Properties on relation (reflexive, symmetric, anti-symmetric and transitive)
So A ={2,4,7,9}
R = {(2,2), (2,4), (2,7), (2,9), (4,7), (4,9), (7,9)}
not reflexive because not all the elements from A are related to one another in R
is symmetric because 2,2 can relate to itself
is anti-symmetric because similarly, 2,2 is symmetric can give 2=2
is transitive because 2,2 can be paired with itself
(2,2) (2,2) ⇒ (2,2)
Is my understanding correct?
discrete-mathematics relations
asked Nov 27 '18 at 19:08
Deus Sued
1
add a comment |
So A ={2,4,7,9}
R = {(2,2), (2,4), (2,7), (2,9), (4,7), (4,9), (7,9)}
not reflexive because not all the elements from A are related to one another in R
is symmetric because 2,2 can relate to itself
is anti-symmetric because similarly, 2,2 is symmetric can give 2=2
is transitive because 2,2 can be paired with itself
(2,2) (2,2) ⇒ (2,2)
Is my understanding correct?
discrete-mathematics relations
asked Nov 27 '18 at 19:08
Deus Sued
1
add a comment |
So A ={2,4,7,9}
R = {(2,2), (2,4), (2,7), (2,9), (4,7), (4,9), (7,9)}
not reflexive because not all the elements from A are related to one another in R
is symmetric because 2,2 can relate to itself
is anti-symmetric because similarly, 2,2 is symmetric can give 2=2
is transitive because 2,2 can be paired with itself
(2,2) (2,2) ⇒ (2,2)
Is my understanding correct?
discrete-mathematics relations
asked Nov 27 '18 at 19:08
Deus Sued
1
So A ={2,4,7,9}
R = {(2,2), (2,4), (2,7), (2,9), (4,7), (4,9), (7,9)}
not reflexive because not all the elements from A are related to one another in R
is symmetric because 2,2 can relate to itself
is anti-symmetric because similarly, 2,2 is symmetric can give 2=2
is transitive because 2,2 can be paired with itself
(2,2) (2,2) ⇒ (2,2)
Is my understanding correct?
discrete-mathematics relations
discrete-mathematics relations
asked Nov 27 '18 at 19:08
Deus Sued
1
asked Nov 27 '18 at 19:08
Deus Sued
1
asked Nov 27 '18 at 19:08
Deus Sued
1
asked Nov 27 '18 at 19:08
Deus Sued
1
asked Nov 27 '18 at 19:08
Deus Sued
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
- indeed not reflexive because e.g. we do not have $4R4$
- not symmetric, because $2R4$ but not $4R2$
- indeed antisymmetric, because for every pair $(a,b)$ that satisfies $aRbwedge bRa$ (the pair $(2,2)$ is the only one here) we also have $a=b$.
- indeed transitive but what you gave as reason for that is not okay. It must be checked that in all cases that we have $aRbwedge bRc$ we also have $aRc$.
answered Nov 27 '18 at 19:20
drhab
97.8k544129
add a comment |
is not reflexive because not all the elements from A are related to one another in R
Correct. Reflexivity requires all elements in A to be self related. $forall xin A:(x,x)in R$.
is symmetric because 2,2 can relate to itself
Incorrect. While $(2,2)$ is its own symmetrical pairing, Symmetry requires that all pairs that are in the relation have a corresponding symmetrical pairing. $forall xforall y: ((x,y)in R to (y,x)in R)$
This is not so. $(2,7)$ is in the relation, but $(7,2)$ is not.
is anti-symmetric because similarly, 2,2 is symmetric can give 2=2
To be precise, anti-symmetry requires that all pairs in R have a cooresponding symmetrical pairing only if they are self-relations, such as for example $(2,2)$. $forall xforall y~(((x,y)in Rland (y,x)in R)to x=y)$
is transitive because 2,2 can be paired with itself (2,2) (2,2) ⇒ (2,2)
Again it is for all cases. You cannot say it is so because one particular case holds. $forall xforall yforall z~((x,y)in Rland (y,z)in Rto(x,z)in R)$
So, therefore you must exhaustively check all possible cases to ensure no counter example exists to satisfy the negation of that; which is $exists xexists yexists z~((x,y)in Rland (y,z)in Rland (x,z)notin R)$
PS: no counter example exists.
Is my understanding correct?
Indications are no.
To ensure a universal statement is satisfied you must ascertain that all cases hold, rather than a single example.
