The Identity Crisis of i: Is i truly unique compared to -i?

[Yayness]

What is i?
A very common way to define the imaginary unit i, is to say that $$i=\sqrt{-1}$$, but -1 has two roots. These are i and -i. Of course you could define i as one of two possible values for z where $$z^{2}=-1$$, but I find that definition kind of incomplete.

To me it seems that i doesn't have any unique properties that -i does not have.
For instance: $$\forall z_1,z_2,z_3\in \mathbb{C} : z_1+z_2=z_3\Longleftrightarrow\overline{z_1}+\overline{z_2}=\overline{z_3}$$, which means that i and -i have the same properties in addition.
The same goes for subtraction, multiplication and division.
Even Euler's formula would still be correct if you exchange every i with -i: $$e^{-i\varphi}=\cos\varphi-i\sin\varphi$$
Also, if you let M be the Mandelbrot set, then $$\forall z\in \mathbb{C} : z\in M \Longleftrightarrow \overline{z}\in M$$.

Of course i has a positive imaginary part and -i has a negative imaginary part, which makes them different, but I'm pretty sure it would look just the same if i and -i switched name somehow, such that the "new i" would have the positive imaginary part. (You have two different numbers with the same properties, and then you pick a random one and choose that its imaginary part shall be positive while the other one shall have a negative imaginary part.)
I guess that would be the difference between i and -i. We gave them that difference.
There's probably a lot more to say about it, other opinions probably. Feel free to post any thoughts or other definitions or anything that has to do with this.

 

[arildno]

With the Hamiltonian formalism, there is a real distinction:

We consider 2-tuples of the form X=(a,b), "a" and "b" being reals.

We set up the following rules of addition&multiplication:
(X=(a,b), Y=(c,d))

X+Y=(a+c, b+d)

X*Y=(a*b-c*d,a*d+c*b)

Now, what we regard as the real number line is re-presented with b=0 (or d=0).

The number "i" is, by definition: i=(0,1), whereas -i=(0,-1)

These are not the same numbers/elements.

 

[HallsofIvy]

What is i?
A very common way to define the imaginary unit i, is to say that $$i=\sqrt{-1}$$, but -1 has two roots. These are i and -i. Of course you could define i as one of two possible values for z where $$z^{2}=-1$$, but I find that definition kind of incomplete.

To me it seems that i doesn't have any unique properties that -i does not have.
For instance: $$\forall z_1,z_2,z_3\in \mathbb{C} : z_1+z_2=z_3\Longleftrightarrow\overline{z_1}+\overline{z_2}=\overline{z_3}$$, which means that i and -i have the same properties in addition.
The same goes for subtraction, multiplication and division.
Even Euler's formula would still be correct if you exchange every i with -i: $$e^{-i\varphi}=\cos\varphi-i\sin\varphi$$
Also, if you let M be the Mandelbrot set, then $$\forall z\in \mathbb{C} : z\in M \Longleftrightarrow \overline{z}\in M$$.

Of course i has a positive imaginary part and -i has a negative imaginary part, which makes them different, but I'm pretty sure it would look just the same if i and -i switched name somehow, such that the "new i" would have the positive imaginary part. (You have two different numbers with the same properties, and then you pick a random one and choose that its imaginary part shall be positive while the other one shall have a negative imaginary part.)
I guess that would be the difference between i and -i. We gave them that difference.
There's probably a lot more to say about it, other opinions probably. Feel free to post any thoughts or other definitions or anything that has to do with this.

You are exactly right- since the complex numbers are NOT an "ordered" field, there is no way to distinguish between "i" and "-i". That is why in more advanced texts, where they want to be more precise and rigorous, they define the complex numbers in quite a different way.

