1
Logic and Proofs
1.1 1.
Propositions and Connectives (a) true (e) false
(b) false (f) false
(c) true (g) false
(d) 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 been assigned. (d) Not a proposition. It would be a proposition if values for x and y had been assigned. (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)
(b)
(c)
(d)
(e)
(f)
P T F
∼P F T
P∧ ∼ P T F
P T F
∼P F T
P∨ ∼ P T T
P T F T F
Q T T F F
∼Q F F T T
P∧ ∼ Q F F T F
P T F T F
Q T T F F
∼Q F F T T
Q∨ ∼ Q T T T T
P T F T F
Q T T F F
∼Q F F T T
P ∧Q T F F F
P T F T F
Q T T F F
P ∧Q T F F F
P ∧ (Q∨ ∼ Q) T F T F (P ∧ Q)∨ ∼ Q T F T T
∼ (P ∧ Q) F T T T 1
1
LOGIC AND PROOFS
(g)
(h)
(i)
(j)
4.
2
∼Q F F T T F F T T
P∨ ∼ Q T F T T T F T T
(P ∨ ∼ Q) ∧ R T F T T F F F F
P T F T F T F T F
Q T T F F T T F F
R T T T T F F F F
P T F T F
Q T T F F
∼P F T F T
P T F T F T F T F
Q T T F F T T F F
R T T T T F F F F
Q∨R T T T T T T F F
P ∧ (Q ∨ R) T F T F T F F F
P T F T F T F T F
Q T T F F T T F F
R T T T T F F F F
P ∧Q T F F F T F F F
P ∧R T F T F F F F F
(a) false (e) false (i) true
∼Q F F T T
∼ P∧ ∼ Q F F F T
(b) true (f) false (j) true
(P ∧ Q) ∨ (P ∧ R) T F T F T F F F (c) true (g) false (k) false
(d) true (h) false (1) false
5. (a) No solution. (b)
P Q P ∨Q Q∨P T T T T F T T T T F T T F F F F Since the third and fourth columns are the same, the propositions are equivalent.
1
LOGIC AND PROOFS
(c)
(d)
(e)
(f)
(g)
3
P Q P ∧Q Q∧P T T T T F T F F T F F F F F F F Since the third and fourth columns are the same, the propositions are equivalent. P Q R Q ∨ R P ∨ (Q ∨ R) P ∨ Q (P ∨ Q) ∨ R T T T T T T T F T T T T T T T F T T T T T F F T T T F T T T F T T T T F T F T T T T T F F F T T T F F F F F F F Since the fifth and seventh columns are the same, the propositions are equivalent. P Q R Q ∧ R P ∧ (Q ∧ R) P ∧ Q (P ∧ Q) ∧ R T T T T T T T F T T T F F F T F T F F F F F F T F F F F T T F F F T F F T F F F F F T F F F F F F F F F F F F F Since the fifth and seventh columns are the same, the propositions are equivalent. P Q R Q ∨ R P ∧ (Q ∨ R) P ∧ Q P ∧ R (P ∧ Q) ∨ (P ∧ R) T T T T T T T T F T T T F F F F T F T T T F T T F F T T F F F F T T F T T T F T F T F T F F F F T F F F F F F F F F F F F F F F Since the fifth and eighth columns are the same, the propositions are equivalent. P T F T F T F T F
Q T T F F T T F F
R T T T T F F F F
Q∧R T T F F F F F F
P ∨ (Q ∧ R) T T T F T F T F
P ∨Q T T T F T T T F
P ∨R T T T T T F T F
(P ∨ Q) ∧ (P ∨ R) T T T F T F T F
1
LOGIC AND PROOFS
4
Since the fifth and eighth columns are the same, the propositions are equivalent. (h) No solution. (i)
P Q P ∨ Q ∼ (P ∨ Q) ∼ P ∼ Q ∼ P ∧ ∼ Q T T T F F F F F T T F T F F T F T F F T F F F F T T T T Since the fourth and eighth columns are the same, the propositions are equivalent.
6.
(a) (c) (e) (g)
equivalent equivalent equivalent not equivalent
7.
(a) ∼ P , true (c) P Q, true
(b) (d) (f) (h)
equivalent equivalent not equivalent not equivalent
(b) 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, Q has 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 is equivalent to R, Q and R have the same truth table. Thus, P and R have the 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, the truth table for P has value true on exactly the same lines that the truth table for Q has value true. Therefore the truth table for ∼ Q has value false on 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 T F T F
Q T T F F
∼P F T F T
∼Q F F T T
P ∧Q T F F F
∼ P∧ ∼ Q F F F T
(P ∧ Q) ∨ (∼ P ∧ ∼ Q) T F F T
(b) ∼ (P ∧ ∼ P ) is a tautology. P T F
∼P F T
P∧ ∼ P F F
∼ (P ∧ ∼ P ) T T
(c) (P ∧ Q) ∨ (∼ P ∨ ∼ Q) is a tautology. P T F T F
Q T T F F
∼P F T F T
∼Q F F T T
P ∧Q T F F F
∼ P∨ ∼ Q F T T T
(P ∧ Q) ∨ (∼ P ∨ ∼ Q) T T T T
(d) (A ∧ B) ∨ (A∧ ∼ B) ∨ (∼ A ∧ B) ∨ (∼ A∧ ∼ B) is a tautology.
1
LOGIC AND PROOFS
A T F T F
B T T F F
∼A F T F T
∼B F F T T
5
A∧B T F F F
A∧ ∼ B F F T F
∼A∧B F T F F
∼ A∧ ∼ B F F F T
(A ∧ B) ∨ (A∧ ∼ B)∨ (∼ A ∧ B) ∨ (∼ A∧ ∼ B) T T T T
(e) (Q∧ ∼ P )∧ ∼ (P ∧ R) is neither. P T F T F T F T F
Q T T F F T T F F
∼P F T F T F T F T
R T T T T F F F F
Q∧ ∼ P F T F F F T F F
P ∧R T F T F F F F F
∼ (P ∧ R) F T F T T T T T
(Q∧ ∼ P )∧ ∼ (P ∧ R) F T F F F T F F
(f) P ∨ [(∼ Q ∧ P ) ∧ (R ∨ Q)] is neither. P T F T F T F T F
10.
Q T T F F T T F F
R T T T T F F F F
∼Q F F T T F F T T
∼Q∧P F F T F F F T F
(a) contradiction (c) tautology
R∨Q T T T T T T F F
[(∼ Q ∧ P ) ∧ (R ∨ Q)] F F T F F F F F
P ∨ [(∼ Q ∧ P ) ∧ (R ∨ Q)] T F T F T F T F
(b) tautology (d) tautology
11. (a) x is not a positive integer. (b) Cleveland will lose the first game and the second game. Or, Cleveland will lose 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 not have 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 a relative 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 does not 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.
1
LOGIC AND PROOFS
13. (a)
i.
ii.
(b)
i.
ii.
iii.
1.2
A T F T F
B T T F F
6
A ∨B F T T F
A B A ∨ B A ∧ B ∼ (A ∧ B) (A ∨ B)∧ ∼ (A ∧ B) T T T T F F F T T F T T T F T F T T F F F F T F Since the final columns of the two tables are identical, the two propositions have the same truth table, thus they are equivalent. A T F T F
B T T F F
A NAND B F T T T
A NOR B F F F T
A B A NANDB A NOR B (A NAND B) ∨ (A NOR B) T T F F F F T T F T T F T F T F F T T T Since the third and last columns are equal, the propositions are equivalent. A B A NAND B A NOR B (A NAND B) ∧ (A NOR B) T T F F F F T T F F T F T F F F F T T T Since the fourth and last columns are equal, the propositions are equivalent.
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.
1
LOGIC AND PROOFS
7
(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 ( in the 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 have three 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 for the Olympic team. Contrapositive: If an athlete records a time that is not 3 minutes and 48 seconds 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 two primes. Consequent: Every odd natural number greater than 5 is the sum of three primes. (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 n elements. Consequent: A × B = mn.
1
LOGIC AND PROOFS
8
(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 a function 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 (e) true
(b) false (f) true
(c) true (g) true
(d) true (h) false
6.
(a) true (e) false
(b) true (f) true
(d) true
(h) true
(i) false
(c) true (g) false (The symbol for helium is He.) (j) true
7. (a)
(b)
(c)
(d)
(e)
P T F T F
Q T T F F
Q∧P T F F F
P T F T F
Q T T F F
∼P F T F T
∼P ⇒Q T T T F
P T F T F
Q T T F F
∼Q F F T T
Q⇔P T F F T
P T F T F
Q T T F F
P ∨Q T T T F
P T F T F T F T F
Q T T F F T T F F
R T T T T F F F F
(k) false
P ⇒ (Q ∧ P ) T T F T
P ∧Q T F F F T F F F
P ∧Q T F F F Q∧R T T F F F F F F
Q⇔P T F F T
(∼ P ⇒ Q) ∨ (Q ⇔ P ) T T T T
∼ Q ⇒ (Q ⇔ P ) T T F T (P ∨ Q) ⇒ (P ∧ Q) T F F T P ∨R T T T T T F T F
(P ∧ Q) ∨ (Q ∧ R) T T F F T F F F
(P ∧ Q) ∨ (Q ∧ R) ⇒ P ∨ R T T T T T T T T
1
LOGIC AND PROOFS
(f)
8. (a)
(b)
(c)
9
P T F T F T F T F T F T F T F T F
Q T T F F T T F F T T F F T T F F
R T T T T F F F F T T T T F F F F
S T T T T T T T T F F F F F F F F
Q⇒S T T T T T T T T F F T T F F T T
Q⇒R T T T T F F T T T T T T F F T T
P T F T F T F T F T F T F T F T F
Q T T F F T T F F T T F F T T F F
R T T T T F F F F T T T T F F F F
S T T T T T T T T F F F F F F F F
(P ∨ Q) ⇒ (S ∨ R) T T T T T T T T T T T T F F F T
P ∨Q T T T F T T T F T T T F T T T F
S∨R T T T T T T T T T T T T F F F F
(Q ⇒ S) ∧ (Q ⇒ R) T T T T F F T T F F T T F F T T
[(Q ⇒ S) ∧ (Q ⇒ R)] ⇒ [(P ∨ Q) ⇒ (S ∨ R)] T T T T T T T T T T T T T T F T
P Q ∼ P P ⇒ Q (∼ P ) ∨ Q T T F T T F T T T T T F F F F F F T T T Since the fourth and fifth columns are the same, the propositions P ⇒ Q and (∼ P ) ∨ Q are equivalent. P Q P ⇒ Q Q ⇒ P P ⇔ Q (P ⇒ Q) ∧ (Q ⇒ P ) T T T T T T F T T F F F T F F T F F F F T T T T Since the fifth and sixth columns are the same, the propositions P ⇔ Q and (P ⇒ Q) ∧ (Q ⇒ P ) are equivalent. P Q ∼ Q P ⇒ Q ∼ (P ⇒ Q) P ∧ ∼ Q T T F T F F F T F T F F T F T F T T F F T T F F Since the fifth and sixth columns are the same, the propositions ∼ (P ⇒ Q) and P ∧ ∼ Q are equivalent.
1
LOGIC AND PROOFS
(d)
(e)
(f)
10
P Q ∼ P ∼ Q P ∧ Q ∼ (P ∧ Q) P T T F F T F F T T F F T T F F T F T F F T T F T Since the sixth, seventh and eighth columns are sitions are equivalent.
⇒∼ Q P ⇒∼ Q F F T T T T T T the same, all three propo-
P Q R Q ⇒ R P ⇒ (Q ⇒ R) P ∧ Q (P ∧ Q) ⇒ R T T T T T T T F T T T T F T T F T T T F T F F T T T F T T T F F F T F F T F F T F T T F F T T F T F F F T T F T Since the fifth and seventh columns are the same, the propositions are equivalent. P T F T F T F T F
Q T T F F T T F F
R T T T T F F F F
Q∧R T T F F F F F F
P ⇒ (Q ∧ R) T T F T F T F T
P ⇒Q T T F T T T F T
P ⇒R T T T T F T F T
(P ⇒ Q) ∧ (P ⇒ R) T T F T F T F T
Since the fifth and eighth columns are the same, the propositions are equivalent. (g)
P T F T F T F T F
Q T T F F T T F F
R T T T T F F F F
P ∨Q T T T F T T T F
P ∨ Q) ⇒ R T T T T F F F T
P ⇒R T T T T F T F T
Q⇒R T T T T F F T T
(P ⇒ R) ∧ (Q ⇒ R) T T T T F F F T
Since the fifth and seventh columns are the same, the propositions are equivalent. 9. (a) yes (d) yes
(b) no (e) no
(c) yes (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)
1
LOGIC AND PROOFS
11
(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 it doesn’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 the symbols D: “The Dolphins make the playoffs” and B: “The Bears win all the rest of their games.” The first translation is preferred, but the speaker may 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 homework first.” It is most likely that a student and parent both interpret this statement as a biconditional, G ⇔ H. (d) Let W be “You win the lottery” and T be “You buy a ticket.” Of the three common interpretations for the word “unless,” only the form ∼ T ⇒∼ W (or, equivalently, W ⇒ T ) makes sense here. 12. (a)
(b)
P Q R P ∨ Q (P ∨ Q) ⇒ R ∼ P ∧ ∼ Q ∼ R ⇒ (∼ P ∧ ∼ Q) T T T T T F T F T T T T F T T F T T T F T F F T F T T T T T F T F F F F T F T F F F T F F T F F F F F F F T T T Since the fifth and seventh columns are the same, (P ∨ Q) ⇒ R and ∼ R ⇒ (∼ P ∧ ∼ Q) are equivalent. P T F T F T F T F
Q T T F F T T F F
R T T T T F F F F
P ∧Q T F F F T F F F
(P ∧ Q) ⇒ R T T T T F T T T
∼Q F F T T F F T T
∼R F F F F T T T T
P∧ ∼ R F F F F T F T F
(P ∧ ∼ R) ⇒∼ Q T T T T F T T T
Since the fifth and ninth columns are the same, the propositions (P ∧ Q) ⇒ R and (P ∧ ∼ R) ⇒∼ Q are equivalent.
1
LOGIC AND PROOFS
(c)
(d)
(e)
(f)
12
P Q R Q ∧ R P ⇒ (Q ∧ R) ∼ Q∨ ∼ R (∼ Q∨ ∼ R) ⇒∼ P T T T T T F T F T T T T F T T F T F F T F F F T F T T T T T F F F T F F T F F T T T T F F F F T F F 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. P Q R Q ∨ R P ⇒ (Q ∨ R) P ∧ ∼ R (P ∧ ∼ R) ⇒ Q T T T T T F T F T T T T F T T F T T T F T F F T T T F T T T F T T T T F T F T T F T T F F F F T F F 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. P Q R P ⇒ Q (P ⇒ Q) ⇒ R P ∧ ∼ Q (P ∧ ∼ Q) ∨ R T T T T T F T F T T T T F T T F T F T T T F F T T T F T T T F T F F F F T F T F F F T F F F T T T F 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. P Q P ⇔ Q ∼ P ∨ Q ∼ Q ∨ P (∼ P ∨ Q) ∧ (∼ Q ∨ P ) T T T T T T F T F T F F T F F F T F F F T T T T Since the third and sixth columns are the same, the propositions P ⇔ Q and (∼ 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 conditional statement will work here.) 14. (a) If 7 is an odd integer, then 6 is an odd integer. (b) This is not possible.
1
LOGIC AND PROOFS
13
(c) This is not possible. (d) If 7 is an odd integer, then 6 is an odd integer. (Any false conditional statement will work here.) 15. (a) Converse: If f (x0 ) = 0, then f has a relative minimum at x0 and is differentiable at x0 . False: f (x) = x3 has first derivative 0 but no minimum at x0 = 0. Contrapositive: If f (x0 ) = 0, then f either has no relative minimum at x0 or is not differentiable at x0 . True. (b) Converse: If n = 2 or n is odd, then n is prime. False: 9 is odd but not prime. 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. True Contrapositive: 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 (d) neither (g) contradiction (j) neither
17. (a)
(b) (e) (h) (k)
tautology tautology tautology tautology
(c) (f) (i) (l)
contradiction neither contradiction neither
P Q P ⇒ Q ∼ P ∼ Q ∼ P ⇒∼ Q T T T F F T F T T T F F T F F F T T F F T T T T Comparison of the third and sixth columns of the truth table shows that P ⇒ 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 of P ⇒ Q is ∼∼ Q ⇒∼∼ P , so the converse and the contrapositive of the inverse are equivalent. The inverse of the contrapositive of P ⇒ Q is also ∼∼ Q ⇒∼∼ P , so it too 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 not beautiful) (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 right triangle∧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)
1
LOGIC AND PROOFS
14
(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 or x 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 or equal to 6. (i) (∃x)(∀y)(x ≤ y) Some integer is less than or equal to every integer, or There is a smallest integer. (j) (∃x)(∀y)(x ≥ y) Some integer is greater than every other integer, or There is a largest integer. (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 number between 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 there is 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.
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(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 number y such that 2y = x. There is a nonzero complex number such that either every product of that number with any complex number is different from π, or there are at least two different complex numbers whose products with the given number are equal 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 The other sentences may be translated as (∀x)[x is a person ⇒ (∃y)(y is a tax ⇒ x dislikes (∃x)[x is a person ⇒ (∀y)(y is a tax ⇒ x dislikes (∃x)[x is a person ⇒ (∃y)(y is a tax ⇒ x dislikes 6. (a) T , U , V and W
(b) T
(c) T , U , V
y)]. y)]. y)]. y)]. (d) T
7. (a) Proof. Let U be any universe. The sentences ∼ (∃x)A(x) is true in U iff (∃x)A(x) is false in U iff the truth set for A(x) is empty iff 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 open sentence 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), which is equivalent to (∀x) ∼ A(x). 8.
(a) false (e) false (i) true
(b) true (f) true (j) false
(c) false (g) false (k) false
(d) true (h) true (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.
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(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 (e) true (i) false
(b) false (f) false (j) true
(c) false (g) true (k) false
(d) false (h) 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 the truth 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) is false in U . (c) Let U be any universe and suppose (∃!x)A(x) is true. Then the truth set for 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 must both be x0 , so y = z. Thus (∃x)A(x) ∧ (∀y)(∀z)(A(y) ∧ A(z) ⇒ y = z) is true. On the other hand, suppose (∃x)A(x) ∧ (∀y)(∀z)(A(y) ∧ A(z) ⇒ y = z) is true in U . Since (∃x)A(x) is true, the truth set for A(x) contains at least one element. Since (∀y)(∀z)(A(y) ∧ A(z) ⇒ y = z) is true, the truth set for A(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 set for 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 the truth set of A(x) ∧ (∀y)(A(y) ⇒ x = y). Then x0 is the only element in the truth 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 (∀)[ > 0 ⇒ (∃δ)(δ > 0 ∧ (∀x)(|x − x0 | < δ ⇒ |f (x) − f (a)| < ] (b) Let the universe be the set R of real numbers, and let f be a function from 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 ⇒ (∀)[ > 0 ⇒ (∀δ)(δ > 0 ∧ (∀x)(|x − x0 | < δ ⇒ |f (x) − f (x0 )| < )] 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 function from R to R. Then lims→a f (x) = L means: (∀)( > 0 ⇒ (∃ δ)[δ > 0 ∧ (∀x)(|x − a| < δ ⇒ |f (x) − L| < )]) (d) A denial of “f is continuous at a” is: (∃ > 0)(∀δ)(δ > 0 ⇒ (∃x)(|x − x0 | < δ ∧ |f (x) − f (a)| ≥ ). A denial of the Mean Value Theorem is: (∃a)(∃b)[a < b ∧ f is continuous (a) )] in [a, b] ∧ f is differentiable on (a, b) ∧ (∀c)(a < c < b ⇒ f (c) = f (b)−f b−a A denial of “ lims→a f (x) = L” is: (∃ )( > 0 ∧ (∀δ)[δ > 0 ⇒ (∃x)(|x − a| < δ ∧ |f (x) − L| ≥ ])
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13. (a) This is not a denial. If the universe has only one element a and P (a) is true, 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) is false, then both the statement and (∃!x)P (x) are false. (e) This statement is not a denial. If the universe has more than one element the statement implies the negation of (∃!x)P (x), but if (∀x)P (x), then both the 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 not zero. (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 subset of C. (d) Suppose the maximum value of the differentiable function f on the closed interval [a, b] occurs at x0 . .. . Thus, either x0 = a or x0 = b or f (x0 ) = 0. Therefore, if the maximum value of the differentiable function f (x) on the closed interval [a, b] occurs at 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 are nonzero. .. . Then A is invertible. Therefore A is invertible whenever all its nonzero entries 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 is invertible.
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(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 has the 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 F for the main connective ⇒. Suppose this connective had the value F. Then the antecedent [(∼ B ⇒ M )∧ ∼ L ∧ (∼ M ∨ L)] must have the value T and the consequent B must have the value F. Then each of ∼ B ⇒ M , ∼ L and ∼ M ∨ L has value T. Then L must have the value F, and since ∼ M ∨ L has value T, M has the value F. Since ∼ B has the value T and M has the value F, ∼ B ⇒ M has the value F. But ∼ B ⇒ M has value T. This contradiction shows that every row 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. By fact (i), if the crime did not take place in the kitchen, then Professor Plum is guilty. Therefore Professor Plum is guilty. (b) Miss Scarlet. The crime did not take place in the library. By fact (iv), the weapon 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 Professor Plum is guilty. Therefore Professor Plum is guilty. (d) Miss Scarlet and Professor Plum. The crime took place at midnight in the conservatory. By fact (ii) Professor Plum is guilty. The crime did not take place 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 that x = 2n and y = 2m. By substitution, x + y = 2n + 2m = 2(n + m). Since x + 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 an integer k such that x = 2k. Then xy = (2k)y = 2(ky). Thus xy is twice the integer ky, so xy is even. (c) Suppose that x and y are even integers. Then there exist integers n and m such that x = 2n and y = 2m. Therefore, xy = 2n · 2m = 4nm. Since nm is an integer, xy is divisible by 4. (d) Suppose that x and y are even integers. Then there are integers k and m such 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 and m 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 is even. (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 an integer, xy is odd.
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(h) Suppose that x is even and y is odd. Then there exist integers n and m such 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 m such 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 conclude that 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 > |ab| = ab = |a||b|. Case 2. If a > 0 and b < 0, then |a| = a and |b| = −b. Also, ab < |ab| = −ab = a(−b) = |a||b|. Case 3. If a < 0 and b > 0, then |a| = −a and |b| = b. Also ab < |ab| = −ab = (−a)b = |a||b|. Case 4. If a < 0 and b < 0, then |a| = −a and |b| = −b. Also ab > |ab| = ab = (−a)(−b) = |a||b|. In every case, |ab| = |a||b|.
0, so 0, so 0, so 0, so
(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
a b
< 0, so | ab | = − ab =
Case 3. Let a < 0 < b. Then
a b
< 0, so | ab | = − ab =
Case 4. Let a < 0 and b < 0. Then
a b
> 0, so | ab | =
a (−b) (−a) b a b
=
|a| |b| . = |a| |b| . (−a) |a| (−b) = |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
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(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. Since a − 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. Then a + 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. By Exercise 5(i) a(a + 1) is even. On the other hand, if a is odd, then by Exercise 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 integer k, b = ka. Since b and a are positive, k must also be positive. Since k is also 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 some integer 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 b divides 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. Then there is a positive integer n such that an = b and a positive integer m such that bm = a. Thus a = bm = (an)m = a(nm). Then nm = 1, so n = 1 and m = 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 = ka and d = jc for some integers k and j. Thus bd = (ka)(jc) = (kj)(ac), and kj is an integer, so ac divides bd. (l) Let a, b, and c be integers. Suppose ab divides c. Then c = k(ab) for some integer 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 integer k such that (ac)k = bc. Thus kac = bc, so that ka = b. Therefore a divides b. 8. (a) Case 1: n is even. Then n = 2k for some natural number k, so n2 + n + 3 = (2k)2 + (2k) + 3 = 4k 2 + 2k + 2 + 1 = 2(2k 2 + k + 1) + 1. Since 2k 2 + 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 = 4k 2 + 4k + 1 + 2k + 1 + 3 = 4k 2 + 6k + 5 = 2(2k 2 + 3k) + 1.
(b) By Exercise 7(d), n2 + n = n(n + 1) is even. Since n2 + n is even and 3 is odd, by Exercise 5(h), n2 + n + 3 is odd.
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√ 2 9. (a) We want to show x+y ≥ xy, which could be derived from ( x+y 2 2 ) ≥ 2 xy, which would follow from (x + y) ≥ 4xy, which would follow from x2 + 2xy + y 2 ≥ 4xy, which would follow from x2 − 2xy + y 2 ≥ 0, which would follow from (x − y)2 ≥ 0. Proof. Suppose x and y are nonnegative real numbers. Then (x − y)2 ≥ 0, 2 so x2 − 2xy + y 2 ≥ 0. Thus x2 + 2xy + y 2 ≥ 4xy, so ( x+y 2 ) ≥ xy. Since √ x and y are nonnegative real numbers, (x + y)2 = x + y and xy is a √ x+y real number. Therefore 2 ≥ xy. We used the fact that x and y are nonnegative in the penultimate sentence in the proof. (b) We want to show a divides 3c, which would follow if a divides c. To show a divides c, we could write c as the difference or sum of two quantities divisible 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 follow if the discriminant b2 − 4ac > 0. Proof. Suppose ab > 0 and bc < 0. Then the product ab2 c < 0. Since b2 ≥ 0, ac < 0. Then −4ac > 0, so b2 − 4ac > 0. Therefore by the discriminant 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. Thus x < −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 = 2ab √ and 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 > C 2 > BC and BC > B 2 , so AC > B 2 . AC is positive, so 4AC > AC. Therefore 4AC > B 2 , so B 2 − 4AC < 0. Thus the graph must be an ellipse. (b) Assume AC < 0. Then −4AC > 0, so B 2 − 4AC > 0. Thus the graph is a hyperbola. Now assume B < C < 4A < 0. Then −B > −C > −4A > 0, so B 2 = (−B)(−B) > (−B)(−C) = BC and BC > 4AC. Thus B 2 −4AC > 0, so the graph is a hyperbola. (c) Assume that the graph is a parabola. Then B 2 − 4AC = 0, so B 2 = 4AC. 2 Assume further that BC = 0. Then C = 0, so A = B 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 are not justified. The steps could be justified either by using the definitions or by referring to previous examples and exercises. (b) C. If a divides both b and c, then there are integers q1 and q2 such that b = 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+ x1 ≥ 2. The proof could be fixed by beginning with the (true) statement that (x − 1)2 ≥ 0 and ending with the conclusion that x + x1 ≥ 2.
