Complex Numbers cannot be Totally Ordered
Theorem
Let $\left({\C, +, \cdot}\right)$ be the field of complex numbers.
There exists no total ordering on $\left({\C, +, \cdot}\right)$ which is compatible with the structure of $\left({\C, +, \cdot}\right)$.
Aiming for a contradiction, suppose there exists a relation $\preceq$ on $\C$ which is ordering compatible with the ring structure of $\C$.
That is:
- $(1): \quad z \ne 0 \implies 0 \prec z \lor z \prec 0$, but not both
- $(2): \quad 0 \prec z_1, z_2 \implies 0 \prec z_1 z_2 \land 0 \prec z_1 + z_2$
By Totally Ordered Ring Zero Precedes Element or its Inverse, $(1)$ can be replaced with:
- $(1'): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both.
As $i \ne 0$, it follows that:
- $0 \prec i$ or $0 \prec -i$
Suppose $0 \prec i$.
Then:
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\(\displaystyle 0\)
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\(\prec\)
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\(\displaystyle i \cdot i\)
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$\quad$ from $(2)$
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$\quad$
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\(\displaystyle \)
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\(=\)
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\(\displaystyle -1\)
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$\quad$ Definition of Complex Number
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$\quad$
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Otherwise, suppose $0 \prec \left({-i}\right)$.
Then:
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\(\displaystyle 0\)
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\(\prec\)
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\(\displaystyle \left({-i}\right) \cdot \left({-i}\right)\)
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$\quad$ from $(2)$
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$\quad$
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\(\displaystyle \)
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\(=\)
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\(\displaystyle -1\)
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$\quad$ Definition of Complex Number
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$\quad$
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Thus by Proof by Cases:
- $0 \prec -1$
Thus it follows that:
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\(\displaystyle 0\)
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\(\prec\)
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\(\displaystyle \left({-1}\right) \cdot \left({-1}\right)\)
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$\quad$ from $(2)$
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$\quad$
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\(\displaystyle \)
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\(=\)
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\(\displaystyle 1\)
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$\quad$
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$\quad$
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Thus both:
- $0 \prec -1$
and:
- $0 \prec 1$
This contradicts hypothesis $(1')$:
- $(1): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both
Hence, by Proof by Contradiction, there can be no such ordering.
$\blacksquare$
Aiming for a contradiction, suppose such a total ordering $\preceq$ exists.
By the definition of a total ordering, $\preceq$ is connected.
That is:
- $0 \preceq i \lor i \preceq 0$
Using Proof by Cases, we will prove that:
- $0 \preceq -1$
- Case 1
Assume that $0 \preceq i$.
By definition of an ordering compatible with the ring structure of $\C$:
- $\forall x, y \in \C: \left({0 \preceq x \land 0 \preceq y}\right) \implies 0 \preceq x \times y$
Substituting $x = i$ and $y = i$ gives:
- $0 \preceq i \times i$
Simplifying:
- $0 \preceq -1$
which is the result required.
- Case 2
Assume that $i \preceq 0$.
By definition of compatibility with addition:
- $\forall x, y, z \in \C: x \preceq y \implies \left({x + z}\right) \preceq \left({y + z}\right)$
Substituting $x = i$, $y = 0$, $z = -i$ gives:
- $i + \left({-i}\right) \preceq 0 + \left({-i}\right)$
Simplifying:
- $0 \preceq -i$
By definition of an ordering compatible with the ring structure of $\C$:
- $\forall x, y \in \C: \left({0 \preceq x \land 0 \preceq y}\right) \implies 0 \preceq x \times y$
Substituting $x = -i$ and $y = -i$ gives:
- $0 \preceq \left({-i}\right) \times \left({-i}\right)$
Simplifying:
- $0 \preceq -1$
This has been demonstrated to follow from both cases, and so by Proof by Cases:
- $0 \preceq -1$
$\Box$
By definition of an ordering compatible with the ring structure of $\C$:
- $\forall x, y \in \C: \left({0 \preceq x \land 0 \preceq y}\right) \implies 0 \preceq x \times y$
Substituting $x = -1$ and $y = -1$:
- $0 \preceq \left({-1}\right) \times \left({-1}\right)$
Simplifying:
- $0 \preceq 1$
By definition of compatibility with addition:
- $\forall x, y, z \in \C: x \preceq y \implies \left({x + z}\right) \preceq \left({y + z}\right)$
Substituting $x = 0$, $y = 1$, $z = -1$ gives:
- $0 + \left({-1}\right) \preceq 1 + \left({-1}\right)$
Simplifying:
- $-1 \preceq 0$
From the definition of ordering:
- $\forall a, b \in \C: \left({a \preceq b \land b \preceq a}\right) \implies a = b$
Substituting $a = -1$ and $b = 0$ gives:
- $-1 = 0$
which is a contradiction.
Hence, from Proof by Contradiction, there can be no such ordering.
$\blacksquare$
From Complex Numbers form Integral Domain, $\left({\C, +, \times}\right)$ is an integral domain.
Aiming for a contradiction, suppose that $\left({\C, +, \times}\right)$ can be ordered.
Thus, by definition, it possesses a positivity property $P$.
Then from Positivity Property induces Total Ordering, let $\le$ be the total ordering induced by $P$.
From Unity of Ordered Integral Domain is Positive:
- $1$ is positive.
Thus by positivity, axiom $3$:
- $-1$ is not positive.
Consider the element $i \in \C$.
By definition of positivity, axiom $3$, either:
- $i$ is positive
or:
- $-i$ is positive.
Suppose $i$ is positive.
Then by Square of Element of Ordered Integral Domain is Positive:
- $i^2 = -1$ is positive.
Similarly, suppose $-i$ is positive.
Then by Square of Element of Ordered Integral Domain is Positive:
- $\left({-i}\right)^2 = -1$ is positive.
In both cases we have that $-1$ is positive.
But it has already been established that $-1$ is not positive.
Hence, by Proof by Contradiction, there can be no such ordering.
$\blacksquare$
The fact that Complex Numbers cannot be Totally Ordered was realized by Leonhard Paul Euler.
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