Complex numbers and their matrix form.












7












$begingroup$


I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane. This thing can be called a vector and be represented as $(a,b), [atext{ }b]^T$ (column vector) or by $amathbf{i}+bmathbf{j}$, where $(mathbf{i},mathbf{j})$ is the stardard basis in $mathbb{R}^2$ Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=sqrt{-1}$.



So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations
$begin{bmatrix} cos(90) & -sin(90) \ sin(90) & cos(90)\ end{bmatrix}$=$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}$ and we find that
$$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}begin{bmatrix} a \ b\ end{bmatrix}=begin{bmatrix} -b \ a\ end{bmatrix}$$
Or, I could choose the complex multiplication way and say,
$i(a+bi)=ai+bi^2=ai-b=-b+ai$



So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?






linear-algebra matrices complex-numbers rotations







share|cite|improve this question











$endgroup$












  • $begingroup$
    For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
    $endgroup$
    – dineshdileep
    Sep 19 '13 at 11:07










  • $begingroup$
    Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
    $endgroup$
    – Incnis Mrsi
    Nov 2 '14 at 7:09


















7












$begingroup$


I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane. This thing can be called a vector and be represented as $(a,b), [atext{ }b]^T$ (column vector) or by $amathbf{i}+bmathbf{j}$, where $(mathbf{i},mathbf{j})$ is the stardard basis in $mathbb{R}^2$ Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=sqrt{-1}$.



So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations
$begin{bmatrix} cos(90) & -sin(90) \ sin(90) & cos(90)\ end{bmatrix}$=$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}$ and we find that
$$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}begin{bmatrix} a \ b\ end{bmatrix}=begin{bmatrix} -b \ a\ end{bmatrix}$$
Or, I could choose the complex multiplication way and say,
$i(a+bi)=ai+bi^2=ai-b=-b+ai$



So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?






linear-algebra matrices complex-numbers rotations







share|cite|improve this question











$endgroup$












  • $begingroup$
    For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
    $endgroup$
    – dineshdileep
    Sep 19 '13 at 11:07










  • $begingroup$
    Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
    $endgroup$
    – Incnis Mrsi
    Nov 2 '14 at 7:09
















7












7








7


5



$begingroup$


I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane. This thing can be called a vector and be represented as $(a,b), [atext{ }b]^T$ (column vector) or by $amathbf{i}+bmathbf{j}$, where $(mathbf{i},mathbf{j})$ is the stardard basis in $mathbb{R}^2$ Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=sqrt{-1}$.



So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations
$begin{bmatrix} cos(90) & -sin(90) \ sin(90) & cos(90)\ end{bmatrix}$=$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}$ and we find that
$$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}begin{bmatrix} a \ b\ end{bmatrix}=begin{bmatrix} -b \ a\ end{bmatrix}$$
Or, I could choose the complex multiplication way and say,
$i(a+bi)=ai+bi^2=ai-b=-b+ai$



So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?






linear-algebra matrices complex-numbers rotations







share|cite|improve this question











$endgroup$




I have a line starting at the origin, and i extend it to a point $(a,b)$ in the plane. This thing can be called a vector and be represented as $(a,b), [atext{ }b]^T$ (column vector) or by $amathbf{i}+bmathbf{j}$, where $(mathbf{i},mathbf{j})$ is the stardard basis in $mathbb{R}^2$ Or it could be seen as a visual representation of a complex number where $(a,b)=a+bi,$ where $i=sqrt{-1}$.



So I want to rotate this vector $(a,b)$ $90$ degrees counter clockwise, so i know I can use my trusty matrix for rotations
$begin{bmatrix} cos(90) & -sin(90) \ sin(90) & cos(90)\ end{bmatrix}$=$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}$ and we find that
$$begin{bmatrix} 0 & -1 \ 1 & 0\ end{bmatrix}begin{bmatrix} a \ b\ end{bmatrix}=begin{bmatrix} -b \ a\ end{bmatrix}$$
Or, I could choose the complex multiplication way and say,
$i(a+bi)=ai+bi^2=ai-b=-b+ai$



So we all know that, but what are some of the advantages and disadvantages to having two things that are completely identical operation in different systems?






linear-algebra matrices complex-numbers rotations




linear-algebra matrices complex-numbers rotations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 0:16









dantopa

6,53942244




6,53942244










asked Sep 19 '13 at 2:37









Irish M PowersIrish M Powers

1155




1155












  • $begingroup$
    For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
    $endgroup$
    – dineshdileep
    Sep 19 '13 at 11:07










  • $begingroup$
    Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
    $endgroup$
    – Incnis Mrsi
    Nov 2 '14 at 7:09




















