Definition of symmetric relation












2












$begingroup$


I know that the relation is symmetric if $forall x forall y xRy implies yRx $.



Consider the set $ A = {{a, b, c, d}}$
and
$R = {{(a, a),(a, b),(a, d),(b, a),(b, b),(c, c),(d, a),(d, d)}}.$



My textbook claims that this relation is symmetric. But what about $(c,d)$ and $(d,c)$ that are not part of the $R$ set? In the definition, it said for all $x$ and $y$, so shouldn't this violate the symmetricity?






discrete-mathematics relations







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$endgroup$








  • 2




    $begingroup$
    There is a typo in your first line. (Or maybe it's not a typo but a mistake - which would explain your confusion about the given example.)
    $endgroup$
    – Stefan Mesken
    Nov 21 '17 at 9:29












  • $begingroup$
    @StefanMesken Thanks a lot! I edited it. It is still confusing me though.
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:37
















2












$begingroup$


I know that the relation is symmetric if $forall x forall y xRy implies yRx $.



Consider the set $ A = {{a, b, c, d}}$
and
$R = {{(a, a),(a, b),(a, d),(b, a),(b, b),(c, c),(d, a),(d, d)}}.$



My textbook claims that this relation is symmetric. But what about $(c,d)$ and $(d,c)$ that are not part of the $R$ set? In the definition, it said for all $x$ and $y$, so shouldn't this violate the symmetricity?






discrete-mathematics relations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    There is a typo in your first line. (Or maybe it's not a typo but a mistake - which would explain your confusion about the given example.)
    $endgroup$
    – Stefan Mesken
    Nov 21 '17 at 9:29












  • $begingroup$
    @StefanMesken Thanks a lot! I edited it. It is still confusing me though.
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:37














2












2








2





$begingroup$


I know that the relation is symmetric if $forall x forall y xRy implies yRx $.



Consider the set $ A = {{a, b, c, d}}$
and
$R = {{(a, a),(a, b),(a, d),(b, a),(b, b),(c, c),(d, a),(d, d)}}.$



My textbook claims that this relation is symmetric. But what about $(c,d)$ and $(d,c)$ that are not part of the $R$ set? In the definition, it said for all $x$ and $y$, so shouldn't this violate the symmetricity?






discrete-mathematics relations







share|cite|improve this question











$endgroup$




I know that the relation is symmetric if $forall x forall y xRy implies yRx $.



Consider the set $ A = {{a, b, c, d}}$
and
$R = {{(a, a),(a, b),(a, d),(b, a),(b, b),(c, c),(d, a),(d, d)}}.$



My textbook claims that this relation is symmetric. But what about $(c,d)$ and $(d,c)$ that are not part of the $R$ set? In the definition, it said for all $x$ and $y$, so shouldn't this violate the symmetricity?






discrete-mathematics relations




discrete-mathematics relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 21 '17 at 14:58







Avocado

















asked Nov 21 '17 at 9:27









AvocadoAvocado

3118




3118








  • 2




    $begingroup$
    There is a typo in your first line. (Or maybe it's not a typo but a mistake - which would explain your confusion about the given example.)
    $endgroup$
    – Stefan Mesken
    Nov 21 '17 at 9:29












  • $begingroup$
    @StefanMesken Thanks a lot! I edited it. It is still confusing me though.
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:37














  • 2




    $begingroup$
    There is a typo in your first line. (Or maybe it's not a typo but a mistake - which would explain your confusion about the given example.)
    $endgroup$
    – Stefan Mesken
    Nov 21 '17 at 9:29












  • $begingroup$
    @StefanMesken Thanks a lot! I edited it. It is still confusing me though.
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:37








2




2




$begingroup$
There is a typo in your first line. (Or maybe it's not a typo but a mistake - which would explain your confusion about the given example.)
$endgroup$
– Stefan Mesken
Nov 21 '17 at 9:29






$begingroup$
There is a typo in your first line. (Or maybe it's not a typo but a mistake - which would explain your confusion about the given example.)
$endgroup$
– Stefan Mesken
Nov 21 '17 at 9:29














$begingroup$
@StefanMesken Thanks a lot! I edited it. It is still confusing me though.
$endgroup$
– Avocado
Nov 21 '17 at 9:37




$begingroup$
@StefanMesken Thanks a lot! I edited it. It is still confusing me though.
$endgroup$
– Avocado
Nov 21 '17 at 9:37










3 Answers
3






active

oldest

votes


















3












$begingroup$

If $(c,d)$ is in the relation, then we check whether $(d,c)$ is in the set.



