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

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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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When $a b < b$ for positive $a, b$

Statement

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

$0 < a b < b ⟺ 0 < a < 1.$

Facts & Assumptions

Given: Elements $a, b$ of an ordered field $F$ with $a > 0$ and $b > 0$ .

[L1]

If $x > 0$ and $y > 0$ then $x y > 0$ : the positive cone is closed under multiplication (Ordered field).

[L2]

$x < y ⟺ x - y < 0 ⟺ y - x > 0$ (Ordered field).

[L3]

For $b > 0$ : $b z < 0 ⟺ z < 0$ ; the sign of $b z$ equals the sign of $z$ (Sign rules for products and monotonicity of multiplication).

[L4]

By distributivity and $b \cdot (- 1) = - b$ , one has $a b - b = b (a - 1)$ (Sign rules for products: $(- a) b = - (a b)$ and $(- a) (- b) = a b$ ).

Proof

technique · direct

1.1

Since $a > 0$ and $b > 0$ , [L1] gives $a b > 0$ .

1.2

By [L4], $a b - b = b (a - 1)$ .

2.1

By [L2], $a b < b ⟺ a b - b < 0$ , and by step 1.2 this is $b (a - 1) < 0$ .

L2step 1.2

2.2

By [L3] with $z = a - 1$ , we have $b (a - 1) < 0 ⟺ a - 1 < 0$ , and by [L2] this is $a < 1$ .

L3L2step 1.2

3.1

Chaining, $a b < b ⟺ a < 1$ ; since $a > 0$ is given and $a b > 0$ by step 1.1, this is exactly $0 < a b < b ⟺ 0 < a < 1$ .

givenstep 1.1step 2.1step 2.2∎

Depends on

Used by

sup(0,1) = 1 and inf(0,1) = 0, with neither attained Example

Dependency tree · two levels

5 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)

When ab<bab<b for positive a,ba,b