The relation $T$ on $mathbb{R}times mathbb{R}$ given by $(x,y)T(a,b)$ iff $x^2+y^2=a^2+b^2$. Sketch the...












0












$begingroup$



The relation $T$ on $mathbb{R}times mathbb{R}$ given by $(x,y)T(a,b)$ iff $x^2+y^2=a^2+b^2$. Sketch the equivalence class of $(1,2)$; of $(4,0)$.




$T$ is an equivalence relation on $mathbb{R}times mathbb{R}$, because:



$T$ is reflexive on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



for all $(x,y) in mathbb{R}times mathbb{R}$, we have



$(x,y)T(x,y) Leftrightarrow x^2+y^2=x^2+y^2$



$T$ is symmetric on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



for all $(x,y),(a,b) in mathbb{R}times mathbb{R}$, we have



$(x,y)T(a,b) Leftrightarrow x^2+y^2=a^2+b^2$



$Leftrightarrow a^2+b^2=x^2+y^2$



$Leftrightarrow (a,b)T(x,y)$



$T$ is transitive on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



for all $(x,y),(a,b),(p,q) in mathbb{R}times mathbb{R}$, we have:



$(x,y)T(a,b) land (a,b)T(p,q) Leftrightarrow x^2+y^2=a^2+b^2 land a^2+b^2=p^2=q^2$



$ Leftrightarrow x^2+y^2+=p^2=q^2$



$ Leftrightarrow (x,y)T(p,q)$.



We have,




  • $(1,2)/ T =[(1,2)]={(x,y)in mathbb{R}times mathbb{R} | (x,y) T (1,2)}={(x,y)in mathbb{R}times mathbb{R} | x^2+y^2=5}$


enter image description here




  • $(4,0)/ T =[(4,0)]={(x,y)in mathbb{R}times mathbb{R} | (x,y) T (4,0)}={(x,y)in mathbb{R}times mathbb{R} | x^2+y^2=16}$


enter image description here



Is that true please?






relations equivalence-relations foundations







share|cite|improve this question









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    0












    $begingroup$



    The relation $T$ on $mathbb{R}times mathbb{R}$ given by $(x,y)T(a,b)$ iff $x^2+y^2=a^2+b^2$. Sketch the equivalence class of $(1,2)$; of $(4,0)$.




    $T$ is an equivalence relation on $mathbb{R}times mathbb{R}$, because:



    $T$ is reflexive on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



    for all $(x,y) in mathbb{R}times mathbb{R}$, we have



    $(x,y)T(x,y) Leftrightarrow x^2+y^2=x^2+y^2$



    $T$ is symmetric on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



    for all $(x,y),(a,b) in mathbb{R}times mathbb{R}$, we have



    $(x,y)T(a,b) Leftrightarrow x^2+y^2=a^2+b^2$



    $Leftrightarrow a^2+b^2=x^2+y^2$



    $Leftrightarrow (a,b)T(x,y)$



    $T$ is transitive on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



    for all $(x,y),(a,b),(p,q) in mathbb{R}times mathbb{R}$, we have:



    $(x,y)T(a,b) land (a,b)T(p,q) Leftrightarrow x^2+y^2=a^2+b^2 land a^2+b^2=p^2=q^2$



    $ Leftrightarrow x^2+y^2+=p^2=q^2$



    $ Leftrightarrow (x,y)T(p,q)$.



    We have,




    • $(1,2)/ T =[(1,2)]={(x,y)in mathbb{R}times mathbb{R} | (x,y) T (1,2)}={(x,y)in mathbb{R}times mathbb{R} | x^2+y^2=5}$


    enter image description here




    • $(4,0)/ T =[(4,0)]={(x,y)in mathbb{R}times mathbb{R} | (x,y) T (4,0)}={(x,y)in mathbb{R}times mathbb{R} | x^2+y^2=16}$


    enter image description here



    Is that true please?






    relations equivalence-relations foundations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      The relation $T$ on $mathbb{R}times mathbb{R}$ given by $(x,y)T(a,b)$ iff $x^2+y^2=a^2+b^2$. Sketch the equivalence class of $(1,2)$; of $(4,0)$.




