Question

?aratati ca radical din 3 nu eate nr rational

Answer 1

$\displaystyle Presupunem \ , \ prin \ reducere \ la \ absurd \ , \ ca \ \sqrt3\in Q \\ \\ \implies \sqrt3=\frac{m}{n} \ , \ m\in Z \ , \ n\in Z \ , \ n\neq 0 \ \Big( consideram \ ca \ fractia \ \frac{m}{n} \ este \ ireductibila\Big) \\ \\ \implies m=\sqrt3n\implies m^2=3n^2 \\ \\ \implies 3|m^2. \\ \\ Deoarece \ 3 \ este \ numar \ prim\implies 3|m\implies m=3p. \\ \\ Inlocuim: \\ \\ 9p^2=n^2\implies 3p=n\implies 3|n \\ \\ , \ adica \ fractia \ \frac{m}{n} \ este \ reductibila \ ; \ contradictie!!$