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Symmetric relation confusion
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My textbook definition on symmetric relation:
A relation R on a set A is called symmetric if(b, a) ∈ R whenever(a, b) ∈ R, for all a, b ∈ A.
Giving a set $A= 1,2,3,4$
and a relation $R1 = (1,1) (1,2)(2,1).$
$R2 = (1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1), (4, 4)$
How come both R1 and R2 are symmetric since they haven't considered other relations such as (1,3), (2,3)etc. Isn't for all a, b ∈ A in the definition means you have to consider every single element in set A?
discrete-mathematics relations
asked Jul 28 at 9:53
stevie lol
306
add a comment |Â
up vote
2
down vote
favorite
My textbook definition on symmetric relation:
A relation R on a set A is called symmetric if(b, a) ∈ R whenever(a, b) ∈ R, for all a, b ∈ A.
Giving a set $A= 1,2,3,4$
and a relation $R1 = (1,1) (1,2)(2,1).$
$R2 = (1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1), (4, 4)$
How come both R1 and R2 are symmetric since they haven't considered other relations such as (1,3), (2,3)etc. Isn't for all a, b ∈ A in the definition means you have to consider every single element in set A?
discrete-mathematics relations
asked Jul 28 at 9:53
stevie lol
306
3
"whenever $(a,b)in R$". With only mathematical symbols, you can write it as $forall a,bin A, ((a,b)in R Rightarrow (b,a)in R)$.
– Suzet
Jul 28 at 9:55
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
My textbook definition on symmetric relation:
A relation R on a set A is called symmetric if(b, a) ∈ R whenever(a, b) ∈ R, for all a, b ∈ A.
Giving a set $A= 1,2,3,4$
and a relation $R1 = (1,1) (1,2)(2,1).$
$R2 = (1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1), (4, 4)$
How come both R1 and R2 are symmetric since they haven't considered other relations such as (1,3), (2,3)etc. Isn't for all a, b ∈ A in the definition means you have to consider every single element in set A?
discrete-mathematics relations
asked Jul 28 at 9:53
stevie lol
306
My textbook definition on symmetric relation:
A relation R on a set A is called symmetric if(b, a) ∈ R whenever(a, b) ∈ R, for all a, b ∈ A.
Giving a set $A= 1,2,3,4$
and a relation $R1 = (1,1) (1,2)(2,1).$
$R2 = (1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1), (4, 4)$
How come both R1 and R2 are symmetric since they haven't considered other relations such as (1,3), (2,3)etc. Isn't for all a, b ∈ A in the definition means you have to consider every single element in set A?
discrete-mathematics relations
asked Jul 28 at 9:53
stevie lol
306
asked Jul 28 at 9:53
stevie lol
306
asked Jul 28 at 9:53
stevie lol
306
asked Jul 28 at 9:53
stevie lol
306
306
3
"whenever $(a,b)in R$". With only mathematical symbols, you can write it as $forall a,bin A, ((a,b)in R Rightarrow (b,a)in R)$.
– Suzet
Jul 28 at 9:55
add a comment |Â
3
"whenever $(a,b)in R$". With only mathematical symbols, you can write it as $forall a,bin A, ((a,b)in R Rightarrow (b,a)in R)$.
– Suzet
Jul 28 at 9:55
3
3
"whenever $(a,b)in R$". With only mathematical symbols, you can write it as $forall a,bin A, ((a,b)in R Rightarrow (b,a)in R)$.
– Suzet
Jul 28 at 9:55
"whenever $(a,b)in R$". With only mathematical symbols, you can write it as $forall a,bin A, ((a,b)in R Rightarrow (b,a)in R)$.
– Suzet
Jul 28 at 9:55
add a comment |Â
4 Answers
4
active
oldest
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up vote
5
down vote
$forall x~(A(x)to B(x))$ is a promise that $B(x)$ will be true whenever $A(x)$ is true. Â No promise is about what may happen when $A(x)$ is false. Â It is only falsified when a counter example, say $c$, can be found where $A(c)$ is true but $B(c)$ is false.
So the criteria for relation to have symmetry when every pair that is in the relation has its corresponding symmetric pair also in the relation. $$forall xforall y~((x,y)in R~to~ (y,x)in R)$$
Although that does in fact mean that for any pair that is not in the relation, then its corresponding symmetric pair will also not be in the relation.$$forall xforall y~((y,x)notin R~to~ (x,y)notin R)$$
The claims are equivalent.
answered Jul 28 at 10:05
Graham Kemp
80k43275
add a comment |Â
up vote
4
down vote
Let us consider $R_1$. Yes, it is symmetric. What this means is: for each $a$ and each $b$ in $A$, if $(a,b)in R_1$, then $(b,a)in R_1$ too. So, for each $a$ and each $b$ in $A$, see whether it belongs to $R_1$. If it doesn't, forget it; there's nothing to deduce from that. Otherwise, you have to check whether $(b,a)$ belongs to $R_1$ too.
