CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27↗ rests on later material (inherited)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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In $R (t)$ the rationals are not dense: no rational lies strictly between $0$ and $1 / t$

Statement refuted

Refuted claim: in every ordered field $F$ the image of $Q$ is dense, that is, for all $x < y$ in $F$ there is a rational $q$ with $x < q \cdot 1_{F} < y$ .

The witness is $R (t)$ with the eventual-sign order (Not every ordered field is Archimedean, The rational function field $R (t)$ ordered by the eventual sign is an ordered field, worked out), and the pair $x = 0$ , $y = 1 / t$ : the interval between them contains no rational at all.

The true statement requires the Archimedean property and is ℚ is dense in every Archimedean ordered field; $R (t)$ is not Archimedean, and this counterexample is exactly the failure that the Archimedean hypothesis rules out.

Facts & Assumptions

Given: The ordered field $R (t)$ with positive cone $P = {f \neq 0 : f (x) > 0 for all sufficiently large real x}$ , and its element $1 / t$ .

[L1]

$R (t)$ is an ordered field and is not Archimedean (Not every ordered field is Archimedean, Archimedean ordered field).

[L2]

$0 < 1 / t$ , and $1 / t < q \cdot 1$ for every rational $q > 0$ (The rational function field $R (t)$ ordered by the eventual sign is an ordered field, worked out).

[L3]

The canonical embedding of $Q$ into an ordered field is an embedding of ordered fields, so $q \cdot 1 > 0$ if and only if $q > 0$ , and $q \cdot 1 \leq 0$ when $q \leq 0$ (The unique embedding of ℚ into an ordered field).

[L4]

$Q$ is dense in every Archimedean ordered field (ℚ is dense in every Archimedean ordered field).

[L5]

In an ordered field the order is total and transitive, exactly one of $u < v$ , $u = v$ , $v < u$ holds, and a positive element has a positive inverse (Ordered field, Inverses of positives are positive, and reciprocation reverses order).

Counterexample

technique · direct

1.1

$R (t)$ is an ordered field, it is not Archimedean, and $0 < 1 / t$ in it.

L1L2L5

1.2

For every rational $q > 0$ one has $1 / t < q \cdot 1$ .

2.1

No rational $q$ satisfies $0 < q \cdot 1 < 1 / t$ : if $q \leq 0$ then $q \cdot 1 \leq 0$ and the left inequality fails, while if $q > 0$ then $1 / t < q \cdot 1$ by step 1.2, so $q \cdot 1 < 1 / t$ fails by trichotomy.

step 1.1step 1.2L3L5

3.1

So $0 < 1 / t$ in $R (t)$ with no rational strictly between them: the image of $Q$ is not dense in $R (t)$ , and the claim is false.

step 1.1step 2.1

4.1

The hypothesis the claim omitted is the Archimedean property, which $R (t)$ lacks and under which the conclusion does hold.

step 1.1L1L4∎

Remarks

What density really needs. Given $0 < x < y$ in an Archimedean field one finds $n$ with $1 / n < y - x$ and then a multiple of $1 / n$ in the gap; the Archimedean property is used precisely to make the mesh $1 / n$ finer than the gap. In $R (t)$ the gap $1 / t - 0$ is smaller than every $1 / n$ , so no mesh built from rationals is ever fine enough.
An element like $1 / t$ is called an infinitesimal: positive, and below every positive rational. A non-Archimedean ordered field always has one, since if $x$ exceeds every canonical natural then $1 / x$ is below every $1 / n$ (Inverses of positives are positive, and reciprocation reverses order). So the failure of density is not special to this field; it happens in every non-Archimedean ordered field, including $R ((t^{- 1}))$ .
Density is not the same as completeness. $Q$ is dense in itself and in $R$ , and $Q$ is not complete. What this counterexample shows is only that density of $Q$ needs the Archimedean property, which is also the hypothesis missing from FALSE: the nested interval property alone implies the least-upper-bound property and FALSE: an ordered field in which every Cauchy sequence converges has the least-upper-bound property.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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

Ordered field (Wikipedia) (standard reference, not scraped)
Dense order (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)

In R(t)R(t) the rationals are not dense: no rational lies strictly between 00 and 1/t1/t