Given a relation $R$, what is the $R^{-1}$, and what is the complement of $R$ on $A$?












0












$begingroup$


For example if I have the following group $A$ and a relation $R$:
$$A= {1,2,3}$$
$$R= {(1,1)(2,2)(3,3)(1,2)(3,2)}$$
$$R^{-1} = ?$$ $$R^C = ;?$$



Thanks.






discrete-mathematics







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$endgroup$








  • 2




    $begingroup$
    Possible duplicate of Inverse Relation (Definition)
    $endgroup$
    – José Carlos Santos
    Jan 20 at 19:03






  • 2




    $begingroup$
    See for $R^{-1}$ Inverse Relation and Complement of Relation.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 20 at 19:07






  • 1




    $begingroup$
    The complement of $R subset Atimes A$ is given by $(Atimes A)setminus R$, i.e., $(Atimes A) cap lnot R$. Also used is the notation: the complement of $R subset Atimes A$ is given by $(Atimes A)-R$.
    $endgroup$
    – jordan_glen
    Jan 20 at 19:15


















0












$begingroup$


For example if I have the following group $A$ and a relation $R$:
$$A= {1,2,3}$$
$$R= {(1,1)(2,2)(3,3)(1,2)(3,2)}$$
$$R^{-1} = ?$$ $$R^C = ;?$$



Thanks.






discrete-mathematics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Possible duplicate of Inverse Relation (Definition)
    $endgroup$
    – José Carlos Santos
    Jan 20 at 19:03






  • 2




    $begingroup$
    See for $R^{-1}$ Inverse Relation and Complement of Relation.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 20 at 19:07






  • 1




    $begingroup$
    The complement of $R subset Atimes A$ is given by $(Atimes A)setminus R$, i.e., $(Atimes A) cap lnot R$. Also used is the notation: the complement of $R subset Atimes A$ is given by $(Atimes A)-R$.
    $endgroup$
    – jordan_glen
    Jan 20 at 19:15
















0












0








0





$begingroup$


For example if I have the following group $A$ and a relation $R$:
$$A= {1,2,3}$$
$$R= {(1,1)(2,2)(3,3)(1,2)(3,2)}$$
$$R^{-1} = ?$$ $$R^C = ;?$$



Thanks.






discrete-mathematics







share|cite|improve this question











$endgroup$




For example if I have the following group $A$ and a relation $R$:
$$A= {1,2,3}$$
$$R= {(1,1)(2,2)(3,3)(1,2)(3,2)}$$
$$R^{-1} = ?$$ $$R^C = ;?$$



Thanks.






discrete-mathematics




discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 18:15









jordan_glen

1




1










asked Jan 20 at 19:00









user3523226user3523226

606




606








  • 2




    $begingroup$
    Possible duplicate of Inverse Relation (Definition)
    $endgroup$
    – José Carlos Santos
    Jan 20 at 19:03






  • 2




    $begingroup$
    See for $R^{-1}$ Inverse Relation and Complement of Relation.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 20 at 19:07






  • 1




    $begingroup$
    The complement of $R subset Atimes A$ is given by $(Atimes A)setminus R$, i.e., $(Atimes A) cap lnot R$. Also used is the notation: the complement of $R subset Atimes A$ is given by $(Atimes A)-R$.
    $endgroup$
    – jordan_glen
    Jan 20 at 19:15
















  • 2




    $begingroup$
    Possible duplicate of Inverse Relation (Definition)
    $endgroup$
    – José Carlos Santos
    Jan 20 at 19:03






  • 2




    $begingroup$
    See for $R^{-1}$ Inverse Relation and Complement of Relation.
    $endgroup$
    – Mauro ALLEGRANZA
    Jan 20 at 19:07






  • 1




    $begingroup$
    The complement of $R subset Atimes A$ is given by $(Atimes A)setminus R$, i.e., $(Atimes A) cap lnot R$. Also used is the notation: the complement of $R subset Atimes A$ is given by $(Atimes A)-R$.
    $endgroup$
    – jordan_glen
    Jan 20 at 19:15










2




2




$begingroup$
Possible duplicate of Inverse Relation (Definition)
$endgroup$
– José Carlos Santos
Jan 20 at 19:03




