第5章习题答æ&iexcl

第5章习题答æ¡

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Unformatted text preview: P151∼ P1531 11 €] 1 21 {-31 2} 1 41 {51 61 71 81 91 101 111 121 131 141 15} 21 €] 1 21 {x| x 1 1 41 {x|ñ x ˆª* µÑ 10< x< 20} } 1 61 {x| x=2k+11 k∈I} * ª 31 Ñ ˆ µñ * ª · ¥ I pˆ> é·ª*} 1 21 * ª 41I pˆ> · ¥ é·ª*} 1 6€] 1 21 1 1 {11 1 }1 ={∅ {1}1 {∅}1 {11 1 }} 1 4 1 1 1 {∅ {a} 1 {∅}} 1 ={∅ {∅} 1 {{a}} 1 {{∅}} 1 {∅ {a}} 1 {∅ {∅}}1 {{a}1 {∅}}1 {∅ {a}1 {∅}}} 71 “ 1 ”`Bµ 11 81 € A∈C 1 21 31 ñ 6ˆµ Ѫ 41 51 A∈C ˆµ ñ Ñ* ª A∈B 1 B∈C 1 A=G1 B=E1 C=F 31 51 ª µ ñ 8ˆ Ñ I1 1€ ª µ ñ1 ˆ Ñ 11 41 61 71 A1 B1ñ C ˆµ Ñ* ª A∈B1 B∈C1 A={a} 1 B={{a}} 1 C={{{a}}} 1 101 1 U={11 21 31 41 5}1 A={11 4}1 B={11 21 5}1 C={21 4}1 1 81 1 1 A1 1 1 1 1 A1 1 1 C1 1 C1 ={∅ {1}1 {4}1 {11 4}}− ∅ {2}1 {4}1 {21 4}} { ={{1}1 {11 4}} ª · 11˜* “å 1 11 1 21 A− B∩C1 =1 A− 1 1 1 A− 1 B C 1 A− 1 1 B=∅ B 11 A− B∩ C1 = A∩ ∼ B∩ C1 = A∩ B∪ ∼ C1 =1 A∩ ∼ B1 1 1 A∩ ∼ C1 =1 A− 1 1 1 A− 1 B C 1 21 A− 1 1 B B =1 A∩ ∼ B1 1 B = A∩ =∅ 121 1 11 1 21 A1 ª B1¥ C p*ˆ> · I é·ª} A− 1 1 1 A− 1 =A B C A− 1 1 1 A− 1 = ∅ B C 11 A− 1 1 1 A− 1 B C B∩ B1 = 1 A∩ ∼ B1 1 1 A∩ ∼ C1 = A∩ B∪ ∼ C1 = A∩ ∼ B∩ C1 = A− B∩ C1 1 1 1 È·ª “ä 21 A− 1 1 1 A− 1 =A 1 B C A∩ B∩C1 =∅ 1 “ 1È·ª ä A− B∩C1 =A A− 1 1 1 A− 1 =A− B∩C1 B C A− B∩ C1 =∅ 1 A− 1 1 1 A− 1 =∅ 1 B C 5.11 1 A⊆ B∩ C1 13. 1 A1 B 1 1 11 1 21 A− B=B1 A · 1 “ B Ȫ ä A− B=B− 1 A 1 · “ B Ȫ A, ä 11 A=B=φ A− B=B 1 B B ⇔ A− ⊆ B∧B⊆ A− B) B B) ⇔( A− ∩ B= A− ∧B∩ ( A− = B B ⇔ A ∩ ~ B ∩ B= A− ∧B ∩A ∩ ~ B = B B= ⇔ A− φ∧ B=φ ⇔ A ⊆ B∧B=φ ⇔ A=B=φ A− B=B B B B ⇔ A− ⊆ B∧ ⊆ A− B B= B ⇔A− − φ∧ B− A− 1 =φ B= ⇔ A− φ∧ B∩~A1 ∪ B∩ B1 =φ B= A= ⇔ A− φ∧ B− φ∧ B=φ B ⇔ A ⊆ B∧ ⊆ A∧B=φ ⇔ A=B=φ (2) A=B1 A− B=B− A B A BA ⇔A− ⊆ B− ∧ − ⊆ A-B B ⇔A∩ ~B⊆ B∩~A ∧ ∩ ~A⊆ A∩ ~B ⇔ A∩ ~B∩B∩~A=A∩ ~B1 ∧ B∩~A∩ A ∩~B=B∩~A1 ⇔A∩ ~B=φ∧B∩ ~A=φ B-A=φ ⇔A-B=φ ∧ B ⇔A⊆ B∧ ⊆ A ⇔A=B 15. 1 A={{φ }1 {{φ }}}1 1 41 1 1 1 1 A1 1 1 1 1 1 A1 =∪ {φ {{φ }}1 {{{φ }}}1 {{φ }1 {{φ }}} ={{φ }1 {{φ }}} =A 16. 