How to use Element-wise Proofs?
$begingroup$
Proof using element wise:
(A ∩ B ∩ C)’ = A’ ∪ B’ ∪ C’
I'm getting some issue proving this question.
calculus algebra-precalculus discrete-mathematics
asked Dec 5 '18 at 19:08
Emmanuel SimonEmmanuel Simon
31
$endgroup$
add a comment |
$begingroup$
Proof using element wise:
(A ∩ B ∩ C)’ = A’ ∪ B’ ∪ C’
I'm getting some issue proving this question.
calculus algebra-precalculus discrete-mathematics
asked Dec 5 '18 at 19:08
Emmanuel SimonEmmanuel Simon
31
$endgroup$
1
$begingroup$
Welcome to stackexchange. Please edit the question to show us what you've tried and where the :"issue" is. Then perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 19:11
add a comment |
$begingroup$
Proof using element wise:
(A ∩ B ∩ C)’ = A’ ∪ B’ ∪ C’
I'm getting some issue proving this question.
calculus algebra-precalculus discrete-mathematics
asked Dec 5 '18 at 19:08
Emmanuel SimonEmmanuel Simon
31
$endgroup$
Proof using element wise:
(A ∩ B ∩ C)’ = A’ ∪ B’ ∪ C’
I'm getting some issue proving this question.
calculus algebra-precalculus discrete-mathematics
calculus algebra-precalculus discrete-mathematics
asked Dec 5 '18 at 19:08
Emmanuel SimonEmmanuel Simon
31
asked Dec 5 '18 at 19:08
Emmanuel SimonEmmanuel Simon
31
asked Dec 5 '18 at 19:08
Emmanuel SimonEmmanuel Simon
31
asked Dec 5 '18 at 19:08
Emmanuel SimonEmmanuel Simon
31
asked Dec 5 '18 at 19:08
Emmanuel SimonEmmanuel Simon
31
31
1
$begingroup$
Welcome to stackexchange. Please edit the question to show us what you've tried and where the :"issue" is. Then perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 19:11
add a comment |
1
$begingroup$
Welcome to stackexchange. Please edit the question to show us what you've tried and where the :"issue" is. Then perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 19:11
1
1
$begingroup$
Welcome to stackexchange. Please edit the question to show us what you've tried and where the :"issue" is. Then perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 19:11
$begingroup$
Welcome to stackexchange. Please edit the question to show us what you've tried and where the :"issue" is. Then perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 19:11
add a comment |
1 Answer
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$begingroup$
By an "element wise" proof I assume you mean the most basic method in set proofs- by considering individual elements in the sets. To prove "X= Y" for sets X and Y, prove both "X is a subset of Y" and "X is a subset of Y". And to prove "X is a subset of Y you start "if x is in X" and use the properties of both X and Y to conclude "then x is in Y".
Here "X" is (A ∩ B ∩ C)'. If x is in (A ∩ B ∩ C)’ then x is not in A ∩ B ∩ C. That, in turn, means that x is not in at least one of A, B, or C. So I would use "cases":
case 1: x is not in A. Then x is in A' so is in A' ∪ B' ∪ C'.
case 2: x is not in B. Then x is in B' so is in A' ∪ B' ∪ C'.
case 3: x is not in C. Then x is in C' so is in A' ∪ B' ∪ C'.
That proves that (A ∩ B ∩ C)'is a subset of A' ∪ B' ∪ C'.
Now do that the other way. If x is in A' ∪ B' ∪ C' then x is in at least one of A', B', C'. Again do "cases":
case 1: x is in A', x is not in A so ...
answered Dec 5 '18 at 19:39
user247327user247327
11.2k1515
$endgroup$
$begingroup$
Thank you for make me understanding "element-wise" proof, this was helpful.
$endgroup$
– Emmanuel Simon
Dec 5 '18 at 20:14
add a comment |
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1 Answer
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$begingroup$
By an "element wise" proof I assume you mean the most basic method in set proofs- by considering individual elements in the sets. To prove "X= Y" for sets X and Y, prove both "X is a subset of Y" and "X is a subset of Y". And to prove "X is a subset of Y you start "if x is in X" and use the properties of both X and Y to conclude "then x is in Y".
Here "X" is (A ∩ B ∩ C)'. If x is in (A ∩ B ∩ C)’ then x is not in A ∩ B ∩ C. That, in turn, means that x is not in at least one of A, B, or C. So I would use "cases":
case 1: x is not in A. Then x is in A' so is in A' ∪ B' ∪ C'.
case 2: x is not in B. Then x is in B' so is in A' ∪ B' ∪ C'.
case 3: x is not in C. Then x is in C' so is in A' ∪ B' ∪ C'.
That proves that (A ∩ B ∩ C)'is a subset of A' ∪ B' ∪ C'.
Now do that the other way. If x is in A' ∪ B' ∪ C' then x is in at least one of A', B', C'. Again do "cases":
case 1: x is in A', x is not in A so ...
answered Dec 5 '18 at 19:39
user247327user247327
11.2k1515
$endgroup$
$begingroup$
Thank you for make me understanding "element-wise" proof, this was helpful.
