Definition D1
b − a means b + (−a).
Proof.
Subtraction is not a new operation but the addition of an inverse.convention
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Definition D2
|a| means a when 0 ≤ a, and −a when a < 0.
Proof.
Trichotomy guarantees that exactly one of the two cases occurs.A1
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Lemma L1
a < b if and only if 0 < b − a.
Proof.
(1)Suppose a < b.assumption
(2)a + (−a) < b + (−a)(1), A3
(3)0 < b − a(2), A6, D1
(4)Conversely, suppose 0 < b − a.assumption
(5)0 + a < (b + (−a)) + a(4), A3, D1
(6)(b + (−a)) + a = b + ((−a) + a) = bA7, A8, A6, A5
(7)0 + a = aA8, A5
(8)a < b(5), (6), (7)
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Lemma L2
(b − a)(b + a) = b² − a²
Proof.
(1)(b + (−a))(b + a) = b(b + a) + (−a)(b + a)A9, A10, D1
(2)b(b + a) = b² + baA9
(3)(−a)(b + a) = (−(ab)) + (−(a²))A9, A11
(4)left side = b² + (ba + (−(ab))) + (−(a²))(1), (2), (3), A7
(5)Since ba = ab, we have ba + (−(ab)) = 0A10, A6
(6)(b − a)(b + a) = b² + 0 + (−(a²)) = b² − a²(4), (5), A5, D1
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(1)If 0 ≤ a, then |a| = a, so |a|² = a².D2
(2)If a < 0, then |a| = −a, so |a|² = (−a)(−a) = a².D2, A12
(3)These are the only two cases, so |a|² = a² always.(1), (2), A1
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(1)If 0 ≤ a, then |a| = a, so 0 ≤ |a|.D2
(2)Suppose a < 0.assumption
(3)a + (−a) < 0 + (−a)(2), A3
(4)The left side is 0 and the right side is −a, so 0 < −a = |a|.(3), A6, A8, A5, D2
(5)These are the only two cases, so 0 ≤ |a| always.(1), (4), A1
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Theorem T1
a² < b² does not follow from a < b.
Proof.
(1)Set a = −3, b = 2. Then −3 < 2.N1
(2)a² = (−3)² = 9 and b² = 2² = 4.N2
(3)Since 4 < 9, we have b² < a².(2), N1
(4)By trichotomy, a² < b² does not hold.(3), A1
(5)Hence a < b alone cannot yield a² < b².(1), (4)
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Theorem T2
If 0 ≤ a and a < b, then a² < b²
Proof.
(1)Suppose 0 ≤ a and a < b.assumption
(2)0 < b − a(1), L1
(3)If 0 = a, then a + b = 0 + b = b and 0 < b, so 0 < a + b(1), A8, A5
(4)If 0 < a, then 0 < b and 0 + a < b + a, so 0 < a < a + b(1), A2, A3, A8, A5
(5)In either case, 0 < a + b(3), (4), A1
(6)0 < (b − a)(b + a)(2), (5), A8, A4
(7)(b − a)(b + a) = b² − a²L2
(8)0 < b² − a²(6), (7), E
(9)a² < b²(8), L1
∎
Corollary C
|a| < |b| if and only if a² < b².
Proof.
(1)Suppose |a| < |b|.assumption
(2)Since 0 ≤ |a|, we have |a|² < |b|²(1), L4, T2
(3)Since |a|² = a² and |b|² = b², we have a² < b²(2), L3, E
(4)Conversely, suppose a² < b².assumption
(5)If |b| < |a|, the same argument gives b² < a²L4, T2, L3
(6)Then a² < b² and b² < a² hold together, contradicting trichotomy.(4), (5), A1
(7)If |a| = |b|, then a² = b², which contradicts (4) and trichotomy.(4), L3, E, A1
(8)By trichotomy, |a| < |b|.(6), (7), A1
∎