Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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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.
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FALSE: ∑k<n+1(−1)kι ⁣(nk)=0\sum_{k<n+1}(-1)^{k}\iota\!\binom{n}{k} = 0 for every n∈Nn \in \mathbb{N}

Statement

FALSE. The statement

∑k<n+1(−1)k ι(nk)=0\displaystyle\sum_{k<n+1}(-1)^{k}\,\iota\binom{n}{k} = 0

for every n∈Nn \in \mathbb{N}.

The claim is what a text whose natural numbers begin at 11 would state truly. In this library N\mathbb{N} contains 00 (The natural numbers N\mathbb{N} (von Neumann)), and the statement acquires a counterexample at its very first index.

Facts & Assumptions

Given: The canonical natural ι\iota (The canonical natural ι(n)=n⋅1F\iota(n) = n \cdot 1_F of a field) and the real finite sum of Finite sums and finite products, by recursion; N\mathbb{N} contains 00 (The natural numbers N\mathbb{N} (von Neumann)).

[L1]

∑k<0uk=0\sum_{k<0}u_k = 0 and ∑k<σ(N)uk=∑k<Nuk+uN\sum_{k<\sigma(N)}u_k = \sum_{k<N}u_k + u_N (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

(00)=1\binom{0}{0} = 1 (The set [A]k[A]^{k} of kk-element subsets and the binomial coefficient (nk):=∣[n]k∣\binom{n}{k} := \lvert [n]^{k}\rvert), and ι(1)=1\iota(1) = 1 (The canonical natural ι(n)=n⋅1F\iota(n) = n \cdot 1_F of a field).

[L3]

a0=1a^{0} = 1 for every real aa, including a=0a = 0; and 0 m=00^{\,m} = 0 for m≥1m \ge 1 (Integer powers ama^m, Multiplication by zero: 0⋅a=00 \cdot a = 0, Field).

[L4]

The true statement: ∑k<n+1(−1)kι(nk)=0\sum_{k<n+1}(-1)^{k}\iota\binom{n}{k} = 0 for n≥1n \ge 1 (∑k<n+1(nk)=2n\sum_{k<n+1}\binom{n}{k} = 2^{n}, and ∑k<n+1(−1)kι ⁣(nk)=0\sum_{k<n+1}(-1)^{k}\iota\!\binom{n}{k} = 0 for n≥1n \ge 1, clause 2), proved from the binomial theorem at x=−1x = -1, y=1y = 1 (The binomial theorem in R\mathbb{R}: (x+y)n=∑k<n+1ι ⁣(nk) xky n−k(x+y)^{n} = \sum_{k<n+1} \iota\!\binom{n}{k}\, x^{k} y^{\,n-k}).

Refutation

technique · direct
1.1

Evaluate the left-hand side at n=0n = 0. The sum runs over k<0+1=1k < 0+1 = 1, so by [L1] it is the single term (−1)0 ι(00)(-1)^{0}\,\iota\binom{0}{0}.

givenL1
2.1

That term is 11: (−1)0=1(-1)^{0} = 1 by [L3], (00)=1\binom{0}{0} = 1 by [L2], and ι(1)=1\iota(1) = 1. So the sum equals 11, not 00, and the displayed statement is false at n=0n = 0.

step 1.1L2L3
3.1

Where the hypothesis is spent in the true version. [L4] obtains the identity by evaluating the binomial theorem at x=−1x = -1, y=1y = 1: the left-hand side becomes (−1+1)n=0 n(-1+1)^{n} = 0^{\,n}, which is 00 only for n≥1n \ge 1, while 00=10^{0} = 1 by [L3]. That single evaluation is the entire difference between the true statement and the false one.

step 2.1L3L4∎

Remarks

  • The convention 00=10^{0} = 1 is not the culprit. It is what makes the binomial theorem itself true at n=0n = 0 and at x=0x = 0, with no exceptional case; the price is that one of its corollaries carries a hypothesis. Changing the convention would move the exception, not remove it.

  • The concrete picture. In Pascal's triangle computed to row 66, with Pascal's rule checked at every interior entry the alternating sums of rows 11 to 66 are all 00 and the alternating sum of row 00 is 11. A reader who computes from row 11 onwards sees only the true pattern.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 86 results over 30 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.

Sources