R ∨ Q)] P ∨ [(∼ Q ∧ P ) ∧ (R ∨ Q)]T T T F F T F TF T T F F T F FT F T T T T T TF F T T F T F FT T F F F T F TF T F F F T F FT F F T T F F TF F F T F F F F

10. (a) contradiction (b) tautology(c) tautology (d) tautology

11. (a) x is not a positive integer.

(b) Cleveland will lose the first game and the second game. Or, Cleveland willlose both games.

(c) 5 < 3

(d) 641,371 is not composite. Or 641,371 is prime.

(e) Roses are not red or violets are not blue.

(f) T is bounded and T is not compact.

(g) M is not odd or M is not one-to-one.

(h) The function.f does not have a positive first derivative at x or does nothave a positive second derivative at x.

(i) The function g does not have a relative maximum at x = 2 (deleted comma)and does not have a relative maximum at x = 4, or else g does not have arelative minimum at x = 3.

(j) z < s or z ≤ t.

(k) R is not transitive or R is reflexive.

(l) If the function g has a relative minimum at x = 2 or x = 4, then g doesnot have a relative minimum at x = 3.

12. (a) [∼ (∼ P )] ∨ [(∼ Q) ∧ (∼ S)]

(b) [Q ∧ (∼ S)]∨ ∼ (P ∧ [Q ∧ (∼ S)]∨ ∼ ((∼ P ∧ Q).

(c) [[P ∧ (∼ Q)] ∨ [(∼ P ) ∧ (∼ R)]] ∨ [(∼ P ) ∧ S]

(d) [(∼ P ) ∨ ([Q ∧ (∼ (∼ P ))] ∧ Q)] ∨ R.


13. (a) i.A B A ©∨ BT T FF T TT F TF F F

ii.A B A ∨ B A ∧ B ∼ (A ∧ B) (A ∨ B)∧ ∼ (A ∧ B)T T T T F FF T T F T TT F T F T TF F F F T F

Since the final columns of the two tables are identical, the two propo-sitions have the same truth table, thus they are equivalent.

(b) i.A B A NAND B A NOR BT T F FF T T FT F T FF F T T

ii.A B A NANDB A NOR B (A NAND B) ∨ (A NOR B)T T F F FF T T F TT F T F TF F T T T

Since the third and last columns are equal, the propositions are equiv-alent.

iii.A B A NAND B A NOR B (A NAND B) ∧ (A NOR B)T T F F FF T T F FT F T F FF F T T T

Since the fourth and last columns are equal, the propositions areequivalent.

1.2 Conditionals and Biconditionals

1. (a) Antecedent: squares have three.Consequent: triangles have four sides.

(b) Antecedent: The moon is made of cheese.Consequent: 8 is an irrational number.

(c) Antecedent: b divides 3.Consequent: b divides 9.

(d) Antecedent: f is differentiable.Consequent: f is continuous.

(e) Antecedent: a is convergent.Consequent: a is bounded.

(f) Antecedent: f if integrable.Consequent: f is bounded.

(g) Antecedent: 1 + 1 = 2.Consequent: 1 + 2 = 3.


(h) Antecedent: the fish bite.Consequent: the moon is full.

(i) Antecedent: An athlete qualifies for the Olympic team.Consequent: The athlete has a time of 3 minutes, 48 seconds or less ( inthe event).

2. (a) Converse: If triangles have four sides, then squares have three sides.Contrapositive: If triangles do not have four sides, then squares do not havethree sides.

(b) Converse: If 8 is irrational, then the moon is made of cheese.Contrapositive: If 8 is rational, then the moon is not made of cheese.

(c) Converse: If b divides 9, then b divides 3.Contrapositive: If b does not divide 9, then b does not divide 3.

(d) Converse: If f is continuous, then f is differentiable.Contrapositive: If f is not continuous, then f is not differentiable.

(e) Converse: If a is bounded, then a is convergent.Contrapositive: If a is not bounded, then a is not convergent.

(f) Converse: If f is bounded, then f is integrable.Contrapositive: If f is not bounded, then f is not integrable.

(g) Converse: If 1 + 2 = 3, then 1 + 1 = 2.Contrapositive: If 1 + 1 �= 2, then 1 + 2 �= 3.

(h) Converse: If the moon is full, then fish will bite.Contrapositive: If the moon is not full, then fish will not bite.

(i) Converse: A time of 3 minutes, 48 seconds or less is sufficient to qualify forthe Olympic team.Contrapositive: If an athlete records a time that is not 3 minutes and 48seconds or less, then that athlete does not qualify for the Olympic team.

3. (a) Q may be either true or false.

(b) Q must be true.

(c) Q must be false.

(d) Q must be false.

(e) Q must be false.

4. (a) Antecedent: A(x) is an open sentence with variable x.Consequent: ∼ (∀x)A(x) is equivalent to (∃x) ∼ A(x).

(b) Antecedent: Every even natural number greater than 2 is the sum of twoprimes.Consequent: Every odd natural number greater than 5 is the sum of threeprimes.

(c) Antecedent: A is a set with n elements.Consequent: P(A) is a set with 2n elements.

(d) Antecedent: S is a subset of N such that 1 ∈ S and, for all n ∈ N, if n ∈ S,then n + 1 ∈ S.Consequent: S = N.

(e) Antecedent: A is a finite set with m elements and B is a finite set with nelements.Consequent: A × B = mn.


(f) Antecedent: R is a partial order for A and B ⊆ A.Consequent: If sup(B) exists, it is unique.

(g) Antecedent: A, B, C, and D are sets, f is a function from A to B, g is afunction from B to C, and h is a function from C to D.Consequent: (h ◦ g) ◦ f = h ◦ (g ◦ f).

(h) Antecedent: A and B are disjoint finite sets.Consequent: A ∪ B is finite and A ∪ B = A + B.

5. (a) true (b) false (c) true (d) true(e) true (f) true (g) true (h) false

6. (a) true (b) true (c) true (d) true(e) false (f) true (g) false (The

symbol forhelium is He.)

(h) true (i) false (j) true (k) false

7. (a)P Q Q ∧ P P ⇒ (Q ∧ P )T T T TF T F TT F F FF F F T

(b)P Q ∼ P ∼ P ⇒ Q Q ⇔ P (∼ P ⇒ Q) ∨ (Q ⇔ P )T T F T T TF T T T F TT F F T F TF F T F T T

(c)P Q ∼ Q Q ⇔ P ∼ Q ⇒ (Q ⇔ P )T T F T TF T F F TT F T F FF F T T T

(d)P Q P ∨ Q P ∧ Q (P ∨ Q) ⇒ (P ∧ Q)T T T T TF T T F FT F T F FF F F F T

(e)P Q R P ∧ Q Q ∧ R P ∨ R (P ∧ Q) ∨ (Q ∧ R) (P ∧ Q) ∨ (Q ∧ R) ⇒ P ∨ RT T T T T T T TF T T F T T T TT F T F F T F TF F T F F T F TT T F T F T T TF T F F F F F TT F F F F T F TF F F F F F F T


(f)P Q R S Q ⇒ S Q ⇒ R P ∨ Q S ∨ R (Q ⇒ S) ∧ (Q ⇒ R)T T T T T T T T TF T T T T T T T TT F T T T T T T TF F T T T T F T TT T F T T F T T FF T F T T F T T FT F F T T T T T TF F F T T T F T TT T T F F T T T FF T T F F T T T FT F T F T T T T TF F T F T T F T TT T F F F F T F FF T F F F F T F FT F F F T T T F TF F F F T T F F T

P Q R S (P ∨ Q) ⇒ (S ∨ R) [(Q ⇒ S) ∧ (Q ⇒ R)] ⇒ [(P ∨ Q) ⇒ (S ∨ R)]T T T T T TF T T T T TT F T T T TF F T T T TT T F T T TF T F T T TT F F T T TF F F T T TT T T F T TF T T F T TT F T F T TF F T F T TT T F F F TF T F F F TT F F F F FF F F F T T

8. (a)P Q ∼ P P ⇒ Q (∼ P ) ∨ QT T F T TF T T T TT F F F FF F T T T

Since the fourth and fifth columns are the same, the propositions P ⇒ Qand (∼ P ) ∨ Q are equivalent.

(b)P Q P ⇒ Q Q ⇒ P P ⇔ Q (P ⇒ Q) ∧ (Q ⇒ P )T T T T T TF T T F F FT F F T F FF F T T T T

Since the fifth and sixth columns are the same, the propositions P ⇔ Qand (P ⇒ Q) ∧ (Q ⇒ P ) are equivalent.

(c)P Q ∼ Q P ⇒ Q ∼ (P ⇒ Q) P∧ ∼ QT T F T F FF T F T F FT F T F T TF F T T F F

Since the fifth and sixth columns are the same, the propositions ∼ (P ⇒ Q)and P∧ ∼ Q are equivalent.


(d)P Q ∼ P ∼ Q P ∧ Q ∼ (P ∧ Q) P ⇒∼ Q P ⇒∼ QT T F F T F F FF T T F F T T TT F F T F T T TF F T T F T T T

Since the sixth, seventh and eighth columns are the same, all three propo-sitions are equivalent.

(e)P Q R Q ⇒ R P ⇒ (Q ⇒ R) P ∧ Q (P ∧ Q) ⇒ RT T T T T T TF T T T T F TT F T T T F TF F T T T F TT T F F F T FF T F F T F TT F F T T F TF F F T T F T


(f)P Q R Q ∧ R P ⇒ (Q ∧ R) P ⇒ Q P ⇒ R (P ⇒ Q) ∧ (P ⇒ R)T T T T T T T TF T T T T T T TT F T F F F T FF F T F T T T TT T F F F T F FF T F F T T T TT F F F F F F FF F F F T T T T


(g)P Q R P ∨ Q P ∨ Q) ⇒ R P ⇒ R Q ⇒ R (P ⇒ R) ∧ (Q ⇒ R)T T T T T T T TF T T T T T T TT F T T T T T TF F T F T T T TT T F T F F F FF T F T F T F FT F F T F F T FF F F F T T T T


9. (a) yes (b) no (c) yes(d) yes (e) no (f) no

10. (a) [(f has a relative minimum at x0)∧(f is differentiable at x0)] ⇒ (f ′(x0) =0)

(b) (n is prime) ⇒ [(n = 2) ∨ (n is odd)]

(c) (x is irrational) ⇒ [(x is real)∧ ∼ (x is rational)]

(d) [(x = 1) ∨ (x = −1)] ⇒ (|x| = 1)

(e) (x0 is a critical point for f) ⇔ [(f ′(x0) = 0) ∨ (f ′(x0) does not exist)]

(f) (S is compact) ⇔ [(S is closed) ∧ (S is bounded)]

(g) (B is invertible) ⇔ (det B �=0)


(h) (6 ≥ n − 3) ⇒ (n > 4) ∨ (n > 10)

(i) (x is Cauchy) ⇒ (x is convergent)

(j) (limx→x0 f(x) = f(x0)) ⇒ (f is continuous at x0)

(k) [(f is differentiable at x0) ∧ (f is strictly increasing at x0)] ⇒ (f ′(x0))

11. (a) Let S be “I go to the store” and R be “It rains.” The preferred translation:is ∼ S ⇒ R (or, equivalently, ∼ R ⇒ S). This could be read as “If itdoesn’t rain, then I go to the store.”The speaker might mean “I go to the store if and only if it doesn’t rain(S ⇒∼ R) or possibly “If it rains, then I don’t go to the store” (R ⇒∼ S).

(b) There are three nonequivalent ways to translate the sentence, using thesymbols D: “The Dolphins make the playoffs” and B: “The Bears win allthe rest of their games.” The first translation is preferred, but the speakermay have intended any of the three.∼ B ⇒∼ D or, equivalently, D ⇒ B∼ D ⇒∼ B or, equivalently, B ⇒ D∼ B ⇔∼ D or, equivalently, B ⇔ D

(c) Let G be “You can go to the game” and H be “You do your homeworkfirst.”It is most likely that a student and parent both interpret this statement asa biconditional, G ⇔ H.

(d) Let W be “You win the lottery” and T be “You buy a ticket.” Of the threecommon interpretations for the word “unless,” only the form ∼ T ⇒∼ W(or, equivalently, W ⇒ T ) makes sense here.

12. (a)P Q R P ∨ Q (P ∨ Q) ⇒ R ∼ P∧ ∼ Q ∼ R ⇒ (∼ P∧ ∼ Q)T T T T T F TF T T T T F TT F T T T F TF F T F T T TT T F T F F FF T F T F F FT F F T F F FF F F F T T T

Since the fifth and seventh columns are the same, (P ∨ Q) ⇒ R and∼ R ⇒ (∼ P∧ ∼ Q) are equivalent.

(b)P Q R P ∧ Q (P ∧ Q) ⇒ R ∼ Q ∼ R P∧ ∼ R (P∧ ∼ R) ⇒∼ QT T T T T F F F TF T T F T F F F TT F T F T T F F TF F T F T T F F TT T F T F F T T FF T F F T F T F TT F F F T T T T TF F F F T T T F T

Since the fifth and ninth columns are the same, the propositions (P ∧Q) ⇒R and (P∧ ∼ R) ⇒∼ Q are equivalent.


(c)P Q R Q ∧ R P ⇒ (Q ∧ R) ∼ Q∨ ∼ R (∼ Q∨ ∼ R) ⇒∼ PT T T T T F TF T T T T F TT F T F F T FF F T F T T TT T F F F T FF T F F T T TT F F F F T FF F F F T T T

Since the fifth and seventh columns are the same, the propositions P ⇒(Q ∧ R) and (∼ Q∨ ∼ R) ⇒∼ P are equivalent.

(d)P Q R Q ∨ R P ⇒ (Q ∨ R) P∧ ∼ R (P∧ ∼ R) ⇒ QT T T T T F TF T T T T F TT F T T T F TF F T T T F TT T F T T T TF T F T T F TT F F F F T FF F F F T F T

Since the fifth and seventh columns are the same, the propositions P ⇒(Q ∨ R) and (P∧ ∼ R) ⇒ Q are equivalent.

(e)P Q R P ⇒ Q (P ⇒ Q) ⇒ R P∧ ∼ Q (P∧ ∼ Q) ∨ RT T T T T F TF T T T T F TT F T F T T TF F T T T F TT T F T F F FF T F T F F FT F F F T T TF F F T F F F

Since the fifth and seventh columns are the same, the propositions (P ⇒Q) ⇒ R and (P∧ ∼ Q) ∨ R are equivalent.

(f)P Q P ⇔ Q ∼ P ∨ Q ∼ Q ∨ P (∼ P ∨ Q) ∧ (∼ Q ∨ P )T T T T T TF T F T F FT F F F T FF F T T T T

Since the third and sixth columns are the same, the propositions P ⇔ Qand (∼ P ∨ Q) ∧ (∼ Q ∨ P ) are equivalent.

13. (a) If 6 is an even integer, then 7 is an odd integer.

(b) If 6 is an odd integer, then 7 is an odd integer.

(c) This is not possible.

(d) If 6 is an even integer, then 7 is an even integer. (Any true conditionalstatement will work here.)

14. (a) If 7 is an odd integer, then 6 is an odd integer.

(b) This is not possible.


(c) This is not possible.

(d) If 7 is an odd integer, then 6 is an odd integer. (Any false conditionalstatement will work here.)

15. (a) Converse: If f ′(x0) = 0, then f has a relative minimum at x0 and isdifferentiable at x0. False: f(x) = x3 has first derivative 0 but no minimumat x0 = 0.Contrapositive: If f ′(x0) �= 0, then f either has no relative minimum at x0or is not differentiable at x0. True.

(b) Converse: If n = 2 or n is odd, then n is prime. False: 9 is odd but notprime.Contrapositive: If n is even and not equal to 2, then n is not prime. True.

(c) Converse: If x is irrational, then x is real and not rational. TrueContrapositive: If x is not irrational, then x is not real or x is rational. True

(d) Converse: If |x| = 1, then x = 1 or x = −1. True.Contrapositive: If |x| �= 1, then x �= 1 and x �= −1. True.

16. (a) tautology (b) tautology (c) contradiction(d) neither (e) tautology (f) neither(g) contradiction (h) tautology (i) contradiction(j) neither (k) tautology (l) neither

17. (a)P Q P ⇒ Q ∼ P ∼ Q ∼ P ⇒∼ QT T T F F TF T T T F FT F F F T TF F T T T T

Comparison of the third and sixth columns of the truth table shows thatP ⇒ Q and ∼ P ⇒∼ Q are not equivalent.

(b) We see from the truth table in part (a) that both propositions P ⇒ Q and∼ P ⇒∼ Q are true only when P and Q have the same truth value.

(c) The converse of P ⇒ Q is Q ⇒ P . The contrapositive of the inverse ofP ⇒ Q is ∼∼ Q ⇒∼∼ P , so the converse and the contrapositive of theinverse are equivalent.The inverse of the contrapositive of P ⇒ Q is also ∼∼ Q ⇒∼∼ P , so ittoo is equivalent to the converse.

1.3 Quantifiers

1. (a) ∼ (∀x)(x is precious⇒ x is beautiful) or (∃ x)(x is precious and x is notbeautiful)

(b) (∀x)(x is precious⇒ x is not beautiful)

(c) (∃ x)(x is isosceles∧x is a right triangle)

(d) (∀x)(x is a right triangle⇒ x is not isosceles) or ∼ (∃ x)(x is a righttriangle∧x is isosceles)

(e) (∀x)(x is honest)∨ ∼ (∃ x)(x is honest)

(f) (∃ x)(x is honest) ∧ (∃ x)(x is not honest)

(g) (∀x)(x �= 0 ⇒ (x > 0 ∨ x < 0))

(h) (∀x)(x is an integer ⇒ (x > −4 ∨ x < 6)) or (∀x ∈ Z)(x > −4 ∨ x < 6)


(i) (∀x)(∃y)(x > y)

(j) (∀x)(∃y)(x < y)

(k) (∀x)(∀y)[(x is an integer ∧y is an integer ∧y > x) ⇒ (∃z)(y > z > x)] or(∀x ∈ Z)(∀y ∈ Z)[y > x ⇒ (∃z)(y > z > x)]

(l) (∃ x)(x is a positive integer and x is smaller than all other positive integers)or (∃ x)(x is a positive integer and (∀y)(y is a positive integer ⇒ x ≤ y))or (∃x ∈ Z)[x > 0 ∧ (∀y ∈ Z)(y > 0 ⇒ y > x)]

(m) (∀x)(∼ (∀y)(x loves y)) or ∼ (∃ x)(∀y)(x loves y) or (∀x)(∃ y)(x does not love y)

(n) (∀x)(∃ y)(x loves y)

(o) (∀x)(x > 0 ⇒ (∃!y)(2y = x)

2. (a) (∀x)(x is precious⇒ x is beautiful)All precious stones are beautiful.

(b) (∃ x)(x is precious∧x is beautiful)There is a beautiful precious stone, or Some precious stones are beautiful.

(c) ∼ (∃ x)(x is isosceles and x is a right triangle) or (∀x)(x is not isosceles orx is not a right triangle) or (∀x)(x is right triangle⇒ x is not isosceles) or(∀x)(x is isosceles⇒ x is not a right triangle).There is no isosceles right triangle.

(d) (∃ x)(x is isosceles∧x is a right triangle)There is an isosceles right triangle.

(e) (∃ x)(x is dishonest) ∧ (∃ x)(x is dishonest)Some people are honest and some people are dishonest.

(f) (∀x)(x is honest) ∨ (∀x)(x is dishonest)All people are honest or no one is honest.

(g) (∃x)(x �= 0 ∧ x is not positive ∧x is not negative)There is a nonzero real number that is neither positive nor negative.

(h) (∃x)(x is an integer ∧x ≤ −4 ∧ x ≥ 6)) or (∃x ∈ Z)(x ≤ −4 ∧ x ≥ 6)There is an integer that is less than or equal to −4 and greater than orequal to 6.

(i) (∃x)(∀y)(x ≤ y)Some integer is less than or equal to every integer, or There is a smallestinteger.

(j) (∃x)(∀y)(x ≥ y)Some integer is greater than every other integer, or There is a largestinteger.

(k) (∃x)(∃y)[x is an integer ∧y is an integer y > x ∧ (∀z)(z ≤ y ∨ x ≤ z)] or(∃x ∈ Z)(∃y ∈ Z)[y > x ∧ (∀z)(z ≤ y ∨ x ≤ z)]There is an integer x and a larger integer y such that there is no real numberbetween them.

(l) (∀x)(x is a positive integer⇒ (∃y)(y is a positive integer) ∧ x > y) or(∀x ∈ Z)[x ≤ 0 ∨ (∃y ∈ Z)(y > 0 ∧ x > y)]. For every positive integer thereis a smaller positive integer.Or, ∼ (∃x)(x is a positive integer∧(∀y)(y is a positive integer⇒ x ≤ y)) or∼ (∃x ∈ Z)[x > 0 ∧ (∀y ∈ Z(y > 0 ⇒ y > x)]There is no smallest positive integer.


(m) (∃ x)(∀y)(x loves y)There is someone who loves everyone.

(n) (∃ x)(∀y)(x does not loves y) or ∼ (∀x)(∃ y)(x loves y).Somebody doesn’t love anyone.

(o) (∃ x)(x > 0∧ ∼ (∃y)(2y = x) ∨ (∃y)(∃z)[y �= z ∧ 2y = x ∧ 2z = x])There is a positive real number x for which there is no unique real numbery such that 2y = x.There is a nonzero complex number such that either every product of thatnumber with any complex number is different from π, or there are at leasttwo different complex numbers whose products with the given number areequal to π.

3. (a) (∃k)(k is an integer ∧x = 2k) or (∃k ∈ Z)(x = 2k)

(b) (∃j)(j is an integer ∧x = 2j + 1) or (∃j ∈ Z)(x = 2j + 1)

(c) (∃k)(k is an integer ∧b = ak) or (∃k ∈ Z)(b = ak)

(d) n �= 1 ∧ (∀m ∈ Z)(m divides n ⇒ (m = 1 ∨ m = n)

(e) n �= 1 ∧ (∃m ∈ Z)(m divides n ∧ (m �= 1 ∨ m �= n)

4. (a) (∀x, y ∈ A)(xRy ⇒ yRx)

(b) (∀x, y, z ∈ A)(xRy ∧ yRz ⇒ xRz)

(c) (∀x, y ∈ A)(f(x) = f(y) ⇒ x = y)

(d) (∀x, y ∈ A)(x · y = y · x)

5. The first interpretation may be translated as(∀x)[x is a person ⇒ (∀y)(y is a tax ⇒ x dislikes y)].The other sentences may be translated as(∀x)[x is a person ⇒ (∃y)(y is a tax ⇒ x dislikes y)].(∃x)[x is a person ⇒ (∀y)(y is a tax ⇒ x dislikes y)].(∃x)[x is a person ⇒ (∃y)(y is a tax ⇒ x dislikes y)].

6. (a) T , U , V and W (b) T (c) T , U , V (d) T

7. (a) Proof. Let U be any universe. The sentences ∼ (∃x)A(x) is true in Uiff (∃x)A(x) is false in Uiff the truth set for A(x) is emptyiff the truth set for ∼ A(x) is U iff (∀x) ∼ A(x) is true in U .

(b) Let A(x) be an open sentence with variable x. Then ∼ A(x) is an opensentence with variable x, so we may apply part (a) of Theorem 1.3.1(b).Thus ∼ (∀x) ∼ A(x) is equivalent to (∃ x) ∼∼ A(x), which is equivalent to(∃ x)A(x). Therefore ∼ (∃ x)A(x) is equivalent to ∼∼ (∀x) ∼ A(x), whichis equivalent to (∀x) ∼ A(x).

8. (a) false (b) true (c) false (d) true(e) false (f) true (g) false (h) true(i) true (j) false (k) false (l) true

9. (a) Every natural number is greater than or equal to 1.

(b) Exactly one real number is both nonnegative and nonpositive.

(c) Every natural number that is prime and different from 2 is odd.

(d) There is exactly one real number whose natural logarithm is 1.


(e) There is no real number whose square is negative.(f) There exists a unique real number whose square is 0.(g) For every natural number, if the number is odd, then its square is odd.

10. (a) true (b) false (c) false (d) false(e) true (f) false (g) true (h) false(i) false (j) true (k) false

11. (a) Let U be any universe and A(x) be an open sentence. Suppose (∃!x)A(x)is true in U . Then the truth set for A(x) has exactly one element, so thetruth set for A(x) is nonempty. Thus (∃x)A(x) is true in U .

(b) Let A(x) be the sentence x2 = 1 and let the universe be the real numbers.Then the truth set for A(x) is {1,−1} so (∃x)A(x) is true but (∃!x)A(x) isfalse in U .

(c) Let U be any universe and suppose (∃!x)A(x) is true. Then the truth setfor A(x) contains exactly one element x0. As in part (a), (∃x)A(x) is true.Suppose u and z are in U and A(y) and A(z) are true. Then u and z mustboth be x0, so y = z. Thus (∃x)A(x) ∧ (∀y)(∀z)(A(y) ∧ A(z) ⇒ y = z) istrue.On the other hand, suppose (∃x)A(x) ∧ (∀y)(∀z)(A(y) ∧ A(z) ⇒ y = z) istrue in U . Since (∃x)A(x) is true, the truth set for A(x) contains at leastone element. Since (∀y)(∀z)(A(y) ∧ A(z) ⇒ y = z) is true, the truth set forA(x) contains only one element Thus (∃!x)A(x) is true in U .

(d) Let U be any universe. Suppose (∃!x)A(x) is true in U . Then the truth setfor A(x) contains exactly one element, x0. Then for every y in U , if A(y)then x0 = y. Thus x0 is in the truth set of A(x) ∧ (∀y)(A(y) ⇒ x = y),so (∃x)[A(x) ∧ (∀y)(A(y) ⇒ x = y)] is true in U . Conversely, suppose(∃x)[A(x) ∧ (∀y)(A(y) ⇒ x = y)] is true in U . Let x0 be an element in thetruth set of A(x)∧ (∀y)(A(y) ⇒ x = y). Then x0 is the only element in thetruth set of A(x). Thus (∃!x)A(x) is true in U.

(e) (∀x)(∼ A(x) ∨ (∃y)(∃z)(A(y) ∧ A(z) ∧ y �= z))

12. (a) f is continuous at a iff (∀ε)[ε > 0 ⇒ (∃δ)(δ > 0 ∧ (∀x)(|x − x0| < δ ⇒|f(x) − f(a)| < ε]

(b) Let the universe be the set R of real numbers, and let f be a functionfrom R to R. The Mean Value Theorem asserts that (∀a)(∀b)([(a <b) ∧ (f is continuous on [a, b] ∧ (f is differentiable on (a, b))]) ⇒

(∃c)[(a < c < b) ∧ (f ′(c) =f(b) − f(a)

b − a)],

where “f is continuous on [a, b]” means:(∀x0)(a ≤ x0 ≤ b ⇒ (∀ε)[ε > 0 ⇒ (∀δ)(δ > 0 ∧ (∀x)(|x − x0| < δ ⇒|f(x) − f(x0)| < ε)] and “f is differentiable on (a, b)” means:(∀x0)(a < x0 < b ⇒ (∃ d)[f ′(x0) = d].

(c) Let the universe be the set R of real numbers, and let f be a functionfrom R to R. Then lims→a f(x) = L means: (∀ε)(ε > 0 ⇒ (∃ δ)[δ >0 ∧ (∀x)(|x − a| < δ ⇒ |f(x) − L| < ε)])

(d) A denial of “f is continuous at a” is: (∃ε > 0)(∀δ)(δ > 0 ⇒ (∃x)(|x − x0| <δ ∧ |f(x) − f(a)| ≥ ε).A denial of the Mean Value Theorem is: (∃a)(∃b)[a < b ∧ f is continuousin [a, b] ∧ f is differentiable on (a, b) ∧ (∀c)(a < c < b ⇒ f ′(c) �= f(b)−f(a)

b−a )]A denial of “ lims→a f(x) = L” is: (∃ ε)(ε > 0∧(∀δ)[δ > 0 ⇒ (∃x)(|x−a| <δ ∧ |f(x) − L| ≥ ε])


13. (a) This is not a denial. If the universe has only one element a and P (a) istrue, then both the statement and (∃!x)P (x) are true.

(b) This is a denial.

(c) This is a denial.

(d) This is not a denial. If the universe has only one element a and P (a) isfalse, then both the statement and (∃!x)P (x) are false.

(e) This statement is not a denial. If the universe has more than one elementthe statement implies the negation of (∃!x)P (x), but if (∀x)P (x), then boththe statement and (∃!x)P (x) are false.

14. For every backwards E, there exists an upside down A! [This is a joke.]

1.4 Basic Proof Methods I

1. (a) Suppose (G, ∗) is a cyclic group....Thus, (G, ∗) is abelian.Therefore, if (G, ∗) is a cyclic group, then (G, ∗) is abelian.

(b) Suppose B is a nonsingular matrix....Thus, the determinant of B is not zero.Therefore, if B is a nonsingular matrix, then the determinant of B is notzero.

(c) Suppose A is a subset of B and B is a subset of C....Thus, A is a subset of C.Therefore, if A is a subset of B and B is a subset of C, then A is a subsetof C.

(d) Suppose the maximum value of the differentiable function f on the closedinterval [a, b] occurs at x0....Thus, either x0 = a or x0 = b or f ′(x0) = 0. Therefore, if the maximumvalue of the differentiable function f(x) on the closed interval [a, b] occursat x0, then either x0 = a or x0 = b or f ′(x0) = 0.

(e) Let A be a diagonal matrix. Suppose all the diagonal entries of A arenonzero....Then A is invertible. Therefore A is invertible whenever all its nonzeroentries are nonzero

2. If A and B are invertible matrices, then AB is invertible.

(a) Suppose that A and B are invertible matrices....Thus, AB is invertible.Therefore, if A and B are invertible matrices, then the product AB isinvertible.


(b) Suppose AB is invertible....Thus, A and B are both invertible.Therefore, if AB is invertible, then A and B are both invertible.

3. One could construct a truth table with 16 rows and observe that every row hasthe value true for the main connective ⇒. However, it is also correct to show,without actually making the truth table, that no row could have the value Ffor the main connective ⇒. Suppose this connective had the value F. Then theantecedent [(∼ B ⇒ M)∧ ∼ L ∧ (∼ M ∨ L)] must have the value T and theconsequent B must have the value F. Then each of ∼ B ⇒ M , ∼ L and ∼ M ∨Lhas value T. Then L must have the value F, and since ∼ M ∨ L has value T, Mhas the value F. Since ∼ B has the value T and M has the value F, ∼ B ⇒ Mhas the value F. But ∼ B ⇒ M has value T. This contradiction shows that everyrow of the truth table has value T, so the propositional form is a tautology.

4. (a) Professor Plum. The crime took place in the library, not the kitchen. Byfact (i), if the crime did not take place in the kitchen, then Professor Plumis guilty. Therefore Professor Plum is guilty.

(b) Miss Scarlet. The crime did not take place in the library. By fact (iv), theweapon was the candlestick. By fact (iii) Miss Scarlet is not innocent.

(c) Professor Plum. The crime was committed at noon with the revolver. By(iii) Miss Scarlet is innocent. By fact (v) either Miss Scarlet or Profes-sor Plum is guilty. Therefore Professor Plum is guilty.

(d) Miss Scarlet and Professor Plum. The crime took place at midnight in theconservatory. By fact (ii) Professor Plum is guilty. The crime did not takeplace in the library. By fact (iv), the weapon was the candlestick. By fact(iii) Miss Scarlet is guilty also.

5. (a) Suppose that x and y are even. Then there are integers n and m such thatx = 2n and y = 2m. By substitution, x + y = 2n + 2m = 2(n + m). Sincex + y is the product of 2 and an integer, x + y is even.

(b) Suppose that x is an even integer, and y is an integer. Then there is aninteger k such that x = 2k. Then xy = (2k)y = 2(ky). Thus xy is twice theinteger ky, so xy is even.

(c) Suppose that x and y are even integers. Then there exist integers n and msuch that x = 2n and y = 2m. Therefore, xy = 2n · 2m = 4nm. Since nmis an integer, xy is divisible by 4.

(d) Suppose that x and y are even integers. Then there are integers k and msuch that x = 2k and y = 2m. Then 3x−5y = 3(2k)−5(2m) = 2(3k−5m).Since 3k − 5m is an integer and 3x − 5y = 2(3k − 5m), 3x − 5y is even.

(e) Suppose that x and y are odd integers. Then there exist integers n andm such that x = 2n + 1 and y = 2m + 1. By substitution, x + y =(2n + 1) + (2m + 1) = 2(n + m + 1). Since x + y is twice an integer,x + y is even.

(f) Then there exist integers k and m such that x = 2k + 1 and y = 2m + 1.Then 3x − 5y = 3(2k + 1) − 5(2m + 1) = 2(3k − 5m − 1). Since 3x − 5y =2(3k − 5m − 1) and 3k − 5m − 1 is an integer, we conclude that 3x − 5y iseven.

(g) Then there are integers k and m such that x = 2k + 1 and y = 2m + 1.Then xy = (2k + 1)(2m + 1) = 2(km + k + m) + 1. Since km + k + m is aninteger, xy is odd.


(h) Suppose that x is even and y is odd. Then there exist integers n and msuch that x = 2n and y = 2m + 1. Therefore, x + y = (2n) + (2m + 1) =2(n + m) + 1. Since n + m is an integer, x + y is odd.

(i) Suppose that x is even and y is odd. Then there exist integers k and msuch that x = 2k and y = 2m + 1. Then xy = (2k)(2m + 1) = 4km + 2k =2(2km+1). Since k+m is an integer and x−y = 2(k+m)+1, we concludethat xy is odd.

6. (a) If a = 0 or b = 0, then |ab| = 0 = |a||b|Otherwise there are four cases.Case 1. If a > 0 and b > 0, then |a| = a and |b| = b. Also, ab > 0, so|ab| = ab = |a||b|.Case 2. If a > 0 and b < 0, then |a| = a and |b| = −b. Also, ab < 0, so|ab| = −ab = a(−b) = |a||b|.Case 3. If a < 0 and b > 0, then |a| = −a and |b| = b. Also ab < 0, so|ab| = −ab = (−a)b = |a||b|.Case 4. If a < 0 and b < 0, then |a| = −a and |b| = −b. Also ab > 0, so|ab| = ab = (−a)(−b) = |a||b|.In every case, |ab| = |a||b|.

(b) Case 1. Let a − b = 0. Then b − a = 0, so |a − b| = 0 = |b − a|.Case 2. Let a−b > 0. Then b−a < 0, so |a−b| = a−b = −(b−a) = |b−a|.Case 3. Let a−b < 0. Then b−a > 0, so |a−b| = −(a−b) = b−a = |b−a|.Thus |a − b| = |b − a| in every case.

(c) If a = 0, then |ab | = 0 = |a|

|b| . Otherwise there are four cases.

Case 1. Let a > 0 and b > 0. Then |ab | = a

b = |a||b| .

Case 2. Let a > 0 > b. Then ab < 0, so |a

b | = −ab = a

(−b) = |a||b| .

Case 3. Let a < 0 < b. Then ab < 0, so |a

b | = −ab = (−a)

b = |a||b| .

Case 4. Let a < 0 and b < 0. Then ab > 0, so |a

b | = ab = (−a)

(−b) = |a||b| .

(d) Case 1: a ≥ 0. There are three subcases.Subcase 1a: b ≥ 0. Then a + b ≥ 0, so |a + b| = a + b = |a| + |b|.Subcase 1b: b < 0 and a ≥ −b. Then a + b ≥ 0, so |a + b| = a + b <a + (−b) = |a| + |b|.Subcase 1c: b < 0 and a < −b. Then a + b < 0, so |a + b| = −(a + b) =−a − b ≤ a − b = |a| + |b|.Case 2: a < 0. There are three subcases.Subcase 2a: b < 0. Then a + b < 0, so |a + b| = −(a + b) = (−a) + (−b) =|a| + |b|.Subcase 2b: −a > b ≥ 0. Then a+b < 0, so |a+b| = −(a+b) = (−a)+(−b) ≤−a + b = |a| + |b|.Subcase 2c: b ≥ −a > 0. Then a+b ≥ 0, so |a+b| = a+b < −a+b = |a|+|b|.In every case, |a + b| ≤ |a| + |b|.

(e) Assume |a| ≤ b. Then there are two cases to consider.Case 1: a ≥ 0. Since b ≥ |a| ≥ 0, we have −b ≤ 0 ≤ a = |a| ≤ b, so−b ≤ a ≤ b.Case 2: a < 0. Then −a = |a| ≤ b, so −b ≤ a < 0 < −a ≤ b and thus−b ≤ a ≤ b.Therefore |a| ≤ b implies −b ≤ a ≤ b


(f) Assume −b ≤ a ≤ b. There are two cases.Case 1: a ≥ 0. Then a = |a|, so |a| ≤ b.Case 2: a < 0. Then −a = |a|, so |a| = −a ≤ −(−b) = b.Thus −b ≤ a ≤ b implies |a| ≤ b.

7. (a) Suppose a is an integer. Then 2a − 1 = 2a − 2 + 1 = 2(a − 1) + 1. Sincea − 1 is an integer, 2a − 1 is odd.

(b) Let a be an integer. Suppose a is even. Then a = 2k for some integer k.Therefore a + 1 = 2k + 1, so a + 1 is odd.

(c) Assume that a is an odd integer. Then for some integer k, a = 2k+1. Thena + 2 = 2k + 3 = 2(a + 1) + 1. Since a + 1 is an integer, a + 2 is odd.

(d) Let a be an integer. If a is even, then by Exercise 7(b) a + 1 is odd. ByExercise 5(i) a(a + 1) is even. On the other hand, if a is odd, then byExercise 5(e) a + 1 is even. Then, again by Exercise 5(i), a(a + 1) is even.

(e) Let a be an integer. Then a = 1 · a, so 1 divides a.

(f) Let a be an integer. Then a = a · 1, so a divides a.

(g) Suppose a and b are positive integers and a divides b. Then for some integerk, b = ka. Since b and a are positive, k must also be positive. Since k isalso an integer, 1 ≤ k. Therefore, a = a · 1 ≤ a · k = b, so a ≤ b.

(h) Let a and b be integers. Suppose that a divides b. Then b = ka for someinteger k, so bc = kac = (kc)a. Since kc is an integer, a divides bc.

(i) Suppose a and b are positive integers and ab = 1. Then a divides 1 and bdivides 1. By part (g), a ≤ 1 and b ≤ 1. But a and b are positive integers,so a = 1 and b = 1.

(j) Let a and b be positive integers. Suppose a divides b and b divides a. Thenthere is a positive integer n such that an = b and a positive integer m suchthat bm = a. Thus a = bm = (an)m = a(nm). Then nm = 1, so n = 1 andm = 1 by part (i). Since n = 1 and an = b, a = b.

(k) Let a, b, and c be integers. Suppose a divides b and c divides d. Then b = kaand d = jc for some integers k and j. Thus bd = (ka)(jc) = (kj)(ac), andkj is an integer, so ac divides bd.

(l) Let a, b, and c be integers. Suppose ab divides c. Then c = k(ab) for someinteger k. Thus c = (kb)a, and kb is an integer, so a divides c.

(m) Let a, b, and c be integers. Suppose ac divides bc. Then there is an integerk such that (ac)k = bc. Thus kac = bc, so that ka = b. Therefore a dividesb.

8. (a) Case 1: n is even. Then n = 2k for some natural number k, so

n2 + n + 3 = (2k)2 + (2k) + 3 = 4k2 + 2k + 2 + 1 = 2(2k2 + k + 1) + 1.

Since 2k2 + k + 1 is an integer, n2 + n + 3 is odd.Case 2: n is odd. Then n = 2k + 1 for some natural number k, so

n2 + n + 3 = (2k + 1)2 + (2k + 1) + 3 = 4k2 + 4k + 1 + 2k + 1 + 3= 4k2 + 6k + 5 = 2(2k2 + 3k) + 1.

(b) By Exercise 7(d), n2 + n = n(n + 1) is even. Since n2 + n is even and 3 isodd, by Exercise 5(h), n2 + n + 3 is odd.


9. (a) We want to show x+y2 ≥ √

xy, which could be derived from (x+y2 )2 ≥

xy, which would follow from (x + y)2 ≥ 4xy, which would follow fromx2 + 2xy + y2 ≥ 4xy, which would follow from x2 − 2xy + y2 ≥ 0, whichwould follow from (x − y)2 ≥ 0.

Proof. Suppose x and y are nonnegative real numbers. Then (x − y)2 ≥ 0,so x2 − 2xy + y2 ≥ 0. Thus x2 + 2xy + y2 ≥ 4xy, so (x+y

2 )2 ≥ xy. Sincex and y are nonnegative real numbers,

√(x + y)2 = x + y and

√xy is a

real number. Therefore x+y2 ≥ √

xy. We used the fact that x and y arenonnegative in the penultimate sentence in the proof.

(b) We want to show a divides 3c, which would follow if a divides c. To showa divides c, we could write c as the difference or sum of two quantitiesdivisible by c.

Proof. Suppose a divides b and a divides b + c. Then using the theorem(proved as an example) on page 34, a divides the difference (b + c) − b = c.Then a divides their difference (b + c) − b = c.

(c) We want to show ax2 + bx+ c = 0 has two real solutions. This would followif the discriminant b2 − 4ac > 0.

Proof. Suppose ab > 0 and bc < 0. Then the product ab2c < 0. Sinceb2 ≥ 0, ac < 0. Then −4ac > 0, so b2 − 4ac > 0. Therefore by thediscriminant test, the equation ax2 + bx + c = 0 has two real solutions.

(d) We want to show 2x + 5 < 11, which would follow from 2x < 6 or x < 3.

Proof. Suppose x3 + 2x2 < 0. Then x2(x + 2) < 0, so x + 2 < 0. Thusx < −2, so x < 3. Therefore 2x < 6, so 2x + 5 < 11.

(e) To show that the triangle is a right triangle, we want to show c2 = a2 + b2.

Proof. Suppose a triangle has sides of length a, b, and c, where c =√

2aband a = b. Then c2 = (

√2ab)2 = 2ab = 2a2 = a2 + a2 = a2 + b2. Therefore

the triangle is a right triangle.

10. (a) Suppose A > C > B > 0. Multiplying by the positive numbers C and B,we have AC > C2 > BC and BC > B2, so AC > B2. AC is positive, so4AC > AC. Therefore 4AC > B2, so B2 − 4AC < 0. Thus the graph mustbe an ellipse.

(b) Assume AC < 0. Then −4AC > 0, so B2 − 4AC > 0. Thus the graph is ahyperbola.Now assume B < C < 4A < 0. Then −B > −C > −4A > 0, soB2 = (−B)(−B) > (−B)(−C) = BC and BC > 4AC. Thus B2−4AC > 0,so the graph is a hyperbola.

(c) Assume that the graph is a parabola. Then B2 − 4AC = 0, so B2 = 4AC.Assume further that BC �= 0. Then C �= 0, so A = B2

4C .

11. (a) F. This proof, while it appears to have the essence of the correct reasoning,has too many gaps. The first “sentence” is incomplete, and the steps arenot justified. The steps could be justified either by using the definitions orby referring to previous examples and exercises.

(b) C. If a divides both b and c, then there are integers q1 and q2 such thatb = aq1 and c = aq2, but q1 and q2 are not necessarily the same number!

(c) C. It looks as if the author of this “proof” assumed that x+ 1x ≥ 2. The proof

could be fixed by beginning with the (true) statement that (x − 1)2 ≥ 0and ending with the conclusion that x + 1

x ≥ 2.


(d) F. This is a proof that if m is odd, then m2 is odd. We cannot prove astatement by proving its converse.

(e) F. Although every statement is correct, the justification is incomplete.Without additional explanation the reader might wonder whether the proofmeans that x2 is always even and x + 1 is always odd. One approach toa correct proof is to use the fact that x2 + x is always even and that theproduct of an integer with an even integer is even. (Exercises 7(d) and5(b).)

1.5 Basic Proof Methods II

1. (a) Suppose (G, ∗) is a not abelian....Thus, (G, ∗) is not a cyclic group.Therefore, if (G, ∗) is a cyclic group, then (G, ∗) is abelian.

(b) Suppose the determinant of B is zero....Thus, B is a singular matrix.Therefore, if B is a nonsingular matrix, then the determinant of B is notzero.

(c) Suppose the set of natural numbers is finite....

Therefore Q (where Q is some proposition)....

Therefore ∼ Q.But Q and ∼ Q is a contradiction.Therefore, the set of natural numbers is not finite.

(d) Suppose x is a real number other than 0. Then x has a multiplicativeinverse, because x · 1

x = 1.Suppose x has another multiplicative inverse z....Then P , where P is some proposition....Then ∼ P .Therefore P and ∼ P , which is a contradiction.We conclude that x has only one multiplicative inverse.

(e) Part 1. Suppose the inverse of the function f from A to B is a functionfrom B to A....Therefore f is one-to-one....Therefore f is onto B.Part 2. Suppose f is one-to-one and onto B....Therefore the inverse of f is a function from B to A.

(f) Part 1. Suppose A is compact....Therefore A is closed and bounded.


Part 2. Suppose A is closed and bounded....Therefore A is compact.

2. If A and B are invertible matrices, then AB is invertible.

(a) Suppose AB is not invertible....Thus, A is not invertible or B is not invertible.Therefore, if A and B are invertible matrices, then AB is invertible.

(b) Suppose A is not invertible or B is not invertible....Thus, AB is not invertible.Therefore, if AB is invertible, then A and B are both invertible.

(c) Suppose both A and B are invertible, and AB is not invertible....Therefore G (where G is some proposition)....Therefore ∼ G.Hence G and ∼ G, which is a contradiction.Therefore, if A and B are invertible matrices, then AB is invertible.

(d) Suppose AB is invertible, and at least one of A or B is not invertible....Therefore G....Therefore ∼ G.Hence G and ∼ G, which is a contradiction.Therefore, if AB is invertible, then A and B are both invertible.

(e) Part 1. Assume A and B are invertible....Therefore AB is invertible.Part 2. Assume AB is invertible....Then A and B are invertible.We conclude that A and B are invertible if and only if AB is invertible.

3. (a) Suppose x + 1 is even (not odd). Then x + 1 = 2k for some integer k. Thenx = 2k − 1 = 2(k − 1) + 1 and k − 1 is an integer, so x is odd. Therefore ifx is even, then x + 1 is odd.

(b) Suppose x + 2 is even (not odd). Then there is an integer m such thatx+2 = 2m, so x = 2m− 2 = 2(m− 1). Since m− 1 is an integer, x is even.Therefore if x is odd, then x + 2 is odd.

(c) Suppose x is even. Then x = 2k for some integer k. Thus x2 = (2k)2 = 4k2

and k2 is an integer, so x2 is divisible by 4. Therefore if x2 is not divisibleby 4, then x is odd.

(d) Suppose x is odd and y is odd. Then x = 2k + 1 and y = 2m + 1 for someintegers k and m. Then 2km+m+k is an integer and 2(2km+m+k)+1 =


4km + 2m + 2k + 1 = (2k + 1)(2m + 1) = xy, so xy is odd. Therefore if xyis even, then x or y is even.

(e) Suppose it is not the case that either x and y are both odd or x and yare both even. Then one of x or y is even and the other is odd. We mayassume that x is even and y is odd. (Otherwise, we could relabel the twointegers.) Then x = 2k and y = 2m + 1 for some integers k and m. Thenx + y = 2k + 2m + 1 = 2(k + m) + 1 and k + m is an integer, so x + y isodd. Therefore if x + y is even then x and y are both odd or both even.

(f) Suppose x and y are not both odd. Then either x or y is even (or both areeven). We may assume x is even. Then x = 2m for some integer m. Thusxy = (2m)y = 2(my), and my is an integer, so xy is even. Therefore if xyis odd then both x and y are odd.

(g) Suppose x is odd. Then x = 2m + 1 for some integer m. Then x2 − 1 =(2m + 1)2 − 1 = 4m2 + 4m = 4m(m + 1). By a previous exercise, m(m + 1)is even, so m(m + 1) = 2k for some integer k. Thus x2 − 1 = 4(2k) = 8k,so 8 divides x2 − 1. Therefore if 8 does not divide x2 − 1, then x is even.

(h) Assume x divides z. Then z = xk for some integer k. Thus yz = y(xk) =x(yk) and yk is an integer, so x divides yz. Therefore if x does not divideyk, then x does not divide z.

4. (a) Suppose x ≥ 0. Then x+2 > 0, and so the product x(x+2) = x2 +2x ≥ 0.Therefore if x2 + 2x < 0,then x < 0.

(b) Suppose x ≤ 2 or x ≥ 3.If x ≤ 2, then x − 2 ≤ 0 and x − 3 ≤ 0, so (x − 2)(x − 3) = x2 − 5x + 6 ≥ 0.If x ≥ 3, then x−3 ≥ 0 and x−2 ≥ 1 > 0, so (x−3)(x−2) = x2−5x+6 ≥ 0.Therefore if x2 − 5x + 6 < 0, then 2 < x and x < 3.

(c) Suppose x ≤ 0. Since x2 ≥ 0, x2 + 1 > 0. Thus the product x(x2 + 1) =x3 + x ≤ 0. Therefore if x3 + x > 0, then x > 0.

5. (a) Suppose (−1, 5) and (5, 1) are both on a circle with center (2, 4). Then theradius of the circle is

√(2 + 1)2 + (4 − 5)2 =

√10 and the radius of the

circle is√

(2 − 5)2 + (4 − 1)2 =√

18. This is impossible. Therefore (−1, 5)and (5, 1) are not both on the circle.

(b) Suppose the circle has radius less than 5 and there is a point (a, b) on thecircle and on the line y = x−6. Then b = a−6 and (a−2)2 +(b−4)2 < 25,so (a − 2)2 + (a − 10)2 < 25. Then 2a2 − 24a + 79 < 0, or 2(a − 6)2 + 7 < 0.But 2(a − 6)2 + 7 ≥ 7, so 2(a − 6)2 + 7 < 0 is impossible. Therefore thecircle does not intersect the line y = x − 6.

(c) Suppose the point (0, 3) is not inside the circle, but (3, 1) is inside thecircle. Then the distance from (2, 4) to (3, 1) is less than the radius andthe distance from (2, 4) to (0, 3) is greater. Therefore (2 − 3)2 + (4 − 1)2 <(2 − 0)2 + (4 − 3)2. But 1 + 9 is not less than 4 + 1. Therefore if (0, 3) isnot inside the circle, then (3, 1) is not inside the circle.

6. (a) Let a and b be positive integers. Suppose a divides b and a > b. Then thereis a natural number k such that b = ak. Since k is a natural number, k ≥ 1.Thus b = ak ≥ a · 1 = a. Thus b ≥ a. This contradicts the assumption thata > b. Therefore if a divides b, then a ≤ b.

(b) Let a and b be positive integers. Suppose ab is odd and that a or b iseven. We may assume a is even. Then a = 2m for some integer m. Thenab = (2m)b = 2(mb). Since mb is an integer, ab is even. Since a numbercannot be both even and odd, this is a contradiction. Therefore if ab is odd,then a and b are both odd.


(c) Suppose a is odd and a+1 is not even. Then a+1 is odd, so a+1 = 2k+1 forsome integer k. Thus a = 2k, so a is even. This contradicts the assumptionthat a is odd. Therefore if a is odd, a + 1 is even.

(d) Suppose a− b is odd and a+ b is even. Then a− b = 2k +1 and a+ b = 2mfor some integers k and m. Then (a − b) + (a + b) = 2a = 2k + 1 + 2m =2(k + m) + 1 is odd, but 2a is even. This is a contradiction. Therefore ifa − b is odd, then a + b is odd.

