Why is A union B also called “A or B”?
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In A union B, the element either belongs to A or B, or A and B right?
So shouldn't it be called A and/or B? Due to this I am unable to solve a problem in my textbook.
elementary-set-theory
asked May 28 at 12:29
user662650user662650
566 bronze badges
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show 1 more comment
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In A union B, the element either belongs to A or B, or A and B right?
So shouldn't it be called A and/or B? Due to this I am unable to solve a problem in my textbook.
elementary-set-theory
asked May 28 at 12:29
user662650user662650
566 bronze badges
$endgroup$
5
$begingroup$
In math, "or" is inclusive.
$endgroup$
– quasi
May 28 at 12:33
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Where is A union B also called "A or B"?
$endgroup$
– Hagen von Eitzen
May 28 at 12:34
2
$begingroup$
When we say "or" in mathematics we mean one thing, the other or both. That is why we can say the union of two sets $A cup B$ is the set whose elements are in $A$ or in $B$ (it can be in both sets).
$endgroup$
– Manuel DaGeo
May 28 at 12:34
$begingroup$
A correct answer is contained in the union of all answers given below. Does that imply that only one answer below is correct?
$endgroup$
– Michael
May 28 at 17:00
$begingroup$
@quasi I honestly think that's the best answer out of all of the ones below.
$endgroup$
– Bladewood
May 28 at 20:38
|
show 1 more comment
$begingroup$
In A union B, the element either belongs to A or B, or A and B right?
So shouldn't it be called A and/or B? Due to this I am unable to solve a problem in my textbook.
elementary-set-theory
asked May 28 at 12:29
user662650user662650
566 bronze badges
$endgroup$
In A union B, the element either belongs to A or B, or A and B right?
So shouldn't it be called A and/or B? Due to this I am unable to solve a problem in my textbook.
elementary-set-theory
elementary-set-theory
asked May 28 at 12:29
user662650user662650
566 bronze badges
asked May 28 at 12:29
user662650user662650
566 bronze badges
asked May 28 at 12:29
user662650user662650
566 bronze badges
asked May 28 at 12:29
user662650user662650
566 bronze badges
asked May 28 at 12:29
user662650user662650
566 bronze badges
566 bronze badges
5
$begingroup$
In math, "or" is inclusive.
$endgroup$
– quasi
May 28 at 12:33
$begingroup$
Where is A union B also called "A or B"?
$endgroup$
– Hagen von Eitzen
May 28 at 12:34
2
$begingroup$
When we say "or" in mathematics we mean one thing, the other or both. That is why we can say the union of two sets $A cup B$ is the set whose elements are in $A$ or in $B$ (it can be in both sets).
$endgroup$
– Manuel DaGeo
May 28 at 12:34
$begingroup$
A correct answer is contained in the union of all answers given below. Does that imply that only one answer below is correct?
$endgroup$
– Michael
May 28 at 17:00
$begingroup$
@quasi I honestly think that's the best answer out of all of the ones below.
$endgroup$
– Bladewood
May 28 at 20:38
|
show 1 more comment
5
$begingroup$
In math, "or" is inclusive.
$endgroup$
– quasi
May 28 at 12:33
$begingroup$
Where is A union B also called "A or B"?
$endgroup$
– Hagen von Eitzen
May 28 at 12:34
2
$begingroup$
When we say "or" in mathematics we mean one thing, the other or both. That is why we can say the union of two sets $A cup B$ is the set whose elements are in $A$ or in $B$ (it can be in both sets).
$endgroup$
– Manuel DaGeo
May 28 at 12:34
$begingroup$
A correct answer is contained in the union of all answers given below. Does that imply that only one answer below is correct?
$endgroup$
– Michael
May 28 at 17:00
$begingroup$
@quasi I honestly think that's the best answer out of all of the ones below.
$endgroup$
– Bladewood
May 28 at 20:38
5
5
$begingroup$
In math, "or" is inclusive.
$endgroup$
– quasi
May 28 at 12:33
$begingroup$
In math, "or" is inclusive.
$endgroup$
– quasi
May 28 at 12:33
$begingroup$
Where is A union B also called "A or B"?
$endgroup$
– Hagen von Eitzen
May 28 at 12:34
$begingroup$
Where is A union B also called "A or B"?