On the other hand, finding a single counter example is all that is needed to prove a universal statement is not satisfied.
answered Nov 27 '18 at 22:09
Graham Kemp
84.7k43378
@DeusSued All elements of the set must be self-related for the relation to be reflexive. Since it does not hold for all, you were correct: the relation is not reflexive.
– Graham Kemp
Nov 29 '18 at 0:29
So antisymmertic (1,2) (2,1) (2,2) because it has (2,2) which gives a=b but if there is (a,b) but not (b,a) it is still antisymmetric
– Deus Sued
Nov 29 '18 at 0:47
(a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) So(1,2) (2,1) then there should be (1,1) for transitive
– Deus Sued
Nov 29 '18 at 0:51
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%2f3016176%2fproperties-on-relation-reflexive-symmetric-anti-symmetric-and-transitive%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
- indeed not reflexive because e.g. we do not have $4R4$
- not symmetric, because $2R4$ but not $4R2$
- indeed antisymmetric, because for every pair $(a,b)$ that satisfies $aRbwedge bRa$ (the pair $(2,2)$ is the only one here) we also have $a=b$.
- indeed transitive but what you gave as reason for that is not okay. It must be checked that in all cases that we have $aRbwedge bRc$ we also have $aRc$.
answered Nov 27 '18 at 19:20
drhab
97.8k544129
add a comment |
- indeed not reflexive because e.g. we do not have $4R4$
- not symmetric, because $2R4$ but not $4R2$
- indeed antisymmetric, because for every pair $(a,b)$ that satisfies $aRbwedge bRa$ (the pair $(2,2)$ is the only one here) we also have $a=b$.
- indeed transitive but what you gave as reason for that is not okay. It must be checked that in all cases that we have $aRbwedge bRc$ we also have $aRc$.
answered Nov 27 '18 at 19:20
drhab
97.8k544129
add a comment |
- indeed not reflexive because e.g. we do not have $4R4$
- not symmetric, because $2R4$ but not $4R2$
- indeed antisymmetric, because for every pair $(a,b)$ that satisfies $aRbwedge bRa$ (the pair $(2,2)$ is the only one here) we also have $a=b$.
- indeed transitive but what you gave as reason for that is not okay. It must be checked that in all cases that we have $aRbwedge bRc$ we also have $aRc$.
answered Nov 27 '18 at 19:20
drhab
97.8k544129
- indeed not reflexive because e.g. we do not have $4R4$
- not symmetric, because $2R4$ but not $4R2$
- indeed antisymmetric, because for every pair $(a,b)$ that satisfies $aRbwedge bRa$ (the pair $(2,2)$ is the only one here) we also have $a=b$.
- indeed transitive but what you gave as reason for that is not okay. It must be checked that in all cases that we have $aRbwedge bRc$ we also have $aRc$.
answered Nov 27 '18 at 19:20
drhab
97.8k544129
answered Nov 27 '18 at 19:20
drhab
97.8k544129
answered Nov 27 '18 at 19:20
drhab
97.8k544129
answered Nov 27 '18 at 19:20
drhab
97.8k544129
97.8k544129
add a comment |
add a comment |
is not reflexive because not all the elements from A are related to one another in R
Correct. Reflexivity requires all elements in A to be self related. $forall xin A:(x,x)in R$.
is symmetric because 2,2 can relate to itself
Incorrect. While $(2,2)$ is its own symmetrical pairing, Symmetry requires that all pairs that are in the relation have a corresponding symmetrical pairing. $forall xforall y: ((x,y)in R to (y,x)in R)$
This is not so. $(2,7)$ is in the relation, but $(7,2)$ is not.
is anti-symmetric because similarly, 2,2 is symmetric can give 2=2
To be precise, anti-symmetry requires that all pairs in R have a cooresponding symmetrical pairing only if they are self-relations, such as for example $(2,2)$. $forall xforall y~(((x,y)in Rland (y,x)in R)to x=y)$
is transitive because 2,2 can be paired with itself (2,2) (2,2) ⇒ (2,2)
Again it is for all cases. You cannot say it is so because one particular case holds. $forall xforall yforall z~((x,y)in Rland (y,z)in Rto(x,z)in R)$
So, therefore you must exhaustively check all possible cases to ensure no counter example exists to satisfy the negation of that; which is $exists xexists yexists z~((x,y)in Rland (y,z)in Rland (x,z)notin R)$
PS: no counter example exists.
