Mixing
Does it make sense to write the complex plane as the cartesian product between $mathbbR$ and $imathbbR$?
Clash Royale CLAN TAG#URR8PPP
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Note: this question is inspired by someone (a teacher long time ago) who drew and labeled the two axes of the standard complex plane as $mathbbR$ and $imathbbR$.
Let $imathbbR = iy$
Does it make sense to write the complex plane $mathbbC$ as:
$mathbbC = mathbbR times imathbbR$?
Doing so, gives me that,
$mathbbC = (x,iy)$
But then again, the complex plane is usually defined as:
$mathbbC = z$
How to resolve these two seemingly different definitions?
elementary-set-theory complex-numbers
asked Aug 1 at 5:50
Roughly Stupid
405
add a comment |Â
up vote
0
down vote
favorite
Note: this question is inspired by someone (a teacher long time ago) who drew and labeled the two axes of the standard complex plane as $mathbbR$ and $imathbbR$.
Let $imathbbR = iy$
Does it make sense to write the complex plane $mathbbC$ as:
$mathbbC = mathbbR times imathbbR$?
Doing so, gives me that,
$mathbbC = (x,iy)$
But then again, the complex plane is usually defined as:
$mathbbC = z$
How to resolve these two seemingly different definitions?
elementary-set-theory complex-numbers
asked Aug 1 at 5:50
Roughly Stupid
405
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Note: this question is inspired by someone (a teacher long time ago) who drew and labeled the two axes of the standard complex plane as $mathbbR$ and $imathbbR$.
Let $imathbbR = iy$
Does it make sense to write the complex plane $mathbbC$ as:
$mathbbC = mathbbR times imathbbR$?
Doing so, gives me that,
$mathbbC = (x,iy)$
But then again, the complex plane is usually defined as:
$mathbbC = z$
How to resolve these two seemingly different definitions?
elementary-set-theory complex-numbers
asked Aug 1 at 5:50
Roughly Stupid
405
Note: this question is inspired by someone (a teacher long time ago) who drew and labeled the two axes of the standard complex plane as $mathbbR$ and $imathbbR$.
Let $imathbbR = iy$
Does it make sense to write the complex plane $mathbbC$ as:
$mathbbC = mathbbR times imathbbR$?
Doing so, gives me that,
$mathbbC = (x,iy)$
But then again, the complex plane is usually defined as:
$mathbbC = z$
How to resolve these two seemingly different definitions?
elementary-set-theory complex-numbers
asked Aug 1 at 5:50
Roughly Stupid
405
edited Aug 1 at 6:04
asked Aug 1 at 5:50
Roughly Stupid
405
asked Aug 1 at 5:50
Roughly Stupid
405
asked Aug 1 at 5:50
Roughly Stupid
405
405
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
It makes perfect sense to define $mathbbC$ in terms of $mathbbR times mathbbR$, with addition defined by
$$(a, b) + (c, d) = (a + c, b + d)$$
and multiplication defined by
$$(a, b)(c, d) = (ac - bd, ad + bc).$$
Structurally, this is absolutely identical to any of the other definitions of complex numbers. If you're happy to abuse notations and notate such pairs of the form $(a, 0)$ by the real number $a$, and denote by $i$ the element $(0, 1)$, then $(a, b) = a + ib$ for any $a, b in mathbbR$.
So, how do we resolve the difference between this and other definitions? Well, we take the view that it doesn't matter. Whether you view complex numbers as matrices of the form $beginpmatrix x & -y \ y & x endpmatrix$, or as elements of the splitting field $mathbbR[x]/langle x^2 + 1 rangle$, or as ordered pairs from the definition above, the fact is that all of them produce an identical structure, and it is very simple to change between them in a way that respects addition and multiplication.
(FYI, I didn't define $mathbbR times imathbbR$, because, before defining complex numbers, it's not clear what $i$ actually is. Sure, you can say it's the square root of $-1$, but that only shows the property it's supposed to satisfy, not what it really is, or what context it can be well-defined.)
answered Aug 1 at 6:00
Theo Bendit
11.7k1841
Aha, you got me completely. I didn't even define imaginary unit before defining the set of complex numbers -.-... Do you have a well-known reference (i.e., some algebra book or complex analysis book) for the matrix representation of complex numbers?