$endgroup$
– Hagen von Eitzen
May 28 at 12:34
2
2
$begingroup$
When we say "or" in mathematics we mean one thing, the other or both. That is why we can say the union of two sets $A cup B$ is the set whose elements are in $A$ or in $B$ (it can be in both sets).
$endgroup$
– Manuel DaGeo
May 28 at 12:34
$begingroup$
When we say "or" in mathematics we mean one thing, the other or both. That is why we can say the union of two sets $A cup B$ is the set whose elements are in $A$ or in $B$ (it can be in both sets).
$endgroup$
– Manuel DaGeo
May 28 at 12:34
$begingroup$
A correct answer is contained in the union of all answers given below. Does that imply that only one answer below is correct?
$endgroup$
– Michael
May 28 at 17:00
$begingroup$
A correct answer is contained in the union of all answers given below. Does that imply that only one answer below is correct?
$endgroup$
– Michael
May 28 at 17:00
$begingroup$
@quasi I honestly think that's the best answer out of all of the ones below.
$endgroup$
– Bladewood
May 28 at 20:38
$begingroup$
@quasi I honestly think that's the best answer out of all of the ones below.
$endgroup$
– Bladewood
May 28 at 20:38
|
show 1 more comment
5 Answers
5
active
oldest
votes
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Because the elements of $A cup B$ are exactly those objects that belong to $A$ or belong to $B$.
answered May 28 at 12:30
Mauro ALLEGRANZAMauro ALLEGRANZA
71.8k5 gold badges51 silver badges122 bronze badges
$endgroup$
1
$begingroup$
I'd like to add [from my limited knowledge] an "or" mostly means an 'inclusive or' [x being in A or in B or even in both [A and B], like stated in the question "and/or"] as opposed to an 'exclusive or' (or XOR) where x shall be in A or B but not in both! en.wikipedia.org/wiki/Exclusive_or
$endgroup$
– nuala
May 28 at 21:12
add a comment
|
$begingroup$
The claim "$p$ or $q$" is true if either $p$ is true, or $q$ is true, or $p$ and $q$ are both true.
That is, the claim "I am human or I am over one meter tall" is true, because at least one of the subclaims is true (in this case, both are true)
answered May 28 at 12:30
5xum5xum
97.6k5 gold badges101 silver badges168 bronze badges
$endgroup$
add a comment
|
$begingroup$
Suppose we are in a university.
The President orders ( for some reason we do not care about) the following thing: I want all math students and all students that play tennis to be present in the great hall tomorrow ( and no other person).
The next day, due to the strong authority of the President, all math students and all students playing tennis are present in the great hall.
Certainly, the President has in front of him all the members of the set : M U T ( that is the union of set M and of set T)
with M = the set of all math students
and T = the set of students playing tennis.
Now, what does the President know of any one of the students present there ( although he has possibly never met anyone of these students before)?
For example, what does the President know about David, one of the students that is there?
The only thing he knows for sure is that :
David is a math student OR David is a student that plays tennis.
Remark : of course, it could be the case David to be both a math student and a tennis player; but of this, the President has absolutely not idea.
In other words he only knows for sure that : one at least of these two propositions is true :
(1) David is a math student
(2) David is a student that plays tennis
and this " at least one" is precisely what defines the inclusive OR operator in logic.
Since the same thing could be said about any student, one can say in general that :
The set : M U T is the set of all x such that (1) x is a math student OR (2) x is a student that plays tennis.
In symbols : M U T = { x | x belongs to M v x belongs to T }
You might say : but there are certainly students in the hall that both are maths students and play tennis.
You would be right, but notice that if a student, say, David is both a math student and a tennis player, this is precisely an excellent reason to say that he is at east one of these two things. So David belongs to the set M U T exactly in the same way as all students that are at least one of these two things : math student OR tennis player.
The OR that defines the union operation is inclusive.
answered May 28 at 14:59
Eleonore Saint JamesEleonore Saint James
1,4231 silver badge18 bronze badges
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2
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Well, at my university the president is a woman.
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– Michael
May 28 at 17:04
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The union of two sets $A,B$ is defined by the ''or'' operation in propositional logic:
$$Acup B = {xmid xin Avee xin B}.$$
answered May 28 at 12:31
WuestenfuxWuestenfux
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A U B are all the elements of set A and set B. Read as 'A or B' (A union B).