In particular, the fact that $(1,3)notin R_1$ is not a problem. On the other hand, if $(1,3)in R_1$ but $(3,1)notin R_1$, then, yes, we would be able to deduce that $R_1$ is not symmetric.
edited Jul 28 at 17:19
Especially Lime
19.1k22252
answered Jul 28 at 9:58
José Carlos Santos
112k1696173
add a comment |Â
up vote
2
down vote
You have to read the word "whenever" in the definition correctly. I suspect you haven't copied the definition exactly as it appears. You are misinterpreting the "for all $a, b in A$".
The right way to say this is
... reflexive if and only if $(a,b) in R$ whenever $(b,a) in R$ .
That does not say $(a,b)$ is in $R$ for all $a$ and $b$.
answered Jul 28 at 10:00
Ethan Bolker
35.7k54199
add a comment |Â
up vote
2
down vote
Also for e.g. $(1,3)$ it is true that $(1,3)in R_1$ implies that $(3,1)in R_1$.
This because $(1,3)in R_1$ is a false statement and "ex falsum sequitur quodlibet" (from a false statement follows whatever you want).
In short: if $p$ is false then $pimplies q$ (or equivalently $neg pvee q$) is true).
Especially a look on $neg pvee q$ makes things more clear. If $p$ is false then of course $neg p$ is true and consequently also $neg pvee q$ is true.
answered Jul 28 at 12:36
Vera
1,696313
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
$forall x~(A(x)to B(x))$ is a promise that $B(x)$ will be true whenever $A(x)$ is true. Â No promise is about what may happen when $A(x)$ is false. Â It is only falsified when a counter example, say $c$, can be found where $A(c)$ is true but $B(c)$ is false.
So the criteria for relation to have symmetry when every pair that is in the relation has its corresponding symmetric pair also in the relation. $$forall xforall y~((x,y)in R~to~ (y,x)in R)$$
Although that does in fact mean that for any pair that is not in the relation, then its corresponding symmetric pair will also not be in the relation.$$forall xforall y~((y,x)notin R~to~ (x,y)notin R)$$
The claims are equivalent.
answered Jul 28 at 10:05
Graham Kemp
80k43275
add a comment |Â
up vote
5
down vote
$forall x~(A(x)to B(x))$ is a promise that $B(x)$ will be true whenever $A(x)$ is true. Â No promise is about what may happen when $A(x)$ is false. Â It is only falsified when a counter example, say $c$, can be found where $A(c)$ is true but $B(c)$ is false.
So the criteria for relation to have symmetry when every pair that is in the relation has its corresponding symmetric pair also in the relation. $$forall xforall y~((x,y)in R~to~ (y,x)in R)$$
Although that does in fact mean that for any pair that is not in the relation, then its corresponding symmetric pair will also not be in the relation.$$forall xforall y~((y,x)notin R~to~ (x,y)notin R)$$
The claims are equivalent.
answered Jul 28 at 10:05
Graham Kemp
80k43275
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$forall x~(A(x)to B(x))$ is a promise that $B(x)$ will be true whenever $A(x)$ is true. Â No promise is about what may happen when $A(x)$ is false. Â It is only falsified when a counter example, say $c$, can be found where $A(c)$ is true but $B(c)$ is false.
So the criteria for relation to have symmetry when every pair that is in the relation has its corresponding symmetric pair also in the relation. $$forall xforall y~((x,y)in R~to~ (y,x)in R)$$
Although that does in fact mean that for any pair that is not in the relation, then its corresponding symmetric pair will also not be in the relation.$$forall xforall y~((y,x)notin R~to~ (x,y)notin R)$$
The claims are equivalent.
answered Jul 28 at 10:05
Graham Kemp
80k43275
$forall x~(A(x)to B(x))$ is a promise that $B(x)$ will be true whenever $A(x)$ is true. Â No promise is about what may happen when $A(x)$ is false. Â It is only falsified when a counter example, say $c$, can be found where $A(c)$ is true but $B(c)$ is false.