(e) Let a, b be positive integers. Assume that a < b and ab < 3. Suppose thata �= 1. Since a is a positive integer, a ≥ 2. And since a < b, b > 3. Thereforeab > 6. This contradicts the assumption that ab < 3.

7. (a) Let a, b, and c be positive integers. Then

a divides b iff b = ak for some integer k

iff bc = (ac)k for some integer k

iff ac divides bc.

(b) Let a and b be positive integers.Part 1. Suppose a = 2 and b = 3. Then a + 1 = 3 divides b = 3 and b = 3divides b + 3 = 6.Part 2. Suppose a + 1 divides b and b divides b + 3. Then b + 3 = bk forsome integer k, so 3 = bk − b = b(k − 1). Therefore b divides 3, so b = 1 orb = 3. Since a+1 divides b, a+1 ≤ b. Thus b �= 1, so b = 3. Since a+1 > 1and a + 1 divides 3, a + 1 = 3. Thus a = 2.

(c) Let a be a positive integer.Part 1. Suppose a is odd. Then a = 2k + 1 for some integer k, soa + 1 = 2k + 2 = 2(k + 1). Since k + 1 is an integer, a + 1 is even.Part 2. Suppose a + 1 is even. Then a + 1 = 2m for some integer m, soa = 2m − 1 = 2(m − 1) + 1. Since m − 1 is an integer, a is odd.

(d) Let a, b, c, d be positive integers.

a + b = c and 2b − a = d iff a = b − c and 2b − a = d

iff a = b − c and 2b − (b − c) = d

iff a = b − c and b + c = d.

8. (a) Suppose m and n have the different parity. Then one is even and the otheris odd. We may assume, without loss of generality, that m is even and nis odd. Then m = 2k and n = 2j + 1 for some integers k and j. Thenm2 −n2=(2k)2−(2j +1)2 = 4k2 −4j2 −4j −1 = 2(2k2 −2j2 −1)+1, whichis odd.

(b) Suppose m2 − n2 is odd.If m is even, then m2 is even. Therefore, since m2 − n2 is odd and m2 iseven, n2 = m2 − (m2 − n2) is odd. From n2 is odd, we conclude that n isodd.If m is odd, then m2 is odd. Therefore, since m2 −n2 is odd and m2 is odd,n2 = m2 − (m2 − n2) is even. From n2 is even, we conclude that n is even.Hence, if m is even, then n is odd, and if m is odd, then n is even. Therefore,m and n have opposite parity.


9. Let n be a natural and suppose to the contrary that nn+1 ≤ n

n+2 . Thenn(n + 2) ≤ n(n + 1), so n + 2 ≤ n + 1, which is impossible. We concludethat n

n+1 > nn+2 .

10. Assume√

5 is a rational number. Then√

5 = pq where p and q are positive

integers, q �= 0, and p and q have no common factors. Then 5 = p2

q2 , so 5q2 = p2.The prime factorization of q2 has an even number of 5’s (twice as many as thefactorization of q), so 5q2 has an odd number of 5’s in its prime factorization.But 5q2 = p2 and p2 has an even number of 5’s in its prime factorization. Thisis impossible. Therefore

√5 is irrational.

11. Suppose x, y, z are real numbers and 0 < x < y < z < 1. Assume thatthe distances from x to y and from y to z are at least 1

2 . That is, assume|x − y| = y − x ≥ 1

2 and |y − z| = z − y ≥ 12 . The total distance from 0 to 1 is 1;

that is (x − 0) + (y − x) + (z − y) + (1 − z) = 1. But x − 0 > 0 and 1 − z > 0, so(x − 0) + (y − x) + (z − y) + (1 − z) > 0 + 1

2 + 12 + 0 = 1. This is a contradiction.

Therefore at least two of x, y, z are within 12 unit from one another.

12. (a) F. This is a proof of the converse of the statement, by contraposition.

(b) A.

(c) F. This seems to be a persuasive “proof” that the sum of two even integersis even, but it assumes that the sum of even numbers is even, which is whatmust be proved.

(d) C. Leaving out the assumption that a divides b and a divides c makesthis proof confusing. If we change the first sentence to “Assume a dividesboth b and c and a does not divide b + c” then we have a correct proof bycontradiction.

1.6 Proofs Involving Quantifiers

1. (a) Choose m = −3 and n = 1. Then 2m + 7n = 1.

(b) Choose m = 1 and n = −1. Then 15m + 12n = 3.

(c) Suppose m and n are integers and 2m + 4n = 7. Then 2 divides 2m and 2divides 4n, so 2 divides their sum 2m+4n. But 2 does not divide 7, so thisis impossible.

(d) Suppose 12m + 15n = 1 for some integers m and n. Then 3 divides the leftside but not the right side, which is impossible.

(e) Let t be an integer. Suppose there exist integers m and n such that15m + 16n = t. Let r and s be the integers 5m and 2n, respectively. Then3r + 8s = 15m + 16n = t.

(f) Suppose 12m + 15n = 1 for some integers m and n, and suppose furtherthat m and n are not both positive. Then 3 divides the left side but not theright, which is impossible. Therefore, if there exist integers m and n suchthat 12m + 15n = 1, then m and n are both positive.

Alternative proof. Let P be the statement “there exist integers m and nsuch that 12m + 15n = 1.” By part (d), P is false. Therefore P implies mand n are both positive.

(g) Let m be an odd integer. Suppose m = 4k + 1 for some integer k. Thenm + 2 = 4k + 3 = 4(k + 1) − 1 = 4j − 1 where j is the integer k + 1.


(h) Suppose m is an odd integer. Then m = 2n + 1 for some integer n. Thenm2 = (2n + 1)2 = 4n2 + 4n + 1 = 4(n2 + n) + 1. But n2 + n = n(n + 1) iseven (Recall Exercise 7(d) of Section 1.4), so n2 + n = 2k for some integerk. Therefore m2 = 4(2k) + 1 = 8k + 1.

(i) Suppose m and n are odd integers. Assume that neither m nor n is of theform 4j − 1 for some integer j. Then m = 4j1 + 1 and n = 4j2 + 1, forsome integers j1 and j2. (Recall the result from Section 1.4 that every oddinteger has either the form 4j + 1 or the form 4j − 1, for some integer j.)Thus mn = (4j1 + 1)(4j2 + 1) = 4(4j1j2 + j1 + j2) + 1. Then mn cannot bewritten in the form 4k − 1, where k is an integer. Therefore, if mn is of theform 4k − 1, then m or n is of the form 4j − 1.

2. (a) Let a, b and c be integers such that c divides a and c divides b. Let x andy be integers. Then there exist integers m and n such that a = cm andb = cn, so ax + by = (cm)x + (cn)y = c(mx + ny) and mx + ny is aninteger, so c divides ax + by.

Alternate proof. Let a, b and c be integers such that c divides a and cdivides b. Let x and y be integers. By a previous exercise, c divides integermultiples of a and b, so c divides ax and by. Then (using another exercise)c divides their sum ax + by.

(b) The proof involves repeated applications of the previous exercise, whichsays that a divides any sum of integer multiples of a.Let a, b and c be integers such that a divides b−1 and a divides c−1. Thenby exercise 2(a) a divides (b − 1)(c − 1) = bc − b − c + 1. Then by Exercise2(a), a divides (bc − b − c + 1) + (b − 1) = bc − c. Then by Exercise 2(a), adivides (bc − c) + (c − 1) = bc − 1.

(c) Let a and b be integers such that b = ka for some integer k. Let n ∈ N.Then bn = (ka)n = knan, so an divides bn.

(d) Let a and c be integers such that a is odd, c > 0, c divides a and c dividesa + 2.By Exercise 2(a), c divides (−1)a+1(a+2), so c divides 2. The only divisorsof 2 are 1 and 2. If c = 2, then 2 divides a, so a is even. But a is odd, soc = 1.

(e) and suppose there exist integers m and n such that am + bn = 1. Alsosuppose that c divides a and c divides b. Then by part (a), c divides am+bn,so c divides 1. Thus c = ±1. Therefore if there exist integers m and n suchthat am+bn = 1, and c �= ±1, then c does not divide a or c does not divideb.

3. Assume that every even natural number greater than 2 is the sum of two primes.Let n be an odd natural number greater than 5. Then n − 3 is an even naturalnumber greater than 2, so by the hypothesis it is the sum of two primes. Let p1and p2 be primes such that n − 3 = p1 + p2. Then n = p1 + p2 + 3. Since p1, p2,and 3 are primes, n is the sum of three primes.

4. (a) False.Counterexample: Let x = 41. Then x2 + x + 41 = 412 + 41 + 41 =41(41 + 1 + 1) = 41(43), which is not prime.

(b) True.

Proof. Let x be a real number. Then y = −x is a real number such thatx + y = 0.


(c) False.Counterexample: Let x = 2 and y = 1. Then yx = 12 = 1�>2 = x.

(d) False.Counterexample: Let a = 6, b = 3 and c = 2. Then a divides bc, but adoesn’t divide b or c.

(e) True.

Proof. Suppose a, b, c, d, j and k are integers such that b − c = ka andc− d = ja. Then b− d = (b− c)+ (c− d) = ka+ ja = (k + j)a, so a dividesb − d.

(f) False.Counterexample: Let x = 1

2 . Then x2 − x = − 14 < 0.

(g) False.Counterexample: Let x = 1

2 . Then 2x =√

2 ≈ 1.415 < 1.5 = x + 1.(h) False.

Counterexample: Let x = 1. Suppose y is a positive real number less thanx. Then 0 < y < 1, and for z = 2 (or any other positive real number)yz < z.

(i) True.

Proof. Let x be a positive real number and choose y = x. Then thestatement [y < x ⇒ (∀z)(yz ≥ z)] is true.

Alternative proof. Let x be a positive real number, and let y = 1. Thenif z is a positive real number, yz = z, so yz ≥ z.

5. (a) Part 1. Suppose x is prime. Then by definition x is not 1 and there is nopositive integer greater than 1 and less than or equal to

√x that divides x.

Part 2. Assume that x > 1 and there is no positive integer greater than 1and less than or equal to that divides x. Suppose x = mn for some naturalnumbers m and n, and m ≤ n. By the hypothesis m = 1 or m >

√x. But

m >√

x implies that n >√

x, from which we conclude that mn > x. Sincethis is impossible, m = 1 and thus n = x. Therefore x is prime.

(b) Suppose p is prime and p �= 3. Then 3 does not divide p, so when p isdivided by 3 the remainder is either 1 or 2. Thus, there is an integer k suchthat p = 3k + 1 or there is an integer k such that p = 3k + 2.If p = 3k+1, then p2+2 = (3k+1)2+2 = (9k2+6k+1)+2 = 9k2+6k+3 =3(3k2 + 2k + 1) so p2 + 2 is divisible by 3.If p = 3k + 2, then p2 + 2 = (3k + 2)2 + 2 = (9k2 + 12k + 4) + 2 =9k2 + 12k + 6 = 3(3k2 + 4k + 2) so p2 + 2 is divisible by 3.

6. (a) Let n be a natural number. Then n ≥ 1, so 1 = nn ≥ 1

n .(b) Choose M = 10. Let n be a natural number greater than m. Then

1n < 1

M = 0.1 < 0.13.(c) Let n be a natural number. Then both 2n and 2n+1 are natural numbers.

Let M = 2n + 1. Then M is a natural number greater than 2n.(d) Let M = 2. Now if n is a natural number, then by part (a) 1

n ≤ 1 < 2.(e) Suppose there is a largest natural number K. Then K + 1 is a natural

number and K + 1 > K. This is a contradiction.(f) Let ε be a positive real number. Then ε

2 is a smaller positive real number.Therefore, for every positive real number there is a smaller positive realnumber.


(g) Let ε > 0 be a real number. Then 1ε is a positive real number and so has

a decimal expression as an integer part plus a decimal part. Let M be theinteger part of 1

ε plus 1. Then M is an integer and M > 1ε .

To prove for all natural numbers n > M that 1n < ε, let n be a natural

number and assume that n > M . Since M > 1ε , we have n > 1

ε . Thus1n < ε. Therefore, for every real number ε > 0, there is a natural numberM such that for all natural numbers n > M , 1

n < ε.(h) Let ε > 0. By part (g), there is M such that 1

n < ε for all n > M . Now ifm > n > M , then 1

n − 1m < 1

n < ε.(i) Let K be 10. Suppose r > K. Then r > 10, so r2 > 100. Therefore

1r2 < 0.01.

(j) Let L = −15 and G = −1. Suppose L < x < G. Then −1.5 < x < −1, so15 > −x > 1. Then 30 > −2x > 2 and 40 > 10 − 2x > 12.

(k) Let M be 51. Then M is an odd integer. Suppose r is a real number andr > M . Then r > 51, so r > 50. Then 2r > 100, so 1

2r < 1100 .

(l) Let x be a natural number. Choose k = −4x+50. Then k is an integer andk < −3.3x + 50, so 3.3x + k < 50.

(m) Let x be 99 and y = 28. Then x+y = 127 < 128. Suppose r > x and s > y.Then r − 50 > 49 and s − 20 > 8. Therefore (r − 50)(s − 20) > 392 > 390.

(n) Let x and y be positive real numbers such that x < y. Then y−x is positive,and 1

y−x is a positive real number. Choose M to be a natural number largerthan 1

y−x . Suppose n is a natural number and n > M . Then n > 1y−x , so

y − x > 1n . That is, 1

n < y − x.

7. (a) F. The false statement referred to is not a denial of the claim.(b) C. Uniqueness has not been shown.(c) F. Listing numerous examples does not constitute a proof.(d) A.(e) F. The proof is correct for Case 2. However, giving examples for Cases 1

and 3 does not prove that the statement is true for all x in those cases.(f) A. Terse, but correct.(g) A. A proof by contraposition would be more natural.(h) F. The “proof” shows that the converse of the claim is true.(i) C. The number 1

2ε may not be a natural number. To correct this error,choose K to be a natural number greater than 1

2ε .(j) A.

1.7 Additional Examples of Proofs

1. (a) Proof. We work both forwards and backwards: From the hypothesis thatis odd we can deduce that 3n is even, from which we can 3n + 1 deducethat n is even. We could reach the conclusion that is 2n+8 divisible by 4 ifwe knew that 4 divides 2n (since 8 is divisible by 4). In turn, the statement2n is divisible by 4 may be derived from the statement that n is divisibleby 2. We combine these steps in the proper order to create the proof.Suppose n is an integer and is odd. Therefore 3n is even, 3n+1 which impliesthat n is even. We are now using properties of even and odd integers thatwe proved earlier, without referencing specific examples or exercises. Sincen is even, n is divisible by 2. Therefore 2n is divisible by 4. Finally since 8is also divisible by is divisible by 4, 2n + 8 is divisible by 4.


(b) Proof. Let a be a real number, a �= 3. The key to the proof is to use thea = 3 idea of “solution” and then work with the resulting equation.Assume that a is a solution to x2 − x − 6 = 0.Then a makes the equation true by the definition of a solution to anequation.Thus a2 − a − 6 = (a − 3)(a + 2) = 0.Then a + 2 = 0, because a − 3 �= 0.Then (a2 + 1)(a + 2) = a3 + 2a2 + a + 3 �= 0.Therefore a is a solution to x3 + 2x2 + x + 3 = 0.

(c) Proof. Assume that a �= 3. Observe that in the proof above, each stepimplies its predecessor. Thus we can modify the given proof to create an iffproof.

a is a solution to x2 − x − 6 = 0iff a2 − a − 6 = (a − 3)(a + 2) = 0.

iff a + 2 = 0. Because a − 3 �= 0.

iff (a2 + 1)(a + 2) = a3 + 2a2 + a + 3 = 0. Because a2 + 1 �= 0.

iff a is a solution to x3 + 2x2 + x + 3 = 0.

(d) Proof. Suppose x2 = 2x + 15 and x > 2. Then (x − 5)(x + 3) = 0. Sincex > 2, x must be 5. Then x−4 and x−3 are positive, so (x−4)/(x−3) > 0.

(e) Proof. Let x and y be real numbers. The statement has the form P ⇒(Q∨R), so it might be proved by assuming P and ∼ Q and deducing R. Inthis case a proof by contrapositive works well.Assume that neither x nor y is irrational. Then both x and y are rational,so they can be written in the form x = p

q and y = rs where p, q, r, and

s are integers, q �= 0, and s �= 0. Therefore, x + y = pq + r

s = ps+qqs . Since

ps+ rq and qs are integers and qs = 0, x+ y is a rational number. We haveshown that if x and y are rational, then x+ y is rational. We conclude thatif x + y is irrational, then either x or y is irrational.

(f) Proof. If we let S be the set of all nonvertical lines in the xy-plane, we cansimplify the symbolic form of the theorem as follows: (∀L1 ∈ S)(∀L2 ∈ S)(L1 and L2 are perpendicular ⇒(slope of L1) · (slope of L2) = −1) >.Let L1 and L2 be nonvertical lines. Suppose L1 and L2 are perpendicular.We now use the fact that the slope of a nonvertical line is tan(α), whereα is the angle of inclination of the line. Let α1 and α2 be the angles ofinclinations of L1 and L2, respectively. See the figure below. We may assumethat α1 > α2. We can make this assumption because the two lines arearbitrary; if α1 < α2 simply interchange the labels of the lines. Since L1and L2 are perpendicular, α1 = α2 + π

2 . Therefore,

tan(α1) = tan(α2 +π

2) = − cot(α2) = − 1

tan(α2).

We use trigonometric identities to rewrite tan(α1). Thus, tan(α1) ·tan(α2) = −1. Since tan(α1) is the slope of L1 and tan(α2) is the slope ofL2, the product of the slopes is −1.


x

y

α2α1

L1 L2

(g) Proof. This is a “non-existence” proof. We could restate the result as“Every point inside the circle is not on the line” and begin a direct proofby assuming that (x, y) is a point inside the circle. We would then have toprove that (x, y) is not on the line. In this instance, a better approach is touse a proof by contradiction. The statement has the form ∼ (∃x)(∃y)((x, y)is inside the circle ∧(x, y) is on the line).Suppose there is a point (a, b) that is inside the circle and on the line. Then(a − 3)2 + b2 < 6 and b = a + 1. We now have two expressions to use.Therefore,

(a − 3)2 + (a + 1)2 < 62a2 − 4a + 10 < 6

a2 − 2a + 5 < 3a2 − 2a + 1 < −1

(a − 1)2 < −1.

This is a contradiction since (a − 1)2 = 0. Thus, no point inside the circleis on the line.

(h) Proof. Proofs that verify equalities or inequalities containing absolute valueexpressions usually involve cases, because of the two-part definition of |x|.The two cases are x − 2 ≥ 0 and x − 2 < 0. The proof in each caseis discovered by working backwards from the desired conclusion. The keysteps are to note that, in the first case, if x ≥ 2, then −6 ≤ x, and, in thesecond case, that if x ≥ 1, then 6

7 ≤ x >.Let x be a real number greater than 1.Case 1. Suppose x − 2 ≥ 0. Then |x − 2| = x − 2. Since x ≥ 2,

−6 ≤ x

3x − 6 ≤ 4x

3(x − 2)x

≤ 4. Remember that x is positive.

Therefore,3|x − 2|

x≤ 4.

Case 2. Suppose x − 2 < 0. Then |x − 2| = −(x − 2). By hypothesis, x ≥ 1.Therefore,

67

≤ x

6 ≤ 7x

6 − 3x ≤ 4x

3[−(x − 2)] ≤ 4x


3[−(x − 2)]x

≤ 4. Remember that x is positive.

Therefore,3|x − 2|

x≤ = 4.

2. (a) Let n be an integer. Then n is either even or odd. If n is even then n = 2kfor some integer k, so that 5n2 +3n+4 = 5(2k)2+3(2k)+4, which is twicethe integer 10k2 +3n+2. If n is odd then n = 2k +1 for some integer k, sothat 5n2 + 3n + 4 = 5(2k + 1)2 + 3(2k + 1) + 4, which is twice the integer10k2 + 13n + 6. In either case, 5n2 + 3n + 4 is even.

(b) Let n be an odd integer. Then n = 2k + 1 for some integer k, so2n2 + 3n + 4 = 2(2k + 1)2 + 3(2k + 1) + 4 = 2(4k2 + 7k + 4) + 1. Since4k2 + 7k + 4 is an integer, 2n2 + 3n + 4 is odd.

(c) Let x be the smallest of five consecutive integers. Then the sum is x+(x+1) + (x + 2) + (x + 3) + (x + 4) = 5x + 10 = 5(x + 2). Since x + 2 is aninteger, the sum is divisible by 5.

(d) Let L1 and L2 be two nonvertical lines such that the product of their slopesis −1. Let α1 and α2 be the inclination angles of L1 and L2 respectively.Since neither line is vertical, the slope of L1 is tan(α1) and the slope of L2is tan(α2).Since tan(α1) tan(α2) = −1, exactly one of these two must be positive andneither can be 0. Suppose without loss of generality that tan(α2) > 0. Then0 < α2 < π

2 . Now tan(α1) = −1tan(α2)

= − cot(α2) = tan(α2 + π2 ), and since

both α1 and α2 + π2 are between π

2 and π, we must have that α1 = α2 + π2 .

Therefore L1 and L2 are perpendicular.(e) Let n be an integer. Then n3 − n = n(n2 − 1) = n(n + 1)(n − 1) is the

product of three consecutive integers. By previous results, if n is even, thenn+1 is odd and if n is odd, then n+1 is even. Therefore, either n or n+1 isdivisible by 2. By the Division Algorithm, the remainder when n is dividedby 3 will be 0, 1, or 2.Case 1. If the remainder is 0, then n is divisible by 3.Case 2. If the remainder is 1, n = 3k+1 for some integer k. Then n−1 = 3kis divisible by 3.Case 3. If the remainder is 2, n = 3k + 2 for some integer k. Thenn + 1 = 3k + 2 + 1 = 3k + 3 is divisible by 3.In all cases, one of n, n − 1, n + 1 has a factor of 3, and either n or n + 1has a factor of 2. Therefore, n3 −n has a factor of 2 ·3 and is divisible by 6.

(f) Let n be an integer. Then (n3 − n)(n + 2) = n(n + 1)(n − 1)(n + 2) is theproduct of four consecutive integers. From part (e) n3 − n is divisible by 6and hence divisible by 3.By previous results, if n is even, then n + 1 is odd and if n is odd, thenn + 1 is even. Therefore, either n or n + 1 is divisible by 2. If n is divisibleby 2, then so is n+2. If n+1 is divisible by 2, then so is n+1− 2 = n− 1.Thus either both n and n+2 are each divisible by 2 or both n−1 and n+1are each divisible by 2.In all cases, n3 − n has a factor of 3, and two terms (either n and n + 2, orn− 1 and n+1) each have a factor of 2. Therefore, (n3 −n)(n+2) has twofactors of 2 and one factor of 3 and therefore is divisible by 2 · 2 · 3 = 12.

3. (a) Suppose the line 2x + ky = 3k has slope 13 . In slope-intercept form the line

has equation y = − 2kx + 3, so − 2

k = 13 . Thus k = −6. Therefore if k �= −6,

then L does not have slope 13 .


(b) Suppose for some real number k that L is parallel to the x-axis. Then Lhas slope 0, so −2

k = 0. This is impossible. Therefore L is not parallel tothe x-axis.

(c) L passes through (1, 4) iff 2(1) + k(4) = 3k, and this happens iff k = −2.

4. (a) Suppose x is rational. Suppose that x + y is also rational. Then there existintegers p and q, q �= 0 such that x = p

q and integers r and s, s �= 0, suchthat x + y = r

s . Then y = (x + y) − x = rs − p

q = rq−pssq . Since rq − ps and

sq are integers and sq �= 0, y is rational. Therefore if x is rational and y isirrational, then x + y is irrational.

(b) Let x = π and y = −π. Then x and y are irrational, and x+ y is irrational.

(c) Let z be a rational number. By part (a), z + π is irrational. Let x = z + πand y = −π. Then x + y = z.

(d) Let z be a rational number and x be an irrational number. Then −x isirrational. Let y = z − x. Then x + y = x + (z − x) = z and z − x isirrational by part (a), so there exists an irrational number y such thatx + y = z.Suppose there is an irrational number w such that x + w = z. Thenw = z − x = y. Therefore the irrational number y such that x + y = zis unique.

5. (a) Assume (x, y) is on the given circle and y �= 0. Then x2 + y2 = r2, so−y2 = x2 − r2 = (x + r)(x − r) and so y

x−r = −x+ry .

Thus the slope of the line passing through (x, y) and (r, 0) is the negativereciprocal of the slope of the line that passes through (x, y) and (−r, 0).Therefore the lines are perpendicular.Note that if y = 0, then x = ±r, so the points for which this argument doesnot apply are (r, 0) and (−r, 0). If (x, y) = (r, 0), then there are many linespassing through (x, y) and (r, 0), only one of which is perpendicular to theline through (x, y) and (−r, 0).

(b) First observe that if y = 0, then the points (x, y), (r, 0), and (−r, 0) are allon the x-axis, so the line through (x, y) and (r, 0) is not perpendicular tothe line through (x, y) and (−r, 0).Suppose y �= 0 and the two lines are perpendicular. Then the slopes of thelines are negative reciprocals, so y

x−r = −x+ry . Thus −y2 = (x+r)(x−r) =

x2 − r2 and so x2 + y2 = r2. But the fact that (x, y) lies inside the circlerequires that x2 + y2 < r2, so this is a contradiction.

6. (a) Then y0 = 6 − x0. The distance between (x0, y0) and (−3, 1) is

D =√

(−3 − x0)2 + (1 − y0)2

=√

(−3 − x0)2 + [1 − (6 − x0)]2

=√

(−3 − x0)2 + (x0 − 5)2

=√

2x20 − 4x0 + 34

=√

2(x20 − 2x0 + 1) + 32

=√[

2 (x0 − 1)2 + 16]

+ 16

>√

16 since[2 (x0 − 1)2 + 16

]> 0


= 4.

Since the distance from (−3, 1) to (x0, y0) is greater than 4, the point(x0, y0) is outside the circle.

(b) 269 is such a number.

(c) By the Extreme Value Theorem, if f does not have a maximum value on[5, 7], then f is not continuous. And if f is not continuous on [5, 7], then fis not differentiable on [5, 7].

(d) Suppose to the contrary that there are real numbers a < b that satisfythis equation. Since f(x) = x3 + 6x − 1 is continuous and differentiableeverywhere, it is certainly continuous on [a, b] and differentiable on (a, b).Thus Rolle’s Theorem asserts the existence of a number c ∈ (a, b) such thatf ′(c) = 3c2 + 6 = 0, which is impossible since 3c2 ≥ 0.

7. (a) (This proof is motivated by working backward from two desired inequalities− one for each possible value of |2x − 1|.)Let x be a nonnegative real number.Case 1: x ≥ 1

2 . In this case 2x−1 ≥ 0, so |2x−1| = 2x−1. Since −1 ≤ 2, wehave 2x − 1 ≤ 2x + 2 = 2(x + 1). Since x + 1 is positive, we have 2x−1

x+1 ≤ 2.

Thus |2x−1|x+1 ≤ 2.

Case 2: 0 < x < 12 . In this case 2x − 1 < 0, so |2x − 1| = 1 − 2x. Since

x > 0, −1 < 4x. Thus 1 − 2x < 2 + 2x = 2(1 + x). Since 1 + x is positive,we have 1−2x

1+x < 2. Therefore |2x−1|x+1 < 2.

(b) If −2 < x < 1, then (x − 1) < 0, (x + 2) > 0, (x − 3) < 0 and (x + 4) > 0,so (x−1)(x+2)

(x−3)(x+4) > 0.

Now if x > 3, then all of the factors are positive, so (x−1)(x+2)(x−3)(x+4) > 0.

8. (a) Suppose (x, y) is inside the first circle. The from the distance formula,|x − 3|2 + |y − 2|2 < 4. Therefore |x − 3|2 < 4 and |y − 2|2 < 4. It followsthat |x − 3| < 2 and |y − 2| < 2, so −2 < x − 3 < 2 and thus 1 < x < 5 and0 < y < 4. Therefore x2 < 25 and y2 < 16, so x2 + y2 < 41. Thus (x, y) isinside the second circle.

(b) Suppose (x, y) is inside the circle. Then as in part (a), |x − 3| < 2 and|y − 2| < 2, so in particular, x − 3 < 2 and −2 < y − 2. Thereforex − 6 < −1 < 0 < y < 3y.

(c) The statement is false. The point (2, 3) is inside the circle (x−3)2+(y−2)2 =4, since (2−3)2+(3−2)2 = 2 < 4. But (2, 3) is not inside (x−5)2+(y+1)2 =25, since (2 − 5)2 + (3 + 1)2 = 25�<25.

9. (a) 310 = 8(38) + 6. The quotient is 38 and remainder is 6.

(b) 36 = 5(7) + 1. The quotient is 7 and remainder is 1.

(c) 36 = −5(−7) + 1. The quotient is -7 and remainder is 1.

(d) −36 = 5(−8) + 4. The quotient is -8 and remainder is 4.

(e) 44 = 7(6) + 2. The quotient is 6 and remainder is 2.

(f) −52 = −8(7) + 4. The quotient is 7 and remainder is 4.

10. (a) Suppose a and b are integers, a > b, and b ≥ 0. Then b = a(0) + b, so whenb is divided by a, the quotient is 0.


(b) Suppose a and b are integers, a > b, and the quotient is 0 when b is dividedby a. Then b = a(0) + r, where the remainder r is ≥ 0. Then r = b, sob ≥ 0.

11. (a) 2, 1,−1,−2 gcd(8, 310) = 2

(b) 1,−1 gcd(−5, 36) = 1

(c) 18, 9, 6, 3, 2, 1,−1,−2,−3,−6,−9,−18gcd(18,−54) = 18

(d) 4, 2, 1,−1,−2,−4 gcd(−8,−52) = 4

12. (a) 2 = (2)12 + (−1)22 and 2 = (−9)12 + 5(22)

(b) −4 = (7)12 + (−4)22 and −4 = (−4)12 + 2(22)

(c) The set of all linear combinations of 12 and 22 is the set of even integers.

13. (a) gcd(13, 15) = 1. 1 = (7)13 + (−6)15

(b) gcd(26, 32) = 2. 2 = (5)26 + (−4)32

(c) gcd(9, 30) = 3. 3 = (7)9 + (−2)30

14. (a) Let a, b, and c be natural numbers and gcd(a, b) = d. Suppose c divides aand c divides b. By Theorem 1.7.3, d is a linear combination of a and b. ByTherorem 1.7.1, c divides every linear combination of a and b. Therefore cdivides d.

(b) Let a and b be natural numbers and gcd(a, b) = d.

i. Suppose a divides b. Then is a common divisor of a and b. No numberlarger than a divides a, so a is the largest common divisor. Thus a = d.

ii. Suppose a = d. Then a is a common divisor of a and b, so a divides b.

(c) Let a, b, and c be natural numbers and gcd(a, b) = d. Suppose a divides bcand d = 1. By Theorem 1.7.3, d is the smallest positive linear combinationof a and b. Therefore there exist integers s and t such that as + bt = 1.Then acs + bct = c. Since a divides acs and a divides bct, a divides theirsum. Thus a divides c.

(d) Let a, b, and c be natural numbers and gcd(a, b) = d. Suppose c divides aand b. Since gcd(a, b) = d, c divides d (by part (a)) and there are integerss and t such that as + bt = d. Then

(c

c

)as +

(c

c

)bt = d.

Thereforea

cs +

b

ct =

d

c.

Since dc is a linear combination of a

c and bc , by Theorem 1.7.1 gcd(a

c , bc )

divides dc . Hence

d

c≥ gcd

(a

c,b

c

).

By Theorem 1.7.3 there exist integers p and q such that

a

cp +

b

cq = gcd(

a

c,b

c).

Therefore

ap + bq = c · gcd(

a

c,b

c

).


Since

c · gcd(

a

c,b

c

)

is a linear combination of a and b, d divides

c · gcd(

a

c,b

c

).

Thus dc divides

gcd(

a

c,b

c

).

Henced

c≤ gcd

(a

c,b

c

).

We conclude thatd

c= gcd

(a

c,b

c

).

(e) Let a and b be natural numbers and gcd(a, b) = d. Let n be a naturalnumber. Since gcd(a, b) = d, there are integers s and t such that as+bt = d.Then n(as + bt) = nd. Therefore (an)s + (bn)t = dn. Since dn is a linearcombination of an and bn, by Theorem 1.7.1 dn ≥ gcd(an, bn).By Theorem 1.7.3 there exist integers p and q such that (an)p + (bn)q =gcd(an, bn). Therefore ap + bq = 1

ngcd(an, bn). Since 1ngcd(an, bn) is a

linear combination of a and b, d divides 1ngcd(an, bn). Thus, dn divides

gcd(an, bn), which implies dn ≤ gcd(an, bn).We conclude that dn ≡ gcd(an, bn).

15. 3, 6, and 10 are relatively prime to 7.

10 is relatively prime to 21.

None is relatively prime to 30.

16. (a) Suppose p is prime and a is any natural number. The only divisors of p are1 and p, and gcd(p, a) divides p, so gcd(p, a) = 1 or p.

i. Assume gcd(p, a) = p. Then p divides a by definition of gcd.ii. Suppose p divides a. Then p is a common divisor of p and a. Since p is

the largest divisor of p it is the largest common divisor of p and a, sogcd(p, a) = p.

(b) Suppose p is prime and a is a natual number.

i. Suppose gcd(p, a) = 1. Then p does not divide a, because otherwise pwould be a divisor of both p and a that is larger than 1.

ii. Suppose p does not divide a. Then the only common divisor of p anda is 1, so gcd(p, a) = 1.

17. Suppose q is a natural number greater than 1 with the property that q dividesa or q divides b whenever q divides ab. Assume q is composite. Then q has adivisor m that is not 1 and not q. That is, q = mn for some integer n, where1 < n < q. Then q divides mn, so by the given property, q divides m or q dividesn. But m and n are less than q, so this is impossible. Therefore q is prime.

18. Suppose a and b are relatively prime nonzero integers and c is an integer. Thengcd(a, b) = 1, so 1 is a linear combination of a and b. That is, 1 = as + bt forsome integers s and t. Then acs + bct = c. Thus x = cs, y = ct is an integersolution for the equation ax + by = c.


19. Let a and b be nonzero integers and d = gcd(a, b). Let m = b/d and n = a/d.Suppose x = s and y = t is a solution for ax + by = c. Then as + bt = c.Now let k be an integer. Then a(s + km) + b(t − kn) = as + bt + akm − btn =c + ak(b/d) − bk(a/d) = c. Therefore, for every integer k, x = s + km andy = t − kn is a solution for ax + by = c.

20. (a) lcm(6, 14) = 42. (b) lcm(10, 35) = 70.(c) lcm(21, 39) = 273. (d) lcm(12, 48) = 48.

21. (a) Part 1. Suppose a divides b. Then b is a common multiple of a and b, socondition (i) is satisfied. Now suppose n is a common multiple. Then bdivides n, so n ≤ b. Therefore condition (ii) is satisfied, so lcm(a, b) = b.Part 2. Suppose m = b. Then b(= m) is a multiple of both a and b, so adivides b.

(b) Part 1. Suppose m = b. Then b is a common multiple of a and b, so adivides b.Part 2. Suppose a divides b. Then b is a common multiple of a and b; andb is the smallest multiple of b, so LCM(a, b) = b.

(c) Suppose d = 1. Since m is a multiple of b, m = bk for some integer k. Bypart (b) m = bk ≤ ab, so k ≤ a. Since m is a multiple of a, a divides bk. Bypart (c) of Exercise 14, a divides k. Thus a ≤ k. Then a = k, so m = ab.

(d) Suppose c divides a and b. Then, since a divides m, c divides m as well, soac , b

c and mc are all natural numbers. Now, since (a

c )c divides (mc )c and ( b

c )cdivides (m

c )c, we have by Exercise 7(m) in Section 1.4 that ac divides m

c andbc divides m

c . That is, mc is a common multiple of a

c and bc . To see that m

cis the least common multiple, let n be another common multiple of a

c andbc . Then nc is a common multiple of a and b. Thus nc divides m. Thereforeby Exercise 7(m), we see that n divides m

c . Therefore lcm(ac , b

c ) = mc .

(e) Let n be a natural number. n divides both an and bn, so by (d),lcm(a, b) =lcm(an

n , bnn ) = 1

n lcm(an, bn). Thus lcm(an, bn) = n·lcm(a, b).

(f) By definition, d divides a and d divides b, so by part (d) lcm(ad , b

d ) =1d lcm(a, b). By Exercise 14(d), gcd(a

d , bd ) = 1. By part (c) lcm(a

d , bd ) = a

d · bd .

Therefore m = d·lcm(ad , b

d ) = d(ad · b

d ) = abd , so d · m = ab.

22. (a) F. The claim is false: 125, 521, 215, and other numbers all work.

(b) A.

(c) F. You cannot prove a statement with an example. Here the “proof” onlyexamines the case where x = π.

(d) A.

(e) C. The second to the last sentence should read, “Then c divides 1.” Thecorrect sentence should be justified by previous exercises.

(f) A.

2 Set Theory

2.1 Basic Concepts of Set Theory

1. (a) {x ∈ N: x < 6} or {x: x ∈ N and x < 6}(b) {x | x ∈ Z and x2 < 17}(c) {x ∈ R: 2 ≤ x ≤ 6} or {x: x ∈ R and 2 ≤ x ≤ 6}.(d) {x | x ∈ R and −1 < x ≤ 9}x < −1}(e) {x | x ∈ R and −5 ≤

2. (a) True (b) True (c) True

3. (a) Suppose that X is a set. If X ∈ X then X is not an ordinary set, so X ∈ X.On the other hand, if X ∈ X, then X is an ordinary set, so X ∈ X. BothX ∈ X and X ∈ X lead to a contradiction. We conclude that the collectionof ordinary sets is not a set.

(b) If the barber is an element of A, then by definition of A the barber does notshave himself. But the barber shaves all men who do not shave themselves,so the barber shaves the barber. Since the barber shaves himself, he is notan element of A. This is a contradiction.On the other hand, if the barber is not an element of A, then he is notamong the men who do not shave themselves. Therefore he shaves himself.This is a contradiction because the barber shaves only men who do notshave themselves.

4. (a) True (b) False (c) True (d) False(e) False (f) False (g) True (h) False(i) False (j) True

5. (a) True (b) True (c) True (d) True(e) False (f) True (g) False (h) False(i) False (j) False (k) True (1) True

6. (a) A = {1, 2} B = {1, 2, 4} C = {1, 2, 5}(b) A = B = C = {1}(c) A = {1, 2, 3} B = {1, 4} C = {1, 2, 3, 5}(d) A = {1} B = {1, 2} C = {3}

7. Assume x /∈ B and A ⊆ B. Suppose x ∈ A. Then since A ⊆ B, x ∈ B. This isimpossible. Therefore x /∈ A.

8. Assume A ⊆ B and B ⊆ C and suppose x ∈ A. Then x ∈ B since A ⊆ B, andthen x ∈ C since B ⊆ C. Therefore A ⊆ C.

9. Assume A ⊆ B, B ⊆ C and C ⊆ A. The last two assumptions and Theorem2.1.1 tell us that B ⊆ A. Since A ⊆ B, A = B by the definition of set equality.The first and the third assumptions along with 2.1.1 tell us that C ⊆ B. SinceB ⊆ C, we have B = C, also by the definition of set equality.

10. Suppose x ∈ X. Then x2−7x+12 = 0, so (x−3)(x−4) = 0. Thus x = 3 or x = 4,so x ∈ Y . If x ∈ Y , then x = 3 or x = 4, so x is a solution to x2 − 7x + 12 = 0.Thus x ∈ Y .

11.

x ∈ iff |x| ≤ 3 and x ∈ Z

iff x = −3 or x = −2 or x = −1 or x = 0 or x = 1 or x = 2 or x = 3iff x ∈ Y

38

2 SET THEORY 39

12.

x ∈ X iff x2 < 30 and x ∈ N

iff x = 1 or x = 2 or x = 3 or x = 4 or x = 5iff x ∈ Y

13. Let a and b be natural numbers.

(a) Suppose a = b. Then by definition of the set aZ, aZ = bZ.

(b) Suppose aZ = bZ. Since a · 1 = a, a ∈ aZ. Thus a ∈ bZ. Therefore thereexists an integer c such that a = bc. Likewise, b ∈ aZ, so there exists aninteger d such that b = ad. Then a = bc = (ad)c = a(dc). Hence dc = 1, soc = 1 or c = −1. Since a and b are both positive, c = 1 and a = b.

14. (a) {{0}, {�}, {�}, {0,�}, {0,�}, {�,�}, X, ∅}(b) {∅, X, {S}, {{S}}}(c) {{∅}, {{a}}, {{b}}, {{a, b}}, {∅, {a}}, {∅, {b}}, {∅, {a, b}}, {{a}, {b}}, {{a}, {a, b}},

{{b}, {a, b}}, {∅, {a}, {b}}, {∅, {a}, {a, b}}, {∅, {b}, {a, b}}, {{a}, {b}, {a, b}}, X, ∅}(d) {�, {�}, {1}, {{2, {3}}}, {�, 1}, {�, {2, {3}}, {1, {2, {3}}, X}

15. (a) false (b) true (c) false (d) true(e) false (f) true (g) true (h) true

16. (a) no proper subsets(b) {∅}, {{∅}}, ∅(c) {1}, {2}, ∅(d) {O}, {�}, {�}, {O,�}, {O,�}, {�,�}, ∅

17. (a) True (b) True (c) True (d) True(e) True (f) True (g) True (h) False(i) True (j) True (k) True (l) False

18. Let A and B be sets.

A = B iff A ⊆ B and B ⊆ A

iff P(A) ⊆ P(B) and P(B) ⊆ P(A) < byTheorem2.1.5. >

iff P(A) = P(B).

19. (a) C. This “proof” shows Y ⊆ X, but not X ⊆ Y .

(b) F. You can’t prove a universal statement by citing an example.

(c) The “proof” asserts that x ∈ C, but fails to justify this assertion witha definite statement that x ∈ A or that x ∈ B. This problem could becorrected by inserting a second sentence “Suppose x ∈ A” and a fourthsentence “Then x ∈ B.”

(d) F. This “proof” is backwards in many respects.

(e) F. The error repeatedly committed in this proof is to say A ⊆ B meansx ∈ A and x ∈ B. The correct meaning of A ⊆ B is that for every x, ifx ∈ A, then X ∈ B.

(f) F. The claim is false. The error is the misunderstanding that x ∈ A impliesx ⊆ A.

(g) F. The claim is false. The error is that the “proof” supposes x ∈ A butshows a different object is in P(A).

2 SET THEORY 40

(h) The proof could be considered correct, but it lacks a statement of the hy-pothesis, helpful explanations and connecting words. How much explanationyou include depends on the presumed level of the reader’s knowledge. Weprefer the use of words, not just symbols.

(i) F. The claim is false. (For example, let A = {1, 2}, B = {1, 2, 4}, C ={1, 2, 5, 6, 7}.)

2.2 Set Operations

1. (a) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} (b) ∅(c) {1, 3, 5, 7, 9} (d) A(e) {3, 9} (f) {1, 2, 3, 5, 6, 7, 8, 9}(g) {1, 5, 7} (h) {1, 5, 7}(i) {1, 5, 7} (j) {0, 3, 4, 6, 9}

2. (a) [2, 8) (b) (1, 8) (c) [3, 6](d) [2, 4) (e) (6, 8) (f) [2, 5](g) [8,∞) (h) (−∞, 3) ∪ [8,∞) (i) ∅(j) (1, 5] ∪ (6, 8)

3. (a) {0,−2,−4,−6,−8, . . .} (b) {1, 3, 5, 7 . . .}(c) � (d) Zi− ∪ {0}(e) Z

+ (f) D(g) {0, 2, 4, 6, 8, . . .} (h) {−1,−3,−5,−7, . . .} ∪ 0 ∪ Z

+

(i) Z

4. A and B are disjoint.

5. C and D are disjoint.

6. (a) A = {x}, B = {x, y}, C = {y}(b) A = B = C = {x}(c) A = B = {x}, C = {x, y}(d) A = {x, z}, B = {y, z}, C = {z}.

(e) A = {x}, B = C = {x, y}(f) A = B = C = {x}. The conditions require that the 3 sets be equal.

7. (a) Suppose x ∈ A. Then x ∈ A or x ∈ B. Therefore x ∈ A ∪ B.

(b) No Solution

(c) Since � is a subset of every set, � ⊆ A ∩ �. By part (h) (proved in thetext) A ∩ � = � ∩ A, and by part (b), � ∩ A ⊆ �. Therefore A ∩ � = �.

(d) Suppose x ∈ A ∪ ∅. Then x ∈ A or x ∈ ∅. Since x ∈ ∅ is false, x ∈ A. Thisshows A ∪ ∅ ⊆ A. By part (a), A ⊆ A ∪ ∅. Therefore A ∪ ∅ = A.

(e) x ∈ A ∩ A iff x ∈ A and x ∈ A iff x ∈ A. Therefore A ∩ A = A.

(f) No Solution

(g) x ∈ A ∪ B iff x ∈ A or x ∈ B iff x ∈ B or x ∈ A iff x ∈ B ∪ A. ThereforeA ∪ B = B ∪ A.

(h) No Solution

(i) x ∈ A − ∅ iff x ∈ A and x /∈ ∅ iff x ∈ A (since x /∈ ∅ is true). ThereforeA − ∅ = A.

(j) x ∈ ∅ − A iff x ∈ ∅ and x /∈ A iff x ∈ ∅ (because x ∈ ∅ is false). Therefore∅ − A = ∅.

2 SET THEORY 41

(k)

x ∈ A ∪ (B ∪ C) iff x ∈ A or x ∈ B ∪ C

iff x ∈ A or (x ∈ B or x ∈ C)iff (x ∈ A or x ∈ B) or x ∈ C

iff (x ∈ A ∪ B) or x ∈ C

iff x ∈ (A ∪ B) ∪ C.

Therefore A ∪ (B ∪ C) = (A ∪ B) ∪ C.

(l)

x ∈ A ∩ (B ∩ C) iff x ∈ A and x ∈ B ∩ C

iff x ∈ A and (x ∈ B and x ∈ C)iff (x ∈ A and x ∈ B) and x ∈ C

iff x ∈ A ∩ B and x ∈ C

iff x ∈ (A ∩ B) ∩ C.

Therefore A ∩ (B ∩ C) = (A ∩ B) ∩ C.

(m) No Solution

(n)

x ∈ A ∪ (B ∩ C) iff x ∈ A or x ∈ B ∩ C

iff x ∈ A or (x ∈ B and x ∈ C)iff (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C)iff x ∈ (A ∪ B) or x ∈ (A ∪ C)iff x ∈ (A ∪ B) ∩ (A ∪ C).

Therefore A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).

(o) Part i. Assume A ⊆ B. By part (a) B ⊆ B∪A and by part (g) B∪A = A∪B.Therefore B ⊆ A∪B. Suppose x ∈ A∪B. Then x ∈ A or x ∈ B. If x ∈ A,then x ∈ B because A ⊆ B. Thus in either case, x ∈ B. This showsA ∪ B = B.Part ii. Assume A ∪ B = B. By part (a) A ⊆ A ∪ B = B, so A ⊆ B.

(p) No Solution

(q) Assume A ⊆ B. Suppose x ∈ A ∪ C. Then x ∈ A or x ∈ C.Case 1: If x ∈ A, then x ∈ B because A ⊆ B. Thus x ∈ B or x ∈ C.Case 2: If x ∈ C, then x ∈ B or x ∈ C.In either case, x ∈ B ∪ C. Therefore A ∪ C ⊆ B ∪ C.

(r) Assume A ⊆ B. Suppose x ∈ A ∩ C, then x ∈ A and x ∈ C. Since x ∈ Aand A ⊆ B, we have x ∈ B. Thus x ∈ B and x ∈ C, so x ∈ B∩C. ThereforeA ∩ C ⊆ B ∩ C.

8. (a) No solution.

(b)

x ∈ A ∪ Ac iff x ∈ A or x ∈ Ac

iff x ∈ A or x /∈ A

iff x ∈ U (because x ∈ A or x /∈ A is a tautology).

Therefore A ∪ Ac = U .

2 SET THEORY 42

(c)

x ∈ A ∩ Ac iff x ∈ Aandx ∈ Ac

iff x ∈ Aandx /∈ A

iff x ∈ ∅ (because x ∈ A and x /∈ A is a contradiction).

Therefore A ∩ Ac = ∅.

(d)

x ∈ A − B iff x ∈ A and x /∈ B

iff x ∈ A and x ∈ Bc

iff x ∈ A ∩ Bc.

Therefore A − B = A ∩ Bc.

(e)

x ∈ (A ∩ B)c iff x /∈ A ∩ B

iff it is not the case that x ∈ A and x ∈ B

iff x /∈ A or x /∈ B

iff x ∈ Ac or x ∈ Bc

iff x ∈ Ac ∪ Bc.

Therefore (A ∩ B)c = Ac ∪ Bc.

(f) Part 1. Assume A ∩ B = ∅. Suppose x ∈ A. If x ∈ B, then x ∈ A ∩ B = ∅,which is impossible. Thus x /∈ B. Therefore x ∈ Bc. This shows A ⊆ Bc.Part 2. Assume A ⊆ Bc. Suppose x ∈ A∩B. Then x ∈ A and x ∈ B. Sincex ∈ A and A ⊆ Bc, x ∈ Bc. Therefore x ∈ B ∩ Bc = ∅. This shows thatA ∩ B ⊆ ∅. Since ∅ ⊆ A ∩ B, we have A ∩ B = ∅.

9. (a) Assume A ⊆ B and that there is an element x in A − B. Then x ∈ A andx ∈ B because A ⊆ B, but x ∈ Bc because x ∈ A − B. This contradictionshows A ⊆ B implies A − B = ∅.Now assume A − B = ∅ and suppose x ∈ A. Then x ∈ B because if x /∈ Bthen x ∈ A − B, which is impossible. Thus A − B = ∅ implies A ⊆ B.

Alternate proof of (a):

A − B = � iff A ∩ Bc = �(by part (d) of Theorem 2.2.2)iff A ⊆ (Bc)c(by part (h) of Theorem 2.2.2)iff A ⊆ B.

(b) Assume A ⊆ B ∪ C and A ∩ B = ∅ and suppose x ∈ A. Then x ∈ B orx ∈ C, but x can’t be in B because A ∩ B = ∅. Thus x ∈ C. ThereforeA ⊆ C.

(c) Suppose C ⊆ A and D ⊆ B. Assume that C and D are not disjoint. Thenthere is an object x ∈ C ∩ D. But then x ∈ C and x ∈ D. Since C ⊆ Aand D ⊆ B, x ∈ A and x ∈ B. Therefore, x ∈ A ∩ B, so A and B are notdisjoint.

(d) Assume A ⊆ B and suppose x ∈ A − C. Then x ∈ B since A ⊆ B andx /∈ C since x ∈ A − C. Thus x ∈ B − C.

2 SET THEORY 43

(e)

x ∈ (A − B) − C iff x ∈ A − B and x /∈ C

iff ∈ A and x /∈ B and x /∈ C

iff x ∈ A and x /∈ C and (x /∈ B or x ∈ C)iff x ∈ A − C and it is not the case that x ∈ B and x /∈ C

iff x ∈ A − C and x /∈ B − C

iff x ∈ (A − C) − (B − C).

(f) Assume A ⊆ B and B ⊆ C. Suppose x ∈ A ∪ B. Then x ∈ A or x ∈ B. Ifx ∈ A, then x ∈ B, so in either case, x ∈ B. Then x ∈ C. Thus A ∪ B ⊆ C.

