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CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-08 (gpt-5.6-terra-codex-subscription)
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Null times divergent has no rule: xk=1/kx_k = 1/k with yk=cky_k = ck gives product limit cc, and with yk=k2y_k = k^2 gives divergence

Statement refuted

That the product 0(+)0 \cdot (+\infty), left undefined by The extended real line R=R{,+}\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}, its order, and the arithmetic that is left undefined, could be given a value compatible with limits: that there is vRv \in \overline{\mathbb{R}} such that for all sequences of reals with xk0x_k \to 0 (Limits and Cauchy sequences of reals) and yk+y_k \to +\infty (Divergence to ++\infty and to -\infty) the products xkykx_k y_k have the single limiting behaviour named by vv.

Equivalently: that knowing a factor is null and the other diverges to ++\infty determines anything at all about the product. It does not, and the two undefined entries in the arithmetic of R\overline{\mathbb{R}} are undefined for exactly this reason.

Facts & Assumptions

Given: The canonical naturals ι(n)=n1R\iota(n) = n \cdot 1_{\mathbb{R}}; the sequence xk:=1/ι(k+1)x_k := 1/\iota(k+1); for a real c>0c > 0 the sequence yk(c):=cι(k+1)y^{(c)}_k := c\,\iota(k+1); and the sequence zk:=ι(k+1)ι(k+1)z_k := \iota(k+1)\,\iota(k+1).

[L1]

Canonical naturals: ι(n)>0\iota(n) > 0 and invertible for n1n \ge 1, ι\iota is strictly increasing, and ι(n)1\iota(n) \ge 1 for n1n \ge 1 (Canonical naturals are positive and strictly increasing, Order on the natural numbers, \le is a linear order on N\mathbb{N}).

[L2]

Archimedean facts: for every real η>0\eta > 0 there is a natural p1p \ge 1 with 1/p<η1/p < \eta, and for every real MM there is a natural p1p \ge 1 with M<ι(p)M < \iota(p); and 0<u<v0 < u < v gives 0<1/v<1/u0 < 1/v < 1/u (For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon, Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L3]

Convergence to a real and divergence to ++\infty; to establish convergence it suffices to produce a threshold for every real ε>0\varepsilon > 0; a constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Divergence to ++\infty and to -\infty).

[L4]

A sequence diverging to ++\infty is unbounded and therefore does not converge to any real (Divergence to ++\infty and to -\infty, Every convergent sequence is bounded); a limit, when it exists, is unique (A sequence has at most one limit).

[L5]

Order and field arithmetic: multiplying an inequality by a positive element preserves it; 0<1<20 < 1 < 2 and 121 \ne 2; u(1/u)=1u \cdot (1/u) = 1 for u0u \ne 0; and the algebra of limits (Sign rules for products and monotonicity of multiplication, Order is preserved by adding a constant and by adding inequalities, The multiplicative identity is positive, Algebra of limits: sums, scalar multiples, products and quotients, Integer powers ama^m, Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · direct
1.1

The sequence xk=1/ι(k+1)x_k = 1/\iota(k+1) is well defined, positive, and converges to 00: given a real ε>0\varepsilon > 0, take a natural p1p \ge 1 with 1/p<ε1/p < \varepsilon; for kpk \ge p we have ι(k+1)>ι(p)>0\iota(k+1) > \iota(p) > 0, hence 0<xk<1/p<ε0 < x_k < 1/p < \varepsilon.

givenL1L2L3
1.2

For every real c>0c > 0 the sequence yk(c)=cι(k+1)y^{(c)}_k = c\,\iota(k+1) diverges to ++\infty: given a real MM, the quotient M/cM/c is real, so there is a natural p1p \ge 1 with M/c<ι(p)M/c < \iota(p), and for kpk \ge p we get ι(k+1)>ι(p)>M/c\iota(k+1) > \iota(p) > M/c, hence yk(c)=cι(k+1)>My^{(c)}_k = c\,\iota(k+1) > M after multiplying by c>0c > 0.

givenL1L2L3L5
1.3

The sequence zk=ι(k+1)ι(k+1)z_k = \iota(k+1)\iota(k+1) diverges to ++\infty: given a real MM, take a natural p1p \ge 1 with M<ι(p)M < \iota(p); for kpk \ge p we have ι(k+1)1\iota(k+1) \ge 1 and ι(k+1)>ι(p)>M\iota(k+1) > \iota(p) > M, so zkι(k+1)>Mz_k \ge \iota(k+1) > M.

givenL1L2L3L5
2.1

For every real c>0c > 0 the product sequence is constant: xkyk(c)=(1/ι(k+1))cι(k+1)=cx_k y^{(c)}_k = \big(1/\iota(k+1)\big)\,c\,\iota(k+1) = c for every kk, so it converges to cc.

step 1.1step 1.2L3L5
2.2

The product with (zk)(z_k) is xkzk=(1/ι(k+1))ι(k+1)ι(k+1)=ι(k+1)x_k z_k = \big(1/\iota(k+1)\big)\iota(k+1)\iota(k+1) = \iota(k+1), which diverges to ++\infty by the argument of step 1.3 with the single factor, and therefore converges to no real number.

step 1.1step 1.3L1L2L3L4L5
3.1

Now take the three pairs (x,y(1))(x, y^{(1)}), (x,y(2))(x, y^{(2)}) and (x,z)(x, z). In each, the first sequence is null and the second diverges to ++\infty, so each pair satisfies the hypotheses of the refuted claim; but the three products converge to 11, converge to 22, and converge to ++\infty in the extended sense. Since 121 \ne 2 and limits are unique, no single vRv \in \overline{\mathbb{R}} describes all three, and the claim is false.

step 2.1step 2.2L4L5L6

Remarks

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