Definition D1
What an extension gains is measured by a single yardstick. Does an equation unsolvable in the previous system become solvable in the new one?
Proof.
(1)View the systems as nested: ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ ⊂ ℂ.convention
(2)'Solvable' means a solution exists within that system.convention
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Lemma L1
0 < 1 < 2 < 4
Proof.
(1)0 < 1A12
(2)Set 2 = 1 + 1 and 4 = 2 + 2.convention
(3)0 + 1 < 1 + 1, that is, 1 < 2(1), (2), A13, A4, A1
(4)0 < 2(1), (3), A14
(5)0 + 2 < 2 + 2, that is, 2 < 4(2), (4), A13, A4, A1
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Theorem T1
In ℤ, x + a = b is solvable for all a, b. In ℕ it may fail to be.
Proof.
(1)Set x = b + (−a).A2
(2)x + a = (b + (−a)) + a(1), E
(3) = b + ((−a) + a)A3
(4)(−a) + a = a + (−a) = 0A4, A2
(5)x + a = b + 0 = b(2), (3), (4), A1
(6)Therefore in ℤ a solution always exists.(5), D1
(7)Suppose there is x in ℕ with x + 2 = 1.assumption
(8)x is a natural number, so 0 < x.(7), A11
(9)0 + 2 < x + 2, that is, 2 < x + 2(8), A13, A4, A1
(10)x + 2 = 1, so 2 < 1.(7), (9), E
(11)1 < 2L1
(12)(10) and (11) violate trichotomy.(10), (11), A15
(13)Unsolvable in ℕ, solvable in ℤ. The additive inverse solves it.(6), (12)
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Theorem T2
In ℚ, ax = b is solvable for all a, b with a ≠ 0. In ℤ it may fail to be.
Proof.
(1)a ≠ 0, so there is c with ac = 1.A6
(2)Set x = cb.convention
(3)ax = a(cb) = (ac)b(2), A7, E
(4)(ac)b = 1 × b = b × 1 = b(1), A8, A5
(5)Therefore in ℚ a solution always exists.(3), (4), E, D1
(6)Suppose there is x in ℤ with 2x = 1.assumption
(7)1 = 2x, so 1 has the form 2k.(6), A10
(8)On the other hand 2 × 0 = 0, so 1 = 2 × 0 + 1.A9, A4, A1
(9)1 would have both forms, violating A10.(7), (8), A10
(10)Unsolvable in ℤ, solvable in ℚ. The multiplicative inverse solves it.(5), (9)
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Theorem T3
In ℝ, x² = 2 is solvable. In ℚ it is not.
Proof.
(1)Set f(x) = x × x.convention
(2)f is continuous.A16
(3)f(1) = 1 × 1 = 1(1), A5
(4)f(2) = 2 × 2 = 4(1), convention
(5)1 < 2 < 4L1
(6)Therefore f(1) < 2 < f(2).(3), (4), (5), E
(7)There is c between 1 and 2 with f(c) = 2.(2), (6), A17
(8)c × c = 2(1), (7)
(9)No rational x satisfies x² = 2.A18
(10)Unsolvable in ℚ, solvable in ℝ. Completeness solves it.(8), (9), D1
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Theorem T4
In ℂ, x² = −1 is solvable. In ℝ it is not.
Proof.
(1)i × i = −1A20
(2)i is an element of ℂ.D1
(3)No real x satisfies x² = −1.A19
(4)Unsolvable in ℝ, solvable in ℂ. i solves it.(1), (2), (3), D1
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