Why is a symmetric relation defined: $forall xforall y( xRyimplies yRx)$ and not $forall xforall y (xRyiff yRx)$?How to prove relation is asymmetric if it is both anti-symmetric and irreflexiveIs an Anti-Symmetric Relation also Reflexive?Graph, Relation $xRy Leftrightarrow$ There is a path between $x$ and $y$ - symmetryCan someone explain antisymmetric versus symmetric relation of sets?Define symmetric relation R on set SProve that if $R$ is a symmetric, transitive relation on $A$ and the domain of $R$ is $A$, then $R$ is reflexive on $A$.If R is symmetric, must $S= (X,Y)in P(A)times P(A) $ be symmetric?Given set A, is the relation A x A always anti symmetric?Definition of symmetric relationCan a relation be transitive when it is symmetric but not reflexive?
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Why is a symmetric relation defined: $forall xforall y( xRyimplies yRx)$ and not $forall xforall y (xRyiff yRx)$?
How to prove relation is asymmetric if it is both anti-symmetric and irreflexiveIs an Anti-Symmetric Relation also Reflexive?Graph, Relation $xRy Leftrightarrow$ There is a path between $x$ and $y$ - symmetryCan someone explain antisymmetric versus symmetric relation of sets?Define symmetric relation R on set SProve that if $R$ is a symmetric, transitive relation on $A$ and the domain of $R$ is $A$, then $R$ is reflexive on $A$.If R is symmetric, must $S=forall xin X exists yin Y (xRy) $ be symmetric?Given set A, is the relation A x A always anti symmetric?Definition of symmetric relationCan a relation be transitive when it is symmetric but not reflexive?
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Why is a symmetric relation defined by $forallxforally(xRy implies yRx)$ and not $forallxforally(xRy iff yRx)$?
(I have only found a couple of sources that defines it with a biconditional)
For example, according to Wolfram:
A relation $R$ on a set $S$ is symmetric provided that for every $x$ and $y$ in $S$ we have $xRy iff yRx$.
But the majority of books defines it the other way.
And I think I agree with the second definition.
Because if we use the first definition with "$implies$", we know the truth table of the implication in particular $P implies Q$ is true when $P$ is false and $Q$ is true. That means in the context of symmetric relation that $(x,y) notin R implies (y,x) in R$ is true.
And the example $A = 1,2,3,4$ with relation $R = (2,1),(3,1),(4,1)$ satisfies the definition because $(x,y) notin R implies (y,x) in R$ is true.
And for me it's weird that this case is considered symmetric.
Or maybe I have a profound confusion with the concept.
I would like that you guys help me clarify. *Sorry for my grammar I'm not a native english speaker.
discrete-mathematics logic definition relations
$endgroup$
add a comment |
$begingroup$
Why is a symmetric relation defined by $forallxforally(xRy implies yRx)$ and not $forallxforally(xRy iff yRx)$?
(I have only found a couple of sources that defines it with a biconditional)
For example, according to Wolfram:
A relation $R$ on a set $S$ is symmetric provided that for every $x$ and $y$ in $S$ we have $xRy iff yRx$.
But the majority of books defines it the other way.
And I think I agree with the second definition.
Because if we use the first definition with "$implies$", we know the truth table of the implication in particular $P implies Q$ is true when $P$ is false and $Q$ is true. That means in the context of symmetric relation that $(x,y) notin R implies (y,x) in R$ is true.
And the example $A = 1,2,3,4$ with relation $R = (2,1),(3,1),(4,1)$ satisfies the definition because $(x,y) notin R implies (y,x) in R$ is true.
And for me it's weird that this case is considered symmetric.
Or maybe I have a profound confusion with the concept.
I would like that you guys help me clarify. *Sorry for my grammar I'm not a native english speaker.
discrete-mathematics logic definition relations
$endgroup$
$begingroup$
Sorry i will correct it
$endgroup$
– Rodrigo Sango
6 hours ago
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It's for ALL x, y. In your example $(2,1)in R not implies (1,2)in R$. But given $A=1,2,3$ and $R=(1,2),(2,1)$ we have $(1,3)in R implies (3,1)in R$ etc.
