Elucubrations on positve, negative & imaginary numbers

[michel123456]

51 minutes ago, studiot said:

I apologise I was rushing when I posted.

I should have made clear what I was doing.

At least you have picked it up correctly.

A is the point (6i, 9i) so the product is -54

B is the point (1i, 54i) so the product is -54

C is the point (54i, 1i) so the product is -54

So my point still stands

What does the -54 mean since there are many different points, each with this value of product?

It is negative.

And I guess all these points are upon an hyperbola. But I don't know if that can have a deeper signification. That was not my goal.

Edited February 21, 2019 by michel123456

[Eise]

Hi Michel,

I think you should see that your way of depicting real numbers and imaginary numbers leads to mathematical nonsense. You cannot treat imaginary numbers as if they are just some additional numbers to the real numbers. If you want to combine the two in one concept, i.e. in complex numbers, you need an independent depiction, in which real numbers and imaginary numbers are two different dimensions. Imaginary numbers do not fit on the real-number line: therefore you cannot say that i is greater or smaller than 1, or even zero. So i is not negative, and it is not positive. -i is just the inverse of i under summation, meaning that -i + i = 0. So complex numbers are more like 2-dimensional vectors: one axis is the real axis, the other axis is the imaginary one.

Now multiplication becomes an interesting phenomenon. If you make a drawing (real axis horizontal, imaginary axis vertical) you can see what happens if you multiply complex numbers. Example:

Take the complex number sqrt(3) + 1.i and calculate the square:

You get: 

• 3 + 2.sqrt(3).i + (i.i)
• 3 + 2.sqrt(3).i - 1
• 2 + 2.sqrt(3).i

What can you say about this calculation in terms of vectors? As you know vectors have a length and an angle. So e.g. the 'length' of i is 1, just as the lengths of 1, -1 or -i.
Length: 

Pythagoras:
Length of sqrt(3) + 1.i:

• sqr(sqrt(3)) + sqr(1) = 3 + 1 = 3 + 1 = 4 => the length is sqrt(4) = 2.

Length of 2 + 2.sqrt(3).i:

• sqr(2) + sqr(2.sqrt(3)) = 4 + 4.3 = 4 + 12  = 16 => the length is sqrt(16) = 4.

So the length of the product is the product of the lengths. And that fits consistently to numbers that lie on the real axis only! No inconcistency.

Now if you would look at the angles (use goniometry) you will discover that with multiplication of 2 complex numbers, you must add the angles.

So what happens when you square e.g. 1 + i? The angle is 45^o, so double the angle of the square: it is 90^o. The length of the 'vector' 1 + i is sqrt(2), so the length of the result will be 2. 90^o means the result lies on the imaginary axis, so the answer is 2i.

Calculation:

• sqr(1 + i) = 1 + 2.i + i.i = 1 + 2.i - 1 = 2i. That fits.

This formalism is used everywhere where complex numbers are used. E.g. Richard Feynman, in his pop-science book about QED,  uses rotating arrows and their vector addition and multiplication  to explain QED. But of course, he really is talking complex numbers, using the vector depiction. The whole building of physics would break down, if this way of treating complex number would be wrong.

Does that help a bit?

 

Edited February 23, 2019 by Eise

[studiot]

37 minutes ago, Eise said:

I think you should see that your way of depicting real numbers and imaginary numbers leads to mathematical nonsense. You cannot treat imaginary numbers as if they are just some additional numbers to the real numbers. If you want to combine the two in one concept

But Michel doesn't want to do this. He want to use only imaginary numbers and that is where his problems start.

Technically by themselves imaginary numbers do not form an 'algebra'.

Of course, both the real numbers and the complex numbers do.

This issue is so fundamental that we normally take it for granted and don't bother to mention it or state it.

We just so used to cracking on with appropriate mathematical object we don't even realise we are automatically selecting them.

Using only only  imaginary numbers doesn't lead to mathematical nonsense, it is just a helluva lot more restrictive than the alternatives - as you rightly observe Michel is finding out.

[michel123456]

6 hours ago, Eise said:

Hi Michel,

I think you should see that your way of depicting real numbers and imaginary numbers leads to mathematical nonsense. You cannot treat imaginary numbers as if they are just some additional numbers to the real numbers. If you want to combine the two in one concept, i.e. in complex numbers, you need an independent depiction, in which real numbers and imaginary numbers are two different dimensions. Imaginary numbers do not fit on the real-number line: therefore you cannot say that i is greater or smaller than 1, or even zero. So i is not negative, and it is not positive. -i is just the inverse of i under summation, meaning that -i + i = 0. So complex numbers are more like 2-dimensional vectors: one axis is the real axis, the other axis is the imaginary one.

