PropositionStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-25

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The sign of a product

Statement

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

$ab > 0 \iff a, b \text{ have the same sign}, \qquad ab < 0 \iff a, b \text{ have opposite signs},$

$ab = 0 \iff a = 0 \text{ or } b = 0.$

Here "same sign" means both positive or both negative, and "opposite signs" means one positive and one negative.

Facts & Assumptions

Given: Elements $a, b$ of an ordered field $F$ .

[L1]

Trichotomy: each $x \in F$ satisfies exactly one of $x > 0$ , $x = 0$ , $x < 0$ (Ordered field).

[L2]

Sign rules: if $x, y > 0$ then $xy > 0$ ; if $x > 0$ and $y < 0$ then $xy < 0$ ; if $x, y < 0$ then $xy > 0$ (Sign rules for products and monotonicity of multiplication).

[L3]

If $xy = 0$ then $x = 0$ or $y = 0$ (A field has no zero divisors: $ab = 0 \Rightarrow a = 0$ or $b = 0$ ).

[L4]

$0 \cdot x = 0$ and $x \cdot 0 = 0$ (Multiplication by zero: $0 \cdot a = 0$ ).

Proof

technique · cases

1.1

If $a = 0$ or $b = 0$ then $ab = 0$ by [L4], and conversely if $ab = 0$ then $a = 0$ or $b = 0$ by [L3]; hence $ab = 0 \iff a = 0$ or $b = 0$ , which is the third biconditional.

L3L4

1.2

For the first two biconditionals assume $a \ne 0$ and $b \ne 0$ ; by trichotomy [L1] each of $a, b$ is then either positive or negative, giving four sign combinations.

2.1

Case $a > 0$ and $b > 0$ (both positive, same sign): $ab > 0$ by [L2].

assume-case posposL2step 1.2

2.2

Case $a < 0$ and $b < 0$ (both negative, same sign): $ab > 0$ by [L2].

assume-case negnegL2step 1.2

2.3

Case $a > 0$ and $b < 0$ (opposite signs): $ab < 0$ by [L2].

assume-case posnegL2step 1.2

2.4

Case $a < 0$ and $b > 0$ (opposite signs): $ab < 0$ by [L2].

assume-case negposL2step 1.2

3.1

By trichotomy [L1] these four cases exhaust every sign combination of the nonzero $a, b$ and are mutually exclusive.

step 2.1step 2.2step 2.3step 2.4cases-exhaustive

4.1

For nonzero $a, b$ we have $ab \ne 0$ by step 1.1, so by trichotomy $ab$ is either $> 0$ or $< 0$ ; from the cases, $ab > 0$ occurs exactly in the same-sign cases 2.1 and 2.2, and $ab < 0$ exactly in the opposite-sign cases 2.3 and 2.4.

step 1.1step 3.1L1

5.1

Therefore $ab > 0 \iff a, b$ have the same sign, and $ab < 0 \iff a, b$ have opposite signs; with step 1.1 all three biconditionals hold.

step 1.1step 4.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 6 results over 4 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.

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)