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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Inverses of positives are positive, and reciprocation reverses order

Statement

Let $F$ be an ordered field (Ordered field) with positive cone $P$ , and let $a, b \in F$ .

If $a > 0$ then $a^{- 1} > 0$ .
If $0 < a < b$ then $0 < b^{- 1} < a^{- 1}$ .

Statement references

Ordered field

Facts & Assumptions

Given: An ordered field $F$ with positive cone $P$ , and elements $a, b \in F$ .

[L1]

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

[L2]

Sign rules: a product of a positive and a negative is negative, a product of two positives is positive, and for $c > 0$ one has $a < b ⟺ a c < b c$ (Sign rules for products and monotonicity of multiplication).

[L3]

$0 < 1$ ; in particular $1 \neq 0$ (The multiplicative identity is positive).

[L4]

$P$ is closed under addition, so $<$ is transitive (Ordered field).

Proof

technique · direct

1.1assume-hypL1L3

Assume $a > 0$ , so $a \neq 0$ and its inverse $a^{- 1}$ exists with $a \cdot a^{- 1} = 1$ ; moreover $a^{- 1} \neq 0$ , since $a^{- 1}$ has $a$ as its inverse while $0$ is non-invertible ( $1 \neq 0$ by L3).

2.1step 1.1L2L3L1

By trichotomy $a^{- 1} \in P$ or $- a^{- 1} \in P$ ; if $- a^{- 1} \in P$ , then $a > 0$ and $a^{- 1} < 0$ give $a \cdot a^{- 1} < 0$ by the sign rules, i.e. $1 < 0$ , contradicting $0 < 1$ ; hence $a^{- 1} \in P$ , i.e. $a^{- 1} > 0$ , proving claim 1.

3.1assume-hypstep 2.1L4L2L1

Assume $0 < a < b$ ; then $0 < b$ by transitivity, so by claim 1 both $a^{- 1} > 0$ and $b^{- 1} > 0$ , and the sign rules give $a^{- 1} b^{- 1} > 0$ .

4.1step 3.1L2algebra

Multiplying $a < b$ by the positive $a^{- 1} b^{- 1}$ via the sign rules gives $a (a^{- 1} b^{- 1}) < b (a^{- 1} b^{- 1})$ ; since $a a^{- 1} = 1$ and $b b^{- 1} = 1$ , this simplifies to $b^{- 1} < a^{- 1}$ .

5.1step 3.1step 4.1∎

Together with $b^{- 1} > 0$ from step 3.1, we conclude $0 < b^{- 1} < a^{- 1}$ , proving claim 2.

Inverses of positives are positive, and reciprocation reverses order

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources