PROOF 94
Prove.
(aᵐ)ⁿ = aᵐⁿ
On the positive reals.
A way to read this proof
First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.
Definition D1
For a natural number n, define a¹ = a, aⁿ⁺¹ = aⁿ × a.
Proof.
(1)This definition fixes the order of multiplication and the parentheses in one way, from the left.convention
(2)It has meaning only when the exponent is a natural number.convention
∎
Lemma L1
For all natural numbers m, n, aᵐ⁺ⁿ = aᵐ aⁿ
Proof.
(1)Let Q(n) be 'aᵐ⁺ⁿ = aᵐ aⁿ'.convention
(2)aᵐ⁺¹ = aᵐ × a = aᵐ a¹, that is, Q(1) holds.D1
(3)Assume Q(k). aᵐ⁺ᵏ = aᵐ aᵏassumption
(4)(m + k) + 1 = m + (k + 1)A3
(5)aᵐ⁺⁽ᵏ⁺¹⁾ = a⁽ᵐ⁺ᵏ⁾⁺¹ = aᵐ⁺ᵏ × a(4), D1, E
(6)aᵐ⁺ᵏ × a = (aᵐ aᵏ) a(3), E
(7)(aᵐ aᵏ) a = aᵐ (aᵏ a)A2
(8)aᵏ a = aᵏ⁺¹D1
(9)aᵐ⁺⁽ᵏ⁺¹⁾ = aᵐ aᵏ⁺¹, that is, Q(k+1) holds.(5), (6), (7), (8)
(10)Therefore Q(n) holds for every natural number n.(2), (9), A1
∎
Theorem T1
For all natural numbers m, n, (aᵐ)ⁿ = aᵐⁿ
Proof.
(1)Let P(n) be '(aᵐ)ⁿ = aᵐⁿ'.convention
(2)(aᵐ)¹ = aᵐD1
(3)m × 1 = m, so (aᵐ)¹ = a^(m×1), that is, P(1) holds.(2), A4, E
(4)Assume P(k). (aᵐ)ᵏ = aᵐᵏassumption
(5)(aᵐ)ᵏ⁺¹ = (aᵐ)ᵏ × aᵐD1
(6) = aᵐᵏ × aᵐ(4), (5), E
(7)aᵐᵏ × aᵐ = aᵐᵏ⁺ᵐL1
(8)mk + m = m(k + 1)A5
(9)(aᵐ)ᵏ⁺¹ = aᵐ⁽ᵏ⁺¹⁾, that is, P(k+1) holds.(6), (7), (8), E
(10)Therefore P(n) holds for every natural number n.(3), (9), A1
∎
Corollary C
Rational exponents are fixed so that T1 is maintained.
Proof.
(1)D1 has meaning only for natural-number exponents, so a^(1/n) has no meaning yet.D1
(2)There is exactly one criterion for fixing it — keep T1 intact up through rational exponents.A6, T1
(3)That requirement gives (a^(1/n))ⁿ = a^((1/n)·n).(2), A6
(4)(1/n) × n = 1N
(5)a¹ = aD1
(6)(a^(1/n))ⁿ = a(3), (4), (5), E
(7)Therefore a^(1/n) must be the number whose n-th power is a.(6)
∎
Remark
Can A2 be removed from this proof?
Check whether the value aⁿ itself wobbles, or whether something else does.