Neutral Element on $(mathbb{C},*,+)$











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I want to show the neutral Element on $$(mathbb{C},*,+)$$



Let the complex Number be defined as:
$$(x,y)$$
Multiplication of complex numbers is defined as:
$$(x_{1}, y_{1}) * (x_{2}, y_{2}):= (x_{1}x_{2}-y_{1}y_{2}, x_{1}y_{2}+x_{2}y_{1})$$
By definition the neutral element is:
$$(e_{x}, e_{y}) *(x,y) = (x,y)$$
$$(x*e_{x}-y*e_{y}, x*e_{y}+y*e_{x})$$
This gives an equation system:
$$x = x*e_{x} -y*e_{y}$$
$$y = x*e_{y} +y*e_{x}$$



Is this the correct approach to finding the neutral Element?



If yes, what is the correct method for solving the equation system?






complex-numbers systems-of-equations







share|cite|improve this question
























  • Of course you can determine the multplicative neutral element by solving the above equation system. But I am sure you already know that the neutral element is $(1,0)$, and in that case it suffices to verify the equation $(1,0) * (x,y) = (x,y)$.
    – Paul Frost
    Nov 19 at 12:40










  • I want to show how to get to the value (1,0)
    – Thomas Christopher Davies
    Nov 19 at 16:05










  • Then simply note that you must have $(e_x,e_y) * (1,0) = (1,0)$. This gives you $1 = e_x$ and $0 = e_y$.
    – Paul Frost
    Nov 19 at 16:17










  • I don't see this being a good derivation
    – Thomas Christopher Davies
    Nov 19 at 17:28















up vote
0
down vote

favorite












I want to show the neutral Element on $$(mathbb{C},*,+)$$



Let the complex Number be defined as:
$$(x,y)$$
Multiplication of complex numbers is defined as:
$$(x_{1}, y_{1}) * (x_{2}, y_{2}):= (x_{1}x_{2}-y_{1}y_{2}, x_{1}y_{2}+x_{2}y_{1})$$
By definition the neutral element is:
$$(e_{x}, e_{y}) *(x,y) = (x,y)$$
$$(x*e_{x}-y*e_{y}, x*e_{y}+y*e_{x})$$
This gives an equation system:
$$x = x*e_{x} -y*e_{y}$$
$$y = x*e_{y} +y*e_{x}$$



Is this the correct approach to finding the neutral Element?



If yes, what is the correct method for solving the equation system?






complex-numbers systems-of-equations







share|cite|improve this question
























  • Of course you can determine the multplicative neutral element by solving the above equation system. But I am sure you already know that the neutral element is $(1,0)$, and in that case it suffices to verify the equation $(1,0) * (x,y) = (x,y)$.
    – Paul Frost
    Nov 19 at 12:40










  • I want to show how to get to the value (1,0)
    – Thomas Christopher Davies
    Nov 19 at 16:05










  • Then simply note that you must have $(e_x,e_y) * (1,0) = (1,0)$. This gives you $1 = e_x$ and $0 = e_y$.
    – Paul Frost
    Nov 19 at 16:17










  • I don't see this being a good derivation
    – Thomas Christopher Davies
    Nov 19 at 17:28













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I want to show the neutral Element on $$(mathbb{C},*,+)$$



Let the complex Number be defined as:
$$(x,y)$$
Multiplication of complex numbers is defined as:
$$(x_{1}, y_{1}) * (x_{2}, y_{2}):= (x_{1}x_{2}-y_{1}y_{2}, x_{1}y_{2}+x_{2}y_{1})$$
By definition the neutral element is:
$$(e_{x}, e_{y}) *(x,y) = (x,y)$$
$$(x*e_{x}-y*e_{y}, x*e_{y}+y*e_{x})$$
This gives an equation system:
$$x = x*e_{x} -y*e_{y}$$
$$y = x*e_{y} +y*e_{x}$$



Is this the correct approach to finding the neutral Element?



If yes, what is the correct method for solving the equation system?






complex-numbers systems-of-equations







share|cite|improve this question















I want to show the neutral Element on $$(mathbb{C},*,+)$$



Let the complex Number be defined as:
$$(x,y)$$
Multiplication of complex numbers is defined as:
$$(x_{1}, y_{1}) * (x_{2}, y_{2}):= (x_{1}x_{2}-y_{1}y_{2}, x_{1}y_{2}+x_{2}y_{1})$$
By definition the neutral element is:
$$(e_{x}, e_{y}) *(x,y) = (x,y)$$
$$(x*e_{x}-y*e_{y}, x*e_{y}+y*e_{x})$$
This gives an equation system:
$$x = x*e_{x} -y*e_{y}$$
$$y = x*e_{y} +y*e_{x}$$



Is this the correct approach to finding the neutral Element?