Is my understanding correct?
Indications are no.
To ensure a universal statement is satisfied you must ascertain that all cases hold, rather than a single example.
On the other hand, finding a single counter example is all that is needed to prove a universal statement is not satisfied.
answered Nov 27 '18 at 22:09
Graham Kemp
84.7k43378
@DeusSued All elements of the set must be self-related for the relation to be reflexive. Since it does not hold for all, you were correct: the relation is not reflexive.
– Graham Kemp
Nov 29 '18 at 0:29
So antisymmertic (1,2) (2,1) (2,2) because it has (2,2) which gives a=b but if there is (a,b) but not (b,a) it is still antisymmetric
– Deus Sued
Nov 29 '18 at 0:47
(a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) So(1,2) (2,1) then there should be (1,1) for transitive
– Deus Sued
Nov 29 '18 at 0:51
add a comment |
is not reflexive because not all the elements from A are related to one another in R
Correct. Reflexivity requires all elements in A to be self related. $forall xin A:(x,x)in R$.
is symmetric because 2,2 can relate to itself
Incorrect. While $(2,2)$ is its own symmetrical pairing, Symmetry requires that all pairs that are in the relation have a corresponding symmetrical pairing. $forall xforall y: ((x,y)in R to (y,x)in R)$
This is not so. $(2,7)$ is in the relation, but $(7,2)$ is not.
is anti-symmetric because similarly, 2,2 is symmetric can give 2=2
To be precise, anti-symmetry requires that all pairs in R have a cooresponding symmetrical pairing only if they are self-relations, such as for example $(2,2)$. $forall xforall y~(((x,y)in Rland (y,x)in R)to x=y)$
is transitive because 2,2 can be paired with itself (2,2) (2,2) ⇒ (2,2)
Again it is for all cases. You cannot say it is so because one particular case holds. $forall xforall yforall z~((x,y)in Rland (y,z)in Rto(x,z)in R)$
So, therefore you must exhaustively check all possible cases to ensure no counter example exists to satisfy the negation of that; which is $exists xexists yexists z~((x,y)in Rland (y,z)in Rland (x,z)notin R)$
PS: no counter example exists.
Is my understanding correct?
Indications are no.
To ensure a universal statement is satisfied you must ascertain that all cases hold, rather than a single example.
On the other hand, finding a single counter example is all that is needed to prove a universal statement is not satisfied.
answered Nov 27 '18 at 22:09
Graham Kemp
84.7k43378
@DeusSued All elements of the set must be self-related for the relation to be reflexive. Since it does not hold for all, you were correct: the relation is not reflexive.
– Graham Kemp
Nov 29 '18 at 0:29
So antisymmertic (1,2) (2,1) (2,2) because it has (2,2) which gives a=b but if there is (a,b) but not (b,a) it is still antisymmetric
– Deus Sued
Nov 29 '18 at 0:47
(a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) So(1,2) (2,1) then there should be (1,1) for transitive
– Deus Sued
Nov 29 '18 at 0:51
add a comment |
is not reflexive because not all the elements from A are related to one another in R
Correct. Reflexivity requires all elements in A to be self related. $forall xin A:(x,x)in R$.
is symmetric because 2,2 can relate to itself
Incorrect. While $(2,2)$ is its own symmetrical pairing, Symmetry requires that all pairs that are in the relation have a corresponding symmetrical pairing. $forall xforall y: ((x,y)in R to (y,x)in R)$
This is not so. $(2,7)$ is in the relation, but $(7,2)$ is not.
is anti-symmetric because similarly, 2,2 is symmetric can give 2=2
To be precise, anti-symmetry requires that all pairs in R have a cooresponding symmetrical pairing only if they are self-relations, such as for example $(2,2)$. $forall xforall y~(((x,y)in Rland (y,x)in R)to x=y)$
is transitive because 2,2 can be paired with itself (2,2) (2,2) ⇒ (2,2)
Again it is for all cases. You cannot say it is so because one particular case holds. $forall xforall yforall z~((x,y)in Rland (y,z)in Rto(x,z)in R)$
So, therefore you must exhaustively check all possible cases to ensure no counter example exists to satisfy the negation of that; which is $exists xexists yexists z~((x,y)in Rland (y,z)in Rland (x,z)notin R)$
PS: no counter example exists.
Is my understanding correct?
Indications are no.
