Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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A positive sequence making all three inequalities of the ratio-to-root chain strict

Example

Let (sk)(s_k) be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing e,oe, o with N\mathbb{N} their disjoint union, and the unique (sk)(s_k) with s0=1s_0 = 1, sσ(k)=sks_{\sigma(k)} = -s_k, which satisfies sk=1|s_k| = 1, se1s \circ e \equiv 1 and so1s \circ o \equiv -1 and define

ak:=2k  when sk=1,ak:=3k  when sk=1.a_k := 2^{-k} \ \text{ when } s_k = 1, \qquad a_k := 3^{-k} \ \text{ when } s_k = -1 .

This interleaves the two geometric sequences 2k2^{-k} and 3k3^{-k}, taking the first at even indices and the second at odd ones. With qk:=ak+1/akq_k := a_{k+1}/a_k and rk:=ak+11/(k+1)r_k := a_{k+1}^{1/(k+1)} as in For ak>0a_k > 0: lim infak+1/aklim infak1/klim supak1/klim supak+1/ak\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k,

lim infkqk=0,lim infkrk=13,lim supkrk=12,lim supkqk=+,\liminf_{k} q_k = 0, \qquad \liminf_{k} r_k = \frac{1}{3}, \qquad \limsup_{k} r_k = \frac{1}{2}, \qquad \limsup_{k} q_k = +\infty,

so the chain of that theorem reads

0  <  13  <  12  <  +0 \;<\; \tfrac{1}{3} \;<\; \tfrac{1}{2} \;<\; +\infty

with all three inequalities strict.

Where each comparison lives. The first two, 0<1/30 < 1/3 and 1/3<1/21/3 < 1/2, are comparisons of real numbers and hold in R\mathbb{R}; they hold in R\overline{\mathbb{R}} as well only because the extended order restricts on R\mathbb{R} to the order of R\mathbb{R} (The extended real line R=R{,+}\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}, its order, and the arithmetic that is left undefined). The third, 1/2<+1/2 < +\infty, is not a comparison in R\mathbb{R} at all: ++\infty is not a real number, and the inequality is the instance of "every real is below the greatest element" in R\overline{\mathbb{R}}. So the outer two values of the chain are of different kinds here, and only the extended line can hold all four at once.

Facts & Assumptions

Given: The alternating sequence (sk)(s_k) with index maps e,oe, o (The even and odd index maps and the alternating sequence: strictly increasing e,oe, o with N\mathbb{N} their disjoint union, and the unique (sk)(s_k) with s0=1s_0 = 1, sσ(k)=sks_{\sigma(k)} = -s_k, which satisfies sk=1|s_k| = 1, se1s \circ e \equiv 1 and so1s \circ o \equiv -1); the sequence aka_k defined above; the ratios qk=ak+1/akq_k = a_{k+1}/a_k; and the roots rk=ak+11/(k+1)r_k = a_{k+1}^{1/(k+1)}.

[L1]

The alternating sequence: sk=1|s_k| = 1, sk+1=sks_{k+1} = -s_k, sej=1s_{e_j} = 1, soj=1s_{o_j} = -1, with ee, oo strictly increasing, so ejje_j \ge j and ojjo_j \ge j; also o0=σ(0)1o_0 = \sigma(0) \ge 1 (The even and odd index maps and the alternating sequence: strictly increasing e,oe, o with N\mathbb{N} their disjoint union, and the unique (sk)(s_k) with s0=1s_0 = 1, sσ(k)=sks_{\sigma(k)} = -s_k, which satisfies sk=1|s_k| = 1, se1s \circ e \equiv 1 and so1s \circ o \equiv -1, A strictly increasing index map satisfies nkkn_k \ge k).

[L3]

The order on R\overline{\mathbb{R}} is total, ++\infty is greatest, every real is <+< +\infty, and the order restricts on R\mathbb{R} to the order of R\mathbb{R} (The extended real line R=R{,+}\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}, its order, and the arithmetic that is left undefined).

[L4]

Powers: 2k=(1/2)k2^{-k} = (1/2)^{k} and 3k=(1/3)k3^{-k} = (1/3)^{k}; xmxm=xm+mx^{m} x^{m'} = x^{m+m'} and (xy)m=xmym(xy)^{m} = x^{m} y^{m} for integer exponents and nonzero bases; xm>0x^{m} > 0 for x>0x > 0; (xn)1/n=x1\big(x^{-n}\big)^{1/n} = x^{-1} for x>0x > 0 and n1n \ge 1 (Integer powers ama^m, Laws of integer exponents, Rational powers ara^r of a positive base, Laws of rational exponents, Existence and uniqueness of nn-th roots: a unique a1/n0a^{1/n} \ge 0 with (a1/n)n=a(a^{1/n})^n = a).

[L5]

Geometric sequences: ρ<1|\rho| < 1 implies ρk0\rho^{k} \to 0, and ρ>1|\rho| > 1 implies ρk+|\rho|^{k} \to +\infty (For r<1|r| < 1 the sequence rkr^k is null, and for r>1|r| > 1 the sequence rk|r|^k diverges to ++\infty, Limits and Cauchy sequences of reals, Divergence to ++\infty and to -\infty).

