Definition D1
For b ≠ 0, a/b means a · b⁻¹.
Proof.
Division is multiplication by an inverse. Without b⁻¹, this notation has no value.convention, A4
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Definition D2
p% means p/100.
Proof.
Percentage is not a new operation but a name for the number divided by 100.convention, D1
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Definition D3
b − a means b + (−a).
Proof.
Subtraction is not a new operation but the addition of an inverse.convention
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Definition D4
The fraction r of a whole A means A × r. The value increased by the fraction r is A + A×r, and the value decreased by it is A − A×r.
Proof.
A ratio is a share when the whole is taken as 1. The word 'of' is translated into multiplication.convention
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Definition D5
For A ≠ 0, the growth rate in changing from A to B is the ratio of the change B − A to A, that is, (B − A)/A.
Proof.
The reference is always the initial value A. Which value sits in the denominator determines the growth rate.convention, D1, D3
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Lemma L1
A + A×r = A(1 + r) and A − A×r = A(1 − r).
Proof.
(1)A(1 + r) = A × 1 + A × rA5
(2)A × 1 = A, so A(1 + r) = A + A×r(1), A3
(3)A(1 − r) = A(1 + (−r)) = A × 1 + A(−r)D3, A5
(4)A(−r) = −(A×r), so A(1 − r) = A + (−(A×r)) = A − A×r(3), A3, A10, D3
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Lemma L2
(1 + r)(1 − r) = 1 − r²
Proof.
(1)(1 + r)(1 − r) = (1 − r)(1 + r)A2
(2)(1 − r)(1 + r) = (1 − r) × 1 + (1 − r)rA5
(3)(1 − r) × 1 = 1 − rA3
(4)(1 − r)r = r(1 − r) = r(1 + (−r)) = r × 1 + r(−r)A2, D3, A5
(5)r × 1 = r and r(−r) = −(r·r) = −r²A3, A10
(6)(1 − r)r = r + (−r²) = r − r²(4), (5), D3
(7)(1 + r)(1 − r) = (1 − r) + (r − r²)(1), (2), (3), (6)
(8)(1 − r) + (r − r²) = 1 + ((−r) + r) + (−r²) = 1 + (−r²) = 1 − r²D3, A6, A7, A9, A8
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Theorem T1
p% of A is A × (p/100), which equals (A × p)/100.
Proof.
(1)Since 100 ≠ 0, 100⁻¹ exists.A4
(2)p% = p/100 = p · 100⁻¹(1), D2, D1
(3)p% of A = A × (p · 100⁻¹)(2), D4
(4)A(p · 100⁻¹) = (Ap)100⁻¹A1
(5)(Ap)100⁻¹ = (A × p)/100D1
(6)p% of A = A × (p/100) = (A × p)/100(3), (4), (5)
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Theorem T2
For A ≠ 0, writing the growth rate from A to B as a percentage gives p = (B − A)/A × 100.
Proof.
(1)Letting r be the growth rate, r = (B − A)/A.D5
(2)Writing r as p% means p/100 = r.D2
(3)Multiplying both sides by 100 gives (p · 100⁻¹)100 = r × 100(2), D1, E
(4)(p · 100⁻¹)100 = p(100⁻¹ · 100) = p × 1 = pA1, A2, A4, A3
(5)p = r × 100 = (B − A)/A × 100(1), (3), (4)
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Theorem T3
Increasing by the fraction r and then decreasing by the same fraction r yields A(1 − r²). If r ≠ 0 and 0 < A, this value is less than A.
Proof.
(1)The increased value is A + A×r = A(1 + r).D4, L1
(2)Calling that value B, the decreased value is B − B×r = B(1 − r).D4, L1
(3)B(1 − r) = (A(1+r))(1 − r) = A((1+r)(1−r)) = A(1 − r²)(1), (2), A1, L2
(4)Since r ≠ 0, we have 0 < r²A11
(5)0 + (1 − r²) < r² + (1 − r²)(4), A12
(6)0 + (1 − r²) = 1 − r² and r² + (1 − r²) = 1D3, A6, A7, A8, A9
(7)1 − r² < 1(5), (6)
(8)If 0 < A, then A(1 − r²) < A × 1 = A(7), A13, A3
(9)Hence the value raised and then lowered by the same fraction is smaller than the initial value.(3), (8)
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Corollary C
Raising A by 20% and then lowering by 20% yields 0.96 × A.
Proof.
(1)20% = 20/100 = 1/5D2, A14
(2)1 + 1/5 = 6/5 = 1.2A15
(3)1 − 1/5 = 4/5 = 0.8A16
(4)The value raised by 20% is A × 1.2, and lowering that by 20% gives (A × 1.2) × 0.8.(1), (2), (3), L1, T3
(5)(A × 1.2) × 0.8 = A(1.2 × 0.8)A1
(6)1.2 × 0.8 = (6/5)(4/5) = 24/25 = 0.96(2), (3), A17
(7)The result is 0.96 × A, and if 0 < A, by T3 it is smaller than A.(4), (5), (6), T3
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