Elucubrations on positve, negative & imaginary numbers

[studiot]

On 2/18/2019 at 8:46 AM, michel123456 said:

Step 3: From the diagram above it comes out that the square of a Real number, positive or negative, is always positive. As shown below, the 2 squares are in a positive area.

Step 4: Then we go to imaginary numbers (Complex numbers), in such a way that negative squares can be handled, as shown below.

Where we see that i^2=-1 (the blue square).

Question 1 is: where is the catch? Why is this diagram wrong? Why -i^2 (the red square) shows negative in the diagram? Although the correct answer is that -i^2=1

Question 2 is the following: what is the sign of 2i? Is it positive or negative? Do we follow the rule of the Real numbers, or the rule of Imaginary numbers?

Question 3 is: how can you combine the Real number diagram with the Imaginary number diagram? Where is the common axis? The only common feature is a point: zero.

Eureka ! The world is saved and we do not need to reinvent the wheel or redefine i and arithmetic works.

It works like this.

 

Michel both your diagrams are correct.

But your interpretation needs a little help.

They are not multiplication diagrams.

To start with your green and yellow squares using real numbers.

 

Quote

As shown below, the 2 squares are in a positive area.

The area of the square is the length of one side times the length of the other.
It is not the product of two coordinates.

You are mixing up corordinates and lengths.
In you opening you have correctly introduced directions showing a positive and negative direction.

length is defined as the difference of two coordinates, taken in a consistent (positive) direction.

length = most positive coordinate minus the least positive coordinate.

So the length of both sides of the green square is (+1 - 0) = +1

So the area of the green square = (1-0)(1-0) = +1

The length of side of the yellow square is [0 - (-1)] = +1

So the area of the yellow square = [0 - (-1)] [0 - (-1)] = (+1) (+1) = +1

Exactly as in your diagram.

So the arithmetic for the real numbers follows the rules of arithmetic for the real numbers.

No suprise.

 

On to your second diagram.

So the length of both sides of the green square is (+i - 0) = +i

So the area of the blue square = (i-0)(i-0) = i² = -1

The length of both sides of side of the yellow square is [0 - (-i)] = +i

So the area of the yellow square = [0 - (-i)] [0 - (-i)] = (+i) (+i) = -1

Exactly as in your diagram.

So the arithmetic for the real numbers follows the rules of arithmetic for the real numbers, provided that i² = -1.

 

What you have done wrong is to say that the length of side the yellow square is -1 and of the the red square is -i.

This effectively reverses the direection of measurement, making your working inconsistent.

This can be even more dramatically brought home by the following diagram.

 

Clearly the area of the large square bdfh is equal to the combined areas of the four smaller squares, A,B,C and D.

But if you count these areas as positive and negative you will find the Area of the large square is exactly zero !

You can confirm this by comparing the measurement of side lengths by your method and by my proper one.

I have not put numbers along the axes, as you can work this out for both 1 and i in order to convice yourself.

 

 

[michel123456]

1 hour ago, studiot said:

Eureka ! The world is saved and we do not need to reinvent the wheel or redefine i and arithmetic works.

It works like this.

 

Michel both your diagrams are correct.

But your interpretation needs a little help.

They are not multiplication diagrams.

To start with your green and yellow squares using real numbers.

 

The area of the square is the length of one side times the length of the other.
It is not the product of two coordinates.

You are mixing up corordinates and lengths.
In you opening you have correctly introduced directions showing a positive and negative direction.

length is defined as the difference of two coordinates, taken in a consistent (positive) direction.

length = most positive coordinate minus the least positive coordinate.

So the length of both sides of the green square is (+1 - 0) = +1

So the area of the green square = (1-0)(1-0) = +1

The length of side of the yellow square is [0 - (-1)] = +1

So the area of the yellow square = [0 - (-1)] [0 - (-1)] = (+1) (+1) = +1

Exactly as in your diagram.

So the arithmetic for the real numbers follows the rules of arithmetic for the real numbers.

No suprise.

 

On to your second diagram.

