Proving well-defineness of function which adds additional variable












2












$begingroup$


I'm trying to solve an exercise from my math lecture. The question is:



Let $k in mathbb{Z}$, $(G, circ) := (mathbb{Z}_{12}, oplus)$, $(H, *) := (mathbb{Z}_{8}, oplus)$ and $varphi: G ni [a]_{12} mapsto [6a + k]_8 in H$.



Prove that $varphi$ is well defined.



Based on another exercise which maps $[a]_{12} mapsto [6a]_8$, I've tried to do this like this:



$$[a]_{12} = [b]_{12} \
iff a equiv bpmod {12} \
iff 12 | b-a \
iff (exists l in mathbb{Z})[12l = b-a]$$



Now insert a new variable, $k$.



$$ iff (exists l,k in mathbb{Z})[12l = (b + k) -(a+k)] \
iff (exists l,k in mathbb{Z})[72l = 6(b+k)-6(a+k)] \
iff (exists l,k in mathbb{Z})[8l = 6(b+k)-6(a+k)] \
iff 8 | (6b+k) - (6a+k) \
iff 6a + k equiv 6b+ k pmod{8} \
iff [6a +k]_8$$



My problem is actually the introducing $k$ part. This does not look right. I think I can simplify $72l$ to $8l$ because 72 is a multiple of 8, but how should I properly introduce $k$ here?



Also, looked at from a set theorey side, if I see $varphi$ as a relation, is it correct that the set would be like this?



$$mathcal{R} = {([0]_{12}, [0+k]_8), ([0]_{12}, [1+k]_8), ldots} $$



i.e. that the $k$ is treated as a constant in the set?






group-theory functions finite-groups







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$endgroup$

















    2












    $begingroup$


    I'm trying to solve an exercise from my math lecture. The question is:



    Let $k in mathbb{Z}$, $(G, circ) := (mathbb{Z}_{12}, oplus)$, $(H, *) := (mathbb{Z}_{8}, oplus)$ and $varphi: G ni [a]_{12} mapsto [6a + k]_8 in H$.



    Prove that $varphi$ is well defined.



    Based on another exercise which maps $[a]_{12} mapsto [6a]_8$, I've tried to do this like this:



    $$[a]_{12} = [b]_{12} \
    iff a equiv bpmod {12} \
    iff 12 | b-a \
    iff (exists l in mathbb{Z})[12l = b-a]$$



    Now insert a new variable, $k$.



    $$ iff (exists l,k in mathbb{Z})[12l = (b + k) -(a+k)] \
    iff (exists l,k in mathbb{Z})[72l = 6(b+k)-6(a+k)] \
    iff (exists l,k in mathbb{Z})[8l = 6(b+k)-6(a+k)] \
    iff 8 | (6b+k) - (6a+k) \
    iff 6a + k equiv 6b+ k pmod{8} \
    iff [6a +k]_8$$



    My problem is actually the introducing $k$ part. This does not look right. I think I can simplify $72l$ to $8l$ because 72 is a multiple of 8, but how should I properly introduce $k$ here?



    Also, looked at from a set theorey side, if I see $varphi$ as a relation, is it correct that the set would be like this?



    $$mathcal{R} = {([0]_{12}, [0+k]_8), ([0]_{12}, [1+k]_8), ldots} $$



    i.e. that the $k$ is treated as a constant in the set?






    group-theory functions finite-groups







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I'm trying to solve an exercise from my math lecture. The question is:



      Let $k in mathbb{Z}$, $(G, circ) := (mathbb{Z}_{12}, oplus)$, $(H, *) := (mathbb{Z}_{8}, oplus)$ and $varphi: G ni [a]_{12} mapsto [6a + k]_8 in H$.



      Prove that $varphi$ is well defined.



      Based on another exercise which maps $[a]_{12} mapsto [6a]_8$, I've tried to do this like this:



      $$[a]_{12} = [b]_{12} \
      iff a equiv bpmod {12} \
      iff 12 | b-a \
      iff (exists l in mathbb{Z})[12l = b-a]$$



      Now insert a new variable, $k$.



      $$ iff (exists l,k in mathbb{Z})[12l = (b + k) -(a+k)] \
      iff (exists l,k in mathbb{Z})[72l = 6(b+k)-6(a+k)] \
      iff (exists l,k in mathbb{Z})[8l = 6(b+k)-6(a+k)] \
      iff 8 | (6b+k) - (6a+k) \
      iff 6a + k equiv 6b+ k pmod{8} \
      iff [6a +k]_8$$



      My problem is actually the introducing $k$ part. This does not look right. I think I can simplify $72l$ to $8l$ because 72 is a multiple of 8, but how should I properly introduce $k$ here?



      Also, looked at from a set theorey side, if I see $varphi$ as a relation, is it correct that the set would be like this?



      $$mathcal{R} = {([0]_{12}, [0+k]_8), ([0]_{12}, [1+k]_8), ldots} $$



      i.e. that the $k$ is treated as a constant in the set?






      group-theory functions finite-groups







      share|cite|improve this question









      $endgroup$




      I'm trying to solve an exercise from my math lecture. The question is:



      Let $k in mathbb{Z}$, $(G, circ) := (mathbb{Z}_{12}, oplus)$, $(H, *) := (mathbb{Z}_{8}, oplus)$ and $varphi: G ni [a]_{12} mapsto [6a + k]_8 in H$.



      Prove that $varphi$ is well defined.



