, while real numbers are as imaginary as imaginary numbers
Topics
About: high school mathematics
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Are Imaginary Numbers Imaginary?
- 2: Does Mathematics Discover or Invent?
- 3: Then, What Do "Real" and "Imaginary" Mean?
- 4: How Complex Numbers Set Models Reality
- 5: Thinking Imaginary Numbers Imaginary Is a Sure Sign of Narrow-Mindedness
- 6: Why We Think of 'Field'
- 7: Why Not Higher-Dimensional Numbers Systems?
Starting Context
- The reader knows the background of this site.
Target Context
- The reader will know how the complex numbers system models the reality.
Orientation
There is an article on becoming a benefactor of humanity by being a conduit of truths
Main Body
1: Are Imaginary Numbers Imaginary?
Special-Student-7-Hypothesizer
Any imaginary number is \(b i\) where \(b\) is any real number and \(i\) is the number such that \(i^2 = - 1\).
There are some (maybe most) people who believe that imaginary numbers are "imaginary".
Special-Student-7-Rebutter
That is a bad name ...
Special-Student-7-Hypothesizer
The objection is natural: "The name is suggesting that they are imaginary ...".
But the name is a remnant of a historical ignorance.
In fact, misnomers are not so rare even in mathematics; "congruent" is one of them.
Special-Student-7-Rebutter
What "ignorance"?
Special-Student-7-Hypothesizer
Probably, the concept of 'imaginary number' was 1st conceived in the attempt to solve cubic equations.
Special-Student-7-Rebutter
"cubic"? instead of quadratic?
Special-Student-7-Hypothesizer
Probably so: you may think that the attempt to solve \(x^2 = - 1\) caused \(i\), but \(x^2 = - 1\) was just "It has no solution" without any further development.
But Cardano's method for cubic equations required \(\omega := (- 1 + \sqrt{3} i) / 2\), which is a cubic root of \(1\): you will see that \(\omega\) is indeed a cubic root of \(1\) by diligently expanding \(\omega^3\) with \(i^2 = - 1\).
But \(i\) at that time was just an expedient: "There is not such a thing, but if we pretend that it existed, ta-da!, we can solve cubic equations.".
Special-Student-7-Rebutter
"ta-da"?
Special-Student-7-Hypothesizer
Anyway, it was thought imaginary, meaning that "it does not really exist.".
Special-Student-7-Rebutter
To be accurate, it does not exist in the real numbers set.
Special-Student-7-Hypothesizer
Certainly, but the people at that time could not imagine beyond the real numbers set, and so, as far as it does not exist in the real numbers set, "it did not exist at all" for them.
Special-Student-7-Rebutter
Just as "\(\sqrt{2}\) did not exist at all" for the people who could not imagine beyond the rational numbers set.
Special-Student-7-Hypothesizer
Nothing was wrong with \(i\), but something was wrong with the people's imagination.
Special-Student-7-Rebutter
In the 1st place, what do "real" and "imaginary" mean?
2: Does Mathematics Discover or Invent?
Special-Student-7-Hypothesizer
In order to think of that, we should think of the question, "Does mathematics discover or invent?".
Special-Student-7-Rebutter
Ah, that is a well-heard question.
Special-Student-7-Hypothesizer
While some people have expressed their own stances, I am pretty sure on that question: mathematics is a model of the reality.
Special-Student-7-Rebutter
So, ..., does it discover or invent?
Special-Student-7-Hypothesizer
Any model itself is a human-made artifact, but mathematics is not something arbitrarily trumped-up but something that aspires to appropriately represent the reality.
Special-Student-7-Rebutter
Cannot a branch of mathematics not aspire so?
Special-Student-7-Hypothesizer
It could, but any branch gets some significant attention only when it represents the reality well, because otherwise, the branch would have no application.
Special-Student-7-Rebutter
So, ..., does mathematics discover or invent?