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(d) F. This is a proof that if m is odd, then m2 is odd. We cannot prove a statement by proving its converse. (e) F. Although every statement is correct, the justification is incomplete. Without additional explanation the reader might wonder whether the proof means that x2 is always even and x + 1 is always odd. One approach to a correct proof is to use the fact that x2 + x is always even and that the product of an integer with an even integer is even. (Exercises 7(d) and 5(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 not zero. (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 multiplicative inverse, because x · x1 = 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 function from 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.
1
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23
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. Then x = 2k − 1 = 2(k − 1) + 1 and k − 1 is an integer, so x is odd. Therefore if x is even, then x + 1 is odd. (b) Suppose x + 2 is even (not odd). Then there is an integer m such that x + 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 = 4k 2 and k 2 is an integer, so x2 is divisible by 4. Therefore if x2 is not divisible by 4, then x is odd. (d) Suppose x is odd and y is odd. Then x = 2k + 1 and y = 2m + 1 for some integers k and m. Then 2km + m + k is an integer and 2(2km + m + k) + 1 =
1
LOGIC AND PROOFS
(e)
(f)
(g)
(h)
24
4km + 2m + 2k + 1 = (2k + 1)(2m + 1) = xy, so xy is odd. Therefore if xy is even, then x or y is even. Suppose it is not the case that either x and y are both odd or x and y are both even. Then one of x or y is even and the other is odd. We may assume that x is even and y is odd. (Otherwise, we could relabel the two integers.) Then x = 2k and y = 2m + 1 for some integers k and m. Then x + y = 2k + 2m + 1 = 2(k + m) + 1 and k + m is an integer, so x + y is odd. Therefore if x + y is even then x and y are both odd or both even. Suppose x and y are not both odd. Then either x or y is even (or both are even). We may assume x is even. Then x = 2m for some integer m. Thus xy = (2m)y = 2(my), and my is an integer, so xy is even. Therefore if xy is odd then both x and y are odd. 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. 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 divide yk, 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 the radius ofthe 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 the circle 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 the circle does not intersect the line y = x − 6. (c) Suppose the point (0, 3) is not inside the circle, but (3, 1) is inside the circle. Then the distance from (2, 4) to (3, 1) is less than the radius and the 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) is not 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 there is 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 that a > 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 is even. We may assume a is even. Then a = 2m for some integer m. Then ab = (2m)b = 2(mb). Since mb is an integer, ab is even. Since a number cannot be both even and odd, this is a contradiction. Therefore if ab is odd, then a and b are both odd.
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25
(c) Suppose a is odd and a+1 is not even. Then a+1 is odd, so a+1 = 2k+1 for some integer k. Thus a = 2k, so a is even. This contradicts the assumption that 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 = 2m for 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 if a − b is odd, then a + b is odd. (e) Let a, b be positive integers. Assume that a < b and ab < 3. Suppose that a = 1. Since a is a positive integer, a ≥ 2. And since a < b, b > 3. Therefore ab > 6. This contradicts the assumption that ab < 3. 7. (a) Let a, b, and c be positive integers. Then a divides b
iff iff iff
b = ak for some integer k bc = (ac)k for some integer k 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 = 3 divides b + 3 = 6. Part 2. Suppose a + 1 divides b and b divides b + 3. Then b + 3 = bk for some integer k, so 3 = bk − b = b(k − 1). Therefore b divides 3, so b = 1 or b = 3. Since a + 1 divides b, a + 1 ≤ b. Thus b = 1, so b = 3. Since a + 1 > 1 and 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, so a + 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, so a = 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 iff iff
a = b − c and 2b − a = d a = b − c and 2b − (b − c) = d a = b − c and b + c = d.
8. (a) Suppose m and n have the different parity. Then one is even and the other is odd. We may assume, without loss of generality, that m is even and n is odd. Then m = 2k and n = 2j + 1 for some integers k and j. Then m2 − n2= (2k)2− (2j + 1)2 = 4k 2 − 4j 2 − 4j − 1 = 2(2k 2 − 2j 2 − 1) + 1, which is odd. (b) Suppose m2 − n2 is odd. If m is even, then m2 is even. Therefore, since m2 − n2 is odd and m2 is even, n2 = m2 − (m2 − n2 ) is odd. From n2 is odd, we conclude that n is odd. 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.
1
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26
n n 9. Let n be a natural and suppose to the contrary that n+1 ≤ n+2 . Then n(n + 2) ≤ n(n + 1), so n + 2 ≤ n + 1, which is impossible. We conclude n n that n+1 > n+2 . √ √ 10. Assume 5 is a rational number. Then 5 = pq where p and q are positive 2
integers, q = 0, and p and q have no common factors. Then 5 = pq2 , so 5q 2 = p2 . The prime factorization of q 2 has an even number of 5’s (twice as many as the factorization of q), so 5q 2 has an odd number of 5’s in its prime factorization. But 5q 2 = p2 and p2 has√an even number of 5’s in its prime factorization. This is impossible. Therefore 5 is irrational. 11. Suppose x, y, z are real numbers and 0 < x < y < z < 1. Assume that the distances from x to y and from y to z are at least 12 . That is, assume |x − y| = y − x ≥ 12 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 + 12 + 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 integers is even, but it assumes that the sum of even numbers is even, which is what must be proved. (d) C. Leaving out the assumption that a divides b and a divides c makes this proof confusing. If we change the first sentence to “Assume a divides both b and c and a does not divide b + c” then we have a correct proof by contradiction.
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 2 divides 4n, so 2 divides their sum 2m + 4n. But 2 does not divide 7, so this is impossible. (d) Suppose 12m + 15n = 1 for some integers m and n. Then 3 divides the left side but not the right side, which is impossible. (e) Let t be an integer. Suppose there exist integers m and n such that 15m + 16n = t. Let r and s be the integers 5m and 2n, respectively. Then 3r + 8s = 15m + 16n = t. (f) Suppose 12m + 15n = 1 for some integers m and n, and suppose further that m and n are not both positive. Then 3 divides the left side but not the right, which is impossible. Therefore, if there exist integers m and n such that 12m + 15n = 1, then m and n are both positive. Alternative proof. Let P be the statement “there exist integers m and n such that 12m + 15n = 1.” By part (d), P is false. Therefore P implies m and n are both positive. (g) Let m be an odd integer. Suppose m = 4k + 1 for some integer k. Then m + 2 = 4k + 3 = 4(k + 1) − 1 = 4j − 1 where j is the integer k + 1.
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27
(h) Suppose m is an odd integer. Then m = 2n + 1 for some integer n. Then m2 = (2n + 1)2 = 4n2 + 4n + 1 = 4(n2 + n) + 1. But n2 + n = n(n + 1) is even (Recall Exercise 7(d) of Section 1.4), so n2 + n = 2k for some integer k. Therefore m2 = 4(2k) + 1 = 8k + 1. (i) Suppose m and n are odd integers. Assume that neither m nor n is of the form 4j − 1 for some integer j. Then m = 4j1 + 1 and n = 4j2 + 1, for some integers j1 and j2 . (Recall the result from Section 1.4 that every odd integer has either the form 4j + 1 or the form 4j − 1, for some integer j.) Thus mn = (4j1 + 1)(4j2 + 1) = 4(4j1 j2 + j1 + j2 ) + 1. Then mn cannot be written in the form 4k − 1, where k is an integer. Therefore, if mn is of the form 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 and y be integers. Then there exist integers m and n such that a = cm and b = cn, so ax + by = (cm)x + (cn)y = c(mx + ny) and mx + ny is an integer, so c divides ax + by. Alternate proof. Let a, b and c be integers such that c divides a and c divides b. Let x and y be integers. By a previous exercise, c divides integer multiples 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, which says 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. Then by exercise 2(a) a divides (b − 1)(c − 1) = bc − b − c + 1. Then by Exercise 2(a), a divides (bc − b − c + 1) + (b − 1) = bc − c. Then by Exercise 2(a), a divides (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 = k n an , so an divides bn . (d) Let a and c be integers such that a is odd, c > 0, c divides a and c divides a + 2. By Exercise 2(a), c divides (−1)a+1(a+2), so c divides 2. The only divisors of 2 are 1 and 2. If c = 2, then 2 divides a, so a is even. But a is odd, so c = 1. (e) and suppose there exist integers m and n such that am + bn = 1. Also suppose 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 such that am + bn = 1, and c = ±1, then c does not divide a or c does not divide b. 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 natural number greater than 2, so by the hypothesis it is the sum of two primes. Let p1 and 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 that x + y = 0.
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28
(c) False. Counterexample: Let x = 2 and y = 1. Then y x = 12 = 1>2 = x. (d) False. Counterexample: Let a = 6, b = 3 and c = 2. Then a divides bc, but a doesn’t divide b or c. (e) True. Proof. Suppose a, b, c, d, j and k are integers such that b − c = ka and c − d = ja. Then b − d = (b − c) + (c − d) = ka + ja = (k + j)a, so a divides b − d. (f) False. Counterexample: Let x = 12 . Then x2 − x = − 14 < 0. (g) False. √ Counterexample: Let x = 12 . Then 2x = 2 ≈ 1.415 < 1.5 = x + 1. (h) False. Counterexample: Let x = 1. Suppose y is a positive real number less than x. 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 the statement [y < x ⇒ (∀z)(yz ≥ z)] is true. Alternative proof. Let x be a positive real number, and let y = 1. Then if 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 no positive 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 1 and less than or equal to that divides x. Suppose x = mn for some√natural numbers ≤ n. By the hypothesis m = 1 or m > x. But √ m and n, and m √ m > x implies that n > x, from which we conclude that mn > x. Since this 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 is divided by 3 the remainder is either 1 or 2. Thus, there is an integer k such that 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 = (9k 2 +6k +1)+2 = 9k 2 +6k +3 = 3(3k 2 + 2k + 1) so p2 + 2 is divisible by 3. If p = 3k + 2, then p2 + 2 = (3k + 2)2 + 2 = (9k 2 + 12k + 4) + 2 = 9k 2 + 12k + 6 = 3(3k 2 + 4k + 2) so p2 + 2 is divisible by 3. 6. (a) Let n be a natural number. Then n ≥ 1, so 1 =
n n
≥
1 n.
(b) Choose M = 10. Let n be a natural number greater than m. Then 1 1 n < 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 be a positive real number. Then 2 is a smaller positive real number. Therefore, for every positive real number there is a smaller positive real number.
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(g) Let > 0 be a real number. Then 1 is a positive real number and so has a decimal expression as an integer part plus a decimal part. Let M be the integer part of 1 plus 1. Then M is an integer and M > 1 . To prove for all natural numbers n > M that n1 < , let n be a natural number and assume that n > M . Since M > 1 , we have n > 1 . Thus 1 n < . Therefore, for every real number > 0, there is a natural number M such that for all natural numbers n > M , n1 < . (h) Let > 0. By part (g), there is M such that 1 < n1 < . m > n > M , then n1 − m
1 n
< for all n > M . Now if
(i) Let K be 10. Suppose r > K. Then r > 10, so r2 > 100. Therefore 1 r 2 < 0.01. (j) Let L = −15 and G = −1. Suppose L < x < G. Then −1.5 < x < −1, so 15 > −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 and 1 1 r > M . Then r > 51, so r > 50. Then 2r > 100, so 2r < 100 . (l) Let x be a natural number. Choose k = −4x + 50. Then k is an integer and k < −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, 1 and y−x is a positive real number. Choose M to be a natural number larger 1 1 , so than y−x . Suppose n is a natural number and n > M . Then n > y−x 1 1 y − x > n . That is, 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. 1 (i) C. The number 2ε may not be a natural number. To correct this error, 1 . choose K to be a natural number greater than 2ε
(j) A.
1.7
Additional Examples of Proofs
1. (a) Proof. We work both forwards and backwards: From the hypothesis that is odd we can deduce that 3n is even, from which we can 3n + 1 deduce that n is even. We could reach the conclusion that is 2n + 8 divisible by 4 if we knew that 4 divides 2n (since 8 is divisible by 4). In turn, the statement 2n is divisible by 4 may be derived from the statement that n is divisible by 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 implies that n is even. We are now using properties of even and odd integers that we proved earlier, without referencing specific examples or exercises. Since n is even, n is divisible by 2. Therefore 2n is divisible by 4. Finally since 8 is also divisible by is divisible by 4, 2n + 8 is divisible by 4.
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30
(b) Proof. Let a be a real number, a = 3. The key to the proof is to use the a = 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 an equation. 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 step implies its predecessor. Thus we can modify the given proof to create an iff proof. a is a solution to x2 − x − 6 = 0 iff a2 − a − 6 = (a − 3)(a + 2) = 0. Because a − 3 = 0.
iff a + 2 = 0. 2
3
2
iff (a + 1)(a + 2) = a + 2a + a + 3 = 0. iff a is a solution to x3 + 2x2 + x + 3 = 0.
Because a2 + 1 = 0.
(d) Proof. Suppose x2 = 2x + 15 and x > 2. Then (x − 5)(x + 3) = 0. Since x > 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. In this 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 = pq and y = rs where p, q, r, and s are integers, q = 0, and s = 0. Therefore, x + y = pq + rs = ps+q qs . Since ps + rq and qs are integers and qs = 0, x + y is a rational number. We have shown that if x and y are rational, then x + y is rational. We conclude that if 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 can simplify 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), where α is the angle of inclination of the line. Let α1 and α2 be the angles of inclinations of L1 and L2 , respectively. See the figure below. We may assume that α1 > α2 . We can make this assumption because the two lines are arbitrary; if α1 < α2 simply interchange the labels of the lines. Since L1 and L2 are perpendicular, α1 = α2 + π2 . Therefore, tan(α1 ) = tan(α2 +
π 1 ) = − cot(α2 ) = − . 2 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 of L2 , the product of the slopes is −1.
1
LOGIC AND PROOFS
y
31
L1
L2
α1
α2
x
(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 proof by assuming that (x, y) is a point inside the circle. We would then have to prove that (x, y) is not on the line. In this instance, a better approach is to use 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
< 6 2a − 4a + 10 < 6 a2 − 2a + 5 < 3 a2 − 2a + 1 < −1 2
(a − 1)2
< −1.
This is a contradiction since (a − 1)2 = 0. Thus, no point inside the circle is on the line. (h) Proof. Proofs that verify equalities or inequalities containing absolute value expressions 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 case is discovered by working backwards from the desired conclusion. The key steps are to note that, in the first case, if x ≥ 2, then −6 ≤ x, and, in the second case, that if x ≥ 1, then 67 ≤ 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) ≤ 4. x 3|x − 2| Therefore, ≤ 4. x
Remember that x is positive.
Case 2. Suppose x − 2 < 0. Then |x − 2| = −(x − 2). By hypothesis, x ≥ 1. Therefore, 6 7 6 6 − 3x 3[−(x − 2)]
≤ x ≤ 7x ≤ 4x ≤ 4x
1
LOGIC AND PROOFS
3[−(x − 2)] x 3|x − 2| Therefore, x
32
≤ 4.
Remember that x is positive.
≤ = 4.
2. (a) Let n be an integer. Then n is either even or odd. If n is even then n = 2k for some integer k, so that 5n2 + 3n + 4 = 5(2k)2 + 3(2k) + 4, which is twice the integer 10k 2 + 3n + 2. If n is odd then n = 2k + 1 for some integer k, so that 5n2 + 3n + 4 = 5(2k + 1)2 + 3(2k + 1) + 4, which is twice the integer 10k 2 + 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, so 2n2 + 3n + 4 = 2(2k + 1)2 + 3(2k + 1) + 4 = 2(4k 2 + 7k + 4) + 1. Since 4k 2 + 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 an integer, the sum is divisible by 5. (d) Let L1 and L2 be two nonvertical lines such that the product of their slopes is −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 L2 is tan(α2 ). Since tan(α1 ) tan(α2 ) = −1, exactly one of these two must be positive and neither can be 0. Suppose without loss of generality that tan(α2 ) > 0. Then −1 = − cot(α2 ) = tan(α2 + π2 ), and since 0 < α2 < π2 . Now tan(α1 ) = tan(α 2) π π both α1 and α2 + 2 are between 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, then n + 1 is odd and if n is odd, then n + 1 is even. Therefore, either n or n+1 is divisible by 2. By the Division Algorithm, the remainder when n is divided by 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 = 3k is divisible by 3. Case 3. If the remainder is 2, n = 3k + 2 for some integer k. Then n + 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 + 1 has 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 the product of four consecutive integers. From part (e) n3 − n is divisible by 6 and hence divisible by 3. By previous results, if n is even, then n + 1 is odd and if n is odd, then n + 1 is even. Therefore, either n or n + 1 is divisible by 2. If n is divisible by 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 + 1 are each divisible by 2. In all cases, n3 − n has a factor of 3, and two terms (either n and n + 2, or n − 1 and n + 1) each have a factor of 2. Therefore, (n3 − n)(n + 2) has two factors 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 = − k2 x + 3, so − k2 = 13 . Thus k = −6. Therefore if k = −6, then L does not have slope 13 .
1
LOGIC AND PROOFS
33
(b) Suppose for some real number k that L is parallel to the x-axis. Then L has slope 0, so −2 k = 0. This is impossible. Therefore L is not parallel to the 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 exist integers p and q, q = 0 such that x = pq and integers r and s, s = 0, such that x + y = rs . Then y = (x + y) − x = rs − pq = rq−ps sq . Since rq − ps and sq are integers and sq = 0, y is rational. Therefore if x is rational and y is irrational, 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 + π and y = −π. Then x + y = z. (d) Let z be a rational number and x be an irrational number. Then −x is irrational. Let y = z − x. Then x + y = x + (z − x) = z and z − x is irrational by part (a), so there exists an irrational number y such that x + y = z. Suppose there is an irrational number w such that x + w = z. Then w = z − x = y. Therefore the irrational number y such that x + y = z is unique. 5. (a) Assume (x, y) is on the given circle and y = 0. Then x2 + y 2 = r2 , so y −y 2 = x2 − r2 = (x + r)(x − r) and so x−r = − x+r y . Thus the slope of the line passing through (x, y) and (r, 0) is the negative reciprocal 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 does not apply are (r, 0) and (−r, 0). If (x, y) = (r, 0), then there are many lines passing through (x, y) and (r, 0), only one of which is perpendicular to the line through (x, y) and (−r, 0). (b) First observe that if y = 0, then the points (x, y), (r, 0), and (−r, 0) are all on the x-axis, so the line through (x, y) and (r, 0) is not perpendicular to the line through (x, y) and (−r, 0). Suppose y = 0 and the two lines are perpendicular. Then the slopes of the y 2 lines are negative reciprocals, so x−r = − x+r y . Thus −y = (x + r)(x − r) = 2 2 2 2 2 x − r and so x + y = r . But the fact that (x, y) lies inside the circle requires that x2 + y 2 < 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 2 (−3 − x0 )2 + [1 − (6 − x0 )] = 2 = (−3 − x0 )2 + (x0 − 5) = 2x20 − 4x0 + 34 = 2(x20 − 2x0 + 1) + 32 2 2 (x0 − 1) + 16 + 16 = √ 2 > 16 since 2 (x0 − 1) + 16 > 0
1
LOGIC AND PROOFS
34
=
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 f is not differentiable on [5, 7]. (d) Suppose to the contrary that there are real numbers a < b that satisfy this equation. Since f (x) = x3 + 6x − 1 is continuous and differentiable everywhere, 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 that f (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 ≥ 12 . In this case 2x−1 ≥ 0, so |2x−1| = 2x−1. Since −1 ≤ 2, we have 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, |2x−1| we have 1−2x 1+x < 2. Therefore 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 follows that |x − 3| < 2 and |y − 2| < 2, so −2 < x − 3 < 2 and thus 1 < x < 5 and 0 < y < 4. Therefore x2 < 25 and y 2 < 16, so x2 + y 2 < 41. Thus (x, y) is inside 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. Therefore x − 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 when b is divided by a, the quotient is 0.
1
LOGIC AND PROOFS
35
(b) Suppose a and b are integers, a > b, and the quotient is 0 when b is divided by a. Then b = a(0) + r, where the remainder r is ≥ 0. Then r = b, so b ≥ 0. 11. (a) 2, 1, −1, −2 (b) 1, −1
gcd(8, 310) = 2
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 a and c divides b. By Theorem 1.7.3, d is a linear combination of a and b. By Therorem 1.7.1, c divides every linear combination of a and b. Therefore c divides 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 number larger 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 bc and d = 1. By Theorem 1.7.3, d is the smallest positive linear combination of 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 their sum. Thus a divides c. (d) Let a, b, and c be natural numbers and gcd(a, b) = d. Suppose c divides a and b. Since gcd(a, b) = d, c divides d (by part (a)) and there are integers s and t such that as + bt = d. Then c c as + bt = d. c c Therefore
a b d s+ t= . c c c
Since dc is a linear combination of ac and cb , by Theorem 1.7.1 gcd( ac , cb ) divides dc . Hence
d a b . ≥ gcd , c c c By Theorem 1.7.3 there exist integers p and q such that a b a b p + q = gcd( , ). c c c c
Therefore ap + bq = c · gcd
a b , c c
.
1
LOGIC AND PROOFS
36
Since c · gcd
a b , c c
is a linear combination of a and b, d divides
a b c · gcd . , c c Thus
d c
divides
gcd
Hence
We conclude that
a b , c c
d ≤ gcd c d = gcd c
.
a b , c c a b , c c
.
.
(e) Let a and b be natural numbers and gcd(a, b) = d. Let n be a natural number. 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 linear combination 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 = n1 gcd(an, bn). Since n1 gcd(an, bn) is a linear combination of a and b, d divides n1 gcd(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 are 1 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, so gcd(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 p would 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 and a is 1, so gcd(p, a) = 1. 17. Suppose q is a natural number greater than 1 with the property that q divides a or q divides b whenever q divides ab. Assume q is composite. Then q has a divisor m that is not 1 and not q. That is, q = mn for some integer n, where 1 < n < q. Then q divides mn, so by the given property, q divides m or q divides n. 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. Then gcd(a, b) = 1, so 1 is a linear combination of a and b. That is, 1 = as + bt for some integers s and t. Then acs + bct = c. Thus x = cs, y = ct is an integer solution for the equation ax + by = c.
1
LOGIC AND PROOFS
37
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 and y = t − kn is a solution for ax + by = c. 20.
(a) lcm(6, 14) = 42. (c) lcm(21, 39) = 273.
(b) lcm(10, 35) = 70. (d) lcm(12, 48) = 48.
21. (a) Part 1. Suppose a divides b. Then b is a common multiple of a and b, so condition (i) is satisfied. Now suppose n is a common multiple. Then b divides 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 a divides b. (b) Part 1. Suppose m = b. Then b is a common multiple of a and b, so a divides b. Part 2. Suppose a divides b. Then b is a common multiple of a and b; and b 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. By part (b) m = bk ≤ ab, so k ≤ a. Since m is a multiple of a, a divides bk. By part (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, so m a m b a b c , c and c are all natural numbers. Now, since ( c )c divides ( c )c and ( c )c m a m divides ( c )c, we have by Exercise 7(m) in Section 1.4 that c divides c and b m m a b m c divides c . That is, c is a common multiple of c and c . To see that c a is the least common multiple, let n be another common multiple of c and b c . Then nc is a common multiple of a and b. Thus nc divides m. Therefore m a b by Exercise 7(m), we see that n divides m c . Therefore lcm( c , c ) = c . (e) Let n be a natural number. n divides both an and bn, so by (d), bn 1 lcm(a, b) =lcm( an n , n ) = 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 , db ) = 1 a b a b a b d lcm(a, b). By Exercise 14(d), gcd( d , d ) = 1. By part (c) lcm( d , d ) = d · d . Therefore m = d·lcm( ad , db ) = d( ad · db ) = ab d , 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” only examines the case where x = π. (d) A. (e) C. The second to the last sentence should read, “Then c divides 1.” The correct sentence should be justified by previous exercises. (f) A.
2
Set Theory
2.1
Basic Concepts of Set Theory
1. (a) (b) (c) (d) (e)
{x ∈ N: x < 6} or {x: x ∈ N and x < 6} {x | x ∈ Z and x2 < 17} {x ∈ R: 2 ≤ x ≤ 6} or {x: x ∈ R and 2 ≤ x ≤ 6}. {x | x ∈ R and −1 < x ≤ 9}x < −1} {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. Both X ∈ X and X ∈ X lead to a contradiction. We conclude that the collection of ordinary sets is not a set. (b) If the barber is an element of A, then by definition of A the barber does not shave 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 not an element of A. This is a contradiction. On the other hand, if the barber is not an element of A, then he is not among the men who do not shave themselves. Therefore he shaves himself. This is a contradiction because the barber shaves only men who do not shave themselves. 4.
(a) True (e) False (i) False
(b) False (f) False (j) True
(c) True (g) True
(d) False (h) False
5.