So the length of both sides of the green square is (+i - 0) = +i

So the area of the blue square = (i-0)(i-0) = i² = -1

The length of both sides of side of the yellow square is [0 - (-i)] = +i

So the area of the yellow square = [0 - (-i)] [0 - (-i)] = (+i) (+i) = -1

Exactly as in your diagram.

So the arithmetic for the real numbers follows the rules of arithmetic for the real numbers, provided that i² = -1.

 

What you have done wrong is to say that the length of side the yellow square is -1 and of the the red square is -i.

This effectively reverses the direection of measurement, making your working inconsistent.

This can be even more dramatically brought home by the following diagram.

 

Clearly the area of the large square bdfh is equal to the combined areas of the four smaller squares, A,B,C and D.

But if you count these areas as positive and negative you will find the Area of the large square is exactly zero !

You can confirm this by comparing the measurement of side lengths by your method and by my proper one.

I have not put numbers along the axes, as you can work this out for both 1 and i in order to convice yourself.

 

 

Right. I understand, thank you for the long explanation.

But that is not my problem.

My problem is : how do you combine the 2 graphs in order to get this one?

1.How do you know that the zone A is positive or negative?

2.How can you combine the 2 previous graphs when they have no common axe?

3. If you state that a point in this zone A is a complex number, how does it come that it is suddenly an addition (1+1i) and not a set of coordinates (1,1i)?

4. How do you know it is an addition in the first place, and not a subtraction?

All those questions.

 

[taeto]

37 minutes ago, michel123456 said:

Right. I understand, thank you for the long explanation.

But that is not my problem.

My problem is : how do you combine the 2 graphs in order to get this one?

3. If you state that a point in this zone A is a complex number, how does it come that it is suddenly an addition (1+1i) and not a set of coordinates (1,1i)?

     The standard construction of \(\mathbb{C}\) indeed introduces complex numbers as pairs \( (x,y  )\) of real numbers, meaning that the real number \(1\) corresponds to \( (1,0) \) and \(i\) to the pair \( (0,1) \) so that \( (x,y) = x\cdot (1,0) + y\cdot (0,1) = x+yi.\) 

     Since squares and square roots feature now and then in this thread, perhaps your question is also related to

en.wikipedia.org/wiki/Square_root#Principal_square_root_of_a_complex_number

Edited February 25, 2019 by taeto

[Eise]

7 minutes ago, michel123456 said:

1.How do you know that the zone A is positive or negative?

It isn't. Only for real numbers positive/negative is defined. It is not for imaginary numbers, and therefore also not for complex numbers. However to imaginary numbers, there is also a number that is its inverse, so that a + a^inv = 0. This inverse is depicted as -a. So in imaginary numbers: 2.i + (-2.i) = 0.i = 0.

9 minutes ago, michel123456 said:

2.How can you combine the 2 previous graphs when they have no common axe?

You can't. Your diagram with imaginary numbers on the x- and y-axis is more or less a (continuous) table of multiplication. The problem is that in the complex plane, you can identify every point with a complex number: e.g. the point defined by the coordinates (1,2) represents the complex number 1 + 2i. However, in your graph with both axes imaginary, you multiply two imaginary numbers, which result in a real number: (x.i).(y.i) = -xy.

20 minutes ago, michel123456 said:

3. If you state that a point in this zone A is a complex number, how does it come that it is suddenly an addition (1+1i) and not a set of coordinates (1,1i)?

Every point that lies in the complex plane (x-axis real numbers, y-axis imaginary numbers) is a complex number. Only the numbers of the x-axis can be defined as real numbers, simply because the imaginary component is 0. Every multiplication on the real axis, yields a real result. (Remember: add the angles when you multiply; the angle is 0^o (or 180^o) for real numbers, so their additions will also always be 0^o (or 180^o), and therefore lie again on the x-axis.) And the addition you mention here is a vector addition: you add the vectors (1,0) and (0,1), which give (1,1). Because the y-axis represents imaginary numbers, you can write this as 1 + 1.i. (In 'normal coordinates you would say that (1,1) means 1.x + 1.y, but again: this is vector addition).

29 minutes ago, michel123456 said:

4. How do you know it is an addition in the first place, and not a subtraction?

Given that we depict complex numbers as coordinates in a plane, it is standard that we call it addition. But again: vector addition.

