Problema cu multimi

alex2008 · Post by **alex2008** » Thu Dec 04, 2008 10:21 pm

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 \( A-\mathb{Q}\neq0 \) .

alex2008 · Post by **alex2008** » Fri Dec 05, 2008 8:40 am

Acolo \( A-\mathb{Q} \) trebuie sa fie diferit de multimea vida .

Post by **Laurian Filip** » Fri Dec 05, 2008 3:55 pm

\( a_1^2=a_0^2+1 \)
\( a_2^2=a_1^2+1 \)
....................
\( a_n^2=a_{n-1}^2+1 \)
_____________ \( \bigoplus \)

\( a_n^2=a_0^2+n \)

pt \( n=2a_0+1 \)
\( a_{2a_0+1}^2=(a_0+1)^2 \) \( \to \) \( a_{2a_0+1} \in \mathbb{Q} \)

mihai++ · Post by **mihai++** » Fri Dec 05, 2008 8:09 pm

cred ca trebuia sa demonstrezi ca are si elemente irationale, nu elemente rationale.

Post by **Laurian Filip** » Fri Dec 05, 2008 8:17 pm

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 \( a_0=0 \) \( \to \) \( a_2=\sqrt{2} \in \mathbb{R}-\mathbb{Q} \)