(a) True (e) False (i) False
(b) True (f) True (j) False
(c) True (g) False (k) True
(d) True (h) False (1) True
6. (a) (b) (c) (d)
A = {1, 2} B = {1, 2, 4} C = {1, 2, 5} A = B = C = {1} A = {1, 2, 3} B = {1, 4} C = {1, 2, 3, 5} A = {1} B = {1, 2} C = {3}
7. Assume x ∈ / B and A ⊆ B. Suppose x ∈ A. Then since A ⊆ B, x ∈ B. This is impossible. Therefore x ∈ / A. 8. Assume A ⊆ B and B ⊆ C and suppose x ∈ A. Then x ∈ B since A ⊆ B, and then x ∈ C since B ⊆ C. Therefore A ⊆ C. 9. Assume A ⊆ B, B ⊆ C and C ⊆ A. The last two assumptions and Theorem 2.1.1 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. Since B ⊆ 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 = 3 iff
x∈Y 38
2
SET THEORY
39
12. iff x2 < 30 and x ∈ N iff x = 1 or x = 2 or x = 3 or x = 4 or x = 5 iff x ∈ Y
x∈X
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 there exists an integer c such that a = bc. Likewise, b ∈ aZ, so there exists an integer d such that b = ad. Then a = bc = (ad)c = a(dc). Hence dc = 1, so c = 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 (e) false
16. (a) (b) (c) (d) 17.
(b) true (f) true
(c) false (g) true
(d) true (h) true
no proper subsets {∅}, {{∅}}, ∅ {1}, {2}, ∅ {O}, {}, {}, {O, }, {O, }, {, }, ∅
(a) True (e) True (i) True
(b) True (f) True (j) True
(c) True (g) True (k) True
(d) True (h) False (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) < byT heorem2.1.5. > 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 with a definite statement that x ∈ A or that x ∈ B. This problem could be corrected by inserting a second sentence “Suppose x ∈ A” and a fourth sentence “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 means x ∈ A and x ∈ B. The correct meaning of A ⊆ B is that for every x, if x ∈ A, then X ∈ B. (f) F. The claim is false. The error is the misunderstanding that x ∈ A implies x ⊆ A. (g) F. The claim is false. The error is that the “proof” supposes x ∈ A but shows 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 hypothesis, helpful explanations and connecting words. How much explanation you include depends on the presumed level of the reader’s knowledge. We prefer 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} (c) {1, 3, 5, 7, 9} (e) {3, 9} (g) {1, 5, 7} (i) {1, 5, 7}
2.
(a) [2, 8) (d) [2, 4) (g) [8, ∞) (j) (1, 5] ∪ (6, 8)
3.
(a) {0, −2, −4, −6, −8, . . .} (c) (e) Z+ (g) {0, 2, 4, 6, 8, . . .} (i) Z
(b) ∅ (d) A (f) {1, 2, 3, 5, 6, 7, 8, 9} (h) {1, 5, 7} (j) {0, 3, 4, 6, 9}
(b) (1, 8) (e) (6, 8) (h) (−∞, 3) ∪ [8, ∞) (b) (d) (f) (h)
(c) [3, 6] (f) [2, 5] (i) ∅
{1, 3, 5, 7 . . .} Zi− ∪ {0} D {−1, −3, −5, −7, . . .} ∪ 0 ∪ 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 the text) A ∩ = ∩ A, and by part (b), ∩ A ⊆ . Therefore A ∩ = . (d) Suppose x ∈ A ∪ ∅. Then x ∈ A or x ∈ ∅. Since x ∈ ∅ is false, x ∈ A. This shows 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. Therefore A ∪ B = B ∪ A. (h) No Solution (i) x ∈ A − ∅ iff x ∈ A and x ∈ / ∅ iff x ∈ A (since x ∈ / ∅ is true). Therefore A − ∅ = A. (j) x ∈ ∅ − A iff x ∈ ∅ and x ∈ / A iff x ∈ ∅ (because x ∈ ∅ is false). Therefore ∅ − A = ∅.
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(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 iff
x ∈ A and x ∈ B ∩ C
iff iff
x ∈ A and (x ∈ B and x ∈ C) (x ∈ A and x ∈ B) and x ∈ C 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 shows A ∪ 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 ∈ A and A ⊆ B, we have x ∈ B. Thus x ∈ B and x ∈ C, so x ∈ B ∩C. Therefore A ∩ 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 .
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42
(c) x ∈ A ∩ Ac
iff
x ∈ Aandx ∈ Ac
iff iff
x ∈ Aandx ∈ /A 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 ∈ B c iff x ∈ A ∩ B c .
Therefore A − B = A ∩ B c . (e) x ∈ (A ∩ B)c
iff x ∈ / A∩B iff it is not the case that x ∈ A and x ∈ B iff iff
x∈ / A or x ∈ /B c x ∈ A or x ∈ B c
iff
x ∈ Ac ∪ B c .
Therefore (A ∩ B)c = Ac ∪ B c . (f) Part 1. Assume A ∩ B = ∅. Suppose x ∈ A. If x ∈ B, then x ∈ A ∩ B = ∅, which is impossible. Thus x ∈ / B. Therefore x ∈ B c . This shows A ⊆ B c . c Part 2. Assume A ⊆ B . Suppose x ∈ A ∩ B. Then x ∈ A and x ∈ B. Since x ∈ A and A ⊆ B c , x ∈ B c . Therefore x ∈ B ∩ B c = ∅. This shows that A ∩ 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 and x ∈ B because A ⊆ B, but x ∈ B c because x ∈ A − B. This contradiction shows A ⊆ B implies A − B = ∅. Now assume A − B = ∅ and suppose x ∈ A. Then x ∈ B because if x ∈ /B then x ∈ A − B, which is impossible. Thus A − B = ∅ implies A ⊆ B. Alternate proof of (a): A − B = iff iff iff
A ∩ B c = (by part (d) of Theorem 2.2.2) A ⊆ (B c )c (by part (h) of Theorem 2.2.2) A ⊆ B.
(b) Assume A ⊆ B ∪ C and A ∩ B = ∅ and suppose x ∈ A. Then x ∈ B or x ∈ C, but x can’t be in B because A ∩ B = ∅. Thus x ∈ C. Therefore A ⊆ C. (c) Suppose C ⊆ A and D ⊆ B. Assume that C and D are not disjoint. Then there is an object x ∈ C ∩ D. But then x ∈ C and x ∈ D. Since C ⊆ A and D ⊆ B, x ∈ A and x ∈ B. Therefore, x ∈ A ∩ B, so A and B are not disjoint. (d) Assume A ⊆ B and suppose x ∈ A − C. Then x ∈ B since A ⊆ B and x∈ / C since x ∈ A − C. Thus x ∈ B − C.
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(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. If x ∈ 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 or x ∈ 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 or x ∈ B ∩ C, so x ∈ A ∪ (B ∩ C). (h) A − B and B are disjoint because (A − B) ∩ B = (A ∩ B C ) ∩ B = A ∩ (B C ∩ 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, then we 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. Therefore D − A ⊆ B − C. (e) Assume A ∪ B ⊆ C ∪ D, A ∩ B = ∅ and C ⊆ A. Suppose x ∈ B. Then x ∈ A ∪ B, so x ∈ C ∪ D. This x ∈ C or x ∈ D. Suppose x ∈ C. Then, since C ⊆ A, so x ∈ A. Then x ∈ A and x ∈ B, so x ∈ A ∩ B. This contradicts the assumption that A ∩ B is empty. We conclude that x ∈ / C, so x ∈ D. Therefore B ⊆ D. A = {1, 2}, B = {1, 3}, C = {2, 3, 4} Let A = {1} and B = {2} = C. Then A ∩ C = ∅ ⊆ B ∩ C, but A ⊆ B. A = {1, 2}, B = {1, 3} 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 = ∅.
11. (a) (b) (c) (d)
12. (a) S ∈ P(A ∪ B)
iff S iff S iff S iff S
⊆A∩B ⊆ A and S ⊆ B[by Exercise 9(c)] ∈ P(A) and S ∈ P(B) ∈ P(A) ∩ P(B).
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(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}} while P(A ∪ B) = {∅, {1}, {2}, {1, 2}}. In general, if A ⊆ B or B ⊆ A, then P(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),
( B×A= { (
{1, 2}, {t}), ({1, 2}, π)} (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. Let b ∈ 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 case A = 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 iff iff iff
(x, y) ∈ A × B and(x, y) ∈ C × D x ∈ A and x ∈ C and y ∈ B and y ∈ D x ∈ A ∩ C and y ∈ B ∩ D (x, y) ∈ (A ∩ C) × (B ∩ D).
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(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 = x and 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∈AB
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 iff
x ∈ A ∪ B and x ∈ / A∩B 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 an acceptable step but not useful in proving A − C ⊆ B − C. After assuming that A ⊆ B, the natural first step for proving that A − C ⊆ B − C is to suppose that x ∈ A − C. (b) C. It is a common error to write something like “Suppose A − C.” The author 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 step that x = x. (h) F. The third sentence contributes nothing to the proof. The fourth sentence makes an assertion without an explanation. What is missing is the proof that if x is a nonempty subset of A − B, then x is a subset of A that is not a subset of B. (i) F. Although a picture may help by suggesting ideas around which a correct proof can be made, a picture alone is rarely sufficient for a proof. Thus a proof 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 and B ⊆ A ∪ B.
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46
Extended Set Operations and Indexed Families of Sets
1. (a) (b)
A = {1, 2, 3, 4, 5, 6, 7, 8};
A = {4, 5} A∈A A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}; A∈A A = ∅ A∈A
A∈A
(c) n∈N An = {5, 6, 10, 11, 12, 15, 16, 17, 18} ∪ {n ∈ N: n ≥ 20}; n∈N An = ∅ (d) n∈N Bn = 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∈N An = ∅ (h) r∈R Ar = [−π, ∞); r∈R Ar = [−π, 0] (i) r∈R Ar = [0, ∞); r∈R Ar = ∅ (j) n∈N Mn = Z; n∈N Mn = {0} 7 1 (k) n≥3 An = (0, 3 ); n≥3 An = [ 3 , 2] (l) n∈Z Cn = R; n∈Z Cn = ∅ (m) n∈Z An = R − Z; n∈Z An = ∅ (n) n∈N Dn = (−∞, 1); n∈N Dn = (−1, 0] (o) A∈A = · − {1}; |A∈A A = ∅
(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). The intersection 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 some A ∈ 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∈A A. 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) is false 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 subset of ∅. 5. (a) Part (a) of Theorem 2.3.2. Let β ∈ Δ. Suppose x ∈ α∈Δ Aα . Then x ∈ Aα for each α ∈ Δ. Since β ∈ Δ, x ∈ Aβ . Therefore α∈Δ Aα ⊆ 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∈Δ Aα . (b) Part (d) of Theorem 2.9. Aα )c iff x∈(
x∈ /
Aα
α∈Δ
iff
there is no α ∈ Δ such that x ∈ Aα
iff iff
for all α ∈ Δ, x ∈ / Aα for all α ∈ Δ, x ∈ (Aα )c
iff
x ∈| (Aα )c .
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6. (a)
x∈B∩
Aα
x ∈ B and x ∈
iff
Aα
α∈Δ
α∈Δ
x ∈ B and x ∈ Aα for some α ∈ Δ x ∈ B ∩ Aα for some α ∈ Δ (B ∩ Aα ) x∈
iff iff iff
α∈Δ
(b) x∈B∪
Aα
iff
x ∈ B or x ∈ Aα for all α ∈ Δ
iff
x∈
α∈Δ
(B ∪ Aα )
α∈Δ
7. (a)
Aα ∩ Bβ
α∈Δ
=
β∈Γ
β∈Γ
=
α∈Δ
β∈Γ
Aα ∩ Bβ Aα ∩ Bβ
α∈Δ
(b)
Aα ∪
α∈Δ
Bβ
=
β∈Γ
β∈Γ
=
Aα ∪ Bβ
α∈Δ
β∈Γ
Aα ∪ Bβ
α∈Δ
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∈Δ Aα ) − B = (|α∈Δ Aα ) ∩ B C = |α∈Δ (Aα ∩ B C ) = |α∈Δ (Aα − B). (d) True. ( α∈Δ Aα ) − B = ( α∈Δ Aα ) ∩ B C = α∈Δ (Aα ∩ B C ) = α∈Δ (Aα − B). 9. (a) Suppose x ∈ α∈Γ Aα . Then x ∈ Aα for someα ∈ Γ. Since Γ ⊆ Δ, α ∈ Δ. Thus x ∈ Aα for some α ∈ Δ. Therefore x ∈ α∈Δ Aα . (b) Suppose x ∈ α∈Δ Aα . Then x ∈ Aα for all x ∈ Δ, so x ∈ Aα for all x ∈ Γ. Therefore x ∈ α∈Γ Aα . 10. (a) Let x ∈ B. For each A ∈ A, B ⊆ A. Thus for each A ∈ A, x ∈ A. Therefore x ∈ A∈A A. (b) X = A∈A A. (c) Suppose x ∈ A∈A A. Then x ∈ A for some A ∈ A. Then x ∈ D because A ⊆ D for all A ∈ A. (d) Y = A∈A A. 11. (a) Let A 1 = {1}, A2 = X, and A = {A1 , A2 }. Then A∈A A = A1 ∩ A2 = {1} and A∈A A = A1 ∪ A2 = X.
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(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 pairwise disjoint and C∈C C = X. 1 1 12. Let A n = (0, 1) − ( 3n , 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 pairwise disjoint. 14. (a) Let C1 and C2 be any two sets in C. Then C1 and C2 are both in A, so if C1 = 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 A and the other comes from B. Without loss of generality, say D1 ∈ A and D ⊆ A and D ⊆ D2 ∈ B. Then 2 A∈A β∈B B by Theorem 2.3.2(b). So, 1 D1 ∩ D2 ⊆ A∈A A ∩ β∈B B = ∅ by Exercise 10(a) in Section 2.2. Thus D1 ∩ D2 = ∅. In any event, we see that D1 = D2 implies D1 ∩ D2 = ∅, so D is pairwise disjoint.
15. (a) x∈
k+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∈
i=1 k
Ai ∪ Ak+1 .
i=1
(b) x∈
k+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∈
i=1 k
Ai ∩ Ak+1 .
i=1
(c) Proof. Let x ∈ i=1 Ai . Then there exists ∞j ∈ N such that k ≤ j ≤ m and x ∈ 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 of m Δ = {1, 2, 3, . . .}. Therefore, by Exercise 9(a), i=k Ai ⊆ i=1 Ai . ∞ x ∈ Ai for every i ∈ N. Thus x ∈ Ai for every (d) Suppose x ∈ i=1 Ai . Then m i, 1 ≤ i ≤ m. Therefore x ∈ i=1 Ai . k (e) Suppose x ∈ i=1 Ai . Then x ∈ Ai for some i ≤ k, so x ∈ Ai for some m i ≤ m. Therefore x ∈ i=1 Ai . m (f) Suppose x ∈ i=1 Ai . Then x ∈ Ai for all i ≤ m, so x ∈ Ai for all i ≤ k. k Therefore x ∈ i=1 Ai .
2
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49
k k 16. (a) Ak ⊆ i=1 Ai by 10(a). On the other hand, i=1 Ai ⊆ Ak by Theorem 2.3.2(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 = − 1i , 1 + 1i for all i ∈ N. (b) Let Ai = (0, 1) for all i ∈ N. 1 (c) Let Ai = [0, 1] − ( i+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 that Aβ ∈ {Aa : α ∈ Δ}. The explanation is that by definition of an indexed family, Δ = ∅. If we allowed Δ = ∅, the claim would be false. (b) C. No connection is made between the first and second sentences. The connection 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.
2.4 1.
∞
n=1 [n, n
+ 1) = [1, ∞).
Mathematical Induction (a) Inductive (d) Not Inductive
(b) Not Inductive (e) Not Inductive
(c) 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 every natural number x. 4.
(a) 24 (d) 56 (g) (n + 2)!
(b) 5040 (e) (n + 2)(n + 1)
(c) 97 (f) (n + 3)(n + 2)
5. (a) Define S as follows: (i) 5 ∈ S (ii) n ∈ S ⇒ n + 5 ∈ S Then 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
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50
(e) Define S as follows: (i) a ∈ S (ii) n ∈ S ⇒ nr ∈ S Then S = {a, ar, ar2 , . . .}. n (f) Define i=1 Ai as follows: 1 n+1 n (i) i=1 Ai = A1 (ii) i=1 Ai = i=1 Ai ∪ An+1 . n (g) Define i=1 Xi as follows: n+1 n 1 (ii) i=1 Xi = i=1 Xi ∪ Xn+1 . (i) i=1 Xi = X1 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 hypothesis 3 + 11 + 19+ · · · +(8n − 5) + [8(n + 1) − 5]
iii. (c) Let i. ii.
iii. (d)
=
(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. By the PMI, the statement is true for every n ∈ N. n S = {n ∈ N: i=1 2i = 2n+1 − 2}. 1 1+1 − 2, so 1 ∈ S. i=1 2i = 2 = 2 n Suppose n ∈ S for some natural number n. Then i=1 2i = 2n+1 − 2. Then n+1 n n+1 = 2n+1 −2+2n+1 = 2(n+1)+1 −2. Therefore i=1 2i+2 i=1 2i = n + 1 ∈ S. By the PMI, S = N.
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 , 1 ∈ S. 2 ii. Assume n ∈ S, for some natural number n. Then 13 + 23 + · · · +n3 + (n + 1)3 = = =
iii. (f) Let i. ii.
n(n + 1) 2 ] + (n + 1)3 2 n2 n2 + 4n + 4 ] + n + 1] = (n + 1)2 [ (n + 1)2 [ 4 4 (n + 1)(n + 2) 2 [ ] . 2 [13 + 23 + . . . + n3 ] + (n + 1)3 = [
Thus n + 1 ∈ S. By the PMI, S = N. n S = {n ∈ N: i=1 (2i − 1)3 = n2 (2n2 − 1)}. 1 3 2 3 i=1 (2i − 1) = 1 = 1 (2 − 1), so 1 ∈ S. Assume n ∈ S. Then n +1
(2i − 1)3
=
n
i=1
(2i − 1)3 + [2(n + 1) − 1]3
i=1
= 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)
1 1 − 1+1 , so statement is true for n = 1. i. 1·2 ii. Assume that the statement is true for some n ∈ N. Then
1 1·2
+ =
n(n + 2) (n + 1)(n + 2)
+ =
1 1 1 1 + + ··· + 2·3 3·4 n(n + 1) (n + 1)(n + 1 + 1) 1 n + n + 1 (n + 1)(n + 2) n2 + 2n + 1 1 = (n + 1)(n + 2) (n + 1)(n + 2) (n + 1)2 n+1 . = (n + 1)(n + 2) n+2
Thus the statement is true for n + 1. iii. By the PMI, the statement is true for all n ∈ N. (h)
1 2 n 1 i. The statement 2! + 3! + · · · + (n+1)! = 1 − (n+1)! is true for n = 1, 1 1 1 because 2! = 2 = 1 − 2! . 1 2 + 3! + ... + ii. Assume the statement is true for some n ∈ N. Then 2! n 1 n+1 = 1 − . Adding to both sides of this equation, we (n+1)! (n+1)! (n+1+1)! have n+1
i=1
n (n + 1)!
=
1−
1 n+1 n+2 n+1 + =1− + (n + 1)! (n + 2)! (n + 2)! (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. 1 1 1 (i) i. The statement is true for n = 1 because i=1 (2i−1)(2i+1) = 1·3 = 1 2·1+1 . ii. Suppose the statement is true for some n ∈ N. We must show that n+1 1 n+1 i=1 (2i−1)(2i+1) = 2(n+1)+1 . By the hypothesis of the induction, n+1
i=1
1 (2i − 1)(2i + 1)
=
n
i=1
= = =
1 1 + (2i − 1)(2i + 1) (2n + 1)(2n + 3)
n 1 + 2n + 1) (2n + 1)(2n + 3) 1 n(2n + 3) + (2n + 1)(2n + 3) (2n + 1)(2n + 3) 2n2 + 3n + 1 (n + 1)(2n + 1) n+1 . = = (2n + 1)(2n + 3) (2n + 1)(2n + 3) 2n + 3
Thus the statement is true for n + 1. iii. By the PMI, the statement is true for every n ∈ N. 1 1 1 (j) i. The statement is true for n = 1, because i=1 (1 − i+1 ) = 1 − 12 = 1+1 . n+1 ii. Suppose that the statement is true for some n ∈ N. Then i=1 (1 − n 1 1 1 1 1 n+2−1 1 ) = (1 − ) i=1 (1 − i+1 ) = (1 − n+2 )( n+1 ) = ( n+2 )( n+1 ) = n+2 i+1 1 n+2 . Thus the statement is true for n + 1. (k) By the PMI the statement is true for every n ∈ N. 1 (l) i. i=1 (2i − 1) = 2(1) − 1 = 1 = (2(1))! 1!21 , so the statement is true for n = 1. n+1 ii. Assume the property holds for some n ∈ N. We must show i=1 (2i − 1) = (2(n+1))! (n+1)!2n+1 . By the hypothesis of induction, the left hand side of the equation is n [2(n + 1) − 1] · i=1 (2i − 1) = (2n + 1) · (2n)! n!2n and the right hand side of the equation is (2n)!(2n+1)(2n+2) . Since these are equal, the property 2(n+1)n!2n is true for n + 1. iii. By the PMI, the statement is true for all n ∈ N. 0 a(r 1 −1) i (m) i. so the statement is true for n = 1. i=0 ar = a = r−1 ii. Assume the statement is true for some n ∈ N. Then n
ari
=
i=0
n−1
i=0
= =
ari + arn =
a(rn − 1) + arn r−1
arn + arn+1 − arn − a a(rn − 1) + arn (r − 1) = r−1 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 true for 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 are divisible 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).
iii. (e) i. ii.
iii. (f) Let i. ii.
By the induction hypothesis (n + 1)(n)(n + 2)(n − 1) is divisible by 12. The expression 4(n)(n + 1)(n + 2) has a factor of 4 and, since the terms n, n + 1, and n + 2 are three consecutive integers, one of them is divisible 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. By the PMI, the statement is true for all n ∈ N. Since 8 divides 52·1 − 1 = 24, the statement is true for n = 1. 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 the hypothesis 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. By the PMI, 8 divides 52n − 1 for every natural number n. S = {n ∈ N: 10n + 3 · 4n+2 + 5 is divisible by 9}. 101 + 3 · 41+2 + 5 = 15 + 3 · 64 = 207 = 9(23), so 1 ∈ S. 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. iii. By the PMI, the statement is true for all n ∈ N. (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 the induction 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 the statement is true for n + 1. iii. By the PMI, the statement is true for all n ∈ N.
(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. 1 (k) i. The statement is true for n = 1, because i=1 i12 = 1 ≤ 2 − 11 . n ii. Suppose that for some n ∈ N, i=1 i12 ≤ 2 − n1 . Then n+1
i=1
1 1 1 + = 2 (n + 1)2 i i2 i=1 n
≤ 2−
1 1 + n (n + 1)2
(n + 1)2 n + n(n + 1)2 n(n + 1)2 n2 + n + 1 = 2− n(n + 1)2 n2 + n 1 . < 2− =2− n(n + 1)2 n+1 =
2−
Thus the statement is true for n + 1. iii. By the PMI the statement is true for every n ∈ N.
2
SET THEORY
(l)
(m)
55
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. iii. By the PMI, the statement is true for all n ∈ N. i.
13 3
+
15 5
ii. Assume
7(1) 15
+
3
n 3
(n + 1)3 3
+
= 1 so the property holds for 1. n5 5
+
7n 15
∈ Z for some n. Then
(n + 1)5 7(n + 1) 1 + = (n3 + 3n2 + 3n + 1) + 5 15 3 1 5 7 4 3 (n + 5n + 10n + 10n2 + 5n + 1) + (n + 1) 15 5 n5 1 7n n3 + n2 + n + + + 5 15 3 3 7 1 +n4 + 2n3 + 2n2 + n + + 5 15 3 n5 7n n + + + n4 + 2n3 + 3n2 + 2n + 1. 3 5 15
+
=
=
3
5
By the hypothesis of induction ( n3 + n5 + 7n 15 ) is an integer, and the remaining terms are integers so the sum is an integer. Thus the property holds for n + 1. iii. By the PMI, the statement is true for all n ∈ N. (n)
d i. dx (x1 ) = 1 = 1 · x0 so the statement is true for 1. ii. Suppose the statement is true for some n ∈ N. Then
d n+1 (x ) dx
d n 1 (x · x ) dx d d n = xn (x) + (x ) · x dx dx = 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)
8.
