Understanding the Ordering of Complex Numbers: Why is it Difficult to Define?

[Spacetectonik]

All,

Could some one tell me please ,why in natural numeric system we are not allowed to move from number one to three? I mean ,is this a properties of the natural numbers to be in sequence?
for instance if we count base on an imaginary numeric system which allows us to shift between numbers on no sequence,is this going to volatile the rule?

Cheers

 

[SteamKing]

Without going all rigorous on you, the natural numbers possesses a certain order which complex numbers lack. By inspection, a sequence of natural numbers a, b, c can be put into a certain order such that a < b < c. For a sampling of random complex numbers, a+bi, c+di, e + fi, these cannot be ordered in a manner similar to the natural numbers.

 

[tadchem]

The "nomenclature" is arbitrary; the arithmetic is physical.
"God made the integers; all else is the work of man." - Leopold Kronecker
Epistemologically the definitions may be as described below:
COMPLEMENTS:
The opposite of "not is" (count "zero") is called "is" (count "one").
COMBINATION:
"One" and "one", combined, are called "two."
EXTENSION:
Every number is obtainable by adding "one" to the next smaller number and/or by removing "one" from the next larger number.
INVERSION:
Removing (de-combining) "one" from "two" leaves "one."
CONTINUATION:
Combining the largest number you have ("two") with "one" produces the next largest number (arbitrarily called "three", in English, but it's properties are separate from its nomenclature).

 

[Spacetectonik]

Thanks ,

Just another stupid question please?!
What do you call a numeric system(if you know any) which works base on the changes on degrees?(changing of the angle?!)

Cheers

PS: Thank to be nice to me king.

 

[tadchem]

Thanks ,

Just another stupid question please?!
What do you call a numeric system(if you know any) which works base on the changes on degrees?(changing of the angle?!)

Polar coordinates.

 

[Spacetectonik]

Polar coordinates.

Thanks, For reply!
But Polar.C is a two dimensional system?! I am sort of thinking about a three dimensional numeric system (degree based),If anyone have heard of such a system! it would not be real number to I reckon.

Cheers

 

[Michael Redei]

Polar coordinates are used in three dimensions as well. One example you surely know begins by describing a place on the surface of the earth, by saying what angle north or south of the equator the place is, plus the angle east or west of the Greenwich meridian. The north/south angle (the "latitude") will be in the range from -90° (south pole) to +90° (north pole), and the east/west angle (the "longitude") will range from -180° to +180°.

If you want to specify any point in space, connect it to the centre of the earth, and you get a meeting point on the Earth's surface. That point's latitude/longitude coordinates, plus the distance from your point in space to the centre of the Earth are the "polar coordinates" of your point.

 

[tadchem]

If you want a coordinate system measured ONLY in degrees, use only the angle from polar coordinates on the circumference of a circle, or spherical coordinates on the *surface* of a sphere - latitude and longitude.
Of course, degrees give you no sense of *scale*. The size (distance) of a degree of longitude will be different at different latitudes.

 

[Diffy]

Without going all rigorous on you, the natural numbers possesses a certain order which complex numbers lack. By inspection, a sequence of natural numbers a, b, c can be put into a certain order such that a < b < c. For a sampling of random complex numbers, a+bi, c+di, e + fi, these cannot be ordered in a manner similar to the natural numbers.

Can you, or someone else explain why this is so?

I'll attempted to define an order, and hopefully someone will tell me why it is bad.

Consider all circles in the complex plane centered at the origin union the origin itself. Call the origin 0.

We define the mapping R for any z = a+bi to be 0 if z = 0
√(a^2 + b^2) for a > 0
-√(a^2 + b^2) for a < 0
and √b^2 for a = 0 b > 0
and - √b^2 for a = 0 b <0

Next we define the mapping T for any z = a+ bi to be 0 if z = 0

T will be the positive angle moving counter clockwise of a complex point with respect to the negative imaginary axis when a is positive.

