Symmetric difference involving three sets: $Atriangle Btriangle C$ [duplicate]

0

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This question already has an answer here:

• 
  Associativity of symmetric difference of sets
  
  3 answers
  
  

I learned that the symmetric difference of sets $A, B$ is given by:
$$Atriangle B=(A-B)cup (B-A).$$

But I encountered the following expression: $Atriangle Btriangle C$, and I don’t understand what it means.

How do I apply the definition of symmetric difference when three sets are involved?

In fact I am stuck on a question

There are three set $A,B,C$.

How can I find the expression that the elements in two of these sets but are not in all three sets?

My answer is $A cup B cup C -[(A-Bcup C)cup (B-Acup C)cup (C-Acup B)+Acap Bcap C]$

But the answer is $Acup B cup (C-A)triangle B triangle C$
seems more simple than mine, how to get that answer?

elementary-set-theory

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edited Jan 17 at 23:25

jordan_glen

1

asked Jan 17 at 2:16

jacksonjackson

1159

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marked as duplicate by Leucippus, user91500, Lord_Farin, José Carlos Santos, Cesareo Jan 17 at 9:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It means $ADelta(BDelta C)$, so it is $(A-(BDelta C))cup ((BDelta C)-A)$.
  $endgroup$
  – Mike Earnest
  Jan 17 at 2:20
  

  
  
  
  

add a comment | 

0

$begingroup$

This question already has an answer here:

• 
  Associativity of symmetric difference of sets
  
  3 answers
  
  

I learned that the symmetric difference of sets $A, B$ is given by:
$$Atriangle B=(A-B)cup (B-A).$$

But I encountered the following expression: $Atriangle Btriangle C$, and I don’t understand what it means.

How do I apply the definition of symmetric difference when three sets are involved?

In fact I am stuck on a question

There are three set $A,B,C$.

How can I find the expression that the elements in two of these sets but are not in all three sets?

My answer is $A cup B cup C -[(A-Bcup C)cup (B-Acup C)cup (C-Acup B)+Acap Bcap C]$

But the answer is $Acup B cup (C-A)triangle B triangle C$
seems more simple than mine, how to get that answer?

elementary-set-theory

share|cite|improve this question

edited Jan 17 at 23:25

jordan_glen

1

asked Jan 17 at 2:16

jacksonjackson

1159

$endgroup$

marked as duplicate by Leucippus, user91500, Lord_Farin, José Carlos Santos, Cesareo Jan 17 at 9:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It means $ADelta(BDelta C)$, so it is $(A-(BDelta C))cup ((BDelta C)-A)$.
  $endgroup$
  – Mike Earnest
  Jan 17 at 2:20
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

This question already has an answer here:

• 
  Associativity of symmetric difference of sets
  
  3 answers
  
  

I learned that the symmetric difference of sets $A, B$ is given by:
$$Atriangle B=(A-B)cup (B-A).$$

But I encountered the following expression: $Atriangle Btriangle C$, and I don’t understand what it means.

How do I apply the definition of symmetric difference when three sets are involved?

In fact I am stuck on a question

There are three set $A,B,C$.

How can I find the expression that the elements in two of these sets but are not in all three sets?

My answer is $A cup B cup C -[(A-Bcup C)cup (B-Acup C)cup (C-Acup B)+Acap Bcap C]$

But the answer is $Acup B cup (C-A)triangle B triangle C$
seems more simple than mine, how to get that answer?

elementary-set-theory

share|cite|improve this question

edited Jan 17 at 23:25

jordan_glen

1

asked Jan 17 at 2:16

jacksonjackson

1159

$endgroup$

This question already has an answer here:

• 
  Associativity of symmetric difference of sets
  
  3 answers
  
  

I learned that the symmetric difference of sets $A, B$ is given by:
$$Atriangle B=(A-B)cup (B-A).$$

But I encountered the following expression: $Atriangle Btriangle C$, and I don’t understand what it means.

How do I apply the definition of symmetric difference when three sets are involved?

In fact I am stuck on a question

There are three set $A,B,C$.

How can I find the expression that the elements in two of these sets but are not in all three sets?

My answer is $A cup B cup C -[(A-Bcup C)cup (B-Acup C)cup (C-Acup B)+Acap Bcap C]$

But the answer is $Acup B cup (C-A)triangle B triangle C$
seems more simple than mine, how to get that answer?