Now multiplication becomes an interesting phenomenon. If you make a drawing (real axis horizontal, imaginary axis vertical) you can see what happens if you multiply complex numbers. Example:

Take the complex number sqrt(3) + 1.i and calculate the square:

You get: 

• 3 + 2.sqrt(3).i + (i.i)
• 3 + 2.sqrt(3).i - 1
• 2 + 2.sqrt(3).i

What can you say about this calculation in terms of vectors? As you know vectors have a length and an angle. So e.g. the 'length' of i is 1, just as the lengths of 1, -1 or -i.
Length: 

Pythagoras:
Length of sqrt(3) + 1.i:

• sqr(sqrt(3)) + sqr(1) = 3 + 1 = 3 + 1 = 4 => the length is sqrt(4) = 2.

Length of 2 + 2.sqrt(3).i:

• sqr(2) + sqr(2.sqrt(3)) = 4 + 4.3 = 4 + 12  = 16 => the length is sqrt(16) = 4.

So the length of the product is the product of the lengths. And that fits consistently to numbers that lie on the real axis only! No inconcistency.

Now if you would look at the angles (use goniometry) you will discover that with multiplication of 2 complex numbers, you must add the angles.

So what happens when you square e.g. 1 + i? The angle is 45^o, so double the angle of the square: it is 90^o. The length of the 'vector' 1 + i is sqrt(2), so the length of the result will be 2. 90^o means the result lies on the imaginary axis, so the answer is 2i.

Calculation:

• sqr(1 + i) = 1 + 2.i + i.i = 1 + 2.i - 1 = 2i. That fits.

This formalism is used everywhere where complex numbers are used. E.g. Richard Feynman, in his pop-science book about QED,  uses rotating arrows and their vector addition and multiplication  to explain QED. But of course, he really is talking complex numbers, using the vector depiction. The whole building of physics would break down, if this way of treating complex number would be wrong.

Does that help a bit?

 

Since i^2 is a thing, I put it into a diagram. That's it.

I put i on the X, i on the Y and I get i^2.

I don't see anything wrong or even weird.

Then I put a sign on the result: it is negative.

Then I put a sign on the reverse (-i) (-i) and it is also negative (as mentioned in the wiki page as I saw afterwards)

Then I suspect that (i) (-i) gives a positive.

note: If you create something like i you should be comfortable to use it anyways.

And then I am asking what is the sign of 3i ? Because 3 is a Real number and multiplication of positives give positive but i is imaginary with the multiplication of positives give negative. There is a conflict IMHO.

Edited February 23, 2019 by michel123456

[taeto]

1 hour ago, michel123456 said:

And then I am asking what is the sign of 3i ? Because 3 is a Real number and multiplication of positives give positive but i is imaginary with the multiplication of positives give negative. There is a conflict IMHO.

There is a conflict with your thinking that \(i\) is "positive"?

Integers, rationals and real numbers have "signs". Complex numbers do not, unless you define a sign as an extra feature. You want to define the "sign" of \( 1- \sqrt{2}i\)? 

Actually when you build the foundations of complex numbers, you just start from, well, \(-1\) has two square roots, so let \(i\) be one of them, then the other one is \(-i\), right? Clearly there is nothing inherently positive or negative, since either one of the two square roots could have chosen to be the "positive" one. Which is a little different from what happens with integers.

Edited February 23, 2019 by taeto

[studiot]

1 hour ago, michel123456 said:

And then I am asking what is the sign of 3i ? Because 3 is a Real number and multiplication of positives give positive but i is imaginary with the multiplication of positives give negative. There is a conflict IMHO

There is no conflict.

nor is the underlined bit always true.

What do you think 0i * 0i makes?

 

This might help.

Humour me and answer this question please.

What is the mains voltage in your house?

 

You might also like to try the following

What is the square of

[math]\frac{{ - 1 + i\sqrt 3 }}{2}[/math]

 

and now the square of

[math]\frac{{ - 1 - i\sqrt 3 }}{2}[/math]

 

Do both and see if you notice something very interesting.

(Note I have put the i before the   [math]{\sqrt 3 }[/math]  to make it clear that the square root sign does not apply to the i.
If I put is after as in 3i it can be confusing as to whether the i is inside or outside the square root.)

But if you really don't like complex numbers you can try another similar type of compound number, but using only real numbers, of the form

[math]\left( {a + b\sqrt 2 } \right)[/math]

 

I await your answer with interest.

 

Edited February 23, 2019 by studiot

[michel123456]

51 minutes ago, taeto said:

There is a conflict with your thinking that i is "positive"?

Integers, rationals and real numbers have "signs". Complex numbers do not, unless you define a sign as an extra feature. You want to define the "sign" of 1−2–√i ? 

