1.5. Complex numbers have no ordering
The complex numbers cannot be given a linear order that is compatible with its field structure.
Recall that a linear order is a total order that satisfies: for any a,b, either a ≤ b or b ≤ a.
A strict ordered ring is a ring that satisfies:
-
if a < b, then a + c < b + c for any c,
-
if a < b and c > 0, then ac < bc.
#check LinearOrder
#check IsStrictOrderedRing
/--
Prove that the complex numbers cannot be given a
linear order that is compatible with its field structure.
-/
theorem not_linearOrderedField_complex
(h1: LinearOrder ℂ) (h2: IsStrictOrderedRing ℂ)
: False := h1:LinearOrder ℂh2:IsStrictOrderedRing ℂ⊢ False
-- In a linear ordered ring, every square is ≥ 0.
have h0_le_I_sq : (0 : ℂ) ≤ I * I := h1:LinearOrder ℂh2:IsStrictOrderedRing ℂ⊢ False
-- `mul_self_nonneg` says `0 ≤ x * x`
-- in any linear ordered ring.
All goals completed! 🐙
-- But in ℂ we know `I * I = -1`.
have h0_le_neg1 : (0 : ℂ) ≤ -1 := h1:LinearOrder ℂh2:IsStrictOrderedRing ℂ⊢ False
All goals completed! 🐙
-- Add 1 to both sides: 0 + 1 ≤ -1 + 1, so 1 ≤ 0.
have h1_le_0 : (1 : ℂ) ≤ 0 := h1:LinearOrder ℂh2:IsStrictOrderedRing ℂ⊢ False
h1:LinearOrder ℂh2:IsStrictOrderedRing ℂh0_le_I_sq:0 ≤ I * Ih0_le_neg1:0 ≤ -1this:1 + 0 ≤ 1 + -1⊢ 1 ≤ 0
-- Simplify both sides.
All goals completed! 🐙
-- In a linear ordered ring we must have 0 < 1.
have h0_lt_1 : (0 : ℂ) < 1 := h1:LinearOrder ℂh2:IsStrictOrderedRing ℂ⊢ False All goals completed! 🐙
-- Combining: 1 < 1, contradiction.
h1:LinearOrder ℂh2:IsStrictOrderedRing ℂh0_le_I_sq:0 ≤ I * Ih0_le_neg1:0 ≤ -1h1_le_0:1 ≤ 0h0_lt_1:0 < 1this:1 < 1⊢ False
All goals completed! 🐙