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– fleablood
6 hours ago
1
$begingroup$
"That means in the context of symmetric relation that (x,y)∉ R ⟹ (y,x)∈ R is true" Only if $x,y$ is actually in $R$. $(2,1)not in R implies (1,2)in R$ is a true statement. But $(3,2)not in R implies (2,3)in R$ is a false statement.
$endgroup$
– fleablood
6 hours ago
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
2 hours ago
add a comment |
$begingroup$
Why is a symmetric relation defined by $forallxforally(xRy implies yRx)$ and not $forallxforally(xRy iff yRx)$?
(I have only found a couple of sources that defines it with a biconditional)
For example, according to Wolfram:
A relation $R$ on a set $S$ is symmetric provided that for every $x$ and $y$ in $S$ we have $xRy iff yRx$.
But the majority of books defines it the other way.
And I think I agree with the second definition.
Because if we use the first definition with "$implies$", we know the truth table of the implication in particular $P implies Q$ is true when $P$ is false and $Q$ is true. That means in the context of symmetric relation that $(x,y) notin R implies (y,x) in R$ is true.
And the example $A = 1,2,3,4$ with relation $R = (2,1),(3,1),(4,1)$ satisfies the definition because $(x,y) notin R implies (y,x) in R$ is true.
And for me it's weird that this case is considered symmetric.
Or maybe I have a profound confusion with the concept.
I would like that you guys help me clarify. *Sorry for my grammar I'm not a native english speaker.
discrete-mathematics logic definition relations
$endgroup$
Why is a symmetric relation defined by $forallxforally(xRy implies yRx)$ and not $forallxforally(xRy iff yRx)$?
(I have only found a couple of sources that defines it with a biconditional)
For example, according to Wolfram:
A relation $R$ on a set $S$ is symmetric provided that for every $x$ and $y$ in $S$ we have $xRy iff yRx$.
But the majority of books defines it the other way.
And I think I agree with the second definition.
Because if we use the first definition with "$implies$", we know the truth table of the implication in particular $P implies Q$ is true when $P$ is false and $Q$ is true. That means in the context of symmetric relation that $(x,y) notin R implies (y,x) in R$ is true.
And the example $A = 1,2,3,4$ with relation $R = (2,1),(3,1),(4,1)$ satisfies the definition because $(x,y) notin R implies (y,x) in R$ is true.
And for me it's weird that this case is considered symmetric.
Or maybe I have a profound confusion with the concept.
I would like that you guys help me clarify. *Sorry for my grammar I'm not a native english speaker.
discrete-mathematics logic definition relations
discrete-mathematics logic definition relations
edited 8 mins ago
Asaf Karagila♦
307k33438769
307k33438769
asked 6 hours ago
Rodrigo SangoRodrigo Sango
1276
1276
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Sorry i will correct it
$endgroup$
– Rodrigo Sango
6 hours ago
$begingroup$
It's for ALL x, y. In your example $(2,1)in R not implies (1,2)in R$. But given $A=1,2,3$ and $R=(1,2),(2,1)$ we have $(1,3)in R implies (3,1)in R$ etc.
$endgroup$
– fleablood
6 hours ago
1
$begingroup$
"That means in the context of symmetric relation that (x,y)∉ R ⟹ (y,x)∈ R is true" Only if $x,y$ is actually in $R$. $(2,1)not in R implies (1,2)in R$ is a true statement. But $(3,2)not in R implies (2,3)in R$ is a false statement.
$endgroup$
– fleablood
6 hours ago
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
2 hours ago
add a comment |
$begingroup$
Sorry i will correct it
$endgroup$
– Rodrigo Sango
6 hours ago
$begingroup$
It's for ALL x, y. In your example $(2,1)in R not implies (1,2)in R$. But given $A=1,2,3$ and $R=(1,2),(2,1)$ we have $(1,3)in R implies (3,1)in R$ etc.