Complex numbers are defined to consist of the set of all ordered pairs of real numbers, (a, b). Addition is defined "coordinate wise"- (a, b)+ (c, d)= (a+ b, c+ d). But multiplication is defined in a different way: (a, b)*(c, d)= (ac- bd, ad+ bc). In that way, (0, 1)*(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0). If we now define "1", for this new system, to be (1, 0) and "i" to mean (0, 1) we have that (a, b)= a+ bi and [itex]i^2= (0, 1)*(0, 1)= (-1, 0)= -1+ 0i= -1.

 

[Hurkyl]

What is iTo me it seems that i doesn't have any unique properties that -i does not have.

This is correct, at least as long as you restrict yourself to properties of complex analysis or similar. That everything has a complex conjugate -- not just numbers but also functions, differential forms, and any other kind of object you might study -- is an important feature of the subject.

e.g. if I am is the imaginary part function, then

$$Im(i) = 1$$​

which distinguishes i from -i. But we also have

$$\overline{Im}(-i) = 1$$​

where $\overline{Im}$ is the complex conjugate of the imaginary part function.

 

[Yayness]

Thanks for the replies!
Yeah, that sounds like a reasonable definition.

 

[bcrowell]

Yayness's original post was correct, and HallsOfIvy was also correct. Arildno and Hurkyl are incorrect.

The number "i" is, by definition: i=(0,1), whereas -i=(0,-1)

This is circular reasoning. We could just as easily have defined i=(0,-1) and -i=(0,1).

e.g. if I am is the imaginary part function, then

$$Im(i) = 1$$​

which distinguishes i from -i. But we also have

$$\overline{Im}(-i) = 1$$​

where $\overline{Im}$ is the complex conjugate of the imaginary part function.

But we could have defined Im(i)=-1 and Im(-i)=1.

-Ben

 

[CRGreathouse]

Arildno and Hurkyl are incorrect.

No, they're both correct.

 

[bcrowell]

No, they're both correct.

I made an argument for why they're incorrect. How about making an argument for why they're correct?

 

[CRGreathouse]

I made an argument for why they're incorrect. How about making an argument for why they're correct?

What, you want the definition of an ordered pair? Seriously?

 

[bcrowell]

What, you want the definition of an ordered pair? Seriously?

No, I didn't ask for a definition of an ordered pair.

 

[disregardthat]

(0,1) is not the definition of i, it's a definition of i which encapsulates the properties we want inside set theory. It's not an argument to say that "we could just as easily" have defined i another way, that is not the point, all we are interested in is that i^2 = -1.

To put it succinctly; we are not motivated by i = (0,1), but i^2 = -1. So it does not matter how we define it as long as this is satisfied.

There are other ways to define i. It could be defined as x (or -x) in the field R[x]/(x^2+1).

 

[bcrowell]

(0,1) is not the definition of i, it's a definition of i which encapsulates the properties we want inside set theory. It's not an argument to say that "we could just as easily" have defined i another way, that is not the point, all we are interested in is that i^2 = 1.

To put it succinctly; are not motivated by i = (0,1), but i^2 = 1. So it does not matter how we define it as long as this is satisfied.

I agree with everything you said. If you're unsatisfied with "we could just as easily," let me put it a different way. The mapping $a+ib \rightarrow a-ib$ is an isomorphism of the complex numbers under the field operations and axioms. This shows that the distinction between i and -i is arbitrary.

 

[arildno]

Yayness's original post was correct, and HallsOfIvy was also correct. Arildno and Hurkyl are incorrect.


This is circular reasoning. We could just as easily have defined i=(0,-1) and -i=(0,1).


But we could have defined Im(i)=-1 and Im(-i)=1.

-Ben

Blather, due to your ignorance of what a definition is.

We could choose to define, say, 1=2+2, but we choose not to.
There is no circular reasoning involved in what I wrote.

 

[Outlined]

So you want a property i has which -i does not have. I can not come up with something that quickly.

For me the complex numbers are just a multiplication of two elements in R². In that case it makes most sense to define i by (0, 1).

 

[arildno]

(0,1) is not the definition of i, it's a definition of i which encapsulates the properties we want inside set theory.

