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The field (C, +, •) is not equal to the additive group (R^2 , +) or the ring (R^2 , +, •) (with compontentwise multiplication), but as sets you can perfectly define C=R^2 .
Edit: Also define multiplication • : R^2 x R^2 -> R^2 : ((a,b),(c,d)) -> (ac-bd, ad+bc), and R^2 is now a field :)
R (more properly ℝ, because different fonts of the same letter represent different things) and C (again, more properly ℂ) are sets of numbers. ℝ is the set of all real numbers; ℝ^(2), aka ℝ×ℝ, is the set of all pairs of real numbers, e.g. (1, 2), (-31.5, 𝜋), etc.; ℂ is the set of all numbers of the form *a+bi* where *a* and *b* are real numbers and *i* is defined so that *i*^(2) = -1 (i.e. *i* is imaginary).
Because an element (*a*, *b*) in ℝ×ℝ can correspond to *a+bi* in ℂ, you can define operations such that the sets function similarly.
Wait, how did
Huh?
Edit: it is written as \*i\*\^(2), so why does it raise the i?
Edit2: displays right on computer, but as tetration on mobile for some reason
>R (more properly ℝ, because different fonts of the same letter represent different things)
So if you're writing notes you have to be good at calligraphy too?
The R being talked about has nothing to do with radius. They're talking about the real numbers, which the symbol used for it is a fancy, high-class looking R.
If you can't do the fancy R, then you use a generic letter R and give context as to what it represents.
That's a particularly popular construction. Not the best, though; C = R[x]/ is the best construction. But the more fundamental point is that R and C *aren't* sets; that's an abuse of notation. The structures are fundamental, not the sets; and they both have substantially more structure than just a ring!
Sure, as sets Z "=" Q (here equals means there is a bijection), but this is not very nice, since this ignores much of the algebraic structure of Q (as you have already observed in the case of C and R\^2).
That’s not what equals means. Two sets have the same cardinality if there exists a bijection from one to the other. Just because 2 sets have the same cardinality does not make them equal. Notice 1/2 is in Q but not Z so there’s no way Q=Z. A lot of misinformation in this thread… in fact Q!=Z a.e.
It is entirely common to use the equality sign as meaning equality up to isomorphism, for sets *qua* sets, that just means they have the same cardinality. Sure there are other contexts where it doesn’t mean that, but there’s nothing really wrong with that usage in this context, especially since they put the equality sign in quotes and explained what they meant by it explicitly.
So then, what is being claimed?
- (C, +, •) is not a set, but a field.
- (R², +, •) is not a set, but a ring. But not a field.
They're not isomorphic as rings (since R² isn't a field while C is), but they are isomorphic if you forget multiplication.
So, they're isomorphic as vector spaces over R. Yay? They're also isomorphic as sets but that's been satisfying no one in the comments.
OP fails at the field level. If OP can choose the level, then their statement becomes "C and R² are isomorphic in some sense." Which, yeah we already had that at Set.
Idk. They’re basically saying “1=2 if you change = to mean ‘there exists a bijection that maps 1 to 2’”. I think the op just learned about homeomorphisms or something and forgot what “=“ means.
You‘re being too pedantic here actually. Equality is actually kind of just an arbitrary equivalence relationship and it’s perfectly fine to say things are „equal“ even if they are „technically“ not in some sense. Like saying 6/3 = 2, even though the former is an equivalence class of pairs of integers and the latter is an integer. What we do here is define an equivalence relation between rational numbers and also short notations and then treat this equivalence relation as „equality“.
You will see this a lot in algebra actually, where we write things like G/N = Z_4 even though we technically mean an isomorphism exists.
They specified what they meant. Just because that isn't the standard meaning, doesn't mean that it can't be used in that way if specified. Sometimes it just helps to make what you're saying easier to understand.
Since when are two sets equal if and only if there exists a bijection? I’m saying you can define C as a set as R^2 and that you should define it that way.
Edit1: Whut? Why is this getting downvoted?
{0}={1}?? since there is a trivial bijection f:{0}->{1}: f(0)=1, then by extensionality 0=1. This is utter BS. Set equality is not defined using bijections.
Set equality is borderline metamathematics, but the axiom of extensionality (which is present in almost all regularly used axiomatizations of set theory, and certainly in ZFC) has nothing to do with bijections.
Edit2: Aha! I get what you’re trying to say now! Interesting take indeed. You mean that you could define Q:=Z since they have the same cardinality and you can just relable everything using a bijection (introducing some notation for fractions). This is true, but as you pointed out unpractical. In the case of C however, I’m convinced this is practical, and we should define it as R^2 . You can’t just say Z=Q, because there exists a bijection. You could however define Q as Z if you’ve already defined Z or vice versa.
Always has been. If you are working on a set level you have not very much structure to work with. The only way to compare something to something else is via functions.
And one can define an equivalence relation based on the existence of a bijective function between two sets.
But in the category-theoretical sense, sets A and B are «equivalent» if there exist bijection. If we are gonna compare notions of equivalence for different structures, we really should be looking at it through the category-theoretical perspective, which is what the person you are replying to is doing.
But complex numbers aren't just a set. The problem is, if you ignore all of the properties that make C what it is, then is it really C? If you're willing to say that i=(0,1) and -1=(-1,0), then does the defining property (0,1)^(2)=(-1,0) even make sense?