Whereas,
A Π B are all the elements which set A and set B have in common. Read as 'A intersection B' (A and B).
answered May 28 at 15:17
Harshit BhardwajHarshit Bhardwaj
112 bronze badges
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
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Because the elements of $A cup B$ are exactly those objects that belong to $A$ or belong to $B$.
answered May 28 at 12:30
Mauro ALLEGRANZAMauro ALLEGRANZA
71.8k5 gold badges51 silver badges122 bronze badges
$endgroup$
1
$begingroup$
I'd like to add [from my limited knowledge] an "or" mostly means an 'inclusive or' [x being in A or in B or even in both [A and B], like stated in the question "and/or"] as opposed to an 'exclusive or' (or XOR) where x shall be in A or B but not in both! en.wikipedia.org/wiki/Exclusive_or
$endgroup$
– nuala
May 28 at 21:12
add a comment
|
$begingroup$
Because the elements of $A cup B$ are exactly those objects that belong to $A$ or belong to $B$.
answered May 28 at 12:30
Mauro ALLEGRANZAMauro ALLEGRANZA
71.8k5 gold badges51 silver badges122 bronze badges
$endgroup$
1
$begingroup$
I'd like to add [from my limited knowledge] an "or" mostly means an 'inclusive or' [x being in A or in B or even in both [A and B], like stated in the question "and/or"] as opposed to an 'exclusive or' (or XOR) where x shall be in A or B but not in both! en.wikipedia.org/wiki/Exclusive_or
$endgroup$
– nuala
May 28 at 21:12
add a comment
|
$begingroup$
Because the elements of $A cup B$ are exactly those objects that belong to $A$ or belong to $B$.
answered May 28 at 12:30
Mauro ALLEGRANZAMauro ALLEGRANZA
71.8k5 gold badges51 silver badges122 bronze badges
$endgroup$
Because the elements of $A cup B$ are exactly those objects that belong to $A$ or belong to $B$.
answered May 28 at 12:30
Mauro ALLEGRANZAMauro ALLEGRANZA
71.8k5 gold badges51 silver badges122 bronze badges
answered May 28 at 12:30
Mauro ALLEGRANZAMauro ALLEGRANZA
71.8k5 gold badges51 silver badges122 bronze badges
answered May 28 at 12:30
Mauro ALLEGRANZAMauro ALLEGRANZA
71.8k5 gold badges51 silver badges122 bronze badges
answered May 28 at 12:30
Mauro ALLEGRANZAMauro ALLEGRANZA
71.8k5 gold badges51 silver badges122 bronze badges
71.8k5 gold badges51 silver badges122 bronze badges
1
$begingroup$
I'd like to add [from my limited knowledge] an "or" mostly means an 'inclusive or' [x being in A or in B or even in both [A and B], like stated in the question "and/or"] as opposed to an 'exclusive or' (or XOR) where x shall be in A or B but not in both! en.wikipedia.org/wiki/Exclusive_or
$endgroup$
– nuala
May 28 at 21:12
add a comment
|
1
$begingroup$
I'd like to add [from my limited knowledge] an "or" mostly means an 'inclusive or' [x being in A or in B or even in both [A and B], like stated in the question "and/or"] as opposed to an 'exclusive or' (or XOR) where x shall be in A or B but not in both! en.wikipedia.org/wiki/Exclusive_or
$endgroup$
– nuala
May 28 at 21:12
1
1
$begingroup$
I'd like to add [from my limited knowledge] an "or" mostly means an 'inclusive or' [x being in A or in B or even in both [A and B], like stated in the question "and/or"] as opposed to an 'exclusive or' (or XOR) where x shall be in A or B but not in both! en.wikipedia.org/wiki/Exclusive_or
$endgroup$
– nuala
May 28 at 21:12
$begingroup$
I'd like to add [from my limited knowledge] an "or" mostly means an 'inclusive or' [x being in A or in B or even in both [A and B], like stated in the question "and/or"] as opposed to an 'exclusive or' (or XOR) where x shall be in A or B but not in both! en.wikipedia.org/wiki/Exclusive_or
$endgroup$
– nuala
May 28 at 21:12
add a comment
|
$begingroup$
The claim "$p$ or $q$" is true if either $p$ is true, or $q$ is true, or $p$ and $q$ are both true.
That is, the claim "I am human or I am over one meter tall" is true, because at least one of the subclaims is true (in this case, both are true)
answered May 28 at 12:30
5xum5xum
97.6k5 gold badges101 silver badges168 bronze badges
$endgroup$
add a comment
|
$begingroup$
The claim "$p$ or $q$" is true if either $p$ is true, or $q$ is true, or $p$ and $q$ are both true.
