Number of relations on a set of $n$ elements [closed]
$begingroup$
http://mathonline.wikidot.com/the-number-of-distinct-relations-on-a-finite-set
From the link above I'm having trouble understanding how from $n^2$ ordered pairs which are either true or false there are a total of $ {2^{n^2}} $, would it not be $2n$?
Also, can someone explain the subset part of the definition, say I have a set ${a,b,c}$, would that mean ${(a,b), (b,a), (c,c)}$ is also a relation? Or is are the subsets only the pairs of elements of ${a,b,c}$?
combinatorics elementary-set-theory relations
edited Dec 31 '18 at 23:49
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 22:10
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
$endgroup$
closed as off-topic by Namaste, Holo, Leucippus, Adrian Keister, Saad Jan 1 at 12:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, Leucippus, Adrian Keister, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
http://mathonline.wikidot.com/the-number-of-distinct-relations-on-a-finite-set
From the link above I'm having trouble understanding how from $n^2$ ordered pairs which are either true or false there are a total of $ {2^{n^2}} $, would it not be $2n$?
Also, can someone explain the subset part of the definition, say I have a set ${a,b,c}$, would that mean ${(a,b), (b,a), (c,c)}$ is also a relation? Or is are the subsets only the pairs of elements of ${a,b,c}$?
combinatorics elementary-set-theory relations
edited Dec 31 '18 at 23:49
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 22:10
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
$endgroup$
closed as off-topic by Namaste, Holo, Leucippus, Adrian Keister, Saad Jan 1 at 12:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, Leucippus, Adrian Keister, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I'm not sure what you mean by an "order relation" in this context but $(a, b)$ and $(b, a)$ in the relation mean that it can't be used to order $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 31 '18 at 22:50
add a comment |
$begingroup$
http://mathonline.wikidot.com/the-number-of-distinct-relations-on-a-finite-set
From the link above I'm having trouble understanding how from $n^2$ ordered pairs which are either true or false there are a total of $ {2^{n^2}} $, would it not be $2n$?
Also, can someone explain the subset part of the definition, say I have a set ${a,b,c}$, would that mean ${(a,b), (b,a), (c,c)}$ is also a relation? Or is are the subsets only the pairs of elements of ${a,b,c}$?
combinatorics elementary-set-theory relations
edited Dec 31 '18 at 23:49
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 22:10
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
$endgroup$
http://mathonline.wikidot.com/the-number-of-distinct-relations-on-a-finite-set
From the link above I'm having trouble understanding how from $n^2$ ordered pairs which are either true or false there are a total of $ {2^{n^2}} $, would it not be $2n$?
Also, can someone explain the subset part of the definition, say I have a set ${a,b,c}$, would that mean ${(a,b), (b,a), (c,c)}$ is also a relation? Or is are the subsets only the pairs of elements of ${a,b,c}$?
combinatorics elementary-set-theory relations
combinatorics elementary-set-theory relations
edited Dec 31 '18 at 23:49
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 22:10
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
edited Dec 31 '18 at 23:49
Maria Mazur
50.3k1361126
asked Dec 31 '18 at 22:10
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
edited Dec 31 '18 at 23:49
Maria Mazur
50.3k1361126
edited Dec 31 '18 at 23:49
Maria Mazur
50.3k1361126
edited Dec 31 '18 at 23:49
Maria Mazur
50.3k1361126
50.3k1361126
asked Dec 31 '18 at 22:10
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
asked Dec 31 '18 at 22:10
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
asked Dec 31 '18 at 22:10
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
536
closed as off-topic by Namaste, Holo, Leucippus, Adrian Keister, Saad Jan 1 at 12:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, Leucippus, Adrian Keister, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Holo, Leucippus, Adrian Keister, Saad Jan 1 at 12:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Holo, Leucippus, Adrian Keister, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I'm not sure what you mean by an "order relation" in this context but $(a, b)$ and $(b, a)$ in the relation mean that it can't be used to order $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 31 '18 at 22:50
add a comment |
$begingroup$
I'm not sure what you mean by an "order relation" in this context but $(a, b)$ and $(b, a)$ in the relation mean that it can't be used to order $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 31 '18 at 22:50
$begingroup$
I'm not sure what you mean by an "order relation" in this context but $(a, b)$ and $(b, a)$ in the relation mean that it can't be used to order $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 31 '18 at 22:50
$begingroup$
I'm not sure what you mean by an "order relation" in this context but $(a, b)$ and $(b, a)$ in the relation mean that it can't be used to order $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 31 '18 at 22:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, it is correct. Every relation is a subset of $Xtimes X$ which has power $n^2$. But the number of subsets in a set with $k$ elements is $2^k$, so in our case $2^{n^2}$.