Actually when you build the foundations of complex numbers, you just start from, well, −1 has two square roots, so let i be one of them, then the other one is −i , right? Clearly there is nothing inherently positive or negative, since either one of the two square roots could have chosen to be the "positive" one. Which is a little different from what happens with integers.

What is the difference?

With the integers the square is always a positive. With i the square is always negative. It is a symmetric situation. I don't understand why I should have to treat the imaginary numbers differently from the real. If real have + and - signs, why not the imaginary ones?

47 minutes ago, studiot said:

Humour me and answer this question please.

What is the mains voltage in your house?



330V



Sorry I can't handle the math scripts.

Edited February 23, 2019 by michel123456

[studiot]

18 minutes ago, michel123456 said:

330V 

Thank you for your answer.

So this mains electricity runs your appliances and aircon and so on, and you would certainly notice the effect if you stuck your finger in the socket.

yet the average voltage in an AC supply is zero.

This curious fact is a real world example of a phenomenon similar to what you are seeing on your plots.

The area under the power curve is also exactly zero over one complete cycle.

 

 

20 minutes ago, michel123456 said:

Sorry I can't handle the math scripts.

 

Don't worry I wasn't looking for you to post the answer, just discover for yourself that the second is the square of the first and the first is the square of the second.

There are many strange things in Maths, which has led some to the occult.

[taeto]

10 hours ago, michel123456 said:

What is the difference?

With the integers the square is always a positive. With i the square is always negative. It is a symmetric situation. I don't understand why I should have to treat the imaginary numbers differently from the real. If real have + and - signs, why not the imaginary ones?

It is asymmetric in that the product of two real numbers is a real number, whereas the product of two imaginary numbers is not an imaginary number.

You could define a new product on the imaginary numbers so that (ix)(iy)=ixy for real numbers x and y. Then the symmetry is restored, and the imaginary numbers form a field isomorphic to the field of real numbers. But I suspect that when you think of multiplying a real number with an imaginary number, then you think of multiplication as it is defined for complex numbers? So if you want to create symmetry, then that is when you end up with a conflict.

[michel123456]

3 hours ago, taeto said:

It is asymmetric in that the product of two real numbers is a real number, whereas the product of two imaginary numbers is not an imaginary number.

You could define a new product on the imaginary numbers so that (ix)(iy)=ixy for real numbers x and y. Then the symmetry is restored, and the imaginary numbers form a field isomorphic to the field of real numbers. But I suspect that when you think of multiplying a real number with an imaginary number, then you think of multiplication as it is defined for complex numbers? So if you want to create symmetry, then that is when you end up with a conflict.

You are a genius! That's it.

ix is not a product, it is a name.

My bad.

[taeto]

30 minutes ago, michel123456 said:

You are a genius! That's it.

I agree with this.

30 minutes ago, michel123456 said:

ix is not a product, it is a name.

We agree here as well, it is only a name. But just to be clear: the point is that at least the imaginary numbers ought to form a self-respecting real vector space, so that when a is a real number and ix is an imaginary number, then the product a(ix) produces the imaginary number i(ax).

[studiot]

1 hour ago, taeto said:

the point is that at least the imaginary numbers ought to form a self-respecting real vector space, so that when a is a real number and ix is an imaginary number, then the product a(ix) produces the imaginary number i(ax).

 

Can't see how that is compatible with the vector space axioms.

Further in all the treatments I have seen, ix an behaves as if it is a perfectly respectable product.

How can ix be 'only a name' ?

How can you do mathematics with 3 * (only a name) or 3i * (only a name) or (only a name)* (a+ib) ?

What on earth does the field set look like?

Edited February 24, 2019 by studiot

[taeto]

50 minutes ago, studiot said:

What on earth does the field set look like?

It is just the same as \(\mathbb{R},\) except instead of \(x\) you have \(ix.\)

Addition is \(ix+iy = i(x+y)\) and the zero element is \(i0,\) etc. The point is that multiplication is "redefined" so that \(ix\cdot iy = i(xy)\) is the rule. It does not say that \(i^2 = -1\) or not, instead it just ignores the \(i.\)

55 minutes ago, studiot said:

How can ix be 'only a name' ?

The way I see it, it means that "i" is a letter and x is a real number. Michel can correct if he sees it differently.

[studiot]

49 minutes ago, taeto said:

The point is that multiplication is "redefined" so that ix⋅iy=i(xy) is the rule.

 

That doesn't work since the product of two imaginary numbers is not an imaginary number.

You can't make (1i *1i) = -1 without the i by that system.

Prefixing an undefined label to every real number certainly retains the field struture of the Reals.

But would be just a pointles as prefixing the label 'real' to every number in the field.

