MAT3253 Complex Variables

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.

LinearOrder.{u_2} (α : Type u_2) : Type u_2#check LinearOrder IsStrictOrderedRing.{u_1} (R : Type u_1) [Semiring R] [PartialOrder R] : Prop#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 ) (h2: IsStrictOrderedRing ) : False := h1:LinearOrder h2:IsStrictOrderedRing False -- In a linear ordered ring, every square is ≥ 0. have h0_le_I_sq : (0 : ) I * I := h1:LinearOrder h2:IsStrictOrderedRing 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 : ) -1 := h1:LinearOrder h2:IsStrictOrderedRing False All goals completed! 🐙 -- Add 1 to both sides: 0 + 1 ≤ -1 + 1, so 1 ≤ 0. have h1_le_0 : (1 : ) 0 := h1:LinearOrder h2:IsStrictOrderedRing False h1:LinearOrder h2:IsStrictOrderedRing h0_le_I_sq:0 I * Ih0_le_neg1:0 -1this:1 + 0 1 + -11 0 -- Simplify both sides. All goals completed! 🐙 -- In a linear ordered ring we must have 0 < 1. have h0_lt_1 : (0 : ) < 1 := h1:LinearOrder h2:IsStrictOrderedRing False All goals completed! 🐙 -- Combining: 1 < 1, contradiction. h1:LinearOrder h2:IsStrictOrderedRing h0_le_I_sq:0 I * Ih0_le_neg1:0 -1h1_le_0:1 0h0_lt_1:0 < 1this:1 < 1False All goals completed! 🐙