why is AxA Transitive, Reflexive, and Symmetric [on hold]
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I just started group-theory. and we have this question but the professor never explained this material, I have no idea how to prove it.
the question as asked by the professor(and Im translating it to English because its in Hebrew):
On a group $A$, is the relation $A times A$ on that group Transitive, Reflexive, and Symmetric?
I know that the identity relation is all of the above. I just don't see the difference between the questions, i.e., I don't see the difference between the identity relation and the relation $A times A$.
calculus sequences-and-series group-theory
edited Nov 16 at 12:41
John Hughes
61.3k24089
asked Nov 16 at 11:45
adi k
11
New contributor
adi k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by 5xum, Derek Holt, Delta-u, Alan Wang, Cameron Buie Nov 16 at 12:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, Derek Holt, Delta-u, Alan Wang
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 3 more comments
up vote
-2
down vote
favorite
I just started group-theory. and we have this question but the professor never explained this material, I have no idea how to prove it.
the question as asked by the professor(and Im translating it to English because its in Hebrew):
On a group $A$, is the relation $A times A$ on that group Transitive, Reflexive, and Symmetric?
I know that the identity relation is all of the above. I just don't see the difference between the questions, i.e., I don't see the difference between the identity relation and the relation $A times A$.
calculus sequences-and-series group-theory
edited Nov 16 at 12:41
John Hughes
61.3k24089
asked Nov 16 at 11:45
adi k
11
New contributor
adi k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by 5xum, Derek Holt, Delta-u, Alan Wang, Cameron Buie Nov 16 at 12:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, Derek Holt, Delta-u, Alan Wang
If this question can be reworded to fit the rules in the help center, please edit the question.
Where is the material?
– Alan Wang
Nov 16 at 11:46
math.meta.stackexchange.com/questions/9959/…
– 5xum
Nov 16 at 11:48
@JohnHughes That is not the relation that I would call $Atimes A$. I'd think that would be the relation of all pairs $(a,b)$ for $a,bin A$.
– user3482749
Nov 16 at 11:53
Right you are! I'll fix this. More coffee needed. :(
– John Hughes
Nov 16 at 12:05
1
Perhaps your question is this: "On any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric." In Mathjax form, that'sOn any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric.You can edit your question to replace it with this more complete and clear and nicely formatted one by clicking "edit" just below your question
– John Hughes
Nov 16 at 12:07
|
show 3 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I just started group-theory. and we have this question but the professor never explained this material, I have no idea how to prove it.
the question as asked by the professor(and Im translating it to English because its in Hebrew):
On a group $A$, is the relation $A times A$ on that group Transitive, Reflexive, and Symmetric?
I know that the identity relation is all of the above. I just don't see the difference between the questions, i.e., I don't see the difference between the identity relation and the relation $A times A$.
calculus sequences-and-series group-theory
edited Nov 16 at 12:41
John Hughes
61.3k24089
asked Nov 16 at 11:45
adi k
11
New contributor
adi k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I just started group-theory. and we have this question but the professor never explained this material, I have no idea how to prove it.
the question as asked by the professor(and Im translating it to English because its in Hebrew):
On a group $A$, is the relation $A times A$ on that group Transitive, Reflexive, and Symmetric?
I know that the identity relation is all of the above. I just don't see the difference between the questions, i.e., I don't see the difference between the identity relation and the relation $A times A$.
calculus sequences-and-series group-theory
calculus sequences-and-series group-theory
edited Nov 16 at 12:41
John Hughes
61.3k24089
asked Nov 16 at 11:45
adi k
11
New contributor
adi k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 16 at 12:41
John Hughes
61.3k24089
asked Nov 16 at 11:45
adi k
11
New contributor
adi k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 16 at 12:41
John Hughes
61.3k24089
edited Nov 16 at 12:41
John Hughes
61.3k24089
edited Nov 16 at 12:41
John Hughes
61.3k24089
61.3k24089
asked Nov 16 at 11:45
adi k
11
New contributor
adi k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 16 at 11:45
adi k
11
asked Nov 16 at 11:45
adi k
11
11
New contributor
adi k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
adi k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
adi k is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by 5xum, Derek Holt, Delta-u, Alan Wang, Cameron Buie Nov 16 at 12:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, Derek Holt, Delta-u, Alan Wang
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by 5xum, Derek Holt, Delta-u, Alan Wang, Cameron Buie Nov 16 at 12:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, Derek Holt, Delta-u, Alan Wang
If this question can be reworded to fit the rules in the help center, please edit the question.