[studiot]

44 minutes ago, michel123456 said:

Right. I understand, thank you for the long explanation.

But that is not my problem.

My problem is : how do you combine the 2 graphs in order to get this one? 

By the two graphs I assume you mean your original pair,

one with both axes real

and the other with both axes imaginary.

The short answer is that since ,as you rightly point out, you have no common axes the only way to make a new graph is to draw a four dimensional one.

No so good huh?

 

44 minutes ago, michel123456 said:

1.How do you know that the zone A is positive or negative?

2.How can you combine the 2 previous graphs when they have no common axe?

3. If you state that a point in this zone A is a complex number, how does it come that it is suddenly an addition (1+1i) and not a set of coordinates (1,1i)?

4. How do you know it is an addition in the first place, and not a subtraction?

All those questions.

Now in drawing this one you have a standard graphical representation of a complex number (not not an imaginary one) with one real axis (the x axis)

and one imaginary axis (the y axis)

 

To answer the other questions,

1)

Zone A is now neither positive nor negative.
Complex numbers can not be positive or negative.
The only use of + and - is to denote direction along the axes.

3)

Points in all four 'zones' are complex numbers.

Thos in zones A and B are of the form x + iy

Those in zones C and D are of the form x - iy

This is the form we use to perform arithmetic with a complex number.
(It is called complex because it is made of more than one part - 2 parts)

There are other forms of presenting or representing the any given complex number.

Your second is used for plotting on diagrams (graphs) and is of conventional plotting representation (x,iy)

The whole plane is called the complex plane.

4)

Answered at the beginning of my answer to question 3

 

I feel we are making definite progress.

 

Note there are several other ways of representing a complex number, each with its own special points.
They are all equivalent.
I will not confuse things by introducing them at this time.

 

Edited February 25, 2019 by studiot

[michel123456]

Ok, lets replace the i axis with something else: oranges.

We get

The x axis is still the real numbers.

The y axis is oranges.

The point a is the set (3,2oranges). It may represent that you have 3packs of 2 oranges each. It means you have 6 oranges (3.2=6)

You don't get 3+2oranges. Even vectorized. And why suddenly vectors?

 

Edited February 25, 2019 by michel123456

[studiot]

49 minutes ago, studiot said:

Points in all four 'zones' are complex numbers.

Thos in zones A and B are of the form x + iy

Those in zones C and D are of the form x - iy

 

I hoped you would spot my non deliberate error here

Very sorry for this I was rushing again.

Those in zone A are of the form +x + iy

Those in zone B are of the form -x + iy

Those in zone C are of the form -x - iy

Those in zone D are of the form +x - iy

Bloody editor just cancelled my previous edit

42 minutes ago, michel123456 said:

The x axis is still the real numbers.

The y axis is oranges. 

The point a is the set (3,2oranges). It may represent that you have 3packs of 2 oranges each. It means you have 6 oranges (3.2=6)

You don't get 3+2oranges. Even vectorized. And why suddenly vectors?

 

What do negative oranges look like?
And what useful properties do positive and negative oranges have , mathematically?
And what does an area of 3 x 2oranges mean?

Perhaps the graph or plane is not the right way to look at counting oranges?

The point of introducing the symbol i is that is is just one extra symbol that has very desirable properties.

 

There is another way of looking at this plot, that I wasn't yet ready to introduce, lest it confuse the issue.

The plot is then not a graph but a tabulation.
It then makes sense.

Do you wish to digress to explore this option?

Edited February 25, 2019 by studiot

[Eise]

2 hours ago, michel123456 said:

You don't get 3+2oranges. Even vectorized. And why suddenly vectors?

Again you are using your depiction as a 'table of multiplication'.

But say you want to keep track of apples and oranges. Apples are on the x-axis, oranges on the y-axis. Then how much would be 3 apples times 2 oranges? Is that meaningful in anyway? It just makes no sense.

However, if you use the graph as a depiction of sums, than it makes sense: the point (3,2) means 3 apples and 2 oranges. And you can see them as vectors: say I have the vectors (1,0) (1 apple) and (2,2) (2 apples and 2 oranges. The vector sum, of course, is (1+2,0+2) = (3,2): 3 apples and 2 oranges. And that is correct.