(g) Suppose x ∈ (A ∪ B) ∩ C. Then x ∈ A ∪ B and x ∈ C. Thus x ∈ A orx ∈ B, and x ∈ C.Case 1. x ∈ A. Then x ∈ A or x ∈ B ∩ C, so x ∈ A ∪ (B ∩ C).Case 2. x ∈ B. Then x ∈ B and x ∈ C, so x ∈ B ∩ C. Then x ∈ A orx ∈ B ∩ C, so x ∈ A ∪ (B ∩ C).

(h) A − B and B are disjoint because (A − B) ∩ B = (A ∩ BC) ∩ B =A ∩ (BC ∩ B) = A ∩ � = �.

10. (a) Assume C ⊆ A and D ⊆ B. Suppose x ∈ C ∩ D. Then x ∈ C and x ∈ D.Since C ⊆ A and D ⊆ B, x ∈ A and x ∈ B. Therefore x ∈ A ∩ B.

(b) Assume C ⊆ A and D ⊆ B. Suppose x ∈ C ∪ D. Then x ∈ C or x ∈ D.Since C ⊆ A, if x ∈ C, then we have x ∈ A. Since D ⊆ B, if x ∈ D, thenwe have x ∈ B. Thus x ∈ A or x ∈ B. Therefore x ∈ A ∪ B.

(c) A = {1, 2}, B = {1, 3}, C = {2, 3}(d) Assume C ⊆ A and D ⊆ B. Suppose x ∈ D − A. Then x ∈ D and x /∈ A.

Since D ⊆ B, x ∈ B. Since C ⊆ A, x /∈ C. Thus x ∈ B − C. ThereforeD − A ⊆ B − C.

(e) Assume A ∪ B ⊆ C ∪ D, A ∩ B = ∅ and C ⊆ A. Suppose x ∈ B. Thenx ∈ A∪B, so x ∈ C ∪D. This x ∈ C or x ∈ D. Suppose x ∈ C. Then, sinceC ⊆ A, so x ∈ A. Then x ∈ A and x ∈ B, so x ∈ A ∩ B. This contradictsthe assumption that A ∩ B is empty. We conclude that x /∈ C, so x ∈ D.Therefore B ⊆ D.

11. (a) A = {1, 2}, B = {1, 3}, C = {2, 3, 4}(b) Let A = {1} and B = {2} = C. Then A ∩ C = ∅ ⊆ B ∩ C, but A ⊆ B.(c) A = {1, 2}, B = {1, 3}(d) Let A = {1, 2} and B = {1}.

Then P(A) − P(B) = {∅, {1}, {2}, {1, 2}} − {∅, {1}} = {{2}, {1, 2}} ⊆{∅, {2}} = P(A − B).

(e) Let A = {1, 2} and B = {1, 3}, and C = {1}.Then A−(B−C) = A−{3} = A, while (A−B)−(A−C) = {2}−{2} = ∅,C = {2, 3}.

(f) Let A = {1, 2}, B = {1}, C = {2}.Then A − (B − C) = {2} but (A − B) − C = ∅.

12. (a)

S ∈ P(A ∪ B) iff S ⊆ A ∩ B

iff S ⊆ A and S ⊆ B[by Exercise 9(c)]iff S ∈ P(A) and S ∈ P(B)iff S ∈ P(A) ∩ P(B).

2 SET THEORY 44

(b) Since A ⊆ A ∪ B, every subset of A is a subset of A ∪ B; thus P(A) ⊆P(A ∪ B). Similarly, P(B) ⊆ P(A ∪ B). Thus P(A) ∪ P(B) ⊆ P(A ∪ B),by Exercise 9(f).

(c) Let A = {2} and B = {1}. Then P(A) ∪ P(B) = {∅, {2}} ∪ {∅, {1}} whileP(A ∪ B) = {∅, {1}, {2}, {1, 2}}. In general, if A ⊆ B or B ⊆ A, thenP(A ∪ B) = P(A) ∪ P(B).

(d) Let A and B be any sets. Since ∅ is a subset of every set, we have∅ ∈ P(A − B). Also ∅ ∈ P(A) and ∅ ∈ P(B). Therefore ∅ /∈ P(A) − P(B).This shows that P(A − B) �⊆ P(A) − P(B).

13. (a)

A × B = { (1, a), (1, e), (1, k), (1, n), (1, r), (3, a), (3, e), (3, k), (3, n),( 3, r), (5, a), (5, e), (5, k), (5, n), (5, r)}B × A

= { (a, 1), (a, 3), (a, 5), (e, 1), (e, 3), (e, 5), (n, 1), (n, 5), (n, 3),( k, 1), (k, 3), (k, 5), (r, 1), (r, 3), (r, 5)}

(b)

A × B = { (1, q), (1, {t}), (1, π), (2, q), (2, {t}), (2, π), ({1, 2}, q),( {1, 2}, {t}), ({1, 2}, π)}

B × A = { (q, 1), (q, 2), (q, {1, 2}), ({t}, 1), ({t}, 2), ({t}, {1, 2}),( π, 1), (π, 2), (π, {1, 2})}

(c)

A × B = { (∅, (∅, {∅})), (∅, {∅}), (∅, ({∅}, ∅)), ({∅}, (∅, {∅})), ({∅}, {∅}),( {∅}, ({∅}, ∅)), ({∅, {∅}}, (∅, {∅})), ({∅, {∅}}, {∅}), ({∅, {∅}},

( {∅}, ∅))}B × A = { ((∅, {∅}), ∅), ((∅, {∅}), {∅}), ((∅, {∅}), {∅, {∅}}), ({∅}, ∅), ({∅}, {∅}),

( {∅}, {∅, {∅}}), (({∅}, ∅), ∅), (({∅}, ∅), {∅}), (({∅}, ∅), {∅, {∅}})}

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

14. Assume that A and B are nonempty sets and that A × B = B × A. Letb ∈ B and a ∈ A. Then (a, b) ∈ A × B = B × A; thus a ∈ B and b ∈ A.This shows A ⊆ B and B ⊆ A; hence A = B. Conversely, if A = B, then(x, y) ∈ A × B iff x ∈ A and y ∈ B iff x ∈ Bandy ∈ A iff (x, y) ∈ B × A.Therefore A × B = B × A.

The statement does not hold if A is empty and B is nonempty. In this caseA �= B, yet A × B = ∅ = B × A.

15. (a) (ART TO COME)

(b) If (x, y) ∈ A × ∅ then y ∈ ∅, which is impossible. Therefore A × ∅ = ∅.

(c)

(x, y) ∈ (A × B) ∩ (C × D) iff (x, y) ∈ A × B and(x, y) ∈ C × D

iff x ∈ A and x ∈ C and y ∈ B and y ∈ D

iff x ∈ A ∩ C and y ∈ B ∩ D

iff (x, y) ∈ (A ∩ C) × (B ∩ D).

2 SET THEORY 45

(d)

(x, y) ∈ (A × B) ∩ (B × A) iff (x, y) ∈ A × B and (x, y) ∈ B × A

iff x ∈ A and x ∈ B and y ∈ B and y ∈ A

iff x ∈ A ∩ B and y ∈ A ∩ B

iff (x, y) ∈ (A ∩ B) × (A ∩ B).

16. (a) A = {1}, B = C = {3}, D = {2}(b) A = C = {1}, B = {2}(c) A = B = C = {1}

17. Suppose (a, b) = (x, y). Then {{a}, {a, b}} = {{x}, {x, y}}. Consequently,{a} = {x} and {a, b} = {x, y}. Therefore a = x and b = y. Conversely, if a = xand b = y, then {a} = {x} and {a, b} = {x, y}. Therefore (a, b) = {{a}, {a, b}} ={{x}, {x, y}} = (x, y).

18. (a) A � B = (A − B) ∪ (B − A) = (B − A) ∪ (A − B) = B � A

(b)

x ∈ A � B iff x ∈ (A − B) ∪ (B − A)iff (x ∈ A and x /∈ B) or (x ∈ B and x /∈ A)iff (x ∈ A or x ∈ B) and (x /∈ A or x /∈ B)iff x ∈ A ∪ B and x /∈ A ∩ B

iff x ∈ (A ∪ B) − (A ∩ B).

(c) A � A = (A − A) ∪ (A − A) = ∅ ∪ ∅ = ∅.(d) A � ∅ = (A − ∅) ∪ (∅ − A) = A ∪ ∅ = A.

19. (a) F. One serious error is the assertion that x ∈ C, which has no justification.The author of this “proof” was misled by supposing x ∈ A, which is anacceptable step but not useful in proving A − C ⊆ B − C. After assumingthat A ⊆ B, the natural first step for proving that A − C ⊆ B − C is tosuppose that x ∈ A − C.

(b) C. It is a common error to write something like “Suppose A − C.” Theauthor must have meant “Suppose x ∈ A − C.”

(c) C. Another common error is to translate “A ⊆ B” as “x ∈ A and x ∈ B.”The second sentence must be deleted and x ∈ B inferred from x ∈ A.

(d) C. The proof that A ∩ B = A is incomplete.(e) F. The claim is false. The statement “x ∈ A and x ∈ ∅ iff x ∈ A” is false.(f) F. The claim is false. The x in A ∩ B may not be the x in B ∩ C.(g) This is probably not the proof most people would write, but it is correct.

Since the sentence “x ∈ A and x /∈ A” is false, it follows from that stepthat x �= x.

(h) F. The third sentence contributes nothing to the proof. The fourth sentencemakes an assertion without an explanation. What is missing is the proofthat if x is a nonempty subset of A − B, then x is a subset of A that is nota subset of B.

(i) F. Although a picture may help by suggesting ideas around which a correctproof can be made, a picture alone is rarely sufficient for a proof. Thus aproof that consists only of Venn diagrams will usually have a grade of F.This “proof” is made better because of the explanation that is included,but the only way to give a complete proof is to show that A ∪ B ⊆ B andB ⊆ A ∪ B.

2 SET THEORY 46

2.3 Extended Set Operations and Indexed Families of Sets

1. (a)⋃

A∈A A = {1, 2, 3, 4, 5, 6, 7, 8};⋂

A∈A A = {4, 5}(b)

⋃A∈A A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};

⋂A∈A A = ∅

(c)⋃

n∈NAn = {5, 6, 10, 11, 12, 15, 16, 17, 18} ∪ {n ∈ N: n ≥ 20};⋂

n∈NAn = ∅

(d)⋃

n∈NBn = N − {1} = B1;

⋂n∈N

Bn = ∅(e)

⋃A∈A A = Z;

⋂A∈A A = {10}

(f)⋃

n≤10 An = {1, 2, 3, . . . , 19};⋂

n≤10 An = ∅(g)

⋃n∈N

An = (0, 1);⋂

n∈NAn = ∅

(h)⋃

r∈RAr = [−π,∞);

⋂r∈R

Ar = [−π, 0]

(i)⋃

r∈RAr = [0,∞);

⋂r∈R

Ar = ∅(j)

⋃n∈N

Mn = Z;⋂

n∈NMn = {0}

(k)⋃

n≥3An = (0, 73 );

⋂n≥3An = [ 13 , 2]

(l)⋃

n∈ZCn = R;

⋂n∈Z

Cn = ∅(m)

⋃n∈Z

An = R − Z;⋂

n∈ZAn = ∅

(n)⋃

n∈NDn = (−∞, 1);

⋂n∈N

Dn = (−1, 0]

(o)⋃

A∈A= · − {1}; |A∈AA = ∅(p) The union is the triangular region bounded by y = 0, x = 0, y = x. The

intersection is the two line segments, y = 0, x = 0

(q) The union is the interior of the square, plus the points (0, 0) and (1, 1). Theintersection is line segment y = x.

2. The families in (b), (l), and (m) are pairwise disjoint.

3. Assume A is a family of sets and B ∈ A. Suppose x ∈ B. Then x ∈ A for someA ∈ A, so x ∈

⋃A∈A A.

4. (a) Let A be the empty family of subsets of R. Then (∀A)(A ∈ A ⇒ x ∈ A)is true for every real number x, because for every set A, the antecedent of(A ∈ A ⇒ x ∈ A) is false. Therefore, for every real number x, x ∈

⋂A∈AA.

Since every A in A is a subset of R,⋂

A∈A A ⊆ R. Therefore⋂

A∈A A = R.

(b) Let A be the empty family of subsets of R. Then (∃A)(A ∈ A∧x ∈ A) isfalse for every x in R. Therefore

⋃A∈A A is empty, so

⋃A∈A A = ∅.

(c) |A∈A A is not a subset of⋃

A∈A A in this example because R is not a subsetof ∅.

5. (a) Part (a) of Theorem 2.3.2. Let β ∈ Δ. Suppose x ∈⋂

α∈ΔAα. Then x ∈ Aα

for each α ∈ Δ. Since β ∈ Δ, x ∈ Aβ . Therefore⋂

α∈ΔAα ⊆ Aβ .Part (b) of Theorem 2.3.2. Let β ∈ Δ and suppose x ∈ Aβ . Then x ∈ Aα

for some α ∈ Δ. Thus x ∈⋃

α∈ΔAα.

(b) Part (d) of Theorem 2.9.

x ∈ (⋃

α∈Δ

Aα)c iff x /∈⋃

Aα

iff there is no α ∈ Δ such that x ∈ Aα

iff for all α ∈ Δ, x /∈ Aα

iff for all α ∈ Δ, x ∈ (Aα)c

iff x ∈| (Aα)c.

2 SET THEORY 47

6. (a)

x ∈ B ∩⋃

α∈Δ

Aα iff x ∈ B and x ∈⋃

α∈Δ

Aα

iff x ∈ B and x ∈ Aα for some α ∈ Δiff x ∈ B ∩ Aα for some α ∈ Δ

iff x ∈⋃

α∈Δ

(B ∩ Aα)

(b)

x ∈ B ∪⋂

α∈Δ

Aα iff x ∈ B or x ∈ Aα for all α ∈ Δ

iff x ∈⋂

α∈Δ

(B ∪ Aα)

7. (a)

( ⋃α∈Δ

Aα

)∩

( ⋃β∈Γ

Bβ

)=

⋃β∈Γ

(( ⋃α∈Δ

Aα

)∩ Bβ

)

=⋃β∈Γ

( ⋃α∈Δ

(Aα ∩ Bβ

))

(b)

( ⋂α∈Δ

Aα

)∪

( ⋂β∈Γ

Bβ

)=

⋂β∈Γ

(( ⋂α∈Δ

Aα

)∪ Bβ

)

=⋂β∈Γ

( ⋂α∈Δ

(Aα ∪ Bβ

))

8. (a) False. B = {0, 1}, Δ = {1, 2}, A1 = {1}, A2 = {2}.

(b) False. B = {1, 2}, Δ = {1, 2}, A1 = {1}, A2 = {2}.

(c) True. (|α∈ΔAα) − B = (|α∈ΔAα) ∩ BC = |α∈Δ(Aα ∩ BC) = |α∈Δ(Aα − B).

(d) True. (⋃

α∈ΔAα)−B = (⋃

α∈ΔAα)∩BC =⋃

α∈Δ(Aα ∩BC) =⋃

α∈Δ(Aα −B).

9. (a) Suppose x ∈⋃

α∈ΓAα. Then x ∈ Aα for some α ∈ Γ. Since Γ ⊆ Δ, α ∈ Δ.Thus x ∈ Aα for some α ∈ Δ. Therefore x ∈

⋃α∈ΔAα.

(b) Suppose x ∈⋂

α∈ΔAα. Then x ∈ Aα for all x ∈ Δ, so x ∈ Aα for all x ∈ Γ.Therefore x ∈

⋂α∈ΓAα.

10. (a) Let x ∈ B. For each A ∈ A, B ⊆ A. Thus for each A ∈ A, x ∈ A. Thereforex ∈

⋂A∈AA.

(b) X =⋂

A∈AA.

(c) Suppose x ∈⋂

A∈AA. Then x ∈ A for some A ∈ A. Then x ∈ D becauseA ⊆ D for all A ∈ A.

(d) Y =⋃

A∈AA.

11. (a) Let A1 = {1}, A2 = X, and A = {A1, A2}. Then⋂

A∈A A = A1 ∩A2 = {1}and

⋃A∈A A = A1 ∪ A2 = X.

2 SET THEORY 48

(b) Let B1 = {1, 2, 3, 4, 5}, B2 = {6, 7, 8, 9, 10}, B3 = {11, 12, 13, 14, 15},B4 = {16, 17, 18, 19, 20} and B = {B1, B2, B3, B4}. Then

⋃B∈B B = X.

(c) Let Ci = {i} for each i ∈ X and C = {Ci: i ∈ X}. Then C is pairwisedisjoint and

⋃C∈C C = X.

12. Let An = (0, 1)− ( 13n , 1− 1

3n ) for all n ∈ N. Then for all m, n ∈ N, An ∩Am �= ∅but

⋂n∈N

An = ∅.

13. Suppose B ⊆ A and A is pairwise disjoint. Let B1 and B2 be any two sets in B.Then B1 and B2 are in A, so if B1 �= B2, then B1 ∩ B2 = ∅. Thus B is pairwisedisjoint.

14. (a) Let C1 and C2 be any two sets in C. Then C1 and C2 are both in A, so ifC1 �= C2 then C1 ∩ C2 = ∅.

(b) A = {{1}} B = {{1, 2}}(c) Suppose that

⋃A∈A A and

⋃β∈B B are disjoint, and let D1 and D2 be

elements of D. Now if both D1 and D2 come from the same family A or B,then D1 �= D2 implies D1 ∩ D2 = ∅. Suppose then that one comes from Aand the other comes from B. Without loss of generality, say D1 ∈ A andD2 ∈ B. Then D1 ⊆

⋃A∈A A and D2 ⊆

⋃β∈B B by Theorem 2.3.2(b). So,

D1 ∩ D2 ⊆⋃

A∈A A ∩⋃

β∈B B = ∅ by Exercise 10(a) in Section 2.2. ThusD1 ∩ D2 = ∅. In any event, we see that D1 �= D2 implies D1 ∩ D2 = ∅, soD is pairwise disjoint.

15. (a)

x ∈k+1⋃i=1

Ai iff x ∈ Ai for some i ∈ {1, . . . , k + 1}

iff x ∈ Ai for some i ∈ {1, . . . , k} or x ∈ Ak+1

iff x ∈k⋃

i=1

Ai ∪ Ak+1.

(b)

x ∈k+1⋂i=1

Ai iff x ∈ Ai for each i ∈ {1, . . . , k + 1}

iff x ∈ Ai for each i ∈ {1, . . . , k} and x ∈ Ak+1

iff x ∈k⋂

i=1

Ai ∩ Ak+1.

(c) Proof. Let x ∈⋃

i=1 Ai. Then there exists j ∈ N such that k ≤ j ≤ m andx ∈ A − j. Since j ∈ N and x ∈ Aj , x ∈

⋃∞i=1 A1.

Alternate Proof. the set of Γ = {k, k + 1, k + 2, . . . , m} is a subset ofΔ = {1, 2, 3, . . .}. Therefore, by Exercise 9(a),

⋃mi=k Ai ⊆

⋃∞i=1 Ai.

(d) Suppose x ∈⋂∞

i=1 Ai. Then x ∈ Ai for every i ∈ N. Thus x ∈ Ai for everyi, 1 ≤ i ≤ m. Therefore x ∈

⋂mi=1 Ai.

(e) Suppose x ∈⋃k

i=1 Ai. Then x ∈ Ai for some i ≤ k, so x ∈ Ai for somei ≤ m. Therefore x ∈

⋃mi=1 Ai.

(f) Suppose x ∈⋂m

i=1 Ai. Then x ∈ Ai for all i ≤ m, so x ∈ Ai for all i ≤ k.Therefore x ∈

⋂ki=1 Ai.

2 SET THEORY 49

16. (a) Ak ⊆⋂k

i=1 Ai by 10(a). On the other hand,⋂k

i=1 Ai ⊆ Ak by Theorem2.3.2(a).

(b) Ai ⊆ A1 for all i ∈ N so⋃∞

i=1 Ai ⊆ A1 by 10(c). On the other hand,A1 ⊆

⋃∞i=1 Ai by 2.3.2(b).

17. (a) Let Ai =[− 1

i , 1 + 1i

)for all i ∈ N.

(b) Let Ai = (0, 1) for all i ∈ N.

(c) Let Ai = [0, 1] − ( 1i+2 , 1 − 1

i+2 ) for all i ∈ N.

18. (a) C. This proof omits an explanation of why there is some β in Δ such thatAβ ∈ {Aa: α ∈ Δ}. The explanation is that by definition of an indexedfamily, Δ �= ∅. If we allowed Δ = ∅, the claim would be false.

(b) C. No connection is made between the first and second sentences. Theconnection that needs to be made is that if x ∈

⋃α∈Δ Aα then x ∈ Aα for

some α ∈ Δ, so x ∈ B because Aα ⊆ B.

(c) F. An example is not a proof of a universal statement.

(d) A.

(e) F. The claim is false.⋃∞

n=1[n, n + 1) = [1,∞).

2.4 Mathematical Induction

1. (a) Inductive (b) Not Inductive (c) Not Inductive(d) Not Inductive (e) Not Inductive (f) Not Inductive

2. (a) Not necessarily true.

(b) True

(c) This is true for an inductive set.

(d) This is true, since 6 ∈ S ⇒ 7 ∈ S ⇒ 8 ∈ S ⇒ 9 ∈ S ⇒ 10 ∈ S ⇒ 11 ∈ S.

(e) False, since {12, 13, 14, . . .} is inductive.

3. (a) The successor properties of N say that x ∈ N implies x + 1 ∈ N.

(b) Since the antecedent is always false, x ∈ ∅ implies x+1 ∈ ∅ is true for everynatural number x.

4. (a) 24 (b) 5040 (c) 97(d) 56 (e) (n + 2)(n + 1) (f) (n + 3)(n + 2)(g) (n + 2)!

5. (a) Define S as follows:(i) 5 ∈ S (ii) n ∈ S ⇒ n + 5 ∈ SThen S = {n: n = 5k for some k ∈ N}

(b) Define S as follows:(i) 11 ∈ S (ii) n ∈ S ⇒ n + 1 ∈ S

Then S = {n: n ∈ N and n > 10}.

(c) A = {n: n = 2k for some k ∈ N} may be defined as(i) 2 ∈ A (ii) x ∈ A ⇒ 2x ∈ A.

(d) Define S as follows:(i) a ∈ S (ii) n ∈ S ⇒ n + d ∈ S.Then S = {a, a + d, a + 2d, . . .}.

2 SET THEORY 50

(e) Define S as follows:(i) a ∈ S (ii) n ∈ S ⇒ nr ∈ S

Then S = {a, ar, ar2, . . .}.

(f) Define⋃n

i=1 Ai as follows:(i)

⋃1i=1 Ai = A1 (ii)

⋃n+1i=1 Ai =

⋃ni=1 Ai ∪ An+1.

(g) Define∏n

i=1 Xi as follows:(i)

∏1i=1 Xi = X1 (ii)

∏n+1i=1 Xi =

∏ni=1 Xi ∪ Xn+1.

6. (a) Proof.

i. 63 = 216 < 720 = 6!, so the statement is true for n = 6.ii. Assume n3 < n! for some n ≥ 6. Then

(n + 1)3 = n3 + 3n2 + 3n + 1< n3 + 3n2 + 3n + n = n3 + 3n2 + 4n

< n3 + 3n2 + n2 = n3 + 4n2

< n3 + n3 = 2n3

< (n + 1)n3

< (n + 1)n! = (n + 1)!.

(b) i. 8(1) − 5 = 3 = 4(1)2 − 1 so the statement is true for n = 1.ii. Assume the property holds for some n. Then by the inductive hypoth-

esis

3 + 11 + 19+ · · · +(8n − 5) + [8(n + 1) − 5]= (4n2 − n) + [8n + 8 − 5]= n2 + 7n + 3 = 4(n2 + 2n + 1) − (n + 1)= 4(n + 1)2 − (n + 1).

So the property is true for n + 1.iii. By the PMI, the statement is true for every n ∈ N.

(c) Let S = {n ∈ N:∑n

i=1 2i = 2n+1 − 2}.

i.∑1

i=1 2i = 2 = 21+1 − 2, so 1 ∈ S.ii. Suppose n ∈ S for some natural number n. Then

∑ni=1 2i = 2n+1 − 2.

Then∑n+1i=1 2i =

∑ni=1 2i+2n+1 = 2n+1−2+2n+1 = 2(n+1)+1−2. Therefore

n + 1 ∈ S.iii. By the PMI, S = N.

(d) i. 1 · 1! = 1 = (1 + 1)! − 1, so the statement is true for 1.ii. Assume the property holds for some n. Then

1 · 1!+ · · · +n · n! + (n + 1) · (n + 1)!= [(n + 1)! − 1] + (n + 1) · (n + 1)!= (n + 1)! + (n + 1) · (n + 1)! − 1= (n + 2)(n + 1)! − 1 = (n + 2)! − 1.

Hence the property holds for n + 1.iii. By the PMI, the statement is true for all n ∈ N.

(e) Let S = {n ∈ N: 13 + 23 + . . . + n3 = [n(n+1)2 ]2}.

2 SET THEORY 51

i. Since 13 = 1 = [1(1+1)2 ]2, 1 ∈ S.

ii. Assume n ∈ S, for some natural number n. Then

13 + 23+ · · · +n3 + (n + 1)3

= [13 + 23 + . . . + n3] + (n + 1)3 = [n(n + 1)

2]2 + (n + 1)3

= (n + 1)2[n2

4+ n + 1] = (n + 1)2[

n2 + 4n + 44

]

= [(n + 1)(n + 2)

2]2.

Thus n + 1 ∈ S.iii. By the PMI, S = N.

(f) Let S = {n ∈ N:∑n

i=1(2i − 1)3 = n2(2n2 − 1)}.

i.∑1

i=1(2i − 1)3 = 13 = 12(2 − 1), so 1 ∈ S.ii. Assume n ∈ S. Then

n+1∑i=1

(2i − 1)3 =n∑

i=1

(2i − 1)3 + [2(n + 1) − 1]3

= n2(2n2 − 1) + (2n + 1)3 = 2n4 − n2 + 8n3 + 12n2 + 6n + 1= 2n4 + 8n3 + 11n2 + 6n + 1 = (n2 + 2n + 1)(2n2 + 4n + 1)= (n + 1)2[2(n + 1)2 − 1].

Thus n + 1 ∈ S.iii. By the PMI, S = N.

(g) i. 11·2 − 1

1+1 , so statement is true for n = 1.ii. Assume that the statement is true for some n ∈ N. Then

11 · 2

+1

2 · 3+

13 · 4

+ · · · 1n(n + 1)

+1

(n + 1)(n + 1 + 1)

=n

n + 1+

1(n + 1)(n + 2)

n(n + 2)(n + 1)(n + 2)

+1

(n + 1)(n + 2)=

n2 + 2n + 1(n + 1)(n + 2)

=(n + 1)2

(n + 1)(n + 2)=

n + 1n + 2

.

Thus the statement is true for n + 1.iii. By the PMI, the statement is true for all n ∈ N.

(h) i. The statement 12! + 2

3! + · · · + n(n+1)! = 1 − 1

(n+1)! is true for n = 1,because 1

2! = 12 = 1 − 1

2! .ii. Assume the statement is true for some n ∈ N. Then 1

2! + 23! + . . . +

n(n+1)! = 1 − 1

(n+1)! . Adding n+1(n+1+1)! to both sides of this equation, we

have

n+1∑i=1

n

(n + 1)!= 1 − 1

(n + 1)!+

n + 1(n + 2)!

= 1 − n + 2(n + 2)!

+n + 1

(n + 2)!

= 1 − 1(n + 2)!

.

Thus the statement is true for n + 1.

2 SET THEORY 52

iii. By the PMI, the statement is true for every n ∈ N.

(i) i. The statement is true for n = 1 because∑1

i=11

(2i−1)(2i+1) = 11·3 =

12·1+1 .

ii. Suppose the statement is true for some n ∈ N. We must show that∑n+1i=1

1(2i−1)(2i+1) = n+1

2(n+1)+1 . By the hypothesis of the induction,

n+1∑i=1

1(2i − 1)(2i + 1)

=n∑

i=1

1(2i − 1)(2i + 1)

+1

(2n + 1)(2n + 3)

=n

2n + 1)+

1(2n + 1)(2n + 3)

=n(2n + 3)

(2n + 1)(2n + 3)+

1(2n + 1)(2n + 3)

=2n2 + 3n + 1

(2n + 1)(2n + 3)=

(n + 1)(2n + 1)(2n + 1)(2n + 3)

=n + 12n + 3

.

Thus the statement is true for n + 1.iii. By the PMI, the statement is true for every n ∈ N.

(j) i. The statement is true for n = 1, because∏1

i=1(1− 1i+1 ) = 1− 1

2 = 11+1 .

ii. Suppose that the statement is true for some n ∈ N. Then∏n+1

i=1 (1 −1

i+1 ) = (1 − 1n+2 )

∏ni=1(1 − 1

i+1 ) = (1 − 1n+2 )( 1

n+1 ) = (n+2−1n+2 )( 1

n+1 ) =1

n+2 . Thus the statement is true for n + 1.

(k) By the PMI the statement is true for every n ∈ N.

(l) i.∏1

i=1(2i − 1) = 2(1) − 1 = 1 = (2(1))!1!21 , so the statement is true for

n = 1.ii. Assume the property holds for some n ∈ N. We must show

∏n+1i=1 (2i − 1) =

(2(n+1))!(n+1)!2n+1 .By the hypothesis of induction, the left hand side of the equation is[2(n + 1) − 1] ·

∏ni=1(2i − 1) = (2n + 1) · (2n)!

n!2n and the right hand sideof the equation is (2n)!(2n+1)(2n+2)

2(n+1)n!2n . Since these are equal, the propertyis true for n + 1.

iii. By the PMI, the statement is true for all n ∈ N.

(m) i.∑0

i=0 ari = a = a(r1−1)r−1 so the statement is true for n = 1.

ii. Assume the statement is true for some n ∈ N. Then

n∑i=0

ari =n−1∑i=0

ari + arn =a(rn − 1)

r − 1+ arn

=a(rn − 1) + arn(r − 1)

r − 1=

arn + arn+1 − arn − a

r − 1

=a(rn+1 − 1)

r − 1.

So the statement is true for n + 1.iii. By the PMI, the statement is true for all n.

7. (a) i. (1)3 + 5(1) + 6 = 12 which is divisible by 3, so the statement is truefor 1.

2 SET THEORY 53

ii. Assume n3 + 5n + 6 is divisible by 3 for some n. Then

(n + 1)3 + 5(n + 1) + 6 = n3 + 3n2 + 3n + 1 + 5n + 5 + 6= (n3 + 5n + 6) + 3n2 + 3n + 6= (n3 + 5n + 6) + 3(n2 + n + 2)

which is divisible by 3, so the property holds for n + 1.iii. By the PMI, the statement is true for all n ∈ N.

(b) i. For n = 1, 41 − 1 = 3 which is divisible by 3.ii. Suppose for some k ∈ N that 4k − 1 is divisible by 3. Then 4k+1 − 1 =

4(4k) − 1 = 4(4k − 1) − 1 + 4 = 4(4k − 1) + 3. Both 3 and 4k − 1 aredivisible by 3, so 4k+1 − 1 is also divisible by 3.

iii. By the PMI, 4n − 1 is divisible by 3 for all n ∈ N.(c) i. 13 − 1 = 0 which is divisible by 6, so the statement is true for n = 1.

ii. Assume that the statement is true for some n ∈ N. Then (n+1)3−(n+1) = n3+3n2+3n+1−n−1 = (n3−n)+3n2+3n = (n3−n)+3n(n+1).By the induction hypothesis n3 − n is divisible by 6. And 3n(n + 1)is divisible by 6 because it has a factor of 3 and is even. Therefore(n + 1)3 − (n + 1) is divisible by 6. Thus the statement is true for n +1.

iii. By the PMI, the statement is true for all n ∈ N.(d) i. (13 −1)(1+2) = 0 which is divisible by 12, so the statement is true for

n = 1.ii. Assume that the statement is true for some n ∈ N. Note that (n3 −

n)(n + 2) = (n + 1)(n)(n + 2)(n − 1). Then

((n + 1)3 − (n + 1))((n + 1) + 2) = ((n + 1)3 − (n + 1))(n + 3)= (n + 1)((n + 1)2 − 1)(n + 3)= (n + 1)(n)(n + 2)(n + 3)= (n + 1)(n)(n + 2)(n − 1)+ 4(n + 1)(n)(n + 2).

By the induction hypothesis (n + 1)(n)(n + 2)(n − 1) is divisible by12. The expression 4(n)(n + 1)(n + 2) has a factor of 4 and, since theterms n, n+1, and n+2 are three consecutive integers, one of them isdivisible by 3. Thus 4(n)(n + 1)(n + 2) is divisible by 12 and therefore((n + 1)3 − (n + 1))((n + 1) + 2) is divisible by 12.Therefore the statement is true for n + 1.

iii. By the PMI, the statement is true for all n ∈ N.(e) i. Since 8 divides 52·1 − 1 = 24, the statement is true for n = 1.

ii. Suppose 8 divides 52n − 1 for some n ∈ N. Then 52(n+1) − 1 =(52n · 25) − 1 = 52n(1) + 52n(24) − 1 = (52n − 1) + 52n(24). By thehypothesis of induction, 8 divides 52n − 1. Since 8 also divides 52n(24),8 divides the sum of these numbers. Thus 8 divides 52(n+1) − 1.

iii. By the PMI, 8 divides 52n − 1 for every natural number n.(f) Let S = {n ∈ N: 10n + 3 · 4n+2 + 5 is divisible by 9}.

i. 101 + 3 · 41+2 + 5 = 15 + 3 · 64 = 207 = 9(23), so 1 ∈ S.ii. Assume n ∈ S. Then 9 divides 10n + 3 · 4n+2 + 5. Then

10n+1 + 3 · 4(n+1)+2 + 5 = 10n+1 + 3 · 4n+3 + 5= 10 · 10n + 4(3 · 4n+2) + 5= (10n + 3 · 4n+2 + 5) + 9 · 10n + 3 · 3 · 4n+2

= (10n + 3 · 4n+2 + 5) + 9(10n + 4n+2).

2 SET THEORY 54

Since 9 divides both terms of the final expression, 9 divides their sum.Thus n + 1 ∈ S.


(g) i. 91 − 1 = 8 which is divisible by 8, so the statement is true for n = 1.ii. Assume that the statement is true for some n ∈ N. Then 9n+1 − 1 =

9n+1 − 9n + 9n − 1 = 9n(9 − 1) + (9n − 1) = 9n(8) + (9n − 1). By theinduction hypothesis 9n − 1 is divisible by 8. It is also true that 9n(8)has a factor of 8. Therefore 9n+1 − 1 is divisible by 8. Therefore thestatement is true for n + 1.


(h) i. 31 = 3 ≥ 1 + 21, so the property is true for n = 1.ii. Assume 3n ≥ 1+2n for some n in N. Then 3n+1 = 3 ·3n ≥ 3(1+2n) =

3 + 3 · 2n ≥ 1 + 2 · 2n = 1 + 2n+1, so the property is true for n + 1.iii. By the PMI, 3n ≥ 1 + 2n for all n in N.

(i) i. The statement is true for n = 1, because 33+1 = 34 = 81 > 64 =(1 + 3)3.

ii. Assume that 3n+3 > (n + 3)3 for some n ∈ N. Then

3(n+1)+3 = 3n+4 = 3 · 3n+3

> 3(n + 3)3 = 3(n3 + 9n2 + 27n + 27)= 3n3 + 27n2 + 81n + 81> n3 + 12n2 + 48n + 64 = (n + 4)3 = [(n + 1) + 1]3.

Therefore the statement is true for n + 1.iii. By the PMI, the statement is true for every n ∈ N.

(j) i. 41+4 = 45 = 1024 > 625 = 54(1+4)4, so the property is true for n = 1.ii. Assume the property is true for some n ∈ N. Then

4(n+1)+4 = 4(4n+4)> 4(n + 4)4 = 4n4 + 64n3 + 384n2 + 1024n + 1024> n4 + 20n3 + 150n2 + 500n + 625 = (n + 5)4.

So the statement holds for n + 1.iii. By the PMI, 4n+4 > (n + 4)4 for all n ∈ N.

(k) i. The statement is true for n = 1, because∑1

i=11i2 = 1 ≤ 2 − 1

1 .ii. Suppose that for some n ∈ N,

∑ni=1

1i2 ≤ 2 − 1

n . Then

n+1∑i=1

1i2

=n∑

i=1

1i2

+1

(n + 1)2≤ 2 − 1

n+

1(n + 1)2

= 2 − (n + 1)2

n(n + 1)2+

n

n(n + 1)2

= 2 − n2 + n + 1n(n + 1)2

< 2 − n2 + n

n(n + 1)2= 2 − 1

n + 1.

Thus the statement is true for n + 1.iii. By the PMI the statement is true for every n ∈ N.

2 SET THEORY 55

(l) i. (1 + x)1 ≥ 1 + 1(x), so the statement is true for n = 1.ii. Assume that the statement is true for some n ∈ N. Then (1 + x)n+1 =

(1 + x)n(1 + x) = (1 + x)n + x(1 + x)n ≥ (1 + nx) + x(1 + x)n ≥(1 + nx) + x(1) = 1 + (n + 1)x.Thus the statement is true for n + 1.


(m) i. 13

3 + 15

5 + 7(1)15 = 1 so the property holds for 1.

ii. Assume n3

3 + n5

5 + 7n15 ∈ Z for some n. Then

(n + 1)3

3+

(n + 1)5

5+

7(n + 1)15

=13(n3 + 3n2 + 3n + 1) +

15(n5 + 5n4 + 10n3 + 10n2 + 5n + 1) +

715

(n + 1)

=(

n3

3+

n5

5+

7n

15

)+ n2 + n +

13

+n4 + 2n3 + 2n2 + n +15

+715

=(

n3

3+

n5

5+

7n

15

)+ n4 + 2n3 + 3n2 + 2n + 1.

By the hypothesis of induction (n3

3 + n5

5 + 7n15 ) is an integer, and the

remaining terms are integers so the sum is an integer. Thus the propertyholds for n + 1.


(n) i. ddx (x1) = 1 = 1 · x0 so the statement is true for 1.

ii. Suppose the statement is true for some n ∈ N. Then

d

dx(xn+1) =

d

dx(xn · x1)

= xn d

dx(x) +

d

dx(xn) · x

= xn + nxn−1 · x

= xn + nxn = (n + 1)xn.

So the statement is true for n + 1.iii. By the PMI, the statement is true for all n.

(o) i. If A has 1 element, then P(A) = {A, ∅} has 2 elements.ii. Let n ∈ N and assume that every set with n elements has 2n subsets.

Let A be a set with have n + 1 elements. Pick an element a ∈ A. ThenA − {a} has n elements, and thus, by the induction hypothesis, has2n subsets. If B is one of these subsets, then both B and B ∪ {a} aresubsets of A. So A must have at least 2n + 2n = 2n+1 subsets. Theseare all the subsets of A, because if a subset of A does not contain athen it is a subset of A − {a}, and otherwise it has the form B ∪ {a}where B is a subset of A − {a}. Therefore A has exactly 2n+1 subsets.

iii. By the PMI, every set with n elements has 2n subsets.

8. i. The statement is true for n = 6, because 63 = 216 < 720 = 6!.ii. Suppose n3 < n! is true for some n ∈ N such that n ≥ 6. Then

(n + 1)3 = n3 + 3n2 + 3n + 1< n3 + 3n2 + 3n + n = n3 + 3n2 + 4n

2 SET THEORY 56

< n3 + 3n2 + n2 = n3 + 4n2

< n3 + n3 = 2n3

< (n + 1)n3

< (n + 1)n! = (n + 1)!

Thus the statement is true for n + 1.iii. By the generalized PMI, the statement is true for all n ≥ 6.

(a)i. 25 = 32 > 25 = 52, so the statement is true for n = 5.ii. Assume the statement is true for some natural number n > 4. Then

2n+1 = 2 ·2n > 2n2 = n2+n2 > n2+4n = n2+2n+2n > n2+2n+1 =(n + 1)2, so the statement is true for n + 1.

iii. By the generalized PMI, the statement is true for all n > 4.

(b) i. Since (5 + 1)! = 6! = 720 > 256 = 28 = 25+3, the statement is true forn = 5.

ii. Assume that (n + 1)! > 2n+3 for some n ≥ 5. Then [(n + 1) + 1]! =(n + 2)(n + 1)! > (n + 2)2n+3 by the induction hypothesis. Since(n+2)2n+3 > 2 ·2n+3 = 2n+4 = 2(n+1)+3 it follows that [(n+1)+1]! >2(n+1)+3.

iii. By the PMI, for all n ≥ 5, (n + 1)! > 2n+3.

(c) i. 4! = 24 and 3(4) = 12, so 4! ≥ 3(4). Thus the statement is true forn = 4.

ii. Assume that the statement is true for some n ∈ N, n ≥ 4. Then(n + 1)! = (n + 1)n! = n! + n(n!) ≥ 3n + n(n!) ≥ 3n + 3 = 3(n + 1).Thus the statement is true for n + 1.

iii. By the PMI, the statement is true for all natural numbers ≥ 4.

(d) i. The statement is true for n = 2, because∏2

i=2i2−1

i2 = 34 = 2+1

2·2 .ii. Suppose the statement is true for some n ∈ N such that n ≥ 2. Then

n+1∏i=2

i2 − 1i2

=(n + 1)2 − 1

(n + 1)2·

n∏i=2

i2 − 1i2

=n2 + 2n

(n + 1)2· n + 1

2n=

n + 22(n + 1)

=(n + 1) + 12(n + 1)

Thus the statement is true for n + 1.iii. By the generalized PMI, the statement is true for all n ≥ 2.

(e) i.∏4

i=11i = 1

112

13

14 = 1

24 ≥ 116 = 2−4.

Thus the statement is true for n = 4.ii. Assume that the statement is true for some n ∈ N, such that n ≥ 4.

Then

n+1∏i=1

1i

=1

n + 1

n∏i=1

1i

≤ 1n + 1

2−n ≤ 122−n = 2−(n+1).

Thus the statement is true for n + 1.iii. By the PMI, the statement is true for all natural numbers ≤ 4.

(f) i. We know from geometry that the sum of the angles of a triangle addup to 180˚. Thus, the statement holds for n = 3.

2 SET THEORY 57

ii. Assume the property holds for some n > 2. Consider a convex polygonwith n + 1 sides. Pick any 3 consecutive vertices and label them A, Band C. Draw a line segment from A to C. Since the original polygonwas convex, it is now split into the triangle ABC and an n-sided convexpolygon. By the induction hypothesis, the angles of the n sided polygonadd up to (n − 2) · 180˚. Thus the sum of the interior angles of theoriginal polygon is (n−2) ·180˚+180˚= [(n+1)−2] ·180˚. Thereforethe property holds for n + 1.

iii. By the generalized PMI, the statement is true for all n > 2.(g) i. Since 1 <

√2, 2 <

√2 + 1, and therefore

√2 < 1 + 1√

2. Thus the

statement is true for n = 2.ii. Assume the statement is true for some n ≥ 2. Then

√n < 1√

1+ 1√

2+

· · · + 1√n. Also,

n < n + 1, son2 < n(n + 1), so

n2

n + 1< n, so

n√n + 1

<√

n, so

√n + 1

n

n + 1<

√n, so

√n + 1

(1 − 1

n + 1

)<

√n, so

√n + 1 − 1√

n + 1<

√n, so

√n + 1 <

√n +

1√n + 1

.

Hence, by the induction hypothesis, the statement is true for n + 1.iii. By the generalized PMI, the statement is true for all n ≥ 2.

9. (a) i. (|1i=1 Ai)c =⋃1

i=1 Aic is A1

c = A1c, which is true.

ii. Assume that the statement is true for some n ∈ N. Then(|n+1i=1 Ai

)c= (|ni=1 Ai ∩ An+1)

c = (|ni=1 Ai)c ∪ An+1

c

=

(n⋃

i=1

Aic

)∪ An+1

c =n+1⋃i=1

Aic

Thus the statement is true for n + 1.iii. By the PMI, the statement is true for all natural numbers.

(b) Let S = {n ∈ N: (⋃n

i=1 Ai)c =|ni=1 Aic}.

i. For n = 1, the expression (⋃1

i=1 Ai)c =|1i=1 Aic is A1

c = A1c, which is

true.ii. Assume that the statement is true for some n ∈ N. Then(⋃

i = 1n+1Ai

)c

=(⋃

i = 1nAi ∪ An+1

)c

=(⋃

i = 1nAi

)c

∩ An+1c

= (| i = 1nAic) ∩ An+1

c =| i = 1n+1Aic.

Thus the statement is true for n + 1.

2 SET THEORY 58


10. (a) It is acceptable to begin with the basis step n = 1, or n = 2. Since there areno line segments if there is just one point, the statement is true if n = 1.Also if n = 2, the number of lines is 1 = 22−2

2 , so the statement is true for2.

(b) Assume the statement is true for some n ≥ 2. Consider n + 1 points in aplane. By the hypothesis of induction, n of them can be joined with n2−n

2distinct line segments. It now takes an additional n segments to join thelast point with each of the first n points. So, the total is

n2 − n

2+n =

n2 − n + 2n

2=

(n2 + 2n + 1) − (n + 1)2

=(n + 1)2 − (n + 1)

2,

and so the statement is true for n + 1.

(c) Thus, by the PMI, the statement is true for all n ∈ N.

11. (a) One disk can be moved to another peg in 1 = 21−1 moves, so the statementis true for n = 1.

(b) Assume the statement true for some n, and suppose we start with n + 1disks. By the induction hypothesis, all but the largest disk can be movedto another peg in 2n − 1 moves. Now move the largest disk to the free pegand then take another 2n −1 moves to move all the other disks on top of it.That makes (2n −1)+1+(2n −1) = 2n+1 −1 moves in all, so the statementis true for n + 1.

(c) By the PMI, the statement is true for all n ∈ N.

12. (a) If there are 3 players x, y and z, and x beats y and y beats z and z beatsx, then they are all top players.

(b) The statement is trivial for n = 1 and n = 2. Assume that for some n ≥ 2,every tournament with n players there has a top player. For n + 1 players,choose any one player s and examine the play between all others. This is atournament with n players, so there is a top player t. Let B be the set ofplayers that t beats.Now if t beats s, then t is top. If not, but some w in B beats s, then t istop. Otherwise s beats t and s beats every player in B. Every other playeris beaten by a player in B, so s is a top player.

13. (a) (a) F. The claim is false. The “proof” fails in the case n + 1 = 2. In thiscase, when either horse is removed from the set, the remaining horse hasthe same color (as itself), [because there is only one horse left] but the twohorses may have different colors.

(b) F. This method, sometimes called “proof by bluffing,” is so common thatit may be a necessary stage for understanding induction proofs, at least forsome students. What is missing is the proof that the statement is true forn + 1.

(c) F. n = 1 is odd, but 11 +1 = 2 is not odd, so the basic step fails. The claimis false.

(d) F. The induction step assumed that n + 1 ∈ S. One must show n ∈ Simplies n + 1 ∈ S.

(e) F. The factorization of xy + 1 is wrong, and there is no reason to believethat x + 1 or y + 1 is prime.

(f) A.

2 SET THEORY 59

2.5 Equivalent Forms of Induction

1. (a) Let S = {n ∈ N: n > 22 and n = 3s+4t for some integers s ≥ 3 and t ≥ 2}.Note that 23 = 3(5) + 4(2), 24 = 3(4) + 4(3) and 25 = 3(3) + 4(4). Thus23, 24, 25 ∈ S. Let m > 22 be a natural number and assume that for allk ∈ {23, 24, . . . , m − 1}, k ∈ S. If m = 23, 24, or 25 we already know m isin S. Otherwise m ≥ 26, so m − 3 ≥ 23. By the hypothesis of induction,m − 3 ∈ S, so m − 3 = 3s + 4t for some integers s and t, where s ≥ 3 andt ≥ 2. Then m = 3(s + 1) + 4t. Thus m ∈ S. By the PCI, the statement istrue for all n ∈ N such that n > 22.

(b) Let S = {n ∈ N: n > 33 and n = 4s+5t for some integers s ≥ 3 and t ≥ 2}.Note that 34 = 4(6) + 5(2), 35 = 4(5) + 5(3), 36 = 4(4) + 5(4), and37 = 4(3) + 5(5). Thus 34, 35, 36, 37 ∈ S.Let m > 33 be a natural number and assume that for all k ∈ {34, 35, . . . , m−1}, k ∈ S. If m = 34, 35, 36, 37 we already know m is in S. Otherwisem ≥ 38, so m − 4 ≥ 34. By the hypothesis of induction, m − 4 ∈ S, som − 4 = 4s + 5t for some integers s and t, where s ≥ 3 and t ≥ 2. Thenm = 4(s + 1) + 5t. Thus m ∈ S.By the PCI, the statement is true for all n ∈ N such that n > 33.

2. It is clear that a1 = 21 and a2 = 22 so the statement is true for m = 1 and form = 2. Assume an = 2n for all natural numbers n ≤ m − 1, for some m ≥ 3.Then

am = 5am−1 − 6am−2

= 5 · 2m−1 − 6 · 2m−2

= (5 · 2 − 6)2m−2 = 4 · 2m−2 = 2m, so the statement is true for m.

Therefore, by the PCI, an = 2n for all n ∈ N.

3. (a) Let a > 0 and T = {n ∈ N: an ≤ 0}. Suppose T �= ∅. Then by the WOP,T has a smallest element, n. Note that n must be greater than 1, becausea1 = a > 0. Thus, n − 1 ∈ N and n − 1 /∈ T , so an−1 > 0. Therefore,because the product of two positive integers is positive, an = a · an−1 > 0.This contradicts the fact that n ∈ T . We conclude that T is empty, soan > 0 for all n ∈ N.

(b) Suppose there were natural numbers a, b such that b = a + b. Then by theWOP, the set T = {n ∈ N: n = a + n} has a least element, b0. And b0 �= 1because 1 is not the successor of any natural number. Now, b0 = a + b0implies (b0 − 1) = a + (b0 − 1) which implies b0 − 1 ∈ T , contradicting theminimality of b0.

(c) Let A = {b ∈ N: (∃a ∈ N)(a2 = 2b2)}. Notice that A �= ∅ iff (∃a, b ∈ N)((a

b )2 = 2) iff√

2 is rational. Suppose A is nonempty. Then bythe WOP, A has a least element b0. Let a0 ∈ N be such that a2

0 = 2b20.

Then a20 is even, which forces a0 to be even. Substituting a0 = 2k, we get

(2k)2 = 2b20, and then b2

0 = 2k2 is even, which forces b0 to be even. Now a02

and b02 are both natural numbers and

(a0

2

)2=

a20

4=

b20

2= 2

(b0

2

)2

,

so b02 ∈ A, contradicting the minimality of b0. Therefore A is empty, so

√2

is irrational.

2 SET THEORY 60

4. (a) Let Rn be the number of rabbit pairs at n months. We already have thatRn = fn for n = 1, 2, 3, 4. Assume Rn = fn for all n up to m − 1 wherem − 1 ≥ 4. Then Rm = (# of pairs from last month)+(# of new pairs).The second term equals the number of breeding pairs, because each breedingpair produces one new pair each month. Now the breeding pairs are exactlythe pairs that were around two months ago, since it takes a month for themto mature. Thus, Rm = Rm−1 + Rm−2 = fm−1 + fm−2 = fm. Thus, by thePCI, Rn = fn for all n ∈ N.

