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On Ordered relations in the set of complex numbers.

The imaginary numbers were defined to solve equations like
x^2+1=0.

The imaginary unit is defined as
i^2= -1. And any other imaginary number can be written as a product of a real number and the imaginary unit i.

Now, going a bit back to the development of the number system, the following problem was encountered:
x+2=1.

As this equation had no solution in the then existing positive natural number system {1, 2, 3, ...} , a negative unit -1 was defined. Further any equation of the form x+a=b (a>b) could be solved and the solution is the product of the negative unit and the difference of a and b(i.e., a~b).

Ordered relations in complex numbers are undefined because one has no idea about the nature and magnitude and behavior of the imaginary unit i.

Similarly there is no physical or intuitive understanding of the number -1.It is as good as or as bad as the square root of negative one. It was defined to be less than zero, and as the positive number multiplied to -1 increased, we said that the negative number's magnitude decreased. And thus an ordered relation was defined on the set of positive and negative integers. There is no absolute thought or truth regarding the negative numbers. It was simply defined to extend the number system to solve the then unsolvable equations with the then existing number system.

Now, my argument is as follows. Is it wrong to define an ordered relation in complex numbers as follows:

i.b > i.c iff b>c ,

when one can define

-b < -c iff b > c > 0 , where the negative unit -1 is equally non-intuitive and not obvious, and there is no absolute truth regarding the behavior of -1?

That is, to be concise, when we have defined a number such as -1, and assigned ordered relations to it which is not obvious, what is wrong in defining similar ordered relations on imaginary numbers?

This was a question that struck my mind. My argument presumably may be an obviously wrong idea.

Thanks in advance.

Note by Samarth M.O.
4 years ago

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Think about how the usual ordering on real numbers is related to addition and multiplication. Would it be possible to extend the ordering to complex numbers in such a way that the properties of multiplication, which you know and love, would continue to hold for all complex numbers? John Smith Staff · 4 years ago

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@John Smith No the complex number cannot be an ordered field because i^2=-1. Samuel Queen · 4 years ago

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There is this system called lexicographical ordering- that is, a+bi<c+di if and only if a<c or a=c and b<d. This order makes the complex plane totally ordered set. However, you cannot have least upper bound for the subset of the complex plane under this order if your subset does not have a bounded imaginary parts. Okay Nho · 4 years ago

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@Okay Nho To add more, there is an axiom called axiom of choice which is equivalent to the statement that you can define an order to any arbitrary set. The problem is that the statement only tells us that an order exists, but does not imply properties of the order. Okay Nho · 4 years ago

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What would it mean to be positive in the complex plane? Is \( i > 0\)? You should run into reductio ad absurdum pretty quickly no matter how you answer that question, but I'll let you have the fun of seeing it for yourself. Eric Edwards · 4 years ago

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