Properties on relation (reflexive, symmetric, anti-symmetric and transitive)












0














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







share|cite|improve this question



























    0














    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







    share|cite|improve this question

























      0












      0








      0







      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







      share|cite|improve this question













      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






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Nov 27 '18 at 19:08









      Deus Sued

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          2 Answers
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          • 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 $aRc$.






          share|cite|improve this answer





























            1















            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.






            share|cite|improve this answer





















            • @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













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            2 Answers
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            2 Answers
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            active

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            1















            • 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 $aRc$.






            share|cite|improve this answer


























              1















              • 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 $aRc$.






              share|cite|improve this answer
























                1












                1








                1







                • 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 $aRc$.






                share|cite|improve this answer













                • 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 $aRc$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 '18 at 19:20









                drhab

                97.8k544129




                97.8k544129























                    1















                    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.






                    share|cite|improve this answer





















                    • @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


















                    1















                    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.






                    share|cite|improve this answer





















                    • @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
















                    1












                    1








                    1







                    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.






                    share|cite|improve this answer













                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    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




















                    • @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




















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