$endgroup$
– fleablood
6 hours ago
1
$begingroup$
"That means in the context of symmetric relation that (x,y)∉ R ⟹ (y,x)∈ R is true" Only if $x,y$ is actually in $R$. $(2,1)not in R implies (1,2)in R$ is a true statement. But $(3,2)not in R implies (2,3)in R$ is a false statement.
$endgroup$
– fleablood
6 hours ago
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
2 hours ago
$begingroup$
Sorry i will correct it
$endgroup$
– Rodrigo Sango
6 hours ago
$begingroup$
Sorry i will correct it
$endgroup$
– Rodrigo Sango
6 hours ago
$begingroup$
It's for ALL x, y. In your example $(2,1)in R not implies (1,2)in R$. But given $A=1,2,3$ and $R=(1,2),(2,1)$ we have $(1,3)in R implies (3,1)in R$ etc.
$endgroup$
– fleablood
6 hours ago
$begingroup$
It's for ALL x, y. In your example $(2,1)in R not implies (1,2)in R$. But given $A=1,2,3$ and $R=(1,2),(2,1)$ we have $(1,3)in R implies (3,1)in R$ etc.
$endgroup$
– fleablood
6 hours ago
1
1
$begingroup$
"That means in the context of symmetric relation that (x,y)∉ R ⟹ (y,x)∈ R is true" Only if $x,y$ is actually in $R$. $(2,1)not in R implies (1,2)in R$ is a true statement. But $(3,2)not in R implies (2,3)in R$ is a false statement.
$endgroup$
– fleablood
6 hours ago
$begingroup$
"That means in the context of symmetric relation that (x,y)∉ R ⟹ (y,x)∈ R is true" Only if $x,y$ is actually in $R$. $(2,1)not in R implies (1,2)in R$ is a true statement. But $(3,2)not in R implies (2,3)in R$ is a false statement.
$endgroup$
– fleablood
6 hours ago
2
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
2 hours ago
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
2 hours ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
For all $x$ and all $y$ make the if and only if unnecessary (albeit perfectly acceptable).
1) $(x,y) in R implies (y,x) in R$ for ALL $x,y in A$
And the statement 2) $(x,y) in R iff (y,x) in R$ are equivalent statements.
If 1) is true and $(x,y) not in R$ then although $(x,y)in Rimplies (y,x)in R$ or $F implies (y,x)in R$ is true, it does not tell us any thing about whether or not $(y,x) in R$. However $(y,x) in R implies (x,y) in Y$ tells us that $(y,x) not in R$. Because $(y,x) in R implies (x,y) in R$ means $(y,x) in R implies F$. An the only thing that implies a false statement is a false statement. So we must have $(y,x) not in R$.
So in your example you have $(1,2)in Rimplies (2,1)in R$ is true but you don't have $(2,1) in R implies (1,2) in R$ as true.
So it isn't symmetric.
=====
Another way to look at it:
If $A = 1,2,3$
Then we will have 9 statments.
By 1) the nine statements are:
$(1,1)in Rimplies (1,1) in R$
$(1,2) in R implies (2,1) in R$
$(1,3) in R implies (3,1) in R$
$(2,1) in R implies (1,2) in R$
... etc... all nine are needed.
With 2) we also have nine statements:
$(1,1)in Riff (1,1) in R$
$(1,2) in R iff (2,1) in R$
$(1,3) in R iff (3,1) in R$
$(2,1) in R iff (1,2) in R$
...etc....
$(1,2) in R iff (2,1) in R$ and $(2,1) in R iff (1,2)in R$ is redundant.
So aesthetically, using definition 2) is .... inefficient.
$endgroup$
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
5 hours ago
1
$begingroup$
"all nine are needed" ... except the three that are tautologies, of course. But it would be more work and less clear to exclude them than to tolerate them.
