How to implement split-complex numbers?












2












$begingroup$


For those who do not know, the split-complex numbers are an analogue to the complex numbers where J is defined such that $J^2=1$ but $Jnepm1$, so they are all of the form $a+bJ$.



By using TagSetDelayed, I tried to define the split-complex numbers as so:



J /: J^2 := 1


If I then type J^2, I get the output 1.
However, if I type J^3, I just get the output J^3. I would like to instead get the output $J$, since $J^3=J^2J=1J=J$.
Is there a better way to implement this number system?






programming algebraic-manipulation algebra







share|improve this question











$endgroup$








  • 1




    $begingroup$
    Might be easier to do this using 2x2 matrix representations.
    $endgroup$
    – Daniel Lichtblau
    3 hours ago










  • $begingroup$
    Ummmm.... isn't $J = -1$?
    $endgroup$
    – David G. Stork
    2 hours ago










  • $begingroup$
    Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
    $endgroup$
    – volcanrb
    1 hour ago


















2












$begingroup$


For those who do not know, the split-complex numbers are an analogue to the complex numbers where J is defined such that $J^2=1$ but $Jnepm1$, so they are all of the form $a+bJ$.



By using TagSetDelayed, I tried to define the split-complex numbers as so:



J /: J^2 := 1


If I then type J^2, I get the output 1.
However, if I type J^3, I just get the output J^3. I would like to instead get the output $J$, since $J^3=J^2J=1J=J$.
Is there a better way to implement this number system?






programming algebraic-manipulation algebra







share|improve this question











$endgroup$








  • 1




    $begingroup$
    Might be easier to do this using 2x2 matrix representations.
    $endgroup$
    – Daniel Lichtblau
    3 hours ago










  • $begingroup$
    Ummmm.... isn't $J = -1$?
    $endgroup$
    – David G. Stork
    2 hours ago










  • $begingroup$
    Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
    $endgroup$
    – volcanrb
    1 hour ago
















2












2








2





$begingroup$


For those who do not know, the split-complex numbers are an analogue to the complex numbers where J is defined such that $J^2=1$ but $Jnepm1$, so they are all of the form $a+bJ$.



By using TagSetDelayed, I tried to define the split-complex numbers as so:



J /: J^2 := 1


If I then type J^2, I get the output 1.
However, if I type J^3, I just get the output J^3. I would like to instead get the output $J$, since $J^3=J^2J=1J=J$.
Is there a better way to implement this number system?






programming algebraic-manipulation algebra







share|improve this question











$endgroup$




For those who do not know, the split-complex numbers are an analogue to the complex numbers where J is defined such that $J^2=1$ but $Jnepm1$, so they are all of the form $a+bJ$.



By using TagSetDelayed, I tried to define the split-complex numbers as so:



J /: J^2 := 1


If I then type J^2, I get the output 1.
However, if I type J^3, I just get the output J^3. I would like to instead get the output $J$, since $J^3=J^2J=1J=J$.
Is there a better way to implement this number system?






programming algebraic-manipulation algebra




programming algebraic-manipulation algebra






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago







volcanrb

















asked 3 hours ago









volcanrbvolcanrb

1684




1684








  • 1




    $begingroup$
    Might be easier to do this using 2x2 matrix representations.
    $endgroup$
    – Daniel Lichtblau
    3 hours ago










  • $begingroup$
    Ummmm.... isn't $J = -1$?
    $endgroup$
    – David G. Stork
    2 hours ago










  • $begingroup$
    Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
    $endgroup$
    – volcanrb
    1 hour ago
















  • 1




    $begingroup$
    Might be easier to do this using 2x2 matrix representations.
    $endgroup$
    – Daniel Lichtblau
    3 hours ago










  • $begingroup$
    Ummmm.... isn't $J = -1$?
    $endgroup$
    – David G. Stork
    2 hours ago










  • $begingroup$
    Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
    $endgroup$
    – volcanrb
    1 hour ago










1




1




$begingroup$
Might be easier to do this using 2x2 matrix representations.
$endgroup$
– Daniel Lichtblau
3 hours ago




$begingroup$
Might be easier to do this using 2x2 matrix representations.
$endgroup$
– Daniel Lichtblau
3 hours ago












$begingroup$
Ummmm.... isn't $J = -1$?
$endgroup$
– David G. Stork
2 hours ago




$begingroup$
Ummmm.... isn't $J = -1$?
$endgroup$
– David G. Stork
2 hours ago












$begingroup$
Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
$endgroup$
– volcanrb
1 hour ago






$begingroup$
Sorry, I forgot to specify that $Jne -1$ either, I've edited the question to fix that
$endgroup$
– volcanrb
1 hour ago












1 Answer
1






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oldest

votes


















4












$begingroup$

Try this:



J /: Power[J, p_Integer?OddQ] := J

J /: Power[J, p_Integer?EvenQ] := 1





J^Range[-10, 10]



{1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}







share|improve this answer









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    1 Answer
    1






    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Try this:



    J /: Power[J, p_Integer?OddQ] := J
    
J /: Power[J, p_Integer?EvenQ] := 1
    

    

    
J^Range[-10, 10]



    {1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}







    share|improve this answer









    $endgroup$


















      4












      $begingroup$

      Try this:



      J /: Power[J, p_Integer?OddQ] := J
      
J /: Power[J, p_Integer?EvenQ] := 1
      

      

      
J^Range[-10, 10]



      {1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}







      share|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Try this:



        J /: Power[J, p_Integer?OddQ] := J
        
J /: Power[J, p_Integer?EvenQ] := 1
        

        

        
J^Range[-10, 10]



        {1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}







        share|improve this answer









        $endgroup$



        Try this:



        J /: Power[J, p_Integer?OddQ] := J
        
J /: Power[J, p_Integer?EvenQ] := 1
        

        

        
J^Range[-10, 10]



        {1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1, J, 1}








        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 3 hours ago









        Henrik SchumacherHenrik Schumacher

        51.1k469146




        51.1k469146






























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