Show $A cap (B cap C) = (A cap B) cap C$ Set algebra approach.
$begingroup$
Show $A cap (B cap C) = (A cap B) cap C$. Using set algebra approach.
I'm just having trouble convincing myself via the algebraic approach that the statement is true.....I know it is true, I convinced myself visually. But trying to work it out through the algebraic approach is leaving me lacking a solution.
Approach:
1) $A cap (B cap C) = (A cap B) cap (A cap C)$
2) $(A cap B) cap C = (A cap C) cap (B cap C)$
What I wanted to do was show those two expressions are equivalent....what am I missing to get my conclusion?
elementary-set-theory
asked Jan 14 at 16:41
dc3rddc3rd
1,51911241
$endgroup$
add a comment |
$begingroup$
Show $A cap (B cap C) = (A cap B) cap C$. Using set algebra approach.
I'm just having trouble convincing myself via the algebraic approach that the statement is true.....I know it is true, I convinced myself visually. But trying to work it out through the algebraic approach is leaving me lacking a solution.
Approach:
1) $A cap (B cap C) = (A cap B) cap (A cap C)$
2) $(A cap B) cap C = (A cap C) cap (B cap C)$
What I wanted to do was show those two expressions are equivalent....what am I missing to get my conclusion?
elementary-set-theory
asked Jan 14 at 16:41
dc3rddc3rd
1,51911241
$endgroup$
1
$begingroup$
What is the "algebraic approach"?
$endgroup$
– MPW
Jan 14 at 16:45
$begingroup$
@MPW maybe wrong choice of words, what I mean is by using the distributive property.
$endgroup$
– dc3rd
Jan 14 at 16:47
$begingroup$
@MPW I expect that what is meant is that the OP wishes to prove the associative property of intersection using "algebraic manipulations" using a proof of the form $Acap (Bcap C)=cdots = cdots = (Acap B)cap C$ while avoiding direct use of what we want to prove (or any results that rely on this) while also (for whatever ill-conceived reason) avoiding rewriting things with words such as $(Acap B)={x~:~xin A~text{and}~xin B}$.
$endgroup$
– JMoravitz
Jan 14 at 16:50
1
$begingroup$
The easiest proof will however be one that uses words instead of only symbols... and will rely on the fact that the logical and is also associative (something which can be shown using a truth table for instance if not already included as an axiom from the start).
$endgroup$
– JMoravitz
Jan 14 at 16:52
$begingroup$
@JMoravitz what I gather from your explanation is that I shouldn't over-complicate things by trying to do symbolic manipulation and should in fact use words to explain why the two expressions are equivalent.
$endgroup$
– dc3rd
Jan 14 at 16:56
add a comment |
$begingroup$
Show $A cap (B cap C) = (A cap B) cap C$. Using set algebra approach.
I'm just having trouble convincing myself via the algebraic approach that the statement is true.....I know it is true, I convinced myself visually. But trying to work it out through the algebraic approach is leaving me lacking a solution.
Approach:
1) $A cap (B cap C) = (A cap B) cap (A cap C)$
2) $(A cap B) cap C = (A cap C) cap (B cap C)$
What I wanted to do was show those two expressions are equivalent....what am I missing to get my conclusion?
elementary-set-theory
asked Jan 14 at 16:41
dc3rddc3rd
1,51911241
$endgroup$
Show $A cap (B cap C) = (A cap B) cap C$. Using set algebra approach.
I'm just having trouble convincing myself via the algebraic approach that the statement is true.....I know it is true, I convinced myself visually. But trying to work it out through the algebraic approach is leaving me lacking a solution.
Approach:
1) $A cap (B cap C) = (A cap B) cap (A cap C)$
2) $(A cap B) cap C = (A cap C) cap (B cap C)$
What I wanted to do was show those two expressions are equivalent....what am I missing to get my conclusion?
elementary-set-theory
elementary-set-theory
asked Jan 14 at 16:41
dc3rddc3rd
1,51911241
asked Jan 14 at 16:41
dc3rddc3rd
1,51911241
asked Jan 14 at 16:41
dc3rddc3rd
1,51911241
asked Jan 14 at 16:41
dc3rddc3rd
1,51911241
asked Jan 14 at 16:41
dc3rddc3rd
1,51911241
1,51911241
1
$begingroup$
What is the "algebraic approach"?
$endgroup$
– MPW
Jan 14 at 16:45
$begingroup$
@MPW maybe wrong choice of words, what I mean is by using the distributive property.
$endgroup$
– dc3rd
Jan 14 at 16:47
$begingroup$
@MPW I expect that what is meant is that the OP wishes to prove the associative property of intersection using "algebraic manipulations" using a proof of the form $Acap (Bcap C)=cdots = cdots = (Acap B)cap C$ while avoiding direct use of what we want to prove (or any results that rely on this) while also (for whatever ill-conceived reason) avoiding rewriting things with words such as $(Acap B)={x~:~xin A~text{and}~xin B}$.
