PROOF 189

Prove.

What each extension loses

On each number system.

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.

What you may use

A1∅ ≠ S ⊆ ℕ ⟹ a least element exists
well-ordering (property of ℕ)
A20 < 1
1 is positive
A3a < b ⟹ a + c < b + c
order and addition
A4no integer n with 0 < n < 1
discreteness (property of ℤ)
A5a < b, 0 < c ⟹ ac < bc
order and multiplication by positives
A60 < a ⟹ 0 < 1/a
the reciprocal of a positive is positive
A7a<b, b<c ⟹ a<c
transitivity of order
A8exactly one of a<b, a=b, b<a
trichotomy of order
A90<a, 0<b ⟹ 0 < ab
order is compatible with multiplication (ordered field)
A10i × i = −1
definition of i (Proof 14)
A11(−a)(−b) = ab
product of negatives (Proof 01)
A12a × 0 = 0
lemma of Proof 01
Fthe computation rules of the four operations
field axioms — associativity, commutativity, distributivity, identities, inverses
Ea = b ⟹ apply the same operation to both sides
properties of equality

T1ℤ has no least element
Theorem 1 — ℕ→ℤ
T2ℚ: 0 < ½ < 1
Theorem 2 — ℤ→ℚ
T3ℂ has no order
Theorem 3 — ℝ→ℂ

Nothing outside this list may be cited.

Theorem T1

In ℕ → ℤ, well-ordering is lost. ℤ has no least element.

Proof.

(1)In ℕ, every nonempty subset has a least element. In particular ℕ itself has one.A1

(2)Suppose ℤ also has a least element m.assumption

(3)m − 1 is also an integer.F

(4)Adding −1 to both sides of 0 < 1 gives −1 < 0.A2, A3, F

(5)Adding m to both sides of −1 < 0 gives m − 1 < m.(4), A3, F

(6)There is an integer smaller than m, so m is not least — a contradiction. ℤ has no least element.(2), (3), (5)

∎

Theorem T2

In ℤ → ℚ, discreteness is lost. The space between 0 and 1 is filled.

Proof.

(1)In ℤ there is no element with 0 < n < 1.A4

(2)Adding 1 to both sides of 0 < 1 gives 1 < 1 + 1 = 2. By transitivity, 0 < 2.A2, A3, A7, convention

(3)0 < 2, so ℚ contains 1/2 and 0 < 1/2.(2), A6, F

(4)Multiplying both sides of 1 < 2 by 1/2 gives 1/2 < 1.(2), (3), A5, F, E

(5)0 < 1/2 < 1 — the property 'nothing lies between 0 and 1' breaks in ℚ.(1), (3), (4)

∎

Theorem T3

In ℝ → ℂ, order is lost. There is no order < on ℂ satisfying A3 and A9.

Proof.

(1)Suppose ℂ has an order < satisfying A3 and A9.assumption

(2)i ≠ 0 — if i = 0, then −1 = i × i = 0 × 0 = 0, contradicting −1 < 0.A10, A12, A2, A3, A8, E

(3)By trichotomy, 0 < i or i < 0.(2), A8

(4)If 0 < i, then 0 < i × i = −1.(3), A9, A10, E

(5)If i < 0, adding −i to both sides gives 0 < −i. Then 0 < (−i)(−i) = i × i = −1.(3), A3, A9, A11, A10, F, E

(6)Either way, 0 < −1. But from 0 < 1 we get −1 < 0, contradicting trichotomy.(4), (5), A2, A3, A8, F

(7)Therefore no such order exists. The order that ℝ had, ℂ cannot have.(6)

∎

Corollary C

Extension is not free.

Proof.

(1)In ℕ → ℤ, well-ordering is lost.T1

(2)In ℤ → ℚ, discreteness is lost.T2

(3)In ℝ → ℂ, order is lost.T3

(4)Each widening exacts a price. Choosing which property to keep is what determines the extension (Proof 10).(1), (2), (3)

∎

Remark

Can A9 be removed from this proof?

Drop the demand that order be compatible with multiplication, and check whether giving ℂ an order is really impossible.