So the criteria for relation to have symmetry when every pair that is in the relation has its corresponding symmetric pair also in the relation. $$forall xforall y~((x,y)in R~to~ (y,x)in R)$$
Although that does in fact mean that for any pair that is not in the relation, then its corresponding symmetric pair will also not be in the relation.$$forall xforall y~((y,x)notin R~to~ (x,y)notin R)$$
The claims are equivalent.
answered Jul 28 at 10:05
Graham Kemp
80k43275
answered Jul 28 at 10:05
Graham Kemp
80k43275
answered Jul 28 at 10:05
Graham Kemp
80k43275
answered Jul 28 at 10:05
Graham Kemp
80k43275
80k43275
add a comment |Â
add a comment |Â
up vote
4
down vote
Let us consider $R_1$. Yes, it is symmetric. What this means is: for each $a$ and each $b$ in $A$, if $(a,b)in R_1$, then $(b,a)in R_1$ too. So, for each $a$ and each $b$ in $A$, see whether it belongs to $R_1$. If it doesn't, forget it; there's nothing to deduce from that. Otherwise, you have to check whether $(b,a)$ belongs to $R_1$ too.
In particular, the fact that $(1,3)notin R_1$ is not a problem. On the other hand, if $(1,3)in R_1$ but $(3,1)notin R_1$, then, yes, we would be able to deduce that $R_1$ is not symmetric.
edited Jul 28 at 17:19
Especially Lime
19.1k22252
answered Jul 28 at 9:58
José Carlos Santos
112k1696173
add a comment |Â
up vote
4
down vote
Let us consider $R_1$. Yes, it is symmetric. What this means is: for each $a$ and each $b$ in $A$, if $(a,b)in R_1$, then $(b,a)in R_1$ too. So, for each $a$ and each $b$ in $A$, see whether it belongs to $R_1$. If it doesn't, forget it; there's nothing to deduce from that. Otherwise, you have to check whether $(b,a)$ belongs to $R_1$ too.
In particular, the fact that $(1,3)notin R_1$ is not a problem. On the other hand, if $(1,3)in R_1$ but $(3,1)notin R_1$, then, yes, we would be able to deduce that $R_1$ is not symmetric.
edited Jul 28 at 17:19
Especially Lime
19.1k22252
answered Jul 28 at 9:58
José Carlos Santos
112k1696173
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let us consider $R_1$. Yes, it is symmetric. What this means is: for each $a$ and each $b$ in $A$, if $(a,b)in R_1$, then $(b,a)in R_1$ too. So, for each $a$ and each $b$ in $A$, see whether it belongs to $R_1$. If it doesn't, forget it; there's nothing to deduce from that. Otherwise, you have to check whether $(b,a)$ belongs to $R_1$ too.
In particular, the fact that $(1,3)notin R_1$ is not a problem. On the other hand, if $(1,3)in R_1$ but $(3,1)notin R_1$, then, yes, we would be able to deduce that $R_1$ is not symmetric.
edited Jul 28 at 17:19
Especially Lime
19.1k22252
answered Jul 28 at 9:58
José Carlos Santos
112k1696173
Let us consider $R_1$. Yes, it is symmetric. What this means is: for each $a$ and each $b$ in $A$, if $(a,b)in R_1$, then $(b,a)in R_1$ too. So, for each $a$ and each $b$ in $A$, see whether it belongs to $R_1$. If it doesn't, forget it; there's nothing to deduce from that. Otherwise, you have to check whether $(b,a)$ belongs to $R_1$ too.
In particular, the fact that $(1,3)notin R_1$ is not a problem. On the other hand, if $(1,3)in R_1$ but $(3,1)notin R_1$, then, yes, we would be able to deduce that $R_1$ is not symmetric.
edited Jul 28 at 17:19
Especially Lime
19.1k22252
answered Jul 28 at 9:58
José Carlos Santos
112k1696173
edited Jul 28 at 17:19
Especially Lime
19.1k22252
edited Jul 28 at 17:19
Especially Lime
19.1k22252
edited Jul 28 at 17:19
Especially Lime
19.1k22252
19.1k22252
answered Jul 28 at 9:58
José Carlos Santos
112k1696173
answered Jul 28 at 9:58
José Carlos Santos
112k1696173
answered Jul 28 at 9:58
José Carlos Santos
112k1696173
112k1696173
add a comment |Â
add a comment |Â
up vote
2
down vote
You have to read the word "whenever" in the definition correctly. I suspect you haven't copied the definition exactly as it appears. You are misinterpreting the "for all $a, b in A$".
