False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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FALSE: the complex field admits an order making it an ordered field

Statement refuted

There is a total order on $C$ compatible with its field operations.

Facts & Assumptions

Given: The proposed ordered-field structure on $C$ .

[F1]

In an ordered field every nonzero square is positive (Squares of nonzero elements are positive).

[F2]

$i^{2} = - 1$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ), and every complex number is uniquely $a + b i$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ), so $i = 0 + 1 i$ is nonzero.

Refutation

technique · contradiction

1.1

Suppose such an ordered-field structure existed. Since $i \neq 0$ , [F1] and [F2] would give $- 1 = i^{2} > 0$ .

assume-contraF1F2

2.1

Also $1 = 1^{2} > 0$ by [F1], so compatibility with addition gives $0 = 1 + (- 1) > 0$ , impossible in a strict order.

F1step 1.1algebra

3.1

Therefore no order compatible with the complex field operations exists.

step 2.1discharge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 32 results over 6 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

R. K. Srivastava, Complex Analysis lecture notes (standard reference, not scraped)