      Based on another exercise which maps $[a]_{12} mapsto [6a]_8$, I've tried to do this like this:



      $$[a]_{12} = [b]_{12} \
      iff a equiv bpmod {12} \
      iff 12 | b-a \
      iff (exists l in mathbb{Z})[12l = b-a]$$



      Now insert a new variable, $k$.



      $$ iff (exists l,k in mathbb{Z})[12l = (b + k) -(a+k)] \
      iff (exists l,k in mathbb{Z})[72l = 6(b+k)-6(a+k)] \
      iff (exists l,k in mathbb{Z})[8l = 6(b+k)-6(a+k)] \
      iff 8 | (6b+k) - (6a+k) \
      iff 6a + k equiv 6b+ k pmod{8} \
      iff [6a +k]_8$$



      My problem is actually the introducing $k$ part. This does not look right. I think I can simplify $72l$ to $8l$ because 72 is a multiple of 8, but how should I properly introduce $k$ here?



      Also, looked at from a set theorey side, if I see $varphi$ as a relation, is it correct that the set would be like this?



      $$mathcal{R} = {([0]_{12}, [0+k]_8), ([0]_{12}, [1+k]_8), ldots} $$



      i.e. that the $k$ is treated as a constant in the set?






      group-theory functions finite-groups




      group-theory functions finite-groups






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 7:32









      HellstormHellstorm

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          $begingroup$

          From what I can tell from your definition of $varphi$, $k$ is indeed a "constant", in the sense that you need to prove that for any fixed $k in mathbb{Z}$ you can define a map $varphi_k$.



          So it is not an "existence of $k$" matter! You fix $k$ in the beginning, and then argue as
          $$[a]_{12} = [b]_{12} \
          iff a equiv bpmod {12} \
          iff 12 | b-a \
          iff (exists l in mathbb{Z})[12l = b-a]
          12l = b-a \
          72l = 6b-6a \ $$
          so, for any fixed $k in mathbb{Z}$
          $$72l = 6b-6a +k - k \
          text{ so }exists m in mathbb{Z} text{ such that } 8m = (6b+k)-(6a+k) \
          text{ so } 8 | (6b+k) - (6a+k) \
          text{ so } 6a + k equiv 6b+ k pmod{8} \
          $$
          and you are done.






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            From what I can tell from your definition of $varphi$, $k$ is indeed a "constant", in the sense that you need to prove that for any fixed $k in mathbb{Z}$ you can define a map $varphi_k$.



            So it is not an "existence of $k$" matter! You fix $k$ in the beginning, and then argue as
            $$[a]_{12} = [b]_{12} \
            iff a equiv bpmod {12} \
            iff 12 | b-a \
            iff (exists l in mathbb{Z})[12l = b-a]
            12l = b-a \
            72l = 6b-6a \ $$
            so, for any fixed $k in mathbb{Z}$
            $$72l = 6b-6a +k - k \
            text{ so }exists m in mathbb{Z} text{ such that } 8m = (6b+k)-(6a+k) \
            text{ so } 8 | (6b+k) - (6a+k) \
            text{ so } 6a + k equiv 6b+ k pmod{8} \
            $$
            and you are done.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              From what I can tell from your definition of $varphi$, $k$ is indeed a "constant", in the sense that you need to prove that for any fixed $k in mathbb{Z}$ you can define a map $varphi_k$.



              So it is not an "existence of $k$" matter! You fix $k$ in the beginning, and then argue as
              $$[a]_{12} = [b]_{12} \
              iff a equiv bpmod {12} \
              iff 12 | b-a \
              iff (exists l in mathbb{Z})[12l = b-a]
              12l = b-a \
              72l = 6b-6a \ $$
              so, for any fixed $k in mathbb{Z}$
              $$72l = 6b-6a +k - k \
              text{ so }exists m in mathbb{Z} text{ such that } 8m = (6b+k)-(6a+k) \
              text{ so } 8 | (6b+k) - (6a+k) \
              text{ so } 6a + k equiv 6b+ k pmod{8} \
              $$
              and you are done.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                From what I can tell from your definition of $varphi$, $k$ is indeed a "constant", in the sense that you need to prove that for any fixed $k in mathbb{Z}$ you can define a map $varphi_k$.



                So it is not an "existence of $k$" matter! You fix $k$ in the beginning, and then argue as
                $$[a]_{12} = [b]_{12} \
                iff a equiv bpmod {12} \
                iff 12 | b-a \
                iff (exists l in mathbb{Z})[12l = b-a]
                12l = b-a \
                72l = 6b-6a \ $$
                so, for any fixed $k in mathbb{Z}$
                $$72l = 6b-6a +k - k \
                text{ so }exists m in mathbb{Z} text{ such that } 8m = (6b+k)-(6a+k) \
                text{ so } 8 | (6b+k) - (6a+k) \
                text{ so } 6a + k equiv 6b+ k pmod{8} \
                $$
                and you are done.






                share|cite|improve this answer









                $endgroup$



                From what I can tell from your definition of $varphi$, $k$ is indeed a "constant", in the sense that you need to prove that for any fixed $k in mathbb{Z}$ you can define a map $varphi_k$.



                So it is not an "existence of $k$" matter! You fix $k$ in the beginning, and then argue as
                $$[a]_{12} = [b]_{12} \
                iff a equiv bpmod {12} \
                iff 12 | b-a \
                iff (exists l in mathbb{Z})[12l = b-a]
                12l = b-a \
                72l = 6b-6a \ $$
                so, for any fixed $k in mathbb{Z}$
                $$72l = 6b-6a +k - k \
                text{ so }exists m in mathbb{Z} text{ such that } 8m = (6b+k)-(6a+k) \
                text{ so } 8 | (6b+k) - (6a+k) \
                text{ so } 6a + k equiv 6b+ k pmod{8} \
                $$
                and you are done.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 10 at 12:04









                AnalysisStudent0414AnalysisStudent0414

                4,376928




                4,376928






























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