The right way to say this is
... reflexive if and only if $(a,b) in R$ whenever $(b,a) in R$ .
That does not say $(a,b)$ is in $R$ for all $a$ and $b$.
answered Jul 28 at 10:00
Ethan Bolker
35.7k54199
add a comment |Â
up vote
2
down vote
You have to read the word "whenever" in the definition correctly. I suspect you haven't copied the definition exactly as it appears. You are misinterpreting the "for all $a, b in A$".
The right way to say this is
... reflexive if and only if $(a,b) in R$ whenever $(b,a) in R$ .
That does not say $(a,b)$ is in $R$ for all $a$ and $b$.
answered Jul 28 at 10:00
Ethan Bolker
35.7k54199
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You have to read the word "whenever" in the definition correctly. I suspect you haven't copied the definition exactly as it appears. You are misinterpreting the "for all $a, b in A$".
The right way to say this is
... reflexive if and only if $(a,b) in R$ whenever $(b,a) in R$ .
That does not say $(a,b)$ is in $R$ for all $a$ and $b$.
answered Jul 28 at 10:00
Ethan Bolker
35.7k54199
You have to read the word "whenever" in the definition correctly. I suspect you haven't copied the definition exactly as it appears. You are misinterpreting the "for all $a, b in A$".
The right way to say this is
... reflexive if and only if $(a,b) in R$ whenever $(b,a) in R$ .
That does not say $(a,b)$ is in $R$ for all $a$ and $b$.
answered Jul 28 at 10:00
Ethan Bolker
35.7k54199
answered Jul 28 at 10:00
Ethan Bolker
35.7k54199
answered Jul 28 at 10:00
Ethan Bolker
35.7k54199
answered Jul 28 at 10:00
Ethan Bolker
35.7k54199
35.7k54199
add a comment |Â
add a comment |Â
up vote
2
down vote
Also for e.g. $(1,3)$ it is true that $(1,3)in R_1$ implies that $(3,1)in R_1$.
This because $(1,3)in R_1$ is a false statement and "ex falsum sequitur quodlibet" (from a false statement follows whatever you want).
In short: if $p$ is false then $pimplies q$ (or equivalently $neg pvee q$) is true).
Especially a look on $neg pvee q$ makes things more clear. If $p$ is false then of course $neg p$ is true and consequently also $neg pvee q$ is true.
answered Jul 28 at 12:36
Vera
1,696313
add a comment |Â
up vote
2
down vote
Also for e.g. $(1,3)$ it is true that $(1,3)in R_1$ implies that $(3,1)in R_1$.
This because $(1,3)in R_1$ is a false statement and "ex falsum sequitur quodlibet" (from a false statement follows whatever you want).
In short: if $p$ is false then $pimplies q$ (or equivalently $neg pvee q$) is true).
Especially a look on $neg pvee q$ makes things more clear. If $p$ is false then of course $neg p$ is true and consequently also $neg pvee q$ is true.
answered Jul 28 at 12:36
Vera
1,696313
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Also for e.g. $(1,3)$ it is true that $(1,3)in R_1$ implies that $(3,1)in R_1$.
This because $(1,3)in R_1$ is a false statement and "ex falsum sequitur quodlibet" (from a false statement follows whatever you want).
In short: if $p$ is false then $pimplies q$ (or equivalently $neg pvee q$) is true).
Especially a look on $neg pvee q$ makes things more clear. If $p$ is false then of course $neg p$ is true and consequently also $neg pvee q$ is true.
answered Jul 28 at 12:36
Vera
1,696313
Also for e.g. $(1,3)$ it is true that $(1,3)in R_1$ implies that $(3,1)in R_1$.
This because $(1,3)in R_1$ is a false statement and "ex falsum sequitur quodlibet" (from a false statement follows whatever you want).
In short: if $p$ is false then $pimplies q$ (or equivalently $neg pvee q$) is true).
Especially a look on $neg pvee q$ makes things more clear. If $p$ is false then of course $neg p$ is true and consequently also $neg pvee q$ is true.
answered Jul 28 at 12:36
Vera
1,696313
edited Jul 28 at 12:42
answered Jul 28 at 12:36
Vera
1,696313
answered Jul 28 at 12:36
Vera
1,696313
answered Jul 28 at 12:36
Vera
1,696313
1,696313
add a comment |Â
add a comment |Â
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3
"whenever $(a,b)in R$". With only mathematical symbols, you can write it as $forall a,bin A, ((a,b)in R Rightarrow (b,a)in R)$.
– Suzet
Jul 28 at 9:55