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. Then A − {a} has n elements, and thus, by the induction hypothesis, has 2n subsets. If B is one of these subsets, then both B and B ∪ {a} are subsets of A. So A must have at least 2n + 2n = 2n+1 subsets. These are all the subsets of A, because if a subset of A does not contain a then 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. 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 for n = 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 .
i. 4! = 24 and 3(4) = 12, so 4! ≥ 3(4). Thus the statement is true for n = 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. 2 2 = 43 = 2+1 (d) i. The statement is true for n = 2, because i=2 i i−1 2 2·2 . ii. Suppose the statement is true for some n ∈ N such that n ≥ 2. Then (c)
n+1 2 i=2
i −1 i2
=
n 2 (n + 1) − 1 i2 − 1 · (n + 1)2 i2 i=2
=
n+2 (n + 1) + 1 n2 + 2n n + 1 = = · (n + 1)2 2n 2(n + 1) 2(n + 1)
Thus the statement is true for n + 1. iii. By the generalized PMI, the statement is true for all n ≥ 2. 4 1 1 (e) i. i=1 1i = 11 12 13 14 = 24 ≥ 16 = 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
n 1 1 1 1 1 = ≤ 2−n ≤ 2−n = 2−(n+1) . n+1 2 i n + 1 i=1 i
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 add up 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 polygon with n + 1 sides. Pick any 3 consecutive vertices and label them A, B and C. Draw a line segment from A to C. Since the original polygon was convex, it is now split into the triangle ABC and an n-sided convex polygon. By the induction hypothesis, the angles of the n sided polygon add up to (n − 2) · 180˚. Thus the sum of the interior angles of the original polygon is (n − 2) · 180˚+180˚= [(n + 1) − 2] · 180˚. Therefore the 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 + √12 . Thus the statement is true for n = 2. √ ii. Assume the statement is true for some n ≥ 2. Then n < √11 + √12 + · · · + √1n . Also, n n2
<
n + 1, so
<
n(n + 1), so
2
n n+1 n √ n+1 √ n n+1 n+1 √ 1 n+1 1− n+1 √ 1 n+1− √ n+1 √ n+1
< < < < < <
n, so √ √ √ √ √
n, so n, so n, so n, so 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. 1 9. (a) i. (|1i=1 Ai )c = i=1 Ai c is A1 c = A1 c , which is true. ii. Assume that the statement is true for some n ∈ N. Then n+1 c c c |i=1 Ai = (|ni=1 Ai ∩ An+1 ) = (|ni=1 Ai ) ∪ An+1 c n n+1 Ai c ∪ An+1 c = Ai c = i=1
iii. (b) Let i. ii.
i=1
Thus the statement is true for n + 1. By the PMI, the statement is true for all natural numbers. n S = {n ∈ N: ( i=1 Ai )c =|ni=1 Ai c }. 1 For n = 1, the expression ( i=1 Ai )c =|1i=1 Ai c is A1 c = A1 c , which is true. Assume that the statement is true for some n ∈ N. Then c c i = 1n+1 Ai = i = 1n Ai ∪ An+1 c = i = 1n Ai ∩ An+1 c =
(| i = 1n Ai c ) ∩ An+1 c =| i = 1n+1 Ai c .
Thus the statement is true for n + 1.
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iii. By the PMI, the statement is true for all n ∈ N. 10. (a) It is acceptable to begin with the basis step n = 1, or n = 2. Since there are no line segments if there is just one point, the statement is true if n = 1. 2 Also if n = 2, the number of lines is 1 = 2 2−2 , so the statement is true for 2. (b) Assume the statement is true for some n ≥ 2. Consider n + 1 points in a 2 plane. By the hypothesis of induction, n of them can be joined with n 2−n distinct line segments. It now takes an additional n segments to join the last point with each of the first n points. So, the total is (n2 + 2n + 1) − (n + 1) (n + 1)2 − (n + 1) n2 − n + 2n n2 − n = = , +n = 2 2 2 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 statement is true for n = 1. (b) Assume the statement true for some n, and suppose we start with n + 1 disks. By the induction hypothesis, all but the largest disk can be moved to another peg in 2n − 1 moves. Now move the largest disk to the free peg and 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 statement is 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 beats x, 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 a tournament with n players, so there is a top player t. Let B be the set of players that t beats. Now if t beats s, then t is top. If not, but some w in B beats s, then t is top. Otherwise s beats t and s beats every player in B. Every other player is 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 this case, when either horse is removed from the set, the remaining horse has the same color (as itself), [because there is only one horse left] but the two horses may have different colors. (b) F. This method, sometimes called “proof by bluffing,” is so common that it may be a necessary stage for understanding induction proofs, at least for some students. What is missing is the proof that the statement is true for n + 1. (c) F. n = 1 is odd, but 11 + 1 = 2 is not odd, so the basic step fails. The claim is false. (d) F. The induction step assumed that n + 1 ∈ S. One must show n ∈ S implies n + 1 ∈ S. (e) F. The factorization of xy + 1 is wrong, and there is no reason to believe that x + 1 or y + 1 is prime. (f) A.
2
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2.5
59
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). Thus 23, 24, 25 ∈ S. Let m > 22 be a natural number and assume that for all k ∈ {23, 24, . . . , m − 1}, k ∈ S. If m = 23, 24, or 25 we already know m is in 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 and t ≥ 2. Then m = 3(s + 1) + 4t. Thus m ∈ S. By the PCI, the statement is true 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), and 37 = 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. Otherwise m ≥ 38, so m − 4 ≥ 34. By the hypothesis of induction, m − 4 ∈ S, so m − 4 = 4s + 5t for some integers s and t, where s ≥ 3 and t ≥ 2. Then m = 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 for m = 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, because a1 = 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, so an > 0 for all n ∈ N. (b) Suppose there were natural numbers a, b such that b = a + b. Then by the WOP, the set T = {n ∈ N: n = a + n} has a least element, b0 . And b0 = 1 because 1 is not the successor of any natural number. Now, b0 = a + b0 implies (b0 − 1) = a + (b0 − 1) which implies b0 − 1 ∈ T , contradicting the minimality of b0 . (c) Let A = {b ∈ N: (∃a√∈ N)(a2 = 2b2 )}. Notice that A = ∅ iff (∃ a, b ∈ N)(( ab )2 = 2) iff 2 is rational. Suppose A is nonempty. Then by the WOP, A has a least element b0 . Let a0 ∈ N be such that a20 = 2b20 . Then a20 is even, which forces a0 to be even. Substituting a0 = 2k, we get (2k)2 = 2b20 , and then b20 = 2k 2 is even, which forces b0 to be even. Now a20 and b20 are both natural numbers and a 2 0
2
a2 b2 = 0 = 0 =2 4 2
b0 2
2 ,
so b20 ∈ A, contradicting the minimality of b0 . Therefore A is empty, so is irrational.
√
2
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60
4. (a) Let Rn be the number of rabbit pairs at n months. We already have that Rn = fn for n = 1, 2, 3, 4. Assume Rn = fn for all n up to m − 1 where m − 1 ≥ 4. Then Rm = (# of pairs from last month)+(# of new pairs). The second term equals the number of breeding pairs, because each breeding pair produces one new pair each month. Now the breeding pairs are exactly the pairs that were around two months ago, since it takes a month for them to mature. Thus, Rm = Rm−1 + Rm−2 = fm−1 + fm−2 = fm . Thus, by the PCI, Rn = fn for all n ∈ N. (b)
f1 = 1 f6 = 8
f2 = 1 f7 = 13
f3 = 2 f8 = 21
f4 = 3 f9 = 34
f5 = 5 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+4 is odd 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 divide fn+1 + fn − fn+1 = fn , so d must be 1. This demonstrates that gcd(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. Thus gcd(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 all n ∈ {1, 2, . . . , m − 1}. Then fm = fm−1 + fm−2 ∈ N by the closure property of 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.
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(c) Let a be a fixed natural number. (We can actually do induction either on a or n here.) Notice fa f1 + fa+1 f2 = fa + fa+1 = fa+2 and that fa f2 + fa+1 f3 = fa + 2fa+1 = fa+1 + fa+2 = fa+3 , so the statement is true for n = 1 and n = 2. Suppose m > 2 and fa fn + fa+1 fn+1 = fa+n+1 for all n ∈ {1, 2, . . . , m − 1}. Then fa fm + fa+1 fm+1
= fa (fm−1 + fm−2 ) + fa+1 (fm + fm−1 ) = (fa fm−1 + fa+1 fm ) + (fa fm−2 + fa+1 fm−1 ) = fa+m + fa+m−1 = fa+m+1 .
Therefore, by the PCI, fa fn + fa+1 fn+1 = fa+n+1 for all n ∈ N. (α+1)−(β+1) α−β −β (d) First notice that f1 = 1 = α−β = αα−β , α−β , and f2 = 1 = α−β = α−β so the statement is true for n = 1 and n = 2. n −β n Suppose that fn = αα−β for all n ∈ {1, 2, . . . , m − 1} and that m > 2. Then 2
fm
2
= fm−1 + fm−2 αm−2 − β m−2 αm−1 − β m−1 + = α−β α−β m−2 m−2 (α + 1) − β (β + 1) α = α−β αm−2 (α2 ) − β m−2 (β 2 ) αm − β m . = = α−β α−β
Therefore, by the PCI, fn =
αn −β n 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 then b − 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 show that 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 member of S. This is impossible, so we must have r < |a|. To complete the proof, we must show that q and r are unique. Suppose there exist integers q, q , r, and r such that b b
= aq + r with 0 ≤ r < |a| and = 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|. Then r − 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 empty because if a > 0 then a(1) + b(0) ∈ S, and if a < 0 then a(−1) + b(0) ∈ S. Thus S is a nonempty subset of N, so by the Well Ordering Principle S has a smallest element.
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9. Suppose n is the smallest positive integer greater than 1 that is not prime and such 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 may be written as: n = p1 p2 p3 . . . pn and n = q1 q2 q3 . . . qm. Since p1 is one of the factors of n, p1 divides n = q1 q2 q3 . . . qm. By Euclid’s Lemma p1 = qj for some 1 < j < m. Then n/p1 = n/qj is a positive integer that is smaller than n and has two different prime factorizations. This is a contradiction. We conclude that every natural number greater than 1 is prime or can be expressed uniquely as a product of primes. 10. Suppose there exists n ∈ N such that some tournament with n players does not have a top player. Let k be the least such number. Then k must be greater than 2. Therefore we can pick a player r out of the set of players, and have at least one player 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 or r 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 = 1 and 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 and assume for some m that {1, 2, . . . , m − 1} ⊆ S. Then m ∈ S because if it 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 a smallest element. (b) Let S be a subset of N such 1 ∈ S and S is inductive. We wish to show that S = N. Assume that S = N and let T = N − S. By the WOP, the nonempty set T has a least element. This least element is not 1, because 1 ∈ S. If the least element is n, then n ∈ T and n − 1 ∈ S. But by the inductive property of 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 by the WOP, T has a smallest element, n. But then {1, . . . , n − 1} ⊆ S (even if 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 assumption that m − 1 is a natural number. In fact, 1 is the smallest natural number n such that 3 does not divide n3 + 2n + 1. There is no contradiction about a 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 (or assumes) 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
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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) (b) (c) (d)
The Combination Rule The Product Rule Combination Rule and Product Rule 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 who got 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 = 72 with units digit 0 and 8 · 8 · 4 = 256 that are even with a nonzero units digit, for a total of 328 that are even. There are 9 · 9 · 8 = 648 three digit positive integers with no repeated digits, and of 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 repeated digits. (b) There are 8 · 8 · 7 · 5 = 2240 odd four digit positive integers with no repeated digits. (c) Among four digit positive integers with no repeated digits, there are 9 · 8 · 7 · 1 = 504 with units digit 0 and 8 · 8 · 7 · 4 = 1792 that are even with a nonzero units digit, for a total of 2296 that are even.
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10. If the bottom right corner square is to be colored the same as the upper left corner, there are 20 choices for the upper left, 1 for the lower right, 19 for the upper 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 painted different 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 + C counts twice exactly the elements of A ∩ C and B ∩ C. Therefore A ∪ B ∪ C = A + B + C − A ∩ C − B ∩ C.
1 12. If k = 1, then there is one task which can be performed in n1 = i=1 ni ways. Now suppose the statement is true for k and there are k + 1 tasks to be done in k sequence. From our hypothesis, the first k tasks can be done in i=1 ni ways. Thinking of this by Theorem 2.6.4 this and the last task can be task, kas one big k+1 done in nk+1 · i=1 ni = i=1 ni ways. 13. (a) There is only 1 permutation of a 1 element set, so the statement is true for n = 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 first ordering A − {x} and then choosing a place to insert x. The first task can be done in n! ways, while the second task can be done in n + 1 ways. Thus the 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 ordering them. 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)! n! so x = (n−r)! . (b) Let n = 1 and 0 ≤ r ≤ 1. Then r = 0 or r = 1, so the number of permutations 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 for n = 1. Assume that n ∈ N and for every r such that 0 ≤ r ≤ n the number of n! permutations of r distinct elements from a set of n elements is (n−r)! . Now suppose there are n + 1 objects in a set and that 0 ≤ r ≤ n + 1. In the special case when r = 0, the number of permutations of r distinct elements (n+1)! from the set is 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 the n! ways to permute r distinct elements hypothesis of induction there are (n−r)! from the set if x is not one of the elements selected. If x is one of the n! ways to permute the other elements, elements selected, we have (n−(r−1))! and then r ways to position x amongst the other elements. Thus there are (n+1)! n! n! (n−r)! + (n−(r−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!
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16. (a) 3! · 4! (b) 5! · 3! There are 5! ways to arrange the digits with the even digits as an unordered block, and then 3! ways to arrange the even digits within the block. (c) 4! · 4! Treating each even and the cluster of all odd digits as 4 distinct objects, there are 4! ways to order them, and then 4! ways to order the cluster of odd digits. (d) 3! · 4! The digits must alternate in the order odd, even, odd, etc. There are 4! 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 first pick such a permutation, and then order the m alike objects. Thus n! = x · m!, n! so x = m! .
18. 19. 20. 21.
If m1 , objects are alike, m2 others are alike, and so forth, up to mk objects alike, permutations. then there are m1 !mn! 2 !···mk ! First choose 4 and then order them: 10 4 · 4! = 5040. 8 8 (a) 19 (b) 11 (c) 11 1 · 2 3 2 · 1 59 5·4 9·8 2 2 = 2 · 2 = 360 (a) i. (a + b)1 = 10 a0 b1 + 11 a1 b0 , so the statement is true for n = 1. ii. Suppose the statement is true for some n ∈ N. Then n+1
n + 1 r n+1−r a b r r=0 n
n + 1 r n+1−r n+1 a b + an+1 =b + r r=1 n
n n + ar bn+1−r + an+1 = bn+1 + r r − 1 r=1 n n
n n r n+1−r ar bn+1−r + an+1 a b + an+1 + = bn+1 + r − 1 r r=1 r=1 n n
n
n n r n−r r−1 n+r−1 n +a a b a b +a =b b + r r−1 r=1 r=1 n n−1
n
n =b ar bn−r + a ar bn−r + an r r r=0 r=0 n n
n
n r n−r r n−r +a a b =b a b r r r=0 r=0 n
n = (a + b) ar bn−r . r r=0 = (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.
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66
(b)
n−1 n−1 + r r−1
= = = = =
(n − 1)! (n − 1)! + r!(n − r − 1)! (r − 1)! [n − 1 − (r − 1)]! (n − 1)! (n − 1)! + r!(n − r − 1)! (r − 1)!(n − r)! 1 1 (n − 1)! + n−r r!(n − r − 1)! r n (n − 1)! r!(n − r − 1)! r(n − r) n! n = r!(n − r)! r
(c) Choose one particular element x a set A of n elements. The number of from subsets 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 those subsets not containing x. We count the of subsets in each collection number and add the results. First, there are n−1 r-element subsets of A that do r not contain x, since each is a subset of A − {x}. Second, there are n−1 r−1 r-element subsets of A that do contain x, because each of these corresponds to the (r − 1)-element subset of A − {x} obtained by removing x from the subset. Thus the sum of the number of subsets in the two collections is n−1 n−1 n + = . r r−1 r 22. (a) a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6 (b) a4 + 8a3 b + 24a2 b2 + 32ab3 + 16b4 (c) 13 3 10 (d) 12 2 ·2 23. (a) The number of ways to select an even number = the number ways to select 0 + 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 odd number of objects. (b) Let A and B be disjoint sets with n and m elements respectively. Then n+m is the number of r-element subsets of A r ∪ B. For each i such that 0 ≤ i ≤ r, we may select i elements from A in ni ways and the remaining r n m m ways. Thus n+m = i=0 i r−i . r − i elements from B in r−i r 1 2n+1 2n+1 1 2n+2 (c) Applying Theorem 2.6.9(c) twice, we have 2 n+1 = 2 [ n+1 + n ] = 2n 2n 2n 2n 2n 2n 1 + + n−1 ]. But n+1 = n−1 , so this expression is 2 [ n+1 + 2nn 2nn equal to n + n+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) (b) (c) (d) (e) (f) (g) (h)
3.
domain R, range R domain R, range [3, ∞) domain [1, ∞), range [0, ∞) domain {x ∈ R | x = 0}, range R+ domain R, range R domain (−2, 2) range {3} domain R, range R domain R, range R
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
4. (a) R1−1 = {(x, y) ∈ R × R | y = x} 67
3
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68
(b) R2−1 = {(x, y) ∈ R × R | y = x−2 −5 } } (c) R3−1 = {(x, y) ∈ R × R | y = x+10 7 √ −1 (d) R4 = {(x, y) ∈ R × R | y = ±√ x − 2 } (e) R5−1 = {(x, y) ∈ R × R | y = ± 25−x } (f) R6−1 = {(x, y) ∈ R × R | y > x − 1} (g) R7−1 = {(x, y) ∈ R × R | y < x+4 3 } 2x (h) R8−1 = {(x, y) ∈ R × R | y = x−2 } (i) R9−1 = {(x, y) ∈ P × P | y is a child of x and x is male} −1 (j) R10 = {(x, y) ∈ P × P | x is a sibling of y} −1 = {(x, y) ∈ P × P | y is loved by x} (k) R11 5.
(a) (c) (e) (g)
R ◦ S = {(3, 5), (5, 2)} T ◦ S = {(2, 1), (3, 1), (3, 4)} S ◦ R = {(1, 5), (2, 4), (5, 4)} R ◦ (S ◦ T ) = {(3, 2)}
(b) (d) (f) (h)
R ◦ T = {(3, 2), (4, 5)} R ◦ R = {(1, 2), (2, 2), (5, 2)} T ◦ T = {(1, 1), (4, 4)} (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} 14x−20 (o) {(x, y) ∈ R × R | y = 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)} (b) R = {(a, c)}
S = {(b, c)}
S = {(c, d)}
R ◦ S = {(b, d)} and S ◦ R = {(a, c)}
(S ◦ R)−1 = {(d, a)} and S −1 ◦ R−1 = ∅
(c) R = {(a, c), (a, b)} S = {(c, d)(d, a)} Then S ◦ R = T ◦ R = {(a, b), (b, d)}
T = {(a, b)(b, d)}
3
RELATIONS AND PARTITIONS
(d) R = {(a,b)}
69
S= {(c,d)}
9. (a) Suppose x ∈Dom(S ◦ R). Then for some z, (x, z) ∈ S ◦ R. Then for some y, (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. Suppose z ∈Rng(S ◦ R). Then for some x, (x, z) ∈ S ◦ R. Then for some y, (x, y) ∈ R and (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)
iff
(S ◦ R)−1 for some x ∈ C and y ∈ A (y, x) ∈ S ◦ R
iff iff
∃z ∈ B such that (y, z) ∈ R and (z, x) ∈ S ∃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} and C = {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 number of ways to select an element of A × B is the number (m) of ways to select an element 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 }. Use induction 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. Then B − {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)}. This means 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) iff x ∈Rng(R). (b) x ∈ b∈B b R iff (∃ b ∈ B)(x ∈ b R) iff x ∈ A and (∃b ∈ B)((x, b) ∈ R) iff x ∈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.
3
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70
(b) C. This “proof” could be corrected by observing that (a, c) ∈ / B ×D implies a∈ / B or c ∈ / D, which implies a ∈ / A or c ∈ / C. This contradicts the fact that 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 is false. (g) F. The statement ”Since (x, z) ∈ R and (y, z) ∈ R, x = y“ is not true in general.
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) (c) (e) (g)
{(x, y) | y = 5x} {(x, y) | x2 + y 2 = 1} {(x, y) | y ≤ x − 1} {(x, y) | x2 + y 2 = 0}
(b) (d) (f) (h)
{(x, y) | y = x or y = x2 } {(x, y) | y = x or x2 + y 2 = 1} {(x, y) | y ≤ x} {(x, y) | y = ±x}
4. (a) Suppose A is nonempty. Let a be an element of A. Then (a, a) is not in the relation ∅, 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) are in ∅, then (x, z) is in ∅. (Both statements are true because their antecedents are false.)
3
RELATIONS AND PARTITIONS
71
5. (a)
i. Let x ∈ R. Then x − x = 0, is a rational number, so x R x. Therefore R 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 and y − 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. The class 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. Thus y R x. iii. Suppose x R y and y R z for some x, y, z ∈ N. Then x has the same digit in the tens place as y and y has the same digit in the tens place as 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, then xz = 1 (in case x = y) or x = z (in case xy = 1). In any case x = z or xz = 1, so x V z. iv. 3/V = {3, 13 }; (− 23 )/V = {− 23 , − 32 }; 0/V = {0}.
(d)
i. Let a ∈ N. Then the prime factorization of a into primes has the same number 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 in its prime factorization that b has, so b has the same number of 2’s in its 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 factorizations of a and b have the same number of 2’s and the prime factorizations of b and c have the same number of 2’s. Thus the prime factorizations 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 + y 2 = x2 + y 2 , so (x, y) T (x, y). ii. Suppose (x, y) T (a, b) for some (x, y), (a, b) ∈ R × R. Then x2 + y 2 = a2 + b2 , so a2 + b2 = x2 + y 2 . 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 + y 2 = a2 + b2 and a2 + b2 = u2 + v 2 . Then x2 + y 2 = u2 + v 2 . Thus (x, y) T (u, v). iv. The equivalence class of (1, 2) is the set {(x, x):√ x2 + y 2 = 5}, which is the 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.
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RELATIONS AND PARTITIONS
72
iv. The elements of {m}/R are {m}, {n}, {p}, {q}, {r}, and {s}. The equivalence 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 equivalence classes contains seven different equivalence classes. (To be precise, 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 possible size 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 class of (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 sin x = sin y, so x T y. ii. Suppose x T y, for some x, y ∈ R. Then sin x = sin y. Therefore sin y = sin x, so y T x. iii. Suppose x T y and y T z, for some x, y, z ∈ R. Then sin x = 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 π2 is { π2 + 2kπ: k ∈ Z}. The class of π4 is {± π4 + 2kπ: k ∈ Z}.
6. (a) Let
p q
∈ Q. Then pq = pg, so pq R pq .
(b) Suppose s p tRq.
p s qRt
for some
p s q, t
∈ Q. Then pt = qs, so sq = tp. Therefore,
(c) Suppose pq R st and st R uv for some pq , st , uv ∈ Q. Then pt = qs and sv = tu. Multiplying by v and then by q, we hav ptv = qsv and qsv = qtu. Therefore pv = qu, so pq R uv . The equivalence class of 23 is { 2k 3k : k ∈ Z − {0}}, which is the set of all rational numbers that reduce to 32 . 7. (a) (b) (c) (d) 8. (a)
Transitive, not reflexive and not symmetric Transitive, reflexive and symmetric Transitive, reflexive and symmetric Reflexive, not transitive and not symmetric 0/ ≡ 5 = {. . . , −10, −5, 0, 5, . . .} 1/ ≡ 5 = {. . . , −9, −4, 1, 6, . . .} 2/ ≡ 5 = {. . . , −8, −3, 2, 7, . . .}
3/ ≡ 5 = {. . . , −7, −2, 3, 8, . . .} 4/ ≡ 5 = {. . . , −6, −1, 4, 9, . . .}
3
RELATIONS AND PARTITIONS
(b)
0/ ≡ 8 1/ ≡ 8 2/ ≡ 8 3/ ≡ 8
73
= {. . . , −16, −8, 0, 8, . . .} = {. . . , −15, −7, 1, 9, . . .} = {. . . , −14, −6, 2, 10, . . .} = {. . . , −13, −5, 3, 11, . . .}
4/ ≡ 8 5/ ≡ 8 6/ ≡ 8 7/ ≡ 8
= {. . . , −12, −4, 4, 12, . . .} = {. . . , −11, −3, 5, 13, . . .} = {. . . , −10, −2, 6, 14, . . .} = {. . . , −9, −1, 7, 15, . . .}
(c) 0/ ≡ 1 = Z (d)
0/ ≡ 7 1/ ≡ 7 2/ ≡ 7 3/ ≡ 7
= {. . . , −14, −7, 0, 7, . . .} = {. . . , −13, −6, 1, 8, . . .} = {. . . , −12, −5, 2, 9, . . .} = {. . . , −11, −4, 3, 10, . . .}
4/ ≡ 7 = {. . . , −10, −3, 4, 11, . . .} 5/ ≡ 7 = {. . . , −9, −2, 5, 12, . . .} 6/ ≡ 7 = {. . . , −8, −1, 6, 13, . . .}
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 ≡ m x, so x ∈ x. (b) Let x ∈ Z. Then x ∈ x by part (a), so x = ∅. (c) Assume x ≡ m y. Let w ∈ x. Then x ≡ m w. By symmetry y ≡ m x and by the transitivity y ≡ m w. Therefore w ∈ y. This shows x ⊆ y. If z ∈ y, then y ≡ m z and by transitivity x ≡ m z. Therefore z ∈ x and y ⊆ x. We conclude that x = y. (d) Assume x = y. By the reflexivity y ≡ Therefore x ≡ m y.
m y,
so y ∈ y. But y = x, so y ∈ x.
(e) Assume x ∩ y = ∅. Then there is some w ∈ Z such that w ∈ x and w ∈ y. Therefore, x ≡ m w and y ≡ m w. By symmetry and transitivity, x ≡ m y. 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 transitive because 5 S 4 and 4 S 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 are symmetric, (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 ∩ S and so R ∩ S is transitive. 13. (a) R is reflexive
iff iff
for all x ∈ A, (x, x) ∈ R 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.
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RELATIONS AND PARTITIONS
74
(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 assume R ◦ 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. By the 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 ∈ A such that (x, a1 ), (a1 , a2 ), . . ., (an , y) ∈ R and there exist b1 , b2 , . . ., bm ∈ A such that (y, b1 ), (b1 , b2 ), . . ., (bm , z) ∈ R. Since x, a1 , a2 , . . ., an , y, b1 , b2 , . . ., bm , z are in A, (x, z) ∈ TR . Therefore TR is transitive. (b) Assume that S is transitive and R ⊆ S. Suppose (x, y) ∈ TR . Then there exist 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 E c be the edge set for the complement of D. Assume that D is a symmetric digraph. Suppose (x, y) ∈ E c . Then (x, y) ∈ / E. Since D is symmetric, it must then be the case that (y, x) ∈ / E. Therefore, (y, x) ∈ E c 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 and y 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 symmetric and transitive, but not reflexive on A. Alternatively, consider the empty relation 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 more complete second sentence.