(b) f1 = 1 f2 = 1 f3 = 2 f4 = 3 f5 = 5f6 = 8 f7 = 13 f8 = 21 f9 = 34 f10 = 55

(c) fn+3 − fn+1 = (fn+2 + fn+1) − fn+1 = fn+2.

5. (a) i. f3 = 2, f4 = 3, f5 = 5, so the statement is true for n = 1.ii. Suppose it is true for n. Then f3(n+1) = f3n+3 = f3n+2 + f3n+1, which

is even since f3n+2 and f3n+1 are odd. Also, f3(n+1)+1 = f3n+4 =f3n+3 + f3n+2, which is odd since f3n+3 is even and f3n+2 is odd.Finally, f3(n+1)+2 = f3n+5 = f3n+4 + f3n+3, which is odd since f3n+4isodd and f3n+3 is even.

(b) i. gcd(1, 1) = 1, so the statement is true for n = 1.ii. Suppose gcd(fn, fn+1) = 1 for some n ∈ N. Suppose also that a

natural number d divides fn+1 and fn+1 + fn. Then d must dividefn+1 + fn − fn+1 = fn, so d must be 1. This demonstrates thatgcd(fn+1, fn+2) =gcd(fn+1, fn+1 + fn) = 1.

iii. Therefore, by the PMI, gcd(fn, fn+1) = 1 for all n ∈ N.

(c) i. Let n ∈ N, and suppose a natural number d divides fn and fn + fn+1.Then d divides fn+1 + fn − fn = fn+1, so by (b), d = 1. Thusgcd(fn, fn+2) =GCD(fn,fn+1 + fn) = 1.

(d) i. In the case of n = 1, the formula is f1 = f1+2 − 1, which is 1 = 2 − 1.Thus the statement is true for n = 1.

ii. Suppose for some k that f1 + f2 + · · · + fk = fk+2 − 1. Then

f1 + f2 + · · · + fk + fk+1 = (f1 + f2 + · · · + fk) + fk+1

= (fk+2 − 1) + fk+1

= (fk+2 + fk+1) − 1= fk+3 − 1.

Therefore the statement is true for k + 1.iii. Thus, by the PMI, f1 + f2 + · · · + fn = fn+2 − 1 for all n ∈ N.

6. (a) f1 = f2 = 1, which is a natural number. Suppose m > 2 and fn ∈ N for alln ∈ {1, 2, . . . , m−1}. Then fm = fm−1 +fm−2 ∈ N by the closure propertyof N. Therefore, by the PCI, fn ∈ N for all n ∈ N.

(b) Notice that f7 = 13 = 4 ·3+1 = 4f4 +f1 and f8 = 21 = 4 ·5+1 = 4f5 +f2,so the statement is true for n = 1 and n = 2. Now assume that fn+6 =4fn+3 + fn for all n ∈ {1, 2, . . . , m − 1} and that m > 2. Then

fm+6 = fm+5 + fm+4

= (4fm+2 + fm−1) + (4fm+1 + fm−2)= 4 (fm+2 + fm+1) + (fm−1 + fm−2)= 4fm+3 + fm.

Therefore, by the PCI, fn+6 = 4fn+3 + fn for all n ∈ N.

2 SET THEORY 61

(c) Let a be a fixed natural number. (We can actually do induction either ona or n here.)Notice faf1 + fa+1f2 = fa + fa+1 = fa+2 and that faf2 + fa+1f3 =fa + 2fa+1 = fa+1 + fa+2 = fa+3, so the statement is true for n = 1and n = 2.Suppose m > 2 and fafn +fa+1fn+1 = fa+n+1 for all n ∈ {1, 2, . . . , m−1}.Then

fafm + fa+1fm+1 = fa(fm−1 + fm−2) + fa+1(fm + fm−1)= (fafm−1 + fa+1fm) + (fafm−2 + fa+1fm−1)= fa+m + fa+m−1 = fa+m+1.

Therefore, by the PCI, fafn + fa+1fn+1 = fa+n+1 for all n ∈ N.

(d) First notice that f1 = 1 = α−βα−β , and f2 = 1 = α−β

α−β = (α+1)−(β+1)α−β = α2−β2

α−β ,so the statement is true for n = 1 and n = 2.Suppose that fn = αn−βn

α−β for all n ∈ {1, 2, . . . , m − 1} and that m > 2.Then

fm = fm−1 + fm−2

=αm−1 − βm−1

α − β+

αm−2 − βm−2

α − β

=αm−2(α + 1) − βm−2(β + 1)

α − β

=αm−2(α2) − βm−2(β2)

α − β=

αm − βm

α − β.

Therefore, by the PCI, fn = αn−βn

α−β for all n ∈ N.

7. Let a and b be integers such that a < 0. We must show the existence of q and r.Let S = {b − ak: k is an integer and b − ak ≥ 0}. If 0 is in S, then a divides b,and we may take q to be the integer b/a and r = 0. Now assume that 0 ∈ S.

It follows from the assumption 0 ∈ S that b �= 0, because otherwise b− a0 = 0 ∈S. The set S is nonempty, because if b > 0 then b − a0 ∈ S, and if b < 0 thenb − a(−2b) = b(1 + 2a) ∈ S.

By the Well Ordering Principle, S has a smallest element, which we may call r.Then r = b−aq for some integer q, so b = aq + r, and r ≥ 0. We must also showthat r < |a| = −a. Suppose r ≥ −a. Then b − a(q − 1) = b − aq + a = r + a ≥ 0,so b − a(q − 1) ∈ S. But b − a(q − 1) < b − aq and b − aq is the smallest memberof S. This is impossible, so we must have r < |a|.To complete the proof, we must show that q and r are unique. Suppose thereexist integers q, q′, r, and r′ such that

b = aq + r with 0 ≤ r < |a| andb = aq′ + r′ with 0 ≤ r′ < |a|.

We may assume without loss of generality that r′ ≥ r. Then aq + r = aq′ + r′

and a(q − q′) = r′ − r. Then a divides r′ − r and 0 ≤ r′ − r ≤ r′ < |a|. Thenr′ − r must be 0, so r′ = r. Since a(q − q′) = 0, q′ = q.

8. Let a and b be nonzero integers. Let S = {ax + by: x, y ∈ Z and ax + by > 0},that is, the set of all positive linear combinations of a and b. Then S is not emptybecause if a > 0 then a(1) + b(0) ∈ S, and if a < 0 then a(−1) + b(0) ∈ S. ThusS is a nonempty subset of N, so by the Well Ordering Principle S has a smallestelement.

2 SET THEORY 62

9. Suppose n is the smallest positive integer greater than 1 that is not prime andsuch that n can be expressed in two different ways as a product of primes .(The order in which the prime factors appear does not matter). Then n maybe written as: n = p1p2p3 . . . pn and n = q1q2q3 . . . qm. Since p1 is one of thefactors of n, p1 divides n = q1q2q3 . . . qm. By Euclid’s Lemma p1 = qj for some1 < j < m. Then n/p1 = n/qj is a positive integer that is smaller than n andhas two different prime factorizations. This is a contradiction. We conclude thatevery natural number greater than 1 is prime or can be expressed uniquely as aproduct of primes.

10. Suppose there exists n ∈ N such that some tournament with n players does nothave a top player. Let k be the least such number. Then k must be greater than2. Therefore we can pick a player r out of the set of players, and have at least oneplayer left. Ignoring the games r played, we have a tournament of k − 1 players,which must have a top player, t. As in Exercise 12(b) of Section 2.4, either t orr must be a top player for the tournament of k players.

11. (a) g1 = 2 g2 = 2 g3 = 4 g4 = 8 g5 = 32

(b) g1 = 2 = 21 = 2f1 , and g2 = 2 = 21 = 2f2 , so the statement is true for n = 1and n = 2. Assume m > 2 and that gn = 2fn for all n ∈ {1, 2, . . . , m − 1}.Then gm = gm−1 · gm−2 = 2fm−1 · 2fm−2 = 2fm−1+fm−2 = 2fm . Therefore,by the PCI, gn = 2fn for n ∈ N.

12. (a) Let T ⊆ N and suppose T has no smallest element. Let S = N − T andassume for some m that {1, 2, . . . , m − 1} ⊆ S. Then m ∈ S because ifit weren’t, m would be the smallest member of T . By the PCI, S = N.Therefore T is empty. We conclude that every nonempty subset of N has asmallest element.

(b) Let S be a subset of N such 1 ∈ S and S is inductive. We wish to show thatS = N. Assume that S �= N and let T = N−S. By the WOP, the nonemptyset T has a least element. This least element is not 1, because 1 ∈ S. If theleast element is n, then n ∈ T and n−1 ∈ S. But by the inductive propertyof S, n − 1 ∈ S implies that n ∈ S. This is a contradiction. Therefore,S = N.

(c) Let S be a subset of N with the property that for all m ∈ N, {1, 2, . . . , m −1} ⊆ S ⇒ m ∈ S. Suppose S �= N and let T = N −S. T is not empty, so bythe WOP, T has a smallest element, n. But then {1, . . . , n − 1} ⊆ S (evenif n = 1) while n /∈ S, contradicting our hypothesis.

13. (a) A.

(b) F. The claim is false. The flaw in the “proof” is the incorrect assumptionthat m − 1 is a natural number. In fact, 1 is the smallest natural numbern such that 3 does not divide n3 + 2n + 1. There is no contradiction abouta smaller natural number because there is no smaller natural number.

(c) F. One cannot know that {1, 2, . . . , x − 1} ⊆ N − T unless one knows (orassumes) that x is the smallest element of T .

(d) F. The PCI condition is misstated.

2.6 Principles of Counting

1. (a) 13 (b) 2 (c) 0 (d) 0

2. (a) A ∩ B = A + B − A ∪ B = 24 + 21 − 37 = 8

2 SET THEORY 63

(b) A − A ∩ B = 24 − 8 = 16

(c) B − A = B ∪ A − A = 37 − 24 = 13

(d) B ∪ C = B + C − B = 21 + 12 = 33

(e) C = B ∪ C − B − C = 33 − 10 = 23

(f) A ∪ C = A + C − A ∩ C = 24 + 23 − 11 = 36

3. {n ∈ N: n < 1,000,000 and n is a square or cube}= {n ∈ N: n < 1,000,000 and n is a square}+{n ∈ N: n < 1,000,000 and n is a cube}−{n ∈ N: n < 1,000,000 and n is a both}= (103 − 1) + (102 − 1) − (10 − 1) = 1,089

4. (a) The Combination Rule(b) The Product Rule(c) Combination Rule and Product Rule(d) Permutation Rule

5. For four sets A, B, C and D,

A ∪ B ∪ C ∪ D = A + B + C + D

− A ∩ B − A ∩ C − A ∩ D − B ∩ C − B ∩ D − C ∩ D

+ A ∩ B ∩ C + A ∩ B ∩ D + A ∩ C ∩ D + B ∩ C ∩ D

− A ∩ B ∩ C ∩ D.

6. Let C = the set of first time campers; M = those who suffered at least on mishap;F = those who fell in the lake; P = those who got poison ivy; L = those whogot lost.

Then M = F + P + L − F ∩ P − F ∩ L − L ∩ P + F ∩ P ∩ L

= 14 + 13 + 16 − 3 − 4 − 8 + 2 = 30.

So the number of lucky ones is C − M = 40 − 30 = 10.

7. (a) 10 · 9 (b) 3 · 2 · 4 (c) 3 · 4 · 5

8. Among three digit positive integers with no repeated digits, there are 9·8·1 = 72with units digit 0 and 8 · 8 · 4 = 256 that are even with a nonzero units digit, fora total of 328 that are even.

There are 9 · 9 · 8 = 648 three digit positive integers with no repeated digits, andof this total there are 8 · 8 · 5 = 320 that are odd, leaving 328 that are even.

9. (a) There are 9 · 9 · 8 · 7 = 4536 four digit positive integers with no repeateddigits.

(b) There are 8 ·8 ·7 ·5 = 2240 odd four digit positive integers with no repeateddigits.

(c) Among four digit positive integers with no repeated digits, there are9 · 8 · 7 · 1 = 504 with units digit 0 and 8 · 8 · 7 · 4 = 1792 that are even witha nonzero units digit, for a total of 2296 that are even.

2 SET THEORY 64

10. If the bottom right corner square is to be colored the same as the upper leftcorner, there are 20 choices for the upper left, 1 for the lower right, 19 for theupper right, and also 19 for the lower left. 20 · 1 · 19 · 19 = 7, 220.

If the bottom right corner square and the upper left corner are to be painteddifferent colors, there are 20 choices for the upper left, 19 for the lower right,and 18 choices for each of the other two corners. 20 · 19 · 18 · 18 = 123,120.

All together there are 7,220 + 123,120 = 130,340 ways to paint the four squares.

11. Assume that A and B are disjoint sets, and C is any set. The sum A + B + Ccounts twice exactly the elements of A ∩ C and B ∩ C.

Therefore A ∪ B ∪ C = A + B + C − A ∩ C − B ∩ C.

12. If k = 1, then there is one task which can be performed in n1 =∏1

i=1 ni ways.Now suppose the statement is true for k and there are k + 1 tasks to be done insequence. From our hypothesis, the first k tasks can be done in

∏ki=1 ni ways.

Thinking of this as one big task, by Theorem 2.6.4 this and the last task can bedone in nk+1 ·

∏ki=1 ni =

∑k+1i=1 ni ways.

13. (a) There is only 1 permutation of a 1 element set, so the statement is true forn = 1.

(b) Assume that for some n ∈ N, every n-element set has n! permutations.Consider an (n + 1)-element set A. Pick x ∈ A. We can order A by firstordering A − {x} and then choosing a place to insert x. The first task canbe done in n! ways, while the second task can be done in n + 1 ways. Thusthe whole operation can be done in n!(n + 1) = (n + 1)! ways.

(c) By the PMI, the statement is true for all n ∈ N.

14. (a) We know that the elements of an n element set can be arranged in n! ways.Let x be the number of ways of selecting r objects from our set and orderingthem. Ordering the remaining n − r objects can be done in (n − r)! ways.Doing one and then the other will order the entire set. Thus n! = x(n − r)!so x = n!

(n−r)! .

(b) Let n = 1 and 0 ≤ r ≤ 1. Then r = 0 or r = 1, so the number ofpermutations of r distinct elements from a set with 1 element is 1 =1!/(1 − 0)! or 1 = 1!/(1 − 1)! respectively. Thus the statement is true forn = 1.Assume that n ∈ N and for every r such that 0 ≤ r ≤ n the number ofpermutations of r distinct elements from a set of n elements is n!

(n−r)! .Now suppose there are n+1 objects in a set and that 0 ≤ r ≤ n+1. In thespecial case when r = 0, the number of permutations of r distinct elementsfrom the set is 1 = (n+1)!

(n+1+0)! and in the special case when r = n + 1, the

number of permutations of r elements from the set is (n+1)! = (n+1)!(n+1−(n+1))! .

Let x be one of the elements of the set, and suppose 1 ≤ r ≤ n. By thehypothesis of induction there are n!

(n−r)! ways to permute r distinct elementsfrom the set if x is not one of the elements selected. If x is one of theelements selected, we have n!

(n−(r−1))! ways to permute the other elements,and then r ways to position x amongst the other elements. Thus there are

n!(n−r)! + n!

(n−(r−1))! = (n+1)!((n+1)−r)! permutations of r distinct elements from

the set of n + 1 elements. Therefore the statement is true for n + 1.

15. 7!

2 SET THEORY 65

16. (a) 3! · 4!

(b) 5! · 3! There are 5! ways to arrange the digits with the even digits as anunordered block, and then 3! ways to arrange the even digits within theblock.

(c) 4! · 4! Treating each even and the cluster of all odd digits as 4 distinctobjects, there are 4! ways to order them, and then 4! ways to order thecluster of odd digits.

(d) 3! · 4! The digits must alternate in the order odd, even, odd, etc. There are4! ways to arrange the odd digits and 3! ways to arrange the even digits.

17. Let x be the number of such permutations. To order all n objects, we could firstpick such a permutation, and then order the m alike objects. Thus n! = x · m!,so x = n!

m! .

If m1, objects are alike, m2 others are alike, and so forth, up to mk objects alike,then there are n!

m1!m2!···mk! permutations.

18. First choose 4 and then order them:(10

4

)· 4! = 5040.

19. (a)(19

3

)(b)

(112

)·(81

)(c)

(111

)·(82

)20.

(52

)(92

)= 5·4

2 · 9·82 = 360

21. (a) i. (a + b)1 =(10

)a0b1 +

(11

)a1b0, so the statement is true for n = 1.

ii. Suppose the statement is true for some n ∈ N. Then

n+1∑r=0

(n + 1

r

)arbn+1−r

= bn+1 +n∑

r=1

(n + 1

r

)arbn+1−r + an+1

= bn+1 +n∑

r=1

((n

r

)+

(n

r − 1

))arbn+1−r + an+1

= bn+1 +n∑

r=1

(n

r

)arbn+1−r + an+1 +

n∑r=1

(n

r − 1

)arbn+1−r + an+1

= b

(bn +

n∑r=1

(n

r

)arbn−r

)+ a

(n∑

r=1

(n

r − 1

)ar−1bn+r−1 + an

)

= b

(n∑

r=0

(n

r

)arbn−r

)+ a

(n−1∑r=0

(n

r

)arbn−r + an

)

= b

(n∑

r=0

(n

r

)arbn−r

)+ a

(n∑

r=0

(n

r

)arbn−r

)

= (a + b)

(n∑

r=0

(n

r

)arbn−r

).

= (a + b)(a + b)n = (a + b)n+1,

therefore the statement is true for n + 1.iii. By the PMI, the statement is true for all n ∈ N.

2 SET THEORY 66

(b)(

n − 1r

)+

(n − 1r − 1

)=

(n − 1)!r!(n − r − 1)!

+(n − 1)!

(r − 1)! [n − 1 − (r − 1)]!

=(n − 1)!

r!(n − r − 1)!+

(n − 1)!(r − 1)!(n − r)!

=(n − 1)!

r!(n − r − 1)!

[1r

+1

n − r

]

=(n − 1)!

r!(n − r − 1)!

[n

r(n − r)

]

=n!

r!(n − r)!=

(n

r

)

(c) Choose one particular element x from a set A of n elements. The number ofsubsets of A with r elements is

(nr

). The collection of r-element subsets may

be divided into two disjoint collections: those subsets containing x and thosesubsets not containing x. We count the number of subsets in each collectionand add the results. First, there are

(n−1

r

)r-element subsets of A that do

not contain x, since each is a subset of A − {x}. Second, there are(n−1r−1

)r-element subsets of A that do contain x, because each of these correspondsto the (r − 1)-element subset of A − {x} obtained by removing x from thesubset. Thus the sum of the number of subsets in the two collections is(

n − 1r

)+

(n − 1r − 1

)=

(n

r

).

22. (a) a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

(b) a4 + 8a3b + 24a2b2 + 32ab3 + 16b4

(c)(13

3

)(d)

(122

)· 210

23. (a) The number of ways to select an even number = the number ways to select0 + the number ways to select 2 +· · ·+ the number of ways to select n−1 =the number of ways to select n + the number of ways to select n−2+ · · ·+the number of ways to select 1 = the number of ways to select an oddnumber of objects.

(b) Let A and B be disjoint sets with n and m elements respectively. Then(n+m

r

)is the number of r-element subsets of A ∪ B. For each i such that

0 ≤ i ≤ r, we may select i elements from A in(ni

)ways and the remaining

r − i elements from B in(

mr−i

)ways. Thus

(n+m

r

)=

∑ri=0

(ni

)(m

r−i

).

(c) Applying Theorem 2.6.9(c) twice, we have 12

(2n+2n+1

)= 1

2 [(2n+1

n+1

)+

(2n+1n

)] =

12 [

( 2nn+1

)+

(2nn

)+

(2nn

)+

( 2nn−1

)]. But

( 2nn+1

)=

( 2nn−1

), so this expression is

equal to(2n

n

)+

( 2nn+1

).

24. (a) A (b) A (c) A

3 Relations and Partitions

3.1 Cartesian Products and Relations

1. (a) Dom(T ) = {1, 2, 3} (b) Rng(T ) = {1, 2, 3, 5, 6}(c) T−1 = {(1, 3), (3, 2), (5, 3), (2, 2), (6, 1), (6, 2), (2, 1)}(d) (T−1)−1 = T

2. (a) domain R, range R

(b) domain R, range [3,∞)(c) domain [1,∞), range [0,∞)(d) domain {x ∈ R | x �= 0}, range R

+

(e) domain R, range R

(f) domain (−2, 2) range {3}(g) domain R, range R

(h) domain R, range R

3. (a) (b)

(c) (d)

(e) (f)

(g) (h)

4. (a) R−11 = {(x, y) ∈ R × R | y = x}

67

3 RELATIONS AND PARTITIONS 68

(b) R−12 = {(x, y) ∈ R × R | y = x−2

−5 }(c) R−1

3 = {(x, y) ∈ R × R | y = x+107 }

(d) R−14 = {(x, y) ∈ R × R | y = ±

√x − 2 }

(e) R−15 = {(x, y) ∈ R × R | y = ±

√5−x2 }

(f) R−16 = {(x, y) ∈ R × R | y > x − 1}

(g) R−17 = {(x, y) ∈ R × R | y < x+4

3 }(h) R−1

8 = {(x, y) ∈ R × R | y = 2xx−2}

(i) R−19 = {(x, y) ∈ P × P | y is a child of x and x is male}

(j) R−110 = {(x, y) ∈ P × P | x is a sibling of y}

(k) R−111 = {(x, y) ∈ P × P | y is loved by x}

5. (a) R ◦ S = {(3, 5), (5, 2)} (b) R ◦ T = {(3, 2), (4, 5)}(c) T ◦ S = {(2, 1), (3, 1), (3, 4)} (d) R ◦ R = {(1, 2), (2, 2), (5, 2)}(e) S ◦ R = {(1, 5), (2, 4), (5, 4)} (f) T ◦ T = {(1, 1), (4, 4)}(g) R ◦ (S ◦ T ) = {(3, 2)} (h) (R ◦ S) ◦ T = {(3, 2)}

6. (a) R1(b) R2(c) {(x, y) ∈ R × R | y = 25x − 8}(d) {(x, y) ∈ R × R | y = −35x + 52}(e) {(x, y) ∈ R × R | y = −5x2 − 8}(f) {(x, y) ∈ R × R | y = 25x2 − 20x + 6}(g) {(x, y) ∈ R × R | y = 16x4 − 40x2 + 27}(h) {(x, y) ∈ R × R | y < −5x + 3}(i) {(x, y) ∈ R × R | y < x2 + 3}(j) {(x, y) ∈ R × R | y < x + 2}(k) {(x, y) ∈ R × R: y > 9x + 16}(l) {(x, y) ∈ R × R | y = −64x4 + 160x2 − 95}(m) {(x, y) ∈ R × R | y = x and x �= 2}(n) {(x, y) ∈ R × R | y = 4x+20

x−2 − 10}(o) {(x, y) ∈ R × R | y = 14x−20

7x−12 }(p) {(x, y) ∈ R × R | y is the paternal grandfather of x}

7. (a) (b)

(c) (d)

(e) (f)

8. (a) R = {(a, b), (c, d)} S = {(b, c)} R ◦ S = {(b, d)} and S ◦ R = {(a, c)}(b) R = {(a, c)} S = {(c, d)} (S ◦ R)−1 = {(d, a)} and S−1 ◦ R−1 = ∅(c) R = {(a, c), (a, b)} S = {(c, d)(d, a)} T = {(a, b)(b, d)}

Then S ◦ R = T ◦ R = {(a, b), (b, d)}


(d) R = {(a,b)} S= {(c,d)}

9. (a) Suppose x ∈Dom(S ◦ R). Then for some z, (x, z) ∈ S ◦ R. Then for somey, (x, y) ∈ R and (y, z) ∈ S. Since (x, y) ∈ R, x ∈Dom(R).

(b) Let R = {(a, b)} and S = {(c, d)}. Then Dom(S ◦R) = ∅ �= {a} = Dom(R).

(c) Rng(S◦R) ⊆ Rng(S) is always true. This may be proved as follows. Supposez ∈Rng(S ◦R). Then for some x, (x, z) ∈ S ◦R. Then for some y, (x, y) ∈ Rand (y, z) ∈ S. Since (y, z) ∈ S, z ∈Rng(S).Let R = {(a, c)} and S = {(b, d)}. Then Rng(S) = {d} ⊆ ∅ =Rng(S ◦ R).

(d) Let R = {(a, a), (b, b)} and S = {(c, c), (d, d)}

10. (a) (a, b) ∈ R for some a ∈ A and b ∈ B iff (b, a) ∈ R−1 iff (a, b) ∈ (R−1)−1.Therefore (R−1)−1 = R.

(b) No solution

(c) No solution

(d)

(x, y) ∈ (S ◦ R)−1 for some x ∈ C and y ∈ A

iff (y, x) ∈ S ◦ R

iff ∃z ∈ B such that (y, z) ∈ R and (z, x) ∈ S

iff ∃z ∈ B such that (z, y) ∈ R−1 and (x, z) ∈ S−1

iff ∃z ∈ B such that (x, z) ∈ S−1 and (z, y) ∈ R−1

iff (x, y) ∈ R−1 ◦ S−1.

11. To see that (A × B) × C = A × (B × C) may be false, let A = {1}, B = {2} andC = {3}.

Then (A × B) × C = {((1, 2), 3)} while A × (B × C) = {(1, (2, 3))}.

12. One method of proof employs the Product Rule (Theorem 2.6.4). The numberof ways to select an element of A × B is the number (m) of ways to select anelement of A times the number (n) of ways to select an element of B.

The statement can also be proved by induction. Assume A = {a1, . . . , am}. Useinduction on B = n.

(a) If n = 0, then B = ∅; therefore A × B = ∅ = 0 = m · 0 = m · n.

(b) Assume that for some t ∈ N, of C is a set with C = t, then A × C = m · t.Let B be any set with t + 1 elements and let x be any element in B. ThenB − {x} = t, so by the induction hypothesis, A × (B − {x}) = A ·B − {x}.Notice that A × B = (A × (B − {x})) ∪ {(a1, x), (a2, x), . . . , (am, x)}. Thismeans A × B = A × (B − {x}) + m = A · (B − {x} + 1) = m · t.

(c) By the PMI, A × B = A · B for all finite sets A and B.

13. (a) x ∈⋃

a∈A Ra iff (∃ a ∈ A) (x ∈ Ra) iff x ∈ B and (∃ a ∈ A)((a, x) ∈ R) iffx ∈Rng(R).

(b) x ∈⋃

b∈B bR iff (∃ b ∈ B)(x ∈ bR) iff x ∈ A and (∃b ∈ B)((x, b) ∈ R) iffx ∈Dom(R).

14. (a, b, c) = (x, y, z) iff ((a, b), c) = ((x, y), z) iff (a, b) = (x, y) and c = z iff a = x,b = y and c = z.

15. (a) F. The statements “x ∈ A×B” and “x ∈ A and x ∈ B” are not equivalent.


(b) C. This “proof” could be corrected by observing that (a, c) /∈ B×D impliesa /∈ B or c /∈ D, which implies a /∈ A or c /∈ C. This contradicts the factthat a ∈ A and c ∈ C. A direct proof would be easier.

(c) F. Division by the set A is not defined.

(d) A.

(e) F. The statement “(x, y) ∈ S ◦ R iff (y, x) ∈ R ◦ S” is not true in general.

(f) C. The proof can be corrected if DOM(R) = A, but otherwise the claim isfalse.

(g) F. The statement ”Since (x, z) ∈ R and (y, z) ∈ R, x = y“ is not true ingeneral.

3.2 Equivalence Relations

1. (a) transitive

(b) reflexive, transitive

(c) reflexive, symmetric, and transitive

(d) transitive

(e) reflexive, transitive

(f) symmetric

(g) reflexive, transitive

(h) symmetric

(i) symmetric

(j) symmetric

(k) symmetric

(l) reflexive, transitive

2. (ART TO COME)

(a) {(1, 1), (2, 2), (2, 3), (3, 1)}(b) {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}(c) {(1, 2), (2, 1)}(d) {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}(e) {(1, 2), (2, 3), (1, 3)}(f) {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}(g) {(1, 2), (2, 1), (2, 2), (1, 1)}(h) {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

3. (a) {(x, y) | y = 5x} (b) {(x, y) | y = x or y = x2}(c) {(x, y) | x2 + y2 = 1} (d) {(x, y) | y = x or x2 + y2 = 1}(e) {(x, y) | y ≤ x − 1} (f) {(x, y) | y ≤ x}(g) {(x, y) | x2 + y2 = 0} (h) {(x, y) | y = ±x}

4. (a) Suppose A is nonempty. Let a be an element of A. Then (a, a) is not in therelation ∅, so ∅ is not reflexive on A.

(b) Let A be any set. If (x, y) is in ∅, then (y, x) is in ∅. If (x, y) and (y, z) arein ∅, then (x, z) is in ∅. (Both statements are true because their antecedentsare false.)


5. (a) i. Let x ∈ R. Then x − x = 0, is a rational number, so x R x. ThereforeR is reflexive on R.

ii. Suppose x R y for some x, y ∈ R. Then x − y ∈ Q. Then y − x =−(x − y) ∈ Q.

iii. Suppose x R y and y R z, for some x, y, z ∈ R. Then x − y ∈ Q andy − z ∈ Q.Then (x−y)+(y−z) = (x−z) ∈ Q, so x R z. Therefore R is transitive.The class of 0 and the class of 1 modulo R are both equal to Q. Theclass of

√2 is the set of all irrational numbers.

(b) i. Let x ∈ N. Then x has the same digit in the tens place as x; so x R x.ii. Suppose x R y, for some x, y ∈ N. Then x has the same digit in the

tens place as y, so y has the same digit in the tens place as x. Thusy R x.

iii. Suppose x R y and y R z for some x, y, z ∈ N. Then x has the samedigit in the tens place as y and y has the same digit in the tens placeas z, so x has the same digit in the tens place as z. Thus x R z.

iv. 106/R contains 8, 202, and 1407. 635/R contains 31, 234, and 3,535.

(c) i. Let x ∈ R. Then x = x, so x V x.ii. Suppose x V y, for some x, y ∈ R. Then x = y or xy = 1. Then y = x

or yx = 1, so y V x.iii. Suppose x V y and y V z for some x, y, z ∈ R. Then x = y or xy = 1

and y = z or yz = 1. If y = z then x = z or xz = 1. If yz = 1, thenxz = 1 (in case x = y) or x = z (in case xy = 1). In any case x = z orxz = 1, so x V z.

iv. 3/V = {3, 13}; (− 2

3 )/V = {− 23 ,− 3

2}; 0/V = {0}.

(d) i. Let a ∈ N. Then the prime factorization of a into primes has the samenumber of 2’s as the prime factorization of a, so a R a.

ii. Suppose a R b for some a, b ∈ N. Then a has the same number of 2’s inits prime factorization that b has, so b has the same number of 2’s inits prime factorization that a has. Therefore b R a.

iii. Assume a R b and b R c for some a, b, c ∈ N. Then the prime factor-izations of a and b have the same number of 2’s and the prime fac-torizations of b and c have the same number of 2’s. Thus the primefactorizations of a and c have the same number of 2’s. Therefore a R c.

iv. 1/R = {1, 3, 5, . . .}, 4/R = {4, 12, 20, . . .}, 72/R = {8, 24, 40, 56, 72, . . .}.

(e) i. Let (x, y) ∈ R × R. Then x2 + y2 = x2 + y2, so (x, y) T (x, y).ii. Suppose (x, y) T (a, b) for some (x, y), (a, b) ∈ R × R. Then x2 + y2 =

a2 + b2, so a2 + b2 = x2 + y2. Therefore (a, b) T (x, y).iii. Suppose (x, y) T (a, b) and (a, b) T (u, v), for some (x, y), (a, b), (u, v) ∈

R × R. Then x2 + y2 = a2 + b2 and a2 + b2 = u2 + v2. Thenx2 + y2 = u2 + v2. Thus (x, y) T (u, v).

iv. The equivalence class of (1, 2) is the set {(x, x): x2 + y2 = 5}, which isthe circle with center at the origin and radius

√5. The class (4, 0)/T

is the circle with center at the origin and radius 4.

(f) i. Let A ⊆ X. Then A R A because A = A.

ii. Suppose A R B for some subsets A and B of X. Then A = B, so B = A.Thus B R A.

iii. Suppose A R B and B R C, for some A, B, C ∈ P(X). Then A = B

and B = C, so A = C. Thus A R C.


iv. The elements of {m}/R are {m}, {n}, {p}, {q}, {r}, and {s}. Theequivalence class {m, n, p, q, r}/R contains the sets {m, n, p, q, r},{m, n, p, q, s}, {m, n, p, r, s}, {m, n, q, r, s}, {m, p, q, r, s}, and {n, p, q, r, s}.There is just one element, X, in X/R. The set P(X)/R of all equiv-alence classes contains seven different equivalence classes. (To be pre-cise, P(X)/R = {∅/R, {m}/R, {m, n}/R, {m, n, p}/R, {m, n, p, q}/R,{m, n, p, q, r}/R, X/R}. There is one equivalence class for each possiblesize of a subset of X.

(g) i. Let (x, y)x ∈ D × R. Then (x, y)P (x, y) because | x − y |=| x − y |.ii. Suppose (x, y)P (z, w). Then | x−y |=| z−w |. Then | z−w |=| x−y |,

so (x, y)P (z, w).iii. Assume that (x, y)P (z, w0 and (z, w)P (u, v). Then | x − y |=| z − w |

and | z − w |=| u − v |. Therefore | x − y |=| u − v |, so (x, y)P (u, v).Some pairs related to (3, 0) are: (4, 1), (−1, 2), (−1,−4) and (1,−2).The equivalence class of (0, 0) is the line y = x. The equivalence classof (1, 0) consists of two lines: y = x + 1 and y = x − 1.

(h) i. Let D be the set of all differentiable functions on R. For all f ∈ D,f R f because f ′ = f ′.

ii. Suppose f, g ∈ D and f R g. Then f ′ = g′, so g′ = f ′, so that g R f .iii. Suppose f , g, h ∈ D and f R g and g R h. Then f ′ = g′ and g′ = h′.

Therefore f ′ = h′ and f R h.iv. x2 + 1, x2 + 2, x2 + 3 ∈ x2/R and 4x3 + 10x + 1, 4x3 + 10x + 2, 4x3 +

10x+3 ∈ (4x3 +10)/R. x3/R = {f ∈ D: f(x) = x3 + c for some c ∈ R}and 7/R = {f ∈ D: f(x) = c for some c ∈ R}.

(i) i. Let x ∈ R. Then sinx = sin y, so x T y.ii. Suppose x T y, for some x, y ∈ R. Then sinx = sin y. Therefore

sin y = sinx, so y T x.iii. Suppose x T y and y T z, for some x, y, z ∈ R. Then sinx = sin y and

sin y = sin z. Therefore sin x = sin z, so y T x.The class of 0 is {0 + kπ: k ∈ Z}. The class of π

2 is {π2 + 2kπ: k ∈ Z}.

The class of π4 is {±π

4 + 2kπ: k ∈ Z}.

6. (a) Let pq ∈ Q. Then pq = pg, so p

q R pq .

(b) Suppose pq R s

t for some pq , s

t ∈ Q. Then pt = qs, so sq = tp. Therefore,st R

pq .

(c) Suppose pq R s

t and st R

uv for some p

q , st ,

uv ∈ Q. Then pt = qs and sv = tu.

Multiplying by v and then by q, we hav ptv = qsv and qsv = qtu. Thereforepv = qu, so p

q Ruv .

The equivalence class of 23 is { 2k

3k : k ∈ Z − {0}}, which is the set of allrational numbers that reduce to 2

3 .

7. (a) Transitive, not reflexive and not symmetric(b) Transitive, reflexive and symmetric(c) Transitive, reflexive and symmetric(d) Reflexive, not transitive and not symmetric

8. (a) 0/ ≡5 = {. . . ,−10,−5, 0, 5, . . .} 3/ ≡5 = {. . . ,−7,−2, 3, 8, . . .}1/ ≡5 = {. . . ,−9,−4, 1, 6, . . .} 4/ ≡5 = {. . . ,−6,−1, 4, 9, . . .}2/ ≡5 = {. . . ,−8,−3, 2, 7, . . .}


(b) 0/ ≡8 = {. . . ,−16,−8, 0, 8, . . .} 4/ ≡8 = {. . . ,−12,−4, 4, 12, . . .}1/ ≡8 = {. . . ,−15,−7, 1, 9, . . .} 5/ ≡8 = {. . . ,−11,−3, 5, 13, . . .}2/ ≡8 = {. . . ,−14,−6, 2, 10, . . .} 6/ ≡8 = {. . . ,−10,−2, 6, 14, . . .}3/ ≡8 = {. . . ,−13,−5, 3, 11, . . .} 7/ ≡8 = {. . . ,−9,−1, 7, 15, . . .}

(c) 0/ ≡1 = Z

(d) 0/ ≡7 = {. . . ,−14,−7, 0, 7, . . .} 4/ ≡7 = {. . . ,−10,−3, 4, 11, . . .}1/ ≡7 = {. . . ,−13,−6, 1, 8, . . .} 5/ ≡7 = {. . . ,−9,−2, 5, 12, . . .}2/ ≡7 = {. . . ,−12,−5, 2, 9, . . .} 6/ ≡7 = {. . . ,−8,−1, 6, 13, . . .}3/ ≡7 = {. . . ,−11,−4, 3, 10, . . .}

9. (a) 5, 10, 15, 20 and −5, −10, −15, −20

(b) 30, 60 and −30, −60

(c) 2, 14, 26 and −10, −22, −34

(d) 3, 23, 43 and −17, −27, −47

(e) 1, 22, 43, and −20, −41, −62

10. (a) Let x ∈ Z. By reflexivityx ≡mx, so x ∈ x.

(b) Let x ∈ Z. Then x ∈ x by part (a), so x �= ∅.

(c) Assume x ≡ my. Let w ∈ x. Then x ≡ mw. By symmetry y ≡ mx andby the transitivity y ≡ mw. Therefore w ∈ y. This shows x ⊆ y. If z ∈ y,then y ≡ mz and by transitivity x ≡ mz. Therefore z ∈ x and y ⊆ x. Weconclude that x = y.

(d) Assume x = y. By the reflexivity y ≡ my, so y ∈ y. But y = x, so y ∈ x.Therefore x ≡ my.

(e) Assume x ∩ y �= ∅. Then there is some w ∈ Z such that w ∈ x andw ∈ y.Therefore, x ≡ mw and y ≡ mw. By symmetry and transitivity, x ≡ my.Then by part (b), x = y.

(f) Suppose x = y. Then x ∈ x by part (a), so x ∈ y. Thus x ∈ x∩y. Therefore,x ∩ y �= ∅.

11. S is not reflexive: 3 does not divide 1 + 1; hence 1 � S 1. Also S is not transitivebecause 5 S 4 and 4S 2, but 5 � S 2.

12. Suppose that R and S are equivalence relations on A.

(a) Since R and S are reflexive, we ahve a R a and a S a for all a ∈ A. Thus(a, a) ∈ R ∩ S.

(b) Suppose (a, b) ∈ R ∩ S. Then (a, b) ∈ R and (a, b) ∈ S. Since R and S aresymmetric, (b, a) ∈ R and (b, a) ∈ S. Therefore (b, a) ∈ R ∩ S.

(c) (a, b) ∈ R and (a, b) ∈ S and (b, c) ∈ R and (b, c) ∈ S. Then (a, c) ∈ R and(a, c) ∈ S, because both R and S are transitive. Therefore (a, c) ∈ R ∩ Sand so R ∩ S is transitive.

13. (a)

R is reflexive iff for all x ∈ A, (x, x) ∈ R

iff IA ⊆ R.

(b) Assume R is symmetric. Then (x, y) ∈ R iff (y, x) ∈ R iff (x, y) ∈ R−1.Therefore R = R−1. Now assume R = R−1. Then (x, y) ∈ R implies(x, y) ∈ R−1, which implies (y, x) ∈ R. Therefore R is symmetric.


(c) Assume R is transitive. Suppose (a, c) ∈ R◦R. Then for some b ∈ A, (a, b) ∈R and (b, c) ∈ R. By transitivity, (a, c) ∈ R; hence R ◦ R ⊆ R. Now assumeR ◦ R ⊆ R. Suppose (x, y) ∈ R and (y, z) ∈ R. Then (x, z) ∈ R ◦ R ⊂ R, so(x, y) ∈ R. Therefore R is transitive.

14. Assume that R is a symmetric, transitive relation on A and Dom(R) = A.Suppose x ∈ A. Then (x, y) ∈ R for some y ∈ R. By symmetry, (y, x) ∈ R. Bythe transitivity, (x, x) ∈ R. Therefore R is reflective on A.

15. (a) Let (x, y) ∈ R ∪ R−1. Then (x, y) ∈ R or (x, y) ∈ R−1. If (x, y) ∈ R, then(y, x) ∈ R−1. Likewise, if (x, y) ∈ R−1, then (y, x) ∈ R. In either case,(y, x) ∈ R ∪ R−1.

(b) Assume that S is a symmetric relation and R ⊆ S. Suppose (x, y) ∈ R−1.Then (y, x) ∈ R ⊆ S. Therefore R−1 ⊆ S.

16. (a) Suppose (x, y) ∈ TR and (y, z) ∈ TR. Then there exist a1, a2, . . ., an ∈ Asuch that (x, a1), (a1, a2), . . ., (an, y) ∈ R and there exist b1, b2, . . ., bm ∈ Asuch that (y, b1), (b1, b2), . . ., (bm, z) ∈ R.Since x, a1, a2, . . ., an, y, b1, b2, . . ., bm, z are in A, (x, z) ∈ TR. ThereforeTR is transitive.

(b) Assume that S is transitive and R ⊆ S. Suppose (x, y) ∈ TR. Then thereexist a1, a2, . . ., an ∈ A such that (x, a1), (a1, a2), . . ., (an, y) ∈ R ⊆ S.Since S transitive, we have (x, y) ∈ S. Therefore TR ⊆ S.

17. (a) Let E be the edge set for D and Ec be the edge set for the complementof D. Assume that D is a symmetric digraph. Suppose (x, y) ∈ Ec. Then(x, y) /∈ E. Since D is symmetric, it must then be the case that (y, x) /∈ E.Therefore, (y, x) ∈ Ec and the complement of D is symmetric.

(b)

18. Assume L is reflexive on A and transitive.

(a) Suppose x ∈ A. Then x L x and x L x because L is reflexive on A, so x R x.

(b) Suppose x R y for some x, y ∈ A. Then x L y and y L x, so y R x.

(c) Suppose x R y and y R z for some x, y, z ∈ A. Then x L y and y Lx andy L z and zL y, so x L z and z L x. Therefore, x R z.

19. (a) F. The statement is false. Consider the relation R = {(1, 5), (5, 1), (2, 6),(6, 2), (1, 1), (2, 2), (5, 5), 6, 6)} on the set A = {1, 2, 3, 4, 5, 6}. R is symmet-ric and transitive, but not reflexive on A. Alternatively, consider the emptyrelation R on a nonempty set A.

(b) F. To show symmetry one must show that (x, y) T (r, s) implies (r, s) T (x, y).

(c) A.

(d) F. This is only an example, not a proof.

(e) A.

(f) F. The last line confuses R ∪ S with R ◦ S. A correct proof requires a morecomplete second sentence.


3.3 Partitions

1. Let X be the set of all female students and Y be the set of all male students. Then{X, Y } is a partition of all university students. Students may also be partitionedby class standings, using sets of all freshmen, sophomores, juniors, seniors, andgraduate students; by major; as in-state and out-of-state students; or by thenumber of credit hours completed.

Note that most of these methods of partitioning may need to be adjusted forspecial circumstances. For example, if students are partitioned by majors, thereprobably should be a partition element for undecided students, and there may bea partition element for unclassified or non-degree-seeking students. Furthermore,one must account for students who are double-majors, perhaps by requiring suchstudents to select a primary major.

2. (a) Not a partition, because P is not pairwise disjoint.

(b) Not a partition of A, because 7 is not an element of an element of P.

(c) Partition of A.

(d) The elements of A are natural numbers, not subsets of N, so A is not apartition of N. Note: {{1, 2, 3, 4}, {n ∈ N: n > 5}} is a partition of N.

(e) Not a partition. P is not a family of subsets of A.

(f) Partition of A.

3. (a) In this partition, there is a partition element for each real number in theinterval [0, 1). All the integers are in the partition set with 0. All realnumbers that can be written as z + 0.071, z ∈ Z, are in the partition setwith 0.071. Each partition set contains exactly one real number in everyinterval [z, z + 1), where z is an integer. If x is any element of a partitionset, the next largest element of that partition set is x+1. The partition setthat contains x is {x + z: z ∈ Z}.

(b) There are 10 equivalence classes. The class 0/R contains, 0, 1, 2, . . . , 9, 100,101, . . . 109, 200, 201, . . . , 209, . . . and all negatives of these numbers. Theclass of 10 modulo R contains all integers that have 1 as the tens digit, andso forth.

(c) There is one partition element for every real number in [−1, 1]. For example,the partition set that contains 0 is {kπ: k ∈ Z} and another partitionelement is{x: sinx = 1

2} = {2kπ + π6 : k ∈ Z} ∪ {2kπ + 5π

6 : k ∈ Z}.

(d) The partition is the set of all sets {x,−x} such that x ∈ R, plus the set{0}.

(e) There are 3 elements in this partition. One element is the set of all pointson the x and y axes; a second element consists of all points in the first andthird quadrants; and the third element contains all points in the secondand fourth quadrants.

(f) In this partition each element is the set of all points on a line with slope 1.

4. {1/R, i/R}, where 1/R = {1,−1} and i/R = {i,−i}.

5. {(1, 1)/S, (i, 1)/S, (−i, 1)/S, (1,−1)/S}, where (1, 1)/S = {(1, 1), (−i, i), (i,−i),(−1,−1)}}, (i, 1)/S = {(i, 1), (−i,−1), (1, i), (−1,−i)}, (−i, 1)/S = {(−i, 1),(i,−1), (1,−i), (−1, i)}, and (1,−1)/S = {(−1, 1), (1,−1), (i, i), (−i,−i)}.

6. (a) x R y iff x has the same number of digits as y.


(b) x R y iff x = y or both x > 2 and y > 2.

(c) x R y iff xy > 0 or x = y = 0.

(d) x R y iff (i) x = y and x ∈ Z or (ii)∫

(x) =∫

(y) and x, y ∈ Z.(Recall that

∫(x) denotes the greatest integer function.)

(e) x R y iff x, y < 3 or x, y ≥ 3.

7. (a) y = −2 − x2

y = −1 − x2

y = −x2

y = 1 − x2

y = 2 − x2

(b) i. Let a ∈ R. Then (0, a) ∈ Aa, so Aa �= ∅.


ii. Suppose a, b ∈ R and a �= b. Assume that (x, y) ∈ Aa ∩ Ab. Thenby definition, y = a − x2 and y = b − x2. Then a = b. This is acontradiction. Therefore Aa ∩ Ab = ∅ when a �= b.

iii. Let (x, y) ∈ R × R. Choose a = y + x2 ∈ R. Then y = a − x2, so(x, y) ∈ Ra. Therefore R × R ⊆

⋃a∈R

Ra. Clearly,⋃

a∈RRa ⊆ R × R.

(c) The point (s, t) in the plane R×R is related to another point (u, v) iff bothpoints are on the same vertical shift of the parabola y = −x2. Alternatively,we could say that (s, t) is related to (u, v) iff there is a real number a suchthat t = a − s2 and v = a − u2.

8. (a) {(1, 2), (2, 1), (3, 4), (4, 3), (3, 5), (5, 3), (4, 5), (5, 4), (1, 1), (2, 2), (3, 3), (4, 4),(5, 5)}

(b) {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 3), (5, 5)}(c) {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (2, 3), (3, 2), (2, 4), (4, 2), (2, 5), (5, 2), (3, 4),

(4, 3), (3, 5), (5, 3), (4, 5), (5, 4)}

9. The two partition elements are D1 = {1, 4, 7} and D2 = {2, 3, 5, 6}. For eachpartition set, the digraph has all possible arcs (including loops) between elementsof the partition set.

10. (a) Suppose x Q y. By definition of Q, there is C ∈ P such that x ∈ C andy ∈ C. Since both y and x belong to C, y Q x. Therefore Q is symmetric.

(b) Let t ∈ A. Since P is a partition of A, A =⋃

X∈P X. Consequently, thereis some C ∈ P so that t ∈ C. Thus t Q t. Therefore Q is reflexive on A.

11. No. Let R be the relation {(1, 1), (1, 2), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}on the set A = {1, 2, 3}, R2 = {1, 2}, and R(3) = {1, 3}. The set A ={{1, 2, 3}, {1, 2}, {1, 3}} is not a partition of A.

12. Let R be a relation on the set A that is reflexive and transitive, but notsymmetric. For example, let A = {1, 2, 3} and define a R b iff a ≥ b. Then R(1) ={1}, R(2) = {1, 2} and R(3) = {1, 2, 3}. The set A = {{1}, {1, 2}, {1, 2, 3}} isnot a partition of A.

13. Let R be symmetric and transitive, but not reflexive on A. For example,let A = {1, 2, 3, 4} and let R = {(1, 1), (2, 2), (1, 2), (2, 1)}. Then R(3) = ∅,so{R(x): x ∈ A} is not a partition of A.

14. (a) Yes. {A} is a partition of A with one element.(b) Yes. {Bc

1, Bc2} is a partition of A, because {Bc

1, Bc2} = {B1, B2}. If B1 = B2,

then B1 = B2 = A; hence Bc1 = Bc

2 = ∅. Thus {Bc1, B

c2} would not be a

partition.(c) No. Let A = {1, 2, 3} and B1 = {1}, B2 = {2} and B3 = {3}. Then

Bc1 = {2, 3} and Bc

2 = {1, 3}, which are neither equal or disjoint. If exactlytwo elements, B1 and B2, of B are equal, then {Bc

1, Bc2, B

c3} = {B3, B3, B1}

is a partition of A. If B1 = B2 = B3, then B1 = A and Bc1 = ∅, so

{Bc1, B

c2, B

c3} is not a partition of A.

15. (a) F. Assume x ∈ y/R. The “proof” shows z /∈ y/R ⇒ z /∈ x/R, which is theconverse of z /∈ x/R ⇒ z /∈ y/R.

(b) A.(c) F. The claim is false, as is the last sentence of part (ii) of the “proof.”(d) A (or C). The proof is complete, because x Q y and y Q x are shown to be

equivalent to equivalent statements. Students who feel that the ideas arenot well connected may give this “proof” a grade of C.


3.4 Ordering Relations

1. (a) Is not antisymmetric because (2, 4) and (4, 2) ∈ R.(b) Is antisymmetric.(c) Is not antisymmetric because (2,−2) and (−2, 2) ∈ R.(d) Is not antisymmetric. 5 ≤ 26 and 6 ≤ 25, so 5 R 6 and 6R 5.(e) Is not antisymmetric because 0 S 1 and 1S 0.(f) Is not antisymmetric because (1, 3) and (3, 1) ∈ R.(g) Is antisymmetric.

2. (a) {(a, a), (c, c)}(b) {(a, a), (b, b), (c, c), (a, c)}(c) {(a, b), (a, c), (c, c)}(d) {(a, b), (b, a), (b, b)}(e) {(a, b), (b, a), (a, c), (c, c)}

3. (a) Assume that R is antisymmetric, x R y, and x �= y. Suppose y R x. Then,since R is antisymmetric, x = y. This contradicts the assumption thatx �= y.