$endgroup$
– Henning Makholm
2 hours ago
add a comment |
$begingroup$
If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$So, why do we choose $(1)$ instead of $(2)$ in general? Because, in general (although not in this case) it is easier to verify the assertion $Aimplies B$ than $Aiff B$. And (again, in general), when we choose between two distinct but equivalent definitions, we usually choose the one which is easier to verify that it holds.
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add a comment |
$begingroup$
If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.
In conclusion, since the implication should hold for all $x,y$, the equivalence already holds.
$endgroup$
add a comment |
$begingroup$
Let it be that $R$ is a symmetric relation.
This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$
Now let it be that $aRb$.
Then we are allowed to conclude that $bRa$.
On the other hand if $bRa$ then also we are conclude that $aRb$.
So apparantly we have:$$aRbiff bRa$$
Proved is now that for a symmetric relation $R$ (based on definition $(1)$) we have:$$forall xforall y[xRyiff yRx]tag2$$
So $(2)$ is a necessary condition for $(1)$.
Next to that it is obvious that $(2)$ is also a sufficient condition for $(1)$ so actually the statements are equivalent.
Both can be used as definition then, but in cases that like that it is good custom to go for the one with less implications.
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add a comment |
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4 Answers
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active
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4 Answers
4
active
oldest
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$begingroup$
For all $x$ and all $y$ make the if and only if unnecessary (albeit perfectly acceptable).
1) $(x,y) in R implies (y,x) in R$ for ALL $x,y in A$
And the statement 2) $(x,y) in R iff (y,x) in R$ are equivalent statements.
If 1) is true and $(x,y) not in R$ then although $(x,y)in Rimplies (y,x)in R$ or $F implies (y,x)in R$ is true, it does not tell us any thing about whether or not $(y,x) in R$. However $(y,x) in R implies (x,y) in Y$ tells us that $(y,x) not in R$. Because $(y,x) in R implies (x,y) in R$ means $(y,x) in R implies F$. An the only thing that implies a false statement is a false statement. So we must have $(y,x) not in R$.
So in your example you have $(1,2)in Rimplies (2,1)in R$ is true but you don't have $(2,1) in R implies (1,2) in R$ as true.
So it isn't symmetric.
=====
Another way to look at it:
If $A = 1,2,3$
Then we will have 9 statments.
By 1) the nine statements are:
$(1,1)in Rimplies (1,1) in R$
$(1,2) in R implies (2,1) in R$
$(1,3) in R implies (3,1) in R$
$(2,1) in R implies (1,2) in R$
... etc... all nine are needed.
With 2) we also have nine statements:
$(1,1)in Riff (1,1) in R$
$(1,2) in R iff (2,1) in R$
$(1,3) in R iff (3,1) in R$
$(2,1) in R iff (1,2) in R$
...etc....
$(1,2) in R iff (2,1) in R$ and $(2,1) in R iff (1,2)in R$ is redundant.
So aesthetically, using definition 2) is .... inefficient.
$endgroup$
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
5 hours ago
1
$begingroup$
"all nine are needed" ... except the three that are tautologies, of course. But it would be more work and less clear to exclude them than to tolerate them.
$endgroup$
– Henning Makholm
2 hours ago
add a comment |
$begingroup$
For all $x$ and all $y$ make the if and only if unnecessary (albeit perfectly acceptable).
1) $(x,y) in R implies (y,x) in R$ for ALL $x,y in A$
And the statement 2) $(x,y) in R iff (y,x) in R$ are equivalent statements.
If 1) is true and $(x,y) not in R$ then although $(x,y)in Rimplies (y,x)in R$ or $F implies (y,x)in R$ is true, it does not tell us any thing about whether or not $(y,x) in R$. However $(y,x) in R implies (x,y) in Y$ tells us that $(y,x) not in R$. Because $(y,x) in R implies (x,y) in R$ means $(y,x) in R implies F$. An the only thing that implies a false statement is a false statement. So we must have $(y,x) not in R$.
