Question

dovediti ca daca x si y sunt doua numere rationale astfel incat √x+√y∈Q,atunci √x∈Q

Answer 1

[tex]\sqrt x+\sqrt y=q\in\mathbb{Q}\Rightarrow\sqrt y=q-\sqrt x\ |^2 \\ y=q+x-2q\sqrt x\Rightarrow q+x-y=2q\sqrt x\\ \Rightarrow \sqrt x=\frac{q+x-y}{2q}\in\mathbb{Q}\\ \text{deoarece in membrul drept avem doar operatii cu numere rationale. } [/tex]
Observatie: Am considerat cazul in care q≠0. Altfel x=y=0 si concluzia re loc.

Answer 2

$E~clasica~problema~aceasta.~Iata~rezolvarea: \\ \\ Pentru~ca~radicalii~sa~fie~de.fi.niti,~trebuie~ca~x,y \geq 0.\\ \\ Daca~x=y=0 \Rightarrow evident! \\ \\ Daca~cel~mult~un~numar~este~egal~cu~0,~rezulta ~ \sqrt{x}+ \sqrt{y} \neq0.\\ \\ x~si~y~sunt~rationale \Rightarrow x-y \in Q. \\ \\ x-y=( \sqrt{x}- \sqrt{y})( \sqrt{x} + \sqrt{y} ) \Rightarrow \sqrt{x} - \sqrt{y}= \frac{x-y}{ \sqrt{x} + \sqrt{y}} .$

$Dar~x-y~si~ \sqrt{x}+ \sqrt{y} ~sunt~rationale \Rightarrow \frac{x-y}{\sqrt{x}+ \sqrt{y}} \in Q \Leftrightarrow \sqrt{x}- \sqrt{y} \in Q. \\ \\ Cum~ \sqrt{x}- \sqrt{y} \in Q~si~ \sqrt{x}+ \sqrt{y} \in Q,~rezulta~ca~ \\ \\ ( \sqrt{x}- \sqrt{y})+( \sqrt{x}+ \sqrt{y})=2 \sqrt{x} \in Q \Rightarrow \boxed{\sqrt{x} \in Q}~.$