If yes, what is the correct method for solving the equation system?






complex-numbers systems-of-equations




complex-numbers systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 19 at 12:31









Cameron Buie

84.8k771155




84.8k771155










asked Nov 19 at 12:24









Thomas Christopher Davies

247




247












  • Of course you can determine the multplicative neutral element by solving the above equation system. But I am sure you already know that the neutral element is $(1,0)$, and in that case it suffices to verify the equation $(1,0) * (x,y) = (x,y)$.
    – Paul Frost
    Nov 19 at 12:40










  • I want to show how to get to the value (1,0)
    – Thomas Christopher Davies
    Nov 19 at 16:05










  • Then simply note that you must have $(e_x,e_y) * (1,0) = (1,0)$. This gives you $1 = e_x$ and $0 = e_y$.
    – Paul Frost
    Nov 19 at 16:17










  • I don't see this being a good derivation
    – Thomas Christopher Davies
    Nov 19 at 17:28


















  • Of course you can determine the multplicative neutral element by solving the above equation system. But I am sure you already know that the neutral element is $(1,0)$, and in that case it suffices to verify the equation $(1,0) * (x,y) = (x,y)$.
    – Paul Frost
    Nov 19 at 12:40










  • I want to show how to get to the value (1,0)
    – Thomas Christopher Davies
    Nov 19 at 16:05










  • Then simply note that you must have $(e_x,e_y) * (1,0) = (1,0)$. This gives you $1 = e_x$ and $0 = e_y$.
    – Paul Frost
    Nov 19 at 16:17










  • I don't see this being a good derivation
    – Thomas Christopher Davies
    Nov 19 at 17:28
















Of course you can determine the multplicative neutral element by solving the above equation system. But I am sure you already know that the neutral element is $(1,0)$, and in that case it suffices to verify the equation $(1,0) * (x,y) = (x,y)$.
– Paul Frost
Nov 19 at 12:40




Of course you can determine the multplicative neutral element by solving the above equation system. But I am sure you already know that the neutral element is $(1,0)$, and in that case it suffices to verify the equation $(1,0) * (x,y) = (x,y)$.
– Paul Frost
Nov 19 at 12:40












I want to show how to get to the value (1,0)
– Thomas Christopher Davies
Nov 19 at 16:05




I want to show how to get to the value (1,0)
– Thomas Christopher Davies
Nov 19 at 16:05












Then simply note that you must have $(e_x,e_y) * (1,0) = (1,0)$. This gives you $1 = e_x$ and $0 = e_y$.
– Paul Frost
Nov 19 at 16:17




Then simply note that you must have $(e_x,e_y) * (1,0) = (1,0)$. This gives you $1 = e_x$ and $0 = e_y$.
– Paul Frost
Nov 19 at 16:17












I don't see this being a good derivation
– Thomas Christopher Davies
Nov 19 at 17:28




I don't see this being a good derivation
– Thomas Christopher Davies
Nov 19 at 17:28










2 Answers
2






active

oldest

votes

















up vote
2
down vote













Your approach is not (quite) correct because you should have $x=xe_x-ye_y$ and $y=xe_y+ye_x$ (just normal multiplication instead of your "*" operation). Also, the equations should hold for any x and y in $mathbb{C}$ and this implies that $e_x=1$ and $e_y=0$






share|cite|improve this answer





















  • But how can I show it?
    – Thomas Christopher Davies
    Nov 19 at 16:05










  • @Thomas Cristopher Davies : Your mistake is that you are confusing “” with the normal multiplication between two real numbers. Your operation is defined so that ($e_x,e_y$)(x,y)= ($e_xx-e_yy,e_xy+e_yx$) and NOT as ($e_x,e_y$)(x,y)= ($e_x*x-e_y*y,e_x*y+e_y*x$). It wouldn’t even make sens for “” to be defined in terms of itself...by the way, “*” is just complex multiplication
    – Sorin Tirc
    Nov 19 at 20:22












  • I would have used the correct dot if I had known that it is required. But you did not show how you derived the neutral element. I want to see how the equation system is being solved.
    – Thomas Christopher Davies
    Nov 20 at 18:11










  • A neutral element is an element $(e_1, e_2)$ such that $(e_1, e_2)*(x,y)=(x,y)$ FOR ANY REAL x and y. Choosing for example $x=1,y=0$ yields $e_1=1,e_2=0$
    – Sorin Tirc
    Nov 20 at 19:22