  • $begingroup$
    For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
    $endgroup$
    – dineshdileep
    Sep 19 '13 at 11:07










  • $begingroup$
    Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
    $endgroup$
    – Incnis Mrsi
    Nov 2 '14 at 7:09


















$begingroup$
For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
$endgroup$
– dineshdileep
Sep 19 '13 at 11:07




$begingroup$
For this particular example, one obvious difference is computation. To use the rotation matrix, you needed 4 trigonometric computations, 4 scalar multiplications and 2 additions. But in the other method, you swapped the entries and changed the sign of one. In general, for any degree of rotation, both will have same amount of computation.
$endgroup$
– dineshdileep
Sep 19 '13 at 11:07












$begingroup$
Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
$endgroup$
– Incnis Mrsi
Nov 2 '14 at 7:09






$begingroup$
Not clear what does the author mean. Multiplication of complex numbers gets two numbers (elements of the field ℂ) and yields a product from the same set. Euclidean 2-vectors are another thing. Although you can “multiply” two rotations and obtain another rotation, there is no multiplication on 2-vectors that gives 2-vectors.
$endgroup$
– Incnis Mrsi
Nov 2 '14 at 7:09












1 Answer
1






active

oldest

votes


















2












$begingroup$

There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21
















2












$begingroup$

There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21














2












2








2





$begingroup$

There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$






share|cite|improve this answer











$endgroup$



There is a homeomorphism between the the complex numbers



$$
color{blue}{a} + color{red}{b}i
$$



and the rotation matrices



$$
left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{a} \
end{array}
right)
$$




Let the Cartesian forms be
$$
z_{1} = color{blue}{a} + color{red}{b}i, quad
z_{2} = color{blue}{c} + color{red}{d}i
$$
and the matrix forms
$$
z_{1} = left(
begin{array}{rc}
color{blue}{a} & color{red}{b} \
-color{red}{b} & color{blue}{c} \
end{array}
right),
quad
z_{2} = left(
begin{array}{rc}
color{blue}{c}& color{red}{d} \
- color{red}{d} & color{blue}{c}\
end{array}
right)
$$



Addition



$$
begin{align}
%
z_{1} +
z_{2} &=
(color{blue}{a}+color{blue}{c}) + (color{red}{b} + color{red}{d} )i \[3pt]
%
&=
left(
begin{array}{rc}
color{blue}{a}+color{blue}{c}& color{red}{b}+ color{red}{d} \
-color{red}{b}- color{red}{d} & color{blue}{a}+color{blue}{c}\
end{array}
right)
end{align}
$$



Multiplication



$$
begin{align}
z_{1}
z_{2} &=
(color{blue}{ac}- color{red}{bd}) +
(color{red}{b}color{blue}{c}+color{blue}{a}color{red}{d})i\[3pt]
%
&=
%
left(
begin{array}{rc}
color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} & color{red}{b}color{blue}{c}+color{blue}{a} color{red}{d} \
- color{red}{b} color{blue}{c}-color{blue}{a} color{red}{d} & color{blue}{a}color{blue}{c}- color{red}{b} color{red}{d} \
end{array}
right)
end{align}
$$



Inversion



$$
begin{align}
%
frac{1}{z} &= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1} left( color{blue}{a} - color{red}{b} i right) \[3pt]
%
&= left( color{blue}{a}^{2} + color{red}{b}^{2} right)^{-1}
left(
begin{array}{cr}
color{blue}{a} & - color{red}{b} \
color{red}{b} & color{blue}{a} \
end{array}
right)
end{align}
$$







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edited Nov 5 '18 at 16:02









amWhy

1




1










answered Apr 24 '17 at 3:39









dantopadantopa

6,53942244




6,53942244








  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21














  • 1




    $begingroup$
    This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
    $endgroup$
    – Starfall
    Apr 24 '17 at 4:04










  • $begingroup$
    @Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
    $endgroup$
    – dantopa
    Apr 24 '17 at 4:21








1




1




$begingroup$
This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
$endgroup$
– Starfall
Apr 24 '17 at 4:04




$begingroup$
This is correct in the sense that these two topological spaces are homeomorphic, but what you (partially) justify in your answer is that these are isomorphic as rings. There is nothing in your answer which says anything about topology.
$endgroup$
– Starfall
Apr 24 '17 at 4:04












$begingroup$
@Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
$endgroup$
– dantopa
Apr 24 '17 at 4:21




$begingroup$
@Starfall: Excellent point; this is the site for mathematical precision and detail. We agree the answer falls short. Hopefully, it is enough for our poster. Thanks for paving the path to completion.
$endgroup$
– dantopa
Apr 24 '17 at 4:21


















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