3
RELATIONS AND PARTITIONS
3.3
75
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 partitioned by class standings, using sets of all freshmen, sophomores, juniors, seniors, and graduate students; by major; as in-state and out-of-state students; or by the number of credit hours completed. Note that most of these methods of partitioning may need to be adjusted for special circumstances. For example, if students are partitioned by majors, there probably should be a partition element for undecided students, and there may be a partition element for unclassified or non-degree-seeking students. Furthermore, one must account for students who are double-majors, perhaps by requiring such students 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 a partition 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 the interval [0, 1). All the integers are in the partition set with 0. All real numbers that can be written as z + 0.071, z ∈ Z, are in the partition set with 0.071. Each partition set contains exactly one real number in every interval [z, z + 1), where z is an integer. If x is any element of a partition set, the next largest element of that partition set is x + 1. The partition set that 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. The class of 10 modulo R contains all integers that have 1 as the tens digit, and so 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 partition element is {x: sin x = 12 } = {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 points on the x and y axes; a second element consists of all points in the first and third quadrants; and the third element contains all points in the second and 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.
3
RELATIONS AND PARTITIONS
(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 = ∅.
76
3
RELATIONS AND PARTITIONS
77
ii. Suppose a, b ∈ R and a = b. Assume that (x, y) ∈ Aa ∩ Ab . Then by definition, y = a − x2 and y = b − x2 . Then a = b. This is a contradiction. Therefore Aa ∩ Ab = ∅ when a = b. 2 iii. Let (x, y) ∈ R × R. Choose a = y + x2 ∈ R. Then y = a − x , so (x, y) ∈ Ra . Therefore R × R ⊆ a∈R Ra . Clearly, a∈R Ra ⊆ R × R. (c) The point (s, t) in the plane R × R is related to another point (u, v) iff both points 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 such that 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 each partition set, the digraph has all possible arcs (including loops) between elements of the partition set. 10. (a) Suppose x Q y. By definition of Q, there is C ∈ P such that x ∈ C and y ∈ 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, there is 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 not symmetric. 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}} is not 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. {B1c , B2c } is a partition of A, because {B1c , B2c } = {B1 , B2 }. If B1 = B2 , then B1 = B2 = A; hence B1c = B2c = ∅. Thus {B1c , B2c } would not be a partition. (c) No. Let A = {1, 2, 3} and B1 = {1}, B2 = {2} and B3 = {3}. Then B c 1 = {2, 3} and B2c = {1, 3}, which are neither equal or disjoint. If exactly two elements, B1 and B2 , of B are equal, then {B1c , B2c , B3c } = {B3 , B3 , B1 } is a partition of A. If B1 = B2 = B3 , then B1 = A and B1c = ∅, so {B1c , B2c , B3c } is not a partition of A. 15. (a) F. Assume x ∈ y/R. The “proof” shows z ∈ / y/R ⇒ z ∈ / x/R, which is the converse 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 are not well connected may give this “proof” a grade of C.
3
RELATIONS AND PARTITIONS
3.4
78
Ordering Relations
1. (a) (b) (c) (d) (e) (f) (g)
Is not antisymmetric because (2, 4) and (4, 2) ∈ R. Is antisymmetric. Is not antisymmetric because (2, −2) and (−2, 2) ∈ R. Is not antisymmetric. 5 ≤ 26 and 6 ≤ 25 , so 5 R 6 and 6 R 5. Is not antisymmetric because 0 S 1 and 1 S 0. Is not antisymmetric because (1, 3) and (3, 1) ∈ R. 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 that x = 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 for some 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. Also b ≤ 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 = x1 x2 . Then α ≤ α, because x1 = x1 and x2 ≤ x2 . (b) Suppose α ≤ β and β ≤ α, where α = x1 x2 and β = y1 y2 . It is not possible that x1 < y1 because β ≤ α implies y1 ≤ x1 . Thus x1 = y1 and x2 ≤ y2 from 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 = x1 x2 , β = y1 y2 and δ = z1 z2 . Then x1 ≤ 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 that x1 < z1 , then α ≤ δ. (The second letters do not need to be compared.)
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79
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 that C = B. On the other hand, if x = C, then for all y ∈ C we have y ∈ B and 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 every element 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}, so c ∈ B. Thus C ⊆ B. This proves C = B or C = B ∪ {x}, so B is an immediate predecessor of B ∪ {x}. 13. (a) Yes. Yes. (b) No. A set consisting of two adjacent congruent squares has no largest element. (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.
3
RELATIONS AND PARTITIONS
(b)
80
Then by Exercise 9(f) of Section 2.2, C ∪ D ⊆ B, so C ∪ D is the least upper 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. By Exercise 9(c) of Section 2.2, S ⊆ C ∩ D, so C ∩ D is the greatest lower bound. 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. ThenB ⊆ W for every B ∈ P. Then (by Exercise 10(c) of Section 2.3) B∈P B ⊆ W . Therefore B is the least upper bound for P. B∈B 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) of Section 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 9 T 10 and 10 T 9. Also T is not transitive because 10 T 8 and 8 T 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 are both even, then t is even and m ≤ n ≤ t, so m V t. Otherwise, m and n 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 5 S 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 ≤ m and 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 = 5 and t = 5, so n S t. If m = 5, then t = 5, so n S t. If t = 5, then n S t by definition of S. Otherwise, m, n, and t are all different from 5 and n ≤ 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 order on N. (d) i. Let n ∈ N. If n = 5, then n ≤ n, so n S n. If n = 5, then 5 S 5 by definition. 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 m S 5. If m = 5, then n = 5 because n S 5. Otherwise n ≤ m and 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 = 5 and 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.
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RELATIONS AND PARTITIONS
81
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 linear order 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 element under the usual ≤ ordering. Let k be this element. Then k V n for every odd natural number in B and k V n for every even natural number in B, so k is the smallest 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 n in 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, then every nonempty subset of A has a smallest element, by definition. If every nonempty 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 a linear ordering that is a well ordering. (b) Suppose R is a well ordering of A. Then every nonempty subset B of A contains a smallest element b. This smallest element b is unique by antisymmetry. Also b is R-related to every element of B because for every x ∈ 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 element that 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 is reflexive on A. ii. Suppose x R y and y R x. Since both x and y are R-related to every element of {x, y}, x = y must be the unique such element. Therefore R is antisymmetric. iii. Suppose x R y and y R z for some x, y, z ∈ A. The set {x, y, z} has a unique element that is R-related to each of its elements. If this element is y then y R x so x = y and x R z. If the unique element is z, then z R y, so y = z and z R x, so y = z = x. Thus x R z. Otherwise, the unique 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 linearly ordered because A is linearly ordered. It only remains to show that every nonempty subset of B has a least element. Suppose C ⊆ B and C = ∅. Since C ⊆ B ⊆ A, we have C ⊆ A and so C has a least element. Therefore B is well-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, so by transitivity, x R b Therefore x ∈ Sb . (b) Suppose a, b ∈ A and Sa ⊆ Sb . By the reflexive property, a R a, so a ∈ 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, so Sa ⊆ Sb . Thus Sa is a predecessor of Sb . Now suppose there exists Sc ∈F such that Sa ⊆ Sc ⊆ Sb . Then a R c and c R b, so a = c or c = b. Thus Sa = Sc or Sc = Sb . Therefore Sa is the immediate predecessor of Sb .
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RELATIONS AND PARTITIONS
82
(d) Suppose B ⊆ A and x is the least upper bound for B. Then b R x for all b ∈ 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. Therefore Sx ⊆ 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 second sentence 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 that if sup(B) exists, then u = sup(B). A correct proof would show that u = sup(B) by showing u has the two supremum properties (u is an upper bound and u R v for all other upper bounds v). (c) F. The claim is false. Consider A = (0, 1) and B = [1, 2]. Then sup(A ∪ B) = 2 = 1 + 2 = sup(A) + sup(B). The error in the proof occurs in 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)
(d) Not possible.
3. (a) 4. (a) (b)
5. (a)
(c)
(e)
(f)
(b)
(c) Not possible.
(d)
and and
(b)
and
(c)
(d)
3
RELATIONS AND PARTITIONS
6.
83
(a) g, f, e, a, b g, f, e, a, d g, h, e, a, b g, h, e, a, d g, c, b, a, e g, c, d, a, e g, c, b, a, d g, c, d, a, b
(b) c, b, a, e, h, g, c c, b, a, e, f, g, c c, d, a, e, f, g, c c, d, a, e, h, g, c c, g, f, e, a, b, c c, g, f, e, a, d, c c, g, h, e, a, d, c c, g, h, e, a, b, c
(c) f, g, h, e, a, b, c, d
(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 vertices with the same degree. Since there are no multiple edges or loops in a simple graph 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 with degree 0, and one vertex w with degree n − 1. But then w is adjacent to every other 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 a path 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 u to 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. Then d(u, v) is the length of a shortest path from u to v, and d(u, v) is the length of a shortest path from v to w. Fix one such path from u to v and one such path from v to w. Traversing the path from u to v and then to w provides a walk from u to w of length d(u, v) + d(v, w). The shortest path from v to w cannot exceed this length. 12. Suppose d(u, r) = n ≥ 2. Then a shortest path from u to r that contains n − 1 vertices other than u and r, and n − 1 ≥ 1. Let w be one of these vertices.
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RELATIONS AND PARTITIONS
84
There can be no shorter path from u to w or from w to r than following the given path, for otherwise there would be a shorter path from u to r. Thus d(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, suppose w is reachable from v and x ∈ C(w). Then x is reachable from w. By symmetry w is reachable from x and by transitivity x is reachable from v. Therefore x ∈ C(v). Thus C(w) ⊆ C(v). Similarly C(v) ⊆ C(w), so C(v) = C(w). (b) Suppose that x is a vertex in both C(v) and C(w). Then x is reachable from v 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 Exercise 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 (c) {(x, y) ∈ A × B: y = x − 2}
(b) {(5, 4), (7, 4)} (d) {(x, y) ∈ A × B: y = 3}
3. The codomain is given in the definition of the function. Another possible codomain 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 − { π2 + 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) (b) (c) (d) (e) (f)
Domain= R − {3}, Range= R − {−1} Domain= R, Range= R Domain= (−π, ∞), Range= (0, ∞) Domain= (−∞, 5], Range= √ (0, ∞) Domain= [3, 5], Range= [ 2, 2] Domain= {−2}, Range= {0}
5. (a) If x = 1 then 2x + y = 2 + y. Since y = 1, 2, 3, or 4, 2 + y 5 only when y = 1. If x = 2 then 2x + y = 4 + y. Since y = 1, 2, 3, or 4, 4 + y 5 only when y = 3. If x = 3 then 2x + y = 6 + y. Since y = 1, 2, 3, or 4, 6 + y 5 only when y = 1. If x = 4 then 2x + y = 8 + y. Since y = 1, 2, 3, or 4, 8 + y 5 only when y = 3. In all cases, for each x ∈ A, there is a unique y such that
is prime and not is prime and not is prime and not is prime and not (x, y) ∈ R.
(b) If x = 1 then 3x + y = 3 + y. Since y = 1, 2, or 3, 3 + y is prime only when y = 2. If x = 2 then 3x + y = 6 + y. Since y = 1, 2, or 3, 6 + y is prime only when y = 1. If x = 3 then 3x + y = 9 + y. Since y = 1, 2, or 3, 9 + y is prime only when y = 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 that Z ⊆ Dom(R), let x ∈ Z. Choose y to be the integer 2 − x2 . Then x2 + y = x2 + (2 − x2 ) = 2. Thus x ∈ Dom(R). We conclude that Dom(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) 8. (a) A 9. 10.
(a) xn = −n (c) xn = 3 +
(d) c
(b) A
(c) ∅
n n+1
(a) f (3) = 3 in Z6 (c) 363 (or any multiple of 3).
(b) xn = (−1)n n (d) xn = n (modulo 3) (b) f (6) = 6 = 0 in Z6 (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 the perspective 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) ∈ ∅, then y = 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 the same way, g ⊆ f . 15. (a) Dom(S)
(b) Rng(S)
16. (a) Let x, y, z ∈ N. i. ii. iii. iv.
By definition of absolute value, d(x, y) = |x − y| ≥ 0 for all x, y ∈ N. d(x, y) = |x − y| = 0 iff x − y = 0 iff x = y. d(x, y) = |x − y| = |y − x| = d(y, x). By the triangle property of absolute value, |x − y| + |y − z| ≥ |x − z|. Thus d(x, y) + d(y, z) ≥ d(x, z).
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
(b)
= 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 as its image. (b) mn. Choose one of the n elements of A, and then one of the m elements of B as the image. 2 (c) m 2 n . Choose 2 elements from A for the first coordinates; each may be assigned any element of B as image. m m m r (d) r=0 (# of functions with r elements in the domain). r=0 r n . This is 18. (a)
i. Let x ∈ A. Since f (x) = f (x), xT x. Thus T is reflexive on A. ii. Let x, y ∈ A and suppose xT y. Then f (x) = f (y), so f (y) = f (x). Thus hT x and T is symmetric.
4
FUNCTIONS
88
iii. Let x, y, z ∈ A and suppose xT y and yT z. Then f (x) = f (y) and f (y) = f (z), so f (x) = f (z). Thus xT z 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 have different 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 the relation 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, then h ∪ 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(sin x) (g ◦ f )(x) = sin(tan x) (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) = (g ◦ f )(x) =
x2 +2 x2 +4 (x+1)2 (x+2)2
+1
(h) (f ◦ g)(x) = 3|x| + 2 (g ◦ f )(x) = |3x + 2| 2x + 1, if (i) (f ◦g)(x) = −2x, if −x + 1, if (7 − 2x)2 , (j) (f ◦g)(x) = (x + 1)2 ,
x ≤ −1 −1 < x < 0 x≥0 if x ≤ 2 if x > 2
(g◦f )(x) =
2x + 2, if x ≤ −2 −x − 1, if −2 < x ≤ 0 −2x, if x > 0
⎧ ⎨ 1 − 4x, if x ≤ − 21 (g◦f )(x) = 2x + 4, if − 12 < x < 3 ⎩ 2 x + 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)] Rng(g ◦ f ) = [−1, 1] (e) Dom(f ◦g) = {k, t, s}, Rng(f ◦g) = {r, l} f) = ∅
Dom(g ◦f ) = R−[ π2 +kπ],
mathrmDom(g ◦f ) = ∅, Rng(g ◦
4
FUNCTIONS
89
(f) Dom(f ◦g) = {1, 2, 4, 5}, Rng(f ◦g) = {2, 5} f ) = {3, 4, 7}
Dom(g◦f ) = {1, 3, 4, 5}, Rng(g◦
(g) Dom(f ◦ g) = R, Rng(f ◦ g) = [ 12 , 1) [1, ∞)
Dom(g ◦ f ) = R − {−2}, Rng(g ◦ f ) =
(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) (−∞, 1) (j) Dom(f ◦ g) = R, Rng(f ◦ g) = [9, ∞)
Dom(g ◦ f ) = R, Rng(g ◦ f ) =
Dom(g ◦ f ) = R, Rng(g ◦ f ) = [3, ∞)
2
3. (a) f (x) = x , g(x) = 3x + 7 f (x) = (x + 7)2 , g(x) = 3x √ (b) f (x) = √x, g(x) = 2x2 − 5 f (x) = 2x − 5, g(x) = x2 (c) f (x) = sin x, 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 (f (f (f
◦ g)([0]) = (f (g([0])) = f (2) = [2] = k([0]). ◦ g)([1]) = (f (g([1])) = f (0) = [0] = k([1]). ◦ g)([2]) = (f (g([2])) = f (2) = [2] = k([2]). ◦ 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−2 5
(b) Not a function. (c) f −1 (x) =
1−2x x−1
(d) Not a function. (e) f −1 (x) = −3 + ln x (f) f −1 (x) = (g) f −1 (x) =
1 1−x x−1 x
(h) f −1 (x) = 3 − x (i) f −1 (x) =
4x 3x+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} (b) {(x, y) ∈ R × R: y = 3}
{(x, y) ∈ R×R: y = x2 }
{(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 and f (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, and A ∩ 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 the other 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) = z and 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 ). Now if 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). Thus h|E = g|E . Now suppose h|E = g|E . We claim that h ∪ g = h ∪ (g|C−E ). To see this, first notice that by Theorem 4.2.5, g|E ∪ g|C−E is a function with domain C. 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 = h ∪ g|C−E .
(since g|E = h|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, so h ∪ g: A ∪ C → B ∪ D.
4
FUNCTIONS
14. (a) Function.
(c) Function.
91
(b) 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, so there 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 = b2 and 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. Therefore f (x) < f (y). (b) Suppose x, y ∈ R and x < y. Then −5x > −5y, so 2−5x > 2−5y. Therefore g(x) > g(y). (c) Suppose 0 ≤ x < y. Then x · x < y · x, and y · x < y · y, so x2 < y 2 . Therefore h(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 the y−1 last inequality by x + 3 and y + 3, we conclude x−1 x+3 < y+3 . Therefore f (x) < f (y). 17. (a) True. Suppose f (x) = ax + b where a > 0, and let c < d. The ac < ad and ac + 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. Suppose this is true, x, y ∈ I and x < y. Then g(x) > g(y) and g(x), g(y) ∈ I, so f (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 . 1 , if x < 0 (e) False. Consider f (x) = x −x, if x ≥ 0. 18. (a) We show that f1 + f2 is a function with domain R. First, f1 + f2 is by definition a relation. For all x ∈ R there is some u ∈ R such that (x, u) ∈ f1 because f1 : R → R and there is some v ∈ R such that (x, v) ∈ f2 because f2 : R → R. Then (x, u + v) ∈ f1 + f2 , so x ∈Dom(f1 + f2 ). It is clear from the definition of f1 + f2 that x ∈Dom(f1 + f2 ) implies x ∈ R, so Dom(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 relation by 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 clear from 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 2 (f + h)(x) = 3x − 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 Exercise 18, 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 ), so cf : R → R. 20. (a) A. (b) C. It is not necessary to refer to the idea of cancellation. The “proof” can be 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 / ). √ √ √ 3 (d) Onto R. Let y ∈ R. Then 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 Intermediate Value Theorem says that f attains every value between 1 and −1.
4
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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, so f (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 that w + 1 > 0 because w ≥ 0, and 2w + 2 ≤ 3w + 6, so 2 ≤ 3w+2 w+1 < 3. This ensures that x ∈ [2, 3). x (l) Onto [0, ∞). First, if x ∈ (1, ∞), then x − 1 > 0 and x > x − 1, so x−1 > 1. w Therefore Rng(f ) ⊆ (1, ∞). Let w ∈ (1, ∞). Then x = w−1 ∈ (1, ∞) and f (x) = w.
2. (a) One-to-one. Suppose f (x) = f (y). Then 21 x + 6 = 12 y + 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 = y 3 , 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 = y 2 + 1, so x2 = y 2 . Since x, y > 0, x = y. (k) One-to-one. Suppose x, z ∈ [2, 3) and f (x) = f (z). Then x = z.
x−2 3−x
=
z−2 3−z ,
so
(l) One-to-one. Suppose x, y ∈ (1, ∞) and f (x) = f (y). Then x = y.
x x−1
=
y y−1 ,
so
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 and f (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 onto Z. (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. onto
onto
5. Suppose f : A −→ B and g: B −→ C. Let c ∈ C. Then there is b ∈ B such that g(b) = c, since g is onto B. Also there is an a ∈ A such that f (a) = b since f is onto A. Thus (g ◦ f )(a) = g(f (a)) = g(b) = c, which proves that g ◦ f is onto C. 1-1 6. Suppose f : A → B, g: B → C and g ◦f : A −→ C. Suppose also that x, y ∈ A and f (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. 1-1 7. (a) Suppose f : A −→ 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
onto
onto
(b) Suppose h: A −→ C, g: B −→ 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)} and g = {(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)}. 1-1 9. (a) Let g(x) = 2 − x and h(x) = x1 . Then g|(−∞,1] : (−∞, 1] −→ [1, ∞) and 1-1 1-1 h|(1,∞) (1, ∞) −→ (0, 1). Since f = g|(−∞,1] ∪ h|(1,∞) , f : R −→ (0, ∞) by Theorem 4.3.5(c). Finally, since there is no real number x such that f (x) = −1, −1 ∈ Rng(a). Therefore f is not onto R. 1-1 (b) Let g(x) = x+4, h(x) = −x and i(x) = x−4. Then g|(−∞,−2] : (−∞, −2] −→ 1-1 1-1 (−∞, 2], h|(−2,2) : (−2, 2) −→ (−2, 2) and i|[2,∞) : [2, ∞) −→ [−2, ∞). Since onto
f = g|(−∞,−2] ∪ h|(−2,2) ∪ i|[2,∞) , f : R −→ 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. For w = 1, choose x = 4w−2 w−1 . Then f (x) = w. w−2 To show that f is one-to-one, suppose f (x) = f (w). Then x−2 x−4 = w−4 for x = 4 and w = 4. Therefore xw − 2w − 4x + 8 = xw − 2x − 4w + 8, from which we derive x = w. We must also consider whether f (x) might be 1 even 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 onto R. 10. Suppose f : R → R and f is increasing. (The proof for decreasing is similar.) Let a, b ∈ R be such that a = b. Without loss of generality, say a < b. Then, since f 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: If f (x) = [1], then [2x] = [1], so 2x ≡8 1. But then 8 divides the odd number 2x − 1, which is impossible. To show that f is an injection, suppose f (x) = f (z). Then 2x =8 2z, so 2x ≡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 50 − n, if n < 50 x(n) = n, 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 n, if n < 50 x(n) = n − 1, if n ≥ 50 13. (a)
n! (n−m)! .
Choosing an image for each of the elements of A, one at a time, we n! choices in all. get n(n − 1) · · · (n − m + 1) = (n−m)!
(b) n! For each element of A, we choose an image from the elements of B that have 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 ways, 2 and then assign each of the remaining n − 1 elements of B as the image of exactly one of the n − 1 remaining n+1 elements of A in (n − 1)! ways. Thus there are n n+1 (n − 1)! = n! functions from A onto B. 2 2 (g) n! For each element of A, choose as image an element of A that has not already been chosen. 14. (a) F. To prove that f is onto we must show that R ⊆Rng(f ). The proof shows only 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
−y −x 1. (a) Let x, y ∈ (2, ∞). Suppose f (x) = f (y). Then x−2 = 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. 2z . From z < −1, we have z + 1 < 0. Thus, Let z ∈ (−∞, −1). Choose t = z+1 z z+1 2z since z < z + 1, z+1 > z+1 = 1 and t = z+1 > 2 · 1 = 2, so so t ∈ (2, ∞).
Then f (t) =
−t t−2
=
2z − z+1 2z z+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 thus z 2z t + 4 < 0. Thus, since z < z + 1, z+1 > z+1 z+1 and t = z+1 > 2 · 1 = 2, so t ∈ (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 . Thus x + 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 as k = 2r q, with nonnegative integer r and odd natural number q. Then q = 2t − 1 for some natural number t. Let s = r + 1. Then (s, t) ∈ NxN and G(s, t) = 2s+2 (2t − 1) = 2r+3 (q) = 8(2r q) = 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. z−3 Let z ∈ (3, ∞). Choose t = ln z−3 2 . Since z > 3, 2 is a positive number ( z−3 ) t and so t ∈ R. Then k(t) = 2e + 3 = 2e 2 + 3 = 2( z−3 2 ) + 3 = z. Thus k is 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 some n, 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 some x, 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 and f (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 some x, 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 and f (−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) for some 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. Then f (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 = y1 iffy = x1 . Therefore h = x1 . To verify that this formula is correct, suppose x ∈ (0, ∞). Then (h ◦ f )(x) = h(f (x)) = h( x1 ) = 11 = x. Since the composite is the x
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
(y, x) ∈ g 4y x= x+2 xy + 2x = 4y y(x − 4) = −2x 2x y= , for x < 4. 4−x
iff iff iff iff iff
Therefore h(x) = x > −2. Then
2x 4−x .
To verify that this formula is correct, suppose
(h ◦ g)(x)
= h(g(x))
4x = h x+2 = =
4x 2 x+2
4−
4x x+2 8x x+2 4(x+2)−4x x+2
= x. Since the composite is the identity function on the domain of g, we conclude that 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 this formula 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 function on the domain of h we conclude that g = h−1 . (d) Let h be the inverse of G. Then (x, y) ∈ h
iff iff iff iff iff
(y, x) ∈ G 5(y − 1) x= y−3 xy − 3x = 5y − 5 xy − 5y = 3x − 5 3x − 5 . y= x−5
Therefore h = 3x−5 x−5 . To verify that this formula is correct, suppose x ∈ (3, ∞). Then (h ◦ f )(x)
= h(f (x))
5(x − 1) = h x−3 5(x−1) 3 x−3 − 5 = 5(x−1) x−3 − 5 =
15x−15−5x+15 x−3 5x−5−5x+15 x−3
=
10x = x. 10
4
FUNCTIONS
98
Since the composite is the identity function on the domain of G, we conclude that h = G−1 . 4. Suppose F : A → B and F −1 is a function. Then by the first part of Theorem 4.4.2 and the fact that (F −1 )−1 = F is a function, F −1 is one-to-one. 1-1 5. (a) Assume that F : A −→ B and G: B → A. Suppose that G = F −1 . Then by onto
part (a) of Theorem 4.4, G ◦ F = IA and F ◦ G = IB . Therefore G ◦ F = IA or F ◦ G = IB . 1-1 1-1 Now suppose G ◦ F = IA or F ◦ G = IB . Since, F : A −→ B, F −1 : B −→ A onto
onto
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 , then G = IA ◦ G = (F −1 ◦ F ) ◦ G = F −1 ◦ (F ◦ G) = F −1 ◦ IB = F −1 . In either case, 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 ∈ A and 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
1-1 1-1 1-1 1-1 8. Suppose f : A −→ B, g: B −→ C, and h: C −→ D. Then g ◦ f : A −→ by onto onto onto onto 1-1 Theorem 4.4.1. and, again by Theorem 4.4.1, h ◦ g ◦ f : A −→ D. Therefore, onto 1-1 (h ◦ g ◦ f )−1 : D −→ A. By Theorem 3.1.3(d), (h ◦ g ◦ f )−1 = ((h ◦ g) ◦ f )−1 = onto
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 necessarily true. (b) A. (c) F. The claim is false. The “proof” assumes that r−1 ◦ s−1 = (r ◦ s)−1 , which is 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] (c) {0} (e) [−3, −2] ∪ [2, 3]
(b) [1,√ 2] ∪√ [5, 17] (d) [− 2, 2 ] (f) [−5, −4] ∪ [−2, 2] ∪ [4, 5]
3.
