Question

Ma poate ajuta cineva?

48c01731916e27f42eced28a52fe7a04.pdf

Answer 1

[tex]c)\ 0\in Q(a)\Rightarrow \exists g\in Q[x] a.i\ g(a)=0\\ Consideram\ g=0 \in Q(X)\\ Analog ~consideram~ g=1~pt~a~demonstra\ ca\ 1\in Q(a)\\ d)\alpha\in Q(a)\Rightarrow\exist g\in Q[X],g(a)=\alpha\\ \beta \in Q(a) \Rightarrow\exist h\in Q[X],h(a)=\beta\\ \alpha+\beta=g(a)+h(a)=(g+h)(a),~deci~ \exists ~polinomul\ g+h\\ \alpha\cdot\beta=g(a)\cdot h(a)=(g\cdot h)(a),~deci~ \exists ~polinomul\ g\cdot h[/tex]
[tex]e) \{p+qa+ra^2|p,q,r\in\mathbb{Q}\}\subset Q(a)~evident \\ Mai\ trebuie\ aratat\ ca\ Q(a)\subset \{p+qa+ra^2|p,q,r\in\mathbb{Q}\} \\ fie\ g\in Q(a),\ daca\ grad(g)\leq2\Rightarrow g\in Q(a)\\ daca \ grad(g)\ \textgreater \ 2\ aplicam\ teorema\ impartirii\ cu\ rest\ la\ f\ (din\ ip):\\ g=f\cdot q+r,\ cu\ grad(r)\leq2\\ g(a)=f(a)\cdot q(a)+r(a)=0\cdot q(a)+(p+qa+ra^2)=p+qa+ra^2\\ \Rightarrow g\in \{p+qa+ra^2|p,q,r\in\mathbb{Q}\}[/tex]
     [tex]f)a\ este\ irational\ si \ f(a)=0\Rightarrow a^3=-3a-3\in R-Q \\ PP\ \exists\ p,q,r \ nenule\ a.i\ ra^2++pa+q=0 \Rightarrow ra^3+pa^2+qa=0\\ Din\ ultimele\ doua\ relatii\ rezulta\ dac\ il\ reducem\ pe \ a^3\\ (pr-3r^3-q^2)a=3r^2+pq\Rightarrow...\Rightarrow(\frac{q}{r})^3+3\frac{q}{r}+3=0\\ Am\ obtinut\ ca\ f\ are\ o\ radacina\ rationala\ CONTRADICTIE \\g) PP\ RA a^{2007}=t\in Q \\ Consideram\ polinomul\ g=X^{2007}-t\\ Aplicam\ teorema\ impartirii\ cu\ rest\ si\ obtinem\ folosind\\ \ punctu \ f) [/tex]
ca g=f*q de unde rezulta ca toate radacinile lui f sunt radacini si ale lui g=X^2007-t,insa g are toate radacinile avand acelasi modul, in timp ce f nu are radacinile de acelasi modul