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PROOF 10

Prove.

What an extension preserves determines the extension

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.

Definition D1

An ordered field is a number system satisfying A1–A8, with a relation < on it satisfying A11–A14.

Proof.

ℚ and ℝ are ordered fields.convention

∎

Definition D2

i is the number satisfying i × i = −1, and ℂ is obtained by adjoining i to ℝ.

Proof.

i is not a number that was found, but a number fixed by this equation.convention

∎

Lemma L1

If a ≠ 0, then 0 < a × a

Proof.

(1)Since a ≠ 0, either 0 < a or a < 0.assumption, A11
(2)If 0 < a, then 0 < a × a.(1), A13
(3)If a < 0, then a + (−a) < 0 + (−a), that is, 0 < −a.(1), A12, A2, A1
(4)Then 0 < (−a)(−a).(3), A13
(5)(−a)(−a) = a × aA10
(6)In either case, 0 < a × a.(2), (4), (5), E

∎

Lemma L2

0 < 1

Proof.

(1)1 ≠ 0A8
(2)0 < 1 × 1(1), L1
(3)1 × 1 = 1A3
(4)0 < 1(2), (3), E

∎

Theorem T1

An ordered field with additive inverses has no least element.

Proof.

(1)Suppose there is a least element m.assumption
(2)0 < 1L2
(3)0 + (−1) < 1 + (−1), that is, −1 < 0(2), A12, A2, A1
(4)m + (−1) < m + 0 = m(3), A12, A1
(5)m + (−1) is also an element of this field.A2, D1
(6)There is an element smaller than m, contradicting (1).(1), (4), (5)
(7)There is no least element. Therefore A15 fails.(6), A15

∎

Theorem T2

An ordered field with multiplicative inverses contains an x with 0 < x < 1.

Proof.

(1)Set 2 = 1 + 1.convention
(2)0 < 1L2
(3)0 + 1 < 1 + 1, that is, 1 < 2(1), (2), A12, A1
(4)0 < 2(2), (3), A14
(5)Since 2 ≠ 0, there is a 2⁻¹ with 2 × 2⁻¹ = 1.(4), A11, A4
(6)If 2⁻¹ = 0, then 1 = 2 × 0 = 0, contradicting A8.(5), A9, A8
(7)0 < 2⁻¹ × 2⁻¹(6), L1
(8)2(2⁻¹ × 2⁻¹) = (2 × 2⁻¹)2⁻¹ = 2⁻¹A5, (5), A3, A6
(9)0 < 2⁻¹(4), (7), (8), A13, E
(10)2⁻¹ + 2⁻¹ = 2⁻¹(1 + 1) = 2⁻¹ × 2 = 1A3, A7, (1), A6, (5)
(11)0 + 2⁻¹ < 2⁻¹ + 2⁻¹, that is, 2⁻¹ < 1(9), (10), A12, A1
(12)0 < 2⁻¹ < 1(9), (11)

∎

Theorem T3

ℂ cannot be an ordered field.

Proof.

(1)Suppose ℂ is an ordered field.assumption, D1
(2)0 < 1L2
(3)0 + (−1) < 1 + (−1), that is, −1 < 0(2), A12, A2, A1
(4)Therefore −1 ≠ 0(3), A11
(5)If i = 0, then i × i = 0, contradicting (4). Therefore i ≠ 0.(4), A9, D2
(6)0 < i × i(5), L1
(7)i × i = −1D2
(8)0 < −1(6), (7), E
(9)(3) and (8) violate A11.(3), (8), A11
(10)ℂ is not an ordered field.(9)

∎

Remark

Can A11 be removed from these proofs?

Take a world where two numbers cannot be compared, and check which lines collapse first.