However, in this case, $(c,d)$ is not in $R$, hence we should not expect $(d,c)$ to be in the relation for it to be symmetric.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh of course. So the interpretation is for all x and y that are in that R set. It doesn't necessarily require the relationship to be between all of the elements in set A?
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:41






  • 1




    $begingroup$
    yes, it is not required.
    $endgroup$
    – Siong Thye Goh
    Nov 21 '17 at 9:42










  • $begingroup$
    Thank you! I am so amazed how quickly the community here helps you out. This is great :)
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:43





















1












$begingroup$

Definition 1: A relation R over set A is symmetric if for all x, y from A the following is true:
(x,y) is in R implies (y,x) is in R.



Definition 2: A relation R over set A is symmetric if for all x, y from A the following is true:
if (x,y) is in R, then (y,x) is in R.



These definitions do not require that every (x,y) has to be in R. They only require that for those (x,y) that are in R, (y,x) has to be in R.



Suppose that no single (x,y), where x is different from y, is in R. In this case R is symmetric.



This is so because proposition



  if (x,y) is in R, then  (y,x) is in R


is automatically true when proposition



  (x,y) is in R


is false (i.e. when (x,y) is not in R).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Right now, your answer is a bit difficult to read, as the mathematics is poorly formatted. It could be vastly improved with a little MathJax code (and, perhaps, one or two fewer newlines).
    $endgroup$
    – Xander Henderson
    Jan 10 '18 at 4:43



















0












$begingroup$

So $(a,c)$ is not included in $mathbb{R}$. So we can not expect $(c,a)$ in $mathbb{R}$.



If $(a,c)$ is included in $mathbb{R}$. So we can expect $(c,a)$ in $mathbb{R}$



And so both are symmetric






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Welcome the Mathematics Stack Exchange. A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. For typesetting your equations, please use MathJax: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – dantopa
    Dec 28 '18 at 5:00











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

If $(c,d)$ is in the relation, then we check whether $(d,c)$ is in the set.



However, in this case, $(c,d)$ is not in $R$, hence we should not expect $(d,c)$ to be in the relation for it to be symmetric.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh of course. So the interpretation is for all x and y that are in that R set. It doesn't necessarily require the relationship to be between all of the elements in set A?
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:41






  • 1




    $begingroup$
    yes, it is not required.
    $endgroup$
    – Siong Thye Goh
    Nov 21 '17 at 9:42










  • $begingroup$
    Thank you! I am so amazed how quickly the community here helps you out. This is great :)
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:43


















3












$begingroup$

If $(c,d)$ is in the relation, then we check whether $(d,c)$ is in the set.



However, in this case, $(c,d)$ is not in $R$, hence we should not expect $(d,c)$ to be in the relation for it to be symmetric.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh of course. So the interpretation is for all x and y that are in that R set. It doesn't necessarily require the relationship to be between all of the elements in set A?
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:41






  • 1




    $begingroup$
    yes, it is not required.
    $endgroup$
    – Siong Thye Goh
    Nov 21 '17 at 9:42










  • $begingroup$
    Thank you! I am so amazed how quickly the community here helps you out. This is great :)
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:43
















3












3








3





$begingroup$

If $(c,d)$ is in the relation, then we check whether $(d,c)$ is in the set.



However, in this case, $(c,d)$ is not in $R$, hence we should not expect $(d,c)$ to be in the relation for it to be symmetric.






share|cite|improve this answer









$endgroup$



If $(c,d)$ is in the relation, then we check whether $(d,c)$ is in the set.



