Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-25
How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Reciprocals and order: 1/r1/r against 11

Statement

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

0<r<1  ⟺  1r>1,r>1  ⟺  0<1r<1,r=1  ⟺  1r=1.0 < r < 1 \iff \tfrac{1}{r} > 1, \qquad r > 1 \iff 0 < \tfrac{1}{r} < 1, \qquad r = 1 \iff \tfrac{1}{r} = 1.

Here 1r\tfrac{1}{r} denotes the multiplicative inverse r−1r^{-1}.

Facts & Assumptions

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

[L1]

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

[L2]

If 0<x<y0 < x < y then 0<y−1<x−10 < y^{-1} < x^{-1}: the reciprocal reverses order on the positives (Inverses of positives are positive, and reciprocation reverses order).

[L3]

0<10 < 1 (The multiplicative identity is positive).

[L4]

In any field, 1−1=11^{-1} = 1 and (x−1)−1=x(x^{-1})^{-1} = x for x≠0x \ne 0 (Field).

Proof

technique · direct
1.1

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

L1
1.2

In particular 1−1=11^{-1} = 1 and (r−1)−1=r(r^{-1})^{-1} = r by uniqueness of inverses.

L4
2.1

If 0<r<10 < r < 1, applying [L2] to 0<r<10 < r < 1 gives 0<1−1<r−10 < 1^{-1} < r^{-1}, that is 1<1r1 < \tfrac{1}{r}.

L2step 1.2
2.2

Conversely, if 1r>1\tfrac{1}{r} > 1, then 0<1<1r0 < 1 < \tfrac{1}{r} by [L3], and applying [L2] gives 0<(r−1)−1<1−10 < (r^{-1})^{-1} < 1^{-1}, that is 0<r<10 < r < 1.

L2L3step 1.2
2.3

If r>1r > 1, then 0<1<r0 < 1 < r by [L3], and applying [L2] gives 0<r−1<1−10 < r^{-1} < 1^{-1}, that is 0<1r<10 < \tfrac{1}{r} < 1.

L2L3step 1.2
2.4

Conversely, if 0<1r<10 < \tfrac{1}{r} < 1, applying [L2] to 0<r−1<10 < r^{-1} < 1 gives 0<1−1<(r−1)−10 < 1^{-1} < (r^{-1})^{-1}, that is 1<r1 < r.

L2step 1.2
2.5

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

step 1.2
3.1

Combining the two directions in each case: 0<r<1  ⟺  1r>10 < r < 1 \iff \tfrac{1}{r} > 1, and r>1  ⟺  0<1r<1r > 1 \iff 0 < \tfrac{1}{r} < 1, and r=1  ⟺  1r=1r = 1 \iff \tfrac{1}{r} = 1.

step 2.1step 2.2step 2.3step 2.4step 2.5∎

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 8 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.

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