Special-Student-7-Hypothesizer
Mathematics models the reality, I am saying.
In fact, I do not think so meaningful to categorize it in "discover" or "invent": it is in "discover" in order to model the reality and is also in "invent" because it is a human-artifact.
3: Then, What Do "Real" and "Imaginary" Mean?
Special-Student-7-Hypothesizer
If "real" means being an entity in the reality, any concept in mathematics is not real, because the concept is an entity in the human-made model, not in the reality.
So, 'real numbers' will be no more real than 'imaginary numbers'.
Special-Student-7-Rebutter
It is not that real numbers exist in the reality.
Special-Student-7-Hypothesizer
But the real numbers set models the reality well.
In fact, the real numbers set models any line well.
Special-Student-7-Rebutter
It is not only spacial lines that the real numbers set models.
Special-Student-7-Hypothesizer
Certainly, for example, also masses of objects are modeled by real numbers.
Anyway, the real numbers set is regarded to be real because it models the reality well.
So, "real" is appropriately understood as modeling the reality well.
So, whether 'imaginary number' is real is the matter of whether it models the reality well.
4: How Complex Numbers Set Models Reality
Special-Student-7-Hypothesizer
The reason why 'imaginary number' was regarded to be non-real is that the people thought that it did not represent anything in the reality.
In fact, what line segment has the i length?
Special-Student-7-Rebutter
But a concept does not need to model a line but to model an entity in the reality.
Special-Student-7-Hypothesizer
In fact, we should think of the complex numbers set, of which the imaginary numbers set is a part.
The complex numbers set models any 2-dimensional plane.
For any 2-dimensional plane, any point is represented by a complex number, \(x + y i\), and any complex number represents a point on the plane; technically, the points set on the plane corresponds to the complex numbers set bijectively.
Then, \(i\) represents the \(0 + 1 i\) point on the plane.
Regarding \(x + y i\) as the vector from the origin to \(x + y i\), \(x + y i = r (cos \theta + sin \theta) i\), where \(r\) is the length of the vector and \(\theta\) is the counter-clock-wise-angle of the vector from the x-axis, as is well known.
Let us define \(e^{\theta i} := cos \theta + sin \theta i\).
Special-Student-7-Rebutter
That definition is possible only because \(e^{0 i} := cos 0 + sin 0 i = 1 = e^0\).
Special-Student-7-Hypothesizer
Well, I could insist that the exponential function for real numbers and the exponential function for imaginary numbers were totally different creatures with the same symbol happened to be shared, but that would be a bad practice, so, yes, we use the same symbol only because the 2 exponential functions are seamlessly connected.
Anyway, that is just a definition, because the exponential map with imaginary number arguments had not been defined as far as we were concerned.
Special-Student-7-Rebutter
The essence of the complex numbers set is not only that the set bijectively corresponds to the plane points set, set-wise.
In fact, what does \(i^2 = - 1\) mean?
Special-Student-7-Hypothesizer
Thinking of \(i^2\) means thinking of multiplication of complex numbers.
As the real numbers set allows the arithmetic operations, the complex numbers set allows the arithmetic operations, \((x + y i) + (x' + y' i) = (x + x') + (y + y') i\); \((x + y i) - (x' + y' i) = (x - x') + (y - y') i\); \((x + y i) (x' + y' i) = (x x' - y y') + (x y' + y x') i\); \((x + y i) / (x' + y' i) = ((x + y i) (x' - y' i)) / (x'^2 + y'^2) = ((x x' + y y') + (- x y' + y x') i) / (x'^2 + y'^2) = (x x' + y y') / (x'^2 + y'^2) + (- x y' + y x') / (x'^2 + y'^2) i\), where \(x' + y' i \neq 0\).
Special-Student-7-Rebutter
An important fact is that that is an expansion of the arithmetic operations for the real numbers set.
