Why Complex Numbers Are Not Ordered

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}.


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.


0 \le -1, a contradiction again.

Therefore, the complex numbers do not possess an order.


Prove the Triangle Inequality Algebraically

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)


|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


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:


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.


(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.

Therefore by (1),

|z_1 + z_2|^2 \le (|z_1|+|z_2|)^2.

By the fact that |z_1 + z_2| \ge 0, |z_1|+|z_2| \ge 0 and (\forall x \ge 0, y \ge 0, x^2 \ge y^2\implies x \ge y),

|z_1+z_2| \le |z_1| + |z_2|.

Exercise-1 Show that \forall x \in \mathbb{R}, |x| \ge x.

Exercise-2 Show that \forall x, y \ge 0, x \ge y \implies x^2 \ge y^2.

Exercise-3 Show that \forall x, y \ge 0, x \ge y \implies \sqrt{x} \ge \sqrt{y}.