That is, the claim "I am human or I am over one meter tall" is true, because at least one of the subclaims is true (in this case, both are true)
answered May 28 at 12:30
5xum5xum
97.6k5 gold badges101 silver badges168 bronze badges
$endgroup$
add a comment
|
$begingroup$
The claim "$p$ or $q$" is true if either $p$ is true, or $q$ is true, or $p$ and $q$ are both true.
That is, the claim "I am human or I am over one meter tall" is true, because at least one of the subclaims is true (in this case, both are true)
answered May 28 at 12:30
5xum5xum
97.6k5 gold badges101 silver badges168 bronze badges
$endgroup$
The claim "$p$ or $q$" is true if either $p$ is true, or $q$ is true, or $p$ and $q$ are both true.
That is, the claim "I am human or I am over one meter tall" is true, because at least one of the subclaims is true (in this case, both are true)
answered May 28 at 12:30
5xum5xum
97.6k5 gold badges101 silver badges168 bronze badges
answered May 28 at 12:30
5xum5xum
97.6k5 gold badges101 silver badges168 bronze badges
answered May 28 at 12:30
5xum5xum
97.6k5 gold badges101 silver badges168 bronze badges
answered May 28 at 12:30
5xum5xum
97.6k5 gold badges101 silver badges168 bronze badges
97.6k5 gold badges101 silver badges168 bronze badges
add a comment
|
add a comment
|
$begingroup$
Suppose we are in a university.
The President orders ( for some reason we do not care about) the following thing: I want all math students and all students that play tennis to be present in the great hall tomorrow ( and no other person).
The next day, due to the strong authority of the President, all math students and all students playing tennis are present in the great hall.
Certainly, the President has in front of him all the members of the set : M U T ( that is the union of set M and of set T)
with M = the set of all math students
and T = the set of students playing tennis.
Now, what does the President know of any one of the students present there ( although he has possibly never met anyone of these students before)?
For example, what does the President know about David, one of the students that is there?
The only thing he knows for sure is that :
David is a math student OR David is a student that plays tennis.
Remark : of course, it could be the case David to be both a math student and a tennis player; but of this, the President has absolutely not idea.
In other words he only knows for sure that : one at least of these two propositions is true :
(1) David is a math student
(2) David is a student that plays tennis
and this " at least one" is precisely what defines the inclusive OR operator in logic.
Since the same thing could be said about any student, one can say in general that :
The set : M U T is the set of all x such that (1) x is a math student OR (2) x is a student that plays tennis.
In symbols : M U T = { x | x belongs to M v x belongs to T }
You might say : but there are certainly students in the hall that both are maths students and play tennis.
You would be right, but notice that if a student, say, David is both a math student and a tennis player, this is precisely an excellent reason to say that he is at east one of these two things. So David belongs to the set M U T exactly in the same way as all students that are at least one of these two things : math student OR tennis player.
The OR that defines the union operation is inclusive.
answered May 28 at 14:59
Eleonore Saint JamesEleonore Saint James
1,4231 silver badge18 bronze badges
$endgroup$
2
$begingroup$
Well, at my university the president is a woman.
$endgroup$
– Michael
May 28 at 17:04
add a comment
|
$begingroup$
Suppose we are in a university.
The President orders ( for some reason we do not care about) the following thing: I want all math students and all students that play tennis to be present in the great hall tomorrow ( and no other person).
The next day, due to the strong authority of the President, all math students and all students playing tennis are present in the great hall.
Certainly, the President has in front of him all the members of the set : M U T ( that is the union of set M and of set T)
with M = the set of all math students
and T = the set of students playing tennis.
Now, what does the President know of any one of the students present there ( although he has possibly never met anyone of these students before)?
For example, what does the President know about David, one of the students that is there?
The only thing he knows for sure is that :
David is a math student OR David is a student that plays tennis.
Remark : of course, it could be the case David to be both a math student and a tennis player; but of this, the President has absolutely not idea.
In other words he only knows for sure that : one at least of these two propositions is true :
(1) David is a math student
(2) David is a student that plays tennis
and this " at least one" is precisely what defines the inclusive OR operator in logic.
Since the same thing could be said about any student, one can say in general that :
The set : M U T is the set of all x such that (1) x is a math student OR (2) x is a student that plays tennis.