Special-Student-7-Hypothesizer
Yes, we have defined the arithmetic operations for the complex numbers set as they are consistent with the arithmetic operations for the real numbers set with the real numbers set as the subset of the complex numbers set.
Special-Student-7-Rebutter
So, the real numbers set sits seamlessly in the complex numbers set.
Special-Student-7-Hypothesizer
And the amazing fact is that \(e^{(\theta + \theta') i} = e^{\theta i} e^{\theta' i}\) holds: \(e^{(\theta + \theta') i} := cos (\theta + \theta') + sin (\theta + \theta') i\) while \(e^{\theta i} e^{\theta' i} = (cos \theta + sin \theta i) (cos \theta' + sin \theta' i) = (cos \theta cos \theta' - sin \theta sin \theta') + (cos \theta sin \theta' + sin \theta cos \theta') i = cos (\theta + \theta') + sin (\theta + \theta') i\).
And the exponential function for complex numbers can be seamlessly defined as \(e^{x + y i} := e^x e^{y i}\), satisfying \(e^{(x + y i) + (x' + y' i)} = e^{x + y i} e^{x' + y' i}\).
Special-Student-7-Rebutter
The definition of the exponential function for complex numbers is just a definition, but the amazing fact is that the function has become seamless with the exponential function for real numbers and satisfies the expected property.
Special-Student-7-Hypothesizer
As each \(x + y i\) is regarded to be a vector, the addition of any 2 complex numbers is the addition of the corresponding 2 vectors, as is obvious.
The product of any 2 complex numbers is the vector whose length is the product of the lengths of the 2 vectors and whose angle is the sum of the angles of the 2 vectors, which is because \((x + y i) (x' + y' i) = r e^{\theta i} r' e^{\theta' i} = r r' e^{(\theta + \theta') i}\).
Special-Student-7-Rebutter
So, the complex numbers set models the plane in the reality well.
Special-Student-7-Hypothesizer
So, the complex numbers set is no less real than the real numbers set.
5: Thinking Imaginary Numbers Imaginary Is a Sure Sign of Narrow-Mindedness
Special-Student-7-Hypothesizer
So, \(i\) is the \((0, 1)\) point on the plane and \(i^2 = - 1\) is reasonably understood as the result of the multiplication: \(i^2 = i i\) has the length, \(1 \times 1 = 1\), and the angle, \(\pi / 2 + \pi / 2 = \pi\), which is nothing but \(- 1\).
So, \(i^2 = - 1\) is no mystery at all.
Special-Student-7-Rebutter
Someone who thinks that \(i\) is imaginary is a 1-dimensional creature that is caged in a line and cannot see or imagine the existence of the world outside the line.
Special-Student-7-Hypothesizer
The word, "imaginary number", was coined at the time when the people did not understand how imaginary numbers modeled the reality, but much time has passed, many things have developed, and you should not be saying like "Imaginary numbers are imaginary." any more.
I heard an argument that wave function in quantum mechanics was not real because it was 'complex numbers'-valued, but that is a sure sign of ignorance.
Special-Student-7-Rebutter
Of course, everyone (especially, we) is ignorant more or less, and it will be fine if you humbly admit your ignorance and continually rectify your misconceptions.
6: Why We Think of 'Field'
Special-Student-7-Hypothesizer
The complex numbers set with the arithmetic operations constitutes a field, as the real numbers set with the arithmetic operations constitutes a field.
Also the rational numbers set with the arithmetic operations is a field, but the integers set with the addition, the subtraction, and the multiplication is not any field but a ring, which is because it does not have any division, which means that \(1 / 2\) does not belong to the integers set.
Special-Student-7-Rebutter
As has been mentioned in What Is a Vector?, it is important to think of the space.
Special-Student-7-Hypothesizer
You may think like "The division operation exists in the integers set, because we can do '1 / 2'", but that is not on the integers set because the result is not in the integers set.