(b) Assume that R is antisymmetric and symmetric, and that Dom(R) = A.Suppose (x, x) ∈ IA. Then x ∈ A, so x ∈Dom(R). Thus (x, y) ∈ R forsome y. Then (y, x) ∈ R by symmetry and x = y by antisymmetry. Thus(x, y) = (x, x), so (x, x) ∈ R. Therefore IA ⊆ R.Suppose (x, y) ∈ R. By symmetry, (y, x) ∈ R. By antisymmetry x = y.Therefore (x, y) = (x, x) ∈ IA. Thus R ⊆ IA. We conclude that R = IA.

4. (a) A = {a, b, c} and R = {(a, a), (b, c), (c, a)}(b) A = {a, b, c, d} and R = {(a, b), (c, d)}

5. (a) Let a ∈ N and choose k = 0. Then a = 2k · a, so a R a.

(b) Assume a R b and b R a. Then b = 2k · a and a = 2n · b for some n,k ≥ 0. Therefore, b = 2k(2n · b); thus, 1 = 2k+n and so k = n = 0.Thus b = 20 · a = a. Therefore R is antisymmetric.

(c) Assume a R b and b R c. Then b = 2k · a and c = 2n · b for some n, k ≥ 0.Then c = 2n ·b = 2n(2k ·a) = 2n+k ·a. Therefore a R c. Thus R is transitive.

6. (a) For each (x, y) ∈ R × R, x ≤ x and y ≤ y, so (x, y) R (x, y).

(b) Suppose (a, b) R (x, y) and (x, y) R (a, b). Then a ≤ x and x ≤ a, so x = a.Similarly, y = b. Therefore (a, b) = (x, y).

(c) Suppose (a, b) R (x, y) and (x, y) R (c, d). Then a ≤ x ≤ c, so a ≤ c. Alsob ≤ y ≤ d, so b ≤ d. Therefore (a, b) R (c, d).

7. R is not a partial order because 22 + 32 ≤ 32 + 22 and that (3 + 2i) R (2 + 3i)because 32 + 22 ≤ 22 + 32, yet (2 + 3i) �= (3 + 2i).

8. (a) Let α ∈ A, where α = x1x2. Then α ≤ α, because x1 = x1 and x2 ≤ x2.

(b) Suppose α ≤ β and β ≤ α, where α = x1x2 and β = y1y2. It is not possiblethat x1 < y1 because β ≤ α implies y1 ≤ x1. Thus x1 = y1 and x2 ≤ y2from the assumption that α ≤ β. Also, from β ≤ α, we have that y2 ≤ x2,so x2 = y2. Therefore α = β.

(c) Suppose α ≤ β and β ≤ δ where α = x1x2, β = y1y2 and δ = z1z2. Thenx1 ≤ y1 ≤ z1, so x1 ≤ z1. In the case that x1 = z1, we have x1 = y1 = z1.Thus x2 ≤ y2 ≤ z2, and so x2 ≤ z2. Therefore α ≤ δ. In the case thatx1 < z1, then α ≤ δ. (The second letters do not need to be compared.)


9.

10. (a) {(a, a), (b, b), (c, c), (c, a), (c, b)}(b) {(a, a), (b, b), (c, c), (d, d), (a, b), (a, c), (b, d), (a, d)}(c) {(a, a), (b, b), (c, c), (d, d), (a, b), (c, d), (b, d), (a, d)}

11. (a) (b)

12. (a) Suppose B − {x} ⊆ C ⊆ B. For any y ∈ B − {x}, we have y ∈ C. If x ∈ C,then for all y ∈ B we have y ∈ C. Therefore, B ⊆ C, which shows thatC = B. On the other hand, if x �= C, then for all y ∈ C we have y ∈ Band y �= x. Thus, C ⊆ B − {x}, which shows that C = B − {x}. Therefore,there is no C different from B and B − {x} such that B − {x} ⊆ C ⊆ B.Hence B − {x} is an immediate predecessor of B.

(b) Clearly B ⊆ B ∪ {x}. Now suppose B ⊆ C ⊆ B ∪ {x}. If x ∈ C, then everyelement of B ∪ {x} other than x is in C because B ⊆ C, and also x ∈ C,so B ∪ {x} ⊆ C. Therefore C = B ∪ {x}.If x /∈ C, then every element c ∈ C is different from x and is in B ∪ {x}, soc ∈ B. Thus C ⊆ B.This proves C = B or C = B ∪ {x}, so B is an immediate predecessor ofB ∪ {x}.

13. (a) Yes. Yes.

(b) No. A set consisting of two adjacent congruent squares has no largestelement.

(c) No. A set consisting of two disjoint squares does not have a lower bound.

(d) No. A set of two disjoint squares has no smallest element.

14. (a) i. C∪D is an upper bound for {C, D} because C ⊆ C∪D and D ⊆ C∪D.Suppose B is an upper bound for {C, D}. Then C ⊆ B and D ⊆ B.


Then by Exercise 9(f) of Section 2.2, C ∪ D ⊆ B, so C ∪ D is the leastupper bound.

ii. C ∩D is a lower bound for {C, D} because C ∩D ⊆ C and C ∩D ⊆ D.Suppose S is a lower bound for {C, D}. Then S ⊆ C and S ⊆ D. ByExercise 9(c) of Section 2.2, S ⊆ C ∩ D, so C ∩ D is the greatest lowerbound.

(b) i.⋃

B∈P B is an upper bound for P, because for all B ∈ P, B ⊆⋃

B∈P B.Suppose W is an upper bound for P. Then B ⊆ W for every B ∈ P.Then (by Exercise 10(c) of Section 2.3)

⋃B∈P B ⊆ W . Therefore⋃

B∈B B is the least upper bound for P.ii.

⋂B∈P B is a lower bound for P because

⋂B∈P B ⊆ B for every B ∈ P.

Suppose V is a lower bound for B ∈ P. Then (by Exercise 10(a) ofSection 2.3) V ⊆

⋂B∈P B. Therefore

⋂B∈P B is the greatest lower

bound for P.

15. (a) T is not a partial order on N because T is not antisymmetric. For example,both 9T 10 and 10T 9. Also T is not transitive because 10 T 8 and 8T 5,but 10 � T 5.

(b) V is a linear order on N.i. Let m ∈ N. Then either m is even and m ≤ m or m is odd and m ≤ m,

so m V m. Thus V is reflexive.ii. Suppose m V n and n V m. Then m and n are both even or both odd,

and m ≤ n and n ≤ m. Therefore m = n. Thus V is antisymmetric.iii. Suppose m V n and n V t. If m is odd and n is even, then t is even, so

m V t. If m and n are odd and t is even, then m V t. If m and n areboth even, then t is even and m ≤ n ≤ t, so m V t. Otherwise, m andn and t are all odd and m ≤ n ≤ t, so m V t. Thus V is transitive.

iv. Suppose m �= n. If exactly one of m, n (say m) is odd, then m V n.Otherwise both are even or both are odd, and one of m or n (say m)is smaller. Then m V n. Thus V is a linear order.

(c) S is a linear order on N.i. Let n ∈ N. If n �= 5, then n ≤ n, so n S n. If n = 5, then 5S 5 as well.

Thus S is reflexive on N.ii. Suppose n S m and m S n for some m, n ∈ N. If n = 5, then m = 5

because 5 S m. If m = 5, then n = 5 because 5 S n. Otherwise n ≤ mand m ≤ n, so n = m. Therefore S is antisymmetric.

iii. Suppose n S m and m S t for some n, m, t ∈ N. If n = 5, then m = 5and t = 5, so n S t. If m = 5, then t = 5, so n S t. If t = 5, then n S tby definition of S. Otherwise, m, n, and t are all different from 5 andn ≤ m ≤ t, so n ≤ t. Thus nS t. Therefore S is transitive.

iv. Let m, n ∈ N. If m = 5, then n S 5. If n = 5, then m S n. Otherwise,either m ≤ n or n ≤ m, so m S n or n S m. Therefore S is a linear orderon N.

(d) i. Let n ∈ N. If n �= 5, then n ≤ n, so n S n. If n = 5, then 5S 5 bydefinition. Thus S is reflexive on N.

ii. Suppose n S m and m S n for some m, n ∈ N. If n = 5, then m = 5because m S 5. If m = 5, then n = 5 because n S 5. Otherwise n ≤ mand m ≤ n, so n = m. Therefore S is antisymmetric.

iii. Suppose n S m and m S t, for some n, m, t ∈ N. If t= 5, then m = 5and n = 5, so n S t. If m = 5, then n = 5, so n S t. If n = 5, the n S t.Otherwise, m, n, and t are all different from 5 and n ≤ m ≤ t, so n ≤ t.Thus n S t. Therefore S is transitive.


iv. Let m, n ∈ N. If m = 5, then m S n. If n = 5, then n S m. Otherwise,either m ≤ n or n ≤ m, so m S n and n S m. Therefore S is a linearorder on N.

16. By Exercise 15(b) V is a linear order on N. Let B be a nonempty subset of N.Let B′ = {n ∈ B: n is odd}. If B′ is nonempty, then B′ has a smallest elementunder the usual ≤ ordering. Let k be this element. Then k V n for every oddnatural number in B and k V n for every even natural number in B, so k is thesmallest element of B under the linear order V . If B contains no odd numbers,let l be the smallest element in B under the usual ≤ ordering. Then every nin B is even and l V n, so l is the smallest element of B under the V ordering.Therefore V is a well ordering of N.

17. (a) Suppose R is a partial order on the set A. If R is a well ordering, thenevery nonempty subset of A has a smallest element, by definition. If everynonempty subset of R has a smallest element, then every two-element subset{x, y} of A has a smallest element, so either x R y or y R x. Thus R is alinear ordering that is a well ordering.

(b) Suppose R is a well ordering of A. Then every nonempty subset B ofA contains a smallest element b. This smallest element b is unique byantisymmetry. Also b is R-related to every element of B because for everyx ∈ B, either b ≤ x or x ≤ b and b is the smallest element of B.Suppose now that every nonempty subset B of A contains a unique elementthat is R-related to every element of B.

i. Let x ∈ A. Then the set {x} has a smallest element and x R x, so R isreflexive on A.

ii. Suppose x R y and y R x. Since both x and y are R-related to everyelement of {x, y}, x = y must be the unique such element. ThereforeR is antisymmetric.

iii. Suppose x R y and y R z for some x, y, z ∈ A. The set {x, y, z} has aunique element that is R-related to each of its elements. If this elementis y then y R x so x = y and x R z. If the unique element is z, thenz R y, so y = z and z R x, so y = z = x. Thus x R z. Otherwise, theunique element is x, so x R z. In any case x R z, so R is transitive.

iv. Suppose x, y ∈ A. Either x or y is the unique smallest element of {x, y},so x R y or x R z.

We conclude that R is a total order that is a well ordering.

18. Let A be any well-ordered set. Let B be any subset of A. Clearly B is linearlyordered because A is linearly ordered. It only remains to show that everynonempty subset of B has a least element. Suppose C ⊆ B and C �= ∅. SinceC ⊆ B ⊆ A, we have C ⊆ A and so C has a least element. Therefore B iswell-ordered.

19. (a) Suppose a, b ∈ A and a R b. If x ∈ Sa, then x R a. Thus x R a and a R b, soby transitivity, x R b Therefore x ∈ Sb.

(b) Suppose a, b ∈ A and Sa ⊆ Sb. By the reflexive property, a R a, soa ∈ Sa ⊆ Sb, hence a ∈ Sb. Therefore a R b.

(c) Let b ∈ A. Suppose a is the immediate predecessor of b. Then a R b, soSa ⊆ Sb. Thus Sa is a predecessor of Sb. Now suppose there exists Sc ∈Fsuch that Sa ⊆ Sc ⊆ Sb. Then a R c and c R b, so a = c or c = b. ThusSa = Sc or Sc = Sb. Therefore Sa is the immediate predecessor of Sb.


(d) Suppose B ⊆ A and x is the least upper bound for B. Then b R x for allb ∈ B, so Sb ⊆ Sx for all b ∈ B. Thus Sx is an upper bound for {Sb: b ∈ B}.Suppose St is an upper bound for {Sb: b ∈ B}. Then Sb ⊆ St for all b ∈ B,so b R t for all b ∈ B. Thus t is an upper bound for B, so x R t. ThereforeSx ⊆ St. This shows that Sx is the least upper bound for {Sb: b ∈ A}.

20. (a) A (or C). The proof is correct. Students who feel that the secondsentence needs more explanation may add that explanation and give this“proof” a grade of C.

F. This proof does not show that sup(B) exists. All it shows is thatif sup(B) exists, then u = sup(B). A correct proof would show thatu = sup(B) by showing u has the two supremum properties (u is an upperbound and u R v for all other upper bounds v).(c) F. The claim is false. Consider A = (0, 1) and B = [1, 2]. Thensup(A ∪ B) = 2 �= 1 + 2 = sup(A) + sup(B). The error in the proof occursin the last sentence: r ≤ t and s ≤ t does not imply r + s ≤ t.

3.5 Graphs

(a)1. (a) d(a) = 1, d(b) = 4, d(c) = d(d) = d(e) = 1. The sum of the degrees is 8(even) and the number of vertices with odd degree is 4 (even).

(b) d(a) = d(e) = 2, d(b) = 4, d(c) = d(d) = 1. The sum of the degrees is 10(even) and the number of vertices with odd degree is 2 (even).

(c) d(a) = 2, d(b) = d(c) = 3, d(d) = d(e) = 2. The sum of the degrees is 12(even) and the number of vertices with odd degree is 0 (even).

2. (a) (b) (c)

(d) Not possible. (e) (f)

3. (a) (b) (c) Not possible. (d)

4. (a) and(b) and and

5. (a) (b) (c) (d)


6. (a) (b)g, f, e, a, b c, b, a, e, h, g, cg, f, e, a, d c, b, a, e, f, g, cg, h, e, a, b c, d, a, e, f, g, cg, h, e, a, d c, d, a, e, h, g, cg, c, b, a, e c, g, f, e, a, b, cg, c, d, a, e c, g, f, e, a, d, cg, c, b, a, d c, g, h, e, a, d, cg, c, d, a, b c, g, h, e, a, b, c

(c) (d)f, g, h, e, a, b, c, d d, a, e, a,b

7. (a) (b)

8. (a) (b) (c) (d)

9. Let G be a simple graph with order n ≥ 2. Suppose there are no two verticeswith the same degree. Since there are no multiple edges or loops in a simplegraph the degree of each vertex is ≤ n − 1. The degree of a vertex must be ≥ 0,so the possible degrees are 0, 1, 2, . . . , n−1. Therefore there is one vertex v withdegree 0, and one vertex w with degree n − 1. But then w is adjacent to everyother vertex, including v, but v is not adjacent to any vertex. This is impossible.Therefore two vertices have the same degree.

10. (a) (b)

11. (a) d(u, v) is the number of edges in a path of minimum length from u to v.Since the graph is connected, there is a path from u to v. The length of apath must be greater than or equal to 0.

(b) If u = v, then d(u, v) = 0 by definition. If u �= v, then every path from uto v has length at least 1, so d(u, v) �= 0.

(c) Let u, v, and w be vertices in G, each reachable from the others. Thend(u, v) is the length of a shortest path from u to v, and d(u, v) is the lengthof a shortest path from v to w. Fix one such path from u to v and one suchpath from v to w. Traversing the path from u to v and then to w providesa walk from u to w of length d(u, v) + d(v, w). The shortest path from v tow cannot exceed this length.

12. Suppose d(u, r) = n ≥ 2. Then a shortest path from u to r that contains n − 1vertices other than u and r, and n − 1 ≥ 1. Let w be one of these vertices.


There can be no shorter path from u to w or from w to r than following thegiven path, for otherwise there would be a shorter path from u to r. Thusd(u, w) + d(w, r) = d(u, r).

13. (a) (b)

(c)

14. (a) Suppose C(v) = C(w). Clearly w is reachable from w, so w ∈ C(w) = C(v).Thus w ∈ C(v) and so w is reachable from v. On the other hand, supposew is reachable from v and x ∈ C(w). Then x is reachable from w. Bysymmetry w is reachable from x and by transitivity x is reachable fromv. Therefore x ∈ C(v). Thus C(w) ⊆ C(v). Similarly C(v) ⊆ C(w), soC(v) = C(w).

(b) Suppose that x is a vertex in both C(v) and C(w). Then x is reachable fromv and x is reachable from w. Then w is reachable from v, via x. Therefore,by part (a), C(v) = C(w). We conclude that if C(v) �= C(w), then C(v)and C(w) have no vertices in common.

15. (a) A.

(b) F. The claim is false. For example, the graph of order 6 shown in Exer-cise 13(c) has no closed path of length 6.

4 Functions

4.1 Functions as Relations

1. (a) Function. Domain: {0,�,�,∩,∪}. Two possible codomains: {0,�,�,∩,∪}and {0,�,�,∩,∪, 1}.(b) Not a function.(c) Function. Domain: {1, 2}. Two possible codomains: {1, 2} and N.(d) Not a function.(e) Not a function.(f) Not a function.(g) Not a function.(h) Function. Domain: {a, b, c}. Two possible codomains: {1, 2} and N.(i) Function. Domain: {a, b, c, d}. Two possible codomains: {2, 3, 4} and N.

2. (a) A × B (b) {(5, 4), (7, 4)}(c) {(x, y) ∈ A × B: y = x − 2} (d) {(x, y) ∈ A × B: y = 3}

3. The codomain is given in the definition of the function. Another possiblecodomain for each function is its range.(a) Domain= R − {1}, Range = R − {0}(b) Domain= R, Range= [5,∞)(c) Domain= N, Range= N − {1, 2, 3, 4, 5}(d) Domain= R − {π

2 + kπ: k ∈ Z}, Range= R

(e) Domain = R, Range= {0, 1}(f) Domain= R, Range= [1,∞)(g) Domain= R − {2}, Range= R − {4}(h) Domain= Z − {2}, Range= Z − {4}

4. (a) Domain= R − {3}, Range= R − {−1}(b) Domain= R, Range= R

(c) Domain= (−π,∞), Range= (0,∞)(d) Domain= (−∞, 5], Range= (0,∞)(e) Domain= [3, 5], Range= [

√2, 2]

(f) Domain= {−2}, Range= {0}

5. (a) If x = 1 then 2x + y = 2 + y. Since y = 1, 2, 3, or 4, 2 + y is prime and not5 only when y = 1.If x = 2 then 2x + y = 4 + y. Since y = 1, 2, 3, or 4, 4 + y is prime and not5 only when y = 3.If x = 3 then 2x + y = 6 + y. Since y = 1, 2, 3, or 4, 6 + y is prime and not5 only when y = 1.If x = 4 then 2x + y = 8 + y. Since y = 1, 2, 3, or 4, 8 + y is prime and not5 only when y = 3.In all cases, for each x ∈ A, there is a unique y such that (x, y) ∈ R.

(b) If x = 1 then 3x + y = 3 + y. Since y = 1, 2, or 3, 3 + y is prime only wheny = 2.If x = 2 then 3x + y = 6 + y. Since y = 1, 2, or 3, 6 + y is prime only wheny = 1.If x = 3 then 3x + y = 9 + y. Since y = 1, 2, or 3, 9 + y is prime only wheny = 2.In all cases, for each x ∈ A, there is a unique y such that (x, y) ∈ R.

(c) i. Suppose x R y and x R z. Then x2 + y = 2 and x2 + z = 2. Therefore,x2 + y = x2 + z and hence y = z.

85

4 FUNCTIONS 86

ii. Since R is defined as a relation on Z, Dom(R) ⊆ Z. To show thatZ ⊆ Dom(R), let x ∈ Z. Choose y to be the integer 2 − x2. Thenx2 + y = x2 + (2 − x2) = 2. Thus x ∈ Dom(R). We conclude thatDom(R) = Z.By (i) and (ii), R is a function on Z.

6. (a) (1, 1) and (1,−1) are both in the relation.

(b) (0, 1) and (0,−1) are both in the relation. Also, the domain is not R.

(c) (1, 0) and (1, 2π) are both in the relation. Also, the domain is not R.

(d) (16, 2) and (16,−2) are both in the domain. Also, the domain is not R.

7. (a) (b)

(c) (d)

8. (a) A (b) Ac (c) ∅

9. (a) xn = −n (b) xn = (−1)nn(c) xn = 3 + n

n+1 (d) xn = n (modulo 3)

10. (a) f(3) = 3 in Z6 (b) f(6) = 6 = 0 in Z6(c) 363 (or any multiple of 3). (d) 1 = {6k + 1, k ∈ Z}

11. (a) Not a function. 0 = 3 in Z3 but f(0) = [0] = [3] = f(3) in Z6.

(b) Function.

(c) Function.

(d) Not a function. 0 = 4 in Z4 but f(0) = [1] = [3] = [9] = f(4) in Z6.

(e) Not a function. 0 = 3 in Z3 but f(0) = [0] = [3] = f(3) in Z4.

(f) Function.

12. f and g are not equal as sets because (−3, 6) ∈ g but (−3, 6) /∈ f . From theperspective of Theorem 4.1.1, f = g because they have different domains.

13. (a) ∅ is a relation on ∅ whose domain is ∅, and if (x, y) ∈ ∅ and (x, z) ∈ ∅, theny = z.

(b) If A = ∅, then Rng(f) = {f(x): x ∈ ∅} = ∅. Also, f ⊆ ∅ × B = ∅, so f = ∅.If Rng(f) = ∅, then f ⊆ A × ∅ = ∅, so f = ∅.Finally, if f = ∅, then A =Dom(f) = ∅ by definition and Rng(f) = ∅.

4 FUNCTIONS 87

14. Suppose (x, y) ∈ f . Then x ∈Dom(f) =Dom(g), so there is a z such that(x, z) ∈ g. But y = f(x) = g(x) = z, so (x, y) ∈ g. Thus f ⊆ g, and in thesame way, g ⊆ f .

15. (a) Dom(S) (b) Rng(S)

16. (a) Let x, y, z ∈ N.

i. By definition of absolute value, d(x, y) = |x − y| ≥ 0 for all x, y ∈ N.ii. d(x, y) = |x − y| = 0 iff x − y = 0 iff x = y.iii. d(x, y) = |x − y| = |y − x| = d(y, x).iv. By the triangle property of absolute value, |x − y| + |y − z| ≥ |x − z|.

Thus d(x, y) + d(y, z) ≥ d(x, z).

(b) i. All values of d(x, y) are either 0 or 1, so d(x, y) ≥ 0.ii. If x = y, then d(x, y) = d(x, x) = d(y, x). If x = y, then d(x, y) = 1 =

d(y, x).iii. If x = z, then d(x, y) + d(y, z) ≥ 0 = d(x, z). Otherwise, there are the

following cases:Case 1: x = y = z. Then d(x, y) + d(y, z) = 2 ≥ 1 = d(x, z).Case 2: x = y = z. Then d(x, y) + d(y, z) = 1 = d(x, z).Case 3: x = y = z. Then d(x, y) + d(y, z) = 1 = d(x, z).

(c) i. d((x, y), (z, w)) =√

(x − z)2 + (y − w)2 ≥ 0.ii. (x, y) = (z, w) iff x = z and y = w iff (x − z)2 and (y − w)2 = 0 iff√

(x − z)2 + (y − w)2 = 0.

iii. d((x, y), (z, w)) =√

(x − z)2 + (y − w)2 =√

(z − x)2 + (w − y)2 =d((z, w), (x, y)).

iv. By the triangle inequality,

d((x, y), (u, v)) + d((u, v), (w, z)) =√

(x − u)2 + (y − v)2

+√

(u − w)2 + (v − z)2

≥√

(x − w)2 + (y − z)2

= d((x, y), (w, z).

(d) i. d((x, y), (z, w)) = |x − z| + |y − w| ≥ 0.ii. (x, y) = (z, w) iff x = z and y = w iff |x − z| = 0 = |y − w| iff

|x − z| + |y − w| = 0.iii. d((x, y), (z, w)) = |x− z|+ |y −w| = |z −x|+ |w −y| = d((z, w), (x, y)).iv. By the triangle property of absolute value, |x − z| ≤ |x − u| + |u − z|

and |y − w| ≤ |y − v| + |v − w|, so d((x, y), (z, w) = |x − z| + |y − w| ≤|x − u| + |u − z| + |y − v| + |v − w| = d((z, w), (x, y)).

17. (a) nm. For each of the n elements of A, choose one of the m elements of B asits image.

(b) mn. Choose one of the n elements of A, and then one of the m elements ofB as the image.

(c)(m2

)n2. Choose 2 elements from A for the first coordinates; each may be

assigned any element of B as image.

(d)∑m

r=0

(mr

)nr. This is

∑mr=0 (# of functions with r elements in the domain).

18. (a) i. Let x ∈ A. Since f(x) = f(x), xTx. Thus T is reflexive on A.ii. Let x, y ∈ A and suppose xTy. Then f(x) = f(y), so f(y) = f(x).

Thus hTx and T is symmetric.

4 FUNCTIONS 88

iii. Let x, y, z ∈ A and suppose xTy and yTz. Then f(x) = f(y) andf(y) = f(z), so f(x) = f(z). Thus xTz and T is transitive.

(b) [0] = {0}, [2] = {−2, 2}, [4] = {−4, 4}.

(c) [0] = {kπ: k ∈ Z}, [π/2] = {π/2 + kπ: k ∈ Z}, [π/4] = {π/4 + 2kπ: k ∈ Z}.

19. (a) F. The claim is false. The two functions are not equal because they havedifferent domains. The “proof” neglects to consider the crucial case x = 0.

(b) A.

(c) F. A visual inspection using the Vertical Line Test does not prove therelation is a function.

(d) F. The claim is false. The “proof” fails to consider two possibilities:(x, y) ∈ h, and (x, z) ∈ g, or (x, y) ∈ g and (x, z) ∈ h. If y = z, thenh ∪ g is not a function.

(e) A.

4.2 Constructions of Functions

1. (a) (f ◦ g)(x) = 17 − 14x(g ◦ f)(x) = −29 − 14x

(b) (f ◦ g)(x) = 4x2 + 8x + 3(g ◦ f)(x) = 2x2 + 4x + 1

(c) (f ◦ g)(x) = sin(2x2 + 1)(g ◦ f)(x) = 2 sin2 x + 1

(d) (f ◦ g)(x) = tan(sinx)(g ◦ f)(x) = sin(tanx)

(e) (f ◦ g)(x) = {(k, r), (t, r), (s, l)}(g ◦ f)(x) = ∅

(f) (f ◦ g)(x) = {(1, 2), (2, 5), (4, 5), (5, 2)}(g ◦ f)(x) = {(1, 7), (3, 4), (4, 3), (5, 3)}

(g) (f ◦ g)(x) = x2+2x2+4

(g ◦ f)(x) = (x+1)2

(x+2)2 + 1

(h) (f ◦ g)(x) = 3|x| + 2(g ◦ f)(x) = |3x + 2|

(i) (f◦g)(x) =

{ 2x + 1, if x ≤ −1−2x, if −1 < x < 0−x + 1, if x ≥ 0

(g◦f)(x) =

{ 2x + 2, if x ≤ −2−x − 1, if −2 < x ≤ 0−2x, if x > 0

(j) (f◦g)(x) ={

(7 − 2x)2, if x ≤ 2(x + 1)2, if x > 2

(g◦f)(x) =

⎧⎨⎩

1 − 4x, if x ≤ − 12

2x + 4, if − 12 < x < 3

x2 + 1, if x ≥ 3

2. (a) Dom(f ◦ g) = R = Rng(f ◦ g) = Dom(g ◦ f) = Rng(g ◦ f)

(b) Dom(f ◦ g) = R = Dom(g ◦ f) Rng(f ◦ g) = Rng(g ◦ f) = [−1,∞)

(c) Dom(f ◦ g) = R, Rng(f ◦ g) = [−1, 1] Dom(g ◦ f) = R, Rng(g ◦ f) = [1, 3]

(d) Dom(f ◦g) = R, Rng(f ◦g) = [tan(−1), tan(1)] Dom(g◦f) = R− [π2 +kπ],

Rng(g ◦ f) = [−1, 1]

(e) Dom(f ◦g) = {k, t, s}, Rng(f ◦g) = {r, l} mathrmDom(g◦f) = ∅, Rng(g◦f) = ∅

4 FUNCTIONS 89

(f) Dom(f◦g) = {1, 2, 4, 5}, Rng(f◦g) = {2, 5} Dom(g◦f) = {1, 3, 4, 5}, Rng(g◦f) = {3, 4, 7}

(g) Dom(f ◦ g) = R, Rng(f ◦ g) = [ 12 , 1) Dom(g ◦f) = R−{−2}, Rng(g ◦f) =[1,∞)

(h) Dom(f ◦ g) = R, Rng(f ◦ g) = [2,∞) Dom(g ◦ f) = R, Rng(g ◦ f) = [0,∞)

(i) Dom(f ◦ g) = R, Rng(f ◦ g) = (−∞, 2) Dom(g ◦ f) = R, Rng(g ◦ f) =(−∞, 1)

(j) Dom(f ◦ g) = R, Rng(f ◦ g) = [9,∞) Dom(g ◦ f) = R, Rng(g ◦ f) = [3,∞)

3. (a) f(x) = x2, g(x) = 3x + 7f(x) = (x + 7)2, g(x) = 3x

(b) f(x) =√

x, g(x) = 2x2 − 5f(x) =

√2x − 5, g(x) = x2

(c) f(x) = sinx, g(x) = |2x + 4|f(x) = sin |x|, g(x) = 2x + 4

4. (a) (k ◦ k)([0]) = (k(k([0])) = k([2]) = [2].(k ◦ k)([1]) = (k(k([1])) = k([0]) = [2].(k ◦ k)([2]) = (k(k([2])) = k([2]) = [2].(k ◦ k)([3]) = (k(k([3])) = k([0]) = [2].

(b) (g ◦ f)(0) = (g(f(0)) = g([2]) = 4 = h(0).(g ◦ f)(1) = (g(f(1)) = g([3]) = 6 = h(1).(g ◦ f)(2) = (g(f(2)) = g([4]) = 0 = h(2).(g ◦ f)(3) = (g(f(3)) = g([5]) = 2 = h(3).(g ◦ f)(4) = (g(f(4)) = g([6]) = 4 = h(4).(g ◦ f)(5) = (g(f(5)) = g([7]) = 6 = h(5).(g ◦ f)(6) = (g(f(6)) = g([0]) = 0 = h(6).(g ◦ f)(7) = (g(f(7)) = g([1]) = 2 = h(7).

(c) (f ◦ g)([0]) = (f(g([0])) = f(2) = [2] = k([0]).(f ◦ g)([1]) = (f(g([1])) = f(0) = [0] = k([1]).(f ◦ g)([2]) = (f(g([2])) = f(2) = [2] = k([2]).(f ◦ g)([3]) = (f(g([3])) = f(0) = [0] = k([3]).

(d) (h ◦ (h(◦h))(0) = h((h(h(0))) = h((h(4)) = h(4) = 4.(h ◦ (h(◦h))(1) = h((h(h(1))) = h((h(6)) = h(0) = 4.(h ◦ (h(◦h))(2) = h((h(h(2))) = h((h(0)) = h(4) = 4.(h ◦ (h(◦h))(3) = h((h(h(3))) = h((h(2)) = h(0) = 4.(h ◦ (h(◦h))(4) = h((h(h(4))) = h((h(4)) = h(4) = 4.(h ◦ (h(◦h))(5) = h((h(h(5))) = h((h(6)) = h(0) = 4.(h ◦ (h(◦h))(6) = h((h(h(6))) = h((h(0)) = h(4) = 4.(h ◦ (h(◦h))(7) = h((h(h(7))) = h((h(2)) = h(0) = 4.

5. (a) f−1(x) = x−25

(b) Not a function.

(c) f−1(x) = 1−2xx−1

(d) Not a function.

(e) f−1(x) = −3 + lnx

(f) f−1(x) = 11−x

(g) f−1(x) = x−1x

(h) f−1(x) = 3 − x

(i) f−1(x) = 4x3x+1

4 FUNCTIONS 90

6. Dom(IB ◦ f) =Dom(f) by Theorem 4.2.1. Suppose x ∈ A. Then (IB ◦ f)(x) =IB(f(x)) = f(x). Therefore IB ◦ f = f .

7. Suppose f : A → B, Rng(f) = C and f−1 is a function.Then Dom(f ◦ f−1) =Dom(f−1) = Rng(f) = C =Dom(IC).Suppose x ∈ C. Then (f ◦ f−1)(x) = f ◦ (f−1(x)) = x = IC(x).

8.

Rng(f |N) = {x ∈ Z: x = 1(mod 3) and x ≤ 1}.

9. (a) {(x, y) ∈ R×R: y = 0 if x < 0 and y = x2 if x ≥ 0} {(x, y) ∈ R×R: y = x2}(b) {(x, y) ∈ R×R: y = 3} {(x, y) ∈ R×R: y = 3 if x ∈ Z and y = 2 if x /∈ Z}(c) {(x, y) ∈ R × R: y = −x} {(x, y) ∈ R × R: y = |x − 1| − 1}

10. Suppose f and g are functions. Suppose (x, y) ∈ f ∩ g. Then (x, y) ∈ f and(x, y) ∈ g, so f(x) = y = g(x). Let A = {x: g(x) = f(x)}. Then x ∈ A,so (x, y) ∈ g|A. Now let (x, y) ∈ g|A. Then in particular, (x, y) ∈ g andf(x) = g(x) = y, so (x, y) ∈ f . Thus (x, y) ∈ f ∩ g. So f ∩ g is a function.

11. Suppose h and g are functions such that Dom(f) = A, Dom(g) = B, andA ∩ B = ∅. Then h and g are relations, so h ∪ g is a relation. If x ∈Dom(h ∪ g),then there is a y such that (x, y) ∈ h or (x, y) ∈ g, so x ∈ A or x ∈ B. On theother hand, if x ∈ A ∪ B, then there is a y such that (x, y) ∈ h or (x, y) ∈ g, so(x, y) ∈ h ∪ g and so x ∈Dom(h ∪ g). Thus Dom(h ∪ g) = A ∪ B.

Now suppose (x, y) and (x, z) ∈ h ∪ g. By the preceding argument, x ∈ A ∪ B,and since A ∩ B = ∅, x can’t be in both A and B. If x ∈ A, then y = h(x) = zand if x ∈ B, then y = g(x) = z. Either way, y = z and so h ∪ g is a function.

12. Dom(f) = D =Dom(i) =Dom(g ◦ i). Suppose x ∈ D. Then f(x) = g|D(x) =g(x) = g(i(x)) = (g ◦ i)(x).

13. Suppose h∪g: A∪C → B ∪D. Then evidently Dom(h|E) = E =Dom(g|E). Nowif x ∈ E, then h|E(x) = h(x) and g|E(x) = g(x). Since (x, h(x)) and (x, g(x))are both in h ∪ g, and h ∪ g is a function, we must have that h(x) = g(x). Thush|E = g|E . Now suppose h|E = g|E . We claim that h ∪ g = h ∪ (g|C−E). To seethis, first notice that by Theorem 4.2.5, g|E ∪ g|C−E is a function with domainC. In fact, it is easy to see that g|E ∪ g|C−E = g. Thus

h ∪ g = h ∪ (g|E ∪ g|C−E)= (h ∪ h|E) ∪ g|C−E (since g|E = h|E)= h ∪ g|C−E . (since h|E ⊆ h)

Now, again by Theorem 4.2.5, h∪(g|C−E) is a function with domain A∪(C−E) =A ∪ C, so h ∪ g is as well. Finally, if (x, y) ∈ h ∪ g, then y ∈ B or y ∈ D, soh ∪ g: A ∪ C → B ∪ D.

4 FUNCTIONS 91

14. (a) Function. (b) Function.

(c) Function. (d) Not a function.

(e) Not a function.

15. (a) f×g is a relation from A×C to B×D since it is a subset of (A×C)×(B×D).

i. Dom(f × g) = A × C. Let (a, c) ∈ A × C. Then a ∈ A and c ∈ C, sothere exists b ∈ B and d ∈ D such that (a, b) ∈ f and (c, d) ∈ g. Thus((a, c), (b, d)) ∈ f × g so (a, c) ⊆Dom(f × g).

ii. Assume ((a, c), (b1, d1)) ∈ f × g and ((a, c), (b2, d2)) ∈ f × g. Then(a, b1) ∈ f and (a, b2) ∈ f and (c, d1) ∈ g and (c, d2) ∈ g. Thus b1 = b2and d1 = d2, and therefore (b1, d1) = (b2, d2).

(b) If (a, c) ∈ A × C, then (f × g)(a, c) = (f(a), g(c)).

16. (a) Suppose x, y ∈ R and x < y. Then 3x < 3y, and 3x− 7 < 3y − 7. Thereforef(x) < f(y).

(b) Suppose x, y ∈ R and x < y. Then −5x > −5y, so 2−5x > 2−5y. Thereforeg(x) > g(y).

(c) Suppose 0 ≤ x < y. Then x ·x < y ·x, and y ·x < y ·y, so x2 < y2. Thereforeh(x) < h(y).

(d) Suppose x and y are in (−3,∞) and x < y. Then 3x < 3y and −y < −x,so 3x − y < 3y − x. Thus xy + 3x − y − 3 < xy + 3y − x − 3 or(x − 1)(y + 3) < (y − 1)(x + 3). Using the fact that x and y are in (−3,∞)we know that x + 3 and y + 3 are positive. Dividing both sides of thelast inequality by x + 3 and y + 3, we conclude x−1

x+3 < y−1y+3 . Therefore

f(x) < f(y).

17. (a) True. Suppose f(x) = ax + b where a > 0, and let c < d. The ac < ad andac + b < ad + b, so f(c) < f(d).

(b) False. Consider f(x) = g(x) = −x on R.

4 FUNCTIONS 92

(c) The statement is true if we assume that g(w) ∈ I for all w ∈ I. Supposethis is true, x, y ∈ I and x < y. Then g(x) > g(y) and g(x), g(y) ∈ I, sof(g(x)) < f(g(y)). Therefore f ◦ g(x) < f ◦ g(y).The statement may be false otherwise. Consider f(x) = −

√x, g(x) = −x

and I = [0, 1]. Then f ◦ g is not defined on I.

(d) False. Consider f(x) = (x + 1)(x − 1)2.

(e) False. Consider f(x) ={

1x , if x < 0−x, if x ≥ 0.

18. (a) We show that f1 + f2 is a function with domain R. First, f1 + f2 is bydefinition a relation. For all x ∈ R there is some u ∈ R such that (x, u) ∈ f1because f1: R → R and there is some v ∈ R such that (x, v) ∈ f2 becausef2: R → R. Then (x, u + v) ∈ f1 + f2, so x ∈Dom(f1 + f2). It is clearfrom the definition of f1 + f2 that x ∈Dom(f1 + f2) implies x ∈ R, soDom(f1 + f2) = R. Now let x ∈ R. Suppose (x, c) and (x, d) are in f1 + f2.Then c = f1(x) + f2(x) = d. Therefore f1 + f2 is a function.We show that f1 · f2 is a function with domain R. First f1 · f2 is a relationby definition. Now let x ∈ R. Then there exits u and v in R such that(x, u) ∈ f1 and (x, v) ∈ f2, so (x, uv) ∈ f1 ·f2, so x ∈Dom(f1 ·f2). It is clearfrom the definition of f1 ·f2, that Dom(f1 ·f2) ⊆ R; hence Dom(f1 ·f2) = R.Now suppose (x, c) and (x, d) are both in f1 ·f2. Then c = f1(x) ·f2(x) = d,so f1 · f2 is a function.

(b) (f + g)(x) = 11 − 5x, (f · g)(x) = −14x2 − 23x + 30(f + h)(x) = 3x2 − 5x + 7, (f · h)(x) = −21x3 + 67x2 − 56x + 12

(c) False. Let f(x) = x, g(x) = 2x + 3, h(x) = −5x.

19. Let g be the constant function defined by g = {(a, c): a ∈ R}. Then by Exercise18, g · f is a function with domain R and g · f = {(a, cd): (a, c) ∈ g and(a, d) ∈ f} = {(a, cd): (a, d) ∈ f} = cf . Also, cd ∈ R for all d ∈Rng(f), socf : R → R.

20. (a) A.

(b) C. It is not necessary to refer to the idea of cancellation. The “proof” canbe corrected by saying simply that IA = IA ◦ f = f .

(c) F. The first “=” is not justified, because g ◦f = f ◦ g is not true in general.

(d) A.

4.3 Functions That Are Onto; One-to-One Functions

1. (a) Onto R. Let w ∈ R. Then 2(w−6) ∈ R and f(2(w−6)) = 12 [2(w−6)]+6 = w.

(b) Onto Z. Let n ∈ Z. Then 1000 − n ∈ Z and f(1000 − n) = −(1000 − n) +1000 = n.

(c) Not onto N × N, since (5, 8) /∈Rng(f).

(d) Onto R. Let y ∈ R. Then 3√

y ∈ R and f( 3√

y ) = ( 3√

y )3 = y.

(e) Not onto R. f(x) ≥ 0 for all x ∈ R, so −1 /∈Rng(f).

(f) Not onto R. f(x) ≥ 0 for all x ∈ R, so −1 /∈Rng(f).

(g) Not onto R. sin x ∈ [−1, 1] for all x ∈ R, so 47 /∈Rng(f).

(h) Onto R. Let x ∈ R. Then (x, 0) ∈ R× R and f(x, 0) = x − 0 = x.

(i) Onto [1,−1]. f is continuous, f(0) = 1 and f(π) = −1, so the IntermediateValue Theorem says that f attains every value between 1 and −1.

4 FUNCTIONS 93

(j) Onto [1,∞). Let y ∈ [1,∞). Then√

y − 1 ∈ R and f(√

y − 1 ) = y.

(k) Onto [0,∞). First, if x ∈ [2, 3), then x − 2 ≥ 0 and 3 − x > 0, sof(x) = x−2

3−x ≥ 0. This shows that Rng(f) ⊆ [0,∞). To see that is onto[0,∞), let w ∈ [0,∞). Choose x = 3w+2

w+1 . Then f(x) = w. Note thatw + 1 > 0 because w ≥ 0, and 2w + 2 ≤ 3w + 6, so 2 ≤ 3w+2

w+1 < 3. Thisensures that x ∈ [2, 3).

(l) Onto [0,∞). First, if x ∈ (1,∞), then x− 1 > 0 and x > x− 1, so xx−1 > 1.

Therefore Rng(f) ⊆ (1,∞). Let w ∈ (1,∞). Then x = ww−1 ∈ (1,∞) and

f(x) = w.

2. (a) One-to-one. Suppose f(x) = f(y). Then 12x + 6 = 1

2y + 6, so x = y.

(b) One-to-one. Suppose f(x) = f(y). Then −x + 1000 = −y + 1000, so x = y.

(c) One-to-one. Suppose f(x) = f(y). Then (x, x) = (x, y) so x = y.

(d) One-to-one. Suppose f(x) = f(y). Then x3 = y3, so x = y.

(e) Not one-to-one. f(1) = f(−1).

(f) One-to-one. Suppose f(x) = f(y)). Then 2x = 2y, so log2(2x) = log2(2y),so x = y.

(g) Not one-to-one. sin(π6 ) = sin(13π

6 ).

(h) Not one-to-one. f(1, 0) = f(2, 1).

(i) Not one-to-one. f(0) = f(2π).

(j) One-to-one. Suppose x, y ∈ [1,∞) and f(x) = f(y). Then x2 + 1 = y2 + 1,so x2 = y2. Since x, y > 0, x = y.

(k) One-to-one. Suppose x, z ∈ [2, 3) and f(x) = f(z). Then x−23−x = z−2

3−z , sox = z.

(l) One-to-one. Suppose x, y ∈ (1,∞) and f(x) = f(y). Then xx−1 = y

y−1 , sox = y.

3. (a) Let a ∈ A. Then IA(a) = a, so a ∈ Rng(IA). Therefore IA maps onto A.

(b) Let f : Z to Zs be the canonical map. Let [x] ∈ Zs. Then x ∈ Z andf(x) = [x]. Thus [x] ∈ Rng(f), so f maps onto Zs.

(c) Let s ∈ Z. Then x ∈ R and∫

x = x. Thus [x] ∈ Rng(int), so f maps ontoZ.

(d) The sequence a is not onto N. The number 3 ∈ N, but there is no n ∈ N

such that an = 2n = 3. Thus 4 /∈ Rng(a).

4. (a) B = {0, 3}, f = {(1, 0), (2, 3), (3, 0), (4, 0)}(b) B = N, f : A → B is the inclusion map.

(c) B = A, f = IA

(d) B = A, f(x) = 1 for all x ∈ A.

5. Suppose f : A onto−→ B and g: B onto−→ C. Let c ∈ C. Then there is b ∈ B such thatg(b) = c, since g is onto B. Also there is an a ∈ A such that f(a) = b since f isonto A. Thus (g ◦ f)(a) = g(f(a)) = g(b) = c, which proves that g ◦ f is onto C.

6. Suppose f : A → B, g: B → C and g◦f : A 1-1−→ C. Suppose also that x, y ∈ A andf(x) = f(y). Since g is a function, (g ◦ f)(x) = g(f(x)) = g(f(y)) = (g ◦ f)(y).Since g ◦ f is one-to-one, x = y. Therefore f is one-to-one.

7. (a) Suppose f : A 1-1−→ B and C ⊆ A. Let x, y ∈ C be such that f |C(x) = f |C(y).Then f(x) = f(y), so x = y.

4 FUNCTIONS 94

(b) Suppose h: A onto−→ C, g: B onto−→ D and A ∩ B = ∅. Then by Theorem 4.2.5,h∪g: A∪B → C ∪D. Suppose x ∈ C ∪D. Then x ∈ C or x ∈ D. If x ∈ C,then there is an a ∈ A such that (h ∪ g)(a) = h(a) = x, since h is onto.Likewise, if x ∈ D, then there is a b ∈ B such that (h ∪ g)(b) = g(b) = x.In either case, there is a y ∈ A ∪ B such that (h ∪ g)(y) = x.

8. (a) Let f : R → R be given by f(x) = 2x and g: R → R be given by g(x) = x2.Then f maps onto R, but g ◦ f is not onto R.

(b) A = {1}, B = {1, 2} = C, f = {(1, 1)} and g = {(1, 2), (2, 1)}.

(c) A = {1}, B = {1, 2}, C = {1}, f = {(1, 1)} and g = {(1, 1), (2, 1)}.

(d) A = {1, 2} = B, C = {1}, f = {(1, 2), (2, 1)} and g = {(1, 1), (2, 1)}.

(e) A = {a, b, c}, B = {1, 2, 3}, C = {x, y, z}, f = {(a, 2), (b, 2), (c, 3)} andg = {(1, x), (2, y), (3, z)}.

(f) A = {1, 2}, B = {a, b, c}, C = {5, 6}, f = {(1, a), (2, b)} and g ={(a, 5), (b, 6), (c, 6)}.

9. (a) Let g(x) = 2 − x and h(x) = 1x . Then g|(−∞,1]: (−∞, 1] 1-1−→ [1,∞) and

h|(1,∞)(1,∞) 1-1−→ (0, 1). Since f = g|(−∞,1] ∪ h|(1,∞), f : R 1-1−→ (0,∞)by Theorem 4.3.5(c). Finally, since there is no real number x such thatf(x) = −1,−1 ∈ Rng(a). Therefore f is not onto R.

(b) Let g(x) = x+4, h(x) = −x and i(x) = x−4. Then g|(−∞,−2]: (−∞,−2] 1-1−→(−∞, 2], h|(−2,2): (−2, 2) 1-1−→ (−2, 2) and i|[2,∞): [2,∞) 1-1−→ [−2,∞). Since

f = g|(−∞,−2] ∪ h|(−2,2) ∪ i|[2,∞), f : R onto−→ R by Theorem 4.3.5(b). Finally,f(4) = f(−4), so f is not one-to-one.

(c) We verify that f maps onto R as follows: 1 ∈Rng(f), because f(4) = 1. Forw = 1, choose x = 4w−2

w−1 . Then f(x) = w.

To show that f is one-to-one, suppose f(x) = f(w). Then x−2x−4 = w−2

w−4 forx = 4 and w = 4. Therefore xw − 2w − 4x + 8 = xw − 2x − 4w + 8, fromwhich we derive x = w. We must also consider whether f(x) might be 1even if x = 4. But if x = 4 and f(x) = x−2

x−4 = 1, then x − 2 = x − 4 and so−2 = −4. This is impossible. Therefore f is one-to-one.

(d) f(10) = f(−7), so f is not one-to-one, and −4 /∈Rng(f), so f is not ontoR.

10. Suppose f : R → R and f is increasing. (The proof for decreasing is similar.) Leta, b ∈ R be such that a = b. Without loss of generality, say a < b. Then, sincef is increasing, f(a) < f(b), so f(a) = f(b). Therefore f is one-to-one.

11. (a) f is not a surjection because [1] has no preimage in Z4. Consider: Iff(x) = [1], then [2x] = [1], so 2x ≡8 1. But then 8 divides the odd number2x − 1, which is impossible.To show that f is an injection, suppose f(x) = f(z). Then 2x =8 2z, so2x ≡8 2z. Therefore 8 divides 2x − 2z, so 4 divides x − z and thus x = z.

(b) Let [y] ∈ Z2. Then y ∈ Z4 and f( y ) = [3y] = [y], since 2 divides 3y − y.Thus f is a surjection.f(0) = f(2). Therefore f is not one-to-one.

(c) Let x and y be in Z6 and suppose x + 1 = y + 1. Then 6 divides (x + 1) −(y + 1), so 6 divides x − y, so x = y. Thus f is one-to-one.Let z ∈ Z6. Then z − 1 ∈ Z6 and f( z − 1 ) = z, so f is onto Z6.

4 FUNCTIONS 95

(d) f(0) = f(2), so f is not one-to-one.1 /∈Rng(f), so f is not onto Z4.

12. (a) x(n) = 1 for all n ∈ N

x(n) = (10 − n)2 + 2 for all n ∈ N

(b) x(n) = n for all n ∈ N

x(n) ={

50 − n, if n < 50n, if n ≥ 50

(c) x(n) = 2n for all n ∈ N

x(n) = n + 10000 for all n ∈ N

(d) x(n) = |n − 5| + 1 for all n ∈ N

x(n) ={

n, if n < 50n − 1, if n ≥ 50

13. (a) n!(n−m)! . Choosing an image for each of the elements of A, one at a time, weget n(n − 1) · · · (n − m + 1) = n!

(n−m)! choices in all.

(b) n! For each element of A, we choose an image from the elements of B thathave not already been chosen.

(c) None. If m > n, there is no one-to-one function from A to B.(d) None. If m < n, there is no function from A onto B.(e) n! For each element of B, choose a pre-image from the elements of A that

have not already been chosen.(f) Since m = n + 1, exactly one element of B has two pre-images. We can

select the element of B in n ways, and the two pre-images in(n+1

2

)ways,

and then assign each of the remaining n − 1 elements of B as the image ofexactly one of the n − 1 remaining elements of A in (n − 1)! ways. Thusthere are n

(n+1

2

)(n − 1)! = n!

(n+1

2

)functions from A onto B.

(g) n! For each element of A, choose as image an element of A that has notalready been chosen.

14. (a) F. To prove that f is onto we must show that R ⊆Rng(f). The proof showsonly that Rng(f) ⊆ R.

(b) F. The claim is false. Dom(f) = [1,∞), so for all x ∈ Dom(f), f(x) ≤ 1.(c) F. The proof shows that Rng(f) ⊆ C, not that C ⊆ Rng(f).(d) A. (A direct proof is easier.)(e) A.(f) A.(g) F. The proof verifies that the relation f is a function.(h) F. The claim if false (see part (i)). One must suppose f(x, y) = f(s, t) for

(x, y) and (s, t) in I × I.(i) A.