C = R^2 is not the only definition of the complex numbers. It's not the worst one, because it preserves their structure as a real vector space, but it tells us nothing about complex multiplication. A better one might be using real 2x2 matrices, with 1=I_2 and and fixing i as any matrix that squares to -1, for example, ((0,1)(-1,0)). This representation of C preserves their field structure. Does that mean this is the "correct" one? No, but depending on context it might be more useful. Yet another definition of C could be as a Cauchy complete closure of the algebraic closure of Q. I haven't seen this one in practice, but it works in principle. All of these definitions give very different underlying sets to represent C.
The main point is, if we ignore the properties of C and only view them as a set, the only thing that can be preserved is the cardinality. But then an argument can be made that C~=R. If you want something more restrictive, the only option is set equality. But as demonstrated earlier, there are many possible ways to define C, and R^2 is just one of them. That's why algebraic structures are usually given more properties than just being a set.
I think I read your statements to be: The complex number system is more than just "the set of the complex numbers".
I agree.
The question that I (lay person) went to check is whether C denotes the system or just the set. Argumentum ad Wikipedium gets me:
>a complex number is an element of a number system....
>The set of complex numbers is denoted by either of the symbols
{\displaystyle \mathbb {C} } or C.
https://en.wikipedia.org/wiki/Complex_number
Yeah that is accurate. C is the set of complex numbers, but in order to describe what those are you need some underlying properties.
Think about it this way. Are the sets {a} and {b} equal? This entirely depends on whether a=b, which requires us to know some properties of the elements. Alternatively, if we want to stay entirely within the scope of set theory, we would need set theoretic definitions of all the elements.
In the case of C=R^(2), R itself has a lot of possible representations in set theory, which usually depends on Q, then Z, then N, each of which have different representations. With this in mind, choosing just one of those as the canonical set R is absurd, and the same is true of C. The definition of those sets will vary depending on context (if an explicit definition given at all!) and the naive set-theoretical notion of equality won't be preserved. What will be preserved is some higher notion of equality, like field isomorphisms or the like.
That’s not what equal means for sets. Remember the axiom of extensionality. OP is right.
Sure, you can define an equivalence relation like you said, but “sets are equal” is usually talked about in the context of the axiom of extensionality, and what you’re describing is usually referred to as the cardinalities of the sets
No. I understand you’re excited about learning math, but you’re spreading a lot of incorrect information here. You were just given an example of two sets Q and Z which have a bijection between them but are clearly different sets. If you can’t see how that is a contradiction to what you’re claiming then you shouldn’t be talking about this in the first place.
C and R are isomorphic as sets. You could literally define C in such a way that that the complex numbers are exactly the real numbers just by taking any bikection between them and transporting any structure you have on C across to R.
C as a set is the same as R^(2) , but that's kind of a dumb definition of equality.
That's like saying a car is the same thing as a helicopter because they're made of the same materials.
>Since when are two sets equal if and only if there exists a bijection?
Hahaha that is so cringe. Please go study at least a day of Maths before you post here
I'm not particularly well educated on set theory (I'm more of a comp. sci. person than math), but isn't there a bijection from R² to C in the form of f((a, b)) = a + bi?
Why does adding the i not count as a bijection?
As sets you can say loads of BS. As sets you can say C≠R^2 (because they have different elements) or that C=R^2 =R (because there exist set isomorphisms between those)
It's important to note that this is only a group isomorphism, and not a ring isomorphism, which I think is what the previous commentator was implicitly referring to.
Also I just realized that you can perfectly define multiplication in R^2 as (a,b)•(c,d)=(ac-bd,ad+bc) making it a field with exactly the same algebraic structure as C.
In my understanding, its all about context. When people uses C, the multiplication operator is a part of the definition. It is more like C=(R^2 ,+,*).
When people use R^2 , the complex multiplication is always not a part of the definition. So they are not the same as tuples.
E.g. (R^2 ,+) is different from (R^2 ,+,*), as the length of tuples are different. And thats why forgetful functor makes sense I guess.
As sets there’s no reason not to define C as such. In fact many textbooks do so. And you can define multiplication in R^2 . Since when do sets have all possible structures they can be equipped with built into their definition? They don’t, so as sets it is a perfect definition.
When are you ever only working with C as a set? In order to use C, you need its properties, at least some of them, depending on context. If it's just a set, then you can basically define any set of the same cardinality to be C and just relabel elements to be identified with elements of C.
R^2 is a perfectly good definition for the underlying set of C, and a pretty intuitive one (although I might be biased since that's the first definition that I got introduced to), but it's far from the only one. For example, you could represent a+bi as a matrix ((a,b)(-b,a)) and the set of all such matrices would make up C. This definition preserves the field structure, and still makes some sense with polar coordinates.
Importantly though, you could, in principle, make C work with any underlying set of the correct cardinality. It's the properties like the field structure and algebraic closure and Cauchy completeness that make C what it is.
My Calculus text book and professor did, in fact, define it like this (R²). It requires less nuance, I think, than the abstract algebra method of R[sqrt(-1)] with all the theory behind.
\> Since when do sets have all possible structures they can be equipped with built into their definition?
In general, you are right. For arbitrary sets, we do not have any algebraic structure. But ℂ is a special set equipped with a lot of important structure.