So in your example you have $(1,2)in Rimplies (2,1)in R$ is true but you don't have $(2,1) in R implies (1,2) in R$ as true.
So it isn't symmetric.
=====
Another way to look at it:
If $A = 1,2,3$
Then we will have 9 statments.
By 1) the nine statements are:
$(1,1)in Rimplies (1,1) in R$
$(1,2) in R implies (2,1) in R$
$(1,3) in R implies (3,1) in R$
$(2,1) in R implies (1,2) in R$
... etc... all nine are needed.
With 2) we also have nine statements:
$(1,1)in Riff (1,1) in R$
$(1,2) in R iff (2,1) in R$
$(1,3) in R iff (3,1) in R$
$(2,1) in R iff (1,2) in R$
...etc....
$(1,2) in R iff (2,1) in R$ and $(2,1) in R iff (1,2)in R$ is redundant.
So aesthetically, using definition 2) is .... inefficient.
$endgroup$
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
5 hours ago
1
$begingroup$
"all nine are needed" ... except the three that are tautologies, of course. But it would be more work and less clear to exclude them than to tolerate them.
$endgroup$
– Henning Makholm
2 hours ago
add a comment |
$begingroup$
For all $x$ and all $y$ make the if and only if unnecessary (albeit perfectly acceptable).
1) $(x,y) in R implies (y,x) in R$ for ALL $x,y in A$
And the statement 2) $(x,y) in R iff (y,x) in R$ are equivalent statements.
If 1) is true and $(x,y) not in R$ then although $(x,y)in Rimplies (y,x)in R$ or $F implies (y,x)in R$ is true, it does not tell us any thing about whether or not $(y,x) in R$. However $(y,x) in R implies (x,y) in Y$ tells us that $(y,x) not in R$. Because $(y,x) in R implies (x,y) in R$ means $(y,x) in R implies F$. An the only thing that implies a false statement is a false statement. So we must have $(y,x) not in R$.
So in your example you have $(1,2)in Rimplies (2,1)in R$ is true but you don't have $(2,1) in R implies (1,2) in R$ as true.
So it isn't symmetric.
=====
Another way to look at it:
If $A = 1,2,3$
Then we will have 9 statments.
By 1) the nine statements are:
$(1,1)in Rimplies (1,1) in R$
$(1,2) in R implies (2,1) in R$
$(1,3) in R implies (3,1) in R$
$(2,1) in R implies (1,2) in R$
... etc... all nine are needed.
With 2) we also have nine statements:
$(1,1)in Riff (1,1) in R$
$(1,2) in R iff (2,1) in R$
$(1,3) in R iff (3,1) in R$
$(2,1) in R iff (1,2) in R$
...etc....
$(1,2) in R iff (2,1) in R$ and $(2,1) in R iff (1,2)in R$ is redundant.
So aesthetically, using definition 2) is .... inefficient.
$endgroup$
For all $x$ and all $y$ make the if and only if unnecessary (albeit perfectly acceptable).
1) $(x,y) in R implies (y,x) in R$ for ALL $x,y in A$
And the statement 2) $(x,y) in R iff (y,x) in R$ are equivalent statements.
If 1) is true and $(x,y) not in R$ then although $(x,y)in Rimplies (y,x)in R$ or $F implies (y,x)in R$ is true, it does not tell us any thing about whether or not $(y,x) in R$. However $(y,x) in R implies (x,y) in Y$ tells us that $(y,x) not in R$. Because $(y,x) in R implies (x,y) in R$ means $(y,x) in R implies F$. An the only thing that implies a false statement is a false statement. So we must have $(y,x) not in R$.
So in your example you have $(1,2)in Rimplies (2,1)in R$ is true but you don't have $(2,1) in R implies (1,2) in R$ as true.
So it isn't symmetric.
=====
Another way to look at it:
If $A = 1,2,3$
Then we will have 9 statments.