T will be the negative angle moving clockwise of a of a complex point with respect to the negative imaginary axis when a is negative.

Now let z1 = a + bi and z2 = c + di

we can say a + bi > c + di

if R(z1) > R(z2) or if R(z1) = R(z2), T(z1) > T(z2)

and a + bi = c + di IFF R(z1) = R(z2) and T(z1) = T(z2)I think this definition of order preserves the order of the real numbers.

 

[Michael Redei]

... and a + bi > c + di IFF R(z1) = R(z2) and T(z1) = T(z2)

Set a=b=c=d=0, then you have z1 = z2, and so R(z1)=R(z2) and T(z1)=T(z2). According to your definition this means z1>z2, but also z2>z1.

Before you try to fix this problem, you might consider this argument:

If you are going to define an ordering at all, you must be able to say of any element z, whether z>0 ("z is positive"), z<0 ("z is negative") or z=0 ("z is zero"). Also, you need the simple rule "minus times minus is plus" to hold for any numbers, or simply "z²>0 for all z≠0". Now look at i, that is one of two complex numbers whose square is 1. (The other one is -i.) Whether i is defined to be positive or negative, i²=-1 is negative, which is impossible for an ordering of all complex numbers.

If you look at this argument, you'll see that it holds for any set of number-like elements that include an element whose square is -1. That's why I explained what I meant by "i", although you knew what that was as a complex number.

 

[jbriggs444]

The ordering given by Diffy appears to be a valid total ordering of the complex numbers _as a set_ which preserves the conventional ordering of the real numbers within that set. That part is no problem.

The difficulty is giving an ordering of the complex numbers _as an ordered field_. i.e. in such a way that the familiar properties of the real or rational numbers with respect to order are preserved.

For instance:

If a > b then a + k > b + k
If a > 0 and b > 0 then ab > 0
If a > 0 then 0 > -a

For a complete ordered field add:

If S is a non-empty set that is bounded above then S has a least upper bound

The classic problem with the complex numbers is that if i is considered to be positive then i*i < 0 which violates the above rules. But if i is considered to be negative then -i*-i < 0 which again violates the above rules.

 

[Diffy]

Set a=b=c=d=0, then you have z1 = z2, and so R(z1)=R(z2) and T(z1)=T(z2). According to your definition this means z1>z2, but also z2>z1.

This was a typo. I editted my original. Thanks!

Before you try to fix this problem, you might consider this argument:

If you are going to define an ordering at all, you must be able to say of any element z, whether z>0 ("z is positive"), z<0 ("z is negative") or z=0 ("z is zero"). Also, you need the simple rule "minus times minus is plus" to hold for any numbers, or simply "z²>0 for all z≠0". Now look at i, that is one of two complex numbers whose square is 1. (The other one is -i.) Whether i is defined to be positive or negative, i²=-1 is negative, which is impossible for an ordering of all complex numbers.

If you look at this argument, you'll see that it holds for any set of number-like elements that include an element whose square is -1. That's why I explained what I meant by "i", although you knew what that was as a complex number.

Thanks for more explanation.

The ordering given by Diffy appears to be a valid total ordering of the complex numbers _as a set_ which preserves the conventional ordering of the real numbers within that set. That part is no problem.

The difficulty is giving an ordering of the complex numbers _as an ordered field_. i.e. in such a way that the familiar properties of the real or rational numbers with respect to order are preserved.

For instance:

If a > b then a + k > b + k
If a > 0 and b > 0 then ab > 0
If a > 0 then 0 > -a

For a complete ordered field add:

If S is a non-empty set that is bounded above then S has a least upper bound

The classic problem with the complex numbers is that if i is considered to be positive then i*i < 0 which violates the above rules. But if i is considered to be negative then -i*-i < 0 which again violates the above rules.

Ahhhh this makes a lot of sense in combination to what Mike said. Thanks!