Question

Salut, am nevoie de ajutor la A2 punctul c)

Answer 1

$M = (-1,1),\,\,x\circ y = \dfrac{x+y}{1+xy}\\ \\\text{Se cere: }\\ \forall x,y \in (-1,1) \Rightarrow x \circ y\in (-1,1)$

|x| < 1 și |y| < 1 ⇒ |x|·|y| < 1 ⇒ |xy| <1

$\Rightarrow -1 < xy < 1$

$\\\left\{\begin{array}{c}x>-1 \Rightarrow 1+x>0\\ y>-1\Rightarrow 1+y>0\end{array}\right| \Rightarrow (1+x)(1+y)>0$

$\Rightarrow 1+y+x+xy>0 \Rightarrow x+y>-1\cdot (\underset{>0}{\underbrace{1+\underset{>-1}{\underbrace{xy}}})} \,\,\Big|:(1+xy) \\ \Rightarrow \dfrac{x+y}{1+xy}>-1\quad (i)$

$\\\left\{\begin{array}{c}x<1 \Rightarrow x-1<0\\ y<1\Rightarrow y-1<0\end{array}\right| \Rightarrow (x-1)(y-1)>0$

$\Rightarrow xy-x-y+1>0\Rightarrow x+y <1+xy \Rightarrow\\ \\ \Rightarrow x+y <(\underset{>0}{\underbrace{1+\underset{>-1}{\underbrace{xy}}})}\Big|:(1+xy)\Rightarrow \dfrac{x+y}{1+xy}<1\quad (ii)$

$\text{Din }(i)\text{ si } (ii) \Rightarrow -1 <\dfrac{x+y}{1+xy}<1\quad (Q.E.D.)$