June 2022

In mathematics, if a structure is a field and has an order $\le$ , two additional axioms need to hold for it to be an ordered field. These axioms express the notion that the ordering is compatible with the field structure:

1) For any three $a,b,c$ elements, $a \le b$ implies $a+c \le b + c.$

2) For any two $a, b$ elements, $0 \le a$ and $0 \le b$ implies $0 \le a\cdot b.$

We can show that such an order does not exist in the set of Complex number.

Let us restate the two axioms:

$\forall a,b,c, (a \le b) \implies a+c \le b+c.\quad\quad\quad(1)$

$\forall a,b, (0 \le a, 0 \le b) \implies 0 \le a\cdot b.\quad\quad\quad(2)$

If $0 \le \bold{i}$ , we write

$0\le \underbrace{\bold{i}}_{a}, 0\le \underbrace{\bold{i}}_{b}$ .

By (2),

$0 \le \underbrace{\bold{i}}_{a}\cdot \underbrace{\bold{i}}_{b}$ .

i.e.,

$0 \le \bold{i}^2 \implies 0 \le -1,$ a contradiction.

Otherwise ( $\bold{i} \le 0$ ), we have

$\underbrace{\bold{i}}_{a} \le \underbrace{0}_{b}$ .

By (1),

$\underbrace{\bold{i}}_{a} + \underbrace{(-\bold{i})}_{c} \le \underbrace{0}_{b} + \underbrace{(-\bold{i})}_{c} \implies 0 \le -\bold{i}.$

We write it as

$0 \le \underbrace{-\bold{i}}_{a}, 0 \le \underbrace{-\bold{i}}_{b}.$

By (2),

$0 \le \underbrace{(-\bold{i})}_{a}\cdot\underbrace{(-\bold{i})}_{b} = \bold{i}^2 = -1.$

i.e.,

$0 \le -1$ , a contradiction again.

Therefore, the complex numbers do not possess an order.

Prove: $\forall z_1, z_2 \in \mathbb{C}, |z_1 + z_2| \le |z_1| + |z_2|.\quad\quad\quad(*)$

Observe that

$|z_1+z_2|^2 = (z_1+z_2)\overline{(z_1+z_2)}$

$= (z_1+z_2)(\overline{z_1}+\overline{z_2})$

$= z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1}+z_2\overline{z_2}$

$= |z_1|^2+z_1\overline{z_2} + z_2\overline{z_1}+|z_2|^2.$

That is

$|z_1+z_2|^2 =|z_1|^2+z_1\overline{z_2} + z_2\overline{z_1}+|z_2|^2, \quad\quad\quad(1)$

or,

$|z_1+z_2| = \sqrt{|z_1|^2+z_1\overline{z_2} + z_2\overline{z_1}+|z_1|^2}.$

It means to prove (*) is to show that

$\sqrt{|z_1|^2+z_1\overline{z_2} + z_2\overline{z_1}+|z_2|^2} \le |z_1| + |z_2|.\quad\quad\quad(2)$

Let’s explore a bit.

Squaring (2), we obtain

$|z_1|^2+z_1\overline{z_2} + z_2\overline{z_1}+|z_2|^2 \le |z_1|^2+2|z_1||z_2|+|z_2|^2$

or,

$z_1\overline{z_2} + z_2\overline{z_1} \le 2|z_1z_2|=2|z_1||z_2|.\quad\quad\quad(3)$

Let $z_1=a+b\bold{i}, z_2=c+d\bold{i}.$

After compute $z_1\overline{z_2} + z_2\overline{z_1}$ using Omega CAS Explorer (See Fig. 1),

Fig. 1

we express (3) as:

$2bd + 2ac \le 2\sqrt{a^2+b^2}\sqrt{c^2+d^2}.$

It simplifies to

$bd + ac \le \sqrt{a^2+b^2}\sqrt{c^2+d^2}.$

If $bd+ac \ge 0$ , then squaring both sides of $\le$ and expanding the right side,

$a^2c^2+2abcd+b^2d^2 \le a^2c^2+a^2d^2+b^2c^2+b^2d^2.$

Simplifying it yields

$2abcd \le a^2c^2+b^2d^2.$

This is true for all $a,b,c,d \in \mathbb{R}.$

We now prove (*) as follows:

Let

$z_1 = a+b\bold{i}, z_2=c+d\bold{i}.$

We have

$2abcd \le a^2d^2 + b^2c^2.$

Adding $a^2c^2, b^2d^2$ on both sides of $\le$ ,

$\underline{a^2c^2} + \underline{b^2d^2} +2abcd \le \underline{a^2c^2}+\underline{b^2d^2}+a^2d^2+b^2c^2.$

i.e.,

$(ac+bd)^2 \le (a^2+b^2)(c^2+d^2).$

Taking the square root on both sides,

$|ac+bd| \le \sqrt{(a^2+b^2)(c^2+d^2)}.$

We know $\forall x \in R, x \le |x|.$ Therefore, $ac+bd \le |ac+bd|.$ And so,

$ac+bd \le \sqrt{a^2+b^2}\sqrt{c^2+d^2}.$

Multiplying $2$ on both isdes,

$2(ac+bd) \le 2(\sqrt{a^2+b^2}\sqrt{c^2+d^2}).$

That is,

$2ac + 2bd \le 2\sqrt{a^2+b^2}\sqrt{c^2+d^2}.$

From Fig. 1: $z_1\overline{z_2} + z_2\overline{z_1}=2ac+2bd$ and, $|z_1|=\sqrt{a^2+b^2}, |z_2| = \sqrt{c^2+d^2}.$ Hence, we have

$z_1\overline{z_2} + z_2\overline{z_1} \le 2|z_1||z_2|.\quad\quad\quad(4)$

Adding $|z_1|^2, |z_2|^2$ on both sides of $\le,$

$\underline{|z_1|^2}+z_1\overline{z_2}+z_2\overline{z_1} + \underline{|z_2|^2} \le \underline{|z_1|^2}+2|z_1||z_2|+\underline{|z_2|^2}\overset{(1)}{=}(|z_1|+|z_2|)^2.$