LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-25

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Squares of nonzero elements are positive

Statement

Let $F$ be an ordered field (Ordered field) with positive cone $P$ . For every $a \in F$ with $a \neq 0$ we have $a^{2} > 0$ , where $a^{2} = a \cdot a$ . In particular, every nonzero square is positive.

Statement references

Ordered field

Facts & Assumptions

Given: An ordered field $F$ with positive cone $P$ , and an element $a \in F$ with $a \neq 0$ .

[L1]

$x > 0 ⟺ x \in P$ , and by trichotomy for $a \neq 0$ exactly one of $a \in P$ , $- a \in P$ holds (Ordered field).

[L2]

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

[L3]

In any field, $(- u) (- v) = u v$ (Sign rules for products: $(- a) b = - (a b)$ and $(- a) (- b) = a b$ ).

Proof

technique · cases

1.1L1

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

2.1assume-case posstep 1.1L2L1

Case $a > 0$ : then $a \in P$ , so closure gives $a \cdot a \in P$ , i.e. $a^{2} > 0$ .

2.2assume-case negstep 1.1L2L3L1

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

3.1cases-exhaustivestep 2.1step 2.2∎

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

Squares of nonzero elements are positive

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources