數學導論

數學導論

學數學

前言

學 學 數學 學 數學 . 學數學

論 . 學 , .

(Logic), (Set) 數 (Function) . , 學

論. 論 學 數學 .

, ,

. , .

, . , 論

, . , .

v

Chapter 1

Basic Logic

學 數學 學 言. “ ” , logic

“ ”. 學 學 ,

, 學 .

, 學 數學 .

數學 論 statement. 2 > 0 statement,

3 < 2 statement x > 0 statement ( x ).

1.1. Connectives

數 statements statement, statements connec-

tives. connectives statement .

1.1.1. And. “and” connective. connective

. P Q statement, P∧ Q P and Q

statement. P∧ Q ? “and” “ ”

, P Q P∧ Q , P Q

, P∧ Q . 2 > 0 and 2 < 7 , 2 > 0

2 > 7 .

truth table connectives statements

. T (true), F (false). true table.

P Q P∧ Q

T T T

T F F

F T F

F F F

Truth table P, Q , P, Q ,

. P T, Q F P∧ Q F.

1

P, Q P∧Q Q∧P . P∧Q

Q∧ P . logically equivalent.

Truth table statements connectives

, (P∧ Q) ∧ R truth table

P Q R P∧ Q (P ∧ Q) ∧ R

T T T T T

T F T F F

F T T F F

F F T F F

T T F T F

T F F F F

F T F F F

F F F F F

Question 1.1. P∧ (Q ∧ R) truth table ?

(P∧ Q) ∧ R P∧ (Q ∧ R) . (P∧ Q) ∧ R P∧ Q

R ; P∧ (Q ∧ R) Q∧ R P . truth table

(P∧ Q) ∧ R P∧ (Q ∧ R) logically equivalent.

1.1.2. Or. P Q statement, P∨ Q P or Q statement.

“or” “ ” . “ ” :

, . “ ” , ;

105 . “ ”

105 , 105

. 數學 , “or” , P Q

P∨ Q ( P Q ). 言 , P Q

, P∨ Q .

, 4 < 5 or 4 < 3 statement , 4 < 5 . 4 > 5 or 4 > 6

statement , . 4 < 5 or 4 > 3 statement

, and , , .

P∨ Q truth table.

P Q P∨ Q

T T T

T F T

F T T

F F F

Question 1.2. P∨ Q Q∨ P logically equivalent? (P∨ Q) ∨ R P∨ (Q ∨ R) logically equivalent?

and, or connectives, . P, Q, R statements

(P∧ Q) ∨ R, (P ∨ Q) ∧ R,... statements.

1.1. Connectives 3

? (P∧ Q) ∨ R (P∧ Q) R . R ,

(P∧ Q) ∨ R , R P, Q , (P∧ Q) ∨ R . ,

(P∧ Q) ∨ R P∧ (Q ∨ R) logically equivalent. P∧ (Q ∨ R)

P Q∨ R . R , Q Q∨ R ,

P P∧ (Q ∨ R) . R , (P∧ Q) ∨ R ,

(P∧ Q) ∨ R P∧ (Q ∨ R) logically equivalent. truth

table logically equivalent.

P Q R P∧ Q (P ∧ Q) ∨ R

T T T T T

T F T F T

F T T F T

F F T F T

T T F T T

T F F F F

F T F F F

F F F F F

P Q R Q∨ R P ∧ (Q ∨ R)

T T T T T

T F T T T

F T T T F

F F T T F

T T F T T

T F F F F

F T F T F

F F F F F

, (P∨ R) ∧ (Q ∨ R) truth table, (P∧ Q) ∨ R (P∨ R) ∧ (Q∨ R) logically equivalent.

P Q R P∨ R Q ∨ R (P ∨ R) ∧ (Q ∨ R)

T T T T T T

T F T T T T

F T T T T T

F F T T T T

T T F T T T

T F F T F F

F T F F T F

F F F F F F

Question 1.3. truth table (P∨ Q) ∧ R (P∧ R) ∨ (Q ∧ R) logically equivalent.

truth table logically equivalent. logic

logical equivalences .

connectives truth table . 論 logical equivalences

, logical equivalences.

“or” 數學 ≥ ≤. 數學 x≥ y x > y or x = y,

4≥ 3 statement or . 4≤ 5, .

1.1.3. If - Then. 數學 connective 學

connective, . P Q statement, P⇒ Q if P

then Q statement, P Q . P⇒ Q 數學

. , 數學 P⇒ Q P, Q (

P, Q ). P, Q statement, x 數 “ ”.

⇒ connective P, Q ( ). 數學

“if x > 3 then x²> 9” statement ( x > 3 x²> 9 statement, if-then , statement). x > 3 x²> 9 .

“if 3 > 2 then 2 is even” statement ( 3 > 2 2 數 ).

P⇒ Q 前, 數學 論 論 .

數學 , if P then Q “ P , Q ”. ( :

statement, , .) ,

if P then Q P , Q . P ,

Q . 數學 論 if P then Q P , Q

“ ”, P . if P then Q

P, Q statements connective , if P then Q

statement, P, Q P⇒ Q .

P⇒ Q Q⇒ P 數學 . 學 P⇒ Q Q⇒ P.

, P⇒ Q P Q , P

Q . if x > 3 then x²> 9, x≤ 3 x²> 9.

Q P . 言 , P⇒ Q Q⇒ P.

if P then Q , P⇒ Q Q⇒ P

equivalent.

Question 1.4. P Q . Q ,

言 P ?

P⇒ Q . 前 數學 , P, Q

statements , P Q , P⇒ Q ,

P⇒ Q . P Q , P⇒ Q ,

P⇒ Q . P , P⇒ Q ? P⇒ Q 論

P , Q , P , Q 前 P⇒ Q ,

P⇒ Q . 2 > 3 2²> 9 , 前

if x > 3 then x²> 9 statement. ,−4 > 3 , (−4)²> 9

, 前 if x > 3 then x²> 9 statement. 言 , P⇒ Q

truth table.

P Q P⇒ Q

T T T

T F F

F T T

F F T

Question 1.5. truth table Q⇒ P P⇒ Q logically equivalent?

(P⇒ Q) ⇒ R P⇒ (Q ⇒ R) logically equivalent?

學 P⇒ Q , “if and only if”

connective .

1.1. Connectives 5

P⇒ Q . if P then Q ,

• Q if P

• P implies Q

• P is sufficient for Q ( P Q )

• Q is necessary for P ( Q P )

• P only if Q ( Q P )

• Q whenever P ( P Q )

1.1.4. If and Only If. P⇒ Q Q⇒ P and , (P⇒ Q) ∧ (Q ⇒ P),

“P if and only if Q”, P⇔ Q .

數學 P⇔ Q . 數學 P⇔ Q P⇒ Q

Q⇒ P. P Q , Q P . P, Q

. 言 , P⇔ Q Q P

Q P ( P Q ).

P⇔ Q “P Q ” ( P Q) .

P⇔ Q . 前 數學 ,

P⇔ Q P Q Q P . . P, Q

. P⇔ Q P Q

. P⇔ Q truth table.

P Q P⇔ Q

T T T

T F F

F T F

F F T

Question 1.6. P⇒ Q Q⇒ P truth table P⇔ Q truth table.

Question 1.7. P⇔ Q Q⇔ P logically equivalent? (P⇔ Q) ⇔ R P⇔ (Q ⇔ R) logically equivalent?

P⇔ Q , 數學 , .

P⇔ Q P , Q , P⇒ Q .

, (P⇒ Q) ∧ (Q ⇒ P) P⇔ Q, P, Q , P⇔ Q ,

P⇒ Q Q⇒ P . P, Q , P⇒ Q .

P Q , Q⇒ P , P⇒ Q P⇔ Q .

P⇒ Q , 導 P⇒ Q, Q ⇒ P P⇔ Q truth table (

equivalent), 前 數學 P⇒ Q Q⇒ P , P Q ,

P⇒ Q .

P⇔ Q . P if and only if Q ,

• P iff Q

• P is equivalent to Q

• P is necessary and sufficient for Q

1.2. Logical Equivalence and Tautology

前 logical equivalence . logical equivalence

導 logical equivalences. Truth table

logical equivalence .

. P, Q statements , P∧ Q Q∧ P

statements ( ), P∧ Q Q∧ P logically

equivalent . P, Q 數 , statement

, P∧ Q P, Q , P∧ Q statement

. , ( ) P, Q ,

connectives “statement form”, P∧ Q Q∧ P

statement forms logically equivalent. “∼” statement forms logically equivalent, (P∧ Q) ∼ (Q ∧ P).

logical equivalence : logically equivalent

statement forms 數 statement form , logical

equivalence. (P∧ Q) ∼ (Q ∧ P), P P⇒ Q

((P⇒ Q) ∧ Q) ∼ (Q ∧ (P ⇒ Q)).

, logically equivalent statement forms truth

table, 數 statement forms

truth table. , 數 ( ) logically equivalent

statement forms , statement forms logically equivalent.

(P∧ Q) ∼ (Q ∧ P) (R∨ S) ∼ (S ∨ R), (P∧ Q) ∼ (Q ∧ P) P R∨ S

, P S∨ R

((R∨ S) ∧ Q) ∼ (Q ∧ (S ∨ R)).

, statement forms A, B logically equivalent B statement form C logically equivalent, A C logically equivalent.

((P∧ Q) ∨ R) ∼ ((Q ∧ P) ∨ R), ((Q∧ P) ∨ R) ∼ (R ∨ (Q ∧ P)), ((P∧ Q) ∨ R) ∼ (R ∨ (Q ∧ P)).

truth table .

truth table statement forms

logically equivalent. logically equivalent “ ” . 前

1.2. Logical Equivalence and Tautology 7

學 logical equivalences, ∧ ∨ ,

(P∧ Q) ∼ (Q ∧ P), (P ∨ Q) ∼ (Q ∨ P) (1.1)

∧ ∨ ,

((P∧ Q) ∧ R) ∼ (P ∧ (Q ∧ R)), ((P ∨ Q) ∨ R) ∼ (P ∨ (Q ∨ R)) (1.2)

∧,∨ ,

((P∧ Q) ∨ R) ∼ ((P ∨ R) ∧ (Q ∨ R)), ((P ∨ Q) ∧ R) ∼ ((P ∧ R) ∨ (Q ∧ R)) (1.3) 導 logical equivalences .

Example 1.2.1. (P∧ Q) ∨ (P ∨ Q) statement form. (1.3) ((P∧ Q) ∨ R) ∼ ((P ∨ R) ∧ (Q ∨ R)), R P∨ Q ,

(P∧ Q) ∨ (P ∨ Q) ∼ ((P ∨ (P ∨ Q)) ∧ (Q ∨ (P ∨ Q))). (1.4) (P∨ (P ∨ Q)) ∼ ((P ∨ P) ∨ Q) (Q∨ (P ∨ Q)) ∼ (Q ∨ (Q ∨ P)) ∼ ((Q ∨ Q) ∨ P)

((P∨ (P ∨ Q)) ∧ (Q ∨ (P ∨ Q)) ∼ (((P ∨ P) ∨ Q) ∧ ((Q ∨ Q) ∨ P)). (1.5) (P∨ P) ∼ P (P∧ P) ∼ P,

(((P∨ P) ∨ Q) ∧ ((Q ∨ Q) ∨ P)) ∼ ((P ∨ Q) ∧ (Q ∨ P)) ∼ (P ∨ Q). (1.6) (1.4), (1.5), (1.6),

((P∧ Q) ∨ (P ∨ Q)) ∼ (P ∨ Q).

statement form truth table , statement form

tautology. . P⇔ P truth table

P P⇔ P

T T

F T

,

P⇔ P tautology.

Question 1.8. P⇒ P tautology? P⇒ (P ⇒ P) tautology?

Tautology , .

logically equivalent. statement forms A, B logically equivalent ,

A, B , A⇔ B . A⇔ B tautology. , A⇔ B

tautology , A, B , truth table. A∼ B.

.

Proposition 1.2.2. A, B statement forms. A B logically equivalent A⇔ B tautology.

前 , A∼ B A⇔ B tautology ( A∼ B A⇔ B tautology), A⇔ B tautology A∼ B. Proposition 1.2.2

A∼ B A⇔ B tautology.

Question 1.9. A, B statement forms. A∼ B A⇒ B tautology?

A⇒ B tautology A∼ B?

Question 1.10. A, B,C statement forms. A⇔ B B⇔ C tautology, A⇔ C tautology?

tautology contradiction ( ). statement form

. contradiction, “not” .

Question 1.11. A, B statement forms.

(1) A tautology, (A∧ B) ∼ B A∨ B tautology.

(2) A contradiction, (A∨ B) ∼ B A∧ B contradiction.

1.3. Not and Contradiction

“not” not equivalences. 前 ,

. , ,

.

Not , statement P, ¬P, not P,

“ P”. P ,¬P . , P ,¬P .

¬P truth table.

P ¬P

T F

F T

. ,

P∼ ¬(¬P). (1.7)

Not P , connectives statement not ,

. ¬(P ∧ Q), (¬P) ∧ (¬Q),

truth table

P Q P∧ Q ¬(P ∧ Q)

T T T F

T F F T

F T F T

F F F T

P Q ¬P ¬Q (¬P) ∧ (¬Q)

T T F F F

T F F T F

F T T F F

F F T T T

, P Q P Q ,¬(P ∧ Q) (¬P) ∧ (¬Q) . ,

truth table,

¬(P ∧ Q) ∼ (¬P) ∨ (¬Q). (1.8)

1.3. Not and Contradiction 9

數學 . 0≤ x ≤ 1, x≤ 1 and

x≥ 0. , x > 1 or x < 0. 數 x P x≤ 1

statement, Q x≥ 0, ¬P, ¬Q x > 1, x < 0. 0≤ x ≤ 1 P∧ Q x > 1 or x < 0 (¬P)∨(¬Q). ¬(P∧Q) (¬P)∨(¬Q) logically equivalent, (¬P) ∧ (¬Q) ( x > 1 and x < 0 ).

statement form logically equivalent not.

(1.8) P, Q ¬P ¬Q ,

¬((¬P) ∧ (¬Q)) ∼ (¬(¬P)) ∨ (¬(¬Q)).

¬(¬P) ∼ P,

¬((¬P) ∧ (¬Q)) ∼ (P ∨ Q).

not,

¬(P ∨ Q) ∼ (¬P) ∧ (¬Q). (1.9)

x≥ 0 , x < 0. P, Q x > 0, x = 0, x≥ 0

P∨ Q. ¬P x≤ 0, ¬Q x̸= 0. (¬P) ∧ (¬Q) x≤ 0 and x ̸= 0, x < 0

x≥ 0 .

(1.7), (1.8), (1.9) 導 not statement forms logical equiva-

lence . (1.8), (1.9) DeMorgan’s laws.

, statement form ¬(P ⇒ Q) logically equivalent ?

P⇒ ¬Q, truth table , P P⇒ Q

P⇒ ¬Q , P . ¬(P ⇒ Q) P⇒ ¬Q

logically equivalent, .

Question 1.12. x≥ 0 ⇒ x ≥ 1 數 x, x≥ 0 ⇒ x < 1

數 x. ?

P⇒ Q P , , .

, A, B statement form A tautology, ¬(A ⇒ B) A⇒ ¬B logically

equivalent. , A , A⇒ B B

.

Question 1.13. x²≥ 0 ⇒ x > 0 數 x, x²≥ 0 ⇒

x≤ 0 數 x. ?

¬(P ⇒ Q) logically equivalent, P⇒ Q.

P⇒ Q P Q . Q , P

. Q , P . Q∨ ¬P statement

form. truth table

P Q ¬P Q ∨ ¬P

T T F T

T F F F

F T T T

F F T T

(P⇒ Q) ∼ (Q ∨ ¬P). (1.10)

(Q∨ ¬P) ∼ ((¬P) ∨ Q) ¬(¬Q) ∼ Q, (P⇒ Q) ∼ ((¬P) ∨ ¬(¬Q)), (1.10) ((¬P) ∨ ¬(¬Q)) ∼ ((¬Q) ⇒ (¬P)),

(P⇒ Q) ∼ ((¬Q) ⇒ (¬P)). (1.11)

P⇒ Q , Q P , .

(1.10), ¬(P ⇒ Q) ∼ ¬(Q ∨ ¬P). DeMorgan’s laws

¬(Q ∨ ¬P) ∼ ((¬Q) ∧ ¬(¬P))

¬(P ⇒ Q) ∼ (P ∧ (¬Q)). (1.12)

(1.10), (1.11), (1.12) “ P Q” 論 logical

equivalences.

(1.10) , statement form logical equivalence

¬,∧,∨ . P⇔ Q ,

(P⇔ Q) ∼ (Q ∨ (¬P)) ∧ (P ∨ (¬Q)). (1.13)

∧,∨ ( (1.3))

(P⇔ Q) ∼ (P ∧ Q) ∨ ((¬P) ∧ (¬Q)). (1.14)

DeMorgan’s laws, (1.7), ∧,∨ ( (1.1),(1.2),

(1.3)), 導 statement form not logical equivalence. (1.13) not

¬(P ⇔ Q) ∼ ((¬Q) ∧ P) ∨ ((¬P) ∧ Q).

, (1.14) Q ¬Q ,

¬(P ⇔ Q) ∼ (P ⇔ ¬Q).

A statement form , ¬A A , A⇔ ¬A truth

table , A⇔ ¬A contradiction. , B statement form

A⇔ B contradiction, A B , B∼ ¬A.

Proposition 1.2.2 .

Proposition 1.3.1. A, B statement forms. ¬A B logically equivalent A⇔ B contradiction.

1.4. Quantifiers 11

1.4. Quantifiers

statement connective not

, statement form . statement,

, . 數學 statement quantifier

( ) , . quantifiers,

.

數學 quantifiers :

• “for all”, “for every” ( ), ∀ .

• “there exists”, “there is” ( , ), ∃ .

• “there is a unique” ( ), ∃! .

∃! , 論 , ∀ ∃.

, 論 quantifiers .

數 數 , 數 數

. quantifiers , .

數. ∀x ∃x, for all x in

R there exists an x inR, 數 .

: ∀x, x²≥ 0. 數 x x²≥ 0.

statement , 數 x , . statement

“∀x, P(x)”. P(x) x ( P(x) x²≥ 0).

x P(x) . statement x ,

. ∀x, x²> 0 (x = 0 ).

, “∃x, P(x)” , x P(x) . statement ,

x P(x) . , ,

, ( there exists ). ∀x, x²> 0

, ∃x, x²> 0 ( x = 1, ).

∀ ∃ , “∀x, P(x)” , “∃x, P(x)” (

x ). . x P(x), x

P(x). ∀ ∃ . “∀x, P(x)” ,

? 前 “∀x, P(x)” x P(x) ,

, x P(x) . ∃x, ¬P(x). 前 ∀x,

x²> 0 , ∃x, x²≤ 0.

學 “∀x, P(x)” “∀x, ¬P(x)”. “∀x, ¬P(x)”

“∀x, P(x)” . “∀x, P(x)” , “∀x, ¬P(x)” .

“∀x, P(x)” “∀x, ¬P(x)”. ∀x, x²> 0 , ∀x, x²≤ 0 ,

∃x, x²≤ 0 . , . 言 logical equivalence

¬(∀x,P(x)) ∼ (∃x,¬P(x)). (1.15)

“∃x, P(x)”, x P(x) . x

P(x), ∀x, ¬P(x). , 學 “∃x, P(x)” “∃x,

¬P(x)”. , x P(x) x P(x).

“∃x, ¬P(x)” “∃x, P(x)”. 言 logical equivalence

¬(∃x,P(x)) ∼ (∀x,¬P(x)). (1.16)

Question 1.14. (1.15) logical equivalence 導 (1.16).

Quantifier 數 , 數 ,

數 數 . 數 , “∀x,∃y,P(x,y)”

statement, P(x, y) x, y . , 數 f (x) x = a

l ( lim_x→af (x) = l) “∀ε > 0,∃δ > 0,0 < |x −a| < δ ⇒ | f (x)−l| < ε”

數 . statement.

(1)∀x,∃y,P(x,y) (2)∃x,∀y,P(x,y) (3)∀x,∀y,P(x,y) (4)∃x,∃y,P(x,y).

(1) : x y P(x, y) . x ,

y, y , x . ∀x,∃y,x + y = 0

statement . x, y x + y = 0. y x

, y =−x. x = 1 y =−1, x = 2 y =−2. x, y ,

.

(2) : x y P(x, y). x ,

y, x , y . ∃x,∀y,x + y = y

statement . x y x + y = y. x

, x = 0. (1) ∀x,∃y,x + y = 0 statement

, ∀x ∃y ∃y,∀x,x + y = 0 statement .

y x x + y = 0. , ,

“∀x,∃y,P(x,y)” “∃y,∀x,P(x,y)” ∀x ∃y ,

.

Question 1.15. ∃x,∀y,x + y = y statement , ∀y,∃x,x + y = y,

? ∀x,∃y,x + y = y ∃y,∀x,x + y = y, ?

Question 1.16. f (x, y), g(x, y) 數 . “∀x,∃y, f (x,y) = 0”

“∃y,∀x,g(x,y) = 0” . f (x, y) = 0 g(x, y) = 0

, x = 101 ?

(3) (4) . (3) x, y P(x, y)

. , (x, y) P(x, y) ,

1.4. Quantifiers 13

∀x ∀y statement. (4) x y

P(x, y). , (x, y) P(x, y) .

∃x ∃y statement. x = 3 , y = 7

P(3, 7) , y = 7 , x = 3 P(x, y) . 言

(3), (4) 數 quantifier , x, y . (3) ∀x,y,

P(x, y), (4) ∃x,y, P(x,y).

數 statement quantifier . (1)

, “∀x,∃y,P(x,y)”. , “∃y,P(x,y)” H(x) .

statement ∀x,H(x). (1.15), ∃x,¬H(x).

(1.16) ¬H(x) ∼ (∀y,¬P(x,y)),

¬(∀x,∃y,P(x,y)) ∼ (∃x,∀y,¬P(x,y)).

¬(∃x,∀y,P(x,y)) ∼ (∀x,∃y,¬P(x,y))

¬(∀x,∀y,P(x,y)) ∼ (∃x,∃y,¬P(x,y))

¬(∃x,∃y,P(x,y)) ∼ (∀x,∀y,¬P(x,y)).

前 , 數 f (x) limx→af (x) = l

∃ε > 0,∀δ > 0,¬(0 < |x − a| < δ ⇒ | f (x) − l| < ε).

(1.12)

¬(0 < |x − a| <δ ⇒ | f (x) − l| < ε) ∼ ((0 < |x − a| < δ) ∧ (| f (x) − l| ≥ ε)).

lim_x→af (x) = l

∃ε > 0,∀δ > 0,(0 < |x − a| < δ) ∧ (| f (x) − l| ≥ ε).

