作业帮证明1.Let A,B,and C be sets.Prove thatA∪包含 (A∪B ∪C).2.Let A,B,
来源：学生作业帮 编辑：作业帮 分类：英语作业 时间：2024/05/20 08:28:58
证明
1.Let A,B,and C be sets.Prove that
A∪包含 (A∪B ∪C).
2.Let A,B,and C be sets.Prove that
(A-C)∩(C -B) = 空集
1.Let A,B,and C be sets.Prove that
A∪包含 (A∪B ∪C).
2.Let A,B,and C be sets.Prove that
(A-C)∩(C -B) = 空集
1、
Pick a∈A∪B ,then a a∈A or a∈B.
there are two cases:
case 1 :a∈A,then a must be a member of
one of A,B,C.that means a a∈A∪B ∪C
case 2:a∈B,similarly discuss.
so in both cases,a must be member of A∪B ∪C
that means A∪B is subset of ∪B ∪C
2.Pick a∈(A-C),a must be in A and not in C.because a is not C,a is not in C-B.
so,for every element a ,a can not be in both (A-C) and (B-C).that means (A-C)∩(C -B) has no element.
therefore,(A-C)∩(C -B) is empty set.
Pick a∈A∪B ,then a a∈A or a∈B.
there are two cases:
case 1 :a∈A,then a must be a member of
one of A,B,C.that means a a∈A∪B ∪C
case 2:a∈B,similarly discuss.
so in both cases,a must be member of A∪B ∪C
that means A∪B is subset of ∪B ∪C
2.Pick a∈(A-C),a must be in A and not in C.because a is not C,a is not in C-B.
so,for every element a ,a can not be in both (A-C) and (B-C).that means (A-C)∩(C -B) has no element.
therefore,(A-C)∩(C -B) is empty set.
英语翻译Let A ,B and C be three points in a plane,such that AB/B
let a,b and c be rational numbers and b=0.24-0.26a,c=0.26-0.
Let is ( )pen friends.A:be B:are C:is
Let a,b and c are raional numbers which satisfy
Let’s ( ) singing and dancing.A.go B.fly C.
Tom,Let,s play__that ball.A.with B./ C.and
急：证明对所有集合A,B,C有（A∩B)∪C=A∩(B∪C),如果C包含于A
任意集合A B C 证明 (A∪B)- (B∪C) = A-B-C
1.Let's ( ) and play games.A.go B.goes C.going 2.There are m
let a,b and c be rational numbers and b=5分之12-5分之13a,c=5分之13
Let a,b and c be ration numbers and b=12/5-13/5a,c=13/5-12/5
Let _____come in A I and my brother B my broher and I C me a