(a) {−5, −3, −1, 1, 3} (c) R (e) [−7, −1)
(b) {z ∈ Z: z < 0 and z is odd} (d) [−2. − 12 ] (f) [−7, −5]
4.
(a) (2, ∞) √ √ √ √ (c) [2 − 3, 3−2 5 ) ∪ ( 3+2 5 , 2 + 3] (e) (−∞, −2] ∪ [2, ∞)
(b) [2, 5.2] (d) ∅ (f) [− 10 3 , −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 ∈ C and b = f (a) or b ∈ D and b = f (a). Thus b ∈ f (C) or b ∈ f (D), so b ∈ 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 iff iff
f (a) ∈ E ∪ F f (a) ∈ E or f (a) ∈ F 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 . Thus a ∈ Dα for every α ∈ Δ. Sinceb = f (a), b ∈f (Dα ) for every α ∈ Δ. Thus b ∈ 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 is an α ∈ δ such that b ∈ f (Dα ). Thus b ∈ α∈δ f (Dα ). On the other hand, suppose b ∈ α∈δ f (Dα ). Thenb ∈ 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 ⎝
⎞ Eβ ⎠ iff
f (a) ∈
β∈Γ
Eβ
β∈Γ
iff iff iff
for all β ∈ Γ, f (a) ∈ Eβ for all β ∈ Γ, a ∈ f −1 (Eβ ) a∈ f −1 (Eβ ). β∈Γ
(d) a ∈ f −1 (
Eβ )
iff
β∈Γ
f (a) ∈
Eβ
β∈Γ
iff iff iff
for some β ∈ Γ, f (a) ∈ Eβ for some β ∈ Γ, a ∈ f −1 (Eβ ) a∈ f −1 (Eβ ). β∈Γ
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. Therefore f (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 and f (a) ∈ / E. Therefore, a ∈ A and a ∈ / f −1 (E). (This is the contrapositive of −1 the statement a ∈ f (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)). Thus there 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)). Therefore E ⊆ 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 − D such that f (a) = b, so obviously b ∈ B. If b ∈ f (D) as well, then by our supposition, 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 suppose / f (A − D). d ∈ f −1 (f (D)). Then f (d) ∈ f (D), so by our assumption f (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 Exercise 10(a). (b) Suppose t ∈ f (X) − f (Y ). Then t ∈ f (X), so there exists x ∈ X such that f (x) = t. We note x ∈ / Y since t = f (x) ∈ / f (Y ). Thus x ∈ X −Y . Therefore t = 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 iff a ∈ f −1 (U − V ). 12. Let X, Y ⊆ A. Then f (X ∩ Y ) ⊆ f (X) ∩ f (Y ) by Theorem 4.5.1(a). Suppose b ∈ f (X) ∩ f (Y ). Then f (x) = b = f (y) for some x ∈ X and y ∈ Y , and x must equal 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). 1-1 13. Suppose f : A −→ B and X ⊆ A. Then f (A) − f (X) ⊆ f (A − X) by Exercise 11(b). Suppose b ∈ f (A − X). Then b = f (a) for some a ∈ A − X . Now if b ∈ 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 is one-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. onto
(b) If f is onto B, then f : P(A) −→ 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 is an 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 same element 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. For every α ∈ Δ there is an x ∈ Dα , but this does not imply that there is an x such that x ∈ Dα for all α.
4
FUNCTIONS
4.6 1.
102
Sequences (a)
5 6 2 3 4 5 , 7 , 9 , 11 , 13
(b) 1, 0, −1, 0, 1
(d)
1 3 7 15 31 2 , 4 , 8 , 16 , 32
(e)
2.
(a) Converges to 12 . (c) Converges to 0. (e) Does not converge.
3.
(a) Does not exist. (d) Does not exist. (g) e2 (j) Does not exist. (m) 0
(c) 1, 12 , 61 ,
1 1 24 , 120
1 1 3 3 15 2, 2, 4, 2, 4
(b) Does not converge. (d) Converges to 1. (b) (e) (h) (k) (n)
0 0 1 e
Does not exist. 1
(c) 45 (f) 35 (i) 0 (1) 0
4. (a) The terms grow larger without bound so the sequence diverges. We prove that the limit is K. Let > 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 if n > M , then xn = 2. 5. (a) The terms grow larger without bound so the sequence diverges. Assume the limit is L. Let = 1 and let N ∈ N. Suppose n is greater than both N and log2 (|L|+1). Then 2n > |L| + 1 so |2n − L| ≥ 2n − |L| > 1 = . (b) The sequence converges to 1. Let > 0. Choose N so that N > 1 . Suppose n > N . Then |xn − 1| = n1 < N1 < . (c) The terms grow larger without bound so the sequence diverges. Assume limn→∞ n2 = L. Let = 1 and choose N ∈ N. Suppose n is greater than both N and L + 1. Then n2 ≥ n > L + 1, so |n2 − L| = n2 − L > 1. (d) The terms are alternately near 12 and − 12 so the sequence diverges. Assume n n the limit is L. Note that for all n ∈ N, |xn | = 2n+1 > 3n = 13 . Let 1 = 3 , and let N ∈ N. If L ≥ 0, then for odd n > N , |xn − L| = n 1 | (−1)n 2n +1 − L| = 2n+1 + L ≥ 3 = . If L < 0, then for even n > N , n n |xn − L| = | 2n +1 − L| = 2n+1 + |L| ≥ 31 = . (e) The sequence converges to 0. Let > 0. Choose N so that N > 1 . Suppose n| n > N . Then |xn − 0| = | cos ≤ n1 < N1 < . Therefore limn→∞ xn = 0. n 1 (f) The sequence converges to 0. Let > 0. Choose N so that N > (2) 2. √ √ 1 1 Suppose n > N . Then |xn − 0| = | n + 1 − n| = √n+1+√n < 2√n < 1 √ 2 N
< .
21 (g) The sequence converges to −7. Let > 0. Choose N so that N > . 2 7(1−n ) 21 Suppose n > N . Then n > 21 , so n2 < . Then |xn − (−7) = | n2 +2 − (−7)| =
21 n2 +2
<
21 n2 +2
<
21 n2
< . Therefore, limn→∞
7(1−n2 ) n2 +2
= −7.
(h) The sequence converges to 0. Choose N such that N > log2 (6/). Suppose n > N N . Then n > log2 (6/), so 2π > 6/. Therefore |6/2π −0| = 6/2π < . (i) The sequence converges to 0. Choose N so that N > 5000/varepsilon. Suppose n > N . Then n! > n > 5000/. Therefore |5000/n! − 0| = 5000/n! < .
4
FUNCTIONS
103
(j) Each term is either −1, 0 or 1, so the sequence diverges. Assume the limit is L. Let = 12 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 + 3π ) = −1. 2 Thus |xn − (L)| ≥ 1 > . = If L < 0, then for n of the form 4k + 1, k ∈ N, xn = sin( 4k+1)π 2 sin(2kπ + π2 ) − 1. Thus, |xn − (L)| ≥ 1 > . (k) The sequence converges to 0. Let > 0. Choose N so that kN > 1 . Suppose n > N . An easy induction argument shows n! ≤ nn−1 , so n−1 |xn − 0| = nn!n ≤ nnn = n1 < N1 < . (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 ln M . Then ln( n2 ) > 1, so ln( n2 )n = n ln( n2 ) > n > ln M , so ( n2 )n > M . Thus x is an unbounded sequence, so x diverges. 6. (a) Let > 0. Then 2 > 0. Since xn → L, there exists N1 ∈ N such that if n > N1 , then |xn − L| < 2 . Likewise, there exists N2 ∈ N such that if n > N2 , then |yn − M | < 2 . Let N3 =max{N1 , N2 } and assume n > N3 . Then |(xn + yn ) − (L + M )| = |(xn − L) + (yn − M )| ≤ |xn − L| + |yn − M | < 2 + 2 = . Therefore, xn + yn → L + M . (b) Let > 0. Choose N1 so that n > N1 implies |xn − L| < 2 and pick N2 so that n > N2 implies |yn − M | < 2 . 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 | ≤ 2 + 2 = . Note: If part (c) has been proved, then part (b) follows by applying part (a) to the sequences xn and −yn . (c) Let > 0. Choose N so that n > N implies |xn − L| < |r| . Now suppose n > N . Then | − xn − (−L)| = | − xn + L| = |xn − L| < . Note: This is a special case of part (d).
(d) Let > 0. Choose N so that N < n implies |xn − L| < n > N . Then |rxn − rL| = |r||xn − L| < .
|r| .
Now suppose
(e) Let > 0. Choose B so that |yn | ≤ B for all n ∈ N. Pick N1 so that for all n > N2 . for all n > N1 , and N2 so that |yn −M | < 2|L| |xn −L| < 2B Let N =max{N1 , N2 }. Suppose n > N . Then |xn yn −LM | = |xn yn −Lyn + Lyn −LM | ≤ |xn −L||yn |+|L||yn −M | ≤ |xn −L|B+|L||yn −M | < 2 + 2 = . (f) Let > 0. Choose N so that n > N implies |xn − L| < . Suppose n > N . Then ||xn | − |L|| ≤ |xn − L| < . 7. (a) Suppose xn → L = 0. Then |xn | → |L| by Exercise 6(f). Thus there exists −|L| < |xn | − |L|, so N such that n > N implies ||xn | − |L|| < |L| 2 2 . Then |L| 2 < |xn | whenever n > N . (b) Suppose xn → L = 0, xn = 0 for all n, and yn → M . By part (a) there exists N1 so that |xn | > |L| 2 for all n > N1 . Let > 0. Then there exists L2 N2 so that n > N2 implies |xn − L| < 4|M | and N3 so that n > N3 implies |yn − M | < |L| 4 . Let N =max{N1 , N2 , N3 }. Suppose n > N . Then yn 1 M = − |Lyn − M xn | xn L |L||xn | 1 ≤ (|Lyn − LM | + |LM − M xn |) |L||xn |
4
FUNCTIONS
104
2 (|L||yn − M | + |M ||L − xn |) L2
L2 2 L2 + < L2 4 4 = .
<
8. (a) x2n =
2 2n ,
x2n+1 = 0
(b) Suppose xn → L. Let > 0. Choose N so that n > N implies |xn − L| < . Then yn = xn+N determines a subsequence such that |yn − L| < for all n ∈ N. (c) Suppose xn → L and (xf (n) ) is a subsequence. Let > 0 and choose N so that n > N implies |xn − L| < . Then f −1 (N ) ∈ N and n > f −1 (N ) implies f (n) > N , so |xf (n) − L| < . (d) Suppose (xf (n) ) and (xg(n) ) are subsequences of (xn ) such that xf (n) → M and xg(n) → L where L = M . Assume further that xn → K. At least one of M or L is not equal to K. Therefore (xn ) has a subsequence that does not converge to K, violating part (c). Thus (xn ) diverges. 9. (a) x1 = 10, x2 = 5, x3 = 52 , x4 = 54 , x5 = 58 , x6 = to 0.
5 16 .
The sequence converges
(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 inequality involving |xn − ( L2 )| + |yn − ( L2 )|is reversed. (b) A. The proof uses Exercise 6(b).
5
Cardinality
5.1
Equivalent Sets; Finite Sets
1-1 1. Reflexive: For any set A, IA : A −→ A, so A ≈ A. onto
Symmetric: For sets A and B, suppose A ≈ B. Then there is a bijection 1-1 1-1 f : A −→ B. Then f −1 : B −→ A, so B ≈ A. onto
onto
Transitive: Let A, B and C be sets such that A ≈ B and B ≈ C. Then there 1-1 1-1 1-1 are bijections f : A −→ B and g: B −→ C, so g ◦ f : A −→ C. Thus A ≈ C. onto
2.
(a) finite (d) infinite (g) finite (j) infinite (m) finite (p) finite
onto
(b) (e) (h) (k) (n)
infinite infinite finite infinite finite
onto
(c) finite (f) finite (i) finite (l) infinite (o) finite
d−c d−c 3. Suppose x, y ∈ (a, b) and f (x) = f (y). Then ( b−a )(x − a) = ( b−a )(y − a), so x = y. Therefore f is one-to-one. {An alternate proof: f is linear function with positive slope, so f is increasing (by Exercise 17(a) of Section 4.2). Therefore f is one-to-one, by Exercise 10 of Section 4.3.
Now suppose y ∈ (c, d). Then
b−a d−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 ∈ B since 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 ) = ∅ implies Dom(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 is finite. This is a contradiction. Thus B is infinite. 7. A ∪ B = (A − B) ∪ (A ∩ B) ∪ (B − A), a union of disjoint sets each of which is 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. 1 8. If A1 is finite, then i=1 Ai = A1 is finite. n Suppose that i=1 Ai is finite whenever A1 , A2 , . . ., An are finite sets for nsome n ∈ N, and consider the finite sets A1 , A2 , . . ., An+1 . By our supposition, i=1 Ai n+1 n is finite, so i=1 Ai = 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) for any (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, so A × 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 B A be the set of all funtions from A to B. Because a function from A to B is a subset of A × B, B A is a subset of P(A × B), the power set of A × B. Since A and B are finite, A × B is finite, and, therefore (by Theorem 2.1.4) P(A × B) is finite. Since B A is a subset of a finite set, B A is finite. Alternate Proof. Assume A and B are finite. If A or B is empty, then the only function from A to B is the empty function; thus B A = {∅} is 1-1 1-1 finite. Suppose g: A −→ Nn and h: B −→ Nm . Define F : B A → Nmn as: onto onto n F (f ) = 1+ i=1 mi−1 (h−1 (f (g −1 (i)))−1) for all f ∈ B A . Then F is a bijection, so B A is finite. 11. (a) Impossible (c) Impossible
(b) For each i ∈ N, let Ai = {5, 6}. (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) is finite by Theorem 5.1.7(a), which contradicts our assumption. Therefore B − A is infinite. 13. Let x ∈ Nr , and define f : Nr − {x} → Nr−1 by: n, if n < x f (n) = n − 1, if n > x for all n ∈ Nr − {x}. Then f is a bijection, so Nr − {x} ≈ Nr−1 . 1-1 14. Let B be a proper subset of a finite set A. Suppose f : A −→ Nk . Then onto 1-1 f |B : B −→ f (B), where f (B) is a proper subset of Nk . Now suppose that B ≈ A, onto 1-1 1-1 and let g: B −→ A. Then f |B ◦ g −1 ◦ f −1 : Nk −→ f (B), so Nk is equivalent to onto onto one of its proper subsets, but this is impossible. 1-1 1-1 1-1 15. Suppose f : S −→ Nm and g: S −→ Nn . Then f ◦ g −1 : Nn −→ Nm and onto onto onto 1-1 g ◦ f −1 : Nm −→ Nn , so by the pigeonhole principle, n ≤ m and m ≤ n. Therefore onto n = 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 is infinite. (b) True. Proof. Suppose A is a set, p ∈ / A and A ≈ A ∪ {p}. If A is finite and has cardinality k, then by Lemma 5.1.4, A ∪ {p} is finite and has cardinality k + 1. Thus A ≈ A ∪ {p} is false. This is a contradiction. Therefore A is infinite. 17. (a) If n = 1, then r ∈ N and r < n is impossible. If n = 2 and r < n, then r = 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. Suppose onto f : Nr −→ Nn+1 . Then (Nn+1 − {f (r)}) ⊆Rng(f |Nr−1 ) and by Lemma 5.1.8, 1-1 onto there is a g: (Nn+1 − {f (r)}) −→ Nn , so g ◦ f |Nr−1 : Nr −→ Nn . This onto contradicts our induction hypothesis. (c) By the PMI, for every n, if r < n there is no function from Nr onto Nn . 1-1 18. (a) Suppose f is one-to-one. Then f : A −→ f (A), so f (A) ≈ A ≈ B and f (A) onto
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 are onto onto finite and equivalent, there are functions h: A −→ Nn and g: B −→ Nn for onto some n ∈ N. Thus g ◦ f ◦ h−1 : Nn −→ 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 = onto g ◦ f ◦ h−1 − {(h(u)), g(f (u))}. Then G: Nn − {g(f (u))} −→ Nn . By Lemma onto 5.1.8, there is a bijection F : Nn−1 → Nn − {y}, so G ◦ F : Nn−1 −→ Nn . This contradicts 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 some n ∈ N ∪ {0}, that any function whose domain has cardinality n has finite range. 1-1 Let f : A → B, A = n + 1, and g: A −→ Nn+1 . Pick (x, y) ∈ f . By Lemma onto 1-1 5.1.8 there is a bijection h: Nn+1 − {g(x)} → Nn , so h ◦ g|A−{x} : A − {x} −→ onto
Nn , so A − {x} = n. By our hypothesis, f |A−{x} has a finite range. Since Rng(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 f is 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 oneto-one function from Nm to Nn by Theorem 4.3.4. But m > n so the conclusion that 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 dates in a year. Assume A = 400 and B = 365. Then N400 ≈ A and B ≈ N65 so there exist bijections f : N400 → A and g: B → N365 . If no two residents have the same birthday, then there is a one-to-one function h: A → B. But then g ◦ h ◦ f : N400 → N365 is one-to-one. This contradicts the Pigeonhole Principle. Therefore, every function from A to B is not one-to-one. Thus, no matter what the birthday assignments may 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 possible sums of the elements of subsets of S, and 210 − 1 = 1,023 nonempty subsets of S, so a least two subsets of S have identical sums for their elements. If the two subsets are not disjoint, delete the common elements from both sets. The result is two disjoint subsets of S whose elements have the same sums.
5
CARDINALITY
108
22. (a) F. f ∪h is not a one-to-one correspondence between A∪B and Nm+n . Some elements of Nm+n (such as 1) have preimages in both A and B, and m + n has no preimages. Furthermore, the “proof” neglects the special case that A 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 but A is infinite. If we assume that B = ∅, the proof is correct. The proof fails when 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 an infinite family of sets.
5.2
Infinite Sets
1. Assume that A is infinite and A ≈ B. Suppose B is finite. This contradicts Theorem 5.1.3. Therefore B is infinite. 2. (a) Define f : N → A by f (n) n1 , for each n ∈ N. The function f is one-to-one 1 because if m, n ∈ N and f (m) = f (n), then m = n1 and so m = n. The function f is onto A because if a ∈ A, then a = k1 for some k ∈ N and f (k) = k1 = 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 + 15 and so m = n. The function f is onto N − N15 because if a ∈ N − N15 , then a > 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). Define f : (0, 0.0005) → (0, 0.001) by f (x) = 2x. The function f is one-to-one because if x, y ∈ (0, 0.0005) and f (x) = f (y), then 2x = 2y and so x = y. The function f is onto (0, 0.001) because if a ∈ (0, 0.001), then a/2 ∈ (0, 0.0005) and f (a/2) = 2(a/2) = a. Therefore, (0, 0.001) is equivalent to a proper subset and is infinite. (d) Note that the interval (0, 1) is a proper subset of (0, ∞). Define f : (0, 1) → x (0, ∞) by f (x) = 1−x . The function f is one-to-one because if x, y ∈ (0, 1) y x . Then x − xy = y − xy and so x = y. and f (x) = f (y), then 1−x = 1−y a is a The function f is onto (0, ∞) because if a ∈ (0, ∞), then t = a+1 a )= real number between 0 and 1 and f (t) = f ( a+1
a a+1
a 1− a+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 f is one-to-one, suppose f (x) = f (y). Then 2x − 1 = 2y − 1, which implies x = 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 x3 = y3 implies x = y. Also, f is onto N because for any n ∈ N, 3n ∈ 3N and f (3n) = n. (c) Define f : 3Z → Z by f (x) = x3 . Then f is one-to-one because x3 = 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-one because x − 6 = y − 6 implies x = y. Also, f is onto N because for any n ∈ 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-one because −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} and f (−n − 12) = −(−n − 12) − 12 = n. x, if x < 5 (f) Define f : N − {5, 6} → N as f (x) = x − 2, if x > 6. Then f is one-to-one. Suppose x, y ∈ N − {5, 6} and f (x) = f (y). If f (x) < 5, then x = f (x) = f (y) = y. If f (x) ≥ 5, then x − 2 = y − 2, so x = y. Also, f is onto. If n < 5, then n ∈ N − {5, 6} and f (n) = n and for any 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 ) ∈ G and π1 (x1 , y1 ) = π1 (x2 , y2 ), then x1 = x2 , so y1 = x11 = x12 = y2 , and thus (x1 , y1 ) = (x2 , y2 ). Also, π1 is onto. Let n ∈ N. Then (n, n1 ) ∈ G and π1 (n, n1 ) = n. (h) Define f : {x ∈ Z: x ≡ 1(mod 5)} → Z by f (x) = x−1 5 . Then f is one-to-one y−1 and onto Z. Suppose x, y ∈ 1/ ≡5 and f (x) = f (y). Then x−1 5 = 5 , so 5z+1−1 = z. x = y. If z ∈ Z, then 5z + 1 ≡ 1(mod 5), and f (5z + 1) = 5 4. (a) Define f : (1, ∞) → (0, 1) by f (x) = x1 . Then f is one-to-one because f is decreasing. Also, f is onto (0, 1) because for any x ∈ (0, 1), x1 ∈ (1, ∞) and f ( x1 ) = 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). 5 + 2x, if x ∈ (0, 12 ) (d) Define f : (0, 1) → [1, 2) ∪ (5, 6) by f (x) = 2x, if x ∈ [ 12 , 1). Then f is onto. Let x ∈ [1, 2) ∪ (5, 6). If x ∈ [1, 2), then 21 x ∈ [ 12 , 1) 1 x−5 and f ( 12 x) = 2( 12 x) = x. If x ∈ (5, 6), then x−5 2 ∈ (0, 2 ) and f ( 2 ) = x−5 5 + 2( 2 ) = x. Also, f is one-to-one. Let x, y ∈ (0, 1) and suppose f (x) = f (y). If f (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. 3 + 6x, if x ∈ (0, 12 ) (e) Define f : (0, 1) → (3, 6) ∪ [10, 20) by f (x) = 20x, if x ∈ [ 12 , 1). 1 Then f is onto. Let x ∈ (3, 6) ∪ [10, 20). If x ∈ (3, 6), then x−3 6 ∈ (0, 2 ) ⊆ x−3 x−3 x 1 (), 1) and f ( 6 ) = 3 + 6( 6 ) = x. If x ∈ [10, 20), then 20 ∈ [ 2 , 1) and x x ) = 20( 20 ) = x. f ( 20 Also, f is one-to-one. Let x, y ∈ (0, 1) and suppose f (x) = f (y). If f (x) ∈ (3, 6), then 3 + 6x = f (x) = f (y) = 3 + 6y, so x = y. If f (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 ⎧ if x ∈ 0 < x ≤ 13 ⎨ 3x f (x) = 3x + 1 if 13 < x ≤ 23 . ⎩ 3x + 2 if 23 < x < 1 The function of f is one-to-one because if , 7 ∈ (0, 1) and f (x) = f (y), then if 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, then if a ∈ (0, 1] then a3 ∈ (0, 13 ] and f ( a3 ) = 3( a3 ) = a; 1 2 a−1 a−1 if a ∈ (2, 3] then a−1 3 ∈ ( 3 , 3 ] and f ( 3 ) = 3( 3 ) + 1 = (a − 1) + 1 = a; 2 a−2 a−2 if a ∈ (4, 5] then a−2 3 ∈ ( 3 , 1) and f ( 3 ) = 3( 3 ) + 2 = (a − 2) + 2 = a; x, if x ∈ /N (g) Define f : R − {0} → R as f (x) = x − 1, if x ∈ N. Then f is onto. Let x ∈ R. If x ∈ R − (N ∪ {0}), then x ∈ R − {0} and f (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). If f (x) ∈ R − (N ∪ {0}), then x = f (x) = f (y) = y. If f (x) ∈ N ∪ {0}, then x + 1 = f (x) = f (y) = y + 1, so x = y. 5. (a) False
(b) True
(c) False
(d) True
(e) True
(f) False
20, if x = 1 2, if x = 10 2x, if x = 1, x = 10. ⎧ 16, if n = 1 ⎪ ⎪ ⎪ 12, if n = 2 ⎪ ⎪ ⎨ 2, if n = 3 (b) Define h: N → E+ by h(n) = 4, if n = 8 ⎪ ⎪ ⎪ ⎪ 6, if n = 6 ⎪ ⎩ 2n, if n = 1, 2, 3, 6, or 8.
6. (a) Define g: N → E+ by g(x) =
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) = (0, 1).