To ensure a universal statement is satisfied you must ascertain that all cases hold, rather than a single example.
On the other hand, finding a single counter example is all that is needed to prove a universal statement is not satisfied.
answered Nov 27 '18 at 22:09
Graham Kemp
84.7k43378
is not reflexive because not all the elements from A are related to one another in R
Correct. Reflexivity requires all elements in A to be self related. $forall xin A:(x,x)in R$.
is symmetric because 2,2 can relate to itself
Incorrect. While $(2,2)$ is its own symmetrical pairing, Symmetry requires that all pairs that are in the relation have a corresponding symmetrical pairing. $forall xforall y: ((x,y)in R to (y,x)in R)$
This is not so. $(2,7)$ is in the relation, but $(7,2)$ is not.
is anti-symmetric because similarly, 2,2 is symmetric can give 2=2
To be precise, anti-symmetry requires that all pairs in R have a cooresponding symmetrical pairing only if they are self-relations, such as for example $(2,2)$. $forall xforall y~(((x,y)in Rland (y,x)in R)to x=y)$
is transitive because 2,2 can be paired with itself (2,2) (2,2) ⇒ (2,2)
Again it is for all cases. You cannot say it is so because one particular case holds. $forall xforall yforall z~((x,y)in Rland (y,z)in Rto(x,z)in R)$
So, therefore you must exhaustively check all possible cases to ensure no counter example exists to satisfy the negation of that; which is $exists xexists yexists z~((x,y)in Rland (y,z)in Rland (x,z)notin R)$
PS: no counter example exists.
Is my understanding correct?
Indications are no.
To ensure a universal statement is satisfied you must ascertain that all cases hold, rather than a single example.
On the other hand, finding a single counter example is all that is needed to prove a universal statement is not satisfied.
answered Nov 27 '18 at 22:09
Graham Kemp
84.7k43378
answered Nov 27 '18 at 22:09
Graham Kemp
84.7k43378
answered Nov 27 '18 at 22:09
Graham Kemp
84.7k43378
answered Nov 27 '18 at 22:09
Graham Kemp
84.7k43378
84.7k43378
@DeusSued All elements of the set must be self-related for the relation to be reflexive. Since it does not hold for all, you were correct: the relation is not reflexive.
– Graham Kemp
Nov 29 '18 at 0:29
So antisymmertic (1,2) (2,1) (2,2) because it has (2,2) which gives a=b but if there is (a,b) but not (b,a) it is still antisymmetric
– Deus Sued
Nov 29 '18 at 0:47
(a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) So(1,2) (2,1) then there should be (1,1) for transitive
– Deus Sued
Nov 29 '18 at 0:51
add a comment |
@DeusSued All elements of the set must be self-related for the relation to be reflexive. Since it does not hold for all, you were correct: the relation is not reflexive.
– Graham Kemp
Nov 29 '18 at 0:29
So antisymmertic (1,2) (2,1) (2,2) because it has (2,2) which gives a=b but if there is (a,b) but not (b,a) it is still antisymmetric
– Deus Sued
Nov 29 '18 at 0:47
(a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) So(1,2) (2,1) then there should be (1,1) for transitive
– Deus Sued
Nov 29 '18 at 0:51
@DeusSued All elements of the set must be self-related for the relation to be reflexive. Since it does not hold for all, you were correct: the relation is not reflexive.
– Graham Kemp
Nov 29 '18 at 0:29
@DeusSued All elements of the set must be self-related for the relation to be reflexive. Since it does not hold for all, you were correct: the relation is not reflexive.
– Graham Kemp
Nov 29 '18 at 0:29
So antisymmertic (1,2) (2,1) (2,2) because it has (2,2) which gives a=b but if there is (a,b) but not (b,a) it is still antisymmetric
– Deus Sued
Nov 29 '18 at 0:47
So antisymmertic (1,2) (2,1) (2,2) because it has (2,2) which gives a=b but if there is (a,b) but not (b,a) it is still antisymmetric
– Deus Sued
Nov 29 '18 at 0:47
(a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) So(1,2) (2,1) then there should be (1,1) for transitive
– Deus Sued
Nov 29 '18 at 0:51
(a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) So(1,2) (2,1) then there should be (1,1) for transitive
– Deus Sued
Nov 29 '18 at 0:51
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3016176%2fproperties-on-relation-reflexive-symmetric-anti-symmetric-and-transitive%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