Yes, the set ${(a,b), (b,a), (c,c)}$ is a subset of $Xtimes X$ and thus a relation (which is symmetric).
answered Dec 31 '18 at 22:15
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
$begingroup$
So {(a,b),(b,a),(c,c)} would be a relation?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 22:31
1
$begingroup$
Yes,and symmetric.
$endgroup$
– Maria Mazur
Dec 31 '18 at 22:32
$begingroup$
Thank you, this all makes sense now.
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 23:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, it is correct. Every relation is a subset of $Xtimes X$ which has power $n^2$. But the number of subsets in a set with $k$ elements is $2^k$, so in our case $2^{n^2}$.
Yes, the set ${(a,b), (b,a), (c,c)}$ is a subset of $Xtimes X$ and thus a relation (which is symmetric).
answered Dec 31 '18 at 22:15
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
$begingroup$
So {(a,b),(b,a),(c,c)} would be a relation?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 22:31
1
$begingroup$
Yes,and symmetric.
$endgroup$
– Maria Mazur
Dec 31 '18 at 22:32
$begingroup$
Thank you, this all makes sense now.
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 23:07
add a comment |
$begingroup$
No, it is correct. Every relation is a subset of $Xtimes X$ which has power $n^2$. But the number of subsets in a set with $k$ elements is $2^k$, so in our case $2^{n^2}$.
Yes, the set ${(a,b), (b,a), (c,c)}$ is a subset of $Xtimes X$ and thus a relation (which is symmetric).
answered Dec 31 '18 at 22:15
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
$begingroup$
So {(a,b),(b,a),(c,c)} would be a relation?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 22:31
1
$begingroup$
Yes,and symmetric.
$endgroup$
– Maria Mazur
Dec 31 '18 at 22:32
$begingroup$
Thank you, this all makes sense now.
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 23:07
add a comment |
$begingroup$
No, it is correct. Every relation is a subset of $Xtimes X$ which has power $n^2$. But the number of subsets in a set with $k$ elements is $2^k$, so in our case $2^{n^2}$.
Yes, the set ${(a,b), (b,a), (c,c)}$ is a subset of $Xtimes X$ and thus a relation (which is symmetric).
answered Dec 31 '18 at 22:15
Maria MazurMaria Mazur
50.3k1361126
$endgroup$
No, it is correct. Every relation is a subset of $Xtimes X$ which has power $n^2$. But the number of subsets in a set with $k$ elements is $2^k$, so in our case $2^{n^2}$.
Yes, the set ${(a,b), (b,a), (c,c)}$ is a subset of $Xtimes X$ and thus a relation (which is symmetric).
answered Dec 31 '18 at 22:15
Maria MazurMaria Mazur
50.3k1361126
edited Dec 31 '18 at 23:26
answered Dec 31 '18 at 22:15
Maria MazurMaria Mazur
50.3k1361126
answered Dec 31 '18 at 22:15
Maria MazurMaria Mazur
50.3k1361126
answered Dec 31 '18 at 22:15
Maria MazurMaria Mazur
50.3k1361126
50.3k1361126
$begingroup$
So {(a,b),(b,a),(c,c)} would be a relation?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 22:31
1
$begingroup$
Yes,and symmetric.
$endgroup$
– Maria Mazur
Dec 31 '18 at 22:32
$begingroup$
Thank you, this all makes sense now.
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 23:07
add a comment |
$begingroup$
So {(a,b),(b,a),(c,c)} would be a relation?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 22:31
1
$begingroup$
Yes,and symmetric.
$endgroup$
– Maria Mazur
Dec 31 '18 at 22:32
$begingroup$
Thank you, this all makes sense now.
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 23:07
$begingroup$
So {(a,b),(b,a),(c,c)} would be a relation?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 22:31
$begingroup$
So {(a,b),(b,a),(c,c)} would be a relation?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 22:31
1
1
$begingroup$
Yes,and symmetric.
$endgroup$
– Maria Mazur
Dec 31 '18 at 22:32
$begingroup$
Yes,and symmetric.
$endgroup$
– Maria Mazur
Dec 31 '18 at 22:32
$begingroup$
Thank you, this all makes sense now.
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 23:07
$begingroup$
Thank you, this all makes sense now.
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 31 '18 at 23:07
add a comment |
$begingroup$
I'm not sure what you mean by an "order relation" in this context but $(a, b)$ and $(b, a)$ in the relation mean that it can't be used to order $a$ and $b$.
$endgroup$
– Mark Bennet
Dec 31 '18 at 22:50