$endgroup$
– JMoravitz
Jan 14 at 16:50
1
$begingroup$
The easiest proof will however be one that uses words instead of only symbols... and will rely on the fact that the logical and is also associative (something which can be shown using a truth table for instance if not already included as an axiom from the start).
$endgroup$
– JMoravitz
Jan 14 at 16:52
$begingroup$
@JMoravitz what I gather from your explanation is that I shouldn't over-complicate things by trying to do symbolic manipulation and should in fact use words to explain why the two expressions are equivalent.
$endgroup$
– dc3rd
Jan 14 at 16:56
add a comment |
1
$begingroup$
What is the "algebraic approach"?
$endgroup$
– MPW
Jan 14 at 16:45
$begingroup$
@MPW maybe wrong choice of words, what I mean is by using the distributive property.
$endgroup$
– dc3rd
Jan 14 at 16:47
$begingroup$
@MPW I expect that what is meant is that the OP wishes to prove the associative property of intersection using "algebraic manipulations" using a proof of the form $Acap (Bcap C)=cdots = cdots = (Acap B)cap C$ while avoiding direct use of what we want to prove (or any results that rely on this) while also (for whatever ill-conceived reason) avoiding rewriting things with words such as $(Acap B)={x~:~xin A~text{and}~xin B}$.
$endgroup$
– JMoravitz
Jan 14 at 16:50
1
$begingroup$
The easiest proof will however be one that uses words instead of only symbols... and will rely on the fact that the logical and is also associative (something which can be shown using a truth table for instance if not already included as an axiom from the start).
$endgroup$
– JMoravitz
Jan 14 at 16:52
$begingroup$
@JMoravitz what I gather from your explanation is that I shouldn't over-complicate things by trying to do symbolic manipulation and should in fact use words to explain why the two expressions are equivalent.
$endgroup$
– dc3rd
Jan 14 at 16:56
1
1
$begingroup$
What is the "algebraic approach"?
$endgroup$
– MPW
Jan 14 at 16:45
$begingroup$
What is the "algebraic approach"?
$endgroup$
– MPW
Jan 14 at 16:45
$begingroup$
@MPW maybe wrong choice of words, what I mean is by using the distributive property.
$endgroup$
– dc3rd
Jan 14 at 16:47
$begingroup$
@MPW maybe wrong choice of words, what I mean is by using the distributive property.
$endgroup$
– dc3rd
Jan 14 at 16:47
$begingroup$
@MPW I expect that what is meant is that the OP wishes to prove the associative property of intersection using "algebraic manipulations" using a proof of the form $Acap (Bcap C)=cdots = cdots = (Acap B)cap C$ while avoiding direct use of what we want to prove (or any results that rely on this) while also (for whatever ill-conceived reason) avoiding rewriting things with words such as $(Acap B)={x~:~xin A~text{and}~xin B}$.
$endgroup$
– JMoravitz
Jan 14 at 16:50
$begingroup$
@MPW I expect that what is meant is that the OP wishes to prove the associative property of intersection using "algebraic manipulations" using a proof of the form $Acap (Bcap C)=cdots = cdots = (Acap B)cap C$ while avoiding direct use of what we want to prove (or any results that rely on this) while also (for whatever ill-conceived reason) avoiding rewriting things with words such as $(Acap B)={x~:~xin A~text{and}~xin B}$.
$endgroup$
– JMoravitz
Jan 14 at 16:50
1
1
$begingroup$
The easiest proof will however be one that uses words instead of only symbols... and will rely on the fact that the logical and is also associative (something which can be shown using a truth table for instance if not already included as an axiom from the start).
$endgroup$
– JMoravitz
Jan 14 at 16:52
$begingroup$
The easiest proof will however be one that uses words instead of only symbols... and will rely on the fact that the logical and is also associative (something which can be shown using a truth table for instance if not already included as an axiom from the start).
$endgroup$
– JMoravitz
Jan 14 at 16:52
$begingroup$
@JMoravitz what I gather from your explanation is that I shouldn't over-complicate things by trying to do symbolic manipulation and should in fact use words to explain why the two expressions are equivalent.
$endgroup$
– dc3rd
Jan 14 at 16:56
$begingroup$
@JMoravitz what I gather from your explanation is that I shouldn't over-complicate things by trying to do symbolic manipulation and should in fact use words to explain why the two expressions are equivalent.
$endgroup$
– dc3rd
Jan 14 at 16:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It depends a bit on how you defined the intersection. But if you defined it as:
$$xin Acap B :iff xin A text{ and } yin B$$
Then you can prove it by using the associative property of the logical "and" which you can show with truth tables.