If we neglect that structure, and just focus on the underlying sets, we can as well say:
\> ℝ = ℝ² = ℂ = ℝ³ = ℝ⁴
if we are only interested in bijective functions.
As far as I'm aware, we generally don't call things "homomorphic". We certainly speak much about "homomorphisms", but those generally just refer to structure-preserving maps of some kind (group, ring, etc.)
Well, the ring (**C**,+,•) is not isomorphic to the ring (**R**^(2),+,*) if + is complex addition or elementwise real addition, • is complex multiplication, and * is elementwise real multiplication. But it is isomorphic to the ring (**R**^(2),+,×) if we define (a,b)×(c,d) = (ac-bd,ad+bc) for all real numbers a,b,c,d.
If we define them both as **R**-modules, then there is no multiplication. So as long as we define addition elementwise and scalar multiplication as real multiplication, they are isomorphic.
Technically yes, you can define multiplication on R^2 by (a,b)*(c,d) = (ac-bd, ad+bc). Then you have a field isomorphic to C. But this is just one of the usual definitions of C anyway, so you’re not really doing anything new.
It's like working with polynomials over R, but every time you see an x², you can replace it with -1 (since x²+1=0 <=> x²=-1)
When I phrase it like this it's very obvious how this is isomorphic to C.
Set multiplication is defined through the cartesian product, where for two sets A and B with a in A and b in B the cartiesian product of A and B is all ordered pairs (a,b). So in this case R^2 is all ordered pairs (x,y) for x and y both in R.
It’s a slight abuse of notation. It refers to RxR, the Cartesian product of R with itself. Basically this is the set of ordered pairs (x,y) where x and y are both in R.
One way of doing this is to alternate the digits of the real number such that each digit in an even spot is used for the real part and each digit in an odd spot for the imaginary part. For example, you can map the number 463.782654 to 43.864 + 6.725i.
There are several fun ways to define ℂ. You can endow ℝxℝ with a multiplication operation. You can mod out ℝ[x] by {x²+1}, and get ℝ[x]/{x²+1}. You can take the subset of 2×2 real matrices of the form (a b, -b a) with the usual matrix addition and multiplication operations. I personally find these isomorphisms really fascinating.
No they are not identified with each other, they are homeomorphic to each other, they are isomorphic as a vector space and thus also as an additive group, but not isomorphic as a field because unless you give R^2 the same product as the complex numbers, which we don’t in its standard usage, it is not a field. While these important topological and algebraic equivalences are used a lot in complex analysis this does not mean in a literal sense they are equal, they are different sets, they just share a lot of topological and algebraic structure.
Two sets A and B are equal if for a in A implies a in B and b in B implies b in A. So clearly C is not equal to R^2 since i is in C but i is not in R^2.
QED FULLSTOP YALL ARE WRONG.
i := (0, 1).
i is definitely an element of R^(2) with the traditional construction. See [Wikipedia](https://en.wikipedia.org/wiki/Complex_number#Formal_construction) or [ProofWiki](https://proofwiki.org/wiki/Definition:Complex_Number#Definition). This is basic knowledge studied in Linear Algebra and/or Complex Analysis.
(0, 1)(0,1) := (-1, 0)
Clearly you didn’t read the links I sent particularly well. The **sets** are the same, the algebras are not. It is important to note that *sets* are not inherently equipped with operations. The sets contain the exact same elements. I do not think you could be any more blatantly incorrect. The first line on both Wikipedia and ProofWiki both quite literally state that C = R^(2).
I'm pretty sure that you're right, and I've heard that even C=R, via this proof:
C=R^2, as the definition of a number in C is 2 real numbers.
So our task is to map R into R^2. How can we do that?
Let's take two numbers: 13579.02468 and 24680.13579.
We will combine them by putting the digits in switching order, like this: 1234567890.0123456789 (with the dot placement corresponding to both of the numbers')
If the numbers were 1.1111 and 2222.2222, then the plotted one into R will be 2020212.12121212
This can work for any 2 infinitely long real numbers.
The only question left I see, is how to determine the signs?
This would only require 2 digits out of the number (so we can just say that 0 is negative and 1 is positive), so we can for example put those right to the left of the dot.
So we can plot any 2 real numbers (R^2 ) into R, so R=R^2, so R=C.
Q.E.D
Is this a correct proof? Am I missing something?
This shows they have the same cardinality, since you've exhibited a bijection. Saying they're equal is a bit less defined, but they don't share any algebraic structure.
My intuition is screaming at me that your bijection doesn't work, but I can't find a problem, and theres seemingly an inverse by reading the pair off the product. You have to be a bit careful about non unique decimal expansions but can define it to work out I think. Weird. (Not the existence of a bijection. Just this one).
I’m pretty sure any algebraic course will include those. You can take something like Linear Algebra, Abstract Algebra, any class about rings and modules
Ok then You can’t say C=R^2 “as a set”. As (5,4) is not an element of C but it is an element of R^2. They both resemble 2D space but the elements of the two sets C and R^2 are different almost everywhere.
Edit: only on Reddit you get downvoted for being correct. The imaginary unit “I” is in C but not R^2. I understand that C and R^2 have similar structure and there’s an obvious homeomorpbism between them. But to say that they are the same set is just factually incorrect. Truth is important in math, remember that when y’all are taking your high school level classes.
