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.

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Denominator exponent is a well-defined integer

Statement

For every nonzero $z \in D$ , the set ${k \in Z : (\sqrt{2})^{k} z \in O}$ is nonempty and bounded below, hence has a least integer.

Facts & Assumptions

Given: $z \neq 0$ has the canonical form $z = (a_{0} + a_{1} ω + a_{2} ω^{2} + a_{3} ω^{3}) / 2^{r}$ with $r \geq 0$ .

[F1]

Canonical dyadic coefficient encoding: the canonical numerator has integer coefficients, is not all even when $r > 0$ , and is nonzero.

Proof

1.1F1givenalgebra

Since $(\sqrt{2})^{2 r} = 2^{r}$ , multiplying $z$ by $(\sqrt{2})^{2 r}$ gives its canonical numerator in $O$ . Hence the exponent set is nonempty.

1.2F1givenchoosealgebra

Suppose $k = - m < 0$ belongs to that set, so $z = (\sqrt{2})^{m} y$ for some $y \in O$ . Then the numerator $a = 2^{r} z = (\sqrt{2})^{m + 2 r} y = 2^{q} (\sqrt{2})^{ϵ} y$ , where $q = ⌊ (m + 2 r) / 2 ⌋$ and $ϵ \in {0, 1}$ . Multiplication by $\sqrt{2}$ has integer coefficients in the power basis, so all coefficients of $a$ are divisible by $2^{q}$ . Choose a nonzero coefficient $a_{j}$ ; its finite $2$ -adic valuation bounds $q$ above, and therefore bounds $m$ above. Thus the negative exponents in the set are bounded below; nonnegative exponents already are.

2.1step 1.1step 1.2algebra∎

A nonempty subset of $Z$ bounded below has a least element. Hence the set of clearing exponents has a least integer; this does not assume that the least exponent is nonnegative.

Remarks

For $z = 0$ the exponent set is all of $Z$ and has no least element; that is why the definition assigns $- \infty$ separately.

Depends on

Used by

Cyclotomic scaled grid and norm-equation solver Algorithm
Smallest square root of two denominator exponent Definition
Clifford+T denominator descent and finite base table Lemma
Finite coefficient bounds for the scaled cyclotomic grid Lemma
Squared entry denominator bound Lemma

Dependency tree · two levels

4 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)