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Problema cu multimi
Posted: Thu Dec 04, 2008 10:21 pm
by alex2008
Fie \( a_0 \) un numar natural . Definim multimea \( A=\{a_0,a_1,a_2, ... ,a_n, ... \} \) , unde \( a_1=\sqrt{a_0^2+1} \) , ... , \( a_{n+1}=\sqrt{a_n^2+1} \) , ...
Aratati ca \( A-\mathb{Q}\neq0 \) .
Posted: Fri Dec 05, 2008 8:40 am
by alex2008
Acolo
\( A-\mathb{Q} \) trebuie sa fie diferit de multimea vida .

Posted: Fri Dec 05, 2008 3:55 pm
by Laurian Filip
\( a_1^2=a_0^2+1 \)
\( a_2^2=a_1^2+1 \)
....................
\( a_n^2=a_{n-1}^2+1 \)
_____________ \( \bigoplus \)
\( a_n^2=a_0^2+n \)
pt \( n=2a_0+1 \)
\( a_{2a_0+1}^2=(a_0+1)^2 \) \( \to \) \( a_{2a_0+1} \in \mathbb{Q} \)
Posted: Fri Dec 05, 2008 8:09 pm
by mihai++
cred ca trebuia sa demonstrezi ca are si elemente irationale, nu elemente rationale.
Posted: Fri Dec 05, 2008 8:17 pm
by Laurian Filip
Ai dreptate, am citit eu gresit.
Daca \( a_0 \neq 0 \) \( \to \) \( a_0^2+1 \) este natural dar nu e patrat perfect, deci \( a_1 \in \mathbb{R}-\mathbb{Q} \)
Daca \( a_0=0 \) \( \to \) \( a_2=\sqrt{2} \in \mathbb{R}-\mathbb{Q} \)