4.4 One-to-One Correspondences and Inverse Functions

1. (a) Let x, y ∈ (2,∞). Suppose f(x) = f(y). Then −xx−2 = −y

y−2 . Thus −x(y−2) =−y(x − 2), since neither x nor y is 2. Therefore, −xy − 2x = −yx − 2y, so−2x = −2y, which implies x = y. Thus f is one-to-one.Let z ∈ (−∞,−1). Choose t = 2z

z+1 . From z < −1, we have z +1 < 0. Thus,since z < z + 1, z

z+1 > z+1z+1 = 1 and t = 2z

z+1 > 2 · 1 = 2, so so t ∈ (2,∞).

Then f(t) = −tt−2 =

− 2zz+1

2zz+1 −2 = −2z

−2 = z. Thus f is onto (−∞,−1).

4 FUNCTIONS 96

(b) Let x, y ∈ (−∞,−4). Suppose g(x) = g(y). Then −|x + 4| = −|y + 4|.Thus −(−(x + 4)) = −(−(y + 4)), since x < −4 and y < −4. Therefore,x + 4 = y + 4, so x = y. Thus g is one-to-one.Let z ∈ (−∞, 0). Choose t = z − 4. From z < 0, we have t < −4 and thust + 4 < 0. Thus, since z < z + 1, z

z+1 > z+1z+1 and t = 2z

z+1 > 2 · 1 = 2, sot ∈ (2,∞). Then g(t) = −|t + 4| = −|(z − 4) + 4| = −|z| = −(−z) = z.Thus g is onto (−∞, 0).

(c) Then 1.25x = 1.25y. Thus x = y and g is one-to-one.Let z ∈ 10N. Then z = 10k, for some natural number k. Choose t = 8k.Then t ∈ 8N and h(t) = 1.25t = 1.25(8k) = 10k = z. Thus h is onto 10N.

(d) Let (x, y), (m, n) ∈ N×N. Suppose G(x, y) = G(m, n). Then 2x+2(2y−1) =2m+2(2n − 1). Since 2y − 1 and 2n − 1 are odd, 2x+2 = 2m+2. Thusx + 2 = m + 2, so x = m. It follows that 2y − 1 = 2n − 1, so y = n.Therefore (x, y) = (m, n), so g is one-to-one.Let z ∈ 8N. Then z = 8k, for some natural number k. Write k ask = 2rq, with nonnegative integer r and odd natural number q. Thenq = 2t− 1 for some natural number t. Let s = r +1. Then (s, t) ∈ NxN andG(s, t) = 2s+2(2t − 1) = 2r+3(q) = 8(2rq) = 8k = z. Thus G is onto 8N.

(e) Let x, y ∈ R. Suppose k(x) = k(y). Then 2ex + 3 = 2ey + 3. Thus ex = ey.Since the exponential function is one-to-one, x = y. Thus k is one-to-one.Let z ∈ (3,∞). Choose t = ln z−3

2 . Since z > 3, z−32 is a positive number

and so t ∈ R. Then k(t) = 2et + 3 = 2e( z−32 ) + 3 = 2( z−3

2 ) + 3 = z. Thus kis onto (3,∞).

2. (a) Let h = {(a, 2), (b, 4), (c, 8), (d, 16), (e, 32), (f, 64)}. In all cases, if h(x) =h(y)then x = y, so h is one-to-one. Rng(h) = {2, 4, 8, 16, 32, 64} so h is onto{2, 4, 8, 16, 32, 64}.

(b) Let f : N → N − {1} be given by f(n) = n + 1. If f(n) = f(m) for somen, m ∈ N, then n + 1 = m + 1, so n = m. Thus f is one-to-one.Let k ∈ N − {1}. Since k = 1, k − 1 is a natural number and f(k − 1) =(k − 1) + 1 = k. Thus k ∈ Rng(f), so f is onto N − {1}.

(c) Let f : (3,∞) → (5,∞) be given by f(x) = x + 2. If f(x) = f(y) for somex, y ∈ (3,∞), then x + 2 = y + 2, so x = y. Thus f is one-to-one.Let z ∈ (5,∞). Since z > 5, z − 2 is a real number greater than 3 andf(z − 2) = (z − 2) + 2 = z. Thus z ∈ Rng(f), so f is onto (5,∞).

(d) Let f : (−∞, 1) → (−1,∞) be given by f(x) = −x. If f(x) = f(y) for somex, y ∈ (−∞, 1), then −x = −y, so x = y. Thus f is one-to-one.Let z ∈ (−1,∞). Since z > −1, −z is a real number less than 1 andf(−z) = −(−z) = z. Thus z ∈ Rng(f), so f is onto (5,∞).

(e) Let f : 12N → 20N be given by f(x) = 5x/3 for x ∈ 12N. If f(x) = f(y) forsome x, y ∈ 12N, then 5x/3 = 5y/3, so x = y. Thus f is one-to-one.Let z ∈ 20N. Since z is a multiple of 20, > 5, 3z/5 is a multiple of 12. Thenf(3z/5) = 5(3z/5)/3 = z. Thus z ∈ Rng(f), so f is onto 20N.

3. (a) Let h be the inverse of f . Then (x, y) ∈ hiff(y, x) ∈ f iffx = 1y iffy = 1

x .Therefore h = 1

x . To verify that this formula is correct, suppose x ∈ (0,∞).Then (h ◦ f)(x) = h(f(x)) = h( 1

x ) = 11x

= x. Since the composite is the

identity function on the domain of f , we conclude that h = f−1.

4 FUNCTIONS 97

(b) Let h be the inverse of g. Then

(x, y) ∈ h iff (y, x) ∈ g

iff x =4y

x + 2iff xy + 2x = 4y

iff y(x − 4) = −2x

iff y =2x

4 − x, for x < 4.

Therefore h(x) = 2x4−x . To verify that this formula is correct, suppose

x > −2. Then

(h ◦ g)(x) = h(g(x))

= h

(4x

x + 2

)

=2 4x

x+2

4 − 4xx+2

=8x

x+24(x+2)−4x

x+2= x.

Since the composite is the identity function on the domain of g, we concludethat h = g−1.

(c) Let g be the inverse of h. Then (x, y) ∈ g iff(y, x) ∈ h iffx = ey+3 iffy +3 =ln(x) iffy = ln(x) − 3. Therefore g(x) = ln(x) − 3. To verify that thisformula is correct, suppose x ∈ R. Then (g ◦ h)(x) = g(h(x)) = g(ex+3) =ln(ex+3) − 3 = (x + 3) − 3 = x. Since the composite is the identity functionon the domain of h we conclude that g = h−1.

(d) Let h be the inverse of G. Then

(x, y) ∈ h iff (y, x) ∈ G

iff x =5(y − 1)y − 3

iff xy − 3x = 5y − 5iff xy − 5y = 3x − 5

iff y =3x − 5x − 5

.

Therefore h = 3x−5x−5 . To verify that this formula is correct, suppose

x ∈ (3,∞). Then

(h ◦ f)(x) = h(f(x))

= h

(5(x − 1)x − 3

)

=3

(5(x−1)

x−3

)− 5

5(x−1)x−3 − 5

=15x−15−5x+15

x−35x−5−5x+15

x−3

=10x

10= x.

4 FUNCTIONS 98

Since the composite is the identity function on the domain of G, we concludethat h = G−1.

4. Suppose F : A → B and F−1 is a function. Then by the first part of Theorem4.4.2 and the fact that (F−1)−1 = F is a function, F−1 is one-to-one.

5. (a) Assume that F : A 1-1−→onto

B and G: B → A. Suppose that G = F−1. Then by

part (a) of Theorem 4.4, G◦F = IA and F ◦G = IB . Therefore G◦F = IA

or F ◦ G = IB .

Now suppose G ◦ F = IA or F ◦ G = IB . Since, F : A 1-1−→onto

B, F−1: B 1-1−→onto

A

by Theorem 4.4.3. If G ◦ F = IA, then G = G ◦ IB = G ◦ (F ◦ F−1) =(G ◦ F ) ◦ F−1 = IA ◦ F−1 = F−1. On the other hand, if F ◦ G = IB , thenG = IA ◦ G = (F−1 ◦ F ) ◦ G = F−1 ◦ (F ◦ G) = F−1 ◦ IB = F−1. In eithercase, G = F−1.

(b) A = {1, 2}, B = {3, 4, 5}, F = {(1, 3), (2, 4)}, G = {(3, 1), (4, 2), (5, 2)}.

6. Let F : A → B and G: B → A. Suppose G◦F = IA and F ◦G = IB . Let x, y ∈ Aand suppose F (x) = F (y). Then F is one-to-one because

x = IA(x) = (G ◦ F )(x) = G(F (x)) = G(F (y)) = (G ◦ F )(y) = IA(y) = y.

Let b ∈ B. Then b = IB(b) = (F ◦ G)(b) = F (G(b)). Therefore b ∈ Rng(F ).Therefore F is onto B.

Let x, y ∈ B and suppose G(x) = G(y). Then G is one-to-one because

x = IB(x) = (F ◦ G)(x) = F (G(x)) = F (G(y)) = (F ◦ G)(y) = IB(y) = y.

Let a ∈ A. Then a = IA(a) = (G ◦ F )(a) = G(F (a)). Therefore a ∈ Rng(G).Therefore G is onto A.

7. (a) f−1 (b) ln ◦f (c) (ln ◦f)−1 = f−1 ◦ ln−1 = f−1 ◦ EXP

8. Suppose f : A 1-1−→onto

B, g: B 1-1−→onto

C, and h: C 1-1−→onto

D. Then g ◦ f : A 1-1−→onto

by

Theorem 4.4.1. and, again by Theorem 4.4.1, h ◦ g ◦ f : A 1-1−→onto

D. Therefore,

(h ◦ g ◦ f)−1: D 1-1−→onto

A. By Theorem 3.1.3(d), (h ◦ g ◦ f)−1 = ((h ◦ g) ◦ f)−1 =

f−1◦(h◦g)−1 = f−1◦g−1◦h−1. Since f−1◦g−1◦h−1 = (h◦g◦f)−1, f−1◦g−1◦h−1

is a one-to-one correspondence from D to A.

9. (a) IC = [1234567]

(b) u = [5624173]

(c) v = [2546173]

(d) w = [2143576]

(e) u ◦ v = [6147532]

(f) v ◦ u = [1756234]

(g) w ◦ w = [1234567] = IC

(h) u ◦ v ◦ w = [1674523]

(i) u−1 = [5374126]

(j) v−1 = [5173246]

(k) (u ◦ v)−1 = [2763514]

(l) v−1 ◦ u−1 = (u ◦ v)−1 = [2763514]

4 FUNCTIONS 99

(m) u−1 ◦ v−1 = [1567342]

10. (a) F. The claim is false. The proof assumes g◦f = f◦g, which is not necessarilytrue.

(b) A.

(c) F. The claim is false. The “proof” assumes that r−1 ◦s−1 = (r◦s)−1, whichis not usually true. The correct inverse of r ◦ s is [42135].

(d) A. The proof would be more aesthetically pleasing if it were shortened.

4.5 Images of Sets

1. (a) f(∅) = ∅, f({1}) = {4}, f({2}) = {4}, f({3}) = {5}, f({1, 2}) ={4}, f({1, 3}) = {4, 5}, f({2, 3}) = {4, 5}, f(A) = {4, 5}

(b) f−1(∅) = ∅, f−1({4}) = {1, 2}, f−1({5}) = {3}. f−1({6}) = ∅,f−1({4, 5}) = A, f−1({5, 6} = {3}, f−1({4, 6}) = {1, 2}, f−1(B) = A

2. (a) [2, 10] (b) [1, 2] ∪ [5, 17](c) {0} (d) [−

√2,

√2 ]

(e) [−3,−2] ∪ [2, 3] (f) [−5,−4] ∪ [−2, 2] ∪ [4, 5]

3. (a) {−5,−3,−1, 1, 3} (b) {z ∈ Z: z < 0 and z is odd}(c) R (d) [−2. − 1

2 ](e) [−7,−1) (f) [−7,−5]

4. (a) (2,∞) (b) [2, 5.2](c) [2 −

√3, 3−

√5

2 ) ∪ ( 3+√

52 , 2 +

√3] (d) ∅

(e) (−∞,−2] ∪ [2,∞) (f) [− 103 ,−2] ∪ [2, 10.1]

5. (a) {54, 108, 216, 162, 324, 648} (b) {(1, 1)}

6. (a) Let a = 2 and D = (−∞, 1). Then a /∈ D, f(a) = 4, and 4 ∈ f(D) = (0,∞).

(b) Let A = (−1, 1) and C = (0, 2). Then f(A∩C) = f(0, 1) = (0, 2) = (0, 4) =f(A) ∩ f(C).

(c) Let D = [1, 2]. Then f(D) = [2, 4] and f−1(f(D)) = [−1, 0] ∪ [1, 2] = D.

(d) Let E = [−2, 2]. Then f−1(E) = [0, 1] and f(f − 1(E)) = (0, 2] = E.

7. (a) No solution.

(b) i. Suppose b ∈ f(C ∪D). Then b = f(a) for some a ∈ C ∪D. Then a ∈ Cand b = f(a) or b ∈ D and b = f(a). Thus b ∈ f(C) or b ∈ f(D), sob ∈ f(C) ∪ f(D).

ii. Suppose b ∈ f(C) ∪ f(D). Then b ∈ f(C) or b ∈ f(D). Thus b = f(a)for some a, and a ∈ C ∪ D. Therefore b ∈ f(C ∪ D).

(c) No solution.

(d)

a ∈ f−1(E ∪ F ) iff f(a) ∈ E ∪ F

iff f(a) ∈ E or f(a) ∈ F

iff a ∈ f−1(E) or a ∈ f−1(F )iff a ∈ f−1(E) ∪ f−1(F )

Therefore f−1(E ∪ F ) = f−1(E) ∪ f−1(F ).

4 FUNCTIONS 100

8. (a) Suppose b ∈ f(⋂

α∈Δ Dα). Then b = f(a0 for some a ∈⋂

α∈Δ Dα. Thusa ∈ Dα for every α ∈ Δ. Since b = f(a), b ∈ f(Dα) for every α ∈ Δ. Thusb ∈

⋂α∈Δ f(Dα). Therefore f(

⋂α∈Δ Dα) ⊆

⋂α∈Δ f(Dα).

(b) Suppose b ∈ f(⋃

α∈δ Dα). Then b = f(a) for some a ∈⋃

α∈δ Dα, so there isan α ∈ δ such that b ∈ f(Dα). Thus b ∈

⋃α∈δ f(Dα). On the other hand,

suppose b ∈⋃

α∈δ f(Dα). Then b ∈ f(Dα) for some α, so there is an a ∈ Dα

such that b = f(a). Since a ∈⋃

α∈δ Dα, b ∈ f(⋃

α∈δ Dα).

(c)

a ∈ f−1

⎛⎝ ⋂

β∈Γ

Eβ

⎞⎠ iff f(a) ∈

⋂β∈Γ

Eβ

iff for all β ∈ Γ, f(a) ∈ Eβ

iff for all β ∈ Γ, a ∈ f−1(Eβ)

iff a ∈⋂β∈Γ

f−1(Eβ).

(d)

a ∈ f−1(⋃β∈Γ

Eβ) iff f(a) ∈⋃β∈Γ

Eβ

iff for some β ∈ Γ, f(a) ∈ Eβ

iff for some β ∈ Γ, a ∈ f−1(Eβ)

iff a ∈⋃β∈Γ

f−1(Eβ).

9. Let f be given by f(x) = x2, D1 = (−∞, 0] and D2 = [0,∞). Then f(D1∩D2) =f({0}) = {0}, but f(D1) ∩ f(D2) = [0,∞).

10. (a) Suppose b ∈ f(f−1(E)). Then there ks a ∈ f−1(E) such that f(a) = b.Since a ∈ f−1(E), f(a) ∈ E. But f(a) = b so b ∈ E. Thereforef(f−1(E)) ⊆ E.

(b) Suppose a ∈ A − f−1(E). Then f(a) ∈ B − E, so a ∈ f−1(B − E).

(c) Suppose a ∈ f−1(B − E). Then f(a) ∈ B − E. That is, f(a) ∈ B andf(a) /∈ E. Therefore, a ∈ A and a /∈ f−1(E). (This is the contrapositive ofthe statement a ∈ f−1(E) ⇒ f(a) ∈ E.) Thus a ∈ A − f−1(E). Therefore,f−1(B − E) ⊆ A − f−1(E).

(d) First, suppose E = f(f−1(E)). Suppose b ∈ E. Then b ∈ f(f−1(E)). Thusthere is a ∈ f−1(E) such that b−f(a), so b ∈Rng(f). Therefore E ⊆Rng(f).Now assume E ⊆Rng(f). We know by part (a) that f(f−1(E)) ⊆ E,so to prove equality we need only E ⊆ f(f−1(E)). Suppose b ∈ E.Then b ∈Rng(f), so b = f(a) for some a ∈ A. Since b = f(a) ∈ E,a ∈ f−1(E). Thus b = f(a) and a ∈ f−1(E) so b ∈ f(f−1(E)). ThereforeE ⊆ f(f−1(E)).

(e) Suppose d ∈ D. Then f(d) ∈ f(D), so d ∈ f−1(f(D)).

(f) Suppose f−1(f(D)) = D, and let b ∈ f(A−D). Then there is an a ∈ A−Dsuch that f(a) = b, so obviously b ∈ B. If b ∈ f(D) as well, then by oursupposition, f−1({b}) ⊆ D, so a ∈ D, a contradiction. Thus b ∈ B − f(D).Now suppose f(A−D) ⊆ B−f(D). By part (c) D ⊆ f−1(f(D)), so supposed ∈ f−1(f(D)). Then f(d) ∈ f(D), so by our assumption f(d) /∈ f(A−D).Then d /∈ A − D. Thus d ∈ D.

4 FUNCTIONS 101

11. (a) Suppose f(X) ⊆ U and let x ∈ X . Then f(x) ∈ f(X) ⊆ U , so x ∈ f−1(U).Now suppose X ⊆ f−1(U). Then f(X) ⊆ f(f−1(U)) ⊆ U , by Exercise10(a).

(b) Suppose t ∈ f(X) − f(Y ). Then t ∈ f(X), so there exists x ∈ X such thatf(x) = t. We note x /∈ Y since t = f(x) /∈ f(Y ). Thus x ∈ X −Y . Thereforet = f(x) ∈ f(X − Y ).

(c) a ∈ f−1(U) − f−1(V ) iff f(a) ∈ U and f(a) /∈ V iff f(a) ∈ U − V iffa ∈ f−1(U − V ).

12. Let X, Y ⊆ A. Then f(X ∩ Y ) ⊆ f(X) ∩ f(Y ) by Theorem 4.5.1(a). Supposeb ∈ f(X)∩ f(Y ). Then f(x) = b = f(y) for some x ∈ X and y ∈ Y , and x mustequal y, since f is one-to-one. Thus x ∈ X ∩ Y and b = f(x), so b ∈ f(X ∩ Y ).

The converse is true. To prove that f is one-to-one, let x = y in A. Then{x} ∩ {y} = ∅ and thus f({x} ∩ {y}) = ∅. By the hypothesis, f({x} ∩ {y}) =f({x}) ∩ f({y}). Thus f(x) = f(y).

13. Suppose f : A 1-1−→ B and X ⊆ A. Then f(A) − f(X) ⊆ f(A − X) by Exercise11(b). Suppose b ∈ f(A − X). Then b = f(a) for some a ∈ A − X . Now ifb ∈ f(X), then b = f(x) for some x ∈ X , so f(x) = f(a) even though x = a.But this is impossible since f is one-to-one, so b ∈ f(A) − f(X).

14. Suppose f(X) = Y . Then X ⊆ f−1(Y ) by Exercise 11(a). Suppose a ∈ f−1(Y ).Then f(a) ∈ Y = f(X), so f(a) = f(x) for some x ∈ X . Thus x = a since f isone-to-one, so a ∈ X . Therefore f−1(Y ) ⊆ X.

Now suppose X = f−1(Y ). Then f(X) ⊆ Y by Exercise 11(a). Suppose b ∈ Y .Since f is onto, there is an a ∈ A such that f(a) = b ∈ Y . Thus a ∈ f−1(Y ) = X.Therefore b = f(a) ∈ f(X), and so Y ⊆ f(X).

15. (a) If f is one-to-one, then the induced function is one-to-one.

(b) If f is onto B, then f : P(A) onto−→ P(B).

16. Suppose x ∈ f(f−1(K)). Then x = f(a) for some a ∈ f−1(K), so x ∈Rng(f)∩K.Therefore f(f−1(K)) ⊆Rng(f)∩K. Now suppose x ∈Rng(f)∩K. Then there isan a ∈ A such that f(a) = x ∈ K, so a ∈ f−1(K). Then x = f(a) ∈ f(f−1(K)).Thus Rng(f) ∩ K ⊆ f(f−1(K)).

17. Suppose x ∈ A. Then f(x) = f(x), so x R x.

Suppose x R y for some x, y ∈ A. Then f(x) = f(y), so f(y) = f(x). Thus y R x.

Suppose x R y and y R z for some x, y, z ∈ A. Then f(x) = f(y) and f(y) = f(z).Therefore f(x) = f(z), so x R z.

Each equivalence class is the set of all elements in A that map to the sameelement in B. The is one equivalence class for each element of Rng(f).

18. (a) F. The claim is false. We cannot conclude x ∈ X from f(x) ∈ f(X).

(b) A.

(c) F. The claim is false. This is an example of the misuse of quantifiers. Forevery α ∈ Δ there is an x ∈ Dα, but this does not imply that there is an xsuch that x ∈ Dα for all α.

4 FUNCTIONS 102

4.6 Sequences

1. (a) 25 , 3

7 , 49 , 5

11 , 613 (b) 1, 0, −1, 0, 1 (c) 1, 1

2 , 16 , 1

24 , 1120

(d) 12 , 3

4 , 78 , 15

16 , 3132 (e) 1

2 , 12 , 3

4 , 32 , 15

4

2. (a) Converges to 12 . (b) Does not converge.

(c) Converges to 0. (d) Converges to 1.(e) Does not converge.

3. (a) Does not exist. (b) 0 (c) 45

(d) Does not exist. (e) 0 (f) 35

(g) e2 (h) 1e (i) 0

(j) Does not exist. (k) Does not exist. (1) 0(m) 0 (n) 1

4. (a) The terms grow larger without bound so the sequence diverges. We provethat the limit is K. Let ε > 0. Choose N = 1. Suppose n ∈ N and n > N .Then n > 1 and cn = K.

(b) A sequence of integers converges to 2 iff there exists M ∈ N such that ifn > M , then xn = 2.

5. (a) The terms grow larger without bound so the sequence diverges. Assume thelimit is L. Let ε = 1 and let N ∈ N. Suppose n is greater than both N andlog2(|L|+1). Then 2n > |L| + 1 so |2n − L| ≥ 2n − |L| > 1 = ε.

(b) The sequence converges to 1. Let ε > 0. Choose N so that N > 1ε . Suppose

n > N . Then |xn − 1| = 1n < 1

N < ε.

(c) The terms grow larger without bound so the sequence diverges. Assumelimn→∞ n2 = L. Let ε = 1 and choose N ∈ N. Suppose n is greater thanboth N and L + 1. Then n2 ≥ n > L + 1, so |n2 − L| = n2 − L > 1.

(d) The terms are alternately near 12 and − 1

2 so the sequence diverges. Assumethe limit is L. Note that for all n ∈ N, |xn| = n

2n+1 > n3n = 1

3 . Letε = 1

3 , and let N ∈ N. If L ≥ 0, then for odd n > N , |xn − L| =| (−1)n2n +1 − L| = n

2n+1 + L ≥ 13 = ε. If L < 0, then for even n > N ,

|xn − L| = | n2n +1 − L| = n

2n+1 + |L| ≥ 13 = ε.

(e) The sequence converges to 0. Let ε > 0. Choose N so that N > 1ε . Suppose

n > N . Then |xn − 0| = | cos n|n ≤ 1

n < 1N < ε. Therefore limn→∞ xn = 0.

(f) The sequence converges to 0. Let ε > 0. Choose N so that N > 1(2ε)2 .

Suppose n > N . Then |xn − 0| = |√

n + 1 −√

n| = 1√n+1+

√n

< 12√

n<

12√

N< ε.

(g) The sequence converges to −7. Let ε > 0. Choose N so that N >√

21ε .

Suppose n > N . Then n >√

21ε , so 21

n2 < ε. Then |xn − (−7) = | 7(1−n2)n2+2 −

(−7)| = 21n2+2 < 21

n2+2 < 21n2 < ε. Therefore, limn→∞

7(1−n2)n2+2 = −7.

(h) The sequence converges to 0. Choose N such that N > log2(6/ε). Supposen > NN . Then n > log2(6/ε), so 2π > 6/ε. Therefore |6/2π−0| = 6/2π < ε.

(i) The sequence converges to 0. Choose N so that N > 5000/varepsilon.Suppose n > N . Then n! > n > 5000/ε. Therefore |5000/n! − 0| =5000/n! < ε.

4 FUNCTIONS 103

(j) Each term is either −1, 0 or 1, so the sequence diverges. Assume the limitis L. Let ε = 1

2 and let N ∈ N. Suppose that n > N .

If L ≥ 0, then for n of the form 4k + 3, k ∈ N, xn = sin (4k+3)π)2 =

sin(2kπ + 3π2 ) = −1.

Thus |xn − (L)| ≥ 1 > ε.

If L < 0, then for n of the form 4k + 1, k ∈ N, xn = sin( 4k+1)π2 =

sin(2kπ + π2 ) − 1. Thus, |xn − (L)| ≥ 1 > ε.

(k) The sequence converges to 0. Let ε > 0. Choose N so that kN > 1ε .

Suppose n > N . An easy induction argument shows n! ≤ nn−1, so|xn − 0| = n!

nn ≤ nn−1

nn = 1n < 1

N < ε.

(l) The term grow larger without bound so the sequence diverges. Let M ∈ N.Choose n so that n is greater than both 2e and lnM . Then ln(n

2 ) > 1, soln(n

2 )n = n ln(n2 ) > n > lnM , so (n

2 )n > M . Thus x is an unboundedsequence, so x diverges.

6. (a) Let ε > 0. Then ε2 > 0. Since xn → L, there exists N1 ∈ N such that if

n > N1, then |xn − L| < ε2 . Likewise, there exists N2 ∈ N such that if

n > N2, then |yn − M | < ε2 . Let N3 =max{N1, N2} and assume n > N3.

Then |(xn +yn)− (L+M)| = |(xn −L)+(yn −M)| ≤ |xn −L|+ |yn −M | <ε2 + ε

2 = ε. Therefore, xn + yn → L + M .

(b) Let ε > 0. Choose N1 so that n > N1 implies |xn − L| < ε2 and pick N2 so

that n > N2 implies |yn − M | < ε2 . Let N3 =max{N1, N2}. If n > N3, then

|(xn − yn) − (L − M)| = |(xn − L) + (M − yn)| ≤ |xn − L| + |M − yn| =|xn − L| + |yn − M | ≤ ε

2 + ε2 = ε. Note: If part (c) has been proved, then

part (b) follows by applying part (a) to the sequences xn and −yn.

(c) Let ε > 0. Choose N so that n > N implies |xn − L| < ε|r| . Now suppose

n > N . Then | − xn − (−L)| = | − xn + L| = |xn − L| < ε. Note: This is aspecial case of part (d).

(d) Let ε > 0. Choose N so that N < n implies |xn − L| < ε|r| . Now suppose

n > N . Then |rxn − rL| = |r||xn − L| < ε.

(e) Let ε > 0. Choose B so that |yn| ≤ B for all n ∈ N. Pick N1 so that|xn−L| < ε

2B for all n > N1, and N2 so that |yn−M | < ε2|L| for all n > N2.

Let N =max{N1, N2}. Suppose n > N . Then |xnyn−LM | = |xnyn−Lyn+Lyn−LM | ≤ |xn−L||yn|+|L||yn−M | ≤ |xn−L|B+|L||yn−M | < ε

2+ ε2 = ε.

(f) Let ε > 0. Choose N so that n > N implies |xn − L| < ε. Suppose n > N .Then ||xn| − |L|| ≤ |xn − L| < ε.

7. (a) Suppose xn → L = 0. Then |xn| → |L| by Exercise 6(f). Thus there existsN such that n > N implies ||xn| − |L|| < |L|

2 . Then −|L|2 < |xn| − |L|, so

|L|2 < |xn| whenever n > N .

(b) Suppose xn → L = 0, xn = 0 for all n, and yn → M . By part (a) thereexists N1 so that |xn| > |L|

2 for all n > N1. Let ε > 0. Then there existsN2 so that n > N2 implies |xn − L| < εL2

4|M | and N3 so that n > N3 implies

|yn − M | < ε|L|4 .

Let N =max{N1, N2, N3}. Suppose n > N . Then∣∣∣∣ yn

xn− M

L

∣∣∣∣ =1

|L||xn| |Lyn − Mxn|

≤ 1|L||xn| (|Lyn − LM | + |LM − Mxn|)

4 FUNCTIONS 104

<2L2 (|L||yn − M | + |M ||L − xn|)

<2L2

(L2

4ε +

L2

4ε

)

= ε.

8. (a) x2n = 22n , x2n+1 = 0

(b) Suppose xn → L. Let ε > 0. Choose N so that n > N implies |xn − L| < ε.Then yn = xn+N determines a subsequence such that |yn − L| < ε for alln ∈ N.

(c) Suppose xn → L and (xf(n)) is a subsequence. Let ε > 0 and choose Nso that n > N implies |xn − L| < ε. Then f−1(N) ∈ N and n > f−1(N)implies f(n) > N , so |xf(n) − L| < ε.

(d) Suppose (xf(n)) and (xg(n)) are subsequences of (xn) such that xf(n) → Mand xg(n) → L where L = M . Assume further that xn → K. At least oneof M or L is not equal to K. Therefore (xn) has a subsequence that doesnot converge to K, violating part (c). Thus (xn) diverges.

9. (a) x1 = 10, x2 = 5, x3 = 52 , x4 = 5

4 , x5 = 58 , x6 = 5

16 . The sequence convergesto 0.

(b) x1 = 1, x2 = 0, x3 = 1, x4 = 0, x5 = 1, x6 = 0. The sequence diverges.

(c) f1 = 1, f2 = 1, f3 = 2, f4 = 3, f5 = 5, f6 = 8. f7 = 13. f8 = 21, f9 = 34,f10 = 55.

(d) The sequence diverges because terms oscillate between 3 and 2/3.

(e) The sequence converges to 4.

10. (a) F. The claim is false. One reason the “proof” fails is that the inequalityinvolving |xn − (L

2 )| + |yn − (L2 )|is reversed.

(b) A. The proof uses Exercise 6(b).

5 Cardinality

5.1 Equivalent Sets; Finite Sets

1. Reflexive: For any set A, IA: A 1-1−→onto

A, so A ≈ A.

Symmetric: For sets A and B, suppose A ≈ B. Then there is a bijectionf : A 1-1−→

ontoB. Then f−1: B 1-1−→

ontoA, so B ≈ A.

Transitive: Let A, B and C be sets such that A ≈ B and B ≈ C. Then thereare bijections f : A 1-1−→

ontoB and g: B 1-1−→

ontoC, so g ◦ f : A 1-1−→

ontoC. Thus A ≈ C.

2. (a) finite (b) infinite (c) finite(d) infinite (e) infinite (f) finite(g) finite (h) finite (i) finite(j) infinite (k) infinite (l) infinite(m) finite (n) finite (o) finite(p) finite

3. Suppose x, y ∈ (a, b) and f(x) = f(y). Then ( d−cb−a )(x − a) = ( d−c

b−a )(y − a), sox = y. Therefore f is one-to-one. {An alternate proof: f is linear function withpositive slope, so f is increasing (by Exercise 17(a) of Section 4.2). Therefore fis one-to-one, by Exercise 10 of Section 4.3.

Now suppose y ∈ (c, d). Then b−ad−c (y − c) + a ∈ (a, b) and f( b−a

d−c (y − c) + a) = y.

4. Suppose (a1, b1), (a2, b2) ∈ A×B and f(a1, b1) = f(a2, b2). Then (h(a1), g(b1)) =(h(a2), g(b2)), so h(a1) = h(a2) and g(b1) = g(b2). Since h and g are one-to-one,this means that a1 = a2 and b1 = b2, so (a1, b1) = (a2, b2). Thus f is one-to-one.

Suppose (c, d) ∈ C × D. Then c = h(a) and d = g(b) for some a ∈ A and b ∈ Bsince h and g are onto. Therefore (a, b) ∈ A×B and f(a, b) = (h(a), g(b)) = (c, d).Thus f is onto C × D.

5. Suppose A ≈ ∅. Then there is a bijection f : A → ∅. But codomain(f) = ∅ impliesDom(f) = ∅. Hence A = ∅.

6. (a) Suppose A is finite. Since A ∩ B ⊆ A, A ∩ B is finite by Theorem 5.1.6.

(b) Suppose A is infinite, A ⊆ B and B is finite. Then by Theorem 5.1.6, A isfinite. This is a contradiction. Thus B is infinite.

7. A ∪ B = (A − B) ∪ (A ∩ B) ∪ (B − A), a union of disjoint sets each of whichis finite by Theorem 5.1.6. By two applications of Theorem 5.1.6, A ∪ B =A − B +A ∩ B +B − A. Now A = (A−B)∪ (A∩B) and B = (B −A)∪ (A∩B),so by Theorem 5.6, A = A − B + A ∩ B and B = B − A + A ∩ B.

Thus A ∪ B = A − B + A ∩ B + B − A + A ∩ B − A ∩ B = A + B − A ∩ B.

8. If A1 is finite, then⋃1

i=1 Ai = A1 is finite.

Suppose that⋃n

i=1 Ai is finite whenever A1, A2, . . ., An are finite sets for somen ∈ N, and consider the finite sets A1, A2, . . ., An+1. By our supposition,

⋃ni=1 Ai

is finite, so⋃n+1

i=1 Ai =⋃n

i=1 Ai ∪ An+1 is finite.

Hence, by the PMI, the statement is true for all n.

9. (a) Define f : A → A × {x} by: f(a) = (a, x). Then f is one-to-one because(a1, x) = (a2, x) implies a1 = a2, and f is onto because f(a) = (a, x) forany (a, x) ∈ A × {x}.

105

5 CARDINALITY 106

(b) Suppose A and B are finite. If A or B is empty, then A × B is empty, soA × B is finite. Now assume that A and B are nonempty. For each b ∈ B,A × {b} ≈ A by part (a) and is therefore a finite set. By Theorem 5.1.7(c),A × B =

⋃b∈B(A × {b}) is finite.

10. Let A and B be finite and BA be the set of all funtions from A to B. Becausea function from A to B is a subset of A × B, BA is a subset of P(A × B), thepower set of A × B. Since A and B are finite, A × B is finite, and, therefore (byTheorem 2.1.4) P(A × B) is finite. Since BA is a subset of a finite set, BA isfinite.

Alternate Proof. Assume A and B are finite. If A or B is empty, thenthe only function from A to B is the empty function; thus BA = {∅} isfinite. Suppose g: A 1-1−→

ontoNn and h: B 1-1−→

ontoNm. Define F : BA → Nmn as:

F (f) = 1+∑n

i=1 mi−1(h−1(f(g−1(i)))−1) for all f ∈ BA. Then F is a bijection,so BA is finite.

11. (a) Impossible (b) For each i ∈ N, let Ai = {5, 6}.(c) Impossible (d) A = {1, 2}, B = {2, 3}

12. Assume A is finite, B is infinite, and B − A is finite. Then B = A ∪ (B − A) isfinite by Theorem 5.1.7(a), which contradicts our assumption. Therefore B − Ais infinite.

13. Let x ∈ Nr, and define f : Nr − {x} → Nr−1 by:

f(n) ={

n, if n < xn − 1, if n > x

for all n ∈ Nr − {x}. Then f is a bijection, so Nr − {x} ≈ Nr−1.

14. Let B be a proper subset of a finite set A. Suppose f : A 1-1−→onto

Nk. Then

f |B : B 1-1−→onto

f(B), where f(B) is a proper subset of Nk. Now suppose that B ≈ A,

and let g: B 1-1−→onto

A. Then f |B ◦ g−1 ◦ f−1: Nk1-1−→onto

f(B), so Nk is equivalent toone of its proper subsets, but this is impossible.

15. Suppose f : S 1-1−→onto

Nm and g: S 1-1−→onto

Nn. Then f ◦ g−1: Nn1-1−→onto

Nm and

g◦f−1: Nm1-1−→onto

Nn, so by the pigeonhole principle, n ≤ m and m ≤ n. Thereforen = m.

16. (a) True.

Proof. Suppose C is infinite and C = A ∪ B. If A and B are both finite,then by Theorem 5.1.7, C is finite. Therefore, at least one of A or B isinfinite.

(b) True.

Proof. Suppose A is a set, p /∈ A and A ≈ A ∪ {p}. If A is finite and hascardinality k, then by Lemma 5.1.4, A ∪ {p} is finite and has cardinalityk + 1. Thus A ≈ A ∪ {p} is false. This is a contradiction. Therefore A isinfinite.

17. (a) If n = 1, then r ∈ N and r < n is impossible. If n = 2 and r < n, thenr = 1. Suppose f : Nr → Nn. Then f is constant, so f is not onto Nn.

5 CARDINALITY 107

(b) Suppose the statement is true for some n > 1. Let r < n + 1. If r = 1,then f : Nr → Nn is not onto, so we may assume 1 < r < n + 1. Supposef : Nr

onto−→ Nn+1. Then (Nn+1 −{f(r)}) ⊆Rng(f |Nr−1) and by Lemma 5.1.8,

there is a g: (Nn+1 − {f(r)}) 1-1−→onto

Nn, so g ◦ f |Nr−1 : Nronto−→ Nn. This

contradicts our induction hypothesis.

(c) By the PMI, for every n, if r < n there is no function from Nr onto Nn.

18. (a) Suppose f is one-to-one. Then f : A 1-1−→onto

f(A), so f(A) ≈ A ≈ B and f(A)

is a subset of the finite set B. Therefore f(A) = B, so f is onto.

(b) Suppose f is not one-to-one. Then A is not empty, and since A and B arefinite and equivalent, there are functions h: A onto−→ Nn and g: B onto−→ Nn forsome n ∈ N. Thus g ◦ f ◦ h−1: Nn

onto−→ Nn but g ◦ f ◦ h−1 is not one-to-one.In particular, for u �= v in A such that f(u) = f(v), h(u) �= h(v) in Nn

but g ◦ f ◦ h−1(h(u)) = g(f(u)) = g(f(v)) = g ◦ f ◦ h−1(h(v)). Let G =g ◦f ◦h−1 −{(h(u)), g(f(u))}. Then G: Nn −{g(f(u))} onto−→ Nn. By Lemma5.1.8, there is a bijection F : Nn−1 → Nn −{y}, so G◦F : Nn−1

onto−→ Nn. Thiscontradicts Exercise 17, so f must be one-to-one.

19. If the domain of a function is empty, then its range is empty. Suppose for somen ∈ N ∪ {0}, that any function whose domain has cardinality n has finite range.Let f : A → B, A = n + 1, and g: A 1-1−→

ontoNn+1. Pick (x, y) ∈ f . By Lemma

5.1.8 there is a bijection h: Nn+1 − {g(x)} → Nn, so h ◦ g|A−{x}: A − {x} 1-1−→onto

Nn, so A − {x} = n. By our hypothesis, f |A−{x} has a finite range. SinceRng(f) =Rng(f |A−{x}) ∪ {y}, Rng(f) is finite, by Lemma 5.1.3. By the PMI,for every function f with finite domain with cardinal number n, the range of fis finite, for every n ≥ 0.

20. Suppose A and B are finite, A = m, B = n, m > n, and f : A → B is one-to-one.Then there exists a one-to-one (and onto) function g: Nm → A and a one-to-one(and onto) function h: B → Nn. See the diagram below. Then h ◦ f ◦ g is a one-to-one function from Nm to Nn by Theorem 4.3.4. But m > n so the conclusionthat h ◦ f ◦ g is a one-to-one contradicts Theorem 5.1.9.

(ART TO COME)

21. (a) Let A be the set of people in Solomeo and B be the set of calendar datesin a year. Assume A = 400 and B = 365. Then N400 ≈ A and B ≈ N65so there exist bijections f : N400 → A and g: B → N365. If no two residentshave the same birthday, then there is a one-to-one function h: A → B. Butthen g ◦ h ◦ f : N400 → N365 is one-to-one.This contradicts the Pigeonhole Principle. Therefore, every function fromA to B is not one-to-one. Thus, no matter what the birthday assignmentsmay be, at least two residents will be maped to the same birthday.

(b) The largest possible sum of 10 elements of N99 is 90 + 91 + · · · + 99 = 945.Let S be a subset of N99 with 10 elements. There are at most 945 possiblesums of the elements of subsets of S, and 210 −1 = 1,023 nonempty subsetsof S, so a least two subsets of S have identical sums for their elements. Ifthe two subsets are not disjoint, delete the common elements from bothsets. The result is two disjoint subsets of S whose elements have the samesums.

5 CARDINALITY 108

22. (a) F. f ∪h is not a one-to-one correspondence between A∪B and Nm+n. Someelements of Nm+n (such as 1) have preimages in both A and B, and m + nhas no preimages. Furthermore, the “proof” neglects the special case thatA or B may be empty.

(b) C. In Case 2, it is not correct that Nk ∪ N1 ≈ Nk+1. In fact, Nk ∪ N1 = Nk.Also, if x ∈ Nk, then S ∪ {x} �≈ Nk ∪ {x} = Nk.

(c) F. The claim is false. Let A = N, and B = ∅. Then A × B = ∅ is finite butA is infinite. If we assume that B �= ∅, the proof is correct. The proof failswhen B = ∅ because it is not possible to choose and element of B.

(d) F. The claim is false. Theorem 5.1.7 does not apply to the union of aninfinite family of sets.

5.2 Infinite Sets

1. Assume that A is infinite and A ≈ B. Suppose B is finite. This contradictsTheorem 5.1.3. Therefore B is infinite.

2. (a) Define f : N → A by f(n) 1n , for each n ∈ N. The function f is one-to-one

because if m, n ∈ N and f(m) = f(n), then 1m = 1

n and so m = n. Thefunction f is onto A because if a ∈ A, then a = 1

k for some k ∈ N andf(k) = 1

k = a. Therefore, A ≈ N and so A is infinite.(b) Define f : N → N − N15 by f(n) = n + 15, for each n ∈ N. The function f

is one-to-one because if m, n ∈ N and f(m) = f(n), then m + 15 = n + 15and so m = n. The function f is onto N − N15 because if a ∈ N − N15, thena > 15 and so a = k +15 for some k ∈ N and f(k) = k +15 = a. Therefore,N ≈ N − N15 and so N − N15 is infinite.

(c) Note that the interval (0, 0.0005) is a proper subset of (0, 0.001). Definef : (0, 0.0005) → (0, 0.001) by f(x) = 2x. The function f is one-to-onebecause if x, y ∈ (0, 0.0005) and f(x) = f(y), then 2x = 2y and sox = y. The function f is onto (0, 0.001) because if a ∈ (0, 0.001), thena/2 ∈ (0, 0.0005) and f(a/2) = 2(a/2) = a. Therefore, (0, 0.001) isequivalent to a proper subset and is infinite.

(d) Note that the interval (0, 1) is a proper subset of (0,∞). Define f : (0, 1) →(0, ∞) by f(x) = x

1−x . The function f is one-to-one because if x, y ∈ (0, 1)and f(x) = f(y), then x

1−x = y1−y . Then x − xy = y − xy and so x = y.

The function f is onto (0,∞) because if a ∈ (0,∞), then t = aa+1 is a

real number between 0 and 1 and f(t) = f( aa+1 ) =

aa+1

1− aa+1

= a. Therefore,(0,∞) is equivalent to a proper subset and is infinite.

3. (a) Let f : N → D+ be given by f(n) = 2n − 1 for each n ∈ N. To see that fis one-to-one, suppose f(x) = f(y). Then 2x − 1 = 2y − 1, which impliesx = y. Also, f is onto D+ since if d is odd, then d = 2r − 1 for some r ∈ N.But then f(r) = d.

(b) Define f : 3N → N by f(x) = x3 . Then f is one-to-one because x

3 = y3

implies x = y. Also, f is onto N because for any n ∈ N, 3n ∈ 3N andf(3n) = n.

(c) Define f : 3Z → Z by f(x) = x3 . Then f is one-to-one because x

3 = y3 implies

x = y. Also, f is onto Z because for any n ∈ Z, 3n ∈ T and f(3n) = n.Thus T ≈ Z ≈ N by Theorem 5.2.2.

(d) Define f : {n ∈ N: n > 6} → N by f(x) = x − 6. Then f is one-to-onebecause x − 6 = y − 6 implies x = y. Also, f is onto N because for anyn ∈ N, n + 6 ∈ {n ∈ N: n > 6} and f(n + 6) = n.

5 CARDINALITY 109

(e) Define f : {n ∈ Z: n < −12} → N by f(x) = −x − 12. Then f is one-to-onebecause −x − 12 = −y − 12 implies x + 12 = y + 12, which implies x = y.Also, f is onto N because for any n ∈ N, −n − 12 ∈ {n ∈ Z, n < −12} andf(−n − 12) = −(−n − 12) − 12 = n.

(f) Define f : N − {5, 6} → N as f(x) ={

x, if x < 5x − 2, if x > 6.

Then f is one-to-one. Suppose x, y ∈ N − {5, 6} and f(x) = f(y). Iff(x) < 5, then x = f(x) = f(y) = y. If f(x) ≥ 5, then x − 2 = y − 2, sox = y. Also, f is onto. If n < 5, then n ∈ N − {5, 6} and f(n) = n and forany n ≥ 5, n + 2 ∈ N − {5, 6} and f(n + 2) = (n + 2) − 2 = n.

(g) Let G = {(x, y) ∈ N × R: xy = 1}. Consider the projection π1: G → N

defined by π1(x, y) = x. Then π1 is one-to-one. If (x1, y1), (x2, y2) ∈ Gand π1(x1, y1) = π1(x2, y2), then x1 = x2, so y1 = 1

x1= 1

x2= y2, and

thus (x1, y1) = (x2, y2). Also, π1 is onto. Let n ∈ N. Then (n, 1n ) ∈ G and

π1(n, 1n ) = n.

(h) Define f : {x ∈ Z: x ≡ 1(mod 5)} → Z by f(x) = x−15 . Then f is one-to-one

and onto Z. Suppose x, y ∈ 1/ ≡5 and f(x) = f(y). Then x−15 = y−1

5 , sox = y. If z ∈ Z, then 5z + 1 ≡ 1(mod 5), and f(5z + 1) = 5z+1−1

5 = z.

4. (a) Define f : (1,∞) → (0, 1) by f(x) = 1x . Then f is one-to-one because f is

decreasing. Also, f is onto (0, 1) because for any x ∈ (0, 1), 1x ∈ (1,∞) and

f( 1x ) = x.

(b) Define f : (a,∞) → (1,∞) by f(x) = x−a+1. Then f is clearly a bijection,so by part (a), (a,∞) ≈ (1,∞) ≈ (0, 1).

(c) Define f : (−∞, b) → (−b, ∞) by f(x) = −x. Then f is clearly a bijection,so by part (b), (−∞, b) ≈ (−b, ≈) ≈ (0, 1).

(d) Define f : (0, 1) → [1, 2) ∪ (5, 6) by f(x) ={

5 + 2x, if x ∈ (0, 12 )

2x, if x ∈ [ 12 , 1).

Then f is onto. Let x ∈ [1, 2) ∪ (5, 6). If x ∈ [1, 2), then 12x ∈ [ 12 , 1)

and f( 12x) = 2( 1

2x) = x. If x ∈ (5, 6), then x−52 ∈ (0, 1

2 ) and f(x−52 ) =

5 + 2(x−52 ) = x.

Also, f is one-to-one. Let x, y ∈ (0, 1) and suppose f(x) = f(y). Iff(x) ∈ [1, 2), then 2x = f(x) = f(2y) = y, so x = y. If f(x) ∈ (5, 6),then 5 + 2x = f(x) = f(y) = 5 + 2y, so x = 5.

(e) Define f : (0, 1) → (3, 6) ∪ [10, 20) by f(x) ={

3 + 6x, if x ∈ (0, 12 )

20x, if x ∈ [ 12 , 1).

Then f is onto. Let x ∈ (3, 6) ∪ [10, 20). If x ∈ (3, 6), then x−36 ∈ (0, 1

2 ) ⊆(), 1) and f(x−3

6 ) = 3 + 6(x−36 ) = x. If x ∈ [10, 20), then x

20 ∈ [ 12 , 1) andf( x

20 ) = 20( x20 ) = x.

Also, f is one-to-one. Let x, y ∈ (0, 1) and suppose f(x) = f(y). Iff(x) ∈ (3, 6), then 3 + 6x = f(x) = f(y) = 3 + 6y, so x = y. Iff(x) ∈ [10, 20), then 20x = f(x) = f(y) = 20y, so x = y.

(f) Let A = (0, 1] ∪ (2, 3] ∪ (4, 5). Define f : (0, 1) → A by

f(x) =

⎧⎨⎩

3x if x ∈ 0 < x ≤ 13

3x + 1 if 13 < x ≤ 2

3 .3x + 2 if 2

3 < x < 1

The function of f is one-to-one because if , 7 ∈ (0, 1) and f(x) = f(y),thenif f(x) ∈ (0, 1] then 3x = f(x) = f(y) = 3y, so x = y;

5 CARDINALITY 110

if f(x) ∈ (2, 3] then 3x + 1 = f(x) = f(y) = 3y + 1, so 3x = 3y and x = y;if f(x) ∈ (4, 5] then 3x + 2 = f(x) = f(y) = 3y + 2, so 3x = 3y and x = y.The function f is onto A because if a ∈ A, thenif a ∈ (0, 1] then a

3 ∈ (0, 13 ] and f(a

3 ) = 3(a3 ) = a;

if a ∈ (2, 3] then a−13 ∈ ( 1

3 , 23 ] and f(a−1

3 ) = 3(a−13 ) + 1 = (a − 1) + 1 = a;

if a ∈ (4, 5] then a−23 ∈ ( 2

3 , 1) and f(a−23 ) = 3(a−2

3 ) + 2 = (a − 2) + 2 = a;

(g) Define f : R − {0} → R as f(x) ={

x, if x /∈ N

x − 1, if x ∈ N.

Then f is onto. Let x ∈ R. If x ∈ R − (N ∪ {0}), then x ∈ R − {0} andf(x) = x. If x ∈ N∪{0}, then x+1 ∈ R−{0}and f(x+1) = x+1−1 = x.Also, f is one-to-one. Let x, y ∈ R − {0} and suppose f(x) = f(y). Iff(x) ∈ R − (N ∪ {0}), then x = f(x) = f(y) = y. If f(x) ∈ N ∪ {0}, thenx + 1 = f(x) = f(y) = y + 1, so x = y.

5. (a) False (b) True (c) False (d) True (e) True (f) False

6. (a) Define g: N → E+ by g(x) =

{ 20, if x = 12, if x = 102x, if x �= 1, x �= 10.

(b) Define h: N → E+ by h(n) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

16, if n = 112, if n = 22, if n = 34, if n = 86, if n = 62n, if n �= 1, 2, 3, 6, or 8.