Special-Student-7-Rebutter
Anyway, why should we introduce such an abstract concept as 'field'?
Special-Student-7-Hypothesizer
A major reason is, that is for the sake of economy.
In fact, as far as a structure is a field, some many conclusions hold for the structure regardless of that it is the rational numbers field, the real numbers field, the complex numbers field, or any other field.
So, it is very uneconomical to state and prove the conclusions individually for each of the various fields.
There seem some people who show sheer rejections to abstract concepts, but do you really want to individually deal with each of the various fields?
If you begin to individually study various fields, probably (unless you are too unintelligent), you will begin to feel studying essentially same things again and again, and probably (unless you are too unintelligent), you will find it unwise.
On the other hand, once you accept the concept of 'field', you can just study 'field' and can make same conclusions for all the fields.
Which do you prefer?
7: Why Not Higher-Dimensional Numbers Systems?
Special-Student-7-Rebutter
The complex numbers system (field) is fine modeling the 2-dimensional plane, but then, a natural question will be "Are there a numbers system for the 3-dimensional space, a numbers system for the 4-dimensional space, etc.?".
Special-Student-7-Hypothesizer
That question is "While the complexes numbers system exists, why not a numbers system like \(x + y i + z j\)? or does it exist?".
The answer is that it does not exist as a field.
Special-Student-7-Rebutter
Why?
Special-Student-7-Hypothesizer
If you define it just as a vectors space, it is of course possible: let \(x + y i + z j\) represent the \((x, y, z)\) point.
But it is not possible to define any set of arithmetic operations on it to make it a field.
Special-Student-7-Rebutter
Why?
Special-Student-7-Hypothesizer
It is just impossible: supposing that it expands the complex numbers field, setting \(i j = a + b i + c j\) where \(a, b, c\) are some real numbers, \(i i j = i (a + b i + c j)\), but the left hand side is \(i^2 j = - j\) and the right hand side is \(a i + b i^2 + c i j = a i - b + c i j = a i - b + c (a + b i + c j) = a c - b + (a + b c) i + c^2 j\), which means that \(a c - b = 0\), \(a + b c = 0\), and \(c^2 = - 1\), which is impossible because \(c^2 = - 1\) is impossible.
Special-Student-7-Rebutter
What if it does not expand the complex numbers field?
Special-Student-7-Hypothesizer
Usually, it is useful only because it expands the complex numbers field, as the complex numbers field is useful only because the complex numbers field has expended the real numbers field.
Special-Student-7-Rebutter
Anyway, what if?
Special-Student-7-Hypothesizer
Anyway, it should be impossible, although I do not show the proof here.
Special-Student-7-Rebutter
Fascinating! Why are the 1-dimensional space and the 2-dimensional space so special to allow the field structures?
Special-Student-7-Hypothesizer
Probably, the concept of 'field' is so special to choose only the 1-dimensional space and the 2-dimensional space.
Special-Student-7-Rebutter
What if the structure is not required to be 'field'?
Special-Student-7-Hypothesizer
In fact, there is 'quaternion' for the 4-dimensional space, which is not any field but an associative division algebra over the real numbers field, where "division" means that each nonzero element has the multiplicative inverse.
Special-Student-7-Rebutter
How is that "associative division algebra" thing different from 'field'?
Special-Student-7-Hypothesizer
That thing is not commutative: \(i j = k\) but \(j i = - k\), for example.
Special-Student-7-Rebutter
But there is no 3-dimensional associative division algebra over the real numbers field?
Special-Student-7-Hypothesizer
Certainly.
Special-Student-7-Rebutter
Fascinating! The 3-dimensional space has been skipped over by the structure.
Special-Student-7-Hypothesizer
It is interesting that while an \(m\)-dimensional Euclidean space and an \(n\)-dimensional Euclidean space do not seem so different as far as vectors space structure is concerned, they can be so different as far as field structure or associative division algebra structure is concerned.
