Definition D1
The stage is an ordered field.
Proof.
A system equipped with the four operations (F) together with an order (A1–A6, A9) is called an ordered field. ℚ and ℝ are examples.convention
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Definition D2
Let A = { q ∈ ℚ : q < 0 or q² < 2 }.
Proof.
(1)That s is an upper bound of A means q ≤ s for every q ∈ A.convention
(2)The least upper bound is the smallest of the upper bounds.convention
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Lemma L1
1 < 2, 0 < 2, 0 < 1/2.
Proof.
(1)Adding 1 to both sides of 0 < 1 gives 1 < 1 + 1 = 2.A1, A2, convention
(2)0 < 1 < 2, so 0 < 2.A1, (1), A4
(3)0 < 2, so 0 < 1/2.(2), A6
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Lemma L2
For 0 < s, setting s′ = (2s+2)/(s+2) gives s′² − 2 = 2(s² − 2)/(s+2)² and s − s′ = (s² − 2)/(s+2).
Proof.
(1)Adding 2 to both sides of 0 < s gives 2 < s + 2. Since 0 < 2, 0 < s + 2 — the denominator is nonzero.A2, L1, A4
(2)s′² − 2 = ((2s+2)² − 2(s+2)²) / (s+2)² — put over a common denominator.(1), F, E
(3)(2s+2)² − 2(s+2)² = (4s² + 8s + 4) − (2s² + 8s + 8) = 2s² − 4 = 2(s² − 2)F, E
(4)Therefore s′² − 2 = 2(s² − 2)/(s+2)² — the same sign as s² − 2.(2), (3), E
(5)s − s′ = (s(s+2) − (2s+2))/(s+2) = (s² − 2)/(s+2).(1), F, E
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Lemma L3
If 0 < u < v, then u² < v².
Proof.
(1)Multiplying both sides of u < v by u (> 0) gives u² < uv.A3
(2)Multiplying both sides of u < v by v (> 0) gives uv < v².A3, A4, F
(3)u² < v².(1), (2), A4
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Theorem T1
Between rationals a < b there is a rational — ℚ is dense.
Proof.
(1)Let c = (a + b)/2. c is rational.convention, F
(2)Adding a to both sides of a < b gives a + a < a + b. Adding b to both sides gives a + b < b + b.A2, F
(3)Multiplying by 1/2 (> 0) gives a < c and c < b.(1), (2), L1, A3, F, E
(4)a < c < b — between any two rationals, however close, there is a rational.(3)
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Theorem T2
A is bounded above, but ℚ contains no least upper bound of A.
Proof.
(1)If 2 ≤ q, then 4 ≤ q² and 2 < 4, so 2 < q² — no such q lies in A. Therefore 2 is an upper bound of A.D2, L1, L3, A2, A4, A5
(2)Let s ∈ ℚ be an upper bound of A.assumption, D2
(3)1² = 1 < 2, so 1 ∈ A. Therefore 1 ≤ s, and in particular 0 < s.D2, L1, A1, A4, F
(4)s² = 2 is impossible.(2), A7
(5)Suppose s² < 2. Then s ∈ A, and s′ − s = (2 − s²)/(s+2) > 0, 2 − s′² = 2(2 − s²)/(s+2)² > 0.assumption, (3), D2, L2, A2, A6, A9, E
(6)That is, s′ ∈ A yet s < s′ — contradicting that s is an upper bound. An upper bound cannot have s² < 2.(2), (5), D2, A2
(7)The only remaining case is 2 < s².(4), (6), A5
(8)Then s′² − 2 = 2(s² − 2)/(s+2)² > 0 and s − s′ = (s² − 2)/(s+2) > 0 — 0 < s′ < s.(3), (7), L2, A2, A6, A9
(9)If q ∈ A with 0 < q, then q² < 2 < s′², so q < s′ (if s′ ≤ q then s′² ≤ q², a contradiction). Each q ≤ 0 also satisfies q < s′. Therefore s′ is an upper bound of A.(8), D2, L3, A4, A5
(10)Every upper bound s admits a smaller upper bound s′. The smallest of the upper bounds is not in ℚ.(8), (9), D2
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Corollary C
Density and completeness are different properties. The ordered field obtained by adding the axiom guaranteeing least upper bounds is ℝ.
Proof.
(1)ℚ is dense (T1). Even so, the place of the least upper bound of A is empty in ℚ (T2).T1, T2
(2)Density does not prevent gaps. Being everywhere close and having no gaps are different properties.(1)
(3)In ℝ, by the completeness axiom, A, being bounded above, has a least upper bound.T2, A8, D1
(4)The same set, the same order — what separates the two is a single axiom. The ordered field that accepts that axiom is ℝ.(1), (3), D1
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