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November 17th, 2011, 10:44 AM   #1
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Complex numbers and ordering

It is thought that complex numbers unlike real numbers don't have ordering. But it seems that even though individual complex numbers don't have ordering, sets of complex numbers do. On the basis of their magnitude.

So, instead of having a complex line, like the real line, there is complex circle that determines the magnitudes of complex numbers. Of course as, by definition, a circle ,unlike a point, has dimensions, the complex circle differentiates between infinite sets of complex numbers.

Is this idea valid?
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November 17th, 2011, 01:14 PM   #2
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Re: Complex numbers and ordering

Quote:
It is thought that complex numbers unlike real numbers don't have ordering.
[color=#000000]
It is not thought, it is so. Take the positive half of the complex plane let us say , meaning , then , but there exists an element with which is impossible for a "positive" element.[/color]
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November 17th, 2011, 01:17 PM   #3
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Re: Complex numbers and ordering

This is trivially valid if you define magnitude in the ordinary way, since real numbers are well-ordered.
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November 17th, 2011, 01:30 PM   #4
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Re: Complex numbers and ordering

Is there any other way to define complex magnitude other than using the pythagorean theorem on the complex plane? I mean any non-equivalent way?
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November 18th, 2011, 11:38 AM   #5
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Re: Complex numbers and ordering

Quote:
Originally Posted by Eureka
It is thought that complex numbers unlike real numbers don't have ordering. But it seems that even though individual complex numbers don't have ordering, sets of complex numbers do. On the basis of their magnitude.

So, instead of having a complex line, like the real line, there is complex circle that determines the magnitudes of complex numbers. Of course as, by definition, a circle ,unlike a point, has dimensions, the complex circle differentiates between infinite sets of complex numbers.

Is this idea valid?
Circles, with center at 0 in the complex plane, lie on numbers with the same absolute value. It sounds to me like you are ordering by the absolute value which is a real number.
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November 19th, 2011, 03:46 PM   #6
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Re: Complex numbers and ordering

Yes, but that's not my point. My point is that complex numbers provide ordering in 2D, that of a plane, that could more simply be done through circles, a thing that they do.
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November 20th, 2011, 11:25 AM   #7
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Re: Complex numbers and ordering

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Originally Posted by Eureka
Yes, but that's not my point. My point is that complex numbers provide ordering in 2D, that of a plane, that could more simply be done through circles, a thing that they do.
[color=#000000]What you are trying prove is wrong, "ordering" does not apply to complex numbers![/color]
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November 21st, 2011, 05:36 AM   #8
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Re: Complex numbers and ordering

First, it is NOT true that the complex number cannot be ordered. We can say that a+ bi< c+ di if a< c or, when a= c, b< d.

What is true is that the complex numbers are not an "ordered field". That is, there is no way to define an order such that if a< b then a+ c< b+ c and if 0< c, then ac< bc.

As for your ordering sets of complex numbers, which of A= {a+ bi| 1< a< 2, 1< b< 2} and C= {a+ bi| -2< a< -1, -2< b< -2} is the smaller?
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November 21st, 2011, 05:47 AM   #9
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Re: Complex numbers and ordering

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Originally Posted by HallsofIvy
First, it is NOT true that the complex number cannot be ordered. We can say that a+ bi< c+ di if a< c or, when a= c, b< d.
[color=#000000]what???????
where did you see that, can you name the book where it is written? Let for example z=a+bi for a,b>0, according to what you are writting you can say that ? Where did you see it?[/color]
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November 21st, 2011, 10:31 PM   #10
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Re: Complex numbers and ordering

Quote:
Originally Posted by HallsofIvy
First, it is NOT true that the complex number cannot be ordered. We can say that a+ bi< c+ di if a< c or, when a= c, b< d.
Isn't that like saying (3, 4) is greater than (2, 7)?
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