1 B={{11 2}1 {21 3}1 {11 3}ª1· φ ç}X *“ 31 ∩ B1 1 41 1 1 ∩ B1 1 31 ∩ B1 =∩ {11 21 3}=1∩2∪ 3 1 41 1 1 ∩ B1 =∪φ =φ 17. “˜ ã ·ª* (A) x x∈A⇔{x}⊆ A ⇔{x}∈P1 A1 ⇔{x}∈P(B) ⇔{x}⊆ B ⇔x∈B 1 24.1 A=B A={01 1}1 B={1ª 1 “2}˜ *·ã 31 B× A1 2 1 1 (A)1 1 (B)1 (B) AB 1 21 A× B1 1 21 A× B={<0,1>,<0,2>,<1,1>,<1,2>} 1 31 B× A1 2 ={<1,0>,<1,1>,<2,0>,<2,1>}× {<1,0>,<1,1>,<2,0>,<2,1>} ={<<1,0>,<1,0>>,<<1,0>,<1,1>>,<<1,0>,<2,0>>,<<1,0>,<2,1>>,<<1,1 >,<1,0>>,<<1,1>,<1,1>>,<<1,1>,<2,0>>,<<1,1>,<2,1>>,<<2,0>,<1,0>>, <<2,0>,<1,1>>,<<2,0>,<2,0>>,<<2,0>,<2,1>>,<<2,1>,<1,0>>,<<2,1>,< 1,1>>,<<2,1>,<2,0>>,<<2,1>,<2,1>>} 26.1 A½1 * Ð B1K C P D ‰ ª´ I 1 1 21 A× 1 31 1 (A∩B)×(C∩D) 1 (A×C)∩(B×D) B∪ C1 =1 A× B1 ∪ A× C1 A×(B1 C)1 (A×B)1 (A×C) x, y y 1ª · “ 1( ì x, y ∈ ( A I B ) × ( C I D ) ⇔ x ∈ A IB ∧ y ∈ C ID ⇔ ( x ∈ A ∧ x ∈ B) ∧ ( y ∈C ∧ y ∈ D) ⇔ ( x ∈ A ∧ y ∈C ) ∧ ( x ∈ B ∧ y ∈ D) ⇔ x , y ∈ A × C ∧ x, y ∈ B × D ⇔ x , y ∈ ( A × C ) I( B × D ) (A∩B)×(C∩D)1 (A×C)∩(B×D) 1 21 1 x, y y x, y ∈ A × ( B U C ) ⇔ x ∈ A ∧ y ∈ B UC ⇔ x ∈ A ∧ ( y ∈ B ∨ y ∈C) ⇔ ( x ∈ A ∧ y ∈ B) ∨ ( x ∈ A ∧ y ∈C ) ⇔ ( x, y ∈ A × B ) ∨ ( x, y ∈ A × C ) ⇔ x, y ∈ ( A × B ) U ( A × C ) ∈ A× ( B UC ) = ( A × B ) U( A × C ) ∈31 1 x, y y x, y ∈ A × B1 − A × C1 ⇔ x, y ∈ A × B1 ∧ x, y ∉ ⇔ x, y ∈ A × B1 ∧ ¬ ⇔ ⇔ A × C1 x, y ∈ A × C1 x∈ A ∧ y∈ B1 ∧ ¬ x∈ A ∧ y∈ C1 x ∈ A ∧ y ∈ B1 ∧ ¬ x∈ A1 ∨ ¬ y∈ C1 ⇔ x∈ A ∧ y∈ B ∧ ¬ x∈ A1 y ∈ C1 ∨ x ∈ A ∧ y B ∧ ¬ y ∈ C1 ⇔ x∈ A ∧ y∈ B ∧ ¬ ⇔ x ∈ A ∧ y ∈ B− 1 C ⇔ x, y ∈ A×(B1 C) 1 27. “] ( ª ·* A×(B1 C)1 (A×B)1 (A×C) A ⊆ C B⊆ D x,y x,y ∈ A×B ⇔ x∈ A ∧ y∈B ⇒ x∈C ∧ y∈D ⇔ A×B ⊆ C×D ∈A ⊆ C B ⊆ D ( ) x,y ∈ C×D X O A×B ⊆ C×D 28.1 ªX ·*“\ A=φ B={a b } C={1 2 } D={x} ⊆ C×D A×B ⊆ C×D\ ª ·*“X A⊆C B⊆D A×B=φ A×B B⊆ D ...
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This note was uploaded on 02/06/2011 for the course CS 343 taught by Professor Zhoujunli during the Spring '08 term at BUPT.

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