Definition of symmetric relation
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
asked Nov 21 '17 at 9:27
Avocado
316
add a comment |
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
asked Nov 21 '17 at 9:27
Avocado
316
2
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.)
– Stefan Mesken
Nov 21 '17 at 9:29
@StefanMesken Thanks a lot! I edited it. It is still confusing me though.
– Avocado
Nov 21 '17 at 9:37
add a comment |
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
asked Nov 21 '17 at 9:27
Avocado
316
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
asked Nov 21 '17 at 9:27
Avocado
316
asked Nov 21 '17 at 9:27
Avocado
316
edited Nov 21 '17 at 14:58
asked Nov 21 '17 at 9:27
Avocado
316
asked Nov 21 '17 at 9:27
Avocado
316
asked Nov 21 '17 at 9:27
Avocado
316
316
2
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.)
– Stefan Mesken
Nov 21 '17 at 9:29
@StefanMesken Thanks a lot! I edited it. It is still confusing me though.
– Avocado
Nov 21 '17 at 9:37
add a comment |
2
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.)
– Stefan Mesken
Nov 21 '17 at 9:29
@StefanMesken Thanks a lot! I edited it. It is still confusing me though.
– Avocado
Nov 21 '17 at 9:37
2
2
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.)
– Stefan Mesken
Nov 21 '17 at 9:29
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.)
– Stefan Mesken
Nov 21 '17 at 9:29
@StefanMesken Thanks a lot! I edited it. It is still confusing me though.
– Avocado
Nov 21 '17 at 9:37
@StefanMesken Thanks a lot! I edited it. It is still confusing me though.
– Avocado
Nov 21 '17 at 9:37
add a comment |
3 Answers
3
active
oldest
votes
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.
answered Nov 21 '17 at 9:39
Siong Thye Goh
99.5k1464117
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?
– Avocado
Nov 21 '17 at 9:41
1
yes, it is not required.
– Siong Thye Goh
Nov 21 '17 at 9:42
Thank you! I am so amazed how quickly the community here helps you out. This is great :)
– Avocado
Nov 21 '17 at 9:43
add a comment |
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).
answered Jan 10 '18 at 4:21
Fernando Sotomayor
111
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).
– Xander Henderson
Jan 10 '18 at 4:43
add a comment |
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
edited Dec 28 '18 at 4:59
dantopa
6,42932042
answered Dec 28 '18 at 4:48
Amar rohith
1
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/…
– dantopa
Dec 28 '18 at 5:00
add a comment |
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3 Answers
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3 Answers
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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.
answered Nov 21 '17 at 9:39
Siong Thye Goh
99.5k1464117
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?
– Avocado
Nov 21 '17 at 9:41
1
yes, it is not required.
– Siong Thye Goh
Nov 21 '17 at 9:42
Thank you! I am so amazed how quickly the community here helps you out. This is great :)
– Avocado
Nov 21 '17 at 9:43
add a comment |
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.
answered Nov 21 '17 at 9:39
Siong Thye Goh
99.5k1464117
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?
– Avocado
Nov 21 '17 at 9:41
1
yes, it is not required.
– Siong Thye Goh
Nov 21 '17 at 9:42
Thank you! I am so amazed how quickly the community here helps you out. This is great :)
– Avocado
Nov 21 '17 at 9:43
add a comment |
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.
answered Nov 21 '17 at 9:39
Siong Thye Goh
99.5k1464117
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.
answered Nov 21 '17 at 9:39
Siong Thye Goh
99.5k1464117
answered Nov 21 '17 at 9:39
Siong Thye Goh
99.5k1464117
answered Nov 21 '17 at 9:39
Siong Thye Goh
99.5k1464117
answered Nov 21 '17 at 9:39
Siong Thye Goh
99.5k1464117
99.5k1464117
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?
– Avocado
Nov 21 '17 at 9:41
1
yes, it is not required.
– Siong Thye Goh
Nov 21 '17 at 9:42
Thank you! I am so amazed how quickly the community here helps you out. This is great :)
– Avocado
Nov 21 '17 at 9:43
add a comment |
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?
– Avocado
Nov 21 '17 at 9:41
1
yes, it is not required.
– Siong Thye Goh
Nov 21 '17 at 9:42
Thank you! I am so amazed how quickly the community here helps you out. This is great :)
– Avocado
Nov 21 '17 at 9:43
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?
– Avocado
Nov 21 '17 at 9:41
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?
– Avocado
Nov 21 '17 at 9:41
1
1
yes, it is not required.
– Siong Thye Goh
Nov 21 '17 at 9:42
yes, it is not required.
– Siong Thye Goh
Nov 21 '17 at 9:42
Thank you! I am so amazed how quickly the community here helps you out. This is great :)
– Avocado
Nov 21 '17 at 9:43
Thank you! I am so amazed how quickly the community here helps you out. This is great :)
– Avocado
Nov 21 '17 at 9:43
add a comment |
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).
answered Jan 10 '18 at 4:21
Fernando Sotomayor
111
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).
– Xander Henderson
Jan 10 '18 at 4:43
add a comment |
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).
answered Jan 10 '18 at 4:21
Fernando Sotomayor
111
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).
– Xander Henderson
Jan 10 '18 at 4:43
add a comment |
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).
answered Jan 10 '18 at 4:21
Fernando Sotomayor
111
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).
answered Jan 10 '18 at 4:21
Fernando Sotomayor
111
answered Jan 10 '18 at 4:21
Fernando Sotomayor
111
answered Jan 10 '18 at 4:21
Fernando Sotomayor
111
answered Jan 10 '18 at 4:21
Fernando Sotomayor
111
111
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).
– Xander Henderson
Jan 10 '18 at 4:43
add a comment |
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).
– Xander Henderson
Jan 10 '18 at 4:43
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).
– Xander Henderson
Jan 10 '18 at 4:43
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).
– Xander Henderson
Jan 10 '18 at 4:43
add a comment |
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
edited Dec 28 '18 at 4:59
dantopa
6,42932042
answered Dec 28 '18 at 4:48
Amar rohith
1
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/…
– dantopa
Dec 28 '18 at 5:00
add a comment |
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
edited Dec 28 '18 at 4:59
dantopa
6,42932042
answered Dec 28 '18 at 4:48
Amar rohith
1
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/…
– dantopa
Dec 28 '18 at 5:00
add a comment |
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
edited Dec 28 '18 at 4:59
dantopa
6,42932042
answered Dec 28 '18 at 4:48
Amar rohith
1
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
edited Dec 28 '18 at 4:59
dantopa
6,42932042
answered Dec 28 '18 at 4:48
Amar rohith
1
edited Dec 28 '18 at 4:59
dantopa
6,42932042
edited Dec 28 '18 at 4:59
dantopa
6,42932042
edited Dec 28 '18 at 4:59
dantopa
6,42932042
6,42932042
answered Dec 28 '18 at 4:48
Amar rohith
1
answered Dec 28 '18 at 4:48
Amar rohith
1
answered Dec 28 '18 at 4:48
Amar rohith
1
1
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/…
– dantopa
Dec 28 '18 at 5:00
add a comment |
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/…
– dantopa
Dec 28 '18 at 5:00
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/…
– dantopa
Dec 28 '18 at 5:00
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/…
– dantopa
Dec 28 '18 at 5:00
add a comment |
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2
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.)
– Stefan Mesken
Nov 21 '17 at 9:29
@StefanMesken Thanks a lot! I edited it. It is still confusing me though.
– Avocado
Nov 21 '17 at 9:37