−x2 +x− 12 x2 (x−1)2
< 0 on (0, 1), so f is decreasing and therefore one-to-one on
Also, limx→1− f (x) = −∞, limx→0+ f (x) = ∞, and f is continuous on (0, 1), so by the Intermediate Value Theorem of Calculus, f attains every real value on the 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). Thus C ≈ R × R ≈ (0, 1). 12. (a) F. W is certainly an infinite subset of N, and D+ is denumerable, but this “proof” claims without justification that every infinite subset of N is denumerable. To show W is denumerable, we need to use another theorem or 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 must be 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 1.
Countable Sets 9 1 x
2. Begin by writing the set K = { 23y : x, y ∈ N} as follows: 21 31
21 32
21 33
21 34
···
2
2 31
2
2 32
2
2 33
2
2 34
···
3
3
3
3
2 31
2 32
2 33
2 34
···
.. .
.. .
.. .
.. .
..
.
Now all the elements of K can be put into a correspondence with N by listing 1 them along the diagonal as in Figure 5.3.1. There is only one number 231 on the 1 first diagonal so 231 corresponds to 1. There are 2 elements of K on the second diagonal, and they correspond to 2 and 3. This correspondence is more natural than the one used to explain Theorem 5.3.1 because it is not necessary to omit any 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 ∪ B is denumerable. Now let B be a set with cardinality n + 1. Let x ∈ B. Then B − {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 − 1 is odd, and h(2n − 1) = f (n) = x. If x ∈ B, then x = g(n) for some n ∈ N. Now 2n 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+1 2 ) = h(x) = y+1 x+1 because f is one-to-one, so x = y. If h(x) ∈ B ). Then = h(y) = f ( y+1 2 2 2 then g( x2 ) = h(x) = h(y) = f ( y2 ). Then x2 = y2 , because g is one-to-one, so x = y. Therefore h is one-to-one. 1-1 5. (a) Let f : N −→ A. Define g: N → A − {x} as: onto f (m), if m < f −1 (x) g(m) = 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 n elements, 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 n elements, then A − B is denumerable.
5
CARDINALITY
6. Let n = 1. Then there is one denumerable set A1 in the family, and which is denumerable.
112 1 i=1
Ai = A1
Suppose that for some k k ∈ N, if there are k pairwise disjoint denumerable sets in the family, then i=1 Ai is denumerable. Consider a family of k + 1 pairwise k disjoint denumerable sets. Since the sets are pairwise disjoint, i=1 Ai and Ak+1 k are disjoint. The set i=1 Ai is denumerable by hypothesis of induction and Ak+1 k k+1 is denumerable, so by Theorem 5.3.6, their union ( i=1 Ai ) ∪ Ak+1 = i=1 Ai is denumerable. By the PMI, the union of any finite number of pairwise disjoint denumerable sets 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 then A ∪ B is denumerable by Theorem 5.3.5. If B − A is denumerable then A ∪ B is denumerable 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 is countable. Now consider a collection A of n + 1 countable sets.Pick A ∈ A. Then A − {A} is of n countable sets and therefore (A − {A}) a collection is countable. Now A = (A − {A}) ∪ A, so by part (b) of Corollary 5.3.9, A is countable. 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 the subset is infinite, it must be denumerable. (b) Suppose A is countable, B is uncountable, and A ⊆ B. Suppose B − A is countable. Then B = A ∪ (B − A) is a union of countable sets, so B is countable. This is a contradiction. We conclude that B − A is uncountable. (c) Q ∩ (1, 2) contains {1 + 21n : n ∈ N}, so it must be infinite. (Q ∈ (1, 2)) ⊆ Q, which is denumerable. Thus Q ∩ (1, 2) is denumerable by part (a). 20 (d) (Q ∩ (1, 2)) ⊆ n=1 (Q ∩ (n, n + 1)), so the union is an infinite subset of Q. 20 Thus n=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 is denumerable. (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 is rational. 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 an integer}. 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 and slope 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 is infinite because it contains the infinite subset { 12 , 31 , 14 , . . .}. 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 m−1 h: i∈N Bi → N by h(x) = ( i=1 ki ) + fm (x) where x ∈ Bm . Then h is a bijection. Suppose x, z ∈ i∈N Bi 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 natural m−1 m−1 number such that i=1 ki < n. Then h(f −1 (n − i=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 disjoint pairwise 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 be written as: .. .f (n) = an1 an2 an3 . . . 0, if aii = 1 Now let b be the sequence b1 b2 b3 . . . where bi = 1, if aii = 0. Then b is not an image of any n ∈ N, so f is not onto. Hence there bijection from N to S. f (1) = a11 a12 a13 . . . f (2) = a21 a22 a23 . . .
.. .
is no
(b) Let n ∈ N. Tn contains the sequence with a block of n consecutive 1’s starting with the first term, the sequence with n consecutive 1’s starting with the second term, etc. Thus Tn is infinite, for every n. Let a ∈ Tn , and suppose k is the first term such that for all i > k, ai = 0. Define f (a) to be
k the product i=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. Therefore Tn is infinite and countable, so Tn is denumerable. (c) By part (b), for every n ∈ N, Tn is denumerable. Therefore, for every n ∈ N, there exists a bijection fn : N → Tn . List the images of each of these functions in rows: f1(1)
f1(2)
f1(3)
f1(4)
f1(5) · · · (elements of T1 )
5
CARDINALITY
f2(1) f2(1) f2(1) .. .
f2(2) f2(2) f2(2) .. .
114
f2(3) f2(3) f2(3) .. .
f2(4) f2(4) f2(4) .. .
f2(5) · · · (elements of T2 ) f2(5) · · · (elements of T3 ) f2(5) · · · (elements of T4 ) .. .
∞ All the elements of T n=1 can be put into a one-to-one correspondence with N by listing them along the diagonal as in Figure 5.3.1. There is no need 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 a bijection. (b) Let g be a bijection from A to N. Let S2 be the set of all 2-element subsets of A. For {x, y} ∈ S2 with x < y, let f ({x, y}) = 2g(x) 3g(y) . The f is a one-to-one function that maps onto Rng(f ). Rng(f ) is an infinite subset of N, 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-element subsets of A. For {x1 , x2 , . . . , xk } ∈ Sk with x1 < x2 < · · · < xk , let
k g(x ) f ({x1 , x2 , . . . , xk }) = i=1 pi i , where pi is the ith prime. Then f is a one-to-one correspondence between Sk and Rng(f ). Since Rng(f ) is an infinite subset of N, Rng(f ) and Sk are denumerable. (d) Since A ≈ N, we may assume that A = N. By Part (c), Sk is denumerable for every k ∈ N. By Theorem 5.3.8, S = k∈N Sk is denumerable. Therefore the 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 the image of 1 will be x. This involves removing the two ordered pairs with second coordinates x and f (1) and replacing them with two other ordered pairs. Let t be the unique element of N such that f (t) = x and define f ∗ = (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 number with 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 with natural 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 that 1-1 1-1 1-1 f : A −→ B and g: B −→ C. Then g ◦ f : A −→ C, so A ≤ C. onto
4. Suppose A ≤ B and A = C. Then there exist functions f and g such that 1-1 1-1 1-1 f : C −→ A and g: A −→ B. Then g ◦ f : C −→ B, so C ≤ B. onto
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 such that 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 functions f and g such that f : A → B and g: B → C. Then g ◦ f is a one-to-one mapping from A to C, so A ≤ C. (d) Suppose A ≤ B and A = B. Then A < B by definition. This shows that A ≤ 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, so A ≤ 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). 1-1 1-1 Now suppose A ≤ B. Let f : A −→ B. Then f : A −→Rng(f ), so Rng(f ) is onto
a subset W of B such that A = W . 7. Suppose there is a set A with the largest cardinal number, A. By Cantor’s Theorem, 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) = (b) (c)
1 14 , f (4) = 19. 1 , g(4) = 20. g(1) = 14 , g(2) = 9, g(3) = 15 1 1 , H(20) = 15 . H(2) = 18 , H(8) = 13 , H(13) = 19
10. B ≤ C and C ≤ A, so B ≤ A. Also A ≤ B, so the Cantor-Schr¨ oder-Bernstein Theorem A = B. Therefore A ≈ B. The same reasoning shows B ≈ C. 11. (a) Any nonzero rational number can be written in exactly one way as a b such that a ∈ Z, b ∈ N, and a and b have no common factors. Define f : Q → N as: ⎧ if a > 0 a ⎨ 2a 3b , = 1, f if a = 0 ⎩ −a b b 5 7 , 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 = W for 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-Schr¨ oder-Bernstein Theorem B = 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 being a set, must be an element of U . Thus P(U ) ⊆ U , so P(U ) ≤ U . By Cantor’s Theorem U < P(U ). Therefore, by Theorem 5.4.4(b), P(U ) < P(U ), which is impossible. 15. (a) Let A be the set of such integers. The inclusion map i: A → Z is oneto-one, so A ≤ Z = N. On the other hand, the map f : N → A given by n f (n) = i=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. On the other hand, define g: (0, 1) × (0, 1) as g(0.a1 a2 a3 . . . , 0.b1 b2 b3 . . .) = 0.a1 b1 a2 b2 a3 b3 . . ., where all decimals are in normalized form. Then g is oneto-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 CantorSchr¨ oder-Bernstein Theorem. 16. (a) Assume there is a bijection from [0, 1] to F. For each a ∈ [0, 1], let the corresponding function in F be fa . Define g: [0, 1] → [0, 1] by g(x) = 0 if fx (x) = 0 1 if fx (x) = 0. Since g ∈ F, g = fb for some b ∈ [0, 1]. If g(b) = 0, then fb (b) = 0, so g(b) = 1. If g(b) = 0, then fb (b) = 0, so g(b) = 0. Thus g(b) = g(b), which is 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-one function from C to B. (b) F. There is no cancellation law for infinite cardinals that would allow us to conclude that (C − B) = 0. The claim is false. (c) F. One must prove that substitution and transitivity properties held for cardinal numbers. As per the definitions of < and ≤, a proof requires consideration of functions. (d) F. If g is not onto B, then some elements of B have no pre-image in A. In this case Dom(f ) = B.
5
CARDINALITY
5.5
117
Comparability of Cardinal Numbers and 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. 1-1 2. (a) Let f : B −→ A and b ∈ B. Define g: A → B by b, if a ∈ / Rng(f ) g(a) = 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 element from each Cb and call it g(b). This defines a function g: B → A for which g(b) ∈ Cb for each b ∈ B. The function g is one-to-one because b1 = b2 implies Cb1 ∩ Cb2 = ∅, which implies that g(b1 ) = g(b2 ). 3. By the Comparability Theorem, we have three cases: 1-1 Case 1: If A < B, then there is an f : A −→ B. 1-1 Case 2: If A= B , then there is an f : A −→ B. onto
Case 3: If A > B, then by Exercise 2, there is an f : A −→ B. onto
4. Let f : A → B. Then f : A −→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 by Theorem 5.5.4. Thus A = N = D ≤ B. By the CSB Theorem, B = A. Thus B ≈ A. 6. Assume that B < C and B/ ≤ A. Suppose C ≤ A. Then by the Comparability Theorem, 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∈N Ai such that for every i ∈ N, F (Ai ) ∈ Ai . Since the Ai ’s are distinct and pairwise disjoint, 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 a denumerable subset B = {an : n ∈ N}. Define g: A → A − {x} by ⎧ if a = x ⎨ a1 , f (a) = ai+1 , if a = ai ∈ B ⎩ a, if a ∈ / B ∪ {x}. Then f is a bijection.
5
CARDINALITY
118
onto
9. Suppose there is a function f : N −→ A. Then by Theorem 5.5.3 A ≤ N, so by Exercise 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. onto Thus, by Exercise 2, there is an f : N −→ A. 10. (a) A. (b) F. The “proof” claims incorrectly that if a set is not denumerable it must be finite. (c) F. The idea of this “proof” is to take out countably many elements, one at a time, until denumerably many elements are left. But if A is uncountable and C is countable, then the set B = A − C of leftover elements will always be 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 detailed explanation. (f) A. (g) A.
6
Concepts of Algebra
6.1
Algebraic Structures
1.
(a) Algebraic structure (c) Algebraic structure (e) Not an algebraic structure (g) Algebraic structure (i) Algebraic structure (k) Algebraic structure
(b) Not an algebraic structure (d) Not an algebraic structure (f) Not an algebraic structure (h) Algebraic structure (j) Not an algebraic structure (1) Not an algebraic structure
2.
(a) not commutative (c) not commutative (e) not an operation (g) not commutative (i) commutative (k) commutative
(b) not an operation (d) not an operation (f) not an operation (h) commutative (j) not an operation (1) not an operation
3.
(a) not associative (c) not associative (e) not an operation (g) not associative (i) associative (k) associative
(b) not an operation (d) not an operation (f) not an operation (h) associative (j) not an operation (1) not an operation
4. (a) a (b) Yes. This is tedious to verify because one must verify that 64 equations of the 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 or z = a, the equation is easily seen to be true. This leaves 27 cases to verify when none of x, y or z is a. For example, (b ◦ c) ◦ b = d ◦ b = c, while b ◦ (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 by cases 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 and finally 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 system with operation *, a does not have an inverse; the elements b and c are their own 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 is the 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 = c b b a a and b ◦ (b ◦ c) = b The element b has two inverses, b and c c c a b 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 the elements, 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 a product of some k factors a1 , a2 , a3 , . . ., ak (k ≤ n) in that order and b2 is a product of the remaining factors ak+1 , . . ., an+1 in that order. Use the induction 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 is an element c of A. Replace a1 ∗ a2 with the element c, so the product has 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 entire product 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 and the first two of these factors are a2 and a3. Then a2 ∗ a3 is an element d 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-associated expression (. . . (a1 ∗ d) . . . ∗ an+1 ). Now re-insert a2 ∗ a − 3 in the place of d, and then substitute (a1 ∗ a2 ) ∗ a3 for a1 ∗ (a2 ∗ a3 ) so the entire product is in left-associated form. 11. (a)
◦ a b c
a a a c
b b b a
c c c b
Both a and b are left identities.
(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 if a ∈ 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 = my for some integers x and y.
6
CONCEPTS OF ALGEBRA
121
(a) Then m divides (a + b) − (c + d) = (a − c) + (b − d) = m(x + y), so a + 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 for classes to emphasize the fact that elements of Zm are equivalence classes. Let x, y, z ∈ Zm . (a) Then (x+m y)+m z = x + y+m z = (x + y) + z = x + (y + z) = x+m y + z = x+m (y+m z), so +m is associative. Also x+m y = x + y = y + x = y+m x, so +m is commutative. The identity is 0, because x+m 0 = x + 0 = x = 0 + x = 0+m x. The inverse of x is −x, because x+m −x = x + (−x) = 0 = (−x) + x = −x+m x. That is −(x) = −x. (b) Then (x·m y)·m z = x · y·m z = (x · y) · z = x · (y · z) = x·m y · z = x·m (y·m z), so ·m is associative. Also x·m y = x · y = y · x = y·m x, so ·m is commutative. The element 1 is the identity because x·m 1 = x · 1 = x = 1 · x = 1·m x. 14. (a)
+ 0 1 2 3 4 5 6 7
0 0 1 2 3 4 5 6 7
1 1 2 3 4 5 6 7 0
2 2 3 4 5 6 7 0 1
3 3 4 5 6 7 0 1 2
4 4 5 6 7 0 1 2 3
(b)
+ 0 1 2 3 4
0 0 1 2 3 4
1 1 2 3 4 0
2 2 3 4 0 1
3 3 4 0 1 2
4 4 0 1 2 3
(c)
· 0 1 2 3 4 5 6 7
0 0 0 0 0 0 0 0 0
1 0 1 2 3 4 5 6 7
2 0 2 4 6 0 2 4 6
3 0 3 6 1 4 7 2 5
4 0 4 0 4 0 4 0 4
(d)
· 0 1 2 3 4
0 0 0 0 0 0
1 0 1 2 3 4
2 0 2 4 1 3
3 0 3 1 4 2
4 0 4 3 2 1
5 5 6 7 0 1 2 3 4
6 6 7 0 1 2 3 4 5
7 7 0 1 2 3 4 5 6
5 0 5 2 7 4 1 6 3
6 0 6 4 2 0 6 4 2
7 0 7 6 5 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.
6
CONCEPTS OF ALGEBRA
122
(d) There are no divisors of zero. 16. (a) F. The claim is false. The proof fails because we cannot assume that the cancellation property holds in every algebraic structure. (b) F. The claim is false. One may premultiply (multiply on the left) or postmultiply (multiply on the right) both sides of an equation by equal quantities. Multiplying one side on the left and the other side on the right does 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)
(c)
· 1 a β
1 1 a β
· 1 −1 i −i
α α β 1 1 1 −1 i −i
β β 1 α −1 −1 1 −i i
We see from the table that {1, α, β} is closed under ·, 1 is the identity, and every element has an inverse. Also, · is associative on C, so it is here too. i i −i −1 1
−i −i i 1 −1
We see from the table that the set is closed under ·, 1 is the identity and each element has an inverse. Also, · is associative.
(d) If A, B ⊆ X, then (A − B) ∪ (B − A) ⊆ A ∪ B ⊆ X, so P(X) is closed under . Also, A ∅ = (A − ∅) ∪ (∅ − A) = A ∪ ∅ = A and ∅ A = (∅ − A) ∪ (A − ∅) = ∅ ∪ A = A, so ∅ is the identity. Each element is 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 w
e e u v w
u u v w e
v v w e u
w w e u v
6
CONCEPTS OF ALGEBRA
3.
◦ e u v w
e e u v w
u u e w v
v v w e u
4. (a) abelian 5.
◦ a b c
a b c a
b a b c
123
w w v u e
(b) abelian c c a b
(c) abelian
This system has no identity and is not associative because (a ◦ a) ◦ c = a ◦ (a ◦ c).
6. (a)
+ 0 1 2 3 4 5
0 0 1 2 3 4 5
1 1 2 3 4 5 0
2 2 3 4 5 0 1
3 3 4 5 0 1 2
4 4 5 0 1 2 3
5 5 1 1 2 3 4
(b)
+ 0 1 2 3 4 5 6
0 0 1 2 3 4 5 6
1 1 2 3 4 5 6 0
2 2 3 4 5 6 0 1
3 3 4 5 6 0 1 2
4 4 5 6 0 1 2 3
5 5 6 0 1 2 3 4
(c)
· 1 2 3 4
1 1 2 3 4
2 2 4 1 3
3 3 1 4 2
4 4 3 2 1
(d)
· 1 2 3 4 5 6
1 1 2 3 4 5 6
2 2 4 6 1 3 5
3 3 6 2 5 1 4
4 4 1 5 2 6 3
5 5 3 1 6 4 2
6 6 5 4 3 2 1
7.
◦ [1 2] [2 1]
[1 2] [1 2] [2 1]
(d) abelian
[2 1] [2 1] [1 2]
6 6 0 1 2 3 4 5
S2 is abelian.
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] −1 −1 −1 −1 −1 −1 −1 −1 −1 9. (a) (a1 a2 a3 )(a−1 3 a2 a1 ) = a1 a2 a3 a3 a2 a1 = a1 a2 ea2 a1 = a1 a2 a2 a1 = −1 −1 −1 −1 −1 −1 −1 −1 a1 ea1 = a1 a1 = e and = (a3 a2 a1 )(a1 a2 a3 ) = a3 a2 a1 a1 a2 a3 = −1 −1 −1 −1 −1 a−1 = a−1 3 a2 a2 a3 = a3 a3 = e. Therefore (a1 a2 a3 ) 3 a2 a1 .
6
CONCEPTS OF ALGEBRA
124
−1 (b) For all n ∈ N, (a1 a2 . . . an )−1 = a − n−1 . . . a−1 2 a1 . −1 The statement is true for n = 1 because a1 = a−1 1 , and it was proved for n = 2 in Theorem 6.2.4. Assume the statement is true for some n > 2. Then
(a1 a2 . . . an an+1 )−1
=
−1
[(a1 a2 . . . an )an+1 ]
−1 = a−1 n+1 (a1 a2 . . . an ) −1 −1 −1 = a−1 n+1 an an−1 . . . a1 −1 −1 −1 = a−1 n+1 an an−1 . . . a1 .
Thus by the PMI, the statement is true for all n ∈ N. 10. Assume that ca = cb. Then a = ea = (c−1 c)a = c−1 (ca) = c−1 (cb) = (c−1 c)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 and ρa (ba−1 = ba−1 a = b. Therefore ρa is onto G. 12. Let a, b ∈ G. Then a(ab)b = a2 b2 = ee = e = (ab)2 = a(ba)b. Then (ab)b = (ba)b by left cancellation and ab = ba by right cancellation. 13.
∗ a b c d
a a b c d
b b a b c
c c d a b
d d c d a
In order to have both cancellation properties for a finite algebraic structure, it is necessary and sufficient that every element occur in every row and column.
14. (a) Suppose G is abelian. Let a, b ∈ G. Then a2 b2 = aabb = abab = (ab)2 . Now suppose a2 b2 = (ab)2 for all a, b ∈ G. Let x, y ∈ G. Then xxyy = x2 y 2 = (xy)2 = xyxy. By cancellation of x (on the left) and right cancellation for y, we have xy = yx. Thus G is abelian. (b) Suppose an bn = (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 a1 b1 = (ab)1 . Assume that ak bk = (ab)k for some k ∈ N. Then ak+1 bk+1 = aak bbk = abak bk = ab(ab)k = (ab)k+1 . By the PMI, an bn = (ab)n for all n ∈ N. 15. (a) Closure. Let a, b ∈ R − {1}. Then a ◦ b = a + b − ab ∈ R, and a ◦ b = 1 iff a + b − ab − 1 = 0 iff (a − 1)(b − 1) = 0 iff a = 1 or b = 1, so a ◦ 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 ◦. a (c) Inverses. If a ∈ R − {1}, then a−1 ∈ R − {1} and a◦
a a−1
a a a2 − a a a2 −a· = + − =0 a−1 a−1 a−1 a−1 a−1 a a a a2 − a a2 +a− ·a= + − =0 a−1 a−1 a−1 a−1 a−1
= a+
a ◦a = a−1
a so a−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).
and
6
CONCEPTS OF ALGEBRA
125
(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) = b and y ∗ a = (b ∗ a−1 ) ∗ a = b. To show uniqueness, suppose u ∈ G and a ∗ u = b. The u = e ∗ u = (a−1 ∗ a) ∗ u = a−1 ∗ b = x, so u = x. Also, if v ∈ 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 < m and xy = m. Then xy = 0(mod m) so Zm − {0} is not closed under multiplication. (b) Let p be prime and let x, y ∈ Zp − {0}. Then xy ∈ Zp and xy = 0 iff p divides xy iff p divides x or p divides y (by Euclid’s Lemma). But x and y are less than p, so this is impossible. Thus Zp − {0} is closed. Furthermore 1x = x1 = x, so 1 is the identity, and associativity follows from associativity for 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 (c) x = 0, 10
(b) x = 0 (d) x = 3, 7, 13, 17
21.
(a) x = 2 (c) x = 6 (e) No solution
(b) x = 4 (d) x = 2, 6 (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. The claim is false. (b) F. The fatal flaw is the use of the undefined division notation.
6
CONCEPTS OF ALGEBRA
126
(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 is corrected 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 is in both H and K, since a and b are both in H and K. Thus ab−1 ∈ H ∩ K, so H ∩ K is a subgroup by Theorem 6.3.3. 5. Suppose ∈ Hα for all α, {Hα : α ∈ Δ}is a family of subgroups of G. Since e 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. Then K is a subset of G and K is a group with the same operation as G, so K is a subgroup of G by definition.
6
CONCEPTS OF ALGEBRA
9.
127
(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
|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
(c)
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 generators of (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. Multiplying both sides of the last equation by y −1 , we have y −1 (ya)y −1 = y −1 (ay)y −1 . Thus (y −1 y)(ay −1 ) = (y −1 a)(yy −1 ), so ay −1 = y −1 a. Therefore, (xy −1 )a = x(y −1 a) = 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. Thus C is not empty. Suppose x, y ∈ C and let z ∈ G. Then (xy −1 )z = x(y −1 z) = x(z −1 y)−1 = x(yz −1 )−1 = (yz −1 )−1 x = (zy −1 )x = z(y −1 x) = z(xy −1 ), so xy −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 by definition. 14. Let a ∈ G and k = {a−1 ha: h ∈ G}. The identity e ∈ H because H is a group, and thus e = a−1 ea ∈ K. Thus K is not empty. Suppose b, c ∈ K. Then b = a−1 h1 a and c = a−1 h2 a for some h1 , h2 ∈ H. Thus bc−1 = (a−1 h1 a)(a−1 h2 a)−1 = −1 −1 (h1 h−1 h1 (aa−1 )h−1 (a−1 h1 a)(a−1 h−1 2 )a. But H is a group, so 2 a = a 2 a) = a −1 −1 h1 h2 ∈ H. Thus bc ∈ K. Therefore, K is a subgroup of G. √
√
15. (a) (α) = {α, α2 , α3 , α4 , α5 , 1}, where α2 = − 12 + i 2 3 , α3 = −1, α4 = − 12 − i 2 3 , √ and α5 = 12 − i 2 3 . (b) α5 16. Let m be a natural number greater than 1. Every k in Zm is the kth multiple of 1, 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 cyclic with generator e. Suppose H = {e}, and let t be the smallest positive integer such that at is in H. (The existence of t is guaranteed by the Well Ordering Principle.) We will show that H = (at ).
6
CONCEPTS OF ALGEBRA
128
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−qt as = a−qt aqt+r = ar ∈ H. Then r = 0, because r < t and t is the smallest positive integer such that at is in H. Therefore s = qt, so x = as ∈ (at ). 18.
(b) a15 (d) a3 , a9 , a21 , a27
(a) 5 (c) a10 , a20
19. (a) C. The proof omits the step of verifying that H ∩ K is nonempty (because the intersection contains the identity.) (b) F. The claim is false. The proof fails because hk −1 x−1 may not be in H, so x(hk −1 x−1 ) may not be in xH.