– Roughly Stupid
Aug 1 at 6:07
@RoughlyStupid I don't. I just came across it in some undergraduate course, from years ago. :-/
– Theo Bendit
Aug 1 at 6:08
That said, it's not long or difficult to verify for yourself!
– Theo Bendit
Aug 1 at 6:09
add a comment |Â
up vote
1
down vote
As points, it makes sense. As additive groups, it makes sense. But there is no natural multiplication on $Bbb Rtimes iBbb R$ because the multiplication on $iBbb R$ isn't closed (it is not a ring, at least not with the standard multiplication). And even if there were, the multiplication in $Bbb C$ is not isomorphic to the natural multiplication of any (non-trivial) direct product of rings.
As for how to reconcile the seemingly different definitions where they do make sense, you apply the "obvious" bijections / isomorphisms.
answered Aug 1 at 6:01
Arthur
98.3k793174
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
It makes perfect sense to define $mathbbC$ in terms of $mathbbR times mathbbR$, with addition defined by
$$(a, b) + (c, d) = (a + c, b + d)$$
and multiplication defined by
$$(a, b)(c, d) = (ac - bd, ad + bc).$$
Structurally, this is absolutely identical to any of the other definitions of complex numbers. If you're happy to abuse notations and notate such pairs of the form $(a, 0)$ by the real number $a$, and denote by $i$ the element $(0, 1)$, then $(a, b) = a + ib$ for any $a, b in mathbbR$.
So, how do we resolve the difference between this and other definitions? Well, we take the view that it doesn't matter. Whether you view complex numbers as matrices of the form $beginpmatrix x & -y \ y & x endpmatrix$, or as elements of the splitting field $mathbbR[x]/langle x^2 + 1 rangle$, or as ordered pairs from the definition above, the fact is that all of them produce an identical structure, and it is very simple to change between them in a way that respects addition and multiplication.
(FYI, I didn't define $mathbbR times imathbbR$, because, before defining complex numbers, it's not clear what $i$ actually is. Sure, you can say it's the square root of $-1$, but that only shows the property it's supposed to satisfy, not what it really is, or what context it can be well-defined.)
answered Aug 1 at 6:00
Theo Bendit
11.7k1841
Aha, you got me completely. I didn't even define imaginary unit before defining the set of complex numbers -.-... Do you have a well-known reference (i.e., some algebra book or complex analysis book) for the matrix representation of complex numbers?
– Roughly Stupid
Aug 1 at 6:07
@RoughlyStupid I don't. I just came across it in some undergraduate course, from years ago. :-/
– Theo Bendit
Aug 1 at 6:08
That said, it's not long or difficult to verify for yourself!
– Theo Bendit
Aug 1 at 6:09
add a comment |Â
up vote
5
down vote
accepted
It makes perfect sense to define $mathbbC$ in terms of $mathbbR times mathbbR$, with addition defined by
$$(a, b) + (c, d) = (a + c, b + d)$$
and multiplication defined by
$$(a, b)(c, d) = (ac - bd, ad + bc).$$
Structurally, this is absolutely identical to any of the other definitions of complex numbers. If you're happy to abuse notations and notate such pairs of the form $(a, 0)$ by the real number $a$, and denote by $i$ the element $(0, 1)$, then $(a, b) = a + ib$ for any $a, b in mathbbR$.
So, how do we resolve the difference between this and other definitions? Well, we take the view that it doesn't matter. Whether you view complex numbers as matrices of the form $beginpmatrix x & -y \ y & x endpmatrix$, or as elements of the splitting field $mathbbR[x]/langle x^2 + 1 rangle$, or as ordered pairs from the definition above, the fact is that all of them produce an identical structure, and it is very simple to change between them in a way that respects addition and multiplication.