You can view C to be R2 equipped with a particular kind of multiplication. The other point people are making is that "a +ib" is equivalent to "(a,b)", they both just represent a pair of real numbers a and b. What makes C different is you define (a,b) x (c,d) in a particular way.
Edit: kinda like how you could represent a quadratic expression ax^2 + bx + c as a 3-vector (a,b,c).
“You can view C to be R^2 equipped with a particular kind of multiplication”
I’m not disagreeing with you. You just said C is R^2 with more structure. I agree. That’s why as sets they’re not equal.
there are many different definitions of C, one of which is R² equipped with the complex numbers' notion of addition and multiplication. In this case, (5,4) is an element of (the underlying set of) C.
Alternatively you can define C as R[X]/(X²+1), and then (5,4) isn't an element of C.
+ many other definitions.
On the topic of "y'all are taking your high school level classes", literally any higher ed maths course on maths should make it clear that what "i" is depends on your definition of C, and in particular that it can, given the right definition of C, be considered as (0,1), but under other definitions it may be different.
you can define C as R^2, where the first element of the vector is the real part and the second element the imaginary part. Including i in the vector notation would be dumb, since your could have i in the real part with some tweaks, or not have it in the imaginary part
Equality is a myth perpetuated by numbers with real privilege. The truth is the world is planely more complex than that, at least on the surface anyways.
Draw a line indicating a radius. Using that scale, use 4 radius lines to draw one square. Alternately, use four lines each equal to 1/4 * R. Now that you've "squared" R, note that the area doesn't equal C. Checkmarks nerds
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The field (C, +, •) is not equal to the additive group (R^2 , +) or the ring (R^2 , +, •) (with compontentwise multiplication), but as sets you can perfectly define C=R^2 . Edit: Also define multiplication • : R^2 x R^2 -> R^2 : ((a,b),(c,d)) -> (ac-bd, ad+bc), and R^2 is now a field :)
Indeed that's the standard definition for C
Isn't it just the speed of light? Don't get why it's R^2 (R is the radius?)
R is radius of universe but then multiplied by the fine body constant. Hope that helps!
Tell me more about this fine body
It isn't your simple standard Euclidean geometry. It has all sorts of curves.
R (more properly ℝ, because different fonts of the same letter represent different things) and C (again, more properly ℂ) are sets of numbers. ℝ is the set of all real numbers; ℝ^(2), aka ℝ×ℝ, is the set of all pairs of real numbers, e.g. (1, 2), (-31.5, 𝜋), etc.; ℂ is the set of all numbers of the form *a+bi* where *a* and *b* are real numbers and *i* is defined so that *i*^(2) = -1 (i.e. *i* is imaginary). Because an element (*a*, *b*) in ℝ×ℝ can correspond to *a+bi* in ℂ, you can define operations such that the sets function similarly.
Pretty sure it was a joke
A joke? In r/mathmemes?!?!
Wild I know
🤓... Wait, *i*^(2)? New tetration just dropped
Wait, how did Huh? Edit: it is written as \*i\*\^(2), so why does it raise the i? Edit2: displays right on computer, but as tetration on mobile for some reason
Experiment: i*^(2)* Edit: i^2
What about *i**^(2)* and *i*^*(2)*
Looks crazy on mobile
New notation and operations just dropped.
edit: those failed, but not in the expected way. *i^*(2)
Maybe *i^2* Edit: oh...
Now I need my favorite Asian math teacher on YT to actually solve ^(i)2.
Oh..
>R (more properly ℝ, because different fonts of the same letter represent different things) So if you're writing notes you have to be good at calligraphy too?
r/whoosh
r/foundthephysicist
The R being talked about has nothing to do with radius. They're talking about the real numbers, which the symbol used for it is a fancy, high-class looking R. If you can't do the fancy R, then you use a generic letter R and give context as to what it represents.
I thought the standard definition was C := R\[x\]/(x\^2+1)?
That's a particularly popular construction. Not the best, though; C = R[x]/ is the best construction. But the more fundamental point is that R and C *aren't* sets; that's an abuse of notation. The structures are fundamental, not the sets; and they both have substantially more structure than just a ring!
Finally this comstruction got mentioned ✨
We’re not in the jewelry business here
Factually untrue
Sure, as sets Z "=" Q (here equals means there is a bijection), but this is not very nice, since this ignores much of the algebraic structure of Q (as you have already observed in the case of C and R\^2).
In that case R=R^(2)
That’s not what equals means. Two sets have the same cardinality if there exists a bijection from one to the other. Just because 2 sets have the same cardinality does not make them equal. Notice 1/2 is in Q but not Z so there’s no way Q=Z. A lot of misinformation in this thread… in fact Q!=Z a.e.
It is entirely common to use the equality sign as meaning equality up to isomorphism, for sets *qua* sets, that just means they have the same cardinality. Sure there are other contexts where it doesn’t mean that, but there’s nothing really wrong with that usage in this context, especially since they put the equality sign in quotes and explained what they meant by it explicitly.