It happens to be THE definition within the Hamiltonian scheme of things, which was what I directly referred to.

That OTHER definitions certainly can be made is..irrelevant.

 

[disregardthat]

It happens to be THE definition within the Hamiltonian scheme of things, which was what I directly referred to.

That OTHER definitions certainly can be made is..irrelevant.

Why is it irrelevant? When speaking of "the definition of i" it is certainly not irrelevant that several definitions can be made.

 

[Outlined]

the question is to find an example in which i behaves different compared to -i. The only thing I can come up with is that Im(i) is multiplicative zero there Im(-i) is not. Not sure whether the starter considers this a valid example.

 

[arildno]

When speaking of "the definition of i" it is certainly not irrelevant that several definitions can be made.

It certainly is, when I made an explicit restriction to the Hamiltonian formalism, in order to be precise.

That other, equally precise, definitions can be made is irrelevant as a criticism.

 

[disregardthat]

It certainly is, when I made an explicit restriction to the Hamiltonian formalism, in order to be precise.

That other, equally precise, definitions can be made is irrelevant as a criticism.

I wasn't referring to your comment, I was referring to bcrowell's post.

 

[rbj]

dunno if i dare jump into this, but the Wikipedia article on the imaginary unit deals with this.

http://en.wikipedia.org/wiki/Imaginary_unit

since the only definition of the imaginary unit is this imaginary number that, when squared, becomes -1, both -i and i have equal claim to the designation. you could go to all of the math and engineering and science lit and change every occurrence of i to -i and vise versa, and all theorems would be just as correct. which one to choose for "i" is a matter of convention, but since i and -i are both imaginary, it doesn't make any difference which one you choose.

the same cannot be said with the real unit, 1, and its negative -1. there is a qualitative difference (the multiplicative identity) between 1 and -1, but there is no qualitative difference between i and -i.

 

[CRGreathouse]

since the only definition of the imaginary unit is this imaginary number that, when squared, becomes -1, both -i and i have equal claim to the designation. you could go to all of the math and engineering and science lit and change every occurrence of i to -i and vise versa, and all theorems would be just as correct. which one to choose for "i" is a matter of convention, but since i and -i are both imaginary, it doesn't make any difference which one you choose.

I'm pretty sure everyone on this thread knows that.

 

[rbj]

I'm pretty sure everyone on this thread knows that.

yeah, you're probably right. i couldn't tell at first.

good to see you CR. it's been a while. i don't hang here much any more.

L8r,

r b-j

 

[disregardthat]

At least 5 posts from 3 different members has been deleted from this thread, may I ask why?

 

[arildno]

At least 5 posts from 3 different members has been deleted from this thread, may I ask why?

If you knew the contents, and PF guidelines, you'd know why.
Those posts have been removed that either contained what was considered personal insults, or posts that were directly referring/replying to such posts.

 

[Yayness]

We could probably post posts that are more relevant to the topic. I like the definition Arildno and HallsofIvy talked about in posts 2 and 3.
About ordered pairs with the following rules for addition and multiplication:
(a,b)+(c,d) = (a+c,b+d)
(a,b)·(c,d) = (ac-bd,ad+bc)
where we define a+bi as (a,b) such that i = (0,1) and -i = (0,-1).
Of course we could have defined a+bi as (a,-b) if we wanted to, and it wouldn't have changed any properties of i and -i, but we choose not to.

 

[arildno]

Of course we could have defined a+bi as (a,-b) if we wanted to, and it wouldn't have changed any properties of i and -i, but we choose not to.

Irrespective of that choice, you'd still have two distinct quantities being additive inverses of each other.

When complex multiplying them with some other complex number, one of those quantities will lead to adding 90 degrees to the product's angular value (relative to the other factor's angular value), whereas multiplying with the other quantity, 90 degrees are subtracted.

Thus, to distinguish between these two quantities is NOT arbitrary, but rather, logically necessary.

