Question

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Answer 1

$f:\mathbb{Q}\to \mathbb{Q},\quad f(x+y) = f(x)+f(y),\quad \forall x,y\in \mathbb{Q}\\ \\ a)\quad f(0+0) = f(0)+f(0) \\ \Leftrightarrow f(0) = 2f(0) \\\Leftrightarrow 2f(0)-f(0) = 0 \\\Leftrightarrow f(0) = 0$

$b) \quad f\Big(x+(-x)\Big) = f(x)+f(-x)\\ \Leftrightarrow f(x-x) = f(x)+f(-x)\\ \Leftrightarrow f(0) = f(x)+f(-x)\\ \Leftrightarrow 0 = f(x)+f(-x)\\ \Leftrightarrow \boxed{f(-x) = -f(x)}$

$c) \quad \text{Daca: }x,y\in \mathbb{Q} \Rightarrow x+y\in \mathbb{Q}\\ \\ f(x_1+x_2+x_3+...+x_n) =\\ = f\Big(x_1+(x_2+x_3+...+x_n)\Big)\\ =f(x_1)+f(x_2+x_3+...+x_n)\\ =f(x_1)+f\Big(x_2+(x_3+x_4+...+x_n)\Big)\\ =f(x_1)+f(x_2)+f(x_3+x_4+...+x_n)\\ \vdots\\ =\boxed{f(x_1)+f(x_2)+f(x_3)+...+f(x_n),\quad \forall n\in \mathbb{N}^*}$

$d)\quad \underline{\text{Inductie matematica:}}\\ \\P(m):\quad f(mx) = mf(x),\quad \forall m\in \mathbb{N}^*,\quad \forall x\in \mathbb{Q}\\ P(1):\quad f(1\cdot x) =1\cdot f(x)\quad (A)\\ P(m+1):\quad \left|\begin{aligned}f\Big((m+1)x\Big) &= f(mx+x)\\&=f(mx)+f(x) \\ &= mf(x)+f(x) \\ &=(m+1)f(x)\quad (A)\end{aligned}\right.\\ \\ \Rightarrow \boxed{f(mx) = mf(x),\quad \forall m\in \mathbb{N}^*,\quad \forall x\in \mathbb{Q}}$

$e)\quad \text{Daca }m = 0:\\ f(mx) = mf(x),\text{ deoarece }f(0\cdot x) = 0\cdot f(x) = 0\\ \\\text{Daca }m< 0\Leftrightarrow -m> 0:\\ f(mx) \overset{\text{din } b)}{=} -f(-mx) \overset{\text{din }d)}{=} -f\Big(\underset{> 0}{\underbrace{(-m)}}\cdot x\Big) =\\ = -\Big(-mf(x)\Big) = mf(x)\\ \Rightarrow f(mx) = mf(x),\quad \forall m\in \mathbb{Q}, x\in \mathbb{Q}$

$\text{Pentru }m \in \mathbb{Q}, n \in \mathbb{Q}^*,\text{avem:}\\\\ \left. \begin{aligned}f\Big(n\cdot \dfrac{m}{n}\Big) &= f(m)=f(m\cdot 1) = mf(1) \\ &=nf\Big(\dfrac{m}{n}\Big)\end{aligned}\right|\Rightarrow mf(1)= nf\Big(\dfrac{m}{n}\Big)\\\\\\\Leftrightarrow f\Big(\dfrac{m}{n}\Big)=\dfrac{m}{n}f(1),\quad\forall\,\dfrac{m}{n}\in \mathbb{Q}\\ \\ \Rightarrow f(x) = xf(1)\\ \\\Rightarrow \boxed{f(x) = ax,\quad \forall x\in \mathbb{Q}}$