So you cannot mix apples and oranges with multiplication, but you can with addition. Same with real and imaginary numbers.

Now the 'only specialty' of imaginary numbers is that 1 orange times 1 orange is exactly one apple less than 0: add one apple and you have 0. A relation that does not exist with real apples and oranges.

[michel123456]

A negative orange may be an orange that you don't have but you owe to someone. Like negative money. Like negative areas that seem to bother you so much: it is an area missing. In real life it may be a part of your property that somebody (the administration, the bank) is asking for. Negatives are not a problem. Negative volumes are those that are produced by the exterior of a building for example.

Now, about apple & oranges, IMHO it is making the things a bit complicated. Since we have the real numbers on the horizontal axis, it cannot be simpler, there is no need to make it more complicated.

My point of view is that you are not allowed to interpret the diagram as you wish. There are rules. When you put orthogonal axes, you must follow the rules. For example, a square (x^2) is geometrically a square. A cube (^3) is a cube (in 3 dimensions). So when the i axis is perpendicular to the x axis, it determines an area (in fact 4 areas). You cannot suddenly change the rules and say that the area is not an area anymore. You cannot change the product with the addition just like that. It is as if I said that the point of coordinates (3,2) equals 5. It is wrong.

And since the imaginary numbers are just regular numbers (that follow a different rule), I don't understand the whole setup.

Sorry for showing such a stubbornness.

[studiot]

17 minutes ago, michel123456 said:

A negative orange may be an orange that you don't have but you owe to someone. Like negative money. Like negative areas that seem to bother you so much: it is an area missing. In real life it may be a part of your property that somebody (the administration, the bank) is asking for. Negatives are not a problem. Negative volumes are those that are produced by the exterior of a building for example.

Now, about apple & oranges, IMHO it is making the things a bit complicated. Since we have the real numbers on the horizontal axis, it cannot be simpler, there is no need to make it more complicated.

My point of view is that you are not allowed to interpret the diagram as you wish. There are rules. When you put orthogonal axes, you must follow the rules. For example, a square (x^2) is geometrically a square. A cube (^3) is a cube (in 3 dimensions). So when the i axis is perpendicular to the x axis, it determines an area (in fact 4 areas). You cannot suddenly change the rules and say that the area is not an area anymore. You cannot change the product with the addition just like that. It is as if I said that the point of coordinates (3,2) equals 5. It is wrong.

And since the imaginary numbers are just regular numbers (that follow a different rule), I don't understand the whole setup.

Sorry for showing such a stubbornness.

 

6 hours ago, studiot said:

There is another way of looking at this plot, that I wasn't yet ready to introduce, lest it confuse the issue.

The plot is then not a graph but a tabulation.
It then makes sense.

Do you wish to digress to explore this option?

 

Stubbornness is not a problem.

Oranges and apples are as they cloud the issue.

[michel123456]

59 minutes ago, studiot said:

 

 

Stubbornness is not a problem.

Oranges and apples are as they cloud the issue.

Tabulation yes but then vectors on tabulation?

A map?

[Eise]

12 hours ago, michel123456 said:

A negative orange may be an orange that you don't have but you owe to someone. Like negative money. Like negative areas that seem to bother you so much: it is an area missing. In real life it may be a part of your property that somebody (the administration, the bank) is asking for. Negatives are not a problem. Negative volumes are those that are produced by the exterior of a building for example.

I have no problem with that. But you have a problem with i. i Is defined as the square root of -1. But i itself is neither positive nor negative. So the same holds for -1.i which is the inverse of i under summation (which does not mean that it is negative! if i itself is nor positive or negative, then -i is neither), and is also a solution to x² = -1:

(-1.i)² = -1² . i² = 1 . -1 = -1

12 hours ago, michel123456 said:

Now, about apple & oranges, IMHO it is making the things a bit complicated. Since we have the real numbers on the horizontal axis, it cannot be simpler, there is no need to make it more complicated.

I think it makes it more complicated: I have no idea what 3 + 2 oranges would be. However I can easily visualise 3 apples and 2 oranges. 