Arbitrariness is restricted to be a second order concern as to which quantity is to be repesented by what symbol.

 

[rbj]

Irrespective of that choice, you'd still have two distinct quantities being additive inverses of each other.

When complex multiplying them with some other complex number, one of those quantities will lead to adding 90 degrees to the product's angular value (relative to the other factor's angular value), whereas multiplying with the other quantity, 90 degrees are subtracted.

Thus, to distinguish between these two quantities is NOT arbitrary, but rather, logically necessary.

Arbitrariness is restricted to be a second order concern as to which quantity is to be represented by what symbol.

and since the two quantities are qualitatively equivalent, it doesn't matter which one you choose for which symbol. nothing is different with the mirror image of the same reality.

 

[arildno]

and since the two quantities are qualitatively equivalent

That is a meaningless expression.

How are two non-zero elements that when added yields zero "qualitatively" equivalent?

 

[rbj]

That is a meaningless expression.

spoken as an expert.

How are two non-zero elements that when added yields zero "qualitatively" equivalent?

you need to look it up in a dictionary. "quality" is not the same as "quantity". i and -i are quantitatively different. but there is no quality, no property that is different about them.

quantum mechanics aside, although clockwise and counterclockwise are opposite, but there is no property (other than human convention) that clockwise has that counterclockwise doesn't have. and vise-versa.

take a look at the Wikipedia article about the imaginary unit. read the section about i and -i. if that doesn't do it for you, then i might recommend a good book on complex variables or even on functional analysis.

other than that they are quantitatively opposite, there is no difference between i and -i. they both have equal claim to squaring to be -1. and there is little else you can say about their properties.

 

[jambaugh]

Some further comments on definitions of i.

There are multiple ways to define i and the complex numbers depending on your application and preferred context. Modern algebra tends to work in terms of free algebras modulo identities. Here is the basic construction:

Begin with the Real number field R. Define the free algebra of polynomials in one variable R where i here, is the variable. This is the algebra of finite degree polynomials such as P(i)=3i^5 - 2i + 7. It is a Real algebra in that we can add elements, multiply them and scale them (multiply them by real numbers).

Now consider the element Z(i)= i^2 + 1. By identifying this with zero we reduce the algebra by imposing the identity equivalent to i^2 = -1. Hence the complex number algebra is the free algebra "modulo Z(i):

C = R \ Z. type-set version: [tex]\mathbb{C}=\mathbb{R} \backslash \mathbf{Z}[/tex]

This is an example of the general method of extension of an algebra and yields a completion of the reals in the sense that all polynomials have roots in this algebra and so factor completely.

Complex conjugation corresponds to the change of variable i -> -i, which is a symmetry of the identity and hence of the algebra. As a symmetry there is no functional difference between i and -i.

------------------
An alternative is to consider the real Lie group U(1) = SO(2) and corresponding Lie algebra. The single generator of the Lie algebra we can call i and the group is generated via g(theta) = exp( i theta). This however only defines the additive property of i not its multiplicative property so it is not the best "definition" of i. Rather we can think of a (real) linear representation (mapping into a real matrix algebra) of the group and its Lie algebra (the above equation becomes a mapping: Rep(g(theta))=exp(i theta) and i becomes the generating matrix in the representation and not the Lie algebra element itself. Hence:

$$\mathbf{i} = \left(\begin{array}{c c} 0 & -1\\ 1 & 0\end{array}\right)$$
This is in the (x,y) basis where our generator rotates x to y to -x to -y. Complex conjugation involves switching this convention or equivalently reversing the direction of rotation in the x-y plane.