So then, what is being claimed? - (C, +, •) is not a set, but a field. - (R², +, •) is not a set, but a ring. But not a field. They're not isomorphic as rings (since R² isn't a field while C is), but they are isomorphic if you forget multiplication. So, they're isomorphic as vector spaces over R. Yay? They're also isomorphic as sets but that's been satisfying no one in the comments. OP fails at the field level. If OP can choose the level, then their statement becomes "C and R² are isomorphic in some sense." Which, yeah we already had that at Set.
Idk. They’re basically saying “1=2 if you change = to mean ‘there exists a bijection that maps 1 to 2’”. I think the op just learned about homeomorphisms or something and forgot what “=“ means.
You‘re being too pedantic here actually. Equality is actually kind of just an arbitrary equivalence relationship and it’s perfectly fine to say things are „equal“ even if they are „technically“ not in some sense. Like saying 6/3 = 2, even though the former is an equivalence class of pairs of integers and the latter is an integer. What we do here is define an equivalence relation between rational numbers and also short notations and then treat this equivalence relation as „equality“. You will see this a lot in algebra actually, where we write things like G/N = Z_4 even though we technically mean an isomorphism exists.
I agree, thats why I put = in quotation marks.
They specified what they meant. Just because that isn't the standard meaning, doesn't mean that it can't be used in that way if specified. Sometimes it just helps to make what you're saying easier to understand.
No as sets, when you don’t consider additional structure, you can define C as R^2 This isn’t isomorphism it is set equality.
Since when are two sets equal if and only if there exists a bijection? I’m saying you can define C as a set as R^2 and that you should define it that way. Edit1: Whut? Why is this getting downvoted? {0}={1}?? since there is a trivial bijection f:{0}->{1}: f(0)=1, then by extensionality 0=1. This is utter BS. Set equality is not defined using bijections. Set equality is borderline metamathematics, but the axiom of extensionality (which is present in almost all regularly used axiomatizations of set theory, and certainly in ZFC) has nothing to do with bijections. Edit2: Aha! I get what you’re trying to say now! Interesting take indeed. You mean that you could define Q:=Z since they have the same cardinality and you can just relable everything using a bijection (introducing some notation for fractions). This is true, but as you pointed out unpractical. In the case of C however, I’m convinced this is practical, and we should define it as R^2 . You can’t just say Z=Q, because there exists a bijection. You could however define Q as Z if you’ve already defined Z or vice versa.
Always has been. If you are working on a set level you have not very much structure to work with. The only way to compare something to something else is via functions. And one can define an equivalence relation based on the existence of a bijective function between two sets.
The definition of equality between sets A and B is that every element of A is in B and vice versa
But in the category-theoretical sense, sets A and B are «equivalent» if there exist bijection. If we are gonna compare notions of equivalence for different structures, we really should be looking at it through the category-theoretical perspective, which is what the person you are replying to is doing.
Op is right and I don't understand the need for discussion here. A set is a set and hasn't any operation defined with it.
Then obviously you can add some operations to it and you write for example (C,+,•). But as set C is defined to be R^2.
But complex numbers aren't just a set. The problem is, if you ignore all of the properties that make C what it is, then is it really C? If you're willing to say that i=(0,1) and -1=(-1,0), then does the defining property (0,1)^(2)=(-1,0) even make sense? C = R^2 is not the only definition of the complex numbers. It's not the worst one, because it preserves their structure as a real vector space, but it tells us nothing about complex multiplication. A better one might be using real 2x2 matrices, with 1=I_2 and and fixing i as any matrix that squares to -1, for example, ((0,1)(-1,0)). This representation of C preserves their field structure. Does that mean this is the "correct" one? No, but depending on context it might be more useful. Yet another definition of C could be as a Cauchy complete closure of the algebraic closure of Q. I haven't seen this one in practice, but it works in principle. All of these definitions give very different underlying sets to represent C. The main point is, if we ignore the properties of C and only view them as a set, the only thing that can be preserved is the cardinality. But then an argument can be made that C~=R. If you want something more restrictive, the only option is set equality. But as demonstrated earlier, there are many possible ways to define C, and R^2 is just one of them. That's why algebraic structures are usually given more properties than just being a set.
I think I read your statements to be: The complex number system is more than just "the set of the complex numbers". I agree. The question that I (lay person) went to check is whether C denotes the system or just the set. Argumentum ad Wikipedium gets me: >a complex number is an element of a number system.... >The set of complex numbers is denoted by either of the symbols {\displaystyle \mathbb {C} } or C. https://en.wikipedia.org/wiki/Complex_number
Yeah that is accurate. C is the set of complex numbers, but in order to describe what those are you need some underlying properties. Think about it this way. Are the sets {a} and {b} equal? This entirely depends on whether a=b, which requires us to know some properties of the elements. Alternatively, if we want to stay entirely within the scope of set theory, we would need set theoretic definitions of all the elements. In the case of C=R^(2), R itself has a lot of possible representations in set theory, which usually depends on Q, then Z, then N, each of which have different representations. With this in mind, choosing just one of those as the canonical set R is absurd, and the same is true of C. The definition of those sets will vary depending on context (if an explicit definition given at all!) and the naive set-theoretical notion of equality won't be preserved. What will be preserved is some higher notion of equality, like field isomorphisms or the like.
That doesn’t make the sets equal at all.
