Can Complex Numbers Be Ordered?

In summary, the complex numbers cannot be ordered in the same way as the real numbers due to the lack of an Archimedean order. However, they can be ordered using a lexicographical order or by using a bijective function with the real numbers. These orders may not have the same properties as the real numbers, such as the axiom of totality or well-ordering.
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Arman777
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Can we order Complex Numbers ? I searched a bit most places says it can but not like the real numbers. I am confused a bit.And I am not sure abouth the truth of those sources.
Thanks
 
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  • #2
We can order every set. But the question is will such an order have any relation to a complex structure or even to linear or topological structure in ##\mathbb{R}^2## :)
 
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  • #3
If I recall correctly, the way it works is that if you try impose an order in the complex numbers, you then lose the field properties.
 
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Arman777 said:
Can we order Complex Numbers ? I searched a bit most places says it can but not like the real numbers. I am confused a bit.And I am not sure abouth the truth of those sources.
Thanks
The complex numbers don't allow an Archimedean order. This is equivalent to the condition that squares are positive, which is not the case for complex numbers, as ##i^2=-1<0##. They allow however an order like the lexicographical order: ##x+iy < u+iv \Longleftrightarrow x < u \,\vee \, (x=u \wedge y < v)##.
 
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If you mean ordering like we think of ordering with real numbers, the answer is no. To demonstrate this, consider this example that uses an informal argument.

If ##\mathbb{C}## is a totally ordered set with properties like the real numbers, we have the axiom of totality, which says that ##\forall x,y \in \mathbb{C}: x \leq y## or ## x \geq y##

Thus, consider ##i \leq 0##, then multiply both sides with ##i##, obtaining ##-1 \geq 0## (multiplying with a number less than ##0## reverses the inequality sign). This is not possible.
Otherwise, assume ##i \geq 0##, then ##-1 \geq 0##. Also impossible.
 
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I understand thanks
 
  • #7
You can find a bijective function between the complex numbers and the real numbers and then use the ordering of the real numbers to "order" the complex numbers. That won't lead to nice properties of the ordering, but it is some sort of order, similar to the one @fresh_42 suggested.

You can even have a well-ordering, which places stronger conditions on an order.
 

Related to Can Complex Numbers Be Ordered?

1. What are complex numbers?

Complex numbers are numbers that consist of a real part and an imaginary part, usually written in the form a + bi, where a and b are real numbers and i is the imaginary unit (i.e. the square root of -1).

2. Why do we need complex numbers?

Complex numbers are useful in representing quantities that involve both real and imaginary components. They are commonly used in fields such as engineering, physics, and mathematics to solve problems that cannot be solved with real numbers alone.

3. Can we order complex numbers?

Yes, we can order complex numbers. They can be compared using the same rules as real numbers, where the real part is compared first, and if they are equal, then the imaginary part is compared.

4. How do we perform operations on complex numbers?

To add or subtract complex numbers, we simply add or subtract the real and imaginary parts separately. To multiply complex numbers, we use the FOIL method (First, Outer, Inner, Last) and simplify using the fact that i squared equals -1. To divide complex numbers, we use the conjugate method.

5. Can we graph complex numbers?

Yes, we can graph complex numbers on a coordinate plane called the complex plane. The horizontal axis represents the real part, while the vertical axis represents the imaginary part. This allows us to visualize and better understand the properties of complex numbers.

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