, ∀ ∃ . , ∀x.

x≥ 3 ⇒ x²≥ 9, statement ∀x,x ≥ 3 ⇒ x²≥ 9. ,

statement 數 x . 數 x, x≥ 3,

x²≥ 9. x < 3, x≥ 3 前 , x≥ 3 ⇒ x²≥ 9

. ∀x,x ≥ 3 ⇒ x²≥ 9 ( P P⇒ Q

, 學 ). ∃x , ∀x ∃x

. 言 , x P(x) Q(x) statement ,

∀x,P(x) ⇒ Q(x) , ∀x P(x)⇒ Q(x) . x

P(x) Q(x) , ∃x. , ∃x ,

∃x,P(x) ⇒ Q(x) , ∃x,P(x) ∧ Q(x) . , ∃x,P(x) ⇒ Q(x)

x P(x)⇒ Q(x) , P(x) x P(x)⇒ Q(x) .

x P(x), ∃x,P(x) ⇒ Q(x) , x

P(x) Q(x) . 3 數 x, x²= 10

∃x,(x > 3) ∧ (x²= 10) ( x =√

10), ∃x,(x > 3) ⇒ (x²= 10) ( x = 2 ).

Question 1.17. f (x, y) 數 . 數 a > 0 f (a, y) = 0

statement, 數學 ? statement .

Chapter 2

Methods of Proof

學 , 學 . .

, ,

. 2.1. IF-Then

數學 P⇒ Q statement. statement,

direct method, contrapositive method contradiction method .

2.1.1. Direct Method. direct method , P

Q . ( , P , Q ). P⇒ Q

, P , . .

Example 2.1.1. p, a, b 數. if p| a and p | b, then p | a + b.

Proof. p| a, p | b, 數 m, n a = pm, b = pn.

a + b = pm + pn = p(m + n).

m + n 數, p| a + b. 

, .

P⇒ Q, P⇒ R R⇒ Q, P⇒ Q .

P⇒ R, P R , R⇒ Q R Q ,

P Q , P⇒ Q. .

Example 2.1.2. a 數 a̸= 1. x, y 數 a^x= a^y, x = y.

Proof. 數 y, a^y̸= 0, a^x= a^y, a^y a^x−y= 1.

數 z, a^z= 1, z = 0, x− y = 0, x = y. 

15

, (a^x= a^y)⇒ (a^x−y= 1), (a^x−y= 1)⇒ (x = y)

(a^x= a^y)⇒ (x = y). , a 數 a̸= 1,

a^z= 1, z = 0. direct method , contradiction

method .

direct method , P ,

Q. proof in cases. .

Example 2.1.3. x 數. if x²− 3x + 2 < 0, then 1 < x < 2.

Proof. x²− 3x + 2 = (x − 1)(x − 2) < 0, 2 , (1) (x− 1) < 0 and (x − 2) > 0;

(2) (x− 1) > 0 and (x − 2) < 0.

(1) x < 1 x > 2. 數 x x < 1 x > 2, (1)

, (2), x > 1 x < 2. 1 < x < 2. 

, , 學 (1), (2)

x²−3x+2 < 0 ? . : x x²−3x+2 < 0,

x (1) (2). (1) (2) , x x²−3x+2 < 0,

x (2). x (2) x²− 3x + 2 < 0, (1),

if 1 < x < 2, then x²− 3x + 2 < 0, if x²− 3x + 2 < 0, then 1 < x < 2. . Question 2.1. x 數.

(1) “If x²− 3x + 2 < 0, then 0 < x < 3.” “If x²− 3x + 2 < 0, then 1.3 < x < 1.7.”

statements ?

(2) “If 0 < x < 3, then x²− 3x + 2 < 0.” “If 1.3 < x < 1.7, then x²− 3x + 2 < 0.”

statements ?

2.1.2. Contrapositive Method. (¬Q) ⇒ (¬P) P⇒ Q statement con- trapositive statement. , P⇒ Q (¬Q) ⇒ (¬P) logically equivalent

( (1.11)). P⇒ Q (¬Q) ⇒ (¬P) . ,

(¬Q) ⇒ (¬P) P⇒ Q .

P⇒ Q , P , ¬Q

contrapositive method. (¬Q) ⇒ (¬P).

. , 導

, statement , contrapositive statement .

contrapositive method . .

Example 2.1.4. x, y 數. if x̸= y, then x³̸= y³.

2.1. IF-Then 17

, f (x) = x³ , .

. contrapositive method , ¬(x³̸= y³) ( x³= y³), ¬(x ̸= y) ( x = y).

Proof. contrapositive method, x³= y³, 0 = x³− y³= (x− y)(x²+ xy + y²).

x²+ xy + y²= (x +1 2y)²+3

4y².

x²+ xy + y²> 0, x = 0 and y = 0 ( x = y). (x− y)(x²+ xy + y²) = 0

x− y = 0, x = y. if x̸= y, then x³̸= y³. 

statement x , x ,

contrapositive method . .

Example 2.1.5. x 數. if x² is even ( 數), then x is even.

Proof. contrapositive method, x 數, x² 數. x 數, x

x = 2n + 1, n 數. x²= (2n + 1)²= 4n²+ 4n + 1 = 2(2n²+ 2n) + 1

數. 

Question 2.2. x, y 數. contrapositive method if x + y is even, then x and y are both even or odd ( 數).

proof in cases , contrapositive method . Example

2.1.3 contrapositive method . , ¬(1 < x < 2) ( x≥ 2 or x≤ 1) ¬(x²− 3x + 2 < 0) ( x²− 3x + 2 ≥ 0). .

Example 2.1.6. x, y, a 數. if xy = a, then x≤√

a or y≤√ a.

Proof. contrapositive method, ¬((x ≤√

a)∨ (y ≤√

a)) ( x >√

a and y >√ a)

¬(xy = a) ( xy ̸= a). x, y, a , x >√

a and y >√

a xy > (√

a)²= a,

xy̸= a. 

Question 2.3. x, y, a 數. if xy = a, then x≥√

a or y≥√

a ?

, Example 2.1.6, ?

2.1.3. Contradiction Method. (1.10) P⇒ Q Q∨ ¬P logically

equivalent. , Q∨ ¬P P⇒ Q. Q∨ ¬P “ ”

, 前 proof in cases, . ¬(P ⇒ Q),

(¬Q) ∧ P logically equivalent ( (1.12)). (¬Q) ∧ P

, P⇒ Q . contradiction method ( ).

Contradiction method , (¬Q) P , 導

statement . (¬Q) P , P⇒ Q.

P ¬Q , 導,

direct method P , contrapositive method ¬Q . ,

direct method contrapositive method 導 ( P 導 Q

¬Q 導 ¬P), 導 “ ” .

P⇒ Q , P ¬Q

, P ¬Q contradiction method.

.

Example 2.1.7. r 數, if r²= 2, then r is irrational ( 數).

數 數, direct method.

contrapositive method r 數, r²̸= 2 . 前

導 , contradiction method .

Proof. , r 數 r²= r, . r 數,

r r = (m/n), m, n 數. m, n 數, 2,

, m, n 數. r²= 2, m²= 2n², Example

2.1.5 m 數. m m = 2m^′, m^′ 數. 4m^′2= 2n²,

n²= 2m^′2. Example 2.1.5 n 數. m, n 數

. if r²= 2, then r is irrational. 

Example 2.1.7 , r = (m/n) 2 ,

. contradiction method “ ”.

Example 2.1.2 , , a̸= 1 數 , z

數 a^z= 1, z = 0. contradiction method statement.