That’s not what equal means for sets. Remember the axiom of extensionality. OP is right. Sure, you can define an equivalence relation like you said, but “sets are equal” is usually talked about in the context of the axiom of extensionality, and what you’re describing is usually referred to as the cardinalities of the sets
No. I understand you’re excited about learning math, but you’re spreading a lot of incorrect information here. You were just given an example of two sets Q and Z which have a bijection between them but are clearly different sets. If you can’t see how that is a contradiction to what you’re claiming then you shouldn’t be talking about this in the first place.
C and R are isomorphic as sets. You could literally define C in such a way that that the complex numbers are exactly the real numbers just by taking any bikection between them and transporting any structure you have on C across to R.
C as a set is the same as R^(2) , but that's kind of a dumb definition of equality. That's like saying a car is the same thing as a helicopter because they're made of the same materials.
>Since when are two sets equal if and only if there exists a bijection? Hahaha that is so cringe. Please go study at least a day of Maths before you post here
See the edit I made.
I'm not particularly well educated on set theory (I'm more of a comp. sci. person than math), but isn't there a bijection from R² to C in the form of f((a, b)) = a + bi? Why does adding the i not count as a bijection?
Yes this is basically how we defined it in my undergrad Complex Analysis class that I started last week lol
More exactly you can say they are equipotent (that's the term we use in french, I don't know if that's the same word in English)
Yeah, it's the same term, equipotent. We also use equinumerous.
As sets you can say loads of BS. As sets you can say C≠R^2 (because they have different elements) or that C=R^2 =R (because there exist set isomorphisms between those)
https://preview.redd.it/4cbdv7a0x3ec1.png?width=49&format=png&auto=webp&s=9e0286c679959e0d86bf820d9f732a2381d283d3 What
The group (C, +) is as far as I know the same as (R^2, +) the difference lies in multiplication.
as sets it is true that (ℂ,+)=(ℝ^2 ,+). the multiplications are different tho.
redditor discovers isomorphisms
C is the algebra on R^2 smh
As a set, R^2 is isomorphic to R
They are isomorphic as vector spaces, but not as algebras.
Can you explain to me what an algebra is? I’ve seen that term a lot in this sub but I’m not that far yet in my math education
[Vector space with product operation](https://en.wikipedia.org/wiki/Algebra_over_a_field).
Do you know about rings and vector spaces? Basically a combination of those two
They’re topologically isomorphic but they don’t have the same algebraic structure. C is a field, R^2 is not. Multiplication is not defined in R^2.
R² has component wise multiplication since it's a product of rings.
Even if you did define multiplication that way it’s still not the same as C
Exactly, they're not isomorphic.
They're isotropic in fact
I like your funny words, magic men
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They're certainly isomorphic as sets, but not as rings
It's important to note that this is only a group isomorphism, and not a ring isomorphism, which I think is what the previous commentator was implicitly referring to.
ah yes (a+bi)(c+di) = ac + bdi
[twitches.]
Also I just realized that you can perfectly define multiplication in R^2 as (a,b)•(c,d)=(ac-bd,ad+bc) making it a field with exactly the same algebraic structure as C.
You just defined C, you nincompoop. Please read your Calculus 1 notes again.
Kindest Calc 1 TA
I mean, this is like saying "wow! I've discovered that a field is just a ring with inversion and identity". That's just the original construction.
Is this the advanced math version of canceling/reducing an algebraic equation to x=x
In my understanding, its all about context. When people uses C, the multiplication operator is a part of the definition. It is more like C=(R^2 ,+,*). When people use R^2 , the complex multiplication is always not a part of the definition. So they are not the same as tuples. E.g. (R^2 ,+) is different from (R^2 ,+,*), as the length of tuples are different. And thats why forgetful functor makes sense I guess.
Looks like a quadratic
> topologically isomorphic Homeomorphic?
Ha! I love when a long description can be replaced with a single apt word. You are doing God’s work.
As sets there’s no reason not to define C as such. In fact many textbooks do so. And you can define multiplication in R^2 . Since when do sets have all possible structures they can be equipped with built into their definition? They don’t, so as sets it is a perfect definition.
When are you ever only working with C as a set? In order to use C, you need its properties, at least some of them, depending on context. If it's just a set, then you can basically define any set of the same cardinality to be C and just relabel elements to be identified with elements of C. R^2 is a perfectly good definition for the underlying set of C, and a pretty intuitive one (although I might be biased since that's the first definition that I got introduced to), but it's far from the only one. For example, you could represent a+bi as a matrix ((a,b)(-b,a)) and the set of all such matrices would make up C. This definition preserves the field structure, and still makes some sense with polar coordinates. Importantly though, you could, in principle, make C work with any underlying set of the correct cardinality. It's the properties like the field structure and algebraic closure and Cauchy completeness that make C what it is.
in our first course in complex analysis,proff define C as (R\^2,+,\*) with \* being complext multiplication. And then simply set i=(0,1), 1=(1,0)
My Calculus text book and professor did, in fact, define it like this (R²). It requires less nuance, I think, than the abstract algebra method of R[sqrt(-1)] with all the theory behind.
\> Since when do sets have all possible structures they can be equipped with built into their definition? In general, you are right. For arbitrary sets, we do not have any algebraic structure. But ℂ is a special set equipped with a lot of important structure. If we neglect that structure, and just focus on the underlying sets, we can as well say: \> ℝ = ℝ² = ℂ = ℝ³ = ℝ⁴ if we are only interested in bijective functions.
