Encyclopedia Cost Cost Trace Rational Exponent Golden Square Has Trace Three
ARTICLE 3 claims 3 theorems
Cost Trace Rational Exponent Golden Square Has Trace Three
The golden ratio's square is the real number that, added to its own reciprocal, gives exactly three, a fact with a surprising consequence for rational arithmetic.
The golden square's trace
The golden ratio is the number usually written as φ, approximately 1.618, the positive solution to the equation x² = x + 1. Its square is about 2.618. The declaration golden_square_has_trace_three states a compact identity: the square of the golden ratio, plus the reciprocal of that square, equals exactly 3. In symbols, if φ² = (3 + √5)/2, then φ² + (φ²)⁻¹ = 3. This is a theorem in the framework's machine-checked library of formal theorems, meaning the equality is proved from the definitions of real arithmetic, not assumed.
The identity is a special case of a broader pattern the library explores. For any real number u greater than 1, the sum u + u⁻¹ is called its trace, a discrete record of the number's size and its reciprocal together. The golden square is the real witness that a trace of exactly three exists. The surrounding theorems show a limit on such traces: no rational number r can satisfy r + r⁻¹ = 3, because that would require a rational square root of five, which does not exist. The golden square's trace is therefore rational in value (the number 3) while its character, the number being traced, is irrational.
This distinction matters for the framework's classification of cost functions, the objects that assign a price to recognition events. The trace equation r + r⁻¹ = 3 has no rational solution, so a cost function with a perfectly rational value at a certain ratio cannot be described by a rational character at that ratio. Asking for such a character asks for something that does not exist. The golden square provides the real, irrational character that does exist, and the theorem records it explicitly.
The declaration does not claim that the golden ratio itself has trace three; the trace of φ is √5, about 2.236, not 3. It does not claim that the golden square is the only number with trace three, nor that the trace equation has any rational solution. It is a precise arithmetic fact, proved in the library, that gives a concrete witness for a structural limit the framework relies on.
THEOREM golden_square_has_trace_three · IndisputableMonolith/Cost/TraceRationalExponent.lean
/-- The real witness behind the previous theorem, recorded so the object is on the page:
the square of the golden ratio has trace exactly three. -/
theorem golden_square_has_trace_three :
((3 + Real.sqrt 5) / 2) + ((3 + Real.sqrt 5) / 2)⁻¹ = 3 := by
have hsq : Real.sqrt 5 ^ 2 = 5 := Real.sq_sqrt (by norm_num)
have hnn : (0 : ℝ) ≤ Real.sqrt 5 := Real.sqrt_nonneg 5
have hne : (3 + Real.sqrt 5) / 2 ≠ 0 := by positivity
field_simp
nlinarith [hsq, hnn]
THEOREM no_rational_character_at_trace_three · IndisputableMonolith/Cost/TraceRationalExponent.lean
/-- **The demand for a carrier-valued character is too strong.** The trace equation
`r + r⁻¹ = 3` has no rational solution. So a cost whose value at the ratio two is the
perfectly rational `1/2` has no rational character at that ratio, and asking the
factorization to produce one asks for something that does not exist. -/
theorem no_rational_character_at_trace_three : ¬ ∃ r : ℚ, r + r⁻¹ = 3 := by
rintro ⟨r, hr⟩
have hr0 : r ≠ 0 := by
intro h
rw [h] at hr
norm_num at hr
have hquad : r ^ 2 - 3 * r + 1 = 0 := by
field_simp at hr
linarith [hr]
exact no_rational_sqrt_five ⟨2 * r - 3, by nlinarith [hquad]⟩
THEOREM no_rational_sqrt_five · IndisputableMonolith/Cost/TraceRationalExponent.lean
/-- No rational number squares to five. Proved by counting the five-adic valuation:
a square has even valuation and five has valuation one. -/
theorem no_rational_sqrt_five : ¬ ∃ s : ℚ, s ^ 2 = 5 := by
haveI : Fact (Nat.Prime 5) := ⟨by norm_num⟩
rintro ⟨s, hs⟩
have hs0 : s ≠ 0 := by
intro h
rw [h] at hs
norm_num at hs
have h1 : padicValRat 5 (s ^ 2) = (2 : ℕ) * padicValRat 5 s :=
padicValRat.pow hs0
have h2 : padicValRat 5 ((5 : ℕ) : ℚ) = 1 := padicValRat.self (by norm_num)
rw [hs] at h1
norm_num at h2
rw [h2] at h1
omega
What this page does not claim
The golden ratio itself has trace three. The golden square is the only real number whose trace is three. The declaration proves anything about cost functions beyond the arithmetic identity.