Assuming this property:
$$ xin Acap(Bcap C) iff xin A text{ and } xin Bcap Ciffdots$$
answered Jan 14 at 16:59
Felix B.Felix B.
694317
$endgroup$
$begingroup$
I think this was the assumption I was missing in my construction, because it all seems clear now. Thanks.
$endgroup$
– dc3rd
Jan 14 at 17:01
add a comment |
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1 Answer
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$begingroup$
It depends a bit on how you defined the intersection. But if you defined it as:
$$xin Acap B :iff xin A text{ and } yin B$$
Then you can prove it by using the associative property of the logical "and" which you can show with truth tables.
Assuming this property:
$$ xin Acap(Bcap C) iff xin A text{ and } xin Bcap Ciffdots$$
answered Jan 14 at 16:59
Felix B.Felix B.
694317
$endgroup$
$begingroup$
I think this was the assumption I was missing in my construction, because it all seems clear now. Thanks.
$endgroup$
– dc3rd
Jan 14 at 17:01
add a comment |
$begingroup$
It depends a bit on how you defined the intersection. But if you defined it as:
$$xin Acap B :iff xin A text{ and } yin B$$
Then you can prove it by using the associative property of the logical "and" which you can show with truth tables.
Assuming this property:
$$ xin Acap(Bcap C) iff xin A text{ and } xin Bcap Ciffdots$$
answered Jan 14 at 16:59
Felix B.Felix B.
694317
$endgroup$
$begingroup$
I think this was the assumption I was missing in my construction, because it all seems clear now. Thanks.
$endgroup$
– dc3rd
Jan 14 at 17:01
add a comment |
$begingroup$
It depends a bit on how you defined the intersection. But if you defined it as:
$$xin Acap B :iff xin A text{ and } yin B$$
Then you can prove it by using the associative property of the logical "and" which you can show with truth tables.
Assuming this property:
$$ xin Acap(Bcap C) iff xin A text{ and } xin Bcap Ciffdots$$
answered Jan 14 at 16:59
Felix B.Felix B.
694317
$endgroup$
It depends a bit on how you defined the intersection. But if you defined it as:
$$xin Acap B :iff xin A text{ and } yin B$$
Then you can prove it by using the associative property of the logical "and" which you can show with truth tables.
Assuming this property:
$$ xin Acap(Bcap C) iff xin A text{ and } xin Bcap Ciffdots$$
answered Jan 14 at 16:59
Felix B.Felix B.
694317
edited Jan 14 at 17:21
answered Jan 14 at 16:59
Felix B.Felix B.
694317
answered Jan 14 at 16:59
Felix B.Felix B.
694317
answered Jan 14 at 16:59
Felix B.Felix B.
694317
694317
$begingroup$
I think this was the assumption I was missing in my construction, because it all seems clear now. Thanks.
$endgroup$
– dc3rd
Jan 14 at 17:01
add a comment |
$begingroup$
I think this was the assumption I was missing in my construction, because it all seems clear now. Thanks.
$endgroup$
– dc3rd
Jan 14 at 17:01
$begingroup$
I think this was the assumption I was missing in my construction, because it all seems clear now. Thanks.
$endgroup$
– dc3rd
Jan 14 at 17:01
$begingroup$
I think this was the assumption I was missing in my construction, because it all seems clear now. Thanks.
$endgroup$
– dc3rd
Jan 14 at 17:01
add a comment |
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1
$begingroup$
What is the "algebraic approach"?
$endgroup$
– MPW
Jan 14 at 16:45
$begingroup$
@MPW maybe wrong choice of words, what I mean is by using the distributive property.
$endgroup$
– dc3rd
Jan 14 at 16:47
$begingroup$
@MPW I expect that what is meant is that the OP wishes to prove the associative property of intersection using "algebraic manipulations" using a proof of the form $Acap (Bcap C)=cdots = cdots = (Acap B)cap C$ while avoiding direct use of what we want to prove (or any results that rely on this) while also (for whatever ill-conceived reason) avoiding rewriting things with words such as $(Acap B)={x~:~xin A~text{and}~xin B}$.
$endgroup$
– JMoravitz
Jan 14 at 16:50
1
$begingroup$
The easiest proof will however be one that uses words instead of only symbols... and will rely on the fact that the logical and is also associative (something which can be shown using a truth table for instance if not already included as an axiom from the start).
$endgroup$
– JMoravitz
Jan 14 at 16:52
$begingroup$
@JMoravitz what I gather from your explanation is that I shouldn't over-complicate things by trying to do symbolic manipulation and should in fact use words to explain why the two expressions are equivalent.
$endgroup$
– dc3rd
Jan 14 at 16:56