Then the group element is represented by:
$$g(\theta)\mapsto e^{\mathbf{i}\theta} = \left(\begin{array}{c c} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{array}\right)=\cos(\theta)\mathbf{1}+ \sin(\theta)\mathbf{i}$$

Considering the subalgebra of the matrix algebra R(2) of 2x2 real matrices generated by these elements we get again an example of the complex algebra with elements:
$$r\cdot e^{\mathbf{i}\theta} = x\mathbf{1}+y\mathbf{i} = \left(\begin{array}{c c} x & -y\\ y & x\end{array}\right)$$
This real linear algebra also happens to be a field (every non-zero element has a reciprocal, no two non-zero elements have zero product)

This constructs the complex numbers as a subalgebra of the real algebra of 2x2 real matrices. A somewhat equivalent construction is to view complex numbers as a Clifford algebra or as a Cayley-Dickson algebra. The Clifford algebra approach applies to spinors, and Fermi-Dirac statistics. The Cayley-Dickson construction can be thought of as sequentially internalizing the process of conjugation as an algebraic element (yielding a new algebra with a new conjugation) and yields increasingly less commutative/associative algebras Reals, Complex #'s, Hamilton's Quaternions, Cayley's Octonian's (the four unique real division algebras possible and thus the four algebras we can think of as "numbers")

Ultimately i is what it does and that is contextual. For example we should consider multiple i's and be more careful about keeping them distinct when doing quantum mechanics. The complex unit of Electro-Magnetism gauge is not quite the same as the complex unit of the phase for the wave-function of a chargeless particle. Nor for that matter is the i of electromagnetism that of the pure hyper-charge i but has a component of the isospin generator of the residual U(1) symmetry when SU(2) weak isospin is broken. These distinct imaginary units have distinct conjugations having physical meaning (swapping matter vs anti-matter, reversing charges, reversing time, inverting space). The fact that we use a common i in the math and can get away with it in the physics has to do with CPT symmetry (an apparently true but not necessarily true symmetry of nature).

 

[arildno]

other than that they are quantitatively opposite,

That is why they are not equivalent, qualitatively or something else.

there is no difference between i and -i. they both have equal claim to squaring to be -1. and there is little else you can say about their properties

So?
Existence of some sort of sameness does not imply equivalence or identity.

 

[jambaugh]

That is why they are not equivalent, qualitatively or something else.

So?
Existence of some sort of sameness does not imply equivalence or identity.

"Equivalence" and identity are not the same thing. Equivalence is "the existence of some sort of sameness" i.e. the existence of an isomorphism. Hence we speak of equivalence relations (between different objects) and equivalence classes. It has a formal meaning in mathematics and that meaning does indeed apply to the conjugation isomorphism in the complex division algebra.

But this argument is silly ****Because**** you and the others have not established within which context you are referring to i and complex numbers. Until you carefully establish such a context questions of "what ____ really means" or "really is defined to be" are subject to reinterpretation and you can argue in circles, never getting anywhere.

 

[arildno]

But this argument is silly ****Because**** you and the others have not established within which context you are referring to i and complex numbers.

I did that in my first post.

 

[disregardthat]

"Equivalence" and identity are not the same thing. Equivalence is "the existence of some sort of sameness"

Isn't "sameness" a bit ambiguous? I prefer the following formal definitions of equivalence and equality: Equivalence is used when objects are equivalent with respect to some defined equivalence relation, may it be an isomorphism, modulo class, etc. Equality however in set theory is set equality, and thus equality is an example of an equivalence. E.g. two numbers are equal because equality refers to some construction of the numbers in set theory. (1=1 because {{}}={{}} )

 

[arildno]

Equivalence is "the existence of some sort of sameness" i.e. the existence of an isomorphism. Hence we speak of equivalence relations (between different objects) and equivalence classes.

Yawn.

Take a field A , like the complex numbers and another one B, for simplicity its mirror image, so that a mapping exists between a+ib in A to a-ib in B.

Here, "i" in A is equivalent to "-i" in B, but "i" and "-i" are not "qualitatively equivalent" within anyone of the systems.

It is precisely because in BOTH fields A and B that elements within each of them retain their uniqueness from other elements that there exists a bijection from A on B or vice versa.