7. (a) c (b) c (c) ℵ0 (d) ℵ0 (e) c (f) c (g) ℵ0

8. (a) A = N ∪ {0}, B = N ∪ {−1},

(b) A = N ∪ {0}, B = {. . . ,−4,−3,−2,−1, 0}(c) A = N ∪ {0}, B = {. . . ,−4,−3,−2,−1, 0}(d) A = N ∪ {0}, B = N ∪ {−1}

9. (a) A = (0, 1) ∪ N, B = (1, 2}(b) A = (0, 1) ∪ N, B = (1, 3)

(c) A = (0, 1) ∪ N, B = (1,∞)

(d) A = (0, 1) ∪ N, B = [0, 1]

10. f ′(x) = −x2+x− 12

x2(x−1)2 < 0 on (0, 1), so f is decreasing and therefore one-to-one on(0, 1).

Also, limx→1− f(x) = −∞, limx→0+ f(x) = ∞, and f is continuous on (0, 1), soby the Intermediate Value Theorem of Calculus, f attains every real value onthe interval (0, 1).

11. Define f : C → R × R by f(a + bi) = (a, b). Then f is one-to-one because(a, b) = (c, d) implies a = c and b = d, which implies a + bi = c + di. Also,f is onto because for any (a, b) ∈ R × R, a + bi ∈ C and f(a + bi) = (a, b). ThusC ≈ R × R ≈ (0, 1).

12. (a) F. W is certainly an infinite subset of N, and D+ is denumerable, butthis “proof” claims without justification that every infinite subset of N isdenumerable. To show W is denumerable, we need to use another theoremor a bijection between W and a denumberable set.

5 CARDINALITY 111

(b) F. This “proof” is based on the false assumption that an infinite set mustbe equivalent to N.

(c) F. The claim is false. A denial of “A and B are infinite” is “A or B is finite.”(d) F. Writing an infinite set as {x1, x2, . . .} is the same as assuming A is

denumerable.(e) A.

5.3 Countable Sets

1. 91

2. Begin by writing the set K = { 2x

3y : x, y ∈ N} as follows:

21

3121

3221

3321

34 · · ·22

3122

3222

3322

34 · · ·23

3123

3223

3323

34 · · ·...

......

.... . .

Now all the elements of K can be put into a correspondence with N by listingthem along the diagonal as in Figure 5.3.1. There is only one number 21

31 on thefirst diagonal so 21

31 corresponds to 1. There are 2 elements of K on the seconddiagonal, and they correspond to 2 and 3. This correspondence is more naturalthan the one used to explain Theorem 5.3.1 because it is not necessary to omitany elements of K. The key idea is that every element of K lies on some diagonal,and that diagonal is reached after counting along finitely many finite diagonals.

3. Suppose A is denumerable and B is finite. If B is empty, then A ∪ B = A

which is denumerable. Suppose for some n ∈ N ∪ {0} that if B = n, then A ∪ Bis denumerable. Now let B be a set with cardinality n + 1. Let x ∈ B. ThenB−{x} is finite with cardinality n, so by the induction hypothesis A∪(B−{x})is denumerable. By Theorem 5.3.4, A∪B = A∪ (B −{x})∪{x} is denumerable.Therefore, by the PMI, A ∪ B is denumerable for every finite set B.

4. Let x ∈ A ∪ B. If x ∈ A, then there is an n ∈ N such that f(n) = x. Now 2n − 1is odd, and h(2n − 1) = f(n) = x. If x ∈ B, then x = g(n) for some n ∈ N. Now2n is even, and h(2n) = g(n) = x. Therefore h is onto A ∪ B.

Let x, y ∈ N be such that h(x) = h(y). If h(x) ∈ A, then f(x+12 ) = h(x) =

h(y) = f(y+12 ). Then x+1

2 = y+12 because f is one-to-one, so x = y. If h(x) ∈ B

then g(x2 ) = h(x) = h(y) = f(y

2 ). Then x2 = y

2 , because g is one-to-one, sox = y. Therefore h is one-to-one.

5. (a) Let f : N 1-1−→onto

A. Define g: N → A − {x} as:

g(m) ={

f(m), if m < f−1(x)f(m + 1), if m ≥ f−1(x).

Then g is a bijection.(b) Let B ⊆ A. If B is empty, then A − B = A is denumerable. Suppose for

some n ∈ N ∪ {0} that if B has n elements, then A − B is denumerable.Now suppose B has n + 1 elements, one of which is x. Then B − {x} has nelements, so A−(B−{x}) is denumerable. Now A−B = A−(B−{x})−{x},so by (a), A − B is denumerable. By the PMI, for all n ∈ N, if B has nelements, then A − B is denumerable.

5 CARDINALITY 112

6. Let n = 1. Then there is one denumerable set A1 in the family, and⋃1

i=1 Ai = A1which is denumerable.

Suppose that for some k ∈ N, if there are k pairwise disjoint denumerable setsin the family, then

⋃ki=1 Ai is denumerable. Consider a family of k + 1 pairwise

disjoint denumerable sets. Since the sets are pairwise disjoint,⋃k

i=1 Ai and Ak+1

are disjoint. The set⋃k

i=1 Ai is denumerable by hypothesis of induction and Ak+1

is denumerable, so by Theorem 5.3.6, their union (⋃k

i=1 Ai) ∪ Ak+1 =⋃k+1

i=1 Ai

is denumerable.

By the PMI, the union of any finite number of pairwise disjoint denumerablesets is denumerable.

7. This result has been proved in these cases:

Case 1. A and B are finite (Theorem 5.1.7)

Case 2. One of A or B is finite and the other is denumerable (Theorem 5.3.5,Exercise 3)

Case 3. A and B are denumerable and disjoint (Theorem 5.3.6)

The only remaining case is

Case 4. A and B are denumerable and not disjoint.

In this case write A∪B as A∪(B−A), a union of disjoint sets. Since B−A ⊆ B,as B −A is either finite or denumerable by Theorem 5.3.2. If B −A is finite thenA ∪ B is denumerable by Theorem 5.3.5. If B − A is denumerable then A ∪ B isdenumerable by Theorem 5.3.6.

8. If A is an empty collection of countable sets, then⋃

A is empty, hence countable.If A contains exactly one countable set A1, then

⋃A = A1, which is countable.

Suppose for some n ∈ N that if A contains n countable sets, then⋃

A iscountable. Now consider a collection A of n + 1 countable sets. Pick A ∈ A.Then A − {A} is a collection of n countable sets and therefore

⋃(A − {A}) is

countable. Now⋃

A =⋃

(A−{A})∪A, so by part (b) of Corollary 5.3.9,⋃

A iscountable. Therefore by the PMI, if A is any finite collection of countable sets,then

⋃A is countable.

9. (a) By Theorem 5.3.2, every subset of a denumerable set is countable. If thesubset is infinite, it must be denumerable.

(b) Suppose A is countable, B is uncountable, and A ⊆ B. Suppose B − A iscountable. Then B = A ∪ (B − A) is a union of countable sets, so B iscountable. This is a contradiction. We conclude that B −A is uncountable.

(c) Q ∩ (1, 2) contains {1 + 12n : n ∈ N}, so it must be infinite. (Q ∈ (1, 2)) ⊆ Q,

which is denumerable. Thus Q ∩ (1, 2) is denumerable by part (a).

(d) (Q ∩ (1, 2)) ⊆⋃20

n=1(Q ∩ (n, n + 1)), so the union is an infinite subset of Q.Thus

⋃20n=1(Q ∩ (n, n + 1)) is denumerable by part (a).

(e) (Q ∩ (1, 2)) is an infinite subset of the union, so⋃

n∈N(Q ∩ (n, n + 1))

is infinite. The union is a subset of the denumerable set Q, so it isdenumerable.

(f) This is another infinite subset of Q and is therefore denumerable by part (a).

10. (a) False. Let A = {1, 2}, B = N.

(b) False. Let A = N, B = R.

5 CARDINALITY 113

(c) True. J = {f : f is a linear function of the form f(x) = x + q, where q isrational.The function H: Q → J , where H(q) is the function given by f(x) = x + q,is a bijection. Therefore, J is denumerable.K = {f : f is a linear function of the form f(x) = 2x + z, where z is aninteger}.The function G: Z → K, where G(z) is the function given by f(x) = x + z,is a bijection. Therefore, K is denumerable.Sets J and K are disjoint because no linear function may have slope 1 andslope 2. Both sets are denumerable, so J ∪ K is denumerable.

(d) True. By Theorem 5.3.1, Q is denumerable and consequently countable.Since Q−Z is a subset of Q, Q−Z is countable by Theorem 5.3.2. Q−Z isinfinite because it contains the infinite subset { 1

2 , 13 , 1

4 , . . .}. Therefore Q−Z

is denumerable.

(e) False. Let A = N, B = N − {1}.

11. For each m ∈ N, there is a bijection fm: Bm → Nkm, where km = Bm. Define

h:⋃

i∈NBi → N by h(x) = (

∑m−1i=1 ki) + fm(x) where x ∈ Bm. Then h is a

bijection. Suppose x, z ∈⋃

i∈NBi and h(x) = h(z). Then fm(x) = fm(z), so

x = z. Therefore h is one-to-one. Suppose n ∈ N. Let m be the smallest naturalnumber such that

∑m−1i=1 ki < n. Then h(f−1(n −

∑m−1i=1 ki)) = n, so h is onto

N.

12. (a) {An: n ∈ N}, where An = {n} for every n ∈ N.

(b) {An: n ∈ N}, where An = {1, 5} for every n ∈ N.

(c) Not possible. By Exercise 11, the union of a denumerable family of disjointpairwise disjoint finite sets is denumerable.

13. (a) S is not finite because the infinite and countable set

{1000 . . . , 0100 . . . , 0010 . . . , . . .} ⊆ S.

Suppose there is a one-to-one function f : N → S. Then the image can bewritten as:

f(1) = a11a12a13 . . . f(2) = a21a22a23 . . ....f(n) = an1an2an3 . . .

...

Now let b be the sequence b1b2b3 . . . where bi ={

0, if aii = 11, if aii = 0.

Then b is not an image of any n ∈ N, so f is not onto. Hence there is nobijection from N to S.

(b) Let n ∈ N. Tn contains the sequence with a block of n consecutive 1’sstarting with the first term, the sequence with n consecutive 1’s startingwith the second term, etc. Thus Tn is infinite, for every n. Let a ∈ Tn, andsuppose k is the first term such that for all i > k, ai = 0. Define f(a) to bethe product

∏ki=1 pai

i , where pi is the ith prime. Then f is one-to-one, so Tn

is equivalent to Rng(f). Since Rng(f) ⊆ N, Rng(f) is countable. ThereforeTn is infinite and countable, so Tn is denumerable.

(c) By part (b), for every n ∈ N, Tn is denumerable. Therefore, for everyn ∈ N, there exists a bijection fn: N → Tn. List the images of each of thesefunctions in rows:

f1(1) f1(2) f1(3) f1(4) f1(5) · · · (elements of T1)

5 CARDINALITY 114

f2(1) f2(2) f2(3) f2(4) f2(5) · · · (elements of T2)f2(1) f2(2) f2(3) f2(4) f2(5) · · · (elements of T3)f2(1) f2(2) f2(3) f2(4) f2(5) · · · (elements of T4)...

......

......

All the elements of T⋃∞

n=1 can be put into a one-to-one correspondencewith N by listing them along the diagonal as in Figure 5.3.1. There is noneed to eliminate duplicate entries because the family of sets {Tn: n ∈ N}is pairwise disjoint.

14. (a) The function f : {B: B ⊆ A and B = 1} → A given by f({a}) = a is abijection.

(b) Let g be a bijection from A to N. Let S2 be the set of all 2-element subsetsof A. For {x, y} ∈ S2 with x < y, let f({x, y}) = 2g(x)3g(y). The f is aone-to-one function that maps onto Rng(f). Rng(f) is an infinite subset ofN, so Rng(f) is denumerable. S2 ≈Rng(f), so S2 is denumerable.

(c) Let g be a bijection from A to N. Let Sk be the set of all k-elementsubsets of A. For {x1, x2, . . . , xk} ∈ Sk with x1 < x2 < · · · < xk, letf({x1, x2, . . . , xk}) =

∏ki=1 p

g(xi)i , where pi is the ith prime. Then f is

a one-to-one correspondence between Sk and Rng(f). Since Rng(f) is aninfinite subset of N, Rng(f) and Sk are denumerable.

(d) Since A ≈ N, we may assume that A = N. By Part (c), Sk is denumerablefor every k ∈ N. By Theorem 5.3.8, S =

⋃k∈N

Sk is denumerable. Thereforethe set S ∪ {∅} of all finite subsets of N is denumerable.

15. (a) C. The proof is valid only when f(1) = x. In the case when f(1) �= x,we need a new function g that is almost the same as f except that theimage of 1 will be x. This involves removing the two ordered pairs withsecond coordinates x and f(1) and replacing them with two other orderedpairs. Let t be the unique element of N such that f(t) = x and definef∗ = (f − {(1, f(1)), (t, x)}) ∪ {(1, x), (t, f(1))}. Now let g(n) = f∗(n + 1)for all n ∈ N.

(b) F. A set that is not denumerable is not necessarily finite.

(c) F. The listing described in the “proof” associates every natural numberwith a number on the first row. Thus the listing is not a bijection.

(d) F. The statement is false. Indexing the elements of the sets A and B withnatural numbers amounts to assuming the sets are denumerable.

(e) F. It is not true that every subset of an uncountable set is uncountable.

5.4 The Ordering of Cardinal Numbers

1. Let n ∈ N. Then Nn ⊆ R, so n = Nn ≤ R =c. Also Nn �≈ R. so n �=c.

2. P(N) = (0, 1) = R < P(R), so by Corollary 5.4.4(b), P(N) < P(R).

3. Suppose A ≤ B and B = C. Then there exist functions f and g such thatf : A 1-1−→ B and g: B 1-1−→

ontoC. Then g ◦ f : A 1-1−→ C, so A ≤ C.

4. Suppose A ≤ B and A = C. Then there exist functions f and g such thatf : C 1-1−→

ontoA and g: A 1-1−→ B. Then g ◦ f : C 1-1−→ B, so C ≤ B.

5 CARDINALITY 115

5. (a) False. Let A = {0, 1} and B = (0, 1).(b) True (c) True (d) True(e) False. Let A = (0, 1) and B = (0, 2).

6. (a) The identity map IA: A → A is one-to-one. Therefore A ≤ A.

(b) Assume that A = B and B = C. Then there exist bijections f and g suchthat f : A → B and g: B → C. Then g ◦ f is a bijection from A to C.Therefore A = C.

(c) Assume that A = B and B = C. Then there exist one-to-one functionsf and g such that f : A → B and g: B → C. Then g ◦ f is a one-to-onemapping from A to C, so A ≤ C.

(d) Suppose A ≤ B and A �= B. Then A < B by definition. This shows thatA ≤ B implies A < B or A = B. Now suppose A < B or A = B.

Case 1: A < B. Then A ≤ B by definition.

Case 2: A = B. Then there is a bijection from A to B which, in particular,is one-to-one, so A ≤ B. Therefore A < B or A=B implies A ≤ B.

(e) Suppose A ⊆ B. Then the inclusion map i from A to B is one-to-one, soA ≤ B.

(f) Suppose there is a subset W of B such that A = W . Then by (d), A ≤ W ,and by (e), W ≤ B, so A ≤ B by (c).

Now suppose A ≤ B. Let f : A 1-1−→ B. Then f : A 1-1−→onto

Rng(f), so Rng(f) is

a subset W of B such that A = W .

7. Suppose there is a set A with the largest cardinal number, A. By Cantor’sTheorem, A < P(A). This is a contradiction.

8. (a) ∅ < {0} < {0, 1} < Q < (0, 1) = [0, 1] = R − N = R < P(R) < P(P(R))

(b) {0, 5} < {0, 3, 5} < P({0, 5}) < [0, 5] = R − {3} = (0, 5) − {3} = R − N <

P((0, 5))

(c) ({0.1}) < Q ∪ {π} = Q < R − {π} = [0, 2] = (0,∞) = R − Z < P(R).

9. (a) f(1) = 13 , f(2) = 8, f(3) = 1

14 , f(4) = 19.

(b) g(1) = 14 , g(2) = 9, g(3) = 1

15 , g(4) = 20.

(c) H(2) = 18 , H(8) = 1

3 , H(13) = 119 , H(20) = 1

15 .

10. B ≤ C and C ≤ A, so B ≤ A. Also A ≤ B, so the Cantor-Schroder-BernsteinTheorem A = B. Therefore A ≈ B. The same reasoning shows B ≈ C.

11. (a) Any nonzero rational number can be written in exactly one way asab such that a ∈ Z, b ∈ N, and a and b have no common factors. Definef : Q → N as:

f(a

b

)=

⎧⎨⎩

2a3b, if a > 01, if a = 05−a7b, if a < 0.

Then f is one-to-one, but not onto N. Let g: N → Q be the inclusion map,which is also one-to-one and not onto.(b) Impossible (c) Impossible (d) Impossible

5 CARDINALITY 116

12. Assume there is such a function. Then A ≤ N, so by Theorem 5.4.1(f), A = Wfor some subset W of N. Therefore A is countable by Corollary 5.3.3.

13. (b) Suppose A ≤ B and B < C. Then B ≤ C, so A ≤ C. Suppose A = C.Then C ≤ B and B ≤ C, so by the Cantor-Schroder-Bernstein TheoremB = C. This contradicts the hypothesis that B < C. Therefore A < C.(d) Suppose A < B and B < C. Then A ≤ B and B < C so by part (b),A < C.

14. Suppose there is a set of all sets U . Then every subset of U , by virtue of beinga set, must be an element of U . Thus P(U) ⊆ U , so P(U) ≤ U . By Cantor’sTheorem U < P(U). Therefore, by Theorem 5.4.4(b), P(U) < P(U), which isimpossible.

15. (a) Let A be the set of such integers. The inclusion map i: A → Z is one-to-one, so A ≤ Z = N. On the other hand, the map f : N → A given byf(n) =

∑ni=0 6 · 10i is also one-to-one. Thus A = N, by the CSBT.

(b) The map f : R → R × R, f(t) = (t, 0) is one-to-one, so R ≤ R × R. Onthe other hand, define g: (0, 1) × (0, 1) as g(0.a1a2a3 . . . , 0.b1b2b3 . . .) =0.a1b1a2b2a3b3 . . ., where all decimals are in normalized form. Then g is one-to-one (but not onto, since 0.1239 /∈Rng(g)), so R × R = (0, 1) × (0, 1) ≤(0, 1) = R. Now by the CSB Theorem R = R × R. (In a letter to a friend,Cantor said of this, “I see it, but I don’t believe it.”)

(c) By Theorem 5.4.1(a), R = (a, b) ≤ A ≤ R, so A = R by the Cantor-Schroder-Bernstein Theorem.

16. (a) Assume there is a bijection from [0, 1] to F . For each a ∈ [0, 1], let thecorresponding function in F be fa. Define g: [0, 1] → [0, 1] by g(x) ={

0 if fx(x) �= 01 if fx(x) = 0.

Since g ∈ F , g = fb for some b ∈ [0, 1]. If g(b) = 0, then fb(b) = 0, sog(b) = 1. If g(b) �= 0, then fb(b) �= 0, so g(b) = 0. Thus g(b) �= g(b), whichis impossible. We conclude that there is no bijection.

(b) For each x ∈ [0, 1], the constant function with range {x} is in F . Thus[0, 1] ≤ F , so F is uncountable.

(c) Since [0, 1] ≤ F and [0, 1] �≈ F , [0, 1] < F

17. (a) F. The main idea in this “proof” is the mistaken assertion that A = C.A correct proof would use a bijection form A to C to create a one-to-onefunction from C to B.

(b) F. There is no cancellation law for infinite cardinals that would allow us toconclude that (C − B) = 0. The claim is false.

(c) F. One must prove that substitution and transitivity properties held forcardinal numbers. As per the definitions of < and ≤, a proof requiresconsideration of functions.

(d) F. If g is not onto B, then some elements of B have no pre-image in A. Inthis case Dom(f) �= B.

5 CARDINALITY 117

5.5 Comparability of Cardinal Numbersand the Axiom of Choice

1. (a) No. Select the odd number in each set.

(b) No. The collection of sets is finite.

(c) No. Select the integer in each set.

(d) Yes.

(e) No. Select the least integer in each set.

(f) No. Select the least element in each set.

(g) Yes.

(h) Yes.

2. (a) Let f : B 1-1−→ A and b ∈ B. Define g: A → B by

g(a) ={

b, if a /∈ Rng(f)f−1(a), if a ∈ Rng(f).

Then g is onto B.

(b) Suppose f : A → B is onto B. Then for each b ∈ B, the set Cb = {a ∈A: f(a) = b} is nonempty. We note that

⋃b∈B Cb = A and if b1 �= b2,

then Cb1 ∩ Cb2 = ∅. By the Axiom of Choice, we may select one elementfrom each Cb and call it g(b). This defines a function g: B → A for whichg(b) ∈ Cb for each b ∈ B. The function g is one-to-one because b1 �= b2implies Cb1 ∩ Cb2 = ∅, which implies that g(b1) �= g(b2).

3. By the Comparability Theorem, we have three cases:

Case 1: If A < B, then there is an f : A 1-1−→ B.

Case 2: If A=B , then there is an f : A 1-1−→ B.

Case 3: If A > B, then by Exercise 2, there is an f : A onto−→ B.

4. Let f : A → B. Then f : A onto−→Rng(f), so by Theorem 5.5.3, Rng(f) ≤ A.

5. Let B ⊆ A with B infinite and A denumerable. Since B ⊆ A, B ≤ A. Since A

is denumerable. A = N. Since B is infinite, B has a denumerable subset D byTheorem 5.5.4. Thus A = N = D ≤ B. By the CSB Theorem, B = A. ThusB ≈ A.

6. Assume that B < C and B/ ≤ A. Suppose C ≤ A. Then by the ComparabilityTheorem, A < C. By transivity (Theorem 5.4.4(d)), A < C.

7. Use the Axiom of Choice to construct a function F : {Ai: i ∈ N} →⋃

i∈NAi

such that for every i ∈ N, F (Ai) ∈ Ai. Since the Ai’s are distinct and pairwisedisjoint, F is one-to-one. Thus Rng(F ) is a denumerable subset of

⋃i∈N

Ai.

8. Let x ∈ A. Then A − {x} is infinite, so by Theorem 5.5.4 A − {x} has adenumerable subset B = {an: n ∈ N}. Define g: A → A − {x} by

f(a) =

⎧⎨⎩

a1, if a = xai+1, if a = ai ∈ Ba, if a /∈ B ∪ {x}.

Then f is a bijection.

5 CARDINALITY 118

9. Suppose there is a function f : N onto−→ A. Then by Theorem 5.5.3 A ≤ N, so byExercise 12 in Section 5.4, A is countable. Suppose now that A is countable.Then by Theorem 5.3.3 A ≈ B, for some B ⊆ N, so A = B ≤ N, and so A ≤ N.Thus, by Exercise 2, there is an f : N onto−→ A.

10. (a) A.

(b) F. The “proof” claims incorrectly that if a set is not denumerable it mustbe finite.

(c) F. The idea of this “proof” is to take out countably many elements, one ata time, until denumerably many elements are left. But if A is uncountableand C is countable, then the set B = A−C of leftover elements will alwaysbe uncountable. (See Exercise 9(b) of Section 5.3.)

(d) F. A − B is not necessarily infinite. In fact, if B = A, then A − B is empty.

(e) A. The ideas are correct. It might be disirable to give more detailedexplanation.

(f) A.

(g) A.

6 Concepts of Algebra

6.1 Algebraic Structures

1. (a) Algebraic structure (b) Not an algebraic structure(c) Algebraic structure (d) Not an algebraic structure(e) Not an algebraic structure (f) Not an algebraic structure(g) Algebraic structure (h) Algebraic structure(i) Algebraic structure (j) Not an algebraic structure(k) Algebraic structure (1) Not an algebraic structure

2. (a) not commutative (b) not an operation(c) not commutative (d) not an operation(e) not an operation (f) not an operation(g) not commutative (h) commutative(i) commutative (j) not an operation(k) commutative (1) not an operation

3. (a) not associative (b) not an operation(c) not associative (d) not an operation(e) not an operation (f) not an operation(g) not associative (h) associative(i) associative (j) not an operation(k) associative (1) not an operation

4. (a) a

(b) Yes. This is tedious to verify because one must verify that 64 equations ofthe form (x◦y)◦z = x◦ (y ◦z) are all true. It helps to observe that if x = a(the identity) then (x ◦ y) ◦ z = y ◦ z = x ◦ (y ◦ z). Similarly, if y = a orz = a, the equation is easily seen to be true. This leaves 27 cases to verifywhen none of x, y or z is a. For example, (b ◦ c) ◦ b = d ◦ b = c, whileb ◦ (c ◦ b) = b ◦ d = c, so the equation is true when x = b, y = c and z = b.

(c) Yes, because the table is symmetric about its main diagonal. To verify bycases that the equation x ◦ y = y ◦ x is true for every choice of x and y,consider first the case that either x or y is a, then the case that x = y andfinally the 3 other cases.

(d) The inverses of a, b, c, d are a, b, c, d, respectively.

(e) No. The product b ◦ c = d is not in B1, so B1 is not closed under ◦.(f) Yes. a ◦ a = a, a ◦ c = c, c ◦ a = c and c ◦ c = a.

(g) {a}, {a, b}, {a, c}, {a, d}, and {a, b, c, d}.

(h) True. In fact for all x ∈ A, x ◦ x = a.

5. (a) c

(b) Yes.

(c) Yes.

(d) The inverses of a, b, c, d are a, d, c, b, respectively.

(e) No. a ∗ b = d

(f) Yes.

(g) {c}, {a, c}, {a, b, c, d}(h) False. a ∗ a = c, b ∗ b = a.

6. (a) ◦, +,×

119

6 CONCEPTS OF ALGEBRA 120

(b) ◦, +

(c) The system with operation ◦ has identity a. The system with operation ∗has identity element b.

(d) In the system with operation◦, a and b are their own inverses. In the systemwith operation *, a does not have an inverse; the elements b and c are theirown inverses.

7. (a) (M, ·) is an algebraic system iff m = n.

(b) (M, +) is an algebraic system for all natural numbers m and n.

8. By associativity, (ac)(db) = a[c(db)] = a[(cd)b]. Since d is the inverse of c and b isthe inverse of a, a[(cd)b] = a[eb] = ab = e. Also (db)(ac) = d[b(ac)] = d[(ba)c] =d[ec] = dc = e. Since (ac)(db) = e = (db)(ac), db is the inverse of ac.

9. (a) Let ◦ be associative and x and y be inverses of a. Then x = x ◦ e =x ◦ (a ◦ y) = (x ◦ a) ◦ y = e ◦ y = y.

(b) Consider ({a, b, c}, ∗) where ∗ is given by the Cayley table:∗ a b c The identity is a.a a b c ∗ is not associative because (b ◦ b) ◦ c = cb b a a and b ◦ (b ◦ c) = bc c a b The element b has two inverses, b and c

10. (a) (a1 ∗ a2) ∗ (a3 ∗ a4) = a1 ∗ (a2 ∗ (a3 ∗ a4)) = a1((a2 ∗ a3) ∗ a4)

(b) Assume that for some natural number n, every product of t elements a1,a2, a3, . . ., at, in that order, is equal to the left-associated product of theelements, for every t ≤ n. Consider a product of n + 1 factors a1, a2, a3,. . ., an+1 in that order. This product has the form b1 ∗ b2, where b1 is aproduct of some k factors a1, a2, a3, . . ., ak (k ≤ n) in that order and b2is a product of the remaining factors ak+1, . . ., an+1 in that order. Use theinduction hypothesis to write b1 and b2 in left-associated form.If k > 1 then there are at least two factors a1 and a2 in b1. Then a1 ∗ a2 isan element c of A. Replace a1 ∗ a2 with the element c, so the producthas n terms. By the hypothesis of induction this product is equal to(. . . (c ∗ a3) . . . ∗ an + 1). Now re-insert a1 ∗ a2 in place of c, so the entireproduct is in left-associated form.If k = 1 and b2 has only one factor, then the product b1 ∗ b2 is just a1 ∗ a2,which is already left-associated. Otherwise, b2 has at least two factors andthe first two of these factors are a2 and a3. Then a2 ∗ a3 is an elementd of A. Replace a2 ∗ a3 with the element d, so the product has n terms.By the hypothesis of induction this product is equal to the left-associatedexpression (. . . (a1 ∗ d) . . . ∗ an+1). Now re-insert a2 ∗ a − 3 in the place ofd, and then substitute (a1 ∗ a2) ∗ a3 for a1 ∗ (a2 ∗ a3) so the entire productis in left-associated form.

11. (a) ◦ a b ca a b c Both a and b are left identities.b a b cc c a b

(b) An element r ∈ A is a right identity for ◦ iff a ◦ r = a for every a ∈ A.

(c) If r is a right identity and l is a left identity, then r = l ◦ r = l. And ifa ∈ A, then a ◦ r = a and r ◦ a = l ◦ a = a, so r is an identity.

12. Assume that a = c(mod m) and b = d(mod m). Then a−c = mx and b−d = myfor some integers x and y.


(a) Then m divides (a + b) − (c + d) = (a − c) + (b − d) = m(x + y), soa + b = c + d(mod m).

(b) Then m divides ab−cd = ab−bc+bc−cd = b(a−c)+c(b−d) = bmx+cmy,so a · b = c · d(mod m).

13. Let m be a natural number. Then (Zm, +m) and (Zm, ·m) are algebraic systems,by Theorem 6.1.2. In the proofs of parts (a) and (b) we use the notation forclasses to emphasize the fact that elements of Zm are equivalence classes. Letx, y, z ∈ Zm.

(a) Then (x+my)+mz = x + y+mz = (x + y) + z = x + (y + z) = x+my + z =x+m(y+mz), so +m is associative. Also x+my = x + y = y + x = y+mx,so +mis commutative. The identity is 0, because x+m0 = x + 0 = x =0 + x = 0+mx. The inverse of x is −x, because x+m−x = x + (−x) = 0 =(−x) + x = −x+mx. That is −(x) = −x.

(b) Then (x·my)·mz = x · y·mz = (x · y) · z = x · (y · z) = x·my · z = x·m(y·mz),so ·m is associative. Also x·my = x · y = y · x = y·mx, so ·mis commutative.The element 1 is the identity because x·m1 = x · 1 = x = 1 · x = 1·mx.

14. (a) + 0 1 2 3 4 5 6 70 0 1 2 3 4 5 6 71 1 2 3 4 5 6 7 02 2 3 4 5 6 7 0 13 3 4 5 6 7 0 1 24 4 5 6 7 0 1 2 35 5 6 7 0 1 2 3 46 6 7 0 1 2 3 4 57 7 0 1 2 3 4 5 6

(b) + 0 1 2 3 40 0 1 2 3 41 1 2 3 4 02 2 3 4 0 13 3 4 0 1 24 4 0 1 2 3

(c) · 0 1 2 3 4 5 6 70 0 0 0 0 0 0 0 01 0 1 2 3 4 5 6 72 0 2 4 6 0 2 4 63 0 3 6 1 4 7 2 54 0 4 0 4 0 4 0 45 0 5 2 7 4 1 6 36 0 6 4 2 0 6 4 27 0 7 6 5 4 3 2 1

(d) · 0 1 2 3 40 0 0 0 0 01 0 1 2 3 42 0 2 4 1 33 0 3 1 4 24 0 4 3 2 1

15. (a) 2, 4, and 6

(b) 2, 3, 4 6, 8, and 10

(c) There are no divisors of zero.


(d) There are no divisors of zero.

16. (a) F. The claim is false. The proof fails because we cannot assume that thecancellation property holds in every algebraic structure.

(b) F. The claim is false. One may premultiply (multiply on the left) orpostmultiply (multiply on the right) both sides of an equation by equalquantities. Multiplying one side on the left and the other side on the rightdoes not always preserve equality.

(c) F. The first sentence of the “proof” is false.

(d) F. The proof makes the assumption that xy �= 0, which may be false.

6.2 Groups

1. (a) {1,−1} is closed under · because 12 = 1 = (−1)2 and −1 · 1 = −1 = 1 · −1.The identity is 1, each element is its own inverse, and · is associative on Z,so it is here as well.

(b) · 1 α β1 1 α β We see from the table that {1, α, β} is closed under ·,a a β 1 1 is the identity, and every element has an inverse.β β 1 α Also, · is associative on C, so it is here too.

(c) · 1 −1 i −i1 1 −1 i −i We see from the table that the set

−1 −1 1 −i i is closed under ·, 1 is the identityi i −i −1 1 and each element has an inverse.

−i −i i 1 −1 Also, · is associative.

(d) If A, B ⊆ X, then (A − B) ∪ (B − A) ⊆ A ∪ B ⊆ X, so P(X) isclosed under . Also, A ∅ = (A − ∅) ∪ (∅ − A) = A ∪ ∅ = A and∅ A = (∅ − A) ∪ (A − ∅) = ∅ ∪ A = A, so ∅ is the identity. Each elementis its own inverse, since A A = (A − A) ∪ (A − A) = ∅ ∪ ∅ = ∅. Finally,if A, B, C ∈ P(X), then

A (B C)= (A − [(B − C) ∪ (C − B)]) ∪ ([(B − C) ∪ (C − B)] − A)= (A − [(B ∪ C) − (B ∩ C)]) ∪ ([B − C) − A] ∪ [(C − B − A])= ([A − (B ∪ C)] ∪ [A ∩ (B ∩ C)]) ∪ ([(B − C) − A] ∪ [(C − A) − B])= [(A − B) − C] ∪ [(B − C) − A] ∪ [(C − A) − B] ∪ [A ∩ B ∩ C]= ([(A − B) − C] ∪ [(B − C) − A]) ∪ (C − [(A ∪ B) − (A ∩ B)])= ([(A − B) − C] ∪ [(B − A) − C]) ∪ (C − [(A ∪ B) − (A ∩ B)])= ([(A − B) ∪ (B − A)] − C) ∪ (C − [(A − B) ∪ (B − A)])= (A B) C,

so A is associative.

2. ◦ e u v we e u v wu u v w ev v w e uw w e u v


3. ◦ e u v we e u v wu u e w vv v w e uw w v u e

4. (a) abelian (b) abelian (c) abelian (d) abelian

5. ◦ a b ca b a c This system has no identity and is notb c b a associative because (a ◦ a) ◦ c �= a ◦ (a ◦ c).c a c b

6. (a) + 0 1 2 3 4 50 0 1 2 3 4 51 1 2 3 4 5 12 2 3 4 5 0 13 3 4 5 0 1 24 4 5 0 1 2 35 5 0 1 2 3 4

(b) + 0 1 2 3 4 5 60 0 1 2 3 4 5 61 1 2 3 4 5 6 02 2 3 4 5 6 0 13 3 4 5 6 0 1 24 4 5 6 0 1 2 35 5 6 0 1 2 3 46 6 0 1 2 3 4 5

(c) · 1 2 3 41 1 2 3 42 2 4 1 33 3 1 4 24 4 3 2 1

(d) · 1 2 3 4 5 61 1 2 3 4 5 62 2 4 6 1 3 53 3 6 2 5 1 44 4 1 5 2 6 35 5 3 1 6 4 26 6 5 4 3 2 1

7. ◦ [1 2] [2 1][1 2] [1 2] [2 1] S2 is abelian.[2 1] [2 1] [1 2]

8. (a) 4! = 24

(b) [3 2 1 4], [1 2 3 4], [2 4 1 3]

(c) [2 1 3 4], [2 4 3 1], [1 2 3 4]

(d) [4 3 2 1] ◦ [3 1 2 4] = [2 4 3 1] �= [4 2 1 3] = [3 1 2 4] ◦ [4 3 2 1]

9. (a) (a1a2a3)(a−13 a−1

2 a−11 ) = a1a2a3a

−13 a−1

2 a−11 = a1a2ea

−12 a−1

1 = a1a2a−12 a−1

1 =a1ea

−11 = a1a

−11 = e and = (a−1

3 a−12 a−1

1 )(a1a2a3) = a−13 a−1

2 a−11 a1a2a3 =

a−13 a−1

2 a2a3 = a−13 a3 = e. Therefore (a1a2a3)−1 = a−1

3 a−12 a−1

1 .


(b) For all n ∈ N, (a1a2 . . . an)−1 = a − n−1 . . . a−12 a−1

1 .The statement is true for n = 1 because a−1

1 = a−11 , and it was proved for

n = 2 in Theorem 6.2.4. Assume the statement is true for some n > 2.Then

(a1a2 . . . anan+1)−1 = [(a1a2 . . . an)an+1]−1

= a−1n+1(a1a2 . . . an)−1

= a−1n+1

(a−1

n a−1n−1 . . . a−1

1

)= a−1

n+1a−1n a−1

n−1 . . . a−11 .

Thus by the PMI, the statement is true for all n ∈ N.

10. Assume that ca = cb. Then a = ea = (c−1c)a = c−1(ca) = c−1(cb) = (c−1c)b =eb = b.

11. Let x, y ∈ G. Suppose ρa(x) = ρa(y). Then xa = ya. By right cancellation,x = y. Therefore ρa is one-to-one. Now let b ∈ G. Then ba−1 ∈ G andρa(ba−1 = ba−1a = b. Therefore ρa is onto G.

12. Let a, b ∈ G. Then a(ab)b = a2b2 = ee = e = (ab)2 = a(ba)b. Then (ab)b = (ba)bby left cancellation and ab = ba by right cancellation.

13. ∗ a b c da a b c d In order to have both cancellation propertiesb b a d c for a finite algebraic structure, it is necessaryc c b a d and sufficient that every element occur in everyd d c b a row and column.

14. (a) Suppose G is abelian. Let a, b ∈ G. Then a2b2 = aabb = abab = (ab)2.Now suppose a2b2 = (ab)2 for all a, b ∈ G. Let x, y ∈ G. Thenxxyy = x2y2 = (xy)2 = xyxy. By cancellation of x (on the left) andright cancellation for y, we have xy = yx. Thus G is abelian.

(b) Suppose anbn = (ab)n for all a, b ∈ G and for all n ∈ N. Then by part (a),G is abelian. Now suppose G is abelian and a, b ∈ G. Then a1b1 = (ab)1.Assume that akbk = (ab)k for some k ∈ N. Then ak+1bk+1 = aakbbk =abakbk = ab(ab)k = (ab)k+1. By the PMI, anbn = (ab)n for all n ∈ N.

15. (a) Closure. Let a, b ∈ R − {1}. Then a ◦ b = a + b − ab ∈ R, anda ◦ b = 1 iff a + b − ab − 1 = 0 iff (a − 1)(b − 1) = 0 iff a = 1 or b = 1, soa ◦ b ∈ R − {1}.

(b) Identity. If a ∈ R − {1}, then a ◦ 0 = a + 0 − a · 0 = a = 0 + a − 0 · a = 0 ◦ a,so 0 is the identity for ◦.

(c) Inverses. If a ∈ R − {1}, then aa−1 ∈ R − {1} and

a ◦ a

a − 1= a +

a

a − 1− a · a

a − 1=

a2 − a

a − 1+

a

a − 1− a2

a − 1= 0 and

a

a − 1◦ a =

a

a − 1+ a − a

a − 1· a =

a

a − 1+

a2 − a

a − 1− a2

a − 1= 0

so aa−1 is the inverse of a.

(d) Associativity. Let a, b, c ∈ R − {1}. Then

(a ◦ b) ◦ c = (a + b − ab) + c − (a + b − ab)c= a + b + c − ab − ac − bc + abc

= a + (b + c − bc) − a(b + c − bc)= a ◦ (b ◦ c).


(e) Commutativity. If a, b ∈ R−{1}, then a◦b = a+b−ab = b+a−ba = b◦a.

16. (a) v; w; e; u.

(b) Let x = a−1 ∗ b and y = b∗a−1. Then x and y are in G and a∗ (a−1 ∗ b) = band y ∗ a = (b ∗ a−1) ∗ a = b. To show uniqueness, suppose u ∈ G anda ∗ u = b. The u = e ∗ u = (a−1 ∗ a) ∗ u = a−1 ∗ b = x, so u = x. Also, ifv ∈ G and v∗a = b, then v = v∗e = v∗(a∗a−1) = (v∗a)∗a−1 = b∗a−1 = y,so v = y.

17. (a) Closure. If a, b ∈ Z, then a # b = a + b + 1 ∈ Z.

(b) Identity. Let a ∈ Z. Then a # −1 = a − 1 + 1 = a = −1 + a + 1 = −1 # a,so −1 is an identity for #.

(c) Inverse. Let a ∈ Z. Then −(a+2) ∈ Z and a # −(a+2) = a− (a+2)+1 =−1 = −(a + 2) + a + 1 = −(a + 2) # a, so −(a + 2) is an inverse for a.

(d) Associativity. Let a, b, c ∈ Z. Then

(a # b) # c = (a + b + 1) + c + 1= a + b + c + 2= a + (b + c + 1) + 1= a # (b # c).

50 # 49 = 100.

18. (a) Let m ∈ N. Suppose there are integers x and y such that 1 < x, y < mand xy = m. Then xy = 0(mod m) so Zm − {0} is not closed undermultiplication.

(b) Let p be prime and let x, y ∈ Zp − {0}. Then xy ∈ Zp and xy = 0 iff pdivides xy iff p divides x or p divides y (by Euclid’s Lemma). But x and yare less than p, so this is impossible. Thus Zp − {0} is closed. Furthermore1x = x1 = x, so 1 is the identity, and associativity follows from associativityfor Z.

(c) Let p be prime and let x ∈ Zp − {0}. Then x and p are relatively prime,so by Theorem 1.7.3, there exist integers r and s such that rx + sp = 1.Then rx = 1(mod p), so the inverse of x in Zp − {0} is r. [More precisely,the inverse is the class of r.]

(d) By parts (b) and (c), Zm − {0} is a group when m is prime, and by part(a) Zm − {0} is not a group when m is not a prime.

19. Let p be a prime natural number. Then (p−1)(p−1) = p2−2p+1 = p(p−2)+1,so (p − 1)(p − 1) = 1(mod p). Thus (p − 1)−1 = (p − 1) in Zp − {0}.

20. (a) x = 0, 4, 8, 12, 16 (b) x = 0(c) x = 0, 10 (d) x = 3, 7, 13, 17

21. (a) x = 2 (b) x = 4(c) x = 6 (d) x = 2, 6(e) No solution (f) x = 3, 7

22. (a) x2 − 1 = 0 (b) x3 − 1 = 0 (c) x4 − 1 = 0

23. (a) F. The 8th “=” is not justified, because it assumes commutativity. Theclaim is false.

(b) F. The fatal flaw is the use of the undefined division notation.


(c) A. The proof is correct, but provides only minimal explanation.

(d) C. We cannot conclude from ab = m(mod m) that ab = m. This error iscorrected in part (e).

(e) A.

(f) F. The expression 1/x has no meaning.

(g) A.

6.3 Subgroups

1. (a) {0}, Z8, {0, 4}, {0, 2, 4, 6}(b) {1}, {1, 6}, {1, 2, 4}, {1, 2, 3, 4, 5, 6}(c) {0}, Z5

(d) {a}, J , {a, b}, {a, c}, {a, d}, {a, e, f}

2. (a) {[1 2 3 4], [2 3 1 4], [3 1 2 4]} and {[1 2 3 4], [1 3 4 2], [1 4 2 3]}(b) {[1 2 3 4], [2 3 4 1], [4 1 2 3], [3 4 1 2]} and {[1 2 3 4], [1 2 4 3], [2 1 3 4], [2 1 4 3]}(c) No. If an element is in a group, then its inverse is in the group.

(d) {[1 2 3 4], [4 2 1 3], [3 2 4 1]}

3. Suppose x ∈ H. Let a be the inverse of x in H and x−1 be the inverse of x in G.By Theorem 6.3.2, the identity of H is e. Thus a = ae = a(xx−1) = (ax)x−1 =ex−1 = x−1.

4. Suppose H and K are subgroups of G. The identity e of G is in both H and K,so e ∈ H ∩ K, and so H ∩ K is not empty. Suppose a, b ∈ H ∩ K. Then ab−1 isin both H and K, since a and b are both in H and K. Thus ab−1 ∈ H ∩ K, soH ∩ K is a subgroup by Theorem 6.3.3.

5. Suppose {Hα: α ∈ Δ} is a family of subgroups of G. Since e ∈ Hα for all α,e ∈

⋂α∈Δ Hα. Thus

⋂α∈Δ Hα is not empty. Suppose a, b ∈

⋂α∈Δ Hα. Then

a, b ∈ Hα for all α, so ab−1 ∈ Hα for all α, and so ab−1 ∈⋂

α∈Δ Hα. Thus⋂α∈Δ Hα is a subgroup by Theorem 6.3.3.

6. In S4 let H = {[1 2 3 4], [2 3 1 4], [3 1 2 4]} and K = {[1 2 3 4], [1 3 4 2], [1 4 2 3]}.

7. (a) Yes. Assume G is abelian and H is a subgroup of G. Suppose x, y ∈ H.Then x, y ∈ G. Therefore xy = yx.

(b) No. {[1 2 3], [3 2 1]} is an abelian subgroup of the nonabelian group S3.

8. Suppose G is a group, H is a subgroup of G, and K is a subgroup of H. ThenK is a subset of G and K is a group with the same operation as G, so K is asubgroup of G by definition.


9. (a) |(123)| = 1|(213)| = 2|(321)| = 2|(132)| = 2|(231)| = 3|(312)| = 3

(b) |0| = 1|1| = 7|2| = 7|3| = 7|4| = 7|5| = 7|6| = 7

(c) |0| = 1|1| = 8|3| = 8|5| = 8|7| = 8|2| = 4|6| = 4|4| = 2

(d) |1| = 1|2| = 3|3| = 6|4| = 3|5| = 6|6| = 2

10. S3 has no generators because it is not a cyclic group. The generators of (Z7, +)are 1, 2, 3, 4, 5 and 6. The generators of (Z8, +) are 1, 3, 5 and 7. The generatorsof (Z7 − {0}, ·) are 3 and 5.

11. Let G be a group and a ∈ G. Then Ca is not empty, because ea = a = ae,so e ∈ Ca. Let x, y ∈ Ca. Then ax = xa and ya = ay. Multiplyingboth sides of the last equation by y−1, we have y−1(ya)y−1 = y−1(ay)y−1.Thus (y−1y)(ay−1) = (y−1a)(yy−1), so ay−1 = y−1a. Therefore, (xy−1)a =x(y−1a) = x(ay−1) = (xa)y−1 = (ax)y−1 = a(xy−1). This shows xy−1 ∈ Ca.Therefore, Ca is a subgroup of G, by Theorem 6.3.3.

12. Let C be the center of the group G. For all x ∈ G, ex = xe, so e ∈ C. ThusC is not empty. Suppose x, y ∈ C and let z ∈ G. Then (xy−1)z = x(y−1z) =x(z−1y)−1 = x(yz−1)−1 = (yz−1)−1x = (zy−1)x = z(y−1x) = z(xy−1), soxy−1 ∈ C. Therefore, by Theorem 6.3.3, C is a subgroup of G.

13. Let a ∈ G. C = {x ∈ G: for all y ∈ G, xy = yx} ⊆ {x ∈ G: xa = ax} = Ca

and C is a group under the same operation as Ca, so C is a subgroup of Ca bydefinition.

14. Let a ∈ G and k = {a−1ha: h ∈ G}. The identity e ∈ H because H is a group, andthus e = a−1ea ∈ K. Thus K is not empty. Suppose b, c ∈ K. Then b = a−1h1aand c = a−1h2a for some h1, h2 ∈ H. Thus bc−1 = (a−1h1a)(a−1h2a)−1 =(a−1h1a)(a−1h−1

2 a) = a−1h1(aa−1)h−12 a = a−1(h1h

−12 )a. But H is a group, so

h1h−12 ∈ H. Thus bc−1 ∈ K. Therefore, K is a subgroup of G.

15. (a) (α) = {α, α2, α3, α4, α5, 1}, where α2 = − 12 + i

√3

2 , α3 = −1, α4 = − 12 − i

√3

2 ,and α5 = 1

2 − i√

32 .

(b) α5

16. Let m be a natural number greater than 1. Every k in Zm is the kth multiple of1, so the element 1 is a generator for (Zm, +). Note that m−1 = −1 (mod m).The element −1 is a generator because every k in Zm is also a multiple of −1,namely the multiple of −1 by −k. Thus m − 1 is a generator.

17. Let H be a subgroup of the cyclic group G = (a). If H = {e}, then H is cyclicwith generator e. Suppose H �= {e}, and let t be the smallest positive integersuch that at is in H. (The existence of t is guaranteed by the Well OrderingPrinciple.) We will show that H = (at).


First, (at) ⊆ H, by the closure property. If x ∈ H, then x = as for some integer s.By the Division Algorithm, s = qt+r for some integers q and r, where 0 ≤ r < t.Since a−qt is in H, a−qtas = a−qtaqt+r = ar ∈ H. Then r = 0, because r < tand t is the smallest positive integer such that at is in H. Therefore s = qt, sox = as ∈ (at).

18. (a) 5 (b) a15

(c) a10, a20 (d) a3, a9, a21, a27

19. (a) C. The proof omits the step of verifying that H ∩ K is nonempty (becausethe intersection contains the identity.)

(b) F. The claim is false. The proof fails because hk−1x−1 may not be in H,so x(hk−1x−1) may not be in xH.

6.4 Operation Preserving Maps

1. (a) Not operation preserving.√

4 + 3 �=√

4 +√

3.

(b) Operation preserving. For all a, b ∈ R+,

√ab =

√a ·

√b.

2. (a) Not operation preserving. (4 + 3)2 �= 42 + 32.

(b) Operation preserving. For all a, b ∈ R, (ab)2 = a2 · b2.

3. (a) Let (a, b), (c, d) ∈ R × R. Then (a, b) ⊗ (c, d) = (ac − bd, ad + bc) ∈ R × R.Thus R × R is closed under the operation ⊗.

(b) Suppose a + bi and c + di are in C and h(a + bi) = h(c + di). Then(a, b) = (c, d), so a = c and b = d. Thus a + bi = c + di. Therefore h isone-to-one. The function h is onto R × R, because for every (a, b) ∈ R × R,(a, b) = h(a + bi).Finally, h((a + bi)(c + di)) = h(ac − bd + (ad + bc)i) = (ac − bd, ad + bc) =(a, b) ⊗ (c, d) = h(a + bi) ⊗ h(c + di). Therefore h is OP.

4. I(f + g) =∫ b

a(f + g)(x) dx =

∫ b

af(x) dx +

∫ b

ag(x) dx = I(f) + I(g).

5. (a) Let x, y ∈ A. Then g ◦ f(x · y) = g(f(x · y)) = g(f(x) ∗ f(y)) =g(f(x)) × g(f(y)) = g ◦ f(x) × g ◦ f(y).

(b) Suppose f−1 is a function. Suppose f(x) and f(y) are in Rng(f). Thenf−1(f(x) ∗ f(y)) = f−1(f(x · y)) = x · y = f−1(f(x)) · f−1(f(y)).

6. (a) Let[

a bc d

]and

[e fg h

]be in M . Then

Det([

a bc d

]·[

e fg h

])= Det

[ae + bg af + bhce + dg cf + dh

]

= (ae + bg)(cf + dh) − (af + bh)(ce + dg)= (ad − bc)(eh − fg)

= Det[

a bc d

]· Det

[e fg h

].

(b) Det([

1 00 0

]+

[0 00 1

])= 1, while Det

[1 00 0

]+Det

[0 00 1

]= 0.


