PropositionStatement: 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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Reciprocals and order: $1 / r$ against $1$

Statement

Let $F$ be an ordered field and let $r \in F$ with $r > 0$ . Then:

$0 < r < 1 ⟺ \frac{1}{r} > 1, r > 1 ⟺ 0 < \frac{1}{r} < 1, r = 1 ⟺ \frac{1}{r} = 1.$

Here $\frac{1}{r}$ denotes the multiplicative inverse $r^{- 1}$ .

Facts & Assumptions

Given: An element $r > 0$ of an ordered field $F$ .

[L1]

If $x > 0$ then $x^{- 1} > 0$ (Inverses of positives are positive, and reciprocation reverses order).

[L2]

If $0 < x < y$ then $0 < y^{- 1} < x^{- 1}$ : the reciprocal reverses order on the positives (Inverses of positives are positive, and reciprocation reverses order).

[L3]

$0 < 1$ (The multiplicative identity is positive).

[L4]

In any field, $1^{- 1} = 1$ and $(x^{- 1})^{- 1} = x$ for $x \neq 0$ (Field).

Proof

technique · direct

1.1

Since $r > 0$ , [L1] gives $r^{- 1} > 0$ , so $\frac{1}{r}$ is positive throughout.

1.2

In particular $1^{- 1} = 1$ and $(r^{- 1})^{- 1} = r$ by uniqueness of inverses.

2.1

If $0 < r < 1$ , applying [L2] to $0 < r < 1$ gives $0 < 1^{- 1} < r^{- 1}$ , that is $1 < \frac{1}{r}$ .

L2step 1.2

2.2

Conversely, if $\frac{1}{r} > 1$ , then $0 < 1 < \frac{1}{r}$ by [L3], and applying [L2] gives $0 < (r^{- 1})^{- 1} < 1^{- 1}$ , that is $0 < r < 1$ .

L2L3step 1.2

2.3

If $r > 1$ , then $0 < 1 < r$ by [L3], and applying [L2] gives $0 < r^{- 1} < 1^{- 1}$ , that is $0 < \frac{1}{r} < 1$ .

L2L3step 1.2

2.4

Conversely, if $0 < \frac{1}{r} < 1$ , applying [L2] to $0 < r^{- 1} < 1$ gives $0 < 1^{- 1} < (r^{- 1})^{- 1}$ , that is $1 < r$ .

L2step 1.2

2.5

If $r = 1$ then $\frac{1}{r} = 1^{- 1} = 1$ ; conversely if $\frac{1}{r} = 1$ then $r = (r^{- 1})^{- 1} = 1^{- 1} = 1$ .

step 1.2

3.1

Combining the two directions in each case: $0 < r < 1 ⟺ \frac{1}{r} > 1$ , and $r > 1 ⟺ 0 < \frac{1}{r} < 1$ , and $r = 1 ⟺ \frac{1}{r} = 1$ .

step 2.1step 2.2step 2.3step 2.4step 2.5∎

Depends on

Used by

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
M. Spivak, Calculus, 4th ed., Ch. 1 (standard reference, not scraped)
University of Innsbruck notes: Ordered fields (standard reference, not scraped)

Reciprocals and order: 1/r1/r against 11