You have not created the extension field gained by adding the special properties available in adding one special symbol with the properties of being the square root of negative 1.

This particular extension field is of course called the complex numbers.

A few posts back, I gave another example adding a less controversial number - the square root of 2.
That creates a proper field.

[taeto]

16 minutes ago, studiot said:

You have not created the extension field gained by adding the special properties available in adding one special symbol with the properties of being the square root of negative 1.

I did not say that it is an extension field. It is an isomorphic field. And yes, it is kind of pointless, except the point is that it is the only way to create the symmetry that Michel postulated.

17 minutes ago, studiot said:

That doesn't work since the product of two imaginary numbers is not an imaginary number.

You can define the product of any two elements of any set to be whatever you like. 

Edited February 24, 2019 by taeto

[michel123456]

1 hour ago, taeto said:

It is just the same as R, except instead of x you have ix.

Addition is ix+iy=i(x+y) and the zero element is i0, etc. The point is that multiplication is "redefined" so that ix⋅iy=i(xy) is the rule. It does not say that i2=−1 or not, instead it just ignores the i.

The way I see it, it means that "i" is a letter and x is a real number. Michel can correct if he sees it differently.

That ruins everything.

The point is to keep i^2=-1

From my understanding if one keeps your ix⋅iy=i(xy )

for x=i and y=i you get i(xy) = i^2

where i(xy) is a notation meaning the multiplication of ix with iy under the i system.

 

Edited February 24, 2019 by michel123456

[taeto]

33 minutes ago, michel123456 said:

From my understanding if one keeps your ix⋅iy=i(xy )

for x=i and y=i you get i(xy) = i^2

where i(xy) is a notation meaning the multiplication of ix with iy under the i system.

We were discussing imaginary numbers, weren't we? So the imaginary numbers ix and iy have imaginary parts x and y that are real numbers. It does not make sense to have imaginary real parts x and y.

[michel123456]

32 minutes ago, taeto said:

We were discussing imaginary numbers, weren't we? So the imaginary numbers ix and iy have imaginary parts x and y that are real numbers. It does not make sense to have imaginary real parts x and y.

You lost me here.

To make things simpler, I imagined that in the imaginery system the rule for multiplication is reversed: the multiplication of 2 positives give a negative. I guess that means that addition gets reversed too.

[taeto]

If was thinking that if ix is an imaginary number, then x is a real number? I am mistaken? What is the range of values that you allow for x, all imaginary numbers, all complex numbers?

[michel123456]

26 minutes ago, taeto said:

If was thinking that if ix is an imaginary number, then x is a real number? I am mistaken? What is the range of values that you allow for x, all imaginary numbers, all complex numbers?

You are not mistaken.

for example i2 is placed at the second interval of the imaginary number line, then comes i3, i4, a.s.o.

the prefix i is simply a statement that the number is imaginary.

that i3.i3=-9

and that i1.i1 = i.i =i^2=-1

[taeto]

2 hours ago, michel123456 said:

The point is to keep i^2=-1

From my understanding if one keeps your ix⋅iy=i(xy )

for x=i and y=i you get i(xy) = i^2

You call ix imaginary when x=i? 

[michel123456]

31 minutes ago, taeto said:

You call ix imaginary when x=i? 

Sorry

I messed it.

when x=1, you get i (unity) on the x axis

when y=1 you get i on the y axis

the product xy, that is to say what you noted i(xy) is equal to i(i^2) = i^2 (the blue square)

Edited February 24, 2019 by michel123456

[taeto]

6 minutes ago, michel123456 said:

Sorry

I messed it.

when x=1, you get i (unity) on the x axis

when y=1 you get i on the y axis

the product xy, that is to say what you noted i(xy) is equal to i(i^2) = i^2 (the blue square)

When x = 1 and y = 1, then the product of the imaginary numbers ix and iy is -xy = -1 when they are interpreted as complex numbers. The product is xy = 1 when you apply the revised product, and that is not i^2.  

[michel123456]

30 minutes ago, taeto said:

When x = 1 and y = 1, then the product of the imaginary numbers ix and iy is -xy = -1 when they are interpreted as complex numbers. The product is xy = 1 when you apply the revised product, and that is not i^2.  

So you are saying that the segments on my diagram are complex numbers, is that correct?

I have lost something, i have to read this thread back again.

The intent was to show imaginary numbers on the axes.

 

[taeto]

The red square in your diagram should be the mirror image of the blue square, e.g. (-i)^2 = i^2, and in general (-ix)(-iy) = (ix)(iy), for real numbers x,y, when you assume complex multiplication. All points in your diagram are pairs (ix,iy), and the sign of the product (ix)(iy) is the exact opposite of the sign of the product xy. The diagram is fine, except for the colors being misleading.