7. (a) Let a + bi, c + di ∈ C. Then

Conj ((a + bi) + (c + di)) = Conj(a + c + (b + d)i)= a + c − (b + d)i= (a − bi) + (c − di)= Conj(a + bi) + Conj(c + di).

(b) Let a + bi, c + di ∈ C. Then

Conj ((a + bi)(c + di)) = Conj(ac − bd + (ad + bc)i)= ac − bd − (ad + bc)i= (a − bi)(c − di)= Conj(a + bi) · Conj(c + di).

8. (a) Let C, D ∈ P(A). Then f(C ∪ D) = f(C) ∪ f(D) by Theorem 4.1.5(b).Therefore, f is an OP mapping.

(b) Let C, D ∈ P(B). Then f−1(C ∩ D) = f−1(C) ∩ f−1(D) by Theorem4.1.5(c).

(c) Let C, D ∈ P(A). Then f−1(C ∪ D) = f−1(C) ∪ f−1(D) by Theorem4.1.5(d).

9. (a) Suppose ◦ is associative on A. Suppose u, v, w ∈ B. Then u = f(x),v = f(y) and w = f(z) for some x, y, z ∈ A. Then

(u ∗ v) ∗ w = [f(x) ∗ f(y)] ∗ f(z)= f ((x ◦ y) ◦ z) = f (x ◦ (y ◦ z)) = f(x) ∗ [f(y) ∗ f(z)] = u ∗ (v ∗ w).

(b) Suppose u = f(x) ∈ B. Then u ∗ f(e) = f(x) ∗ f(e) = f(x ◦ e) = f(x) =f(e ◦ x) = f(e) ∗ f(x) = f(e) ∗ u. Thus f(e) is the identity for B.

(c) We show that f(x−1) satisfies the condition of being the inverse of f(x).f(x−1) ∗ f(x) = f(x−1 ◦ x) = f(e) = f(x ◦ x−1) = f(x) ∗ f(x−1). Thereforef(x−1) = f−1(x).

10. (a) Let G = {e, g} and H = {e, h} be two groups with operations ◦ and ·and identities e and e. By Theorem 6.4.3, if f is to be an OP map fromG to H, then f(e) = e. This means that f(g) must be h. The functionf = {(e, e ), (g, h)} is one-to-one and maps G to H, and (by checkingfour cases) is easily seen to be OP. In the case where we must verify thatf(g ◦ g) = f(g) · f(g), we first note that g ◦ g = e and h · h = e. Thenf(g ◦ g) = f(e) = e = h · h = f(g) · f(g).

(b) Let G = {e, a, b} and H = {e, c, d} be two groups with identities e ande. Then f = {(e, e ), (a, c), (b, d)} is one-to-one and maps G onto H. Weobserve that in (G, ◦), a ◦ b �= a ◦ e and a ◦ b �= e ◦ b (by cancellation),so a ◦ b = e. Similarly each of c and d is the inverse of the other in(H, ·). Now we verify that f is OP by checking 9 cases. For example,f(a ◦ b) = f(e) = e = c · d = f(a) · f(b).

(c) Let G be the group in Exercise 2 of Section 6.2 Suppose f is an isomorphismfrom G to Z4. Then 1 = f(x), where x is one of u, v, or w. Thenf(x2) = f(x) + f(x) = 1 + 1 = 2. But x2 = e, so f(x2) = 0. This is acontradiction.

11. (a) Let 3a and 3b be in 3Z. Then f(3a + 3b) = 4(3a + 3b) = 4(3a) + 4(3b) =f(3a) + f(3b).


(b) 12Z

12. Not a homomorphism. 3, 6 ∈ 3Z and f(3 + 6) = 12, while g(3) + g(6) = 15.

13. g(0 + n) = g(n) = a ◦ g(n) = g(0) ◦ g(n) for all n ∈ Z6g(1 + 1) = g(2) = c = b ◦ b = g(1) ◦ g(1)g(1 + 2) = g(3) = a = b ◦ c = g(1) ◦ g(2)g(1 + 3) = g(4) = b = b ◦ a = g(1) ◦ g(3)g(1 + 4) = g(5) = c = b ◦ b = g(1) ◦ g(4)g(1 + 5) = g(0) = a = b ◦ c = g(1) ◦ g(5)g(2 + 2) = g(4) = b = c ◦ c = g(2) ◦ g(2)g(2 + 3) = g(5) = c = c ◦ a = g(2) ◦ g(3)g(2 + 4) = g(0) = a = c ◦ b = g(2) ◦ g(4)g(2 + 5) = g(1) = b = c ◦ c = g(2) ◦ g(5)g(3 + 3) = g(0) = a = a ◦ a = g(3) ◦ g(3)g(3 + 4) = g(1) = b = a ◦ b = g(3) ◦ g(4)g(3 + 5) = g(2) = c = a ◦ c = g(3) ◦ g(5)g(4 + 4) = g(2) = c = b ◦ b = g(4) ◦ g(4)g(4 + 5) = g(3) = a = b ◦ c = g(4) ◦ g(5)g(5 + 5) = g(4) = b = c ◦ c = g(5) ◦ g(5)

This covers all the cases since both groups are abelian.

14. (a) Suppose x = y in Z18. Then 18 divides x− y. Therefore 6 divides x− y andso 24 divides 4(x − y) = 4x − 4y. Thus f(x) = [4x] = [4y] = f(y) in Z24, sof is well defined.Now let x, y ∈ Z18. Then f(x +18 y) = [4(x + y)] = [4x + 4y] = [4x] +24 [4y] = f(x) +24 f(y).

(b) Rng(f) = {[0] , [4] , [8] , [12] , [16] , [20]}.[0] [4] [8] [12] [16] [20]

[0] [0] [4] [8] [12] [16] [20][4] [4] [8] [12] [16] [20] [0][8] [8] [12] [16] [20] [0] [4][12] [12] [16] [20] [0] [4] [8][16] [16] [20] [0] [4] [8] [12][20] [20] [0] [4] [8] [12] [16]

15. (a) Suppose x = y in Z15. Then 15 divides x − y, so 3 divides x − y, so 12divides 4x − 4y and so f(x) = [4x] = [4y] = f(y) in Z24. Thus f is well-defined. Now let x, y ∈ Z15. Then f(x +15 y) = [4(x + y)] = [4x + 4y] =[4x] +12 [4y] = f(x)+12f(y).

(b) Rng(f) = { [0] , [4] , [8] }[0] [4] [8]

[0] [0] [4] [8][4] [4] [8] [0][8] [8] [0] [4]

16. Let e be the identity in G. Then f(e) = i, so e ∈ker(f). Thus ker(f) is not empty.Suppose a, b ∈ker(f). Then f(a◦b−1) = f(a)∗f(b−1) = f(a)∗f−1(b) = i∗i−1 = i,so a ◦ b−1 ∈ker(f). Thus ker(f) is a subgroup of G.

17. Define f : Z4 → {1,−1, i,−i} by setting f(n) = in . Then f is a bijection andfor m, n ∈ Z4, f(m + n) = im+n = imin = f(m) · f(n).

18. S3 is not isomorphic to (Z6, +). The homomorphic image of an abelian group isabelian; (Z6, +) is abelian but S3 is not.


19. (a) Let (G, ·) be a group. Then the identity function IA maps A one-to-one andonto A. Suppose x, y ∈ A. Then IA(x · y) = x · y = IA(x) · IA(y).

(b) Suppose (G, ·) is isomorphic to (H, ∗). Then there exists f : G1–1,Onto,OP−→ H.

The f−1 is a function that maps H one-to-one and onto G. By Exercise5(b), f−1 is OP. Therefore (H, ∗) is isomorphic to (G, ·).

(c) Suppose (G, ·) is isomorphic to (H, ∗) and (H, ∗) is isomorphic to (K, ⊗).Then there exist f : G

1–1,Onto,OP−→ H and g : H1–1,Onto,OP−→ K. Then the

composite function maps G one-to-one and onto K. By Exercise 5(a), g ◦ fis OP. Therefore (G, ·) is isomorphic to (K, ⊗).

20. For each group G, we form the group H = {θa : a ∈ G} of left translations. Asshown in Theorem 6.4.5, H is a permutation group isomorphic to G.

(a) H = {θ0, θ1, θ2} where θ0 = [0 12], θ1 = [1 2 0], and θ2 = [2 0 1].

(b) H = {θ0, θ1, θ2, θ3, θ4} where θ0 = [0 1 2 3 4], .θ1 = [1 2 3 4 0],θ2 = [2 3 4 0 1], θ3 = [3 4 0 1 2], and θ4 = [4 0 1 2 3].

(c) H = {θa : a ∈ R} where θa : R → R is given by θa(r) = a + r.

21. (a) F. One cannot show that the claim holds for every pair of points in R × R

by giving one example.

(b) F. One must begin by supposing a and b are elements of G. The serious erroris the omission of operation symbols in each step. The “proof” should say:g◦f(a∗b) = g(f(a∗b) = g(f(a)·f(b)) = g(f(a))⊗g(f(b)) = g◦f(a)⊗g◦f(b).

6.5 Exercise Solutions

1. (a) Not a ring. (N, +) is not an abelian group because it has no identity.

(b) Not a ring. The set is not closed under addition.

(c) Ring.

(d) Not a ring. The set is not closed under multiplication.

2. The set Z[√

2] is closed under the operations. The identity is 0 = 0 + 0√

2,and the inverse of a + b

√2 is −a + (−b)

√2. The commutative, associative, and

distributive properties are inherited from R.

3. Let a, b, c ∈ Zm. Then (b +m c )·m a = b + c·m a = (b + c)a = ba + ca =ba +m ca = (b·m a ) +m (c·m a )

4. Z × Z is closed under ⊕ because a + c and b + d are in Z, and closed under⊗ because ac and bd are in Z. The addition operation is commutative because(a + c, b + d) = (c + a, d + b) and associative because ((a + c) + e, (b + d) + f) =(a+(c+e), b+(d+f)). The additive identity is (0, 0) and (−a,−b) is the inverse of(a, b) because (a, b) ⊕ (−a,−b) = (0, 0). The operation ⊗ is associative because((a, b) ⊗ (c, d)) ⊗ (e, f) = (ac, bd) ⊗ (e, f) = ((ac)e, (bd)f) = (a(ce), b(df)) =(a, b) ⊗ (ce, df) = (a, b) ⊗ ((c, d) ⊗ (e, f))

The operation ⊗ distributes over ⊕ because (a, b) ⊗ (c, d) ⊕ (e, f)) = (a, b) ⊗(c + e, d + f) = (a(c + e), b(d + f)) = (ac + ae, bd + bf) = (ac, bd) ⊕ (ae, bf) =((a, b) ⊗ (c, d)) ⊕ ((a, b) ⊗ (e, f)).

5. By definition F(R) is closed under the operations. The identity is the constantfunction f0 given by f0(x) = 0, for all x ∈ R. The inverse of f is the function−f given by −f(x) = −(f(x)), for all x ∈ R. The associative, commutative anddistributive properties all follow immediately from the properties for R.


6. First b + (−a) is a solution, because (b + (−a)) + a = b + ((−a) + a) = b + 0 = b.The solution is unique, because if x + a = b, then x = x + 0 = x + (a + (−a)) =(x + a) + (−a) = b + (−a), so x = b + (−a).

7. (a) (a · 0) + 0 = a · 0 = a · (0 + 0) = a · 0 + a · 0. By cancellation, 0 = a · 0.

(b) (−a) · b + a · b = ((−a) + a) · b = 0 · b = 0. Therefore (−a) · b is the additiveinverse of a · b.

(c) (a − b) · c = (a + (−b)) · c = a · c + (−b) · c = a · c + (−bc) = a · c − b · c.

8. (a) The nonempty set {0} is closed under + and ·. The identity is 0 and theinverse of 0 is 0. The associative, commutative and distributive propertiesare inherited from R.

(b) Suppose T is a nonempty subset of R that is closed under subtraction andmultiplication . Then by the subgroup test, (T, +) is a group, and is abelianbecause (R, +) is abelian. The associative property for and the distributiveproperty are inherited from the ring R.

9. Suppose a and b are in 3Z. Then a = 3k and b = 3m for some k, m ∈ Z. Thena − b = 3k − 3m = 3(k − m) ∈ 3Z. Therefore 3Z is a subring of Z.

10. (a) h(4 · 2) = h(8) = 24, but h(4) · h(2) = 12 · 6 = 72, so h does not preservemultiplication.

(b) h(4 · 5) = h(2) = 4, but h(4) · h(5) = 8 · 10 = 2 · 4 = 8 = 2, so h does notpreserve multiplication.

(c) Let p(x) = anxn + . . . + a1x + a0 and q(x) = bnxn + . . . + b1x + b0 be inZ[x]. Then g((p + q)(x)) = g(p(x) + q(x)) = g((an + bn)xn + . . . + (a1 +b1)x + (a0 + b0)) = (p + q)(0) = a0 + b0 = p(0) + q(0) = g(p(x)) + g(q(x)).Also, g(pq(x)) = g(pq(0)) = a0b0 = p(0) · q(0) = g(p(x)) · g(q(x)).

11. Suppose pq and r

s are in Q. Then f(pq + r

s ) = f(ps+rqqs ) = (ps + rq, qs) = (p, q) ⊕

(r, s) = f(pq )⊕f( r

s ) and f(pq · r

s ) = f(prqs ) = (pr, qs) = (p, q)⊗(r, s) = f(p

q )⊗f( rs ).

12. Suppose 0 has a multiplicative inverse x. Then x · 0 = 1. But x · 0 = 0, so 0 = 1.This contradicts the definition of an integral domain.

13. (a) Let m ∈ N. If m is composite then m = st for some natural numbers s, tsuch that 1 < s, t < m. Then in Zm we have st = 0, so s and t are divisorsof 0. Now suppose m is prime and a and b are divisors of zero in Zm. Thenm divides ab. By Euclid’s Lemma, m divides a or m divides b. But thisis impossible, because a and b are less then m. We conclude that if m isprime, then Zm has no divisors of zero.

(b) We have already verified that for all m > 1, (Zm.+, ·) is a commutativering with unity. Thus (Zm.+, ·) is a field iff (Zm.+, ·) has no zero divisorsiff m is prime.

14. Let a, b, c ∈ R. Assume b · a = c · a and a �= 0. Then b · a − c · a = (b − c) · a = 0.Since a �= 0 and R has no divisors of zero, b − c = 0. Therefore b = c.

15. (a) Suppose a, b, c ∈ R. Then (a + b) · c = c · (a + b) [because · is commutative]and c · (a + b) = c · a + c · b = a · c + b · c.

(b) Suppose a, b ∈ R, a · b = 0, and a �= 0. Then a has a multiplicative inversea−1 in R−{0}, and a−1 · (a ·b) = a−1 ·0 = 0. But a−1 · (a ·b) = (a−1 ·a) ·b =1 · b = b . Therefore b = 0.

16. (a) F. The statement is false. (b) A.

7 Concepts of Analysis

7.1 Completeness of the Real Numbers

1. (a)√

10, 4, 11, 43 (b) 13 , 1, 2, 3

(c) 5+√

212 , 400, 10100, 2222

(d) 1,√

2, π, e(e) −1, − 1

2 , 0, 5 (f) No upper bound exists.(g) 11,044, 11.5, 12.0, 98.6

2. (a) −√

10 (b) 0(c) No lower bound exists (d) −3(e) No lower bound exists (f) 0(g) 10−11

3. (a) sup: 1, inf: 0 (b) sup: 2, inf: 1(c) sup: does not exist, inf: 0 (d) sup: 1.5, inf: -2(e) sup: 1, inf: 1

3 (f) sup:√

10, inf: −√

10(g) sup: 5, inf: −1 (h) sup: 1, inf: −1(i) sup: does not exist, inf: 0 (j) sup: does not exist, inf: does not

exist

4. (a) Suppose A is bounded above and B ⊆ A. Then there exists a real numberu such that for all x ∈ A, x ≤ u. Since B ⊆ A, if b ∈ B, then b ∈ A and sob ≤ u. Therefore B is bounded above by u.

(b) Suppose A is bounded below and B ⊆ A. Then there exists a real numberv such that for all x ∈ A, v ≤ x. Since B ⊆ A, if b ∈ B, then b ∈ A and sov ≤ b. Therefore B is bounded below by v.

(c) Suppose A is bounded above and B is bounded above. Then there exist realnumbers u and v such that for all a ∈ A, a ≤ u and for all b ∈ B, b ≤ v.Let w = maxu, v. Then for all a ∈ A, a ≤ w and for all b ∈ B, b ≤ w.Therefore for all x ∈ A ∪ B, x ≤ w. Hence A ∪ B is bounded above.

(d) Suppose A is bounded below and B is bounded below. Then there existreal numbers u and v such that for all a ∈ A, u ≤ a and for all b ∈ B,v ≤ b. Let w = minu, v. Then for all a ∈ A, w ≤ a and for all b ∈ B, w ≤ b.Therefore for all x ∈ A ∪ B, w ≤ x. Hence A ∪ B is bounded below.

5. (a) Suppose x < y. Then if a ∈ A, a ≤ x < y, so y is an upper bound for a.

(b) Suppose x ∈ A. Suppose also that t is another upper bound for A. ThenThen x ≤ t, since x ∈ A. Thus x = supA.

6. (a) Suppose that b is an upper bound for A and assume that Ac also has anupper bound, c. Let x ∈ R be a number greater than b or c. Then x �∈ Aand x �∈ Ac. This is a contradiction.

7. A = Z

8. (a) Let x and y be least upper bounds for A. Then x and y are upper boundsfor A. Since y is an upper bound and x is a least upper bound, x ≤ y. Sincex is an upper bound and y is a least upper bound, y ≤ x. Thus x = y.

(b) Let x and y be greatest lower bounds for A. Since y is a lower bound andx is a greatest lower bound, y ≤ x. Since x is a lower bound and y is agreatest lower bound, x ≤ y. Thus x = y.

9. (a) Let x = sup(A) and y = sup(B). If a ∈ A, then a ∈ B, so a ≤ y. Thus y isan upper bound for A. Therefore x ≤ y.

1

7 CONCEPTS OF ANALYSIS 2

(b) Let x = inf(A) and y = inf(B). If a ∈ A, then a ∈ B, so y ≤ a Thus y is alower bound for A. Therefore y ≤ x.

10. Let A be a subset of an ordered field F . Then

y = inf(A) iff (1) (∀ε > 0)(x ∈ A ⇒ x > g−ε) and (2) (∀ε > 0)(∃y ∈ A)(y < g+ε).

Proof. First suppose g = inf(A). Let ε > 0. Then x ≥ g > g − ε for all x ∈ A,which establishes property (1). To verify property (2), suppose there were no ysuch that y < g + ε. Then g + ε is a lower bound for A, which is greater thanthe greatest lower bound of A. This is a contradiction.

Suppose now thatg satisfies (1) and (2). To show that g is a lower bound for A,assume there is a y ∈ A such that y < g. If we let ε = g−y

2 , then y < g − ε, whichviolates (1). Thus g is a lower bound for A.

Now assume that there is another lower bound t such that g < t. If we letε = t − g, then (2) says that there is some y ∈ A such that y < g + ε, so t is nota lower bound. Therefore g is the greatest lower bound of A.

11. (a) (0, 4) (b) {x ∈ Q: x ∈ (0, 4)}(c) impossible (d) {5, 6}

12. {x ∈ Q: x2 < 2}

13. (a) Let s = sup(A), B = {u: u is an upper bound for A}. Then B is boundedbelow (by any element of A), so inf(B) exists. Let t = inf(B). We showthat s = t.

i. To show t ≤ s we note that since s = sup(A), s is an upper bound forA. Thus s ∈ B. Therefore t ≤ s.

ii. To show s ≤ t we will show t is an upper bound for A. If t is not anupper bound for A, then there exists a ∈ A with a > t. Let ε = a−t

2 .Since t = inf(b) and t < t+ε, there exist u ∈ B such that u < t+ε. Buttε < a. Therefore, u < a. This is a contradiction, since u is an upperbound for a.

(b) Let i = inf(a) and C = {l: l is a lower bound of A}. C is bounded aboveby elements of A, so sup C exists. Let j = sup(C). We show that i = y.

i. Since i = inf(A), i is a lower bound for A. Thus i ∈ C. Since j = sup(c),i < j.

ii. We claim j is a lower bound for A. If not, there is a ∈ A such thata < j. Let ε = j−a

2 . Since j = sup(C) and ε > 0, there is l ∈ C suchthat l > j − ε, But j − ε > a. Therefore, l > a, which contradicts thefact that l is a lower bound for a.

14. (a) Let m = max{sup(A), sup(B)}.

i. Since A ⊆ A ∪ B, sup(A) ≤ sup(A ∪ B). Also, B ⊆ A ∪ B implessup(A) ≤ sup(A∪B). Thus m = max{sup(A), sup(B)} ≤ sup(AcupB).

ii. It suffices to show m is an upper bound for A ∪ B. Let x ∈ A ∪ B. Ifx ∈ A, then x ≤ sup(A) ≤ m. If x ∈ B, then x ≤ sup(B) ≤ m. This mis an upper bound for A ∪ B. Hence sup(A ∪ B) ≤ m.

(b) If inf(A) and inf(B) exist, then inf(A ∪ B) exists and inf(A ∪ B) =min{inf(A), inf(B)}.


Proof. Let x = inf(A), y = inf B, and z = min{x, y}. If a ∈ A ∪ B, thena ∈ A, in which case z ≤ x ≤ a, or a ∈ B, in which case z ≤ y ≤ a. So z isa lower bound for A ∪ B. If t is another lower bound for A ∪ B, then t is alower bound for both A and B, so t ≤ x and t ≤ y, so therefore t ≤ z.

15. (a) Let A = {1, 2, 3, 4} and B = {1, 32 , 5

2 , 72}. Then sup(A ∩ B) = 1.

(b) If sup(A) and sup(B) exist, then sup(A ∩ B) exists and sup(A ∩ B) ≤min{sup(A), sup(B)}.

Proof. A∩B ⊆ A and A is bounded above, so A∩B is bounded above. Thussup(A ∩ B) exists. By Exercise 9(a), sup(A ∩ B) ≤ min{sup(A), sup(B)}.

16. (a) Let A = {13 , 1

2 , 1} and B = {−1, 0, 1}. Then inf(A ∩ B) = 1.

(b) If inf(A) and inf(B) exist, then inf(A ∩ B) exists and inf(A ∩ B) ≥max{inf(A), inf(B)}.

Proof. A∩B is bounded below because A is bounded and A∩B ⊆ A. There-fore inf(A∩B) exists. By Exercise 9(b), inf(A∩B) ≥ max{inf(A), inf(B)}.

17. Suppose there is a positive real number r such that for all integers K, 1K ≥ r.

Then for all K, 1 ≥ Kr. Therefore the set W = {nr: n ∈ N} is bounded aboveby 1. By the completeness property for R, sup(W ) exists. Let t = sup(W ). Thent − r < t, so t − r is not an upper bound for W . Thus there exists mr ∈ W suchthat mr > t − r. Then mr + r > t, so (m + 1)r > t. But (m1)r is in W and t isan upper bound for W . so this is impossible.

18. For the irrational number z, let B = {x ∈ Q: x < z}.

19. (a) Suppose F is a complete ordered field. Let B be a nonempty subset of Fthat has a lower bound v. Let A = {−x: x ∈ B}. Then A is a subset of Fand since v ≤ b for all b ∈ B, −v ≥ −b for all b ∈ B; that is, −v ≥ a for alla ∈ A. Thus A is bounded above by −v and so s = sup(A) exists. We nowshow −s is inf(B).Since s = sup(A), s is an upper bound for A. Therefore, a ≤ s for all a ∈ A.Thus −a ≥ −s for all a ∈ A; that is, b ≥ −s for all b ∈ B. Hence −s is alower bound for B.Let w be a lower bound for B. Then w ≤ b for all b ∈ B. Therefore −w ≥ −bfor all b ∈ B; that is, −w ≥ a for all a ∈ A. Thus, −w is an upper boundfor A and so −w ≥ s. Therefore w ≤ −s, which proves −s = inf(B).

(b) Suppose F is a field with the property that every nonempty subset of Fthat has a lower bound in F has an infimum in F . Let A be a nonemptysubset of A that is bounded above by u. To show F is complete, we showthat sup(A) exists. Let B = {−x: x ∈ A}. Then B is a subset of F andsince a ≤ u for all a ∈ A, −a ≥ −u for all a ∈ A; that is, b ≥ −u for allb ∈ B. Thus B is bounded below by −u and so z = inf(B) exists. We nowshow that −z is sup(A).Since z = inf(B), z is a lower bound for B. Therefore, z ≤ b for all b ∈ B.Thus −z ≥ −b for all b ∈ B; that is, −z ≥ a for all a ∈ A. Hence −z is anupper bound for A.Let t be an upper bound for A. Then a ≤ t for all a ∈ A. Therefore −a ≥ −tfor all a ∈ A; that is, b ≥ −t for all b ∈ B. Thus, −t is a lower bound for Band so z ≥ −t. Therefore −z ≤ t, which proves −z = sup(A).

20. (a) By the trichotomy property, at least one of the properties must be true.Suppose x < y and y < z. Then x < x, which is a contradiction. Suppose


x < y and x = y. Then y < y, another contradiction. Finally, x = y andy < x leads to x < x as well. Thus at most one of x < y, x = y, and y < xcan be true.

(b) Assume x < 0. Then 0 = x + (−x) < 0 + (−x) = −x, so −x > 0.

(c) It is helpful to prove first that (−1)(−1) = 1.(−1)(−1) = (−1)(−1 + 0) = (−1)(−1 + 1 + (−1)) = (−1)(−1) + (−1) +(−1)(−1). Adding the inverse of (−1)(−1) to each side of this equation, wehave 0 = (−1)(−1) + (−1). Since 1 + (−1) = 0, we have (−1)(−1) = 1.Now 0 �= 1, and by part (a), exactly one of 0 < 1, 0 = 1, 1 < 0 istrue. Suppose 1 < 0. Then by part (b), 0 < −1. By travsitivity 1 < −1.Multiplying by −1, we have −1 < 1, which contradicts 1 < −1. Therefore0 < 1.

(d) By part (c), 0 < 1. Thus −1 = 0 + (−1) < 1 + (−1) = 0.

(e) Assume x < y. Then 0 = −x + x < −x + y, so −y < −x + y − y = −x.

(f) We first prove two lemmas.

i. For all x ∈ F , 0 · x = 0.

Proof. 0 ·x = 0 ·x+x+(−x) = 0 · · ·x+1 ·x+(−x) = (0+1)x+(−x) =x + (−x) = 0.

ii. For all x ∈ F , −1 · x = −x.

Proof. −1·x = −1·x+x+(−x) = −1·x+1·x+(−x) = (−1+1)x+−x =0 · x + (−x) = 0 + (−x) = −x.

Now assume x < y and z > 0. Then 0 < −z, so −xy = (−1)xz =x(−1)z = x(−z) < y(−z) = y(−1)z = (−1)yz = −yz. Therefore, bypart (c), xy > yz.

(g) 1 = 1 + 0 = 1 + (−1) · (0) = 1 + (−1) · [(−1) + 1] = 1 + (−1) · (−1) +(−1) · (1) = (−1) · (−1) + 1 · 1 + (−1) · 1 = (−1) · (−1) + 1 · 1 + (−1) · 1 =(−1) ·(−1)+[1+(−1)] ·1 = (−1) ·(−1)+0 ·1 = (−1) ·(−1)+0 = (−1) ·(−1).

(h) 0 = 0 · x + −(0 · x) = (0 + 0) · x + −(0 · x) = (0 · x + 0 · x) + −(0 · x) =0 · x + [0 · x + −(0 · x)] = 0 · x + 0 = 0 · x.

(i) −x = −x+0− = x+(0 ·x) = −x+ [1+ (−1)] ·x = −x+ [1 ·x+(−1) ·x] =−x + [x + (−1) · x] = (−x + x) + (−1) · x = 0 + (−1) · x = (−1) · x.

21. (a) F. The claim is true, but y = i + ε2 might not be in A.

(b) F. The claim is false. A = {0} is a counterexample.

(c) A.

(d) F. The claim is false. The “proof” assumes without justification that f isincreasing, but the claim is false even for increasing functions. The fatalflaw is the assumption that m ∈ Dom(f).

(e) A.

7.2 The Heine–Borel Theorem

1. (a) x = 9.5, δ = 2.5 (b) x = 3.825, δ = 0.025 (c) x = 6.0235, δ = 0.0005

2. (a) N (x1, δ1) ∩ N (x1 ∩ δ2) = N (x1, δ3), where δ3 = min{δ1, δ2}.

(b) If |x2 − x1| < 2δ1, then N (x1, δ1) ∩ \(x2, δ1) = N(

x1+x22 , δ1 − |x1+x2|

2

). If

|x2 − x1| ≥ 2δ1, then N (x1, δ1) ∩ \(x2, δ1) = ∅.


(c) N (x1, δ1) ∩ N (x2, δ2) = {x ∈ R: |x1 − x| < δ1 and |x2 − x| < δ2}. This setmay be empty.

3. limx→∞ f(x) = L if for all ε > 0 there is a δ > 0 such that if x ∈ N (a, δ) thenf(x) ∈ N (L, ε).

4. (a) (−1, 1) (b) (−1, 1) (c) ∅(d) ∅ (e) ∅ (f) ∅(g) R − N (h) R − ({ 1

3k : k ∈ �} ∪{∅})

(i)⋃

n∈N(n + 0.1, n +

0.2)

5. (a) open (b) open (c) neither (d) neither (e) open(f) closed (g) open (h) open (i) closed (j) neither

6. (a,∞) =⋃

n∈N(a, a + n) and (−∞, a) =

⋃n∈N

(a − n, a), so by Theorem 7.2.1and Theorem 7.2.2(a), (a,∞) and (−∞, a) are open sets.

7. (a) [a,∞)c = (−∞, a) and (−∞, a]c = (a,∞). Since their complements areopen, the closed rays are closed.

(b) [a, b]c = (−∞, a) ∪ (b, ∞) so by solution 6 and Theorem 7.2.2(a), [a, b] is aclosed set.

8. Suppose A is a nonempty collection of closed subsets of R. Then for eachA ∈ A, Ac is open. By Theorem 7.2.2(a),

⋃A∈A Ac is open. Therefore⋂

A∈A A = ((⋂

A∈A A)c)c = (⋃

A∈A Ac)c is closed.

Supposed A is a finite nonempty collection of closed subsets of R. Then foreach A ∈ A, Ac is open. By Theorem 7.2.2(b)

⋂A∈A Ac is open. Therefore⋃

A∈A A = ((⋃

A∈A A)c)c = (⋂

A∈A Ac)c is closed.

(a)(b)⋃

x∈(0,1){x} is not closed.

9. (a) Let A be an open set. Then A − {x} = A ∩ [(−∞, x) ∪ (x,∞)], which isopen by Theorem 7.2.2.

(b) Let A be open and B be closed. Then Bc is open and so A − B = A ∩ Bc

is open.

(c) Let A be open and B be closed. Then (B − A)c = (B ∩ Ac)c = Bc ∪ A,which is open, so B − A is closed.

10. Designate the set of interior points of A and∫

(A). If∫

(A) is empty, then it’sopen. Otherwise, let x ∈

∫(A). Since x is an interior point of A, there is a

δ1 such the N (x, δ1) lies in A. (We will show N (x, δ1) ⊆∫

(A) as well.) Picky ∈ N (x, δ1). Since y ∈ N (x, δ1) is an open set in R, there is a δ2 such thatN (y, δ2) ⊆ N (x, δ1) ⊆ A. Thus y is an interior point of A, or in other words,y ∈

∫(A). Therefore N (x, δ1) ⊆

∫(A), so x is an interior point of

∫(A). This

shows that∫

(A) is open.

11. (a) (i) 2, 5 (ii) 0, 1 (iii) 3, 5, 6 (iv) Q

(b) Suppose x is a boundary point of A. Then x is not an interior point of A,because every δ-neighborhood of x intersects Ac. Also, x is not an interiorpoint of Ac, because every δ-neighborhood of x intersects A. Now suppose xis not an interior point of A and not an interior point of Ac. Let δ > 0. ThenN (x, δ) �⊆ A, so N (x, δ) ∩ Ac �= ∅, and N (x, δ) �⊆ Ac, so N (x, δ) ∩ A �= ∅.

(c) First, suppose A is open and x is a boundary point of A. Then by part (b),x is not an interior point of A. But every point of A is an interior point ofA, so x �∈ A. Now suppose A is a set that contains none of its boundarypoints, and suppose x ∈ A. Then x is not a boundary point of A, so by


part (b), x is an interior point of A or an interior point of Ac. Since x isnot in Ac, x is not an interior point of Ac. Therefore x is an interior pointof A. Therefore A is open.

(d) First note that the boundary of A is the same set as the boundary of Ac.Now A is closed iff Ac is open iff Ac contains none of its boundary pointsiff A contains alll of its boundary points.

12. Suppose A is closed and A ∩ N (x, δ) �= ∅ for all δ > 0. Then x is not an interiorpoint of Ac. But Ac is open, so the interior of Ac is Ac. Therefore x ∈ A.

13. (a) not compact (not bounded) (b) compact(c) compact (d) not compact (not closed, not

bounded)(e) compact (f) compact(g) not compact (not closed) (h) not compact (not closed)

14. (a) A = (0, 1) and C = {( 1n , 2): n ∈ N}

(b) A = N and C = {(0, n + 1):n ∈ N}(c) A = (0, 1), B = [01, ], C = (0, 2], D = [0, 2].

15. (a) Let A and B be compacts subsets of R. Suppose {Oα: α ∈ Δ} is a coverfor A ∪ B. Then {Oα: α ∈ Δ} is a cover for A and a cover for B. Thusthere are finite sets Δ1 and Δ2 such that {Oα: α ∈ Δ1} is a subcover for Aand {Oα: α ∈ Δ2} is a subcover for B. Then {Oα: α ∈ Δ1 ∪ Δ2} is a finitesubcover for A ∪ B. Thus A ∪ B is compact.

(b) Let A and B be compacts subsets of R. By the Heine-Borel Theorem, A isclosed and bounded and B is closed and bounded. By Exercise 8, A ∩ B isclosed. Since A ∩ B ⊆ B, A ∩ B is bounded. Therefore, by the Heine-BorelTheorem, A ∩ B is compact.

(c) Let A and B be compacts subsets of R. By the Heine-Borel Theorem, A isclosed and bounded and B is closed and bounded. By Exercise 8, A ∩ B isclosed. By Exercise 4 of Section 7.1, A ∪ B is bounded. Therefore, by theHeine-Borel Theorem, A ∪ B is compact.

16. (a) Let x ∈ S. Since limn→∞n+22n = 0 and 21/n > 1 for all n, there is an n such

that x ∈(

n+22n , 21/n

). Thus (0, 1] ⊆ ∪C∈IC.

(b) No.(c) S is not compact (not closed).

17. L {Aα: α ∈ Δ} be a family of compact sets. Then ∪αinΔAα is closed (byExercise 8(a)). Let β ∈ Δ. Then ∪αinΔAα ⊆ Aβ is bounded. By the Heine-Borel Theorem, ∪αinΔAα is compact.

18. Let A = {{x}: x ∈ R} or {[−n, n]:n ∈ N} or{

[ 1n+1 , 1]:n ∈ N

}.

19. (a) C. The idea is correct, but δ must be chosen to be x − a.(b) C. One must begin with an arbitrary open cover of A ∪ B.(c) C. With the addition of O∗ to the cover {Oα: α ∈ Δ} we are assured

that there is a finite subcover of {O∗} ∪ {Oα: αinΔ}, but not necessarilya subcover of {Oα: αinΔ}. Since O∗ = R − B is useless in a cover of B, itcan be deleted from the subcover.

(d) A.(e) F. The claim is false. It is not sufficient to show that one cover has a finite

subcover.


7.3 The Bolzano–Weierstrass Theorem

1. (a) Let x be 7 and suppose δ > 0. If δ > 1, then 6 is a point of N(x, δ) thatis in [3, 7) and is distinct from 7. If δ ≤ 1, then 7 − δ

2 is a point of N(x, δ)that is in [3, 7) and is distinct from 7.

(b) Let A = { 1+(−1)n

n : n ∈ N}. Suppose δ > 0. Let n be an even natural number

greater than 2/δ. Then 2/n < δ. Since n is even,∣∣∣ 1+(−1)n

n

∣∣∣ = 2/n < δ. Thus1+(−1)n

n is in N (0, δ) and is distinct from 0.

(c) Let A = {(1 + 1n )n: n ∈ N}. Let δ > 0. Since limn→∞(1 + 1

n )n = e, thereexists a natural number N such that for all n > N , (1 + 1

n )n − e < δ. Then(1 + 1

N+1 )N+1 is an element of A and N (e, δ). Therefore A ∩ N (e, δ) �= ∅.Therefore, e is an accumulation point for A.

2. (a) N (b) { 1n : n ∈ N}

(c) {(−1)n(n+1n ): n ∈ N} (d) {m − 1

n : m, n ∈ N}(e) Q

3. (a) { 12} (b) ∅ (c) ∅ (d) {0} (e) [0, 1]

(f) [3, 7] (g) {0, 2} (h) ∅ (i) [0, 1] (j) {0}(k) [−1, 1] (l) [0, 1] (m) N (n) {0}

4. {0, 1}

5. Since z is an upper bound for A and z /∈ A. Let δ > 0. Then z − δ < sup(A), sothere is an x ∈ A such that z − δ < x < z (by Theorem 7.1.1). Thus x ∈ N (x, δ)and x is distinct from z.

6. (a) Suppose A ⊆ B ⊆ R and x ∈ A′. Let δ > 0. Then there is an a ∈ A distinctfrom x such that a ∈ N (x, δ). Since a ∈ B, there is an element of B distinctfrom x in N (x, δ). Therefore x ∈ B′.

(b) No. [0, 1]′ ⊆ (0, 1)′, but [0, 1] �⊆ (0, 1).

7. (a) First notice that A ⊆ A ∪ B and B ⊆ A ∪ B, so by Exercise 6(a),A′ ⊆ (A ∪ B)′ and B′ ⊆ (A ∪ B)′. Thus A′ ∪ B′ ⊆ (A ∪ B)′. Now supposethere is an x /∈ A′∪B′. Then there exist δ1 and δ2, both greater than 0, suchthat (N (x, δ1)−{x})∩A = 0 = (N (x, δ2)−{x})∩B. So for δ = min{δ1, δ2},we have (N (x, δ) − {x}) ∩ (A ∪ B) = ∅. Thus x /∈ (A ∪ B)′.

(b) A ∩ B ⊆ A and A ∩ B ⊆ B, so by Exercise 6(a), (A ∩ B)′ ⊆ A′ and(A ∩ B)′ ⊆ B′. Thus (A ∩ B)′ ⊆ A′ ∩ B′.

(c) A = [0, 1) and B = (1, 2]. (A ∩ B)′ = ∅ and A′ ∩ B′ = {1}.

8. (a) Suppose B is closed and A ⊆ B. Let x ∈ A′. Then every neighborhood ofx meets (has a nonempty intersection with) A, so every neighborhood of xmeets the closed set B. By Lemma 7.2.4, x ∈ B.

(b) We first show that (A′) ⊆ A′. Let x ∈ {A′}′ and let δ > 0. Then there isa point y ∈ A′ such that y ∈ N (x, δ). If x = y, then x ∈ A′. If x �= y,let δ1 = δ − x − y. Then δ1 > 0 because y ∈ N (x, δ) and δ1 < δ becausex �= y. Since y ∈ A′ there exists z ∈ A such that z ∈ N (x, δ1). But thenz ∈ N (x, δ) because

x − z < x − y + y − z < x − y + δ1 = x − y + (δ − x − y) = δ.

Therefore x ∈ A′. We conclude that (A′)′ ⊆ A′.


From Exercise 7(a) and the above,

(A ∪ A′)′ = A′ ∪ (A′)′ ⊆ A′ ∪ A′ = A′ ⊆ A ∪ A′.

Since (A ∪ A′)′ ⊆ A ∪ A′, the set A ∪ A′ is closed by Theorem 7.3.2.

9. (a) Suppose x is an interior point of A. Then there exists δ′ such that N (x, δ′) ⊆A. Let δ > 0. Let δ′′ = min(δ, δ′); Then N (x, δ′′) ⊆ A and x+ δ′′

2 is a pointof A distinct from x in N (x, δ′′) ⊆ N (x, δ).

(b) False. An accumulation point for a set need not even be in the set, whereasinterior points of A are always in A.

(c) Assume S is open. The interior of S is a subset of S′, by part (a). Since Xis a subset of the interior of S, S ⊆ S′.

(d) False. [0, 1] ⊆ [0, 1]′, but [0, 1] is not open

10. (a) no accumulationpoints

(b) has at least one ac-cumulation point

(c) has at least one ac-cumulation point

(d) may have no accu-mulation points

(e) has one accumula-tion point

(f) has accumulationpoints

(g) has at least one ac-cumulation point

11. Let A ⊆ R. Let x ∈ (A′)c. Then x /∈ A′. Therefore, there exists ε > 0 such thatN (x, ε) ∩ A = ∅. Thus N (x, ε) ⊆ Ac.

To show that x ∈ (Ac)′, let δ > 0. Choose δ1 = min(δ, ε). Then N (x, δ1) ⊆N (x, ε) ⊆ Ac. Thus N (x, δ1) ∩ Ac �= ∅. Hence N (x, δ) ∩ Ac �= ∅, so x ∈ (Ac)′.

12. Let δ > 0. Then by Theorem 7.3.1, N (x, δ) contains an infinite number of pointsof A, so N (x, δ) contains an infinite number of points of A−F . Thus x ∈ (A−F )′.

13. (a) F. Part (ii) uses the false deduction (∀x)(P (x)∨Q(x)) (∀x)P (x)∨(∀x)Q(x).(b) F. The errors occurs at the third “iff.”(c) F. (Bc)′ need not be a subset of (B′)c.(d) A.(e) F. The proof assumes (incorrectly) that if an infinite set of real numbers

has an accumulation point, it must be bounded.

7.4 The Bounded Monotone Sequence Theorem

1. (a) Bounded below (b) Bounded(c) Bounded (d) Bounded below(e) Bounded (f) Bounded(g) Not bounded (h) Not bounded(i) Bounded (j) Not bounded(k) Bounded (l) Bounded

2. (a) xn = (−1)n (b) xn = n (c) xn = (−1)n

n(d) xn = (−1)n (e) xn = π

2

3. Pick B so that |yn| ≤ B or all n ∈ mathbbN . Let ε > 0. Choose N so that n > Nimplies |xn| ≤ ε

B . Now suppose n > N . Then |xnyn − 0| = |xn| |yn| ≤ |xn|B < ε.

4. Suppose |xn| ≤ B for all n ∈ N. Then −B ≤ xn ≤ B for all n ∈ N,so x is bounded. Now suppose x is bounded. Then there eist B1 and B2 sothat B1 ≤ xn ≤ B2 for all n ∈ N. Then |xn| ≤ B for all n ∈ N, whereB = max{|B1|, |B2|}.


5. (a) The sequence x given by xn = n+2n is decreasing. Suppose that m, n are

natural numbers and m < n. Then

1m

>1n

,

2m

>2n

,

1 +2m

> 1 +2n

,

m + 2m

>n + 2

n.

(b) The sequence y given by yn = 2−n is decreasing. Suppose that m, n arenatural numbers and m < n. Then n−m > 0 and thus 2n−m > 1. Thereforeym = 2−m = 2−n2n−m > 2−n = yn.

(c) The sequence x given by xn = (n − 2)(n − 5)2 is neither increasing nordecreasing because the first four terms are x1 = −16, x2 = 0, x3 = 4,x4 = 2.

(d) The sequence y given by yn = 10n! is decreasing. Suppose that m, n are

natural numbers and m < n. Then n! = (n − m)!(m!) > m!, so n!m! > 1.

Then 10m! = n!

m! · 10n! > 10

n! .

(e) The sequence x given by xn = 2n−5n+3 is increasing. Suppose that m, n are

natural numbers and m < n. Then

m + 3 < n + 31

m + 3>

1n + 3

−11m + 3

<−11n + 3

2 − 11m + 3

< 2 − 11n + 3

2m − 5m + 3

<2n − 5n + 3

(f) The sequence y given by yn = n!nn is decreasing. We will compare consecutive

terms. For n ∈ N,

yn =n!nn

=n!(n + 1)nn(n + 1)

=(n + 1)!

nn(n + 1)=

(n + 1)!(n + 1)n(n + 1)

=(n + 1)!

(n + 1)n+1 = yn+1

(g) The sequence x given by xn =√

n + 1 is increasing. We first note that asimple proof by contradiction shows that if t is a real number such that0 < t < 1, then 0 <

√t < 1. Suppose that m, n are natural numbers and

m < n. Then m+1 < n+1, which implies 0 < m+1n+1 < 1 and 0 <

√m+1n+1 < 1.

Then√

m + 1 =√

(m + 1)n+1n+1 =

√n + 1

√m+1n+1 <

√n + 1.

6. Assume x is a bounded, decreasing sequence.

If {xn: n ∈ mathbbN} is finite, then let L = min{xn: n ∈ mathbbN}. For someN ∈ N, xN = L and since x is decreasing, xn = L for all n > N . THereforexn → L.


Supposed {xn: n ∈ N} is infinite. Then by the Bolzano-Weierstrass Theorem,{xn: n ∈ N} must have an accumulation point, L. Wd claim L ≤ xn for alln ∈ N If there exists N such that xN < L, then xn < L for all n > N . ByExercise 12 of Section 7.3, L is an accumulation point of {xn : n ≥ N}. But forδ = |xn − L|, N(L, δ contains no points of {xn: n ≥ N}. This is a contradiction.Thus xn ≥ L for all n. Now show that xn → L. Let ε > 0. Since L is anaccumultion point of {xn: n ∈ N}, there exists M ∈ N such that xM ∈ N (L, ε).Suppose n > M . Then L − ε < L ≤ xn ≤ xM < L + ε, so |xn − L| < ε.

7. Assume k, n ∈ N and k ≤ n. We see from the binomial expansion of (n + 1)k−1

reveals that (n + 1)k−1 ≤ nk. Thus

1k!

n(n − 1) . . . [n − (k − 1)]nk

≤ 1k!

n(n − 1) . . . [n − (k − 1)](n + 1)k−1

=1k!

(n + 1)n(n − 1) . . . [n − (k − 1)](n + 1)k

8. Let x be a bounded increasing sequence. Then {xn: n ∈ N} is bounded above.By the completeness property, sup{xn: n ∈ N} exists. Let s = sup{xn: n ∈ N}.Let ε > 0. Then by part (b) of Theorem 7.1.1, there is an N ∈ N such thatxn > s− ε. Suppose n > N . Then s− ε < xN ≤ xn < s+ ε, so |xn − s| < ε. Thusxn → s.

9. Suppose x is a bounded sequence. Then there exists B such that for every n ∈ N,|xn| ≤ B. Let y be a subsequence of x, and n ∈ N. Then yn = xf(n) for someincreasing function f , and f(n) ∈ N, so |yn| = |xf(n)| ≤ B. Thus y is bounded.

10. (a) Suppose x is a Cauchy sequence and pick N ∈ N so that m, n > N implies|xn −xm| < 1. Then |xn| ≤ |xn −xN |+ |xN | < 1+ |xN | for all n ≥ N . Thusx is bounded by max{|xk| + 1 : 1 ≤ k ≤ N}.

(b) Suppose xn → L. Let ε > 0. Pick N so that n > N implies |xn − L| < ε2 .

Suppose m, n > N . Then |xm − xn| ≤ |xn − L| + |xn − L| < ε2 + ε

2 = ε.

11. (a) Let n = 1. First observe that a1 − b1 = x+y2 − √

xy = 12 (x − 2

√xy + y) =

12 (

√x − √

y)2 > 0, because x > y. Thus a1 > b1. We now compute:

a1 − a2 = a1 − 12(a1 + b1) =

12(a1 − b1) > 0;

b2 − b1 =√

a1b1 − b1 =√

b1(√

a1 −√

b1 ) > 0; and

a2 − b2 =12(a1 + b1) −

√a1b1 =

12(a1 − 2

√a1b1 + b1) =

12(√

a1 −√

b1 )2 > 0.

Therefore, a1 > a2 > b2 > b1, so the statement is true for n = 1.Now assume the statement is true for some n ∈ N. That is, assumean > an+1 > bn+1 > bn. Then

an+1 − an+2 =12(an+1 − bn+1) > 0;

bn+2 − bn+1 =√

bn+1(√

an+1 −√

bn+1 ) > 0; and

an+2 − bn+2 =12(an+1 − 2

√an+1bn+1 + bn+1) =

12(√

an+1 −√

bn+1 )2 > 0.

Therefore, an+1 > an+2 > bn+2 > bn+1, so the statement is true for n + 1.By the PMI, the statement is true for every n ∈ N.


(b) The sequence an converges because it is monotone decreasing and boundedby a. The sequence bn is convergent because it is monotone increasing andis also bounded by a.

(c) Let limn→∞ an = L and limn→∞ bn = M . Then limn→∞ an+1 = limn→∞an+bn

2 ,so L = L+M

2 . Therefore 2L = L + M , so L = M .

12. (a) A. The proof makes use of Exercise 6(c) of Section 4.6.

(b) F (or C). A correct proof could be made using the fact that the subsequencex3, x4, x5, . . . converges, but the Bounded Monotone Sequence Theoremdoes not apply to the sequence x.

7.5 Equivalents of Completeness

1. Let x and y be such that yn → s and xn − yn → 0. By Exercise 6(a) in Section4.6, xn = (yn + (xn − yn)) → s + 0 = s.

2. Assume xn → s and t < s. Then s − t > 0. Pick N so that |xn − s| < s − t forall n ≥ N . Then xn > t for all n ≥ N .

3. (a) {x ∈ Q: 49 ≤ x2 ≤ 50}(b) The set {7, 7.07, 7.071, . . .} of rational approximations to

√50 is a bounded

infinite subset of Q ∩ [7, 8} that has no accumulation point.

(c) The sequence whose terms are the successive decimal approximations to√50 has no limit in Q.

4. (a) Let A = (R − Q) ∩ [3, 4]. Then A is closed and bounded but not compact.

(b) {π, π − 0.1, π − 0.14, π − 0.141, . . .}

(c) {4√

n−1n : n ∈ N and n ≥ 3}

5. (a) (i) {0} (ii) [2, 4] (iii) ∅ (iv) ∅(b) Suppose An = [an, bn] is a sequence of closed intervals such that An+1 ⊆ An

for all n ∈ N. Let n ∈ N. If m ≤ n, then am ≤ an ≤ bn and if m > n thenam ≤ bm ≤ bn. Thus am ≤ bn for all m ∈ N. Therefore a = sup{an: n ∈ N}exists and a ≤ bn. Since this is true for all n ∈ N, b = inf{bn: n ∈ N} existsand a ≤ b. Thus

⋂An = [a, b] �= ∅.

6. (a) C. This is a promising idea, but the limit L is not necessarily the sup ofA. For example, if [0, 2] ⊆ A, the process described might produce thesequence xn = n

n+1 , but xn → 1 and sup(A) ≥ 2.

(b) F. The claim is correct, but there is little that is correct in this proof.For instance, the upper bound a0 for A may be negative, in which case Bwould have to be defined differently. More seriously, there is no connectionbetween being an accumulation point and being an upper bound for a set.

Introduction to Advanced Mathematics (7E) Solutions Manual.pdf

Documents

Transcript of Introduction to Advanced Mathematics (7E) Solutions Manual.pdf