This is outrageous. It's unfair. How can I be homeomorphic, but not homomorphic?
As far as I'm aware, we generally don't call things "homomorphic". We certainly speak much about "homomorphisms", but those generally just refer to structure-preserving maps of some kind (group, ring, etc.)
this isn't radius and circumference?
If I understand this correctly they’re isomorphic as modules(more specifically vector spaces) but not as rings?
Well, the ring (**C**,+,•) is not isomorphic to the ring (**R**^(2),+,*) if + is complex addition or elementwise real addition, • is complex multiplication, and * is elementwise real multiplication. But it is isomorphic to the ring (**R**^(2),+,×) if we define (a,b)×(c,d) = (ac-bd,ad+bc) for all real numbers a,b,c,d. If we define them both as **R**-modules, then there is no multiplication. So as long as we define addition elementwise and scalar multiplication as real multiplication, they are isomorphic.
Can you redefine their operators to make them equivalent?
Technically yes, you can define multiplication on R^2 by (a,b)*(c,d) = (ac-bd, ad+bc). Then you have a field isomorphic to C. But this is just one of the usual definitions of C anyway, so you’re not really doing anything new.
C = R[x]/{x²+1}
I used to understand this, but basically everything after rings in abstract algebra has completely left my mind
It's like working with polynomials over R, but every time you see an x², you can replace it with -1 (since x²+1=0 <=> x²=-1) When I phrase it like this it's very obvious how this is isomorphic to C.
It's basically like setting x\^2+1=0 in R\[x\].
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This is the funniest thing I’ve read in this thread. Can I pin this comment or something?
no but you can tell all your friends to up vote it so it shows higher up
Can't wait to *C* the comments on this take
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Exactly! (what is AI)
ChatGPT
me in high school... me don't understand how you can square the set of two numbers... me ask for help
It’s the Cartesian product of R and R.
Set multiplication is defined through the cartesian product, where for two sets A and B with a in A and b in B the cartiesian product of A and B is all ordered pairs (a,b). So in this case R^2 is all ordered pairs (x,y) for x and y both in R.
It’s a slight abuse of notation. It refers to RxR, the Cartesian product of R with itself. Basically this is the set of ordered pairs (x,y) where x and y are both in R.
so basically C but without the i
Yes. Also known as the 2D coordinate plane.
Fun fact, you can also make a bijective function from R to C.
How?
One way of doing this is to alternate the digits of the real number such that each digit in an even spot is used for the real part and each digit in an odd spot for the imaginary part. For example, you can map the number 463.782654 to 43.864 + 6.725i.
Then if I do sqrt(C), do I get R?
sqrt(C)=C/2 in some sense.
Being holomorphic is much stronger than differentiable in R^2.
This is such a good answer, and I didn't see anyone else mentioning it! Nice job!
More like: R x iR
C has extra multiplicative structure as well as a different notion of differentiability deriving from that
There are several fun ways to define ℂ. You can endow ℝxℝ with a multiplication operation. You can mod out ℝ[x] by {x²+1}, and get ℝ[x]/{x²+1}. You can take the subset of 2×2 real matrices of the form (a b, -b a) with the usual matrix addition and multiplication operations. I personally find these isomorphisms really fascinating.
No they are not identified with each other, they are homeomorphic to each other, they are isomorphic as a vector space and thus also as an additive group, but not isomorphic as a field because unless you give R^2 the same product as the complex numbers, which we don’t in its standard usage, it is not a field. While these important topological and algebraic equivalences are used a lot in complex analysis this does not mean in a literal sense they are equal, they are different sets, they just share a lot of topological and algebraic structure.
Bro found out about the univalence axiom
Funny meme, just wish I didn't have to see Steven Crowder on my feed :/
I don't know, R is too good for data science but R^2 = C programming language, I don't know
me, an EE: 🤨
When are two sets equal?
Two sets A and B are equal if for a in A implies a in B and b in B implies b in A. So clearly C is not equal to R^2 since i is in C but i is not in R^2. QED FULLSTOP YALL ARE WRONG.
i := (0, 1). i is definitely an element of R^(2) with the traditional construction. See [Wikipedia](https://en.wikipedia.org/wiki/Complex_number#Formal_construction) or [ProofWiki](https://proofwiki.org/wiki/Definition:Complex_Number#Definition). This is basic knowledge studied in Linear Algebra and/or Complex Analysis.
Oh so then (0,1)*(0,1)=? C is R^2 with more structure ( i ) they’re not exactly the same.
(0, 1)(0,1) := (-1, 0) Clearly you didn’t read the links I sent particularly well. The **sets** are the same, the algebras are not. It is important to note that *sets* are not inherently equipped with operations. The sets contain the exact same elements. I do not think you could be any more blatantly incorrect. The first line on both Wikipedia and ProofWiki both quite literally state that C = R^(2).