By 1) the nine statements are:
$(1,1)in Rimplies (1,1) in R$
$(1,2) in R implies (2,1) in R$
$(1,3) in R implies (3,1) in R$
$(2,1) in R implies (1,2) in R$
... etc... all nine are needed.
With 2) we also have nine statements:
$(1,1)in Riff (1,1) in R$
$(1,2) in R iff (2,1) in R$
$(1,3) in R iff (3,1) in R$
$(2,1) in R iff (1,2) in R$
...etc....
$(1,2) in R iff (2,1) in R$ and $(2,1) in R iff (1,2)in R$ is redundant.
So aesthetically, using definition 2) is .... inefficient.
answered 5 hours ago
fleabloodfleablood
73k22789
73k22789
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
5 hours ago
1
$begingroup$
"all nine are needed" ... except the three that are tautologies, of course. But it would be more work and less clear to exclude them than to tolerate them.
$endgroup$
– Henning Makholm
2 hours ago
add a comment |
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
5 hours ago
1
$begingroup$
"all nine are needed" ... except the three that are tautologies, of course. But it would be more work and less clear to exclude them than to tolerate them.
$endgroup$
– Henning Makholm
2 hours ago
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
5 hours ago
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
5 hours ago
1
1
$begingroup$
"all nine are needed" ... except the three that are tautologies, of course. But it would be more work and less clear to exclude them than to tolerate them.
$endgroup$
– Henning Makholm
2 hours ago
$begingroup$
"all nine are needed" ... except the three that are tautologies, of course. But it would be more work and less clear to exclude them than to tolerate them.
$endgroup$
– Henning Makholm
2 hours ago
add a comment |
$begingroup$
If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$So, why do we choose $(1)$ instead of $(2)$ in general? Because, in general (although not in this case) it is easier to verify the assertion $Aimplies B$ than $Aiff B$. And (again, in general), when we choose between two distinct but equivalent definitions, we usually choose the one which is easier to verify that it holds.
$endgroup$
add a comment |
$begingroup$
If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$So, why do we choose $(1)$ instead of $(2)$ in general? Because, in general (although not in this case) it is easier to verify the assertion $Aimplies B$ than $Aiff B$. And (again, in general), when we choose between two distinct but equivalent definitions, we usually choose the one which is easier to verify that it holds.
$endgroup$
add a comment |
$begingroup$
If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$So, why do we choose $(1)$ instead of $(2)$ in general? Because, in general (although not in this case) it is easier to verify the assertion $Aimplies B$ than $Aiff B$. And (again, in general), when we choose between two distinct but equivalent definitions, we usually choose the one which is easier to verify that it holds.
$endgroup$
If $A$ is a set and $R$ is a binary relation defined on $A$ (that is, $R$ is a subset of $Atimes A$), then the usual definition of symmetry (as far as $R$ is concerned) is$$(forall xin A)(forall yin A):xmathrel Ryimplies ymathrel Rx.tag1$$And this is equivalent to$$(forall xin A)(forall yin A):xmathrel Ryiff ymathrel Rx.tag2$$So, why do we choose $(1)$ instead of $(2)$ in general? Because, in general (although not in this case) it is easier to verify the assertion $Aimplies B$ than $Aiff B$. And (again, in general), when we choose between two distinct but equivalent definitions, we usually choose the one which is easier to verify that it holds.
answered 6 hours ago
José Carlos SantosJosé Carlos Santos
169k23132237
169k23132237
add a comment |
add a comment |
$begingroup$
If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.
In conclusion, since the implication should hold for all $x,y$, the equivalence already holds.
$endgroup$
add a comment |
$begingroup$
If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.
In conclusion, since the implication should hold for all $x,y$, the equivalence already holds.
$endgroup$
add a comment |
$begingroup$
If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.
In conclusion, since the implication should hold for all $x,y$, the equivalence already holds.
$endgroup$
If $xRy implies yRx$ for all $x$ and all $y$, then we can choose $x := tildey$ and $y := tildex$ and get $tildeyRtildex implies tildexRtildey$ or, equivalently, $yRx implies xRy$, which is $impliedby$.