This question already has an answer here:

• 
  Associativity of symmetric difference of sets
  
  3 answers
  
  

elementary-set-theory

elementary-set-theory

share|cite|improve this question

edited Jan 17 at 23:25

jordan_glen

1

asked Jan 17 at 2:16

jacksonjackson

1159

share|cite|improve this question

edited Jan 17 at 23:25

jordan_glen

1

asked Jan 17 at 2:16

jacksonjackson

1159

share|cite|improve this question

share|cite|improve this question

edited Jan 17 at 23:25

jordan_glen

1

edited Jan 17 at 23:25

jordan_glen

1

edited Jan 17 at 23:25

jordan_glen

1

1

asked Jan 17 at 2:16

jacksonjackson

1159

asked Jan 17 at 2:16

jacksonjackson

1159

asked Jan 17 at 2:16

jacksonjackson

1159

1159

marked as duplicate by Leucippus, user91500, Lord_Farin, José Carlos Santos, Cesareo Jan 17 at 9:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Leucippus, user91500, Lord_Farin, José Carlos Santos, Cesareo Jan 17 at 9:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It means $ADelta(BDelta C)$, so it is $(A-(BDelta C))cup ((BDelta C)-A)$.
  $endgroup$
  – Mike Earnest
  Jan 17 at 2:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It means $ADelta(BDelta C)$, so it is $(A-(BDelta C))cup ((BDelta C)-A)$.
  $endgroup$
  – Mike Earnest
  Jan 17 at 2:20
  

  
  
  
  

1

1

$begingroup$
It means $ADelta(BDelta C)$, so it is $(A-(BDelta C))cup ((BDelta C)-A)$.
$endgroup$
– Mike Earnest
Jan 17 at 2:20

$begingroup$
It means $ADelta(BDelta C)$, so it is $(A-(BDelta C))cup ((BDelta C)-A)$.
$endgroup$
– Mike Earnest
Jan 17 at 2:20

add a comment | 

1 Answer
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Color a venn diagram. Count how many of $A$, $B$ and $C$ that any element of $(ADelta B)Delta C$ belong to. You will quickly see this operation is associative and be able to see what $A_1Delta A_2 cdots A_n$ is.

share|cite|improve this answer

answered Jan 17 at 2:27

ncmathsadistncmathsadist

42.8k260103

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

$begingroup$

Color a venn diagram. Count how many of $A$, $B$ and $C$ that any element of $(ADelta B)Delta C$ belong to. You will quickly see this operation is associative and be able to see what $A_1Delta A_2 cdots A_n$ is.

share|cite|improve this answer

answered Jan 17 at 2:27

ncmathsadistncmathsadist

42.8k260103

$endgroup$

add a comment | 

0

$begingroup$

Color a venn diagram. Count how many of $A$, $B$ and $C$ that any element of $(ADelta B)Delta C$ belong to. You will quickly see this operation is associative and be able to see what $A_1Delta A_2 cdots A_n$ is.

share|cite|improve this answer

answered Jan 17 at 2:27

ncmathsadistncmathsadist

42.8k260103

$endgroup$

add a comment | 

0

0

0

$begingroup$

Color a venn diagram. Count how many of $A$, $B$ and $C$ that any element of $(ADelta B)Delta C$ belong to. You will quickly see this operation is associative and be able to see what $A_1Delta A_2 cdots A_n$ is.

share|cite|improve this answer

answered Jan 17 at 2:27

ncmathsadistncmathsadist

42.8k260103

$endgroup$

Color a venn diagram. Count how many of $A$, $B$ and $C$ that any element of $(ADelta B)Delta C$ belong to. You will quickly see this operation is associative and be able to see what $A_1Delta A_2 cdots A_n$ is.

share|cite|improve this answer

answered Jan 17 at 2:27

ncmathsadistncmathsadist

42.8k260103

share|cite|improve this answer

share|cite|improve this answer

answered Jan 17 at 2:27

ncmathsadistncmathsadist

42.8k260103

answered Jan 17 at 2:27

ncmathsadistncmathsadist

42.8k260103

answered Jan 17 at 2:27

ncmathsadistncmathsadist

42.8k260103

42.8k260103

add a comment | 

add a comment | 

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