Can complex numbers be ordered in the way of Pythagorean triples?

Question

This post did not quite seem to answer my question. If we have $\quad a+bi\in\mathbb{C}\quad$ can ordering be done the way of Pythagorean triples $\quad A^2+B^2=C^2?\quad$

There are infinite triples for each of $\quad (C-B),\space (C-A),\space\text{and}\space (B\pm A)\quad$ but these may be further ordered as sets of sets

$ (C-B)=(2n-1)^2\in\big\{1,3,5,\cdots\big\} $

$ (C=A)=2k^2\in\big\{2,8,18,\cdots\big\} $

\begin{align*} &(B\pm A)=P:\\ &P=p_1,\cdots ,p_n),\space p_k\equiv \pm 1 \text{ (mod }8) \in\big\{1,7,17, \cdots\big\} \end{align*}

For each of $(A\text{ or } B\space\text{ or } \space C),\space$ there are $2^{n-1}\space$ primitive triples where $\space n\space $ is the number of unique prime factors of $\space(A\text{ or } B\space\text{ or } \space C)\space $ and these may be of vastly different sizes. Perhaps these may be ordered by the pairs of natural numbers $\space(n,k)\space$ that generate them in this formula

\begin{align*} A=(2n-1)^2+ & 2(2n-1)k \tag{a} \\ B= \qquad\qquad\quad & 2(2n-1)k+ ]] 2k^2 \tag{b} \\ C=(2n-1)^2+ & 2(2n-1)k+ 2k^2 \tag{c} \end{align*}

where $\space n\space $ is a set number and $\space k\space$ is the ordinal triple within each set. For example, $C=65\space$ represents triples $\space f(4,1)=(63,16,65)\quad f(5,4)=(33,56,65).\space$ where $\space(63,16,65)\space$ is the first member of $Set_4$ and $\space(33,56,65)\space$ is the fourth member of $Set_5.$

Can $\space (4+1i),\space (5+4i)\space$ be treated as natural ordering of these complex numbers or are their other considerations such as the size of the resulting squares or products to consider?

Answer 1

There are an infinite number of ways to order the complex numbers as a set.

There is absolutely no way to order the complex numbers as a field.

I didn't bother to read your explanation of the ordering of pythogorean triples but it seems similar to "lexigraphic ordering".

For two complex number as numbers $w,z$ we can define the lexigraphic order as:

1. If $Re(w) < Re(z)$ then $w \prec z$
2. If $Re(w) = Re(z)$ but $Im(w) < Im(z)$ then $w\prec z$.
3. Otherwise $w \not \prec z$.

(In other words: If $w = a+bi; z = c+di; a,b,c,d \in \mathbb R$ to compare $w$ to $z$ compare $a$ to $c$. If $a = c$ then compare $b$ to $d$.)

That satisfies everything we need to have an ordered set. To be an order on a set we need:

1. Given complex $w$ and $z$ then exactly one of the following is true: Either $w \prec z$ or $z\prec w$ or $w = z$.

2. if $w\prec z$ and $z\prec \alpha$ then it always follows that $w \prec \alpha$.

It's easy to see that those are both always true. So $\prec$ is an order on $\mathbb C$ as a set.

......

But it does not satisfy an order on a field.

To be an order on a field we need to be able to do math on these numbers and have the resulting inequalities be preserved. We need:

3. If $w \prec z$ then for every $a$ we have $w + a \prec z+a$.

This is actually true.

and

4. If $w \prec z$ and $0 \prec a$ then $aw \prec az$.

This fails. This fails big time!

Let $w = 4 - 6i$ and $z = 5-3i$ and $a = i$.

Then $4 - 6i \prec 5-3i$ and $0\prec i$ but $(4-6i)i = 6 + 4i \not\prec 3 + 5i = (5-3i)i$.

.....

Now having an order as a field is impossible.

Why. Consider any $w$ in a field. We either have $w > 0$ or $w = 0$ or $w< 0$.

If $w > 0$ then $w^2 = w\times w > 0$. If $w=0$ then $w^2 = 0$. And if $w < 0$ then $0 =w +(-w) < 0 + (-w) =-w$ and so $w^2 = (-w)\cdot (-w) > 0$.

So we have $w^2 \ge 0$ and we never have $w^2 < 0$.

But in the complex we have $1^2 = 1$ so we would have to have $1^2 = 1 > 0$. Then we'd have to have $0 < 1$ so $-1 =0+(-1) < 1 +(-1) = 0$ so $-1 < 0$.

But we also have And we have $i^2 = -1$ so we would HAVE to have $i^2 = -1 > 0$. So we have BOTH $-1 < 0$ AND $-1 > 0$. That's utterly impossible.

So the complex numbers can not be ordered as a field.