In conclusion, since the implication should hold for all $x,y$, the equivalence already holds.
answered 6 hours ago
Viktor GlombikViktor Glombik
1,2352528
1,2352528
add a comment |
add a comment |
$begingroup$
Let it be that $R$ is a symmetric relation.
This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$
Now let it be that $aRb$.
Then we are allowed to conclude that $bRa$.
On the other hand if $bRa$ then also we are conclude that $aRb$.
So apparantly we have:$$aRbiff bRa$$
Proved is now that for a symmetric relation $R$ (based on definition $(1)$) we have:$$forall xforall y[xRyiff yRx]tag2$$
So $(2)$ is a necessary condition for $(1)$.
Next to that it is obvious that $(2)$ is also a sufficient condition for $(1)$ so actually the statements are equivalent.
Both can be used as definition then, but in cases that like that it is good custom to go for the one with less implications.
$endgroup$
add a comment |
$begingroup$
Let it be that $R$ is a symmetric relation.
This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$
Now let it be that $aRb$.
Then we are allowed to conclude that $bRa$.
On the other hand if $bRa$ then also we are conclude that $aRb$.
So apparantly we have:$$aRbiff bRa$$
Proved is now that for a symmetric relation $R$ (based on definition $(1)$) we have:$$forall xforall y[xRyiff yRx]tag2$$
So $(2)$ is a necessary condition for $(1)$.
Next to that it is obvious that $(2)$ is also a sufficient condition for $(1)$ so actually the statements are equivalent.
Both can be used as definition then, but in cases that like that it is good custom to go for the one with less implications.
$endgroup$
add a comment |
$begingroup$
Let it be that $R$ is a symmetric relation.
This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$
Now let it be that $aRb$.
Then we are allowed to conclude that $bRa$.
On the other hand if $bRa$ then also we are conclude that $aRb$.
So apparantly we have:$$aRbiff bRa$$
Proved is now that for a symmetric relation $R$ (based on definition $(1)$) we have:$$forall xforall y[xRyiff yRx]tag2$$
So $(2)$ is a necessary condition for $(1)$.
Next to that it is obvious that $(2)$ is also a sufficient condition for $(1)$ so actually the statements are equivalent.
Both can be used as definition then, but in cases that like that it is good custom to go for the one with less implications.
$endgroup$
Let it be that $R$ is a symmetric relation.
This according to the first mentioned definition:$$forall xforall y[xRyimplies yRx]tag1$$
Now let it be that $aRb$.
Then we are allowed to conclude that $bRa$.
On the other hand if $bRa$ then also we are conclude that $aRb$.
So apparantly we have:$$aRbiff bRa$$
Proved is now that for a symmetric relation $R$ (based on definition $(1)$) we have:$$forall xforall y[xRyiff yRx]tag2$$
So $(2)$ is a necessary condition for $(1)$.
Next to that it is obvious that $(2)$ is also a sufficient condition for $(1)$ so actually the statements are equivalent.
Both can be used as definition then, but in cases that like that it is good custom to go for the one with less implications.
answered 6 hours ago
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
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$begingroup$
Sorry i will correct it
$endgroup$
– Rodrigo Sango
6 hours ago
$begingroup$
It's for ALL x, y. In your example $(2,1)in R not implies (1,2)in R$. But given $A=1,2,3$ and $R=(1,2),(2,1)$ we have $(1,3)in R implies (3,1)in R$ etc.
$endgroup$
– fleablood
6 hours ago
1
$begingroup$
"That means in the context of symmetric relation that (x,y)∉ R ⟹ (y,x)∈ R is true" Only if $x,y$ is actually in $R$. $(2,1)not in R implies (1,2)in R$ is a true statement. But $(3,2)not in R implies (2,3)in R$ is a false statement.
$endgroup$
– fleablood
6 hours ago
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
2 hours ago