False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

How statement and proof provenance work

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

FALSE: $\sum_{k < n + 1} (- 1)^{k} ι ⁣ (\binom{n}{k}) = 0$ for every $n \in N$

Statement

FALSE. The statement

$\sum_{k < n + 1} (- 1)^{k} ι (\binom{n}{k}) = 0$

for every $n \in N$ .

The claim is what a text whose natural numbers begin at $1$ would state truly. In this library $N$ contains $0$ (The natural numbers $N$ (von Neumann)), and the statement acquires a counterexample at its very first index.

Facts & Assumptions

Given: The canonical natural $ι$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field) and the real finite sum of Finite sums and finite products, by recursion; $N$ contains $0$ (The natural numbers $N$ (von Neumann)).

[L1]

$\sum_{k < 0} u_{k} = 0$ and $\sum_{k < σ (N)} u_{k} = \sum_{k < N} u_{k} + u_{N}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L2]

$(\binom{0}{0}) = 1$ (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ), and $ι (1) = 1$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L3]

$a^{0} = 1$ for every real $a$ , including $a = 0$ ; and $0^{m} = 0$ for $m \geq 1$ (Integer powers $a^{m}$ , Multiplication by zero: $0 \cdot a = 0$ , Field).

[L4]

The true statement: $\sum_{k < n + 1} (- 1)^{k} ι (\binom{n}{k}) = 0$ for $n \geq 1$ ( $\sum_{k < n + 1} (\binom{n}{k}) = 2^{n}$ , and $\sum_{k < n + 1} (- 1)^{k} ι ⁣ (\binom{n}{k}) = 0$ for $n \geq 1$ , clause 2), proved from the binomial theorem at $x = - 1$ , $y = 1$ (The binomial theorem in $R$ : $(x + y)^{n} = \sum_{k < n + 1} ι ⁣ (\binom{n}{k}) x^{k} y^{n - k}$ ).

Refutation

technique · direct

1.1

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

givenL1

2.1

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

step 1.1L2L3

3.1

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

step 2.1L3L4∎

Remarks

The convention $0^{0} = 1$ is not the culprit. It is what makes the binomial theorem itself true at $n = 0$ and at $x = 0$ , 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 $6$ , with Pascal's rule checked at every interior entry the alternating sums of rows $1$ to $6$ are all $0$ and the alternating sum of row $0$ is $1$ . A reader who computes from row $1$ onwards sees only the true pattern.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

47 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

Binomial coefficient (Wikipedia) (standard reference, not scraped)
Binomial theorem (Wikipedia) (standard reference, not scraped)
Pascal's triangle (Wikipedia) (standard reference, not scraped)

FALSE: ∑k<n+1(−1)kι ⁣(nk)=0∑k<n+1​(−1)kι(kn​)=0 for every n∈Nn∈N