LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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.

Squared entry denominator bound

Statement

For nonzero $z \in D$ , write $s = sde (z)$ . There is $m \in {0, 1}$ such that $sde (∣ z ∣^{2}) = 2 s - m$ . In particular, if $sde (∣ z ∣^{2}) \leq 3$ , then $sde (z) \leq 2$ .

Facts & Assumptions

Given: $O = Z [ω]$ , $\sqrt{2} = ω - ω^{3}$ , and the exponent is the least integer power placing an element in $O$ .

[F1]

Square root of two divisibility criterion in the eighth cyclotomic ring: for $x = a + b ω + c ω^{2} + d ω^{3}$ , $\sqrt{2} ∣ x$ iff $a + c$ and $b + d$ are even, and $x \overset{ˉ}{x} = P + \sqrt{2} Q$ with $P \equiv (a + c) + (b + d)$ and $Q \equiv (a + c) (b + d) (m o d 2)$ .

Proof

1.1givenF1algebra

Put $s = sde (z)$ and $x = (\sqrt{2})^{s} z \in O$ . Minimality of $s$ implies $\sqrt{2} ∤ x$ , since otherwise $x = \sqrt{2} y$ with $y \in O$ would make $(\sqrt{2})^{s - 1} z = y \in O$ .

2.1F1step 1.1algebra

Write $x = a + b ω + c ω^{2} + d ω^{3}$ and set $A = a + c$ , $B = b + d$ . By [F1], $\sqrt{2} ∤ x$ means $(A, B)$ is not $(0, 0)$ modulo $2$ . The two congruences $P \equiv A + B$ and $Q \equiv A B$ show that $P, Q$ are not both even: among the three nonzero pairs $(A, B) \in {0, 1}^{2}$ , the pair $(A + B, A B)$ is never $(0, 0)$ . For $P + \sqrt{2} Q$ , divisibility by $(\sqrt{2})^{2} = 2$ is equivalent to both coefficients $P, Q$ being even. Hence $x \overset{ˉ}{x}$ is divisible by $\sqrt{2}$ at most once.

3.1step 2.1algebra

Let $m \in {0, 1}$ be the number of factors $\sqrt{2}$ dividing $x \overset{ˉ}{x}$ ; write $x \overset{ˉ}{x} = (\sqrt{2})^{m} y$ with $y \in O$ and $\sqrt{2} ∤ y$ . Since $∣ z ∣^{2} = (x \overset{ˉ}{x}) / (\sqrt{2})^{2 s}$ , its least denominator exponent is $2 s - m$ : multiplying by that power gives $y$ , while any smaller exponent leaves a denominator because $\sqrt{2} ∤ y$ . Thus $sde (∣ z ∣^{2}) = 2 s - m$ .

4.1step 3.1givenalgebra∎

The hypothesis gives $2 s - m \leq 3$ and $m \leq 1$ , so $2 s \leq 4$ and $s \leq 2$ .

Remarks

The zero case is excluded in the statement; for $z = 0$ both denominator exponents have the stipulated value $- \infty$ and the implication also holds.

Depends on

Used by

Dependency tree · two levels

7 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

Kliuchnikov, Maslov, and Mosca, Practical approximation of single-qubit unitaries by single-qubit quantum circuits (standard reference, not scraped)