Example 2.1.8. a̸= 1 數. z 數 a^z= 1, z = 0.

, z̸= 0 a^z= 1 . ?

statement a = 1 前 ( ), a̸= 1.

Proof. contradiction method, z̸= 0 a^z= 1. z̸= 0, 1/z , (a^z)^1/z= a a^z= 1,

a = (a^z)^1/z= 1^1/z= 1.

a̸= 1 , a^z= 1, z = 0. 

2.1.4. If and Only If . P⇔ Q P⇒ Q Q⇒ P.

, 導 P⇒ Q ,

P⇔ Q. , ,

, P⇔ Q. , 導

. P⇔ Q P⇒ Q Q⇒ P

.

2.2. Existence and Uniqueness 19

a 數 “a² 數 ⇔ a 數”, ⇐ .

a = 2n, n 數, a²= (2n)²= 4n² 數. 導 ,

. a² 數, a² 4n² ? ⇒

導 ( Example 2.1.5 ).

(P⇒ Q) ∼ ((¬Q) ⇒ (¬P)) (Q⇒ P) ∼ ((¬P) ⇒ (¬Q)) P⇔ Q (¬P) ⇔

(¬Q) logically equivalent. 學 P⇔ Q , (¬P) ⇔

(¬Q). , . a, b 數,

“ab is even⇔ a is even or b is even” “ab is odd⇔ a and b are odd”

. , , .

學 (¬Q) ⇒ (¬P) P⇒ Q. ,

, . statement ,

前 . , .

, ?

數學 statement:

The following are equivalent. (1) P; (2) Q; (3) R.

( P, Q, R , ).

P⇔ Q, Q ⇔ R and R ⇔ P.

, “ ” ⇒ .

P⇒ Q, Q ⇒ R and R ⇒ P

. Q⇒ P , Q⇒ R R⇒ P , R⇒ Q , R⇒ P

P⇒ Q . P⇒ R , P⇒ Q Q⇒ R . ,

,

R⇒ Q, Q ⇒ P and P ⇒ R.

, P⇔ R ,

P⇔ Q and Q ⇔ R

. P⇒ R , P⇒ Q Q⇒ R , R⇒ P , R⇒ Q

Q⇒ P . , , 導 , statement

. 前 statement 導 statement, .

2.2. Existence and Uniqueness

Existence , uniqueness . 數

學 . , existence uniqueness .

. , . ,

論 existence uniqueness .

2.2.1. Existence. existence . constructive

method . nonconstructive method

論 導 , .

數 x 6x²−x−1 = 0. 6x²−x−1 (2x−1)(3x+1)

x = 1/2 ( x =−1/3) 數 6x²− x − 1 = 0. construct

method. 數 f (x) = 6x²− x − 1, f (0) =−1 < 0 f (1) = 4 > 0.

數 數 數 , f (x) = 0 0 < x < 1

, . , x

6x²− x − 1 = 0, nonconstructive method. .

Example 2.2.1. there exists irrational numbers a, b such that a^b is rational.

Proof. Constructive Method: a =√

2 b = log₂9. a 數.

b 數. m, n 數 log₂3 = n/m.

2^m= 3ⁿ, 3 數 數, . b = log₂9 數.

a^b= 2^12log29= 2^log23= 3,

數 a, b a^b 數.

Nonconstructive Method: a^′=√

2 b^′=√

2. a^′, b^′ 數.

c = (a^′)^b′. c 數, a =√

2, b =√

2 . c 數, a = c, b =√

2,

a^b= (√ 2

√2

)

√2=√ 2²= 2.

數 a, b a^b 數. 

nonconstructive method √

2

√2

數.

a = b =√

2 a =√

2

√2

, b =√

2 ,

, nonconstructive method. √

2

√2

數, a =√

2

√2

, b =√

2 constructive method. √

2

√2

數 (

). , , nonconstructive

method .

Question 2.4. , construct method there exists

irrational numbers a, b such that a^b is rational.

constructive method ,

, .

, 學 導 ,

. 學 , . 導 ,

, 導 “ ”. , 導 “ ”

, , , , (

2.2. Existence and Uniqueness 21

P Q ). ,

, . .

Example 2.2.2. 數 x √

3− 2x = x − 2?

, 3− 2x = x²− 4x + 4, x²− 2x + 1 = 0. x = 1.

, x = 1. x = 1 , 數

. x = 1 , 1 =−1 . 數 x √

3− 2x = x − 2.

nonconstructive method , . 前

數 數 x 6x²− x − 1 = 0.

, .

, pigeonhole principle ( ). “Dirichlet’s drawer principle”, pigeonhole principle .

Theorem 2.2.3 (Pigeonhole Principle). n 數. n n

. , .

Proof. , constructive method. , statement

,

n n , n

. . 

pigeonhole principle . ,

, . 6 數, 數 5

數 . 6 數 6 , 0 4 . 5

數 0 0 , 數 1 1 , . 數 6

數 5, , . 數

5 數 .

. 16 數, 4

數 5 數. 數 5 數 0, 1, 2, 3, 4 3 ,

數 15 , 16 數 .

Theorem 2.2.3 , .

Theorem 2.2.4. k, n 數. n kn .

, k .

Question 2.5. Theorem 2.2.4

, 數 數 ( ).

數 數 .

數 數 . ,

.

2.2.2. Uniqueness. 前 .

. , ,

. Example 2.2.2 x x = 1 ,

. , .

. 前 ,

. ,

. R² .

Example 2.2.5. R² −→

O R² −→

− V

→V +−→ O =−→

V , −→

O .

(1) : −→

O = (x, y), −→

V = (a, b)∈ R². −→

O −→

V +−→ O =−→

V , (a, b) + (x, y) = (a + x, b + y) = (a, b). a + x = a, b + y = b, x = 0, y = 0. −→

O , (0, 0).

(2) : −→

O ,−→

Q ∈ R² −→ O ̸=−→

Q −→

V ∈ R²,

−

→V +−→ O =−→

V (2.1)

−

→V +−→ Q =−→

V (2.2)

−

→Q =−→

V (2.1) −→

Q +−→ O =−→

Q . −→

O =−→

V (2.2)

−

→O +−→ Q =−→

O . −→

Q +−→ O =−→

O +−→ Q , −→

Q =−→

O . −→

O ̸=−→

Q ,

.

Question 2.6. −→

V ∈ R² : R² −→

− W

→V +−→ W =−→

O , −→

W .

Example 2.2.5 , −→

O , −→

O = (0, 0).

, −→

O = (0, 0) , . .

. , .

數學 , .

.

Example 2.2.6. 數 r , r³= 3, 數 r .

Proof. Example 2.1.4 , x, y 數 x̸= y, x³̸= y³.

r∈ R , r³= 3 s̸= r 數 s³= 3, Example 2.1.4

3 = s³̸= r³= 3. 數 s s³= 3. 

Example 2.2.6 x̸= y , x, y ,

. . , Example 2.2.6

數 r r³= 3, . ,

數 ( 數 f (x) = x³ ) .

3^1/3 r³= 3 數 r.

2.3. Mathematical Induction 23

2.3. Mathematical Induction

“數學 ”. 數學 數

axiom ( ), 論 . 數學

. 數學 , , ,

.

數學 論 axiom ( ). 數學

前, well-ordering principle. ,

前 . well-ordering principle

. , .

well-ordering “ ” ,

數 , .

principle ( ) ! , 數 1 , .

. ,

, .

, 數學 .

Theorem 2.3.1 (Mathematical Induction). statement (1) P(1)

(2) k∈ N P(k) , P(k + 1)

數 n, P(n) .

Proof. 數 n P(n) , .

(1), (2) 數 n, P(n) .

數 n, P(n) , 數 n, P(n) .