However, in this case, $(c,d)$ is not in $R$, hence we should not expect $(d,c)$ to be in the relation for it to be symmetric.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 21 '17 at 9:39









Siong Thye GohSiong Thye Goh

103k1468119




103k1468119












  • $begingroup$
    Oh of course. So the interpretation is for all x and y that are in that R set. It doesn't necessarily require the relationship to be between all of the elements in set A?
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:41






  • 1




    $begingroup$
    yes, it is not required.
    $endgroup$
    – Siong Thye Goh
    Nov 21 '17 at 9:42










  • $begingroup$
    Thank you! I am so amazed how quickly the community here helps you out. This is great :)
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:43




















  • $begingroup$
    Oh of course. So the interpretation is for all x and y that are in that R set. It doesn't necessarily require the relationship to be between all of the elements in set A?
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:41






  • 1




    $begingroup$
    yes, it is not required.
    $endgroup$
    – Siong Thye Goh
    Nov 21 '17 at 9:42










  • $begingroup$
    Thank you! I am so amazed how quickly the community here helps you out. This is great :)
    $endgroup$
    – Avocado
    Nov 21 '17 at 9:43


















$begingroup$
Oh of course. So the interpretation is for all x and y that are in that R set. It doesn't necessarily require the relationship to be between all of the elements in set A?
$endgroup$
– Avocado
Nov 21 '17 at 9:41




$begingroup$
Oh of course. So the interpretation is for all x and y that are in that R set. It doesn't necessarily require the relationship to be between all of the elements in set A?
$endgroup$
– Avocado
Nov 21 '17 at 9:41




1




1




$begingroup$
yes, it is not required.
$endgroup$
– Siong Thye Goh
Nov 21 '17 at 9:42




$begingroup$
yes, it is not required.
$endgroup$
– Siong Thye Goh
Nov 21 '17 at 9:42












$begingroup$
Thank you! I am so amazed how quickly the community here helps you out. This is great :)
$endgroup$
– Avocado
Nov 21 '17 at 9:43






$begingroup$
Thank you! I am so amazed how quickly the community here helps you out. This is great :)
$endgroup$
– Avocado
Nov 21 '17 at 9:43













1












$begingroup$

Definition 1: A relation R over set A is symmetric if for all x, y from A the following is true:
(x,y) is in R implies (y,x) is in R.



Definition 2: A relation R over set A is symmetric if for all x, y from A the following is true:
if (x,y) is in R, then (y,x) is in R.



These definitions do not require that every (x,y) has to be in R. They only require that for those (x,y) that are in R, (y,x) has to be in R.



Suppose that no single (x,y), where x is different from y, is in R. In this case R is symmetric.



This is so because proposition



  if (x,y) is in R, then  (y,x) is in R


is automatically true when proposition



  (x,y) is in R


is false (i.e. when (x,y) is not in R).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Right now, your answer is a bit difficult to read, as the mathematics is poorly formatted. It could be vastly improved with a little MathJax code (and, perhaps, one or two fewer newlines).
    $endgroup$
    – Xander Henderson
    Jan 10 '18 at 4:43
















1












$begingroup$

Definition 1: A relation R over set A is symmetric if for all x, y from A the following is true:
(x,y) is in R implies (y,x) is in R.



Definition 2: A relation R over set A is symmetric if for all x, y from A the following is true:
if (x,y) is in R, then (y,x) is in R.



These definitions do not require that every (x,y) has to be in R. They only require that for those (x,y) that are in R, (y,x) has to be in R.



Suppose that no single (x,y), where x is different from y, is in R. In this case R is symmetric.



This is so because proposition



  if (x,y) is in R, then  (y,x) is in R


is automatically true when proposition



  (x,y) is in R


is false (i.e. when (x,y) is not in R).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Right now, your answer is a bit difficult to read, as the mathematics is poorly formatted. It could be vastly improved with a little MathJax code (and, perhaps, one or two fewer newlines).
    $endgroup$
    – Xander Henderson
    Jan 10 '18 at 4:43














1












1








1





$begingroup$

Definition 1: A relation R over set A is symmetric if for all x, y from A the following is true:
(x,y) is in R implies (y,x) is in R.



Definition 2: A relation R over set A is symmetric if for all x, y from A the following is true:
if (x,y) is in R, then (y,x) is in R.



These definitions do not require that every (x,y) has to be in R. They only require that for those (x,y) that are in R, (y,x) has to be in R.



Suppose that no single (x,y), where x is different from y, is in R. In this case R is symmetric.



This is so because proposition



  if (x,y) is in R, then  (y,x) is in R


is automatically true when proposition



  (x,y) is in R


is false (i.e. when (x,y) is not in R).






share|cite|improve this answer









$endgroup$



Definition 1: A relation R over set A is symmetric if for all x, y from A the following is true:
(x,y) is in R implies (y,x) is in R.