$endgroup$
– Emmanuel Simon
Dec 5 '18 at 20:14
add a comment |
$begingroup$
By an "element wise" proof I assume you mean the most basic method in set proofs- by considering individual elements in the sets. To prove "X= Y" for sets X and Y, prove both "X is a subset of Y" and "X is a subset of Y". And to prove "X is a subset of Y you start "if x is in X" and use the properties of both X and Y to conclude "then x is in Y".
Here "X" is (A ∩ B ∩ C)'. If x is in (A ∩ B ∩ C)’ then x is not in A ∩ B ∩ C. That, in turn, means that x is not in at least one of A, B, or C. So I would use "cases":
case 1: x is not in A. Then x is in A' so is in A' ∪ B' ∪ C'.
case 2: x is not in B. Then x is in B' so is in A' ∪ B' ∪ C'.
case 3: x is not in C. Then x is in C' so is in A' ∪ B' ∪ C'.
That proves that (A ∩ B ∩ C)'is a subset of A' ∪ B' ∪ C'.
Now do that the other way. If x is in A' ∪ B' ∪ C' then x is in at least one of A', B', C'. Again do "cases":
case 1: x is in A', x is not in A so ...
answered Dec 5 '18 at 19:39
user247327user247327
11.2k1515
$endgroup$
$begingroup$
Thank you for make me understanding "element-wise" proof, this was helpful.
$endgroup$
– Emmanuel Simon
Dec 5 '18 at 20:14
add a comment |
$begingroup$
By an "element wise" proof I assume you mean the most basic method in set proofs- by considering individual elements in the sets. To prove "X= Y" for sets X and Y, prove both "X is a subset of Y" and "X is a subset of Y". And to prove "X is a subset of Y you start "if x is in X" and use the properties of both X and Y to conclude "then x is in Y".
Here "X" is (A ∩ B ∩ C)'. If x is in (A ∩ B ∩ C)’ then x is not in A ∩ B ∩ C. That, in turn, means that x is not in at least one of A, B, or C. So I would use "cases":
case 1: x is not in A. Then x is in A' so is in A' ∪ B' ∪ C'.
case 2: x is not in B. Then x is in B' so is in A' ∪ B' ∪ C'.
case 3: x is not in C. Then x is in C' so is in A' ∪ B' ∪ C'.
That proves that (A ∩ B ∩ C)'is a subset of A' ∪ B' ∪ C'.
Now do that the other way. If x is in A' ∪ B' ∪ C' then x is in at least one of A', B', C'. Again do "cases":
case 1: x is in A', x is not in A so ...
answered Dec 5 '18 at 19:39
user247327user247327
11.2k1515
$endgroup$
By an "element wise" proof I assume you mean the most basic method in set proofs- by considering individual elements in the sets. To prove "X= Y" for sets X and Y, prove both "X is a subset of Y" and "X is a subset of Y". And to prove "X is a subset of Y you start "if x is in X" and use the properties of both X and Y to conclude "then x is in Y".
Here "X" is (A ∩ B ∩ C)'. If x is in (A ∩ B ∩ C)’ then x is not in A ∩ B ∩ C. That, in turn, means that x is not in at least one of A, B, or C. So I would use "cases":
case 1: x is not in A. Then x is in A' so is in A' ∪ B' ∪ C'.
case 2: x is not in B. Then x is in B' so is in A' ∪ B' ∪ C'.
case 3: x is not in C. Then x is in C' so is in A' ∪ B' ∪ C'.
That proves that (A ∩ B ∩ C)'is a subset of A' ∪ B' ∪ C'.
Now do that the other way. If x is in A' ∪ B' ∪ C' then x is in at least one of A', B', C'. Again do "cases":
case 1: x is in A', x is not in A so ...
answered Dec 5 '18 at 19:39
user247327user247327
11.2k1515
answered Dec 5 '18 at 19:39
user247327user247327
11.2k1515
answered Dec 5 '18 at 19:39
user247327user247327
11.2k1515
answered Dec 5 '18 at 19:39
user247327user247327
11.2k1515
11.2k1515
$begingroup$
Thank you for make me understanding "element-wise" proof, this was helpful.
$endgroup$
– Emmanuel Simon
Dec 5 '18 at 20:14
add a comment |
$begingroup$
Thank you for make me understanding "element-wise" proof, this was helpful.
$endgroup$
– Emmanuel Simon
Dec 5 '18 at 20:14
$begingroup$
Thank you for make me understanding "element-wise" proof, this was helpful.
$endgroup$
– Emmanuel Simon
Dec 5 '18 at 20:14
$begingroup$
Thank you for make me understanding "element-wise" proof, this was helpful.
$endgroup$
– Emmanuel Simon
Dec 5 '18 at 20:14
add a comment |
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$begingroup$
Welcome to stackexchange. Please edit the question to show us what you've tried and where the :"issue" is. Then perhaps we can help.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 19:11