P(n) 數 n . , well-ordering

principle , 數 m P(m) . (1) P(1) ,

m̸= 1. m 1 數. m− 1 數 m− 1 < m, m

P(m) 數 P(m− 1) . (2) , P(m− 1) ,

P((m− 1) + 1) = P(m) . P(m) . 數 n,

P(n) , 數 n, P(n) . 

數學 , (1) P(1) , (2) k = 1 ,

P(1) P(2) . k = 2 , P(2) P(3) .

. P(1) . (2) P(k) P(k + 1)

. P(k) , P(k) ,

P(k) P(k + 1) . P(k) P(k + 1)

, 數學 . f (x) = x²+ x + 41.

x = 1 f (1) = 43 數. x = 2, f (2) = 47 數. f (3) = 53, f (4) = 61,

f (5) = 71 數. x 數 n f (n) 數.

x = 39 數. , f (k)

數 f (k + 1) 數, 數學 . , x = 40 ,

f (40) = 40²+ 40 + 41 = 40(40 + 1) + 41 41 , 數.

數學 .

Example 2.3.2. a, b 數, 數學 : 數 n

aⁿ− bⁿ a− b 數.

Proof. aⁿ− bⁿ (a− b)(aⁿ⁻¹+ aⁿ⁻²b +··· + bⁿ⁻¹), aⁿ− bⁿ a− b

數. 數學 . n = 1 a− b.

a− b 數, . a^k− b^k a− b 數, a^k+1− b^k+1

a− b 數. a^k+1− b^k+1 a^k− b^k . a^k+1− b^k+1

a^k− b^k ,

a^k+1− b^k+1= aa^k− bb^k= aa^k− ab^k+ ab^k− bb^k= a(a^k− b^k) + (a− b)b^k.

a^k− b^k a− b 數, a^k− b^k (a− b)m, m 數.

a^k+1− b^k+1= a(a− b)m + (a − b)b^k= (a− b)(am + b^k). a^k+1− b^k+1 a− b 數.

數學 , 數 n, aⁿ− bⁿ a− b 數 

, 數學 , . P

Q , 前 2.1 . P(k)

P(k + 1) , P(1), P(2) , P(2), P(3) , . P(k)

P(k + 1) . , P(k) P(k + 1),

. .

Example 2.3.3. 數學 n 數 . 論

, 論 .

: 數, .

: k 數 , k + 1 數 . k + 1

數, 數 a, k 數 . k 數

b. 數, 數 a . k 數

, a = b. 數學 , n 數 .

. 數 k− 1 數 .

k = 1 , , 數 a 數 ,

a = b. 論 k≥ 2 , P(k) P(k + 1) , k = 1

P(k) P(k + 1) . 數學 數 k,

P(k) , P(k + 1) . 論 .

2.3. Mathematical Induction 25

n = 1 , n≥ 5 , 2ⁿ> n². 數學

1 , . , Theorem 2.3.1 論,

數學 .

Corollary 2.3.4 (Extended Mathematical Induction). m 數. state- ment

(1) P(m)

(2) k≥ m 數 P(k) , P(k + 1)

m 數 n P(n) .

Proof. Q(n) = P(m + n− 1). P(m) , Q(1) = P(m) . k∈ N

Q(k) , m + k− 1 ≥ m 數, P(m + k− 1) = Q(k) , (2)

P(m + k) = Q(k + 1) . , k∈ N Q(k) , Q(k + 1) . Theorem

2.3.1 n∈ N, Q(n) = P(m + n − 1) . n m 數 ,

P(n) . 

“extended mathematical induction” Corollary , Theo-

rem 2.3.1 . m = 1 Corollary 2.3.4 Theorem 2.3.1,

Theorem 2.3.1 Corollary 2.3.4 equivalent . Example 2.3.5. 數 n, n≥ 5 , 2ⁿ> n².

Proof. extended mathematical induction m = 5 P(n) 2ⁿ> n² . n = 5 , 2⁵= 32 > 25 = 5², P(5) . k≥ 5 數 2^k> k². 2^k+1= 2× 2^k,

2^k > k² 2^k+1 > 2k². (k + 1)²= k²+ 2k + 1 2k²> k²+ 2k + 1, 2^k+1> (k + 1)², P(k + 1) . 2k² > k²+ 2k + 1 k²− 2k > 1, k²− 2k = k(k − 2), k≥ 5 k²− 2k > 1. k≥ 5 數 P(k) , P(k + 1) , extended mathematical induction (Corollary 2.3.4)

5 數 n P(n), 2ⁿ> n² . 

數學 P(k) P(k + 1) .

數 , 前 . P(1)

P(2) , P(2) P(3) , P(1) , P(2)

前 , P(1) . P(3) P(4) , P(1), P(2)

, 數學 .

Corollary 2.3.6 (Strong Mathematical Induction). m 數. statement

(1) P(m)

(2) k≥ m 數 P(m), P(m + 1), . . . , P(k− 1),P(k) , P(k + 1)

m 數 n P(n) .

Proof. m 數 n, Q(n) = P(m)∧P(m+1)∧···∧P(n−1)∧P(n). P(m) , Q(m) = P(m) . k≥ m 數 Q(k) , P(m), . . . , P(k−1),P(k)

, (2) P(k + 1) . P(m), P(m + 1), . . . , P(k− 1),P(k) , Q(k + 1) = P(m)∧ P(m + 1) ∧ ··· ∧ P(k) ∧ P(k + 1) . , k≥ m 數

Q(k) , Q(k + 1) . Corollary 2.3.4 m 數 n

Q(n) . Q(n) P(m), P(m + 1), . . . , P(n− 1),P(n) , P(n)

, n m 數 , P(n) . 

Corollary 2.3.4 Corollary 2.3.6. P(k)⇒ P(k + 1) P(m), P(m + 1), . . . , P(k− 1),P(k) ⇒ P(k + 1) ,

Corollary 2.3.6 Corollary 2.3.4. strong mathematical induction extended mathematical induction .

Question 2.7. Corollary 2.3.6 Corollary 2.3.4.

前 extended mathematical induction mathematical induction

, 數學 .

. strong mathematical induction .

Example 2.3.7. 1 數 數 .

學 . n 1 數, n 數, n

數 . n 數, n n 1 數 ,

數 .

. , “ ”

( 數 ) . 數學 ,

. , k 數

, k + 1 數 , strong mathematical

induction .

Proof. n = 2 , 2 數, . k≥ 2 i = 2, 3, . . . , k, i

數 . k + 1 . k + 1 數 ,

k + 1 數 . k + 1 = ab, 1 < a, b≤ k. 前 a, b

數 , k + 1 = ab 數 . strong

mathematical induction 1 數 數 . 

數學 , P(k)

( P(m), P(m + 1), . . . , P(k) ) P(k + 1) . 導 k

. Example 2.3.3 , “ k 數

k + 1 數 ” k≥ 2 . k = 1 ,

2.3. Mathematical Induction 27

( Example 2.3.3 ).

, k , k ,

, . , 數學 ,

, .

Example 2.3.8. Fibonacci sequence{F0, F₁, F₂, . . .}, F₀= 0, F₁= 1 i≥ 2, Fi Fi= F_i−1+ F_i−2. Fn< 2ⁿ⁻², for n≥ 4.

