The complex numbers not having a total ordering is not so strange when one thinks about how they are defined. C is a vector space of dimension 2 over the field R, equipped with an additional multiplication operator. In general, vector spaces of dimension 2 cannot be totally ordered, much less of higher dimensions. It is only possible to order sets if they can form a vector spaces of dimension 1 over a simple enough field, or dimension 0, which is not interesting to consider.
Fun fact: because this construction of the complex numbers defines C as a vector space over R with a product satisfying the field axioms, you can also form a Hilbert space with this vector space. Given the complex product as defined in the video, and given the complex conjugate operator *, which is an involution on C, you can define the inner product of two complex numbers x, y as = (x*·y + x·y*)/2. Since C is complete with respect to the metric induced by absolute value sqrt(z*·z), which is induced by this inner product, this makes (C, ) a Hilbert space of dimension 2 over R. This is part of what makes it so useful for complex analysis.
@@angelmendez-rivera351 I forget where and to whom you recommended this channel to, but at some point you suggested that someone might find this channel's series on real analysis interesting. Thank you for doing so, as that is the reason I found this channel, and thank you for these interesting asides you wrote for the Start Learning Mathematics series.
Isn't it a bit weird to use the distributive law on the short notation to justify the definition of the multiplication ? I would assume that if we could find a way to define the multiplication in a different manner on the vector notation, which also satisfies distributivity, same results on purely real numbers, and i*i = -1, we wouldn't be writing complex numbers with this short notation ? Do you think there are other ways to define the multiplication on R^2 that would satisfy all the properties ? In this case, why do we choose this definition instead of the others ?
The distributive "property" of complex multiplication is an axiom of complex arithmetic, it is not a theorem. You cannot prove that complex multiplication intrinsically satisfies distributivity. You postulate that multiplication satisfies distributivity as an axiom, and then find explicit well-definitions that satisfy this axiom, as well as all the other field axioms. The whole point of the complex numbers is to extend the real arithmetic operators to complex arithmetic operators, while still preserving the real results as a special case, so the axioms for the real numbers are also axioms for the complex numbers. If you do not do it this way, then you are not defining an extension of real arithmetic, you are defining an unrelated arithmetic. The point of extensions is to preserve the operations of the original arithmetic you started with. Since real multiplication satisfies distributivity, any suitable extension of the real numbers should also satisfy distributivity. Otherwise, it is by definition not actually an extension.
@@angelmendez-rivera351 I think my main point was not really about proving that complex multiplication always satisfies distributivity. As you said, it's an axiom we postulate. It was more about the explicit definition that is given in the video, and that is shown to satisfies the axioms. Are there other definitions that could also be used ? If so, what would motivate choosing this one ? Because in the video, I had the impression that the distributivity argument was used as a way to justify that the current definition was the only way to perform complex multiplication, instead of just being a property that we would like to have, which sounded weird to me at first since I don't see why it might not be possible to define other kinds of complex arithmetic.
@@Admo120104 Well, why would you not choose this definition? If it satisfies the axioms, then why does anything else even matter to you? Besides, the definition is the most intuitive of them all, and it captures best the notion that we are really working with an algebraic field extension of R. I imagine there are equivalent definitions, and sure, you can choose those if you want, but you are just end up falling back on this other characterization 99% of the time, because the other characterizations are genuinely not very useful to work with. Working with complex numbers is not very different than working with elements of Q(sqrt(2)), which is the field extension of Q containing all the elements of the form a + b·sqrt(2). If you do a multiplication of two elements x(0) + y(0)·sqrt(2) and x(1) + y(1)·sqrt(2), in this set, you get [x(0)·x(1) + 2·y(0)·y(1)] + [x(0)·y(1) + x(1)·y(0)]·sqrt(2). Notice how analogous this definition of multiplication in Q[sqrt(2)] is to multiplication in C, except 2 is replaced with -1, completely capturing the fact that i = sqrt(-1) is what we want. The difference is that Q[sqrt(2)] is a subfield of R, and so it does not need a completely separate formalization as a vector space Q^2 equipped with a special product, but C needs to be rigorously axiomatized as an extension of R. This is done in a way that emulates other field extensions, like in the case above, but more directly associating it with vector notation. And it would not matter which definition you start with: you would always be able to prove that this is an equivalent characterization, and so the definition in the video is the only one even worth looking at ever. Again, I have no idea why you would even think of using an alternate definition. Most mathematicians would not give it a second thought. Of course, there is another formalization you can do using matrices instead, and this is the only other definition of multiplication that is useful, but for most applications, this is not as natural to work with. The advantage is that, when complex numbers are treated as matrices of a special kind, their tranpose is their complex conjugate, their modulus is their determinant, and their trace is twice the real part, while complex multiplication is just ordinary matrix multiplication. It makes the arithmetic easier, but it is not as convenient in other applications.