(FYI, I didn't define $mathbbR times imathbbR$, because, before defining complex numbers, it's not clear what $i$ actually is. Sure, you can say it's the square root of $-1$, but that only shows the property it's supposed to satisfy, not what it really is, or what context it can be well-defined.)
answered Aug 1 at 6:00
Theo Bendit
11.7k1841
Aha, you got me completely. I didn't even define imaginary unit before defining the set of complex numbers -.-... Do you have a well-known reference (i.e., some algebra book or complex analysis book) for the matrix representation of complex numbers?
– Roughly Stupid
Aug 1 at 6:07
@RoughlyStupid I don't. I just came across it in some undergraduate course, from years ago. :-/
– Theo Bendit
Aug 1 at 6:08
That said, it's not long or difficult to verify for yourself!
– Theo Bendit
Aug 1 at 6:09
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
It makes perfect sense to define $mathbbC$ in terms of $mathbbR times mathbbR$, with addition defined by
$$(a, b) + (c, d) = (a + c, b + d)$$
and multiplication defined by
$$(a, b)(c, d) = (ac - bd, ad + bc).$$
Structurally, this is absolutely identical to any of the other definitions of complex numbers. If you're happy to abuse notations and notate such pairs of the form $(a, 0)$ by the real number $a$, and denote by $i$ the element $(0, 1)$, then $(a, b) = a + ib$ for any $a, b in mathbbR$.
So, how do we resolve the difference between this and other definitions? Well, we take the view that it doesn't matter. Whether you view complex numbers as matrices of the form $beginpmatrix x & -y \ y & x endpmatrix$, or as elements of the splitting field $mathbbR[x]/langle x^2 + 1 rangle$, or as ordered pairs from the definition above, the fact is that all of them produce an identical structure, and it is very simple to change between them in a way that respects addition and multiplication.
(FYI, I didn't define $mathbbR times imathbbR$, because, before defining complex numbers, it's not clear what $i$ actually is. Sure, you can say it's the square root of $-1$, but that only shows the property it's supposed to satisfy, not what it really is, or what context it can be well-defined.)
answered Aug 1 at 6:00
Theo Bendit
11.7k1841
It makes perfect sense to define $mathbbC$ in terms of $mathbbR times mathbbR$, with addition defined by
$$(a, b) + (c, d) = (a + c, b + d)$$
and multiplication defined by
$$(a, b)(c, d) = (ac - bd, ad + bc).$$
Structurally, this is absolutely identical to any of the other definitions of complex numbers. If you're happy to abuse notations and notate such pairs of the form $(a, 0)$ by the real number $a$, and denote by $i$ the element $(0, 1)$, then $(a, b) = a + ib$ for any $a, b in mathbbR$.
So, how do we resolve the difference between this and other definitions? Well, we take the view that it doesn't matter. Whether you view complex numbers as matrices of the form $beginpmatrix x & -y \ y & x endpmatrix$, or as elements of the splitting field $mathbbR[x]/langle x^2 + 1 rangle$, or as ordered pairs from the definition above, the fact is that all of them produce an identical structure, and it is very simple to change between them in a way that respects addition and multiplication.
(FYI, I didn't define $mathbbR times imathbbR$, because, before defining complex numbers, it's not clear what $i$ actually is. Sure, you can say it's the square root of $-1$, but that only shows the property it's supposed to satisfy, not what it really is, or what context it can be well-defined.)
answered Aug 1 at 6:00
Theo Bendit
11.7k1841
answered Aug 1 at 6:00
Theo Bendit
11.7k1841
answered Aug 1 at 6:00
Theo Bendit
11.7k1841
answered Aug 1 at 6:00
Theo Bendit
11.7k1841
11.7k1841
Aha, you got me completely. I didn't even define imaginary unit before defining the set of complex numbers -.-... Do you have a well-known reference (i.e., some algebra book or complex analysis book) for the matrix representation of complex numbers?
– Roughly Stupid
Aug 1 at 6:07
@RoughlyStupid I don't. I just came across it in some undergraduate course, from years ago. :-/
– Theo Bendit
Aug 1 at 6:08
That said, it's not long or difficult to verify for yourself!