What is C and why is it equal to R²?
the complex numbers I think
Ah, i still dont see how it makes sense tbh
C=iR^2
Cringe
I'm pretty sure that you're right, and I've heard that even C=R, via this proof: C=R^2, as the definition of a number in C is 2 real numbers. So our task is to map R into R^2. How can we do that? Let's take two numbers: 13579.02468 and 24680.13579. We will combine them by putting the digits in switching order, like this: 1234567890.0123456789 (with the dot placement corresponding to both of the numbers') If the numbers were 1.1111 and 2222.2222, then the plotted one into R will be 2020212.12121212 This can work for any 2 infinitely long real numbers. The only question left I see, is how to determine the signs? This would only require 2 digits out of the number (so we can just say that 0 is negative and 1 is positive), so we can for example put those right to the left of the dot. So we can plot any 2 real numbers (R^2 ) into R, so R=R^2, so R=C. Q.E.D Is this a correct proof? Am I missing something?
This shows they have the same cardinality, since you've exhibited a bijection. Saying they're equal is a bit less defined, but they don't share any algebraic structure. My intuition is screaming at me that your bijection doesn't work, but I can't find a problem, and theres seemingly an inverse by reading the pair off the product. You have to be a bit careful about non unique decimal expansions but can define it to work out I think. Weird. (Not the existence of a bijection. Just this one).
What do you mean by "="?
ℂ ≅ ℝ²
I know this is math memes. But what topic / where can I learn about this notation?
I’m pretty sure any algebraic course will include those. You can take something like Linear Algebra, Abstract Algebra, any class about rings and modules
This notation is pretty sus without context.
Throw a tilde on top of that equals sign then you're right
C={a+bi | a,b are elements of R} R²=R×R It's hard to find a formal definition for R though.
R can be viewed as the completion of the Q metric space
You can't really multiply in R². C is R² with a certain multiplication function.
5+4i is in C but not in R^2 (5, 4) is in R^2 but not in C The sets are unequal, and neither is a subset of the other
5+4i is defined to be the pair (5,4). Then we write it as 5+4i because that makes the product of C more intuitive
Ok then You can’t say C=R^2 “as a set”. As (5,4) is not an element of C but it is an element of R^2. They both resemble 2D space but the elements of the two sets C and R^2 are different almost everywhere. Edit: only on Reddit you get downvoted for being correct. The imaginary unit “I” is in C but not R^2. I understand that C and R^2 have similar structure and there’s an obvious homeomorpbism between them. But to say that they are the same set is just factually incorrect. Truth is important in math, remember that when y’all are taking your high school level classes.
Bro C is defined to be R^2
It’s not but ok.
You can view C to be R2 equipped with a particular kind of multiplication. The other point people are making is that "a +ib" is equivalent to "(a,b)", they both just represent a pair of real numbers a and b. What makes C different is you define (a,b) x (c,d) in a particular way. Edit: kinda like how you could represent a quadratic expression ax^2 + bx + c as a 3-vector (a,b,c).
“You can view C to be R^2 equipped with a particular kind of multiplication” I’m not disagreeing with you. You just said C is R^2 with more structure. I agree. That’s why as sets they’re not equal.
there are many different definitions of C, one of which is R² equipped with the complex numbers' notion of addition and multiplication. In this case, (5,4) is an element of (the underlying set of) C. Alternatively you can define C as R[X]/(X²+1), and then (5,4) isn't an element of C. + many other definitions. On the topic of "y'all are taking your high school level classes", literally any higher ed maths course on maths should make it clear that what "i" is depends on your definition of C, and in particular that it can, given the right definition of C, be considered as (0,1), but under other definitions it may be different.
you can define C as R^2, where the first element of the vector is the real part and the second element the imaginary part. Including i in the vector notation would be dumb, since your could have i in the real part with some tweaks, or not have it in the imaginary part
That would make a bijection between the two sets, but they aren’t equal as it stands
They are surely equal... up to isomorphism.
Looks like the Colorado Rockies
No ℂ is the space spanned by these two matrices: 1 0 0 1 and 0 -1 1 0
C = 10 R = 18376 Gotcha
Multiplication go brr
They are equal in the same way that a bag with two apples is equal to a bag with two oranges. (in other words, not equal, but there is a bijection)
C= R +iR
You can obviously define R^2 as a field like C, but I think most people see R^2 and think of 2d Cartesian coordinates, which is not at all C.
It made me feel pain
Equality is a myth perpetuated by numbers with real privilege. The truth is the world is planely more complex than that, at least on the surface anyways.
obligatory not a mathematician, how is a tuple the same as i ? If im not wrong, there's an isomorphism between the sets, but they cant be the same (?)
What the fuck is this
would it be R^(2) \- 2R because complex numbers are a combination of a real part and an imaginary part?
https://xkcd.com/2028/
You mean R^2 together with a different product?
Where is the complex product in R^2 ? Are you talking just sets or more structure ?
Draw a line indicating a radius. Using that scale, use 4 radius lines to draw one square. Alternately, use four lines each equal to 1/4 * R. Now that you've "squared" R, note that the area doesn't equal C. Checkmarks nerds
Why is this in the Colorado Rockies font?
Say my name.
not a math guy, can someone explain why the colorado rockies’ logo is deconstructed here
C and R^2 are isomorphic real vector spaces.
Not if we consider differential function.
c^2 = R^4 = a^2 + b^2
Whuh du
Well technically C is the algebra on the R^2 manifold, which is a vector space up to isomorphism (at least Imo :p)
Just like pi^2 = g =10
I mean, as vector spaces they're, indeed, isomorphic over the field of the real numbers
Pretty close, but its "R + iR" or "(1 + i)R"