Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-24
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FALSE: the rationals are complete

Statement

False claim: every Cauchy sequence of rationals converges to a rational (where (an)→q(a_n) \to q means (an−q)(a_n - q) is null).

This is precisely the defect the construction of R\mathbb{R} repairs.

Facts & Assumptions

Given: The decimal truncations of 2\sqrt 2, built below.

[A1]

In N\mathbb{N}: for each nn there is a largest natural kk with k2≤2⋅102nk^2 \le 2 \cdot 10^{2n} (only finitely many candidates, since k>2⋅10nk > 2 \cdot 10^n already fails); and n<10nn < 10^n.

[L1]

Ordered-field arithmetic in Q\mathbb{Q} (The rationals form a totally ordered field, Absolute value and the triangle inequality).

[L2]

No rational squares to 22 (FALSE: some rational number squares to 2).

[L3]

A constant sequence is null only if the constant is 00; sums of null sequences are null; a Cauchy multiple of a null sequence is null (Null sequence, Null sequences form an ideal).

[L4]

Archimedean property, so 10−n10^{-n} falls below any positive rational (The rationals are Archimedean).

Refutation

technique · direct
1.1

For each nn let knk_n be the largest natural with kn2≤2⋅102nk_n^2 \le 2 \cdot 10^{2n}, and set sn=kn/10ns_n = k_n / 10^n; then sn2≤2<(sn+10−n)2s_n^2 \le 2 < (s_n + 10^{-n})^2, and sn≤2s_n \le 2 (since kn2≤2⋅102n<(2⋅10n)2k_n^2 \le 2\cdot 10^{2n} < (2 \cdot 10^n)^2).

A1L1
2.1

(sn)(s_n) is Cauchy: for m≥nm \ge n, (10m−nkn)2≤2⋅102m(10^{m-n} k_n)^2 \le 2 \cdot 10^{2m} gives 10m−nkn≤km10^{m-n} k_n \le k_m, so sn≤sms_n \le s_m; and sm2≤2<(sn+10−n)2s_m^2 \le 2 < (s_n + 10^{-n})^2 forces sm<sn+10−ns_m < s_n + 10^{-n}; hence 0≤sm−sn<10−n0 \le s_m - s_n < 10^{-n}, and 10−n10^{-n} eventually falls below any ε>0\varepsilon > 0.

step 1.1L1L4
2.2

(sn2−2)(s_n^2 - 2) is null: 0≤2−sn2<(sn+10−n)2−sn2=10−n(2sn+10−n)≤5⋅10−n0 \le 2 - s_n^2 < (s_n + 10^{-n})^2 - s_n^2 = 10^{-n}(2 s_n + 10^{-n}) \le 5 \cdot 10^{-n}.

step 1.1L1L4
3.1

If (sn)(s_n) converged to a rational qq, then (sn−q)(s_n - q) would be null; since (sn+q)(s_n + q) is Cauchy, (sn2−q2)=(sn−q)(sn+q)(s_n^2 - q^2) = (s_n - q)(s_n + q) would be null; adding the null (2−sn2)(2 - s_n^2), the constant 2−q22 - q^2 would be null, forcing q2=2q^2 = 2.

step 2.1step 2.2L3L1
4.1

No rational squares to 22, so (sn)(s_n) is a Cauchy sequence of rationals with no rational limit: the claim is refuted.

step 3.1step 2.1L2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 35 results over 16 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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