Question

Fie multimile G = {x/x=15a+7,a e N } si H = {x/x=27b+8,b e N}. Demonstrati ca G n H=O

Answer 1

$G = \{x|x=15a + 7, \ a \in \mathbb{N}\} \\H = \{x|x=27b + 8, \ b \in \mathbb{N}\}\\\\m, n \in \mathbb{N}\\Fie \ g \in G => g = 15n + 7;\\Presupunem \ prin \ asburd \ ca \ g \in H =>\\g = 27n + 8 = 15m + 7\\27n + 8 = 15m + 7\\27n + 1 = 15m => 15 | 27n + 1, \forall n \in \mathbb{N}.\\M_{27} - multiplii \ lui \ 27;\\M_{27} = \{0, 27, 54, 81, ... \}\\M_{27} + 1 - numerele \ cu \ restul \ 1 \ la \ impartirea \ cu \ 27.$

$M_27 + 1 = \{1, 28, 55, 82 ..\}, \ dupa \ cum \ vezi \ 15 \ nu \ e \ un \ factor\\=> ca \ ipoteza \ initiala \ e \ falsa => ca \ daca \ g \in G => g \notin H \ si \ vice-versa => G \cap H = \O$