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 $y$ - symmetryDefine 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 relationName for relation property: If $xRy$ and $xRz$ and $x not =y$, then $yRz$.Can a relation be transitive when it is symmetric but not reflexive?
How much character growth crosses the line into breaking the character
Could solar power be utilized and substitute coal in the 19th Century
A Permanent Norse Presence in America
Melting point of aspirin, contradicting sources
What does this horizontal bar at the first measure mean?
Is it possible to use .desktop files to open local pdf files on specific pages with a browser?
Is it possible to have a strip of cold climate in the middle of a planet?
Greco-Roman egalitarianism
Journal losing indexing services
THT: What is a squared annular “ring”?
Can we have a perfect cadence in a minor key?
How can Trident be so inexpensive? Will it orbit Triton or just do a (slow) flyby?
Is it improper etiquette to ask your opponent what his/her rating is before the game?
Why do IPv6 unique local addresses have to have a /48 prefix?
Can somebody explain Brexit in a few child-proof sentences?
Can I sign legal documents with a smiley face?
Confusion on Parallelogram
Does having a TSA Pre-Check member in your flight reservation increase the chances that everyone gets Pre-Check?
Are all species of CANNA edible?
Why has "pence" been used in this sentence, not "pences"?
Do the concepts of IP address and network interface not belong to the same layer?
Can I use my Chinese passport to enter China after I acquired another citizenship?
How can "mimic phobia" be cured or prevented?
How to color a curve
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 $y$ - symmetryDefine 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 relationName for relation property: If $xRy$ and $xRz$ and $x not =y$, then $yRz$.Can a relation be transitive when it is symmetric but not reflexive?
$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
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
yesterday
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
yesterday
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
yesterday
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 yesterday
Asaf Karagila♦
307k33438770
307k33438770
asked yesterday
Rodrigo SangoRodrigo Sango
1326
1326
$begingroup$
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
yesterday
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
yesterday
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
yesterday
add a comment |
$begingroup$
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
yesterday
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
yesterday
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
yesterday
$begingroup$
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
yesterday
$begingroup$
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
yesterday
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
yesterday
$begingroup$
"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
yesterday
2
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
yesterday
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
yesterday
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
yesterday
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
yesterday
$begingroup$
Well, none of them are needed if you state it as all $(x,y)$ but that includes, for each $(x,y)$, $(y,x)$ as well so... well, why say more than you need to.
$endgroup$
– fleablood
yesterday
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 $xRy implies yRx$ 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.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3159427%2fwhy-is-a-symmetric-relation-defined-forall-x-forall-y-xry-implies-yrx-and%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
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
yesterday
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
yesterday
$begingroup$
Well, none of them are needed if you state it as all $(x,y)$ but that includes, for each $(x,y)$, $(y,x)$ as well so... well, why say more than you need to.
$endgroup$
– fleablood
yesterday
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
yesterday
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
yesterday
$begingroup$
Well, none of them are needed if you state it as all $(x,y)$ but that includes, for each $(x,y)$, $(y,x)$ as well so... well, why say more than you need to.
$endgroup$
– fleablood
yesterday
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 yesterday
fleabloodfleablood
73.1k22790
73.1k22790
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
yesterday
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
yesterday
$begingroup$
Well, none of them are needed if you state it as all $(x,y)$ but that includes, for each $(x,y)$, $(y,x)$ as well so... well, why say more than you need to.
$endgroup$
– fleablood
yesterday
add a comment |
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
yesterday
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
yesterday
$begingroup$
Well, none of them are needed if you state it as all $(x,y)$ but that includes, for each $(x,y)$, $(y,x)$ as well so... well, why say more than you need to.
$endgroup$
– fleablood
yesterday
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
yesterday
$begingroup$
Got it. Thank you, fleablood.
$endgroup$
– Rodrigo Sango
yesterday
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
yesterday
$begingroup$
"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
yesterday
$begingroup$
Well, none of them are needed if you state it as all $(x,y)$ but that includes, for each $(x,y)$, $(y,x)$ as well so... well, why say more than you need to.
$endgroup$
– fleablood
yesterday
$begingroup$
Well, none of them are needed if you state it as all $(x,y)$ but that includes, for each $(x,y)$, $(y,x)$ as well so... well, why say more than you need to.
$endgroup$
– fleablood
yesterday
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 yesterday
José Carlos SantosJosé Carlos Santos
170k23132238
170k23132238
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 yesterday
Viktor GlombikViktor Glombik
1,2372528
1,2372528
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 yesterday
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3159427%2fwhy-is-a-symmetric-relation-defined-forall-x-forall-y-xry-implies-yrx-and%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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
yesterday
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
yesterday
2
$begingroup$
The definitions are equivalent.
$endgroup$
– PyRulez
yesterday