TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)

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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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The rationals form a totally ordered field

Statement

The relation of Order on the rationals is well defined and makes the field $Q$ (The rationals form a field) a totally ordered field: the order is total, $x \leq y$ implies $x + z \leq y + z$ , and $0 < x$ , $0 < y$ imply $0 < x y$ .

Facts & Assumptions

Given: Rationals $x = [(a, b)]$ , $y = [(c, d)]$ , $z = [(e, f)]$ with $b, d, f > 0$ .

[L1]

$Z$ is a totally ordered commutative ring; positives are closed under products (The integers form a totally ordered ring).

Proof

technique · direct

1.1L1algebra

Order-scaling in $Z$ : for $p > 0$ , if $u < v$ then $0 < (v - u) p$ , so $u p < v p$ ; conversely if $u p \leq v p$ and $v < u$ then $v p < u p$ , impossible; hence $u \leq v$ iff $u p \leq v p$ .

1.2given

Suppose $(a, b) \sim (a^{'}, b^{'})$ and $(c, d) \sim (c^{'}, d^{'})$ with $b^{'}, d^{'} > 0$ , i.e. $a b^{'} = a^{'} b$ and $c d^{'} = c^{'} d$ ; suppose also $a d \leq c b$ .

1.3L1

Totality: $a d \leq c b$ or $c b \leq a d$ in $Z$ , so $x \leq y$ or $y \leq x$ .

1.4L1

Antisymmetry: $a d \leq c b$ and $c b \leq a d$ give $a d = c b$ , i.e. $x = y$ as classes.

1.5L1

Positive products: if $0 < x$ and $0 < y$ then $0 < a$ and $0 < c$ , so $x y = [(a c, b d)]$ has $a c > 0$ and $b d > 0$ , hence $0 < x y$ .

1.6given

For transitivity, let $z = [(e, f)]$ with $f > 0$ and suppose additionally $y \leq z$ , i.e. $c f \leq e d$ .

2.1step 1.1step 1.2L1

Scaling the hypothesis $a d \leq c b$ by $b^{'} d^{'} > 0$ : $(a d) (b^{'} d^{'}) \leq (c b) (b^{'} d^{'})$ .

2.2step 1.2L1

Rearranging both sides with $a b^{'} = a^{'} b$ and $c d^{'} = c^{'} d$ : $(a d) (b^{'} d^{'}) = (a b^{'}) (d d^{'}) = (a^{'} b) (d d^{'}) = (a^{'} d^{'}) (b d)$ and $(c b) (b^{'} d^{'}) = (c d^{'}) (b b^{'}) = (c^{'} d) (b b^{'}) = (c^{'} b^{'}) (b d)$ .

2.3step 1.1step 1.2step 1.6L1

Transitivity: from $a d \leq c b$ and $c f \leq e d$ , scaling by $f > 0$ and $b > 0$ gives $(a f) d = (a d) f \leq (c b) f = (c f) b \leq (e d) b = (e b) d$ ; cancelling $d > 0$ via order-scaling, $a f \leq e b$ , i.e. $x \leq z$ .

2.4step 1.1L1

Compatibility with addition: $x + z \leq y + z$ reads $(a f + e b) (d f) \leq (c f + e d) (b f)$ , which expands to $(a d) f^{2} + (e b) (d f) \leq (c b) f^{2} + (e d) (b f)$ ; the second terms are equal, so this is $(a d) f^{2} \leq (c b) f^{2}$ , equivalent by order-scaling with $f^{2} > 0$ to $a d \leq c b$ , i.e. $x \leq y$ .

3.1step 2.1step 2.2step 1.1L1

Combining: $(a^{'} d^{'}) (b d) \leq (c^{'} b^{'}) (b d)$ with $b d > 0$ , so $a^{'} d^{'} \leq c^{'} b^{'}$ by order-scaling: the order is independent of representatives.

4.1step 3.1step 1.3step 1.4step 2.3step 2.4step 1.5∎

The order is well defined, total, compatible with addition, and positives are closed under multiplication: $Q$ is a totally ordered field.

The rationals form a totally ordered field

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources