[Paper Review] Approximation to real numbers by cubic algebraic integers II
This paper establishes that the optimal exponent for approximating transcendental real numbers by cubic algebraic integers is (3 + √5)/2 ≈ 2.618, not 3 as previously conjectured. Using extremal real numbers defined via recurrence relations in GL₂(ℤ), the authors construct a class of transcendental numbers for which the approximation measure is sharp, proving that the exponent γ² = ((1+√5)/2)² is best possible via a detailed analysis of continued fraction-like sequences and Diophantine approximation properties.
It has been conjectured for some time that, for any integer n\ge 2, any real number ε>0 and any transcendental real number ξ, there would exist infinitely many algebraic integers αof degree at most n with the property that |ξ-α| < H(α)^{-n+ε}, where H(α) denotes the height of α. Although this is true for n=2, we show here that, for n=3, the optimal exponent of approximation is not 3 but (3+\sqrt{5})/2 = 2.618...
Motivation & Objective
- To resolve the conjecture that the optimal approximation exponent for transcendental reals by cubic algebraic integers is 3.
- To demonstrate that the actual optimal exponent is (3 + √5)/2 ≈ 2.618, less than 3.
- To construct explicit examples of extremal real numbers satisfying the sharp approximation bound.
- To prove that the exponent γ² = ((1+√5)/2)² is best possible using recurrence structures in GL₂(ℤ) and Diophantine approximation techniques.
Proposed method
- Define the height H(α) of an algebraic integer α as the maximum absolute value of coefficients in its minimal polynomial over ℤ.
- Characterize extremal real numbers via sequences (xₖ) in ℤ³ satisfying growth, approximation, and determinant bounds (Proposition 2.1).
- Use recurrence relations in GL₂(ℤ) to define a class E(M) of extremal numbers with specific matrix M = [[a,1],[-1,0]] for positive integers a.
- Apply Proposition 2.2 to show that if {xₖ,₀ξ³} is bounded below, then |ξ − α| ≥ cH(α)^−γ² for all cubic algebraic integers α.
- Analyze the sequence {xₖ,₀ξ³} modulo 1 using recurrence formulas from Lemma 2.6 to prove it has positive lower bounds.
- Leverage Proposition 9.2 of [6] to show {xₖ,₀ξ³} ≥ c₉Xₖ^−1/γ³, ensuring the lower bound condition in Proposition 2.2 is met.
Experimental results
Research questions
- RQ1Is the conjectured exponent n = 3 optimal for approximating transcendental reals by cubic algebraic integers?
- RQ2Can the exponent be improved below 3, and if so, what is the true optimal value?
- RQ3Do extremal real numbers constructed via Fibonacci-like continued fractions satisfy the sharp approximation bound?
- RQ4What conditions ensure that the sequence {xₖ,₀ξ³} remains bounded away from zero modulo 1?
- RQ5Can the recurrence structure in GL₂(ℤ) be used to construct explicit examples of numbers achieving the optimal exponent?
Key findings
- The optimal exponent for approximation of transcendental reals by cubic algebraic integers is τ₃ = ((1+√5)/2)² = (3+√5)/2 ≈ 2.618, not 3.
- There exists a transcendental real number ξ and a constant c₁ > 0 such that |ξ − α| ≥ c₁H(α)^−γ² for all cubic algebraic integers α.
- The class Ea of extremal numbers defined via M = [[a,1],[-1,0]] for positive integers a satisfies the sharp approximation bound with exponent γ².
- For ξ ∈ Ea, the sequence {xₖ,₀ξ³} modulo 1 has at most three non-zero accumulation points and is bounded below by a positive constant for large k.
- The proof relies on showing that {xₖ,₀ξ³} ≥ c₉Xₖ^−1/γ³ and using this to verify the hypothesis of Proposition 2.2.
- The result implies that the conjecture τₙ = n for all n ≥ 2 fails for n = 3, as τ₃ = γ² < 3.
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This review was created by AI and reviewed by human editors.