In symbols : M U T = { x | x belongs to M v x belongs to T }
You might say : but there are certainly students in the hall that both are maths students and play tennis.
You would be right, but notice that if a student, say, David is both a math student and a tennis player, this is precisely an excellent reason to say that he is at east one of these two things. So David belongs to the set M U T exactly in the same way as all students that are at least one of these two things : math student OR tennis player.
The OR that defines the union operation is inclusive.
answered May 28 at 14:59
Eleonore Saint JamesEleonore Saint James
1,4231 silver badge18 bronze badges
$endgroup$
2
$begingroup$
Well, at my university the president is a woman.
$endgroup$
– Michael
May 28 at 17:04
add a comment
|
$begingroup$
Suppose we are in a university.
The President orders ( for some reason we do not care about) the following thing: I want all math students and all students that play tennis to be present in the great hall tomorrow ( and no other person).
The next day, due to the strong authority of the President, all math students and all students playing tennis are present in the great hall.
Certainly, the President has in front of him all the members of the set : M U T ( that is the union of set M and of set T)
with M = the set of all math students
and T = the set of students playing tennis.
Now, what does the President know of any one of the students present there ( although he has possibly never met anyone of these students before)?
For example, what does the President know about David, one of the students that is there?
The only thing he knows for sure is that :
David is a math student OR David is a student that plays tennis.
Remark : of course, it could be the case David to be both a math student and a tennis player; but of this, the President has absolutely not idea.
In other words he only knows for sure that : one at least of these two propositions is true :
(1) David is a math student
(2) David is a student that plays tennis
and this " at least one" is precisely what defines the inclusive OR operator in logic.
Since the same thing could be said about any student, one can say in general that :
The set : M U T is the set of all x such that (1) x is a math student OR (2) x is a student that plays tennis.
In symbols : M U T = { x | x belongs to M v x belongs to T }
You might say : but there are certainly students in the hall that both are maths students and play tennis.
You would be right, but notice that if a student, say, David is both a math student and a tennis player, this is precisely an excellent reason to say that he is at east one of these two things. So David belongs to the set M U T exactly in the same way as all students that are at least one of these two things : math student OR tennis player.
The OR that defines the union operation is inclusive.
answered May 28 at 14:59
Eleonore Saint JamesEleonore Saint James
1,4231 silver badge18 bronze badges
$endgroup$
Suppose we are in a university.
The President orders ( for some reason we do not care about) the following thing: I want all math students and all students that play tennis to be present in the great hall tomorrow ( and no other person).
The next day, due to the strong authority of the President, all math students and all students playing tennis are present in the great hall.
Certainly, the President has in front of him all the members of the set : M U T ( that is the union of set M and of set T)
with M = the set of all math students
and T = the set of students playing tennis.
Now, what does the President know of any one of the students present there ( although he has possibly never met anyone of these students before)?
For example, what does the President know about David, one of the students that is there?
The only thing he knows for sure is that :
David is a math student OR David is a student that plays tennis.
Remark : of course, it could be the case David to be both a math student and a tennis player; but of this, the President has absolutely not idea.
In other words he only knows for sure that : one at least of these two propositions is true :
(1) David is a math student
(2) David is a student that plays tennis
and this " at least one" is precisely what defines the inclusive OR operator in logic.
Since the same thing could be said about any student, one can say in general that :
The set : M U T is the set of all x such that (1) x is a math student OR (2) x is a student that plays tennis.
In symbols : M U T = { x | x belongs to M v x belongs to T }
You might say : but there are certainly students in the hall that both are maths students and play tennis.
You would be right, but notice that if a student, say, David is both a math student and a tennis player, this is precisely an excellent reason to say that he is at east one of these two things. So David belongs to the set M U T exactly in the same way as all students that are at least one of these two things : math student OR tennis player.
The OR that defines the union operation is inclusive.
answered May 28 at 14:59
Eleonore Saint JamesEleonore Saint James
1,4231 silver badge18 bronze badges
edited May 29 at 16:14
answered May 28 at 14:59
Eleonore Saint JamesEleonore Saint James
1,4231 silver badge18 bronze badges
answered May 28 at 14:59
Eleonore Saint JamesEleonore Saint James
1,4231 silver badge18 bronze badges
answered May 28 at 14:59
Eleonore Saint JamesEleonore Saint James
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1,4231 silver badge18 bronze badges
2
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Well, at my university the president is a woman.