  • I know the definition of a neutral element, but you did not show how you got to the neutral element. Well, don't bother, I will post tomorrow the solution
    – Thomas Christopher Davies
    Nov 21 at 11:07


















up vote
0
down vote



accepted










The answer is
$$mathrm{I}quad x*e_x-y*e_y = x quad|*x$$
$$mathrm{II}quad x*e_y+y*e_x = x quad|*y$$
$$mathrm{I+II} quad x^2*e_x+y^2*e_x = x^2 + y^2$$
$$Leftrightarrow quad e_x * (x^2 + y^2) = x^2 + y^2 quad | :(x^2 + y^2 )$$
$$Leftrightarrow quad e_x = 1$$
Putting the solution $e_x = 1 $ into $mathrm{I}$
$$x*1-y *e_y = x quad |-x$$
$$Leftrightarrow quad-y*e_y=0 quad|:(-y)$$
$$Leftrightarrow quad e_y=0 $$
Thus the neutral element is $(1,0)$






share|cite|improve this answer





















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    2 Answers
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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

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    up vote
    2
    down vote













    Your approach is not (quite) correct because you should have $x=xe_x-ye_y$ and $y=xe_y+ye_x$ (just normal multiplication instead of your "*" operation). Also, the equations should hold for any x and y in $mathbb{C}$ and this implies that $e_x=1$ and $e_y=0$






    share|cite|improve this answer





















    • But how can I show it?
      – Thomas Christopher Davies
      Nov 19 at 16:05










    • @Thomas Cristopher Davies : Your mistake is that you are confusing “” with the normal multiplication between two real numbers. Your operation is defined so that ($e_x,e_y$)(x,y)= ($e_xx-e_yy,e_xy+e_yx$) and NOT as ($e_x,e_y$)(x,y)= ($e_x*x-e_y*y,e_x*y+e_y*x$). It wouldn’t even make sens for “” to be defined in terms of itself...by the way, “*” is just complex multiplication
      – Sorin Tirc
      Nov 19 at 20:22












    • I would have used the correct dot if I had known that it is required. But you did not show how you derived the neutral element. I want to see how the equation system is being solved.
      – Thomas Christopher Davies
      Nov 20 at 18:11










    • A neutral element is an element $(e_1, e_2)$ such that $(e_1, e_2)*(x,y)=(x,y)$ FOR ANY REAL x and y. Choosing for example $x=1,y=0$ yields $e_1=1,e_2=0$
      – Sorin Tirc
      Nov 20 at 19:22












    • I know the definition of a neutral element, but you did not show how you got to the neutral element. Well, don't bother, I will post tomorrow the solution
      – Thomas Christopher Davies
      Nov 21 at 11:07















    up vote
    2
    down vote













    Your approach is not (quite) correct because you should have $x=xe_x-ye_y$ and $y=xe_y+ye_x$ (just normal multiplication instead of your "*" operation). Also, the equations should hold for any x and y in $mathbb{C}$ and this implies that $e_x=1$ and $e_y=0$






    share|cite|improve this answer





















    • But how can I show it?
      – Thomas Christopher Davies
      Nov 19 at 16:05










    • @Thomas Cristopher Davies : Your mistake is that you are confusing “” with the normal multiplication between two real numbers. Your operation is defined so that ($e_x,e_y$)(x,y)= ($e_xx-e_yy,e_xy+e_yx$) and NOT as ($e_x,e_y$)(x,y)= ($e_x*x-e_y*y,e_x*y+e_y*x$). It wouldn’t even make sens for “” to be defined in terms of itself...by the way, “*” is just complex multiplication
      – Sorin Tirc
      Nov 19 at 20:22












    • I would have used the correct dot if I had known that it is required. But you did not show how you derived the neutral element. I want to see how the equation system is being solved.
      – Thomas Christopher Davies
      Nov 20 at 18:11










    • A neutral element is an element $(e_1, e_2)$ such that $(e_1, e_2)*(x,y)=(x,y)$ FOR ANY REAL x and y. Choosing for example $x=1,y=0$ yields $e_1=1,e_2=0$
      – Sorin Tirc
      Nov 20 at 19:22