[L6]

Order arithmetic: 0<10 < 1, so 0<1/3<1/2<1<2<30 < 1/3 < 1/2 < 1 < 2 < 3; multiplying an inequality by a positive element preserves it; reciprocals reverse the order; the order is total; t=1|t| = 1 forces t=1t = 1 or t=1t = -1 (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Inverses of positives are positive, and reciprocation reverses order, Basic properties of the absolute value, Absolute value in an ordered field, Ordered field, Complete ordered field (least-upper-bound property)).

[L7]

The order on N\mathbb{N} is total, so any two indices have a common upper bound (Order on the natural numbers, \le is a linear order on N\mathbb{N}).

[L8]

The chain lim infkqklim infkrklim supkrklim supkqk\liminf_k q_k \le \liminf_k r_k \le \limsup_k r_k \le \limsup_k q_k (For ak>0a_k > 0: lim infak+1/aklim infak1/klim supak1/klim supak+1/ak\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k).

Verification

technique · direct
1.1

Each sks_k is 11 or 1-1, so aka_k is well defined, and ak>0a_k > 0 for every kk since positive powers of positive bases are positive.

givenL1L4L6
1.2

For every nNn \in \mathbb{N} there are indices k,knk, k' \ge n with sk=1s_k = 1 and sk=1s_{k'} = -1, namely k=enk = e_n and k=onk' = o_n; and there are indices l,lnl, l' \ge n with sl+1=1s_{l+1} = 1 and sl+1=1s_{l'+1} = -1, namely l=ej1l = e_j - 1 and l=oj1l' = o_j - 1 for any jn+1j \ge n+1, these being natural numbers because ejj1e_j \ge j \ge 1 and ojj1o_j \ge j \ge 1, and satisfying lj1nl \ge j - 1 \ge n and lj1nl' \ge j-1 \ge n.

givenL1L7
1.3

Since sk+1=sks_{k+1} = -s_k, the ratios are qk=3(k+1)/2k=31(2/3)kq_k = 3^{-(k+1)}/2^{-k} = 3^{-1}(2/3)^{k} when sk=1s_k = 1, and qk=2(k+1)/3k=21(3/2)kq_k = 2^{-(k+1)}/3^{-k} = 2^{-1}(3/2)^{k} when sk=1s_k = -1; in both cases qk>0q_k > 0.

givenL1L4L6
1.4

Likewise the roots are rk=(2(k+1))1/(k+1)=21r_k = \big(2^{-(k+1)}\big)^{1/(k+1)} = 2^{-1} when sk+1=1s_{k+1} = 1, and rk=(3(k+1))1/(k+1)=31r_k = \big(3^{-(k+1)}\big)^{1/(k+1)} = 3^{-1} when sk+1=1s_{k+1} = -1.

givenL1L4
2.1

By steps 1.2 and 1.4 the tail range of (rk)(r_k) at every index nn is exactly {1/2,1/3}\{1/2, 1/3\}, whose least upper bound is 1/21/2 and greatest lower bound 1/31/3, since 1/3<1/21/3 < 1/2 and both belong to the set. Hence lim supkrk=1/2\limsup_k r_k = 1/2 and lim infkrk=1/3\liminf_k r_k = 1/3.

step 1.2step 1.4L2L3L6
2.2

lim supkqk=+\limsup_k q_k = +\infty. Fix nn and a real MM. Since 3/2>1|3/2| > 1, the sequence (3/2)k(3/2)^{k} diverges to ++\infty, so there is KK with 21(3/2)k>M2^{-1}(3/2)^{k} > M for all kKk \ge K; taking jj at least as large as both nn and KK and putting k:=ojk := o_j, we get kjnk \ge j \ge n and sk=1s_k = -1, hence qk=21(3/2)k>Mq_k = 2^{-1}(3/2)^{k} > M. So no real bounds the tail range of (qk)(q_k) above, its least upper bound in R\overline{\mathbb{R}} is ++\infty for every nn, and lim supkqk\limsup_k q_k is the greatest lower bound of {+}\{+\infty\}, namely ++\infty.

step 1.2step 1.3L2L3L5L6L7
2.3

lim infkqk=0\liminf_k q_k = 0. Fix nn. All qkq_k are positive, so 00 is a lower bound of the tail range. If >0\ell > 0 were a lower bound, then, since 2/3<1|2/3| < 1 gives (2/3)k0(2/3)^{k} \to 0 and hence 31(2/3)k<3^{-1}(2/3)^{k} < \ell for all kKk \ge K for some KK, taking jj at least as large as both nn and KK and putting k:=ejk := e_j would give kjnk \ge j \ge n, sk=1s_k = 1 and qk=31(2/3)k<q_k = 3^{-1}(2/3)^{k} < \ell, contradicting that \ell is a lower bound. So every lower bound is 0\le 0 and the greatest lower bound of each tail range is 00; hence lim infkqk\liminf_k q_k is the least upper bound of {0}\{0\}, namely 00.

step 1.2step 1.3L2L3L5L6L7
3.1

Collecting the four values, the chain [L8] reads 01/31/2+0 \le 1/3 \le 1/2 \le +\infty, and each inequality is strict: 0<1/30 < 1/3 and 1/3<1/21/3 < 1/2 hold in R\mathbb{R} and therefore in R\overline{\mathbb{R}}, while 1/2<+1/2 < +\infty holds because ++\infty is the greatest element of R\overline{\mathbb{R}} and is distinct from every real. So no two of the four quantities coincide.

step 2.1step 2.2step 2.3L3L6L8

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