Given two sets A,B,the symmetric difference of A and B is de
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Given two sets A,B,the symmetric difference of A and B is denoted by
A△B and defined as A△B = (A − B) ∪ (B − A).
For very sets A,B,C,prove that:
A△(B△C) = (A△B)△C
翻译:对于集合A、B,有:A△B = (A − B) ∪ (B − A).证明:A△(B△C) = (A△B)△C
A△B and defined as A△B = (A − B) ∪ (B − A).
For very sets A,B,C,prove that:
A△(B△C) = (A△B)△C
翻译:对于集合A、B,有:A△B = (A − B) ∪ (B − A).证明:A△(B△C) = (A△B)△C
我们先来严格证明:
A△(B△C)=(A-B△C)∪(B△C-A)=[A-(B-C)∪(C-B)]∪[(B-C)∪(C-B)-A]
={[A-(B-C)]∩[A-(C-B)]}∪{[(B-C)-A]∪[(C-B)-A]}
={[(A-B)∪A∩B∩C)]∩[(A-C)∪A∩B∩C)]}∪{[(B-C∪A)]∪[C-(B∪A)]}
=[(A-B)∩(A-C)]∪A∩B∩C∪{[(B-C∪A)]∪[C-(B∪A)]}
=[A-(B∪C)]∪A∩B∩C∪[(B-C∪A)]∪[(C-B)∩(C-A)]
={[A-(B∪C)]∪[(B-C∪A)]}∪{A∩B∩C∪[(C-B)∩(C-A)]}
={[(A-B)-C)]∪[(B-A)-C)]}∪{[A∩B∩C∪(C-B)]∩[A∩B∩C∪(C-A)]}
={[(A-B)∪(B-A)-C]}∪{[C-(B-A)]∩[C-(A-B)]}
=(A△B-C)∪[C-(B-A)∪(A-B)]
=(A△B-C)∪[C-A△B]
=(A△B)△C
显然代数证明不太好理解,但配合上几何来说明,就会清晰很多.
如图三个圆分别表示三个集合A,B,C,每个单独集合部分用一个数字表示
如图有A={1,2,3,4},B={1,2,5,6},C={1,3,6,7}
则有B△C={2,3,5,7}=> A△(B△C)={1,4}∪{5,7}={1,4,5,7}
同样A△B={3,4,5,6}=> (A△B)△C={4,5}∪{1,7}={1,4,5,7}
∴A△(B△C) = (A△B)△C
同样可以通过图形来理解上面的证明过程,这样要容易的多.
再问: You are awesome!! But, 从第二步到第三步[(B-C)∪(C-B)-A]问什么等于[(B-C)-A]∪[(C-B)-A]
再答: 很高兴能对你有所帮助
A△(B△C)=(A-B△C)∪(B△C-A)=[A-(B-C)∪(C-B)]∪[(B-C)∪(C-B)-A]
={[A-(B-C)]∩[A-(C-B)]}∪{[(B-C)-A]∪[(C-B)-A]}
={[(A-B)∪A∩B∩C)]∩[(A-C)∪A∩B∩C)]}∪{[(B-C∪A)]∪[C-(B∪A)]}
=[(A-B)∩(A-C)]∪A∩B∩C∪{[(B-C∪A)]∪[C-(B∪A)]}
=[A-(B∪C)]∪A∩B∩C∪[(B-C∪A)]∪[(C-B)∩(C-A)]
={[A-(B∪C)]∪[(B-C∪A)]}∪{A∩B∩C∪[(C-B)∩(C-A)]}
={[(A-B)-C)]∪[(B-A)-C)]}∪{[A∩B∩C∪(C-B)]∩[A∩B∩C∪(C-A)]}
={[(A-B)∪(B-A)-C]}∪{[C-(B-A)]∩[C-(A-B)]}
=(A△B-C)∪[C-(B-A)∪(A-B)]
=(A△B-C)∪[C-A△B]
=(A△B)△C
显然代数证明不太好理解,但配合上几何来说明,就会清晰很多.
如图三个圆分别表示三个集合A,B,C,每个单独集合部分用一个数字表示
如图有A={1,2,3,4},B={1,2,5,6},C={1,3,6,7}
则有B△C={2,3,5,7}=> A△(B△C)={1,4}∪{5,7}={1,4,5,7}
同样A△B={3,4,5,6}=> (A△B)△C={4,5}∪{1,7}={1,4,5,7}
∴A△(B△C) = (A△B)△C
同样可以通过图形来理解上面的证明过程,这样要容易的多.
再问: You are awesome!! But, 从第二步到第三步[(B-C)∪(C-B)-A]问什么等于[(B-C)-A]∪[(C-B)-A]
再答: 很高兴能对你有所帮助
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