    • I know the definition of a neutral element, but you did not show how you got to the neutral element. Well, don't bother, I will post tomorrow the solution
      – Thomas Christopher Davies
      Nov 21 at 11:07













    up vote
    2
    down vote










    up vote
    2
    down vote









    Your approach is not (quite) correct because you should have $x=xe_x-ye_y$ and $y=xe_y+ye_x$ (just normal multiplication instead of your "*" operation). Also, the equations should hold for any x and y in $mathbb{C}$ and this implies that $e_x=1$ and $e_y=0$






    share|cite|improve this answer












    Your approach is not (quite) correct because you should have $x=xe_x-ye_y$ and $y=xe_y+ye_x$ (just normal multiplication instead of your "*" operation). Also, the equations should hold for any x and y in $mathbb{C}$ and this implies that $e_x=1$ and $e_y=0$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 19 at 12:29









    Sorin Tirc

    89710




    89710












    • But how can I show it?
      – Thomas Christopher Davies
      Nov 19 at 16:05










    • @Thomas Cristopher Davies : Your mistake is that you are confusing “” with the normal multiplication between two real numbers. Your operation is defined so that ($e_x,e_y$)(x,y)= ($e_xx-e_yy,e_xy+e_yx$) and NOT as ($e_x,e_y$)(x,y)= ($e_x*x-e_y*y,e_x*y+e_y*x$). It wouldn’t even make sens for “” to be defined in terms of itself...by the way, “*” is just complex multiplication
      – Sorin Tirc
      Nov 19 at 20:22












    • I would have used the correct dot if I had known that it is required. But you did not show how you derived the neutral element. I want to see how the equation system is being solved.
      – Thomas Christopher Davies
      Nov 20 at 18:11










    • A neutral element is an element $(e_1, e_2)$ such that $(e_1, e_2)*(x,y)=(x,y)$ FOR ANY REAL x and y. Choosing for example $x=1,y=0$ yields $e_1=1,e_2=0$
      – Sorin Tirc
      Nov 20 at 19:22












    • I know the definition of a neutral element, but you did not show how you got to the neutral element. Well, don't bother, I will post tomorrow the solution
      – Thomas Christopher Davies
      Nov 21 at 11:07


















    • But how can I show it?
      – Thomas Christopher Davies
      Nov 19 at 16:05










    • @Thomas Cristopher Davies : Your mistake is that you are confusing “” with the normal multiplication between two real numbers. Your operation is defined so that ($e_x,e_y$)(x,y)= ($e_xx-e_yy,e_xy+e_yx$) and NOT as ($e_x,e_y$)(x,y)= ($e_x*x-e_y*y,e_x*y+e_y*x$). It wouldn’t even make sens for “” to be defined in terms of itself...by the way, “*” is just complex multiplication
      – Sorin Tirc
      Nov 19 at 20:22












    • I would have used the correct dot if I had known that it is required. But you did not show how you derived the neutral element. I want to see how the equation system is being solved.
      – Thomas Christopher Davies
      Nov 20 at 18:11










    • A neutral element is an element $(e_1, e_2)$ such that $(e_1, e_2)*(x,y)=(x,y)$ FOR ANY REAL x and y. Choosing for example $x=1,y=0$ yields $e_1=1,e_2=0$
      – Sorin Tirc
      Nov 20 at 19:22












    • I know the definition of a neutral element, but you did not show how you got to the neutral element. Well, don't bother, I will post tomorrow the solution
      – Thomas Christopher Davies
      Nov 21 at 11:07
















    But how can I show it?
    – Thomas Christopher Davies
    Nov 19 at 16:05




    But how can I show it?
    – Thomas Christopher Davies
    Nov 19 at 16:05












    @Thomas Cristopher Davies : Your mistake is that you are confusing “” with the normal multiplication between two real numbers. Your operation is defined so that ($e_x,e_y$)(x,y)= ($e_xx-e_yy,e_xy+e_yx$) and NOT as ($e_x,e_y$)(x,y)= ($e_x*x-e_y*y,e_x*y+e_y*x$). It wouldn’t even make sens for “” to be defined in terms of itself...by the way, “*” is just complex multiplication
    – Sorin Tirc
    Nov 19 at 20:22






    @Thomas Cristopher Davies : Your mistake is that you are confusing “” with the normal multiplication between two real numbers. Your operation is defined so that ($e_x,e_y$)(x,y)= ($e_xx-e_yy,e_xy+e_yx$) and NOT as ($e_x,e_y$)(x,y)= ($e_x*x-e_y*y,e_x*y+e_y*x$). It wouldn’t even make sens for “” to be defined in terms of itself...by the way, “*” is just complex multiplication
    – Sorin Tirc
    Nov 19 at 20:22














    I would have used the correct dot if I had known that it is required. But you did not show how you derived the neutral element. I want to see how the equation system is being solved.
    – Thomas Christopher Davies
    Nov 20 at 18:11