12 hours ago, michel123456 said:

My point of view is that you are not allowed to interpret the diagram as you wish. There are rules. When you put orthogonal axes, you must follow the rules. For example, a square (x^2) is geometrically a square. A cube (^3) is a cube (in 3 dimensions). So when the i axis is perpendicular to the x axis, it determines an area (in fact 4 areas).

So can you draw me an area spanned by the complex number 3 + 2i? According to you the area should be 3 * 2i = 6i. What would be the meaning of that?

12 hours ago, michel123456 said:

You cannot suddenly change the rules and say that the area is not an area anymore. You cannot change the product with the addition just like that.

The rules were changed when introducing the y-axis as imaginary numbers.

12 hours ago, michel123456 said:

And since the imaginary numbers are just regular numbers (that follow a different rule), I don't understand the whole setup.

I would say this is a contradiction: something cannot be 'just a regular number' and then follow a different rule.

Imaginary numbers are not just regular numbers. But you can extend the set of real numbers with them, but then you must define e.g. multiplication in a way that still gives the same results when applied to reals only. This can be shown in the complex plane. 

[michel123456]

50 minutes ago, Eise said:

I think it makes it more complicated: I have no idea what 3 + 2 oranges would be.

But that is exactly what you are doing when additioning real & complex numbers. You get 3+2i = 3 + 2 oranges.

52 minutes ago, Eise said:

So can you draw me an area spanned by the complex number 3 + 2i? According to you the area should be 3 * 2i = 6i. What would be the meaning of that?

6 oranges.

53 minutes ago, Eise said:

The rules were changed when introducing the y-axis as imaginary numbers.

That is my question: why do you change the rules?

As I have shown, the rules are similar, with the difference of sign. For i numbers, the product of 2 positives give a negative. But still, the diagram shows a product, not an addition.

[studiot]

14 hours ago, michel123456 said:

Tabulation yes but then vectors on tabulation?

A map?

Firstly I would avoid using vectors as analogies.

The geometric vectors Eise is talking about are not the same as complex numbers.

The two possess some common features but there are also differences.

If the goal is to model one with the other, obviously this can only be done for models only concerned with the common features.
However the differences are important here.
In particular there is only one rule of multiplication for complex numbers, but there are two quite different forms of multiplication with vectors and neither correspond to the multiplication of complex numbers.

This, of course, is why we have both of these.

 

As regards the second question.

Map, yes, tabulation, yes.

Let us just use the integers for a minute.

Using these we can construct 'pigeonholes'

There is even a theorem called 'the pigeonhole principle'.

https://en.wikipedia.org/wiki/Pigeonhole_principle

(For once the Wiki article is accessible to those without a triple Phd in maths.)

We can regard the coordinates as identifying a particular pigeonhole, rather than a length.

It does not matter what 'size' each pigeonhole has or even that they are square.

One pigeonhole is one pigeonhole.

And we can put anything we like into it.

So we could put the product of the x and y coordinate in there.

This does not mean that this product equals an area.

In fact in this model there is no such quantity as 'area'.

However our pigeonhole array forms the multiplication table or tabulation I referred to earlier.

Note also that Wiki goes further and discusses the case where some pigeonsholes have multiple occupancy.

We want to specify for our purposes that each pigeonhole has exactly one thing in it.

 

This can be developed further to give meaning to your idea of crossed imaginary axes.

 

[Eise]

1 hour ago, michel123456 said:

But that is exactly what you are doing when additioning real & complex numbers. You get 3+2i = 3 + 2 oranges.

Yes, I do, but as vector addition. Under addition the sum of two complex numbers is the sum of the 2 components of a complex number:

(1 + i) + (5 -3i) = 6 - 2i. So unless the imaginary parts happen to cancel (1 + i) + (5 - i) = 6 + 0i, the sum of 2 complex numbers is again a complex number.

Multiplication with a real number is also the same as multiplication of a vector with a real:

2 * (5 -3i) = (2*5 - 2*3i) = 10 -6i.

Only when multiplying 2 complex numbers, we get a new definition, that is not defined for 'normal' vector products (inner- or outer-product):

(1 + i) * (5 -3i) = 1*5 -1*3i + 5i - i*3i = 5 + 2i + 3 = 8 + 2i.