, Fk+1 Fk Fk−1 , Fk Fk+1.

strong mathematical induction . F2= F1+ F0= 1, F3= F2+ F1= 1 + 1 = 2, n = 4 , F₄= F₃+ F₂ = 2 + 1 = 3, F₄= 3 < 2⁴⁻²= 4 . k≥ 4

4≤ i ≤ k, Fi < 2ⁱ⁻². Fk+1 = Fk+ Fk−1. Fk < 2^k−2 F_k−1< 2^(k−1)−2= 2^k−3 F_k+1< 2^(k+1)−2= 2^k−1. k = 4 , i = k− 1

4≤ i ≤ k, Fk−1< 2^k−3 ( Fk−1= F₃= 2 = 2⁴⁻³).

k = 4 , F_k+1= F₅= F4+ F3= 5 < 2⁵⁻²= 8, . Proof. F4= 3 < 2⁴⁻², F5= 5 < 2⁵⁻².

k≥ 5 i = 4, 5, . . . , k Fi< 2ⁱ⁻². 4≤ k − 1 ≤ k 4≤ k ≤ k, F_k< 2^k−2, F_k−1< 2^(k−1)−1= 2^k−3,

Fk+1= F_k+ F_k−1< 2^k−2+ 2^k−3= 2^k−3(2 + 1) < 4× 2^k−3= 2^k−1= 2^(k+1)−2.

數學 F_n< 2ⁿ⁻², for n≥ 4. 

, 20 數 4 5 數 .

n > 20, 數 l, m n = 4l + 5m. 學 1 = 5− 4,

k = 4l + 5m, k + 1 = 4l + 5m + (5− 4) = 4(l − 1) + 5(m + 1). ,

數 4 數 5 數 , 4 5 數

. , k = 4l + 5m, k + 4 = 4(l + 1) + 5m,

21, 22, 23, 24 4 5 數 , 數學

.

Example 2.3.9. n 數 n > 20, l, m 數 n = 4l + 5m.

前 , n≥ 0 數 , 21 + 4n, 22 + 4n, 23 + 4n, 24 + 4n

4l + 5m, l.m 數 . 20 數 .

21 = 4× 4 + 5 × 1, . k≥ 0 21 + 4k = 4l + 5m, l.m 數.

21 + 4(k + 1) = (21 + 4k) + 4 = 4l + 5m + 4 = 4(l + 1) + 5m, . n≥ 0 數 ,

21 + 4n 4l + 5m, l.m 數 . 22 = 4× 3 + 5 × 2,

23 = 4× 2 + 5 × 3 24 = 4× 1 + 5 ×4, 數學 . 數學

, strong mathematical induction.

Proof. 21 = 4× 4 + 5 × 1, 22 = 4 × 3 + 5 × 2, 23 = 4 × 2 + 5 × 3 24 = 4× 1 + 5 × 4

n = 21, 22, 23, 24 . k≥ 24 21≤ i ≤ k 數 i,

數 l, m i = 4l + 5m. k + 1 , k + 1 = (k− 3) + 4, i = k− 3 21≤ i ≤ k, 數 l, m k− 3 = 4l + 5m, k + 1 = 4l + 5m + 4 = 4(l + 1) + 5m.

數學 , n 20 數, l, m 數 n = 4l + 5m. 

Question 2.8. 數 4 數 5 數 .

數學 數學 . 數 ,

數 數學 . 數 ,

數 數學 . 數 , row

數 column 數 數學 .

Chapter 3

Set

論 數學 論 .

論. 論, .

3.1. Basic Definition

, .

數學 , .

, . ( set)

, , ( element).

set, A, B, S , set . 數學

. : N 數 , Z 數

,Q 數 , 數 R . x

S , x∈ S , x belongs to S ( x S). x

S , x̸∈ S .

數學 , set element element.

S, x, x∈ S . ,

, . S ={1,2,3}

3 , 1, 2 3. S , 1∈ S,

4̸∈ S. , set-builder notation

. {x : P(x)} , : x

x , : P(x) x P(x).

{x : P(x)} P(x) x .

前, , .

Definition 3.1.1. A, B . B element A element, B A

subset ( ), B is contained in ( ) A, B⊆ A. B⊆ A A⊆ B,

A, B equal, A = B. B⊆ A B̸= A, B A proper subset, B⊂ A.

29

subset proper set “⊆” “⊂”, , .

B⊆ A, B x, A , 數學

“x∈ B ⇒ x ∈ A”. A = B “x∈ B ⇔ x ∈ A”.

, , ,

. :

Example 3.1.2. A ={1,2,2}, B = {1,2,3}, C = {3,3,1,2}, D = {n ∈ N : 1 ≤ n ≤ 3}, E ={2,4}.

A 1, 2 , 1∈ B 2∈ B, A⊆ B. 3∈ B 3̸∈ A, B

A, A⊂ B. B = C.

x∈ B, x∈ N 1≤ x ≤ 3, x∈ D. B⊆ D. , x∈ D

x∈ N 1≤ x ≤ 3, x = 1, 2, 3 B , D⊆ B, B = D.

1∈ B 1̸∈ E, B E subset. , 4∈ E 4̸∈ B, E

B subset.

A, B sets, B A subset , B* A . B⊆ A

A* B, B⊂ A.

Question 3.1. P(x), Q(x) statement form. P ={x : P(x)} Q ={x : Q(x)}

:

(1) P⊆ Q P(x)⇒ Q(x).

(2) P = Q P(x)⇔ Q(x).

, . ,

subset , universal set (

). 論 數, R universal set.

x∈ R . universal set

, a, b 數 , universal set Q 論 ax + b = 0 . 論

ax²+ b = 0, universal set R 數 C . ,

, universal set .

universal set , 論 set universal set subset.

empty set ( ). ,

/0 . /0 ?

, “ x∈ /0 ” .

operations, /0

. universal set empty set, .

Proposition 3.1.3. X universal set A set. A⊆ X /0⊆ A.

3.1. Basic Definition 31

Proof. X universal set, A X subset, A⊆ X. , /0⊆ A

x∈ /0 x∈ A. x∈ /0 , P P⇒ Q

“x∈ /0 ⇒ x ∈ A” /0⊆ A. 

Question 3.2. Question X universal set. universal set ?

empty set ?

數學 , ,

論 subset .

Proposition 3.1.4. A, B,C sets, .

(1) A⊆ A.

(2) A⊆ B B⊆ C, A⊆ C.

Proof. (1) x∈ A, x∈ A, A⊆ A.

(2) x∈ A, A⊆ B x∈ B. B⊆ C x∈ C. 言 , x∈ A x∈ C,

A⊆ C. 

Question 3.3. Proposition 3.1.4 A = B B = C, A = C.

Question 3.4. A, B,C sets, ?

(1) A⊂ B B⊆ C, A⊆ C.

(2) A⊆ B B⊆ C, A⊂ C.

(3) A⊂ B B⊂ C, A⊆ C.

(4) A⊂ B B⊆ C, A⊂ C.

, A = B A⊆ B B⊆ A .

, 數 , 學

. .

Example 3.1.5. A ={(x,y) ∈ R²: x²− x = y = 2} B ={(2,2),(−1,2)}. A = B.

Proof. (x, y)∈ A, x²− x = 2 y = 2, x = 2 x =−1 y = 2.

(x, y)∈ A, (x, y) = (2, 2) (x, y) = (−1,2). x∈ B, A⊆ B. (x, y)∈ B, (x, y) = (2, 2) (x, y) = (−1,2) x²−x = y = 2, (x, y)∈ A. B⊆ A,

A = B 

Question 3.5. A ={x ∈ R :√

x = x− 2}, B = {1}, C = {4}, D = {1,4}. A, B,C, D .

Venn diagrams . ,

universal set, ( )

set. X set A.

..

A .

X

Venn diagrams A, B .

A B.... X

. A

. B

.

X

. A

.

B .

X

A, B , A, B

, A⊆ B.

Venn diagrams ,

. Proposition 3.1.4 (2) A⊆ B B⊆ C, A⊆ C

.

A...

B .

C .

X

Venn diagrams , .

Question 3.6. A, B,C sets. A⊆ B. B C ,

Venn diagrams. A C ? B C ,

Venn diagrams, A C ? , A,C

Venn diagrams, B,C .