    I would have used the correct dot if I had known that it is required. But you did not show how you derived the neutral element. I want to see how the equation system is being solved.
    – Thomas Christopher Davies
    Nov 20 at 18:11












    A neutral element is an element $(e_1, e_2)$ such that $(e_1, e_2)*(x,y)=(x,y)$ FOR ANY REAL x and y. Choosing for example $x=1,y=0$ yields $e_1=1,e_2=0$
    – Sorin Tirc
    Nov 20 at 19:22






    A neutral element is an element $(e_1, e_2)$ such that $(e_1, e_2)*(x,y)=(x,y)$ FOR ANY REAL x and y. Choosing for example $x=1,y=0$ yields $e_1=1,e_2=0$
    – Sorin Tirc
    Nov 20 at 19:22














    I know the definition of a neutral element, but you did not show how you got to the neutral element. Well, don't bother, I will post tomorrow the solution
    – Thomas Christopher Davies
    Nov 21 at 11:07




    I know the definition of a neutral element, but you did not show how you got to the neutral element. Well, don't bother, I will post tomorrow the solution
    – Thomas Christopher Davies
    Nov 21 at 11:07










    up vote
    0
    down vote



    accepted










    The answer is
    $$mathrm{I}quad x*e_x-y*e_y = x quad|*x$$
    $$mathrm{II}quad x*e_y+y*e_x = x quad|*y$$
    $$mathrm{I+II} quad x^2*e_x+y^2*e_x = x^2 + y^2$$
    $$Leftrightarrow quad e_x * (x^2 + y^2) = x^2 + y^2 quad | :(x^2 + y^2 )$$
    $$Leftrightarrow quad e_x = 1$$
    Putting the solution $e_x = 1 $ into $mathrm{I}$
    $$x*1-y *e_y = x quad |-x$$
    $$Leftrightarrow quad-y*e_y=0 quad|:(-y)$$
    $$Leftrightarrow quad e_y=0 $$
    Thus the neutral element is $(1,0)$






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      The answer is
      $$mathrm{I}quad x*e_x-y*e_y = x quad|*x$$
      $$mathrm{II}quad x*e_y+y*e_x = x quad|*y$$
      $$mathrm{I+II} quad x^2*e_x+y^2*e_x = x^2 + y^2$$
      $$Leftrightarrow quad e_x * (x^2 + y^2) = x^2 + y^2 quad | :(x^2 + y^2 )$$
      $$Leftrightarrow quad e_x = 1$$
      Putting the solution $e_x = 1 $ into $mathrm{I}$
      $$x*1-y *e_y = x quad |-x$$
      $$Leftrightarrow quad-y*e_y=0 quad|:(-y)$$
      $$Leftrightarrow quad e_y=0 $$
      Thus the neutral element is $(1,0)$






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        The answer is
        $$mathrm{I}quad x*e_x-y*e_y = x quad|*x$$
        $$mathrm{II}quad x*e_y+y*e_x = x quad|*y$$
        $$mathrm{I+II} quad x^2*e_x+y^2*e_x = x^2 + y^2$$
        $$Leftrightarrow quad e_x * (x^2 + y^2) = x^2 + y^2 quad | :(x^2 + y^2 )$$
        $$Leftrightarrow quad e_x = 1$$
        Putting the solution $e_x = 1 $ into $mathrm{I}$
        $$x*1-y *e_y = x quad |-x$$
        $$Leftrightarrow quad-y*e_y=0 quad|:(-y)$$
        $$Leftrightarrow quad e_y=0 $$
        Thus the neutral element is $(1,0)$






        share|cite|improve this answer












        The answer is
        $$mathrm{I}quad x*e_x-y*e_y = x quad|*x$$
        $$mathrm{II}quad x*e_y+y*e_x = x quad|*y$$
        $$mathrm{I+II} quad x^2*e_x+y^2*e_x = x^2 + y^2$$
        $$Leftrightarrow quad e_x * (x^2 + y^2) = x^2 + y^2 quad | :(x^2 + y^2 )$$
        $$Leftrightarrow quad e_x = 1$$
        Putting the solution $e_x = 1 $ into $mathrm{I}$
        $$x*1-y *e_y = x quad |-x$$
        $$Leftrightarrow quad-y*e_y=0 quad|:(-y)$$
        $$Leftrightarrow quad e_y=0 $$
        Thus the neutral element is $(1,0)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 5 at 18:34









        Thomas Christopher Davies

        247




        247






























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