We can check the 'lengths' of the complex numbers (use Pythagoras)

1 + i :  sqrt(1² + 1²) =  sqrt(2)

5 -3i: sqrt(5² + 3²) = sqrt(34)

Product of their lengths = sqrt(68).

8 + 2i: sqrt(8² + 2²) = sqrt(68) => correct.

I am too lazy for the angles, but you can bet on it that the sum of the angles of 1 + i and 5 -3i equals the angle of 8 + 2i in the complex plane.

1 hour ago, michel123456 said:

6 oranges

And that does not work. An area should be at least oranges². 

1 hour ago, michel123456 said:

That is my question: why do you change the rules?

Because i cannot be treated as a real number!

1 hour ago, michel123456 said:

As I have shown, the rules are similar, with the difference of sign.

Well, that is not similar to the rules for reals at all:

• Every square of a real number is positive
• The reals can be ordered. The combination of reals and imaginary numbers cannot.

1 hour ago, michel123456 said:

For i numbers, the product of 2 positives give a negative. But still, the diagram shows a product, not an addition.

Again, there are no positive imaginary numbers. Positive/Negative is only defined for reals. (would you prefer 100i dollars or -100i dollars?; what if I replace i with -i everywhere. Would you notice?))

And again, your two diagrams are just tables of multiplication, like such a one:

 

[taeto]

1 hour ago, studiot said:

In particular there is only one rule of multiplication for complex numbers, but there are two quite different forms of multiplication with vectors and neither correspond to the multiplication of complex numbers.

You are being hurried again.

The set \(\mathbb{C}\) of complex numbers equipped with addition and multiplication is the extension field \(\mathbb{R}(i)\) of the real numbers, and it is a vector space over \(\mathbb{R}.\) Just as every extension field of a field \(\mathbb{F}\) is a vector space over \(\mathbb{F},\) for the appropriately defined product that extends the product in \(\mathbb{F}.\) 

As candidates for multiplications in \(\mathbb{C}\) as a real vector space you may be thinking of the dot product \( (a+bi)\cdot(c+di) = ac+bd \) and a cross product \(\times\) (?) each of which is not the product that works to make \(\mathbb{C} \) a field.

Edited February 26, 2019 by taeto

[studiot]

1 hour ago, taeto said:

You are being hurried again.

The set C of complex numbers equipped with addition and multiplication is the extension field R(i) of the real numbers, and it is a vector space over R. Just as every extension field of a field F is a vector space over F, for the appropriately defined product that extends the product in F.  

As candidates for multiplications in C as a real vector space you may be thinking of the dot product (a+bi)⋅(c+di)=ac+bd and a cross product × (?) each of which is not the product that works to make C a field.

Not really.

Addition and multiplication are both closed in the complex plane.

Neither form of 'vector' multiplication is closed in the plane.

In fact there is no axiom at all in the vector space space list for multiplication of a vector by a vector.

This is why you can say that C is a vector space, and a field, but the planar space spanned by a pair of non parallel coplanar geometric vectors is not.

Note I also said geometric vectors.

 

However what I said stands

3 hours ago, studiot said:

The two possess some common features but there are also differences.

 

 

I think that geometric (pointy) vectors are confusing enough here, without going to the full complexity

[taeto]

3 hours ago, studiot said:

In fact there is no axiom at all in the vector space space list for multiplication of a vector by a vector.

We can even say that there is no theorem that states that a multiplication which turns the vector space into a field is at all possible in general.

 

3 hours ago, studiot said:

This is why you can say that C is a vector space, and a field, but the planar space spanned by a pair of non parallel coplanar geometric vectors is not.

Note I also said geometric vectors.

I cannot make sense of that comment.

How do you define "geometric vector"? 

If you restrict even down to Euclidean geometry without the Cantor-Dedekind property, you still have that the set of 2-dimensional vectors \(\mathbb{Q}^2\) forms a field, with multiplication defined as for \(\mathbb{Q}(i)\), the rational complex numbers. 