“∈” ( ) “⊆” ( ) . “∈”

, “⊆” . A⊆ B B⊆ C A⊆ C

, A∈ B B∈ C A∈ C.

A ={1}, B ={ {1}}

, C ={{

{1}}}

.

A∈ B B∈ C, A̸∈ C.

3.2. Set Operations 33

3.2. Set Operations

set operation, .

operations, intersection, union set difference, set

operations .

3.2.1. Intersection and Union. intersection union.

Definition 3.2.1. A, B sets. A∩B = {x : x ∈ A and x ∈ B} the intersec- tion of A and B (A B ). A∪ B = {x : x ∈ A or x ∈ B} the union of A and

B (A B ).

A∩ B A, B , A∪ B A, B

. .

Example 3.2.2. A ={1,2,3}, B = {2,4,6}. 2 A B,

A∩ B = {2}. 1, 3 B A, A B 1, 3

A∪ B. 4, 6 A∪ B. 2 A B A

B , 2 A∪ B. 數 A B ,

A∪ B = {1,2,3,4,6}.

. A∩ B = {2} ⊆ A =

{1,2,3} B ={2,4,6} ⊆ A ∪ B = {1,2,3,4,6}. , x∈ A ∩ B, x∈ A x∈ B,

x A x B,

(A∩ B) ⊆ A and (A ∩ B) ⊆ B. (3.1)

A∩ B , A, B disjoint. ,

A, B disjoint . x∈ A, x A B, x∈ A∪B

.

A⊆ (A ∪ B) and B ⊆ (A ∪ B) (3.2)

Question 3.7. (A∩ A) = A (A∪ A) = A.

, .

Proposition 3.2.3. A, B,C, D sets A⊆ B C⊆ D.

(A∩C) ⊆ (B ∩ D) and (A ∪C) ⊆ (B ∪ D).

Proof. A⊆ B, x∈ A x∈ B. C⊆ D, x∈ C x∈ D. x∈ A ∩C, x∈ A x∈ C. x∈ B x∈ D. (A∩C) ⊆ (B ∩ D). , x∈ A ∪C,

x∈ A x∈ C. x∈ A x∈ B, x∈ C x∈ D. x∈ A ∪C

x∈ B ∪ D. (A∪C) ⊆ (B ∪ D). 

, A⊆ B A⊆ D , C = A Proposition 3.2.3 (A∩ A) ⊆ (B ∩ D). (A∩ A) = A, A⊆ (B ∩ D). , A⊆ B C⊆ B ,

(A∪C) ⊆ (B ∪ B). (B∪ B) = B, (A∪C) ⊆ B. .

Proposition 3.2.3 導 , corollary ( ) .

Corollary 3.2.4. A, B,C, D, E sets.

(1) A⊆ B A⊆ C, A⊆ (B ∩C).

(2) D⊆ A E⊆ A, (D∪ E) ⊆ A.

Question 3.8. Corollary 3.2.4, 導 Proposition 3.2.3.

. .

Proposition 3.2.5. A, B sets. equivalent.

(1) A⊆ B.

(2) (A∩ B) = A.

(3) (A∪ B) = B.

Proof. (1)⇔ (2) (1)⇔ (3).

(1)⇔ (2): A⊆ B, (A∩ B) = A. (3.1) (A∩ B) ⊆ A,

A⊆ (A ∩ B). A⊆ A A⊆ B, Corollary 3.2.4 A⊆ (A ∩ B).

(1)⇒ (2). , (3.1) (A∩ B) ⊆ B. A = (A∩ B)

A⊆ B, (2)⇒ (1).

(1)⇔ (3): A⊆ B, (A∪ B) = B. (3.2) B⊆ (A ∪ B),

(A∪ B) ⊆ B. A⊆ B B⊆ B, Corollary 3.2.4, (A∪ B) ⊆ B.

(1)⇒ (3). , (3.2) A⊆ (A ∪ B). (A∪ B) = B

A⊆ B, (3)⇒ (1). 

Definition 3.2.1 “ ” “and” , “ ” “or” .

: (1) A∩ B = B ∩ A.

(2) A∪ B = B ∪ A.

(3) (A∩ B) ∩C = A ∩ (B ∩C).

(4) (A∪ B) ∪C = A ∪ (B ∪C).

(3) , ,

A∩ B ∩C. (4), ,

A∪ B ∪C.

3.2. Set Operations 35

∧,∨ ,

((P∧ Q) ∨ R) ∼ ((P ∨ R) ∧ (Q ∨ R)), ((P ∨ Q) ∧ R) ∼ ((P ∧ R) ∨ (Q ∧ R)), .

Proposition 3.2.6. A, B,C sets,

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

Proof. (A∩ B) ⊆ A C⊆ C Proposition 3.2.3 ((A∩ B) ∪C) ⊆ (A ∪C).

((A∩B)∪C) ⊆ (B∪C). Corollary 3.2.4 ((A∩B)∪C) ⊆ (A∪C)∩(B∪C).

x∈ (A∪C)∩(B∪C) x∈ A∪C x∈ B∪C. proof in cases, x∈ C

x̸∈ C . x∈ C, x∈ (A∩B)∪C. x̸∈ C, x∈ A∪C x∈ B∪C,

x∈ A x∈ B, x∈ A ∩ B. x∈ (A ∩ B) ∪C. ,

x∈ (A ∪C) ∩ (B ∪C) x∈ (A ∩ B) ∪C, ((A∪C) ∩ (B ∪C)) ⊆ (A ∩ B) ∪C.

((A∩ B) ∪C) = (A ∪C) ∩ (B ∪C).

((A∪ B) ∩ C) = (A ∩ C) ∪ (B ∩ C) , A⊆ (A ∪ B) C⊆ C Proposition 3.2.3 (A∩ C) ⊆ ((A ∪ B) ∩ C), (B∩ C) ⊆ ((A ∪ B) ∩ C).

Corollary 3.2.4 (A∩C)∪(B∩C) ⊆ ((A∪B)∩C). , x∈ (A∪B)∩C, x∈ A∪B

x∈C. x∈ A∪B, x∈ A x∈ B. x∈ A , x∈ C, x∈ A∩C.

x∈ (A∩C)∪(B∩C). , x∈ B , x∈ B∩C. x∈ (A∩C)∪(B∩C),

((A∪ B) ∩C) ⊆ (A ∩C) ∪ (B ∩C). ((A∪ B) ∩C) = (A ∩C) ∪ (B ∩C)  Question 3.9. Proposition 3.2.5 (1)⇒ (2) Proposition 3.2.6 Proposition 3.2.5 (2)⇒ (3).

3.2.2. Set Difference. set difference.

Definition 3.2.7. A, B sets, A\ B = {x : x ∈ A and x ̸∈ B}, the set difference of B in A (B A ). X universal set, A^c= X\ A = {x : x ̸∈ A}

the complement of A (A ).

A^c {x : x ∈ X and x ̸∈ A}, X universal set,

X , x∈ X x̸∈ A.

, . Q , Q^c= /0, R ,

Q^c 數 .

, A\B = A∩B^c, A\B B\A

. (A\ B)∩ (B \ A) = /0. A∩ A^c= /0 B∩ B^c= /0,

(A\ B) ∩ (B \ A) = (A ∩ B^c)∩ (B ∩ A^c) = (A∩ A^c)∩ (B ∩ B^c) = /0.

Example 3.2.8. X ={1,2,3,4,5,6},A = {1,2,3},B = {2,4,6}. 1, 3∈ A 1̸∈ B 3̸∈ B, 1, 3∈ A \ B. 2∈ A 2∈ B, 2̸∈ A \ B. A\ B = {1,3}.