Mixing
Neutral Element on $(mathbb{C},*,+)$
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 23 hours ago
Cameron Buie
84.4k771155
asked 23 hours ago
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 23 hours ago
Cameron Buie
84.4k771155
asked 23 hours ago
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
23 hours ago
I want to show how to get to the value (1,0)
– Thomas Christopher Davies
19 hours ago
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
19 hours ago
I don't see this being a good derivation
– Thomas Christopher Davies
18 hours ago
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 23 hours ago
Cameron Buie
84.4k771155
asked 23 hours ago
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 23 hours ago
Cameron Buie
84.4k771155
asked 23 hours ago
Thomas Christopher Davies
247
edited 23 hours ago
Cameron Buie
84.4k771155
asked 23 hours ago
Thomas Christopher Davies
247
edited 23 hours ago
Cameron Buie
84.4k771155
edited 23 hours ago
Cameron Buie
84.4k771155
edited 23 hours ago
Cameron Buie
84.4k771155
84.4k771155
asked 23 hours ago
Thomas Christopher Davies
247
asked 23 hours ago
Thomas Christopher Davies
247
asked 23 hours ago
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
23 hours ago
I want to show how to get to the value (1,0)
– Thomas Christopher Davies
19 hours ago
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
19 hours ago
I don't see this being a good derivation
– Thomas Christopher Davies
18 hours ago
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
23 hours ago
I want to show how to get to the value (1,0)
– Thomas Christopher Davies
19 hours ago
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
19 hours ago
I don't see this being a good derivation
– Thomas Christopher Davies
18 hours ago
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
23 hours ago
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
23 hours ago
I want to show how to get to the value (1,0)
– Thomas Christopher Davies
19 hours ago
I want to show how to get to the value (1,0)
– Thomas Christopher Davies
19 hours ago
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
19 hours ago
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
19 hours ago
I don't see this being a good derivation
– Thomas Christopher Davies
18 hours ago
I don't see this being a good derivation
– Thomas Christopher Davies
18 hours ago
add a comment |
1 Answer
1
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 23 hours ago
Sorin Tirc
4888
But how can I show it?
– Thomas Christopher Davies
19 hours ago
@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
15 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
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 23 hours ago
Sorin Tirc
4888
But how can I show it?
– Thomas Christopher Davies
19 hours ago
@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
15 hours ago
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 23 hours ago
Sorin Tirc
4888
But how can I show it?
– Thomas Christopher Davies
19 hours ago
@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
15 hours ago
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 23 hours ago
Sorin Tirc
4888
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 23 hours ago
Sorin Tirc
4888
answered 23 hours ago
Sorin Tirc
4888
answered 23 hours ago
Sorin Tirc
4888
answered 23 hours ago
Sorin Tirc
4888
4888
But how can I show it?
– Thomas Christopher Davies
19 hours ago
@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
15 hours ago
add a comment |
But how can I show it?
– Thomas Christopher Davies
19 hours ago
@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
15 hours ago
But how can I show it?
– Thomas Christopher Davies
19 hours ago
But how can I show it?
– Thomas Christopher Davies
19 hours ago
@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
15 hours ago
@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
15 hours ago
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004861%2fneutral-element-on-mathbbc%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
23 hours ago
I want to show how to get to the value (1,0)
– Thomas Christopher Davies
19 hours ago
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
19 hours ago
I don't see this being a good derivation
– Thomas Christopher Davies
18 hours ago