Definition 2: A relation R over set A is symmetric if for all x, y from A the following is true:
if (x,y) is in R, then (y,x) is in R.



These definitions do not require that every (x,y) has to be in R. They only require that for those (x,y) that are in R, (y,x) has to be in R.



Suppose that no single (x,y), where x is different from y, is in R. In this case R is symmetric.



This is so because proposition



  if (x,y) is in R, then  (y,x) is in R


is automatically true when proposition



  (x,y) is in R


is false (i.e. when (x,y) is not in R).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 '18 at 4:21









Fernando SotomayorFernando Sotomayor

111




111












  • $begingroup$
    Right now, your answer is a bit difficult to read, as the mathematics is poorly formatted. It could be vastly improved with a little MathJax code (and, perhaps, one or two fewer newlines).
    $endgroup$
    – Xander Henderson
    Jan 10 '18 at 4:43


















  • $begingroup$
    Right now, your answer is a bit difficult to read, as the mathematics is poorly formatted. It could be vastly improved with a little MathJax code (and, perhaps, one or two fewer newlines).
    $endgroup$
    – Xander Henderson
    Jan 10 '18 at 4:43
















$begingroup$
Right now, your answer is a bit difficult to read, as the mathematics is poorly formatted. It could be vastly improved with a little MathJax code (and, perhaps, one or two fewer newlines).
$endgroup$
– Xander Henderson
Jan 10 '18 at 4:43




$begingroup$
Right now, your answer is a bit difficult to read, as the mathematics is poorly formatted. It could be vastly improved with a little MathJax code (and, perhaps, one or two fewer newlines).
$endgroup$
– Xander Henderson
Jan 10 '18 at 4:43











0












$begingroup$

So $(a,c)$ is not included in $mathbb{R}$. So we can not expect $(c,a)$ in $mathbb{R}$.



If $(a,c)$ is included in $mathbb{R}$. So we can expect $(c,a)$ in $mathbb{R}$



And so both are symmetric






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Welcome the Mathematics Stack Exchange. A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. For typesetting your equations, please use MathJax: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – dantopa
    Dec 28 '18 at 5:00
















0












$begingroup$

So $(a,c)$ is not included in $mathbb{R}$. So we can not expect $(c,a)$ in $mathbb{R}$.



If $(a,c)$ is included in $mathbb{R}$. So we can expect $(c,a)$ in $mathbb{R}$



And so both are symmetric






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Welcome the Mathematics Stack Exchange. A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. For typesetting your equations, please use MathJax: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – dantopa
    Dec 28 '18 at 5:00














0












0








0





$begingroup$

So $(a,c)$ is not included in $mathbb{R}$. So we can not expect $(c,a)$ in $mathbb{R}$.



If $(a,c)$ is included in $mathbb{R}$. So we can expect $(c,a)$ in $mathbb{R}$



And so both are symmetric






share|cite|improve this answer











$endgroup$



So $(a,c)$ is not included in $mathbb{R}$. So we can not expect $(c,a)$ in $mathbb{R}$.



If $(a,c)$ is included in $mathbb{R}$. So we can expect $(c,a)$ in $mathbb{R}$



And so both are symmetric







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 4:59









dantopa

6,61442245




6,61442245










answered Dec 28 '18 at 4:48









Amar rohithAmar rohith

1




1












  • $begingroup$
    Welcome the Mathematics Stack Exchange. A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. For typesetting your equations, please use MathJax: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – dantopa
    Dec 28 '18 at 5:00


















  • $begingroup$
    Welcome the Mathematics Stack Exchange. A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. For typesetting your equations, please use MathJax: math.meta.stackexchange.com/questions/5020/…
    $endgroup$
    – dantopa
    Dec 28 '18 at 5:00
















$begingroup$
Welcome the Mathematics Stack Exchange. A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. For typesetting your equations, please use MathJax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– dantopa
Dec 28 '18 at 5:00




$begingroup$
Welcome the Mathematics Stack Exchange. A quick tour of the site (math.stackexchange.com/tour) will help you get the most of your time here. For typesetting your equations, please use MathJax: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– dantopa
Dec 28 '18 at 5:00


















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