      $T$ is an equivalence relation on $mathbb{R}times mathbb{R}$, because:



      $T$ is reflexive on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



      for all $(x,y) in mathbb{R}times mathbb{R}$, we have



      $(x,y)T(x,y) Leftrightarrow x^2+y^2=x^2+y^2$



      $T$ is symmetric on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



      for all $(x,y),(a,b) in mathbb{R}times mathbb{R}$, we have



      $(x,y)T(a,b) Leftrightarrow x^2+y^2=a^2+b^2$



      $Leftrightarrow a^2+b^2=x^2+y^2$



      $Leftrightarrow (a,b)T(x,y)$



      $T$ is transitive on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



      for all $(x,y),(a,b),(p,q) in mathbb{R}times mathbb{R}$, we have:



      $(x,y)T(a,b) land (a,b)T(p,q) Leftrightarrow x^2+y^2=a^2+b^2 land a^2+b^2=p^2=q^2$



      $ Leftrightarrow x^2+y^2+=p^2=q^2$



      $ Leftrightarrow (x,y)T(p,q)$.



      We have,




      • $(1,2)/ T =[(1,2)]={(x,y)in mathbb{R}times mathbb{R} | (x,y) T (1,2)}={(x,y)in mathbb{R}times mathbb{R} | x^2+y^2=5}$


      enter image description here




      • $(4,0)/ T =[(4,0)]={(x,y)in mathbb{R}times mathbb{R} | (x,y) T (4,0)}={(x,y)in mathbb{R}times mathbb{R} | x^2+y^2=16}$


      enter image description here



      Is that true please?






      relations equivalence-relations foundations







      share|cite|improve this question









      $endgroup$





      The relation $T$ on $mathbb{R}times mathbb{R}$ given by $(x,y)T(a,b)$ iff $x^2+y^2=a^2+b^2$. Sketch the equivalence class of $(1,2)$; of $(4,0)$.




      $T$ is an equivalence relation on $mathbb{R}times mathbb{R}$, because:



      $T$ is reflexive on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



      for all $(x,y) in mathbb{R}times mathbb{R}$, we have



      $(x,y)T(x,y) Leftrightarrow x^2+y^2=x^2+y^2$



      $T$ is symmetric on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



      for all $(x,y),(a,b) in mathbb{R}times mathbb{R}$, we have



      $(x,y)T(a,b) Leftrightarrow x^2+y^2=a^2+b^2$



      $Leftrightarrow a^2+b^2=x^2+y^2$



      $Leftrightarrow (a,b)T(x,y)$



      $T$ is transitive on $mathbb{R}times mathbb{R}$, and we can proving this as the following:



      for all $(x,y),(a,b),(p,q) in mathbb{R}times mathbb{R}$, we have:



      $(x,y)T(a,b) land (a,b)T(p,q) Leftrightarrow x^2+y^2=a^2+b^2 land a^2+b^2=p^2=q^2$



      $ Leftrightarrow x^2+y^2+=p^2=q^2$



      $ Leftrightarrow (x,y)T(p,q)$.



      We have,




      • $(1,2)/ T =[(1,2)]={(x,y)in mathbb{R}times mathbb{R} | (x,y) T (1,2)}={(x,y)in mathbb{R}times mathbb{R} | x^2+y^2=5}$


      enter image description here




      • $(4,0)/ T =[(4,0)]={(x,y)in mathbb{R}times mathbb{R} | (x,y) T (4,0)}={(x,y)in mathbb{R}times mathbb{R} | x^2+y^2=16}$


      enter image description here



      Is that true please?






      relations equivalence-relations foundations




      relations equivalence-relations foundations






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      asked Nov 30 '18 at 7:48









      DimaDima

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          Yes, every thing is true. Your solution is fine !






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          • $begingroup$
            Thank you so much prof.
            $endgroup$
            – Dima
            Nov 30 '18 at 7:50











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          $begingroup$

          Yes, every thing is true. Your solution is fine !






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much prof.
            $endgroup$
            – Dima
            Nov 30 '18 at 7:50
















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          $begingroup$

          Yes, every thing is true. Your solution is fine !






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you so much prof.
            $endgroup$
            – Dima
            Nov 30 '18 at 7:50














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          $begingroup$

          Yes, every thing is true. Your solution is fine !






          share|cite|improve this answer









          $endgroup$



          Yes, every thing is true. Your solution is fine !







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 7:50









          FredFred

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          44.4k1845












          • $begingroup$
            Thank you so much prof.
            $endgroup$
            – Dima
            Nov 30 '18 at 7:50


















          • $begingroup$
            Thank you so much prof.
            $endgroup$
            – Dima
            Nov 30 '18 at 7:50
















          $begingroup$
          Thank you so much prof.
          $endgroup$
          – Dima
          Nov 30 '18 at 7:50




          $begingroup$
          Thank you so much prof.
          $endgroup$
          – Dima
          Nov 30 '18 at 7:50


















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