PROOF 88
Prove.
The complex numbers cannot be ordered by size
On the complex numbers ℂ.
A way to read this proof
First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.
Definition D1
An ordered field is a number system with a relation < on top of the field operations, where < satisfies all of A1 through A4.
Proof.
All four are required: trichotomy, transitivity, addition and order, multiplication and order.A1, A2, A3, A4
∎
Lemma L1
0 < 1
Proof.
(1)1 ≠ 0A11
(2)Suppose 1 < 0.assumption, A1
(3)1 + (−1) < 0 + (−1)(2), A3
(4)1 + (−1) = 0 and 0 + (−1) = −1.A6, A7, A5
(5)Therefore 0 < −1.(3), (4), E
(6)0 < (−1)(−1)(5), A4
(7)(−1)(−1) = 1 × 1 = 1A9, A8
(8)Therefore 0 < 1.(6), (7), E
(9)1 < 0 and 0 < 1 violate trichotomy.(2), (8), A1
(10)Since 1 ≠ 0, 0 < 1.(1), (9), A1
∎
Lemma L2
−1 < 0
Proof.
(1)0 < 1L1
(2)0 + (−1) < 1 + (−1)(1), A3
(3)0 + (−1) = −1 and 1 + (−1) = 0.A7, A5, A6
(4)Therefore −1 < 0.(2), (3), E
∎
Theorem T
ℂ cannot be an ordered field.
Proof.
(1)Suppose ℂ is an ordered field.assumption, D1
(2)i × i = −1A12
(3)−1 < 0L2
(4)If i = 0, then i × i = 0, and by (3) −1 ≠ 0, so i ≠ 0.(2), (3), A10, A1
(5)Therefore 0 < i or i < 0.(4), A1
(6)If 0 < i, then 0 < i × i.(5), A4
(7)If i < 0, then i + (−i) < 0 + (−i), that is, 0 < −i.(5), A3, A6, A7, A5
(8)Then 0 < (−i)(−i).(7), A4
(9)(−i)(−i) = i × iA9
(10)In either case 0 < i × i.(5), (6), (8), (9), E
(11)i × i = −1, so 0 < −1.(2), (10), E
(12)−1 < 0 and 0 < −1 violate trichotomy.(3), (11), A1
(13)No such order exists on ℂ.(12)
∎
Remark
Can A4 be removed from this proof?
Remove just one of the four axioms. Check again whether placing i on either side still produces a contradiction.