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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-25
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Squares of nonzero elements are positive

Statement

Let FF be an ordered field (Ordered field) with positive cone PP. For every aFa \in F with a0a \neq 0 we have a2>0a^2 > 0, where a2=aaa^2 = a \cdot a. In particular, every nonzero square is positive.

Facts & Assumptions

Given: An ordered field FF with positive cone PP, and an element aFa \in F with a0a \neq 0.

[L1]

x>0    xPx > 0 \iff x \in P, and by trichotomy for a0a \neq 0 exactly one of aPa \in P, aP-a \in P holds (Ordered field).

[L2]

PP is closed under multiplication: if u,vPu, v \in P then uvPuv \in P (axiom O2 of Ordered field).

Proof

technique · cases
1.1

Since a0a \neq 0, trichotomy leaves exactly two possibilities: aPa \in P or aP-a \in P.

L1
2.1

Case a>0a > 0: then aPa \in P, so closure gives aaPa \cdot a \in P, i.e. a2>0a^2 > 0.

assume-case posstep 1.1L2L1
2.2

Case a<0a < 0: then aP-a \in P, so closure gives (a)(a)P(-a)(-a) \in P; since (a)(a)=aa=a2(-a)(-a) = a \cdot a = a^2 by L3, we get a2Pa^2 \in P, i.e. a2>0a^2 > 0.

assume-case negstep 1.1L2L3L1
3.1

The cases a>0a > 0 and a<0a < 0 exhaust all of a0a \neq 0 by trichotomy, and both yield a2>0a^2 > 0.

cases-exhaustivestep 2.1step 2.2

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 4 results over 3 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