f(q_1 + q_2) = f(q_1) + f(q_2) \\
f(0) = f(0 + 0) = 2 f(0) \Rightarrow f(0) = 0 \\
f(q - q) = f(q) + f(-q) = 0 \Rightarrow f(-q) = -f(q)\\
\\
f(q_1 \cdot q_2) = f(q_1) \cdot f(q_2) \\
f(1) = f(1 \cdot 1) = f(1) \cdot f(1) \Rightarrow f(1) = 1 $ jer je izomorfizam pa $ f(1) \neq f(0) = 0 \\
\\
\forall n \in N \\
f(1) = f \underbrace{\left(\frac{1}{n} + \frac{1}{n} + \ldots + \frac{1}{n}\right)}_n = \underbrace{f\left(\frac{1}{n}\right) + f\left(\frac{1}{n}\right) + \ldots + f\left(\frac{1}{n}\right)}_n = n \cdot f\left(\frac{1}{n}\right)\\ 
\Longrightarrow f\left(\frac{1}{n}\right) = \frac{1}{n} \\
\\
q = \frac{m}{n} \in Q^+
\Rightarrow f(q) = f \underbrace{\left(\frac{1}{n} + \frac{1}{n} + \ldots + \frac{1}{n}\right)}_m = \underbrace{\frac{1}{n} + \frac{1}{n} + \ldots + \frac{1}{n}}_m = \frac{m}{n} = q\\
q \in Q^- \Rightarrow -q \in Q^+ \Rightarrow f(-q) = -q \Rightarrow f(q) = - f(-q) = q