数学 离散数学

离散数学:第一次作业

[leftmargin=1.5cm] (1) (2) (3) (4) (5) (6) (7) (8)。

本页目录7 节

题目 P97/3 #

确定下列个命题的真假性:

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  • (1) ∅⊆∅\varnothing \subseteq \varnothing;
  • (2) ∅∈∅\varnothing \in \varnothing;
  • (3) ∅⊆{∅}\varnothing \subseteq \{\varnothing\};
  • (4) ∅∈{∅}\varnothing \in \{\varnothing\};
  • (5) {a,b}⊆{a,b,c,{a,b,c}}\{a,b\} \subseteq \{a,b,c,\{a,b,c\}\};
  • (6) {a,b}∈{a,b,c,{a,b,c}}\{a,b\} \in \{a,b,c,\{a,b,c\}\};
  • (7) {a,b}⊆{a,b,{{a,b,c}}}\{a,b\} \subseteq \{a,b,\{\{a,b,c\}\}\};
  • (8) {a,b}∈{a,b,{{a,b,c}}}\{a,b\} \in \{a,b,\{\{a,b,c\}\}\};
答案

(1) 真 (2) 假 (3) 真 (4) 真 (5) 真 (6) 假 (7) 真 (8) 假

题目 P97/4 #

对任意集合 A,B,CA,B,C,确定下列命题的真假性:

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  • (1) 如果 A∉B∧B∉CA\not\in B \wedge B \not\in C,则 A∉CA\not\in C;
  • (2) 如果 A∈B∧B∉CA\in B \wedge B \not\in C,则 A∉CA\not\in C;
  • (3) 如果 A⊆B∧B∉CA\subseteq B \wedge B \not\in C,则 A∉CA\not\in C.
答案

(1) 假 (2) 假 (3) 假

题目 P97/5 #

对任意集合 A,B,CA,B,C,确定下列命题的真假性:

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  • (1) 如果 A∈B∧B⊆CA\in B \wedge B \subseteq C,则 A∈CA\in C;
  • (2) 如果 A∈B∧B⊆CA\in B \wedge B \subseteq C,则 A⊆CA\subseteq C;
  • (3) 如果 A⊆B∧B∈CA\subseteq B \wedge B \in C,则 A∈CA\in C.
答案

(1) 真 (2) 假 (3) 假

题目 P98/6 #

求下列集合的幂集:

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  • (1) {a,b,c}\{a,b,c\};
  • (2) {a,{b,c}}\{a,\{b,c\}\};
  • (3) {∅}\{\varnothing\};
  • (4) {∅,{∅}}\{\varnothing,\{\varnothing\}\}.
答案

(1) {∅,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}\{\varnothing,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},\{a,b,c\}\};

(2) {∅,{a},{{b,c}},{a,{b,c}}}\{\varnothing,\{a\},\{\{b,c\}\},\{a,\{b,c\}\}\};

(3) {∅,{∅}}\{\varnothing,\{\varnothing\}\};

(4) {∅,{∅},{{∅}},{∅,{∅}}}\{\varnothing,\{\varnothing\},\{\{\varnothing\}\},\{\varnothing,\{\varnothing\}\}\}.

题目 P98/8 #

设 A,B,CA,B,C 是集合, 证明:

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  • (1) (A\B)\C=A\(B∪C)(A \backslash B)\backslash C = A\backslash(B \cup C);
  • (2) (A\B)\C=(A\C)\(B\C)(A \backslash B)\backslash C = (A\backslash C)\backslash(B \backslash C);
  • (3) (A\B)\C=(A\C)\B(A \backslash B)\backslash C = (A \backslash C)\backslash B;
证明

\

我们记 D=A∪B∪CD=A\cup B\cup C 为全集.

(1) (A\B)\C=(A∩B′)∩C′=A∩(B′∩C′)=A∩(B∪C)′=A\(B∪C)(A\backslash B) \backslash C = (A\cap B')\cap C' = A \cap (B' \cap C') = A\cap(B \cup C)' = A\backslash (B \cup C);

(2) (A\B)\C=(A∩B′)∩C′=(A∩C′)∩(B′∪C)=(A\C)∩(B∩C′)′=(A\C)∩(B\C)′=(A\C)\(B\C)(A\backslash B) \backslash C = (A\cap B')\cap C' = (A \cap C') \cap (B' \cup C) = (A \backslash C)\cap(B \cap C')' \\= (A \backslash C)\cap (B \backslash C)' = (A \backslash C)\backslash (B \backslash C);

(3) (A\B)\C=(A∩B′)∩C′=(A∩C′)∩B′=(A\C)∩B′=(A\C)\B(A\backslash B) \backslash C = (A\cap B')\cap C' = (A \cap C') \cap B' = (A \backslash C)\cap B'= (A \backslash C)\backslash B.

题目 P98/9 #

设 A,BA,B 是集合 XX 的子集, 证明:

A⊆B⇔A′∪B=X⇔A∩B′=∅A \subseteq B \Leftrightarrow A' \cup B = X \Leftrightarrow A\cap B' = \varnothing

先证明 A⊆B⇔A′∪B=XA \subseteq B \Leftrightarrow A' \cup B = X

证明

"⇒""\Rightarrow": A′∪B=A′∪(A∪B)=A′∪A∪B=X∪B=XA'\cup B = A' \cup (A \cup B) = A'\cup A \cup B = X \cup B = X

"⇐""\Leftarrow" A′∪B=X⇒(X\A′)⊆B⇒A⊆BA' \cup B = X \Rightarrow (X \backslash A') \subseteq B \Rightarrow A \subseteq B

再证明 A⊆B⇔A∩B′=∅A \subseteq B \Leftrightarrow A\cap B' = \varnothing

证明

"⇒""\Rightarrow": A∩B′=A∩(X\B)=(A∩X)\B=A\B=∅A\cap B' = A \cap (X \backslash B) = (A \cap X) \backslash B = A \backslash B = \varnothing

"⇐""\Leftarrow" A∩B′=∅⇒A⊆(X\B′)⇒A⊆BA \cap B' = \varnothing \Rightarrow A \subseteq (X \backslash B') \Rightarrow A \subseteq B

题目 P98/10 #

对于任意集合 A,B,CA,B,C, 下列各式是否成立, 为什么?

  • (1) A∪B=A∪C⇒B=CA \cup B = A\cup C \Rightarrow B = C;
  • (2) A∩B=A∩C⇒B=CA \cap B = A \cap C \Rightarrow B = C.
答案
  • (1) 不成立, 例如取 A={1,2},B={1},C={2}A=\{1,2\},B=\{1\},C=\{2\}.
  • (2) 不成立, 例如取 A={1},B={1,2},C={1,3}A=\{1\},B=\{1,2\},C=\{1,3\}

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