Where is the material?
– Alan Wang
Nov 16 at 11:46
math.meta.stackexchange.com/questions/9959/…
– 5xum
Nov 16 at 11:48
@JohnHughes That is not the relation that I would call $Atimes A$. I'd think that would be the relation of all pairs $(a,b)$ for $a,bin A$.
– user3482749
Nov 16 at 11:53
Right you are! I'll fix this. More coffee needed. :(
– John Hughes
Nov 16 at 12:05
1
Perhaps your question is this: "On any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric." In Mathjax form, that'sOn any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric.You can edit your question to replace it with this more complete and clear and nicely formatted one by clicking "edit" just below your question
– John Hughes
Nov 16 at 12:07
|
show 3 more comments
Where is the material?
– Alan Wang
Nov 16 at 11:46
math.meta.stackexchange.com/questions/9959/…
– 5xum
Nov 16 at 11:48
@JohnHughes That is not the relation that I would call $Atimes A$. I'd think that would be the relation of all pairs $(a,b)$ for $a,bin A$.
– user3482749
Nov 16 at 11:53
Right you are! I'll fix this. More coffee needed. :(
– John Hughes
Nov 16 at 12:05
1
Perhaps your question is this: "On any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric." In Mathjax form, that'sOn any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric.You can edit your question to replace it with this more complete and clear and nicely formatted one by clicking "edit" just below your question
– John Hughes
Nov 16 at 12:07
Where is the material?
– Alan Wang
Nov 16 at 11:46
Where is the material?
– Alan Wang
Nov 16 at 11:46
math.meta.stackexchange.com/questions/9959/…
– 5xum
Nov 16 at 11:48
math.meta.stackexchange.com/questions/9959/…
– 5xum
Nov 16 at 11:48
@JohnHughes That is not the relation that I would call $Atimes A$. I'd think that would be the relation of all pairs $(a,b)$ for $a,bin A$.
– user3482749
Nov 16 at 11:53
@JohnHughes That is not the relation that I would call $Atimes A$. I'd think that would be the relation of all pairs $(a,b)$ for $a,bin A$.
– user3482749
Nov 16 at 11:53
Right you are! I'll fix this. More coffee needed. :(
– John Hughes
Nov 16 at 12:05
Right you are! I'll fix this. More coffee needed. :(
– John Hughes
Nov 16 at 12:05
1
1
Perhaps your question is this: "On any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric." In Mathjax form, that's
On any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric. You can edit your question to replace it with this more complete and clear and nicely formatted one by clicking "edit" just below your question– John Hughes
Nov 16 at 12:07
Perhaps your question is this: "On any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric." In Mathjax form, that's
On any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric. You can edit your question to replace it with this more complete and clear and nicely formatted one by clicking "edit" just below your question– John Hughes
Nov 16 at 12:07
|
show 3 more comments
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Where is the material?
– Alan Wang
Nov 16 at 11:46
math.meta.stackexchange.com/questions/9959/…
– 5xum
Nov 16 at 11:48
@JohnHughes That is not the relation that I would call $Atimes A$. I'd think that would be the relation of all pairs $(a,b)$ for $a,bin A$.
– user3482749
Nov 16 at 11:53
Right you are! I'll fix this. More coffee needed. :(
– John Hughes
Nov 16 at 12:05
1
Perhaps your question is this: "On any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric." In Mathjax form, that's
On any set $A$, there's a relation consisting of all pairs $(p, q)$ for $p in A, q in A$, i.e., the relation is $A times A$. Show that this relation is transitive, reflexive, and symmetric.You can edit your question to replace it with this more complete and clear and nicely formatted one by clicking "edit" just below your question– John Hughes
Nov 16 at 12:07