– Theo Bendit
Aug 1 at 6:09
add a comment |Â
Aha, you got me completely. I didn't even define imaginary unit before defining the set of complex numbers -.-... Do you have a well-known reference (i.e., some algebra book or complex analysis book) for the matrix representation of complex numbers?
– Roughly Stupid
Aug 1 at 6:07
@RoughlyStupid I don't. I just came across it in some undergraduate course, from years ago. :-/
– Theo Bendit
Aug 1 at 6:08
That said, it's not long or difficult to verify for yourself!
– Theo Bendit
Aug 1 at 6:09
Aha, you got me completely. I didn't even define imaginary unit before defining the set of complex numbers -.-... Do you have a well-known reference (i.e., some algebra book or complex analysis book) for the matrix representation of complex numbers?
– Roughly Stupid
Aug 1 at 6:07
Aha, you got me completely. I didn't even define imaginary unit before defining the set of complex numbers -.-... Do you have a well-known reference (i.e., some algebra book or complex analysis book) for the matrix representation of complex numbers?
– Roughly Stupid
Aug 1 at 6:07
@RoughlyStupid I don't. I just came across it in some undergraduate course, from years ago. :-/
– Theo Bendit
Aug 1 at 6:08
@RoughlyStupid I don't. I just came across it in some undergraduate course, from years ago. :-/
– Theo Bendit
Aug 1 at 6:08
That said, it's not long or difficult to verify for yourself!
– Theo Bendit
Aug 1 at 6:09
That said, it's not long or difficult to verify for yourself!
– Theo Bendit
Aug 1 at 6:09
add a comment |Â
up vote
1
down vote
As points, it makes sense. As additive groups, it makes sense. But there is no natural multiplication on $Bbb Rtimes iBbb R$ because the multiplication on $iBbb R$ isn't closed (it is not a ring, at least not with the standard multiplication). And even if there were, the multiplication in $Bbb C$ is not isomorphic to the natural multiplication of any (non-trivial) direct product of rings.
As for how to reconcile the seemingly different definitions where they do make sense, you apply the "obvious" bijections / isomorphisms.
answered Aug 1 at 6:01
Arthur
98.3k793174
add a comment |Â
up vote
1
down vote
As points, it makes sense. As additive groups, it makes sense. But there is no natural multiplication on $Bbb Rtimes iBbb R$ because the multiplication on $iBbb R$ isn't closed (it is not a ring, at least not with the standard multiplication). And even if there were, the multiplication in $Bbb C$ is not isomorphic to the natural multiplication of any (non-trivial) direct product of rings.
As for how to reconcile the seemingly different definitions where they do make sense, you apply the "obvious" bijections / isomorphisms.
answered Aug 1 at 6:01
Arthur
98.3k793174
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As points, it makes sense. As additive groups, it makes sense. But there is no natural multiplication on $Bbb Rtimes iBbb R$ because the multiplication on $iBbb R$ isn't closed (it is not a ring, at least not with the standard multiplication). And even if there were, the multiplication in $Bbb C$ is not isomorphic to the natural multiplication of any (non-trivial) direct product of rings.
As for how to reconcile the seemingly different definitions where they do make sense, you apply the "obvious" bijections / isomorphisms.
answered Aug 1 at 6:01
Arthur
98.3k793174
As points, it makes sense. As additive groups, it makes sense. But there is no natural multiplication on $Bbb Rtimes iBbb R$ because the multiplication on $iBbb R$ isn't closed (it is not a ring, at least not with the standard multiplication). And even if there were, the multiplication in $Bbb C$ is not isomorphic to the natural multiplication of any (non-trivial) direct product of rings.
As for how to reconcile the seemingly different definitions where they do make sense, you apply the "obvious" bijections / isomorphisms.
answered Aug 1 at 6:01
Arthur
98.3k793174
edited Aug 1 at 6:06
answered Aug 1 at 6:01
Arthur
98.3k793174
answered Aug 1 at 6:01
Arthur
98.3k793174
answered Aug 1 at 6:01
Arthur
98.3k793174
98.3k793174
add a comment |Â
add a comment |Â
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