用中文解答下题Let A = {a, b, c} and B = P(A). 1) Prove that [B; ∪ , ∩ ,~] is a Boolean algebra. 2) Write out the operation tables for the Boolean algebra.
时间: 2023-06-17 08:05:29 浏览: 45
1) 首先需要证明[B; ∪ , ∩ ,~]满足Boolean代数的四个公理:结合律、交换律、分配律和互补律。
结合律:对于任意的A,B,C∈B,有(A∪B)∪C=A∪(B∪C)和(A∩B)∩C=A∩(B∩C)。这是因为集合的并、交运算符都满足结合律。
交换律:对于任意的A,B∈B,有A∪B=B∪A和A∩B=B∩A。这是因为集合的并、交运算符都满足交换律。
分配律:对于任意的A,B,C∈B,有A∪(B∩C)=(A∪B)∩(A∪C)和A∩(B∪C)=(A∩B)∪(A∩C)。这是因为集合的并、交运算符都满足分配律。
互补律:对于任意的A∈B,有A∪~A=A和A∩~A=∅。这是因为集合的补集~A定义为A的所有元素的补集,即~A={x∣x∉A},A∪~A=A∩~A=A的所有元素的集合,因此满足互补律。
由此可知[B; ∪ , ∩ ,~]是一个Boolean代数。
2) 接下来是[B; ∪ , ∩ ,~]的操作表:
并集 ∪:
| | a | b | c |
|---|---|---|---|
| a | a | a | a |
| b | a | b | b |
| c | a | b | c |
交集 ∩:
| | a | b | c |
|---|---|---|---|
| a | a | ∅ | ∅ |
| b | ∅ | b | ∅ |
| c | ∅ | ∅ | c |
补集 ~:
| | a | b | c |
|---|---|---|---|
| | b,c | a,c | a,b |
其中,∅代表空集。
相关问题
解答下题Let A = {a, b, c} and B = P(A). 1) Prove that [B; ∪ , ∩ ,~] is a Boolean algebra. 2) Write out the operation tables for the Boolean algebra.
1) To prove that [B; ∪ , ∩ ,~] is a Boolean algebra, we need to show that it satisfies the following axioms:
- Closure under union and intersection: For any two sets X, Y ∈ B, X ∪ Y and X ∩ Y ∈ B.
- Associativity of union and intersection: For any three sets X, Y, Z ∈ B, (X ∪ Y) ∪ Z = X ∪ (Y ∪ Z) and (X ∩ Y) ∩ Z = X ∩ (Y ∩ Z).
- Commutativity of union and intersection: For any two sets X, Y ∈ B, X ∪ Y = Y ∪ X and X ∩ Y = Y ∩ X.
- Distributivity of intersection over union and vice versa: For any three sets X, Y, Z ∈ B, X ∩ (Y ∪ Z) = (X ∩ Y) ∪ (X ∩ Z) and X ∪ (Y ∩ Z) = (X ∪ Y) ∩ (X ∪ Z).
- Identity elements of union and intersection: There exist two sets 0, 1 ∈ B such that for any set X ∈ B, X ∪ 0 = X and X ∩ 1 = X.
- Complementation: For any set X ∈ B, there exists a set ¬X ∈ B such that X ∪ ¬X = 1 and X ∩ ¬X = 0.
All of these axioms can be easily verified for [B; ∪ , ∩ ,~], therefore it is a Boolean algebra.
2) The operation tables for the Boolean algebra [B; ∪ , ∩ ,~] are:
| ∪ | a b c ∅
--|----------------
a | a a a a
b | a b b b
c | a b c c
∅ | a b c ∅
| ∩ | a b c ∅
--|----------------
a | a b c ∅
b | b b ∅ ∅
c | c ∅ c ∅
∅ | ∅ ∅ ∅ ∅
| ~ | a b c ∅
--|------------------------
| ∅ {b,c} {a,c} {a,b}
a | {b,c} ∅ {b} {c}
b | {a,c} {a} ∅ {c}
c | {a,b} {a} {b} ∅
∅ | {a,b,c} {a,b,c} {a,b,c} ∅
Note: ∅ represents the empty set, and {} represents the set with no elements.
Prove that∥p∥1 := |a| + |b| + |c|defines a norm on V .
根据向量范数的定义,要证明∥p∥1是V上的范数,需要满足以下三个条件:
1.非负性:对于任意的p∈V,有∥p∥1≥0,并且当且仅当p=0时,有∥p∥1=0。
2.齐次性:对于任意的p∈V和任意的实数α,有∥αp∥1=|α|∥p∥1。
3.三角不等式:对于任意的p,q∈V,有∥p+q∥1≤∥p∥1+∥q∥1。
首先证明非负性和齐次性:
1.对于任意的p=(a,b,c)∈V,有∥p∥1=|a|+|b|+|c|≥0,并且当且仅当p=0时,有∥p∥1=0。因为|a|、|b|和|c|都是非负数,所以它们的和也是非负数,当且仅当a=b=c=0时,和为0。
2.对于任意的p=(a,b,c)∈V和任意的实数α,有∥αp∥1=|αa|+|αb|+|αc|=|α||a|+|α||b|+|α||c|=|α|(|a|+|b|+|c|)=|α|∥p∥1。
接下来证明三角不等式:
对于任意的p=(a1,b1,c1)和q=(a2,b2,c2)∈V,有p+q=(a1+a2,b1+b2,c1+c2),所以
∥p+q∥1=|a1+a2|+|b1+b2|+|c1+c2|
≤(|a1|+|a2|)+(|b1|+|b2|)+(|c1|+|c2|)
=|a1|+|b1|+|c1|+|a2|+|b2|+|c2|
=∥p∥1+∥q∥1
因此,∥p∥1 := |a| + |b| + |c|是V上的范数。