$begingroup$
Possible duplicate of Inverse Relation (Definition)
$endgroup$
– José Carlos Santos
Jan 20 at 19:03




2




2




$begingroup$
See for $R^{-1}$ Inverse Relation and Complement of Relation.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 19:07




$begingroup$
See for $R^{-1}$ Inverse Relation and Complement of Relation.
$endgroup$
– Mauro ALLEGRANZA
Jan 20 at 19:07




1




1




$begingroup$
The complement of $R subset Atimes A$ is given by $(Atimes A)setminus R$, i.e., $(Atimes A) cap lnot R$. Also used is the notation: the complement of $R subset Atimes A$ is given by $(Atimes A)-R$.
$endgroup$
– jordan_glen
Jan 20 at 19:15






$begingroup$
The complement of $R subset Atimes A$ is given by $(Atimes A)setminus R$, i.e., $(Atimes A) cap lnot R$. Also used is the notation: the complement of $R subset Atimes A$ is given by $(Atimes A)-R$.
$endgroup$
– jordan_glen
Jan 20 at 19:15












2 Answers
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1












$begingroup$

$$ R^{-1} = {(x,y) mid (y,x) in R}$$
$$ R^C = { (x,y) in A times A mid (x,y) notin R}$$
so with



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



we get



$$R^{-1} = {(1,1),(2,2),(3,3),(2,1),(2,3)}$$
$$R^C = {(2,1),(2,3),(1,3),(3,1)}$$



Also, there was no significance of calling $A$ a group. Calling it the set
$$ A = {1,2,3}$$
is enough for this problem.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    $$begin{align} Atimes A &= {(x, y)mid (xin A) land (yin A)} \ &= {(1, 1), (1,2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3. 2), (3, 3)}end{align}$$



    $$Rsubset Atimes A = {(1,1),(2,2),(3,3),(1,2),(3,2)}tag{given}$$



    $$R^{-1} = {(y, x) mid (x, y) in R } = {(1, 1), (2, 2), (3, 3), (2, 1), (2, 3)}tag{inverse of $R$}$$



    $$begin{align} R^C subset Atimes A &= {(x, y) in Atimes A mid (x, y)notin R}\ &= {(1, 3), (2, 1), (2, 3), (3, 1)}tag{complement of $R$}end{align}$$




    Note that $R^{-1}$ represents the inverse relation of $R$ on $A$.



    Also, $R^C subset Atimes A$ represents the complement of the relation $R$ on $A$, and that $R cup R^{C} = Atimes A$.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes









      1












      $begingroup$

      $$ R^{-1} = {(x,y) mid (y,x) in R}$$
      $$ R^C = { (x,y) in A times A mid (x,y) notin R}$$
      so with



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



      we get



      $$R^{-1} = {(1,1),(2,2),(3,3),(2,1),(2,3)}$$
      $$R^C = {(2,1),(2,3),(1,3),(3,1)}$$



      Also, there was no significance of calling $A$ a group. Calling it the set
      $$ A = {1,2,3}$$
      is enough for this problem.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        $$ R^{-1} = {(x,y) mid (y,x) in R}$$
        $$ R^C = { (x,y) in A times A mid (x,y) notin R}$$
        so with



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



        we get



        $$R^{-1} = {(1,1),(2,2),(3,3),(2,1),(2,3)}$$
        $$R^C = {(2,1),(2,3),(1,3),(3,1)}$$



        Also, there was no significance of calling $A$ a group. Calling it the set
        $$ A = {1,2,3}$$
        is enough for this problem.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          $$ R^{-1} = {(x,y) mid (y,x) in R}$$
          $$ R^C = { (x,y) in A times A mid (x,y) notin R}$$
          so with



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



          we get



          $$R^{-1} = {(1,1),(2,2),(3,3),(2,1),(2,3)}$$
          $$R^C = {(2,1),(2,3),(1,3),(3,1)}$$



          Also, there was no significance of calling $A$ a group. Calling it the set
          $$ A = {1,2,3}$$
          is enough for this problem.






          share|cite|improve this answer











          $endgroup$



          $$ R^{-1} = {(x,y) mid (y,x) in R}$$
          $$ R^C = { (x,y) in A times A mid (x,y) notin R}$$
          so with



