A comparison scheme for complex numbers (or vectors in general)

Question

Below I am stating a possible ordering scheme for complex numbers and n-dimensional vectors in general. I am vaguely aware that such kind of thigs exist in Mathematics, however I have put it as a number code towards the end, which I am not sure has been done earlier. I would like to know that overall my idea is mathematically acceptable, or I am missing some fundamental mathematical aspects?

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On a number line, we compare real numbers by their position (left, equal or right). In essence, comparison between real numbers $(<,= \text{or} >)$ is just the comparison of their relative positions on the number line.

Now, going to complex numbers, we say that it is not possible to compare two complex numbers mathematically, like we do for real numbers. This is because they are made up of two components.

But, if we go back to the real number line, we observe that we are simply comparing the position of the points representing those numbers on the number line.

Extending this idea to the complex plane, we can have a positional comparison of two complex numbers.

A complex number has to be in one of the nine different relative positions with respect to another complex number. There is no other possibility. Just as two points can have only three relative positions on a number line, they can have only nine relative positions in the plane. These are as follows:

• Same place (1)
• Directly above/below (2, 3)
• Directly left/right (4, 5)
• To the right and above (6)
• To the right and below (7)
• To the left and above (8)
• To the left and below (9)

These nine positions correspond to nine combinations of component-wise inequalities for two complex numbers.

Let us say we have two complex numbers:

• $a + ib$
• $c+ id$

Then the above nine possibilities correspond to the nine situations below:

• a=c, b=d (1)
• a=c with (b$<$d or b$>$d) (2,3)
• (a$>$c or a$<$c) with b=d (4,5)
• a$>$b, c$>$d (6)
• a>b, c<d (7)
• a<b, c>d (8)
• a<b, c<d (9)

Now complex numbers are just vectors in the form (a, b). So this idea can be extended to higher dimensions, vectors having $n$ components. The number of relative positions in n dimensions will be $3^n$.

Now, if we assign < as 0, = as 1, > as 2, then for a pair of complex numbers, we can have a unique doublet comparing component-wise. e.g. if a$<$b and c$>$d, we can say z1 [<>] z2, or convert it to a ternary number z1 [02] z2, or convert that ternary to decimal: z1 [2] z2.

We can circle the central number on paper. This final decimal number will exactly indicate the relative position between two complex numbers. The same idea can be extended to n-dimensional vectors.

Answer 1

The thing is, any set can be ordered. There are infinitely many possible orderings of the complex numbers. Sometimes, which ordering to work with in a given situation depends on the needs.

I'm addressing what you quote "it is not possible to compare two complex numbers mathematically like we do for real numbers".

What people who say this mean is that, with complex numbers as a field (the algebraic structure it constitutes with addition and multiplication), there is no order relation on $\mathbb C$ that is compatible with the algebraic structure, contrary to $\mathbb R$ (look up for ordered field).

Indeed, imagine you have one such ordering on $\mathbb C$. You would want it to satify the following compatibility properties for any complex numbers:

• Compatibility with addition: $x<y\implies x+z<y+z$;
• Compatibility with multiplication by nonnegative numbers: $(x<y \,\text{and}\,0\le z)\implies x+z\le y+z$;
• Total ordering: for any distinct $x,y\in \mathbb C$, either $x<y$ or $y<x$.

From compatibility with addition, it follows that $(a<b\,\text{and}\,c<d)\implies a+c<b+d)$. Indeed, $a+c<b+c<b+d$.

One key property: such an ordering must satisfy $-1<0$. Indeed, if $0<-1$, then $0\times(-1)<(-1)^2$, i.e., $0<1$, thus $0+0<(-1)+1$, i.e., $0<0$, absurd.

Another consequence is that $x<y\implies -x>-y$. Indeed, if $x<y$ and $-x\le -y$, then $x-x<y-y$, i.e., $0<0$, absurd.

Now, the order being total, you have either $0<i$ or $i<0$.

1. Assume $0<i$. This means $i$ is positive. Thus $0\times i<i^2$, i.e., $0<-1$, which is false.
2. Assume $i<0$. Then $-i>0$. Again $(-i)^2>0\times (-i)$, i.e., $-1>0$, which is false.

This show that no matter what ordering you choose for $\mathbb C$, you will not have nice compatibility properties with the algebraic operations on $\mathbb C$.