[studiot]

53 minutes ago, taeto said:

If you restrict even down to Euclidean geometry without the Cantor-Dedekind property, you still have that the set of 2-dimensional vectors Q2 forms a field, with multiplication defined as for Q(i) , the rational complex numbers. 

 

Yes, and funnily enough my pigeonholing was approaching this from a different direction.

But the vectors of Q² or R² are not the planar geometric vectors Eise and I were talking about.

These do not form a field as they do not satisfy the field axiom (not the non existant vector space axiom) of completeness under multiplication.

 

[taeto]

1 hour ago, studiot said:

But the vectors of Q² or R² are not the planar geometric vectors Eise and I were talking about.

These do not form a field as they do not satisfy the field axiom (not the non existant vector space axiom) of completeness under multiplication.

I wish that I knew what it is you guys are talking about . Some pointer to the concept of "geometric vector" would seem in order, assuming you are not just making things up .

Look, both \(\mathbb{Q}^2\) and \(\mathbb{R}^2\) do form fields, even if you restrict the possible addition operations to be exactly the same as they are when the same sets are viewed as vector spaces over \(\mathbb{Q}\) and \(\mathbb{R}\), respectively. You seem to argue that they are not fields, because the vector space axioms do not require them to be. You have things backwards. First you define what you mean by the set \(\mathbb{Q}^2\) of pairs of rational numbers. Then you define addition and scalar multiplication to prove that the set forms a vector space under those operations. Nothing more is required. Then, if you so desire, you commence to prove that the set also forms a field, given an appropriate description of a multiplication operator.  

Compare with the assertion "The set \(\{0,1,2,3,4,5\}\) forms a field." How does your assertion \(\mathbb{Q}^2\) is not a field compare to that?

Edited February 26, 2019 by taeto

[studiot]

Is the dot product or the cross product of co-planar vectors  a vector in the same plane?

Is the product of one complex number with another, yet another complex number?

 

You mentioned symmetry a while back.

There is an interesting application of symmetry in this thread.
Perhaps you can think of something better?

 

 

Edited February 26, 2019 by studiot

[michel123456]

22 hours ago, Eise said:

Only when multiplying 2 complex numbers, we get a new definition, that is not defined for 'normal' vector products (inner- or outer-product):

(1 + i) * (5 -3i) = 1*5 -1*3i + 5i - i*3i = 5 + 2i + 3 = 8 + 2i.

We can check the 'lengths' of the complex numbers (use Pythagoras)

1 + i :  sqrt(1² + 1²) =  sqrt(2)

5 -3i: sqrt(5² + 3²) = sqrt(34)

Product of their lengths = sqrt(68).

8 + 2i: sqrt(8² + 2²) = sqrt(68) => correct.

How do you put this on a graph?

[Eise]

2 hours ago, michel123456 said:

How do you put this on a graph?

 

So the (Pythagorean) length (1 + i) times the length of (5 -3i) is the length of (8 + 2i), as described in my previous posting.

And the angle of c is a + b (where b is negative in this example. Or somewhere between 270^o and 360^o, that doesn't matter.)

Edited February 27, 2019 by Eise

[michel123456]

I expected to see an orthogonal triangle somewhere.

Got it.

Edited February 27, 2019 by michel123456

[michel123456]

 

To me

1. multiplication has a very strong relationship with orthogonality.

2. Multiplication has also a relation with units (when you multiply meters with meters, you get m^2 which is different from m, when you multiply seconds with sec., you get s^2 which is also different, etc.) To me, it is bizarre to multiply length by length & still obtain a length.

Yes I understand somehow the result as stated by Eise above, but geometrically I don't get how the multiplication 2 "lengths" give a "length" and not a "length squared"

So in the below graph, I have reproduced the multiplication of 2+i with 5-3i. I have reported the "lengths" on both orthogonal axis, obtaining points α & β* The product gives the yellow rectangle (which correctly is an area, not a "length"). The area gives arithmetically the same number with "lenght" L.

The result is 8.2462 for the area & 8.2462 for L (0,8+2i).

Angle c does not come out immediately. You must report point β on point γ in order to get the correct angle. The point 8+2i does not come out from the construction.

I don't know If it gives any insight.

* the projection could have been made on other orthogonal axes, the report on i is for simplicity.