What you can show is that as long as multiplication is "bilinear," ie distributive and (5x) * y = 5(x*y), then it is uniquely determined by 1*1, 1*i, i*1, and i*i. This is because every complex number can be written in the form a*1 + b*i, so (a1+bi)(c1+di) = (a1+bi)c1 + (a1+bi)di = (a1)*(c1) + (bi)(c1) + (a1)(di) =(bi)(di) = (ac)(1*1) + (bc)*(i*1) + (ad)(1*i) +(bd)(i*i). So, it isn't enough to simply specify i*i = -1 and distributivity, but it is enough to specify bilinearity, i*i= -1, and that 1 acts as the multiplicative identity.
@@matthewma11hew Bilinearity is implied by distributivity and commutativity, as this is a vector space, and both of these are part of the field axioms, along with the existence of the multiplicative identity. So "specifying" those is completely redundant. The only axiom that is not a field axiom is that i^2 = -1. In other words, C := R(i), or even more explicitly, C := R[X]/(X^2 + 1). This is what the definition of complex multiplication captures, but this is being presented without having the video appeal to quotient fields and field extensions directly, because that does require some higher level concepts that would be inappropriate to talk about prior to a video like this.
![Start Learning Complex Numbers 3 | Absolute Value, Conjugate, Argument [dark version]](http://i.ytimg.com/vi/g83_37nW8zg/mqdefault.jpg)
![Start Learning Complex Numbers 3 | Absolute Value, Conjugate, Argument [dark version]](/img/tr.png)
6:22
Just blew my mind, it makes so much sense
Very good Insights! Please consider doing a series on Geometric Algebra too...
These videos make complex numbers totally intuitive! Hurrah! 😃
The complex numbers not having a total ordering is not so strange when one thinks about how they are defined. C is a vector space of dimension 2 over the field R, equipped with an additional multiplication operator. In general, vector spaces of dimension 2 cannot be totally ordered, much less of higher dimensions. It is only possible to order sets if they can form a vector spaces of dimension 1 over a simple enough field, or dimension 0, which is not interesting to consider.
Fun fact: because this construction of the complex numbers defines C as a vector space over R with a product satisfying the field axioms, you can also form a Hilbert space with this vector space. Given the complex product as defined in the video, and given the complex conjugate operator *, which is an involution on C, you can define the inner product of two complex numbers x, y as = (x*·y + x·y*)/2. Since C is complete with respect to the metric induced by absolute value sqrt(z*·z), which is induced by this inner product, this makes (C, ) a Hilbert space of dimension 2 over R. This is part of what makes it so useful for complex analysis.
@@angelmendez-rivera351 I forget where and to whom you recommended this channel to, but at some point you suggested that someone might find this channel's series on real analysis interesting. Thank you for doing so, as that is the reason I found this channel, and thank you for these interesting asides you wrote for the Start Learning Mathematics series.
@@delta3244 Thank you. And I am glad you enjoy this series.
When is Start learning Abstract Algebra?
I hope he will get there.
You are fantastic!
I recently read that generalizations of the complex field lose more properties , so octonions are non associative isn't that amazing
Will we see squared root of complex numbers?
Of course!
Isn't it a bit weird to use the distributive law on the short notation to justify the definition of the multiplication ? I would assume that if we could find a way to define the multiplication in a different manner on the vector notation, which also satisfies distributivity, same results on purely real numbers, and i*i = -1, we wouldn't be writing complex numbers with this short notation ? Do you think there are other ways to define the multiplication on R^2 that would satisfy all the properties ? In this case, why do we choose this definition instead of the others ?
