Divisor de 7
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This document proves by induction that for all natural numbers x, the number 23x - 1 is divisible by 7. It first shows the base case of x=0 is true, as 23*0 - 1 = 0, which is divisible by 7. It then assumes the inductive hypothesis that 23k - 1 is divisible by 7 for some natural number k. It performs algebraic manipulations to show that if 23k - 1 is divisible by 7, then 23(k+1) - 1 must also be divisible by 7. This completes the inductive proof that for any natural number x, the number 23x - 1 is divisible by 7.
Divisor de 7
- 1. Prova por indu¸˜o: ca ∀x(x ∈ N )(div7(23x − 1)). a) Base da indu¸˜o ( div7(23×0 − 1)): ca 1 ∀x(div7(x) ↔ ∃y(y ∈ N )(7y = x)) Defini¸˜o div7(x) ca 2 23×0 − 1 = 0 Teorema 3 7×0=0 Teorema 4 7 × 0 = 23×0 − 1 2,3, = s 5 0∈N Teorema 6 ∃y(y ∈ N )(7y = 23×0 − 1) 4,5 I ∃ 3×0 7 div7(2 − 1) 6,1 MP b) Passo indutivo ( div7(23k − 1) → div7(23(k+1) − 1)): 1 ∀x(div7(x) ↔ ∃y(y ∈ N )(7y = x) Defini¸˜o div7(x) ca 3k 2 div7(2 − 1) Hip´tese o 3 ∃y(y ∈ N )(7y = 23k − 1) 2, 3, MP 3k 4 7a = 2 −1 4, E∃ 5 7a + 1 = 23k 5, x + 1 = x + 1 3k 6 (7a + 1) × 8 = 2 ×8 6, 8x = 8x 7 (7a + 1) × 8 = 23k × 23 7, 23 = 8 8 (7a + 1) × 8 = 23k+3 8, xy × xz = xy+z 9 (7a + 1) × 8 = 23×(k+1) 9, Distributiva ×/+ 3×(k+1) 10 8(7a) + 8 = 2 10, Distributiva ×/+ 11 8(7a) + 7 = 23×(k+1) − 1 11, x − 1 = x − 1 3×(k+1) 12 7(8a) + 7 = 2 −1 12, Comutativa × 13 7(8a + 1) = 23×(k+1) − 1 13, Distributiva ×/+ 3×(k+1) 14 ∃y(7y = 2 − 1) 14, I∃ 15 div7(23×(k+1) − 1) 15, 1, MP 16 div7(23k − 1) → div7(23×(k+1) − 1) 2, 16, I → 1