6.4
Operation Preserving Maps
√ 4 + 3. √ √ √ (b) Operation preserving. For all a, b ∈ R+ , ab = a · b.
1. (a) Not operation preserving.
√
4 + 3 =
√
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 is one-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. b b b 4. I(f + g) = a (f + g)(x) dx = a f (x) dx + a g(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 ). Then f −1 (f (x) ∗ f (y)) = f −1 (f (x · y)) = x · y = f −1 (f (x)) · f −1 (f (y)). a b e f 6. (a) Let and be in M . Then c d g h a b e f ae + bg af + bh Det · = Det c d g h ce + dg cf + dh = (ae + bg)(cf + dh) − (af + bh)(ce + dg) = (ad − bc)(eh − f g) a b e f = Det · Det . c d g h (b) Det
1 0
0 0
+
0 0
0 1
= 1, while Det
1 0
0 0
+ Det
0 0
0 1
= 0.
6
CONCEPTS OF ALGEBRA
129
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 Theorem 4.1.5(c). (c) Let C, D ∈ P(A). Then f −1 (C ∪ D) = f −1 (C) ∪ f −1 (D) by Theorem 4.1.5(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 ). Therefore f (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 from G to H, then f (e) = e¯. This means that f (g) must be h. The function f = {(e, e ), (g, h)} is one-to-one and maps G to H, and (by checking four cases) is easily seen to be OP. In the case where we must verify that f (g ◦ g) = f (g) · f (g), we first note that g ◦ g = e and h · h = e. Then f (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 and e. Then f = {(e, e ), (a, c), (b, d)} is one-to-one and maps G onto H. We observe 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 isomorphism from G to Z4 . Then 1 = f (x), where x is one of u, v, or w. Then f (x2 ) = f (x) + f (x) = 1 + 1 = 2. But x2 = e, so f (x2 ) = 0. This is a contradiction. 11. (a) Let 3a and 3b be in 3Z. Then f (3a + 3b) = 4(3a + 3b) = 4(3a) + 4(3b) = f (3a) + f (3b).
6
CONCEPTS OF ALGEBRA
130
(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 ∈ Z6 g(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 and so 24 divides 4(x − y) = 4x − 4y. Thus f (x) = [4x] = [4y] = f (y) in Z24 , so f 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] [0] [0] [4] [8] [12] [16] [4] [8] [12] [16] [20] [4] [8] [8] [12] [16] [20] [0] [4] [12] [12] [16] [20] [0] [16] [16] [20] [0] [4] [8] [20] [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 12 divides 4x − 4y and so f (x) = [4x] = [4y] = f (y) in Z24 . Thus f is welldefined. Now let x, y ∈ Z15 . Then f (x + 15 y) = [4(x + y)] = [4x + 4y] = [4x] +12 [4y] = f (x)+12 f (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 and for m, n ∈ Z4 , f (m + n) = im+n = im in = f (m) · f (n). 18. S3 is not isomorphic to (Z6 , +). The homomorphic image of an abelian group is abelian; (Z6 , +) is abelian but S3 is not.
6
CONCEPTS OF ALGEBRA
131
19. (a) Let (G, ·) be a group. Then the identity function IA maps A one-to-one and onto 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 : G −→ H. The f −1 is a function that maps H one-to-one and onto G. By Exercise 5(b), f −1 is OP. Therefore (H, ∗) is isomorphic to (G, ·). 1–1,Onto,OP
(c) Suppose (G, ·) is isomorphic to (H, ∗) and (H, ∗) is isomorphic to (K, ⊗). 1–1,Onto,OP 1–1,Onto,OP −→ H and g : H −→ K. Then the Then there exist f : G composite function maps G one-to-one and onto K. By Exercise 5(a), g ◦ f is OP. Therefore (G, ·) is isomorphic to (K, ⊗). 20. For each group G, we form the group H = {θa : a ∈ G} of left translations. As shown 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 error is 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 constant function 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 and distributive properties all follow immediately from the properties for R.
6
CONCEPTS OF ALGEBRA
132
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 additive inverse 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 the inverse of 0 is 0. The associative, commutative and distributive properties are inherited from R. (b) Suppose T is a nonempty subset of R that is closed under subtraction and multiplication . Then by the subgroup test, (T, +) is a group, and is abelian because (R, +) is abelian. The associative property for and the distributive property 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. Then a − 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 preserve multiplication. (b) h(4 · 5) = h(2) = 4, but h(4) · h(5) = 8 · 10 = 2 · 4 = 8 = 2, so h does not preserve multiplication. (c) Let p(x) = an xn + . . . + a1 x + a0 and q(x) = bn xn + . . . + b1 x + b0 be in Z[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)) = a0 b0 = p(0) · q(0) = g(p(x)) · g(q(x)). 11. Suppose pq and rs are in Q. Then f ( pq + rs ) = f ( ps+rq qs ) = (ps + rq, qs) = (p, q) ⊕ pr p r p r (r, s) = f ( q )⊕f ( s ) and f ( q · s ) = f ( qs ) = (pr, qs) = (p, q)⊗(r, s) = f ( pq )⊗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, t such that 1 < s, t < m. Then in Zm we have st = 0, so s and t are divisors of 0. Now suppose m is prime and a and b are divisors of zero in Zm . Then m divides ab. By Euclid’s Lemma, m divides a or m divides b. But this is impossible, because a and b are less then m. We conclude that if m is prime, then Zm has no divisors of zero. (b) We have already verified that for all m > 1, (Zm .+, ·) is a commutative ring with unity. Thus (Zm .+, ·) is a field iff (Zm .+, ·) has no zero divisors iff 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 inverse a−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 1.
2.
3.
Completeness of the Real Numbers (a)
√
10, 4, 11, 43
√ 5+ 21 , 400, 10100 , 2 −1, − 12 , 0, 5
(b)
1 3,
1, 2, 3 √ (d) 1, 2, π, e (f) No upper bound exists.
2
22
2 (c) (e) (g) 11,044, 11.5, 12.0, 98.6 √ (a) − 10 (c) No lower bound exists (e) No lower bound exists (g) 10−11
(b) 0 (d) −3 (f) 0
(a) sup: 1, inf: 0 (c) sup: does not exist, inf: 0 (e) sup: 1, inf: 13 (g) sup: 5, inf: −1 (i) sup: does not exist, inf: 0
(b) sup: 2, inf: 1 (d) sup: √ 1.5, inf: -2 √ (f) sup: 10, inf: − 10 (h) sup: 1, inf: −1 (j) sup: does not exist, inf: does not exist
4. (a) Suppose A is bounded above and B ⊆ A. Then there exists a real number u such that for all x ∈ A, x ≤ u. Since B ⊆ A, if b ∈ B, then b ∈ A and so b ≤ u. Therefore B is bounded above by u. (b) Suppose A is bounded below and B ⊆ A. Then there exists a real number v such that for all x ∈ A, v ≤ x. Since B ⊆ A, if b ∈ B, then b ∈ A and so v ≤ b. Therefore B is bounded below by v. (c) Suppose A is bounded above and B is bounded above. Then there exist real numbers 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 exist real 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. Then Then x ≤ t, since x ∈ A. Thus x = sup A. 6. (a) Suppose that b is an upper bound for A and assume that Ac also has an upper bound, c. Let x ∈ R be a number greater than b or c. Then x ∈ A and 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 bounds for A. Since y is an upper bound and x is a least upper bound, x ≤ y. Since x 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 and x is a greatest lower bound, y ≤ x. Since x is a lower bound and y is a greatest 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 is an 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 a lower bound for A. Therefore y ≤ x. 10. Let A be a subset of an ordered field F . Then y = inf(A) iff (1) (∀ > 0)(x ∈ A ⇒ x > g−) and (2) (∀ > 0)(∃y ∈ A)(y < g+).
Proof. First suppose g = inf(A). Let > 0. Then x ≥ g > g − for all x ∈ A, which establishes property (1). To verify property (2), suppose there were no y such that y < g + . Then g + is a lower bound for A, which is greater than the 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 = g−y 2 , then y < g − , which violates (1). Thus g is a lower bound for A. Now assume that there is another lower bound t such that g < t. If we let = t − g, then (2) says that there is some y ∈ A such that y < g + , so t is not a lower bound. Therefore g is the greatest lower bound of A. 11.
(a) (0, 4) (c) impossible
(b) {x ∈ Q: x ∈ (0, 4)} (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 bounded below (by any element of A), so inf(B) exists. Let t = inf(B). We show that s = t. i. To show t ≤ s we note that since s = sup(A), s is an upper bound for A. 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 an upper bound for A, then there exists a ∈ A with a > t. Let = a−t 2 . Since t = inf(b) and t < t+, there exist u ∈ B such that u < t+. But t < a. Therefore, u < a. This is a contradiction, since u is an upper bound for a. (b) Let i = inf(a) and C = {l: l is a lower bound of A}. C is bounded above by 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 that a < j. Let = j−a 2 . Since j = sup(C) and > 0, there is l ∈ C such that l > j − , But j − > a. Therefore, l > a, which contradicts the fact 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 imples sup(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. If x ∈ A, then x ≤ sup(A) ≤ m. If x ∈ B, then x ≤ sup(B) ≤ m. This m is 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)}.
7
CONCEPTS OF ANALYSIS
3
Proof. Let x = inf(A), y = inf B, and z = min{x, y}. If a ∈ A ∪ B, then a ∈ A, in which case z ≤ x ≤ a, or a ∈ B, in which case z ≤ y ≤ a. So z is a lower bound for A ∪ B. If t is another lower bound for A ∪ B, then t is a lower 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, 23 , 52 , 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. Thus sup(A ∩ B) exists. By Exercise 9(a), sup(A ∩ B) ≤ min{sup(A), sup(B)}. 16. (a) Let A = { 13 , 21 , 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. Therefore inf(A ∩ B) exists. By Exercise 9(b), inf(A ∩ B) ≥ max{inf(A), inf(B)}. 1 17. Suppose there is a positive real number r such that for all integers K, K ≥ r. Then for all K, 1 ≥ Kr. Therefore the set W = {nr: n ∈ N } is bounded above by 1. By the completeness property for R, sup(W ) exists. Let t = sup(W ). Then t − r < t, so t − r is not an upper bound for W . Thus there exists mr ∈ W such that mr > t − r. Then mr + r > t, so (m + 1)r > t. But (m1 )r is in W and t is an 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 F that has a lower bound v. Let A = {−x: x ∈ B}. Then A is a subset of F and since v ≤ b for all b ∈ B, −v ≥ −b for all b ∈ B; that is, −v ≥ a for all a ∈ A. Thus A is bounded above by −v and so s = sup(A) exists. We now show −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 a lower bound for B. Let w be a lower bound for B. Then w ≤ b for all b ∈ B. Therefore −w ≥ −b for all b ∈ B; that is, −w ≥ a for all a ∈ A. Thus, −w is an upper bound for 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 F that has a lower bound in F has an infimum in F . Let A be a nonempty subset of A that is bounded above by u. To show F is complete, we show that sup(A) exists. Let B = {−x: x ∈ A}. Then B is a subset of F and since a ≤ u for all a ∈ A, −a ≥ −u for all a ∈ A; that is, b ≥ −u for all b ∈ B. Thus B is bounded below by −u and so z = inf(B) exists. We now show 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 an upper bound for A. Let t be an upper bound for A. Then a ≤ t for all a ∈ A. Therefore −a ≥ −t for all a ∈ A; that is, b ≥ −t for all b ∈ B. Thus, −t is a lower bound for B and 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
7
CONCEPTS OF ANALYSIS
4
x < y and x = y. Then y < y, another contradiction. Finally, x = y and y < x leads to x < x as well. Thus at most one of x < y, x = y, and y < x can 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, we have 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 is true. Suppose 1 < 0. Then by part (b), 0 < −1. By travsitivity 1 < −1. Multiplying by −1, we have −1 < 1, which contradicts 1 < −1. Therefore 0 < 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, by part (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 +
2
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 is increasing, but the claim is false even for increasing functions. The fatal flaw 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 }. 2 , δ1 − (b) If |x2 − x1 | < 2δ1 , then N (x1 , δ1 ) ∩ \(x2 , δ1 ) = N x1 +x 2 |x2 − x1 | ≥ 2δ1 , then N (x1 , δ1 ) ∩ \(x2 , δ1 ) = ∅.
|x1 +x2 | 2
. If
7
CONCEPTS OF ANALYSIS
5
(c) N (x1 , δ1 ) ∩ N (x2 , δ2 ) = {x ∈ R: |x1 − x| < δ1 and |x2 − x| < δ2 }. This set may be empty. 3. limx→∞ f (x) = L if for all > 0 there is a δ > 0 such that if x ∈ N (a, δ) then f (x) ∈ N (L, ). 4.
(a) (−1, 1) (d) ∅ (g) R − N
(b) (−1, 1) (e) ∅ 1 (h) R − ({ 3k : k ∈ } ∪ {∅})
(c) ∅ (f) ∅ (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 7.2.1 and Theorem 7.2.2(a), (a, ∞) and (−∞, a) are open sets. 7. (a) [a, ∞)c = (−∞, a) and (−∞, a]c = (a, ∞). Since their complements are open, the closed rays are closed. (b) [a, b]c = (−∞, a) ∪ (b, ∞) so by solution 6 and Theorem 7.2.2(a), [a, b] is a closed set. 8. Suppose A is a nonempty collection of closedsubsets of R. Then for each c A Theorem 7.2.2(a), A∈A Ac is open. Therefore open. By ∈ A, A is c c c c A∈A A = (( A∈A A) ) = ( A∈A A ) is closed. Supposed A is a finite nonempty collection of closed subsets of R. Then for c c each A ∈ A, A is open. By Theorem 7.2.2(b) A∈A A is open. Therefore c c c c A∈A A = (( A∈A A) ) = ( A∈A A ) is closed. (b) (a) x∈(0,1) {x} is not closed. 9. (a) Let A be an open set. Then A − {x} = A ∩ [(−∞, x) ∪ (x, ∞)], which is open by Theorem 7.2.2. (b) Let A be open and B be closed. Then B c is open and so A − B = A ∩ B c is open. (c) Let A be open and B be closed. Then (B − A)c = (B ∩ Ac )c = B c ∪ 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’s open. 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.) Pick y ∈ N (x, δ1 ). Since y ∈ N (x, δ1 ) is an open set in R, there is a δ2 such that N (y,δ2 ) ⊆ N (x, δ1 ) ⊆ A. Thus y is an interior point of A, or in other words, ) ⊆ (A), so x is an interior point of (A). This y ∈ (A). Therefore N (x, δ 1 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 interior point of Ac , because every δ-neighborhood of x intersects A. Now suppose x is not an interior point of A and not an interior point of Ac . Let δ > 0. Then N (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 of A, so x ∈ A. Now suppose A is a set that contains none of its boundary points, and suppose x ∈ A. Then x is not a boundary point of A, so by
7
CONCEPTS OF ANALYSIS
6
part (b), x is an interior point of A or an interior point of Ac . Since x is not in Ac , x is not an interior point of Ac . Therefore x is an interior point of 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 points iff 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 interior point of Ac . But Ac is open, so the interior of Ac is Ac . Therefore x ∈ A. 13.
(a) not compact (not bounded) (c) compact (e) compact (g) not compact (not closed)
(b) compact (d) not compact (not closed, not bounded) (f) compact (h) not compact (not closed)
14. (a) A = (0, 1) and C = {( n1 , 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 cover for A ∪ B. Then {Oα : α ∈ Δ} is a cover for A and a cover for B. Thus there are finite sets Δ1 and Δ2 such that {Oα : α ∈ Δ1 } is a subcover for A and {Oα : α ∈ Δ2 } is a subcover for B. Then {Oα : α ∈ Δ1 ∪ Δ2 } is a finite subcover for A ∪ B. Thus A ∪ B is compact. (b) Let A and B be compacts subsets of R. By the Heine-Borel Theorem, A is closed and bounded and B is closed and bounded. By Exercise 8, A ∩ B is closed. Since A ∩ B ⊆ B, A ∩ B is bounded. Therefore, by the Heine-Borel Theorem, A ∩ B is compact. (c) Let A and B be compacts subsets of R. By the Heine-Borel Theorem, A is closed and bounded and B is closed and bounded. By Exercise 8, A ∩ B is closed. By Exercise 4 of Section 7.1, A ∪ B is bounded. Therefore, by the Heine-Borel Theorem, A ∪ B is compact. 1/n > 1 for all n, there is an n such 16. (a) Let x ∈ S. Since limn→∞ n+2 2n = 0 and 2 n+2 1/n that x ∈ 2n , 2 . Thus (0, 1] ⊆ ∪C∈I C. (b) No. (c) S is not compact (not closed).
17. L {Aα : α ∈ Δ} be a family of compact sets. Then ∪αinΔ Aα is closed (by Exercise 8(a)). Let β ∈ Δ. Then ∪αinΔ Aα ⊆ Aβ is bounded. By the HeineBorel Theorem, ∪αinΔ Aα is compact.
1 , 1]: n ∈ N . 18. Let A = {{x}: x ∈ R} or {[−n, n]: n ∈ N} or [ n+1 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 necessarily a subcover of {Oα : αinΔ}. Since O∗ = R − B is useless in a cover of B, it can 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
CONCEPTS OF ANALYSIS
7.3
7
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) that is in [3, 7) and is distinct from 7. If δ ≤ 1, then 7 − 2δ is a point of N (x, δ) that is in [3, 7) and is distinct from 7. n
(b) Let A = { 1+(−1) : n ∈ N}. Suppose δ > 0. Let n be an even n natural number 1+(−1)n greater than 2/δ. Then 2/n < δ. Since n is even, n = 2/n < δ. Thus 1+(−1)n n
is in N (0, δ) and is distinct from 0.
(c) Let A = {(1 + n1 )n : n ∈ N}. Let δ > 0. Since limn→∞ (1 + n1 )n = e, there exists a natural number N such that for all n > N , (1 + n1 )n − e < δ. Then (1 + N1+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 (c) {(−1)n ( n+1 n ): n ∈ N} (e) Q
3.
(a) { 12 } (f) [3, 7] (k) [−1, 1]
(b) ∅ (g) {0, 2} (l) [0, 1]
(b) { n1 : n ∈ N} (d) {m − n1 : m, n ∈ N} (c) ∅ (h) ∅ (m) N
(d) {0} (i) [0, 1] (n) {0}
(e) [0, 1] (j) {0}
4. {0, 1} 5. Since z is an upper bound for A and z ∈ / A. Let δ > 0. Then z − δ < sup(A), so there 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 distinct from x such that a ∈ N (x, δ). Since a ∈ B, there is an element of B distinct from 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 suppose there is an x ∈ / A ∪B . Then there exist δ1 and δ2 , both greater than 0, such that (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 of x meets (has a nonempty intersection with) A, so every neighborhood of x meets 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 is a 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 because x = y. Since y ∈ A there exists z ∈ A such that z ∈ N (x, δ1 ). But then z ∈ 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 .
7
CONCEPTS OF ANALYSIS
8
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 + δ2 is a point of 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, whereas interior 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 X is a subset of the interior of S, S ⊆ S . (d) False. [0, 1] ⊆ [0, 1] , but [0, 1] is not open 10.
(a) no accumulation points (d) may have no accumulation points (g) has at least one accumulation point
(b) has at least one accumulation point (e) has one accumulation point
(c) has at least one accumulation point (f) has accumulation points
/ A . Therefore, there exists > 0 such that 11. Let A ⊆ R. Let x ∈ (A )c . Then x ∈ c N (x, ) ∩ A = ∅. Thus N (x, ) ⊆ A . To show that x ∈ (Ac ) , let δ > 0. Choose δ1 = min(δ, ). Then N (x, δ1 ) ⊆ N (x, ) ⊆ 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 points of 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. (B c ) 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 (c) Bounded (e) Bounded (g) Not bounded (i) Bounded (k) Bounded
2.
(a) xn = (−1)n (d) xn = (−1)n
(b) Bounded (d) Bounded below (f) Bounded (h) Not bounded (j) Not bounded (l) Bounded (b) xn = n (e) xn = π2
(c) xn =
(−1)n n
3. Pick B so that |yn | ≤ B or all n ∈ mathbbN . Let > 0. Choose N so that n > N implies |xn | ≤ B . Now suppose n > N . Then |xn yn − 0| = |xn | |yn | ≤ |xn |B < . 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 so that B1 ≤ xn ≤ B2 for all n ∈ N. Then |xn | ≤ B for all n ∈ N, where B = max{|B1 |, |B2 |}.
7
CONCEPTS OF ANALYSIS
9
5. (a) The sequence x given by xn = n+2 n is decreasing. Suppose that m, n are natural numbers and m < n. Then 1 m 2 m 2 1+ m m+2 m
> >
1 , n 2 , n 2 , n n+2 . n
> 1+ >
(b) The sequence y given by yn = 2−n is decreasing. Suppose that m, n are natural numbers and m < n. Then n−m > 0 and thus 2n−m > 1. Therefore ym = 2−m = 2−n 2n−m > 2−n = yn . (c) The sequence x given by xn = (n − 2)(n − 5)2 is neither increasing nor decreasing because the first four terms are x1 = −16, x2 = 0, x3 = 4, x4 = 2. (d) The sequence y given by yn = 10 n! is decreasing. Suppose that m, n are n! > 1. natural numbers and m < n. Then n! = (n − m)!(m!) > m!, so m! 10 10 10 n! Then m! = m! · n! > n! . (e) The sequence x given by xn = 2n−5 n+3 is increasing. Suppose that m, n are natural numbers and m < n. Then m+3 1 m+3 −11 m+3 11 2− m+3 2m − 5 m+3 (f) The sequence y given by yn = terms. For n ∈ N, yn =
n! nn
<
n+3 1 > n+3 −11 < n+3 11 n+3 2n − 5 n+3
< 2− <
is decreasing. We will compare consecutive
n!(n + 1) n! (n + 1)! (n + 1)! (n + 1)! = n = yn+1 = n = = nn n (n + 1) n (n + 1) (n + 1)n (n + 1) (n + 1)n+1
√ (g) The sequence x given by xn = n + 1 is increasing. We first note that a simple proof by contradiction shows that if t is a real number such that √ 0 < t < 1, then 0 < t < 1. Suppose that m, n are natural numbers and m+1 m+1 m < n. Then m+1 < n+1, which implies 0 < n+1 < 1 and 0 < n+1 < 1. √ √ √ n+1 m+1 Then m + 1 = (m + 1) n+1 = n + 1 n+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 some N ∈ N, xN = L and since x is decreasing, xn = L for all n > N . THerefore xn → L.
7
CONCEPTS OF ANALYSIS
10
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 all n ∈ N If there exists N such that xN < L, then xn < L for all n > N . By Exercise 12 of Section 7.3, L is an accumulation point of {xn : n ≥ N }. But for δ = |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 > 0. Since L is an accumultion point of {xn : n ∈ N}, there exists M ∈ N such that xM ∈ N (L, ). Suppose n > M . Then L − < L ≤ xn ≤ xM < L + , so |xn − L| < . 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 1 n(n − 1) . . . [n − (k − 1)] k! nk
≤ =
1 n(n − 1) . . . [n − (k − 1)] k! (n + 1)k−1 1 (n + 1)n(n − 1) . . . [n − (k − 1)] k! (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 > 0. Then by part (b) of Theorem 7.1.1, there is an N ∈ N such that xn > s − . Suppose n > N . Then s − < xN ≤ xn < s + , so |xn − s| < . Thus xn → 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 some increasing 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 . Thus x is bounded by max{|xk | + 1 : 1 ≤ k ≤ N }. (b) Suppose xn → L. Let > 0. Pick N so that n > N implies |xn − L| < 2 . Suppose m, n > N . Then |xm − xn | ≤ |xn − L| + |xn − L| < 2 + 2 = . √ √ xy = 21 (x − 2 xy + y) = 11. (a) Let√n = 1. First observe that a1 − b1 = x+y 2 − √ 1 y)2 > 0, because x > y. Thus a1 > b1 . We now compute: 2( x − a1 − a2 b2 − b1 a2 − b2
1 1 = a1 − (a1 + b1 ) = (a1 − b1 ) > 0;
2
2√ a1 b1 − b1 = b1 ( a1 − b1 ) > 0; and =
1 √ 1 1 = (a1 + b1 ) − a1 b1 = (a1 − 2 a1 b1 + b1 ) = ( a1 − b1 )2 > 0. 2 2 2
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, assume an > an+1 > bn+1 > bn . Then an+1 − an+2 bn+2 − bn+1 an+2 − bn+2
1 (an+1 − bn+1 ) > 0; 2
√ = bn+1 ( an+1 − bn+1 ) > 0; and
1 1 √ = (an+1 − 2 an+1 bn+1 + bn+1 ) = ( an+1 − bn+1 )2 > 0. 2 2 =
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.
7
CONCEPTS OF ANALYSIS
11
(b) The sequence an converges because it is monotone decreasing and bounded by a. The sequence bn is convergent because it is monotone increasing and is also bounded by a. (c) Let limn→∞ an = L and limn→∞ bn = M . Then limn→∞ an+1 = limn→∞ so L = L+M 2 . Therefore 2L = L + M , so L = M .
an +bn , 2
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 subsequence x3 , x4 , x5 , . . . converges, but the Bounded Monotone Sequence Theorem does 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 Section 4.6, 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 for all 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−1 n : 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 then am ≤ 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} exists and a ≤ b. Thus An = [a, b] = ∅. 6. (a) C. This is a promising idea, but the limit L is not necessarily the sup of A. For example, if [0, 2] ⊆ A, the process described might produce the n sequence xn = 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 B would have to be defined differently. More seriously, there is no connection between being an accumulation point and being an upper bound for a set.
