Composition of relation and transitive closure Announcing the arrival of Valued Associate...

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Composition of relation and transitive closure



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Need assistance determining whether these relations are transitive or antisymmetric (or both?)Transitive closure of these relations on ${1,2,3,4}$?Finding the smallest relation that is reflexive, transitive, and symmetricSmallest relation for reflexive, symmetry and transitivityHow to determine whether a given relation on a finite set is transitive?Symmetric closure of the reflexive closure of the transitive closure of a relationTransitive Closure, $R ={(0,0), (0,3), (1,0), (1,2), (2,0), (3,2)}$Equivalence Relation, transitive relationAm I correct about the transitive closure of this relation?binary relation that is both symmetric and irreflexive












1












$begingroup$


Am I right in the following?



Let $$A = {1, 2, 3, 4, 5}$$ and consider the following relation on A:
$$R = {(1, 2),(2, 3),(3, 4),(4, 5),(5, 1)}$$



a)



Here I am to find the composition of R on R. I got this:



$$R^2={(1,3),(2,4),(3,5),(4,1)}$$



b)



$$R={(1,2),(2,3),(3,4),(4,5),(5,1)}$$



The transitive closure is



$$R={(1,2),(2,3),(1,3)(3,4),(4,5),(4,1),(3,5),(5,1)}$$



EDIT1: I added (4,1), because we have (4,5),(5,1), which would be (a,b),(b,c) and the transitive closure of those would be (4,1).










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Am I right in the following?



    Let $$A = {1, 2, 3, 4, 5}$$ and consider the following relation on A:
    $$R = {(1, 2),(2, 3),(3, 4),(4, 5),(5, 1)}$$



    a)



    Here I am to find the composition of R on R. I got this:



    $$R^2={(1,3),(2,4),(3,5),(4,1)}$$



    b)



    $$R={(1,2),(2,3),(3,4),(4,5),(5,1)}$$



    The transitive closure is



    $$R={(1,2),(2,3),(1,3)(3,4),(4,5),(4,1),(3,5),(5,1)}$$



    EDIT1: I added (4,1), because we have (4,5),(5,1), which would be (a,b),(b,c) and the transitive closure of those would be (4,1).










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Am I right in the following?



      Let $$A = {1, 2, 3, 4, 5}$$ and consider the following relation on A:
      $$R = {(1, 2),(2, 3),(3, 4),(4, 5),(5, 1)}$$



      a)



      Here I am to find the composition of R on R. I got this:



      $$R^2={(1,3),(2,4),(3,5),(4,1)}$$



      b)



      $$R={(1,2),(2,3),(3,4),(4,5),(5,1)}$$



      The transitive closure is



      $$R={(1,2),(2,3),(1,3)(3,4),(4,5),(4,1),(3,5),(5,1)}$$



      EDIT1: I added (4,1), because we have (4,5),(5,1), which would be (a,b),(b,c) and the transitive closure of those would be (4,1).










      share|cite|improve this question









      $endgroup$




      Am I right in the following?



      Let $$A = {1, 2, 3, 4, 5}$$ and consider the following relation on A:
      $$R = {(1, 2),(2, 3),(3, 4),(4, 5),(5, 1)}$$



      a)



      Here I am to find the composition of R on R. I got this:



      $$R^2={(1,3),(2,4),(3,5),(4,1)}$$



      b)



      $$R={(1,2),(2,3),(3,4),(4,5),(5,1)}$$



      The transitive closure is



      $$R={(1,2),(2,3),(1,3)(3,4),(4,5),(4,1),(3,5),(5,1)}$$



      EDIT1: I added (4,1), because we have (4,5),(5,1), which would be (a,b),(b,c) and the transitive closure of those would be (4,1).







      discrete-mathematics






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      asked Mar 25 at 17:52









      EventhorizonEventhorizon

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          0












          $begingroup$

          Not exactly.



          a) You missed $(5,2)$, otherwise it's ok.



          b) Try to prove that all pairs are in the transitive closure.






          share|cite|improve this answer









          $endgroup$














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            1 Answer
            1






            active

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            active

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            active

            oldest

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            0












            $begingroup$

            Not exactly.



            a) You missed $(5,2)$, otherwise it's ok.



            b) Try to prove that all pairs are in the transitive closure.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Not exactly.



              a) You missed $(5,2)$, otherwise it's ok.



              b) Try to prove that all pairs are in the transitive closure.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Not exactly.



                a) You missed $(5,2)$, otherwise it's ok.



                b) Try to prove that all pairs are in the transitive closure.






                share|cite|improve this answer









                $endgroup$



                Not exactly.



                a) You missed $(5,2)$, otherwise it's ok.



                b) Try to prove that all pairs are in the transitive closure.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Mar 25 at 22:05









                BerciBerci

                62.1k23776




                62.1k23776






























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