Verify this page
Every tagged claim above names its theorem. To check one yourself rather than trust this page, elaborate the source module with Lean 4 and audit its axiom basis:
$ lake env lean IndisputableMonolith/Cost/TraceRationalExponent.lean
expected axiom basis: [propext, Classical.choice, Quot.sound] (the Lean kernel's standard three; no RS-specific axioms)
A page whose claims cannot be reproduced this way does not ship. In production, every anchor links to the exact declaration in the public source release, and this block carries the build receipt for the page itself.
Derived articles
This page is generated by a question-recursion engine: the questions its answers raise become the next pages. The current agenda, with open targets marked red:
- What is the trace of the golden ratio itself?
- How does the absence of a rational character at trace three constrain the framework's cost classification?
- What role does the six exponentials theorem play in the broader exponent classification?
MACHINE LAYER · GROUNDED CLAIM TABLE · CLICK TO EXPAND
THEOREM golden_square_has_trace_three · IndisputableMonolith/Cost/TraceRationalExponent.lean
/-- The real witness behind the previous theorem, recorded so the object is on the page: the square of the golden ratio has trace exactly three. -/ theorem golden_square_has_trace_three : ((3 + Real.sqrt 5) / 2) + ((3 + Real.sqrt 5) / 2)⁻¹ = 3 := by have hsq : Real.sqrt 5 ^ 2 = 5 := Real.sq_sqrt (by norm_num) have hnn : (0 : ℝ) ≤ Real.sqrt 5 := Real.sqrt_nonneg 5 have hne : (3 + Real.sqrt 5) / 2 ≠ 0 := by positivity field_simp nlinarith [hsq, hnn]The square of the golden ratio, plus the reciprocal of that square, equals exactly 3. golden_square_has_trace_three · IndisputableMonolith/Cost/TraceRationalExponent.leanTHEOREM no_rational_character_at_trace_three · IndisputableMonolith/Cost/TraceRationalExponent.lean
/-- **The demand for a carrier-valued character is too strong.** The trace equation `r + r⁻¹ = 3` has no rational solution. So a cost whose value at the ratio two is the perfectly rational `1/2` has no rational character at that ratio, and asking the factorization to produce one asks for something that does not exist. -/ theorem no_rational_character_at_trace_three : ¬ ∃ r : ℚ, r + r⁻¹ = 3 := by rintro ⟨r, hr⟩ have hr0 : r ≠ 0 := by intro h rw [h] at hr norm_num at hr have hquad : r ^ 2 - 3 * r + 1 = 0 := by field_simp at hr linarith [hr] exact no_rational_sqrt_five ⟨2 * r - 3, by nlinarith [hquad]⟩No rational number r can satisfy r + r⁻¹ = 3. no_rational_character_at_trace_three · IndisputableMonolith/Cost/TraceRationalExponent.leanTHEOREM no_rational_sqrt_five · IndisputableMonolith/Cost/TraceRationalExponent.lean
/-- No rational number squares to five. Proved by counting the five-adic valuation: a square has even valuation and five has valuation one. -/ theorem no_rational_sqrt_five : ¬ ∃ s : ℚ, s ^ 2 = 5 := by haveI : Fact (Nat.Prime 5) := ⟨by norm_num⟩ rintro ⟨s, hs⟩ have hs0 : s ≠ 0 := by intro h rw [h] at hs norm_num at hs have h1 : padicValRat 5 (s ^ 2) = (2 : ℕ) * padicValRat 5 s := padicValRat.pow hs0 have h2 : padicValRat 5 ((5 : ℕ) : ℚ) = 1 := padicValRat.self (by norm_num) rw [hs] at h1 norm_num at h2 rw [h2] at h1 omegaNo rational number squares to five. no_rational_sqrt_five · IndisputableMonolith/Cost/TraceRationalExponent.lean