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



          we get



          $$R^{-1} = {(1,1),(2,2),(3,3),(2,1),(2,3)}$$
          $$R^C = {(2,1),(2,3),(1,3),(3,1)}$$



          Also, there was no significance of calling $A$ a group. Calling it the set
          $$ A = {1,2,3}$$
          is enough for this problem.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 21 at 19:10

























          answered Jan 21 at 5:19









          MetricMetric

          1,23649




          1,23649























              1












              $begingroup$

              $$begin{align} Atimes A &= {(x, y)mid (xin A) land (yin A)} \ &= {(1, 1), (1,2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3. 2), (3, 3)}end{align}$$



              $$Rsubset Atimes A = {(1,1),(2,2),(3,3),(1,2),(3,2)}tag{given}$$



              $$R^{-1} = {(y, x) mid (x, y) in R } = {(1, 1), (2, 2), (3, 3), (2, 1), (2, 3)}tag{inverse of $R$}$$



              $$begin{align} R^C subset Atimes A &= {(x, y) in Atimes A mid (x, y)notin R}\ &= {(1, 3), (2, 1), (2, 3), (3, 1)}tag{complement of $R$}end{align}$$




              Note that $R^{-1}$ represents the inverse relation of $R$ on $A$.



              Also, $R^C subset Atimes A$ represents the complement of the relation $R$ on $A$, and that $R cup R^{C} = Atimes A$.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                $$begin{align} Atimes A &= {(x, y)mid (xin A) land (yin A)} \ &= {(1, 1), (1,2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3. 2), (3, 3)}end{align}$$



                $$Rsubset Atimes A = {(1,1),(2,2),(3,3),(1,2),(3,2)}tag{given}$$



                $$R^{-1} = {(y, x) mid (x, y) in R } = {(1, 1), (2, 2), (3, 3), (2, 1), (2, 3)}tag{inverse of $R$}$$



                $$begin{align} R^C subset Atimes A &= {(x, y) in Atimes A mid (x, y)notin R}\ &= {(1, 3), (2, 1), (2, 3), (3, 1)}tag{complement of $R$}end{align}$$




                Note that $R^{-1}$ represents the inverse relation of $R$ on $A$.



                Also, $R^C subset Atimes A$ represents the complement of the relation $R$ on $A$, and that $R cup R^{C} = Atimes A$.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  $$begin{align} Atimes A &= {(x, y)mid (xin A) land (yin A)} \ &= {(1, 1), (1,2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3. 2), (3, 3)}end{align}$$



                  $$Rsubset Atimes A = {(1,1),(2,2),(3,3),(1,2),(3,2)}tag{given}$$



                  $$R^{-1} = {(y, x) mid (x, y) in R } = {(1, 1), (2, 2), (3, 3), (2, 1), (2, 3)}tag{inverse of $R$}$$



                  $$begin{align} R^C subset Atimes A &= {(x, y) in Atimes A mid (x, y)notin R}\ &= {(1, 3), (2, 1), (2, 3), (3, 1)}tag{complement of $R$}end{align}$$




                  Note that $R^{-1}$ represents the inverse relation of $R$ on $A$.



                  Also, $R^C subset Atimes A$ represents the complement of the relation $R$ on $A$, and that $R cup R^{C} = Atimes A$.






                  share|cite|improve this answer











                  $endgroup$



                  $$begin{align} Atimes A &= {(x, y)mid (xin A) land (yin A)} \ &= {(1, 1), (1,2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3. 2), (3, 3)}end{align}$$



                  $$Rsubset Atimes A = {(1,1),(2,2),(3,3),(1,2),(3,2)}tag{given}$$



                  $$R^{-1} = {(y, x) mid (x, y) in R } = {(1, 1), (2, 2), (3, 3), (2, 1), (2, 3)}tag{inverse of $R$}$$



                  $$begin{align} R^C subset Atimes A &= {(x, y) in Atimes A mid (x, y)notin R}\ &= {(1, 3), (2, 1), (2, 3), (3, 1)}tag{complement of $R$}end{align}$$




                  Note that $R^{-1}$ represents the inverse relation of $R$ on $A$.



                  Also, $R^C subset Atimes A$ represents the complement of the relation $R$ on $A$, and that $R cup R^{C} = Atimes A$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 20 at 19:41

























                  answered Jan 20 at 19:31









                  jordan_glenjordan_glen

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