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– Michael
May 28 at 17:04
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2
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Well, at my university the president is a woman.
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– Michael
May 28 at 17:04
2
2
$begingroup$
Well, at my university the president is a woman.
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– Michael
May 28 at 17:04
$begingroup$
Well, at my university the president is a woman.
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– Michael
May 28 at 17:04
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The union of two sets $A,B$ is defined by the ''or'' operation in propositional logic:
$$Acup B = {xmid xin Avee xin B}.$$
answered May 28 at 12:31
WuestenfuxWuestenfux
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The union of two sets $A,B$ is defined by the ''or'' operation in propositional logic:
$$Acup B = {xmid xin Avee xin B}.$$
answered May 28 at 12:31
WuestenfuxWuestenfux
11.7k2 gold badges6 silver badges17 bronze badges
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$begingroup$
The union of two sets $A,B$ is defined by the ''or'' operation in propositional logic:
$$Acup B = {xmid xin Avee xin B}.$$
answered May 28 at 12:31
WuestenfuxWuestenfux
11.7k2 gold badges6 silver badges17 bronze badges
$endgroup$
The union of two sets $A,B$ is defined by the ''or'' operation in propositional logic:
$$Acup B = {xmid xin Avee xin B}.$$
answered May 28 at 12:31
WuestenfuxWuestenfux
11.7k2 gold badges6 silver badges17 bronze badges
answered May 28 at 12:31
WuestenfuxWuestenfux
11.7k2 gold badges6 silver badges17 bronze badges
answered May 28 at 12:31
WuestenfuxWuestenfux
11.7k2 gold badges6 silver badges17 bronze badges
answered May 28 at 12:31
WuestenfuxWuestenfux
11.7k2 gold badges6 silver badges17 bronze badges
11.7k2 gold badges6 silver badges17 bronze badges
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$begingroup$
A U B are all the elements of set A and set B. Read as 'A or B' (A union B).
Whereas,
A Π B are all the elements which set A and set B have in common. Read as 'A intersection B' (A and B).
answered May 28 at 15:17
Harshit BhardwajHarshit Bhardwaj
112 bronze badges
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Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
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– dantopa
May 28 at 15:36
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$begingroup$
A U B are all the elements of set A and set B. Read as 'A or B' (A union B).
Whereas,
A Π B are all the elements which set A and set B have in common. Read as 'A intersection B' (A and B).
answered May 28 at 15:17
Harshit BhardwajHarshit Bhardwaj
112 bronze badges
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$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
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– dantopa
May 28 at 15:36
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$begingroup$
A U B are all the elements of set A and set B. Read as 'A or B' (A union B).
Whereas,
A Π B are all the elements which set A and set B have in common. Read as 'A intersection B' (A and B).
answered May 28 at 15:17
Harshit BhardwajHarshit Bhardwaj
112 bronze badges
$endgroup$
A U B are all the elements of set A and set B. Read as 'A or B' (A union B).
Whereas,
A Π B are all the elements which set A and set B have in common. Read as 'A intersection B' (A and B).
answered May 28 at 15:17
Harshit BhardwajHarshit Bhardwaj
112 bronze badges
edited May 30 at 15:06
answered May 28 at 15:17
Harshit BhardwajHarshit Bhardwaj
112 bronze badges
answered May 28 at 15:17
Harshit BhardwajHarshit Bhardwaj
112 bronze badges
answered May 28 at 15:17
Harshit BhardwajHarshit Bhardwaj
112 bronze badges
112 bronze badges
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
May 28 at 15:36
add a comment
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$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
May 28 at 15:36
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
May 28 at 15:36
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
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– dantopa
May 28 at 15:36
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5
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In math, "or" is inclusive.
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– quasi
May 28 at 12:33
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Where is A union B also called "A or B"?
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– Hagen von Eitzen
May 28 at 12:34
2
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When we say "or" in mathematics we mean one thing, the other or both. That is why we can say the union of two sets $A cup B$ is the set whose elements are in $A$ or in $B$ (it can be in both sets).
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– Manuel DaGeo
May 28 at 12:34
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A correct answer is contained in the union of all answers given below. Does that imply that only one answer below is correct?
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– Michael
May 28 at 17:00
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@quasi I honestly think that's the best answer out of all of the ones below.
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– Bladewood
May 28 at 20:38