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lemma

phi_squared_bounds

proved
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module
IndisputableMonolith.Constants
domain
Constants
line
107 · github
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IndisputableMonolith.Constants on GitHub at line 107.

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 104
 105/-- φ² is between 2.5 and 2.7.
 106    φ² = φ + 1 ≈ 2.618 (exact: (3 + √5)/2). -/
 107lemma phi_squared_bounds : (2.5 : ℝ) < phi^2 ∧ phi^2 < 2.7 := by
 108  rw [phi_sq_eq]
 109  have h1 := phi_gt_onePointFive
 110  have h2 := phi_lt_onePointSixTwo
 111  constructor <;> linarith
 112
 113/-! ### Fibonacci power identities for φ -/
 114
 115/-- Key identity: φ³ = 2φ + 1 (Fibonacci recurrence).
 116    φ³ = φ × φ² = φ(φ + 1) = φ² + φ = (φ + 1) + φ = 2φ + 1. -/
 117lemma phi_cubed_eq : phi^3 = 2 * phi + 1 := by
 118  calc phi^3 = phi * phi^2 := by ring
 119    _ = phi * (phi + 1) := by rw [phi_sq_eq]
 120    _ = phi^2 + phi := by ring
 121    _ = (phi + 1) + phi := by rw [phi_sq_eq]
 122    _ = 2 * phi + 1 := by ring
 123
 124/-- Key identity: φ⁴ = 3φ + 2 (Fibonacci recurrence).
 125    φ⁴ = φ × φ³ = φ(2φ + 1) = 2φ² + φ = 2(φ + 1) + φ = 3φ + 2. -/
 126lemma phi_fourth_eq : phi^4 = 3 * phi + 2 := by
 127  calc phi^4 = phi * phi^3 := by ring
 128    _ = phi * (2 * phi + 1) := by rw [phi_cubed_eq]
 129    _ = 2 * phi^2 + phi := by ring
 130    _ = 2 * (phi + 1) + phi := by rw [phi_sq_eq]
 131    _ = 3 * phi + 2 := by ring
 132
 133/-- Key identity: φ⁵ = 5φ + 3 (Fibonacci recurrence).
 134    φ⁵ = φ × φ⁴ = φ(3φ + 2) = 3φ² + 2φ = 3(φ + 1) + 2φ = 5φ + 3. -/
 135lemma phi_fifth_eq : phi^5 = 5 * phi + 3 := by
 136  calc phi^5 = phi * phi^4 := by ring
 137    _ = phi * (3 * phi + 2) := by rw [phi_fourth_eq]