Neutral Element on $(mathbb{C},*,+)$
up vote
0
down vote
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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
edited Nov 19 at 12:31
Cameron Buie
84.8k771155
asked Nov 19 at 12:24
Thomas Christopher Davies
247
add a comment |
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
edited Nov 19 at 12:31
Cameron Buie
84.8k771155
asked Nov 19 at 12:24
Thomas Christopher Davies
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
add a comment |
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
edited Nov 19 at 12:31
Cameron Buie
84.8k771155
asked Nov 19 at 12:24
Thomas Christopher Davies
247
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
edited Nov 19 at 12:31
Cameron Buie
84.8k771155
asked Nov 19 at 12:24
Thomas Christopher Davies
247
edited Nov 19 at 12:31
Cameron Buie
84.8k771155
asked Nov 19 at 12:24
Thomas Christopher Davies
247
edited Nov 19 at 12:31
Cameron Buie
84.8k771155
edited Nov 19 at 12:31
Cameron Buie
84.8k771155
edited Nov 19 at 12:31
Cameron Buie
84.8k771155
84.8k771155
asked Nov 19 at 12:24
Thomas Christopher Davies
247
asked Nov 19 at 12:24
Thomas Christopher Davies
247
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
add a comment |
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
add a comment |
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 $e_y=0$
answered Nov 19 at 12:29
Sorin Tirc
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
add a comment |
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 $(1,0)$
answered Dec 5 at 18:34
Thomas Christopher Davies
247
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 $e_y=0$
answered Nov 19 at 12:29
Sorin Tirc
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
add a comment |
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 $e_y=0$
answered Nov 19 at 12:29
Sorin Tirc
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
add a comment |
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 $e_y=0$
answered Nov 19 at 12:29
Sorin Tirc
89710
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 $e_y=0$
answered Nov 19 at 12:29
Sorin Tirc
89710
answered Nov 19 at 12:29
Sorin Tirc
89710
answered Nov 19 at 12:29
Sorin Tirc
89710
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
add a comment |
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
add a comment |
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 $(1,0)$
answered Dec 5 at 18:34
Thomas Christopher Davies
247
add a comment |
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 $(1,0)$
answered Dec 5 at 18:34
Thomas Christopher Davies
247
add a comment |
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 $(1,0)$
answered Dec 5 at 18:34
Thomas Christopher Davies
247
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 $(1,0)$
answered Dec 5 at 18:34
Thomas Christopher Davies
247
answered Dec 5 at 18:34
Thomas Christopher Davies
247
answered Dec 5 at 18:34
Thomas Christopher Davies
247
answered Dec 5 at 18:34
Thomas Christopher Davies
247
247
add a comment |
add a comment |
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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