The distributive "property" of complex multiplication is an axiom of complex arithmetic, it is not a theorem. You cannot prove that complex multiplication intrinsically satisfies distributivity. You postulate that multiplication satisfies distributivity as an axiom, and then find explicit well-definitions that satisfy this axiom, as well as all the other field axioms. The whole point of the complex numbers is to extend the real arithmetic operators to complex arithmetic operators, while still preserving the real results as a special case, so the axioms for the real numbers are also axioms for the complex numbers. If you do not do it this way, then you are not defining an extension of real arithmetic, you are defining an unrelated arithmetic. The point of extensions is to preserve the operations of the original arithmetic you started with. Since real multiplication satisfies distributivity, any suitable extension of the real numbers should also satisfy distributivity. Otherwise, it is by definition not actually an extension.
@@angelmendez-rivera351 I think my main point was not really about proving that complex multiplication always satisfies distributivity. As you said, it's an axiom we postulate. It was more about the explicit definition that is given in the video, and that is shown to satisfies the axioms. Are there other definitions that could also be used ? If so, what would motivate choosing this one ? Because in the video, I had the impression that the distributivity argument was used as a way to justify that the current definition was the only way to perform complex multiplication, instead of just being a property that we would like to have, which sounded weird to me at first since I don't see why it might not be possible to define other kinds of complex arithmetic.
@@Admo120104 Well, why would you not choose this definition? If it satisfies the axioms, then why does anything else even matter to you? Besides, the definition is the most intuitive of them all, and it captures best the notion that we are really working with an algebraic field extension of R. I imagine there are equivalent definitions, and sure, you can choose those if you want, but you are just end up falling back on this other characterization 99% of the time, because the other characterizations are genuinely not very useful to work with. Working with complex numbers is not very different than working with elements of Q(sqrt(2)), which is the field extension of Q containing all the elements of the form a + b·sqrt(2). If you do a multiplication of two elements x(0) + y(0)·sqrt(2) and x(1) + y(1)·sqrt(2), in this set, you get [x(0)·x(1) + 2·y(0)·y(1)] + [x(0)·y(1) + x(1)·y(0)]·sqrt(2). Notice how analogous this definition of multiplication in Q[sqrt(2)] is to multiplication in C, except 2 is replaced with -1, completely capturing the fact that i = sqrt(-1) is what we want. The difference is that Q[sqrt(2)] is a subfield of R, and so it does not need a completely separate formalization as a vector space Q^2 equipped with a special product, but C needs to be rigorously axiomatized as an extension of R. This is done in a way that emulates other field extensions, like in the case above, but more directly associating it with vector notation. And it would not matter which definition you start with: you would always be able to prove that this is an equivalent characterization, and so the definition in the video is the only one even worth looking at ever. Again, I have no idea why you would even think of using an alternate definition. Most mathematicians would not give it a second thought.
Of course, there is another formalization you can do using matrices instead, and this is the only other definition of multiplication that is useful, but for most applications, this is not as natural to work with. The advantage is that, when complex numbers are treated as matrices of a special kind, their tranpose is their complex conjugate, their modulus is their determinant, and their trace is twice the real part, while complex multiplication is just ordinary matrix multiplication. It makes the arithmetic easier, but it is not as convenient in other applications.
What you can show is that as long as multiplication is "bilinear," ie distributive and (5x) * y = 5(x*y), then it is uniquely determined by 1*1, 1*i, i*1, and i*i. This is because every complex number can be written in the form a*1 + b*i, so (a1+bi)(c1+di) = (a1+bi)c1 + (a1+bi)di = (a1)*(c1) + (bi)(c1) + (a1)(di) =(bi)(di) = (ac)(1*1) + (bc)*(i*1) + (ad)(1*i) +(bd)(i*i). So, it isn't enough to simply specify i*i = -1 and distributivity, but it is enough to specify bilinearity, i*i= -1, and that 1 acts as the multiplicative identity.
@@matthewma11hew Bilinearity is implied by distributivity and commutativity, as this is a vector space, and both of these are part of the field axioms, along with the existence of the multiplicative identity. So "specifying" those is completely redundant. The only axiom that is not a field axiom is that i^2 = -1. In other words, C := R(i), or even more explicitly, C := R[X]/(X^2 + 1). This is what the definition of complex multiplication captures, but this is being presented without having the video appeal to quotient fields and field extensions directly, because that does require some higher level concepts that would be inappropriate to talk about prior to a video like this.
First
Second
@@brightsideofmaths no i am first
@@mrbeast8743 And I am second.
@@brightsideofmaths Oh ok