B Inequalities of complex number

[Einstein's Cat]

I am under the impression that the following cannot be stated,
a < b, if the a term is a complex number and the b term is either a natural number or a complex number, or any other type of number for that matter.

Firstly am I correct? Secondly, if I am, does there exist a theorem of some sort that proves this statement here? Finally, can this be generalised to all inequalities?

 

[PeroK]

I am under the impression that the following cannot be stated,
a < b, if the a term is a complex number and the b term is either a natural number or a complex number, or any other type of number for that matter.

Firstly am I correct? Secondly, if I am, does there exist a theorem of some sort that proves this statement here? Finally, can this be generalised to all inequalities?

The complex numbers form a plane not a line, so are not ordered. You cannot line them all up in order, in other words.

 

[Einstein's Cat]

The complex numbers form a plane not a line, so are not ordered. You cannot line them all up in order, in other words.

Yet say in the plane there are two points, a and b (or two complex numbers) and they are both points on the same line. Could not one say that the point further from another point on the line (point c) towards the positive quadrant is greater than the other point?
Analogous to the point (6,6) being further from the origin than point (1,1) on the line y = x; therefore 6 > 1.

 

[PeroK]

Yet say in the plane there are two points, a and b (or two complex numbers) and they are both points on the same line. Could not one say that the point further from another point on the line (point c) towards the positive quadrant is greater than the other point?
Analogous to the point (6,6) being further from the origin than point (1,1) on the line y = x; therefore 6 > 1.

At a basic level, the complex numbers are not ordered, in the sense that there is not a total order. The real question is what properties you expect $<$ to have. See, for example:

https://en.wikipedia.org/wiki/Total_order

You could order the complex numbers by magnitude, by real part, by imaginary part, but in all cases you get many different numbers that cannot be put in order. They are all the "same" when it comes to comparing them.

You can also do it in ways that not all numbers are "comparable". For example, if you only compare numbers with the same imaginary part, then you have a partial order. That means that you can't compare $2 + i$ with $3 - i$, but you could say that $2 + i < 3 + i$.

See:

https://en.wikipedia.org/wiki/Partially_ordered_set

 

[mfb]

You can define ">" for complex numbers in many ways, but none of those definitions is useful enough to be used widely.

 

[fresh_42]

The short answer is: $\; a > 0 \,\wedge \, b>0 \Rightarrow ab>0$ is one of the defining conditions for an ordering on a field. (Definition in my copy of van der Waerden's Algebra I)

The shorter is: All squares have to be positive in ordered fields.

The shortest is: $\; i^2 < 0$.

 

[FactChecker]

I am under the impression that the following cannot be stated,
a < b, if the a term is a complex number and the b term is either a natural number or a complex number, or any other type of number for that matter.

Firstly am I correct? Secondly, if I am, does there exist a theorem of some sort that proves this statement here? Finally, can this be generalised to all inequalities?

With the meaning of '<' that you are probably thinking of, you are correct.

When you ask about "generalizing to all inequalities", there are more general uses of the symbol '<' in what are called 'preorders' that can be applied to the complex numbers. For instance, a < b iff |a| < |b| is a preorder. That would put two complex numbers in the same equivalance class if their modulus was equal. (see https://en.wikipedia.org/wiki/Preorder ) There are many other preorders that can be defined on the complex numbers. I assume that this was not what you had in mind. It does not satisfy the antisymetric property: a ≤ b and b ≤ a implies a = b.