Definition D1
ℂ is the set of all pairs (a, b) of reals, with two operations defined as follows.
Proof.
(1)(a, b) + (c, d) = (a + c, b + d)convention
(2)(a, b) × (c, d) = (ac − bd, ad + bc)convention
(3)Identify (a, 0) with the real number a, and write (0, 1) as i.convention
(4)Then (a, b) = a + bi.(1), (2), (3)
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Lemma L1
For all a, a × 0 = 0
Proof.
(1)0 + 0 = 0A5
(2)a(0 + 0) = a × 0(1), E
(3)a×0 + a×0 = a×0(2), A1
(4)a×0 + a×0 = a×0 + 0(3), A5
(5)a × 0 = 0(4), A7
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Lemma L2
For all a, b, a(−b) = −(ab)
Proof.
(1)ab + a(−b) = a(b + (−b))A1
(2) = a × 0A6
(3) = 0L1
(4)The inverse of ab is unique, so a(−b) = −(ab).(3), A6
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Lemma L3
Under the multiplication of D1, i × i = −1
Proof.
(1)(0,1) × (0,1) = (0×0 − 1×1, 0×1 + 1×0)D1
(2)0 × 0 = 0 and 0 × 1 = 1 × 0 = 0L1, A3
(3)1 × 1 = 1A4
(4)Therefore (0,1) × (0,1) = (0 + (−1), 0) = (−1, 0)(1), (2), (3), A5
(5)(−1, 0) is the real number −1 and (0,1) is i, so i × i = −1.(4), D1
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Lemma L4
If distributivity, associativity, and commutativity are to hold and i × i = −1 is required, the multiplication rule of D1 is the only one.
Proof.
(1)(a + bi)(c + di) = a(c + di) + (bi)(c + di)A1
(2) = ac + a(di) + (bi)c + (bi)(di)A1
(3) = ac + (ad)i + (bc)i + (bd)(i × i)A2, A3
(4)(bd)(i × i) = (bd)(−1) = −(bd)L3, L2, A4
(5) = (ac − bd) + (ad + bc)i(3), (4), A1
(6)This is the same as the multiplication rule of D1.(5), D1
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Theorem T1
No real number x satisfies x² = −1.
Proof.
(1)0 < 1A11
(2)Adding −1 to both sides, −1 < 0.(1), A8, A6, A5
(3)Suppose there is an x with x² = −1.assumption
(4)If x = 0 then x² = 0, and by (2), 0 ≠ −1, so x ≠ 0.L1, (2), A10
(5)If 0 < x, then 0 = 0 × x < x × x, so 0 < x².A9, L1, A3
(6)If x < 0, adding −x to both sides gives 0 < −x.A8, A6, A5
(7)From x < 0 and 0 < −x, x(−x) < 0 × (−x) = 0.(6), A9, L1, A3
(8)x(−x) = −(x × x), so −x² < 0.(7), L2
(9)Adding x² to both sides, 0 < x².(8), A8, A6, A5
(10)x is nonzero, so (5) or (9) applies. Either way, 0 < x².A10, (4), (5), (9)
(11)But x² = −1 and −1 < 0.(3), (2)
(12)0 < x² and x² < 0 cannot stand together.A10, (10), (11)
(13)No such x exists.(12)
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Corollary C
ℂ has no order satisfying A8–A11.
Proof.
(1)Suppose ℂ has such an order.assumption
(2)T1 uses only the axioms of an ordered field, so it holds in every field carrying an order.T1, L1, L2
(3)Therefore in ℂ too, x² = −1 has no solution.(1), (2)
(4)But i is an element of ℂ and i × i = −1.L3
(5)Contradiction. No such order exists.(3), (4)
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