I have two models of real numbers, $\mathbb{R}$ and $\mathbb{R'}$, and a function $f:\mathbb{R} \rightarrow \mathbb{R'}$ such that $f(x+y)=f(x)+f(y)$, $f(xy)=f(x)f(y)$ and $f \not\equiv 0$.
I have to show that $f$ is a bijective mapping and also that $f$ preserves the order.
I've been able to prove that $f(0)=0'$, $f(1)=1'$ and that $f:\mathbb{Q} \rightarrow \mathbb{Q'}$ is bijective and preserves the order, but not on all real numbers.
I am not sure if this reasoning is right: if $f(x_0)=0'$ and $x_0 \neq 0$, then $f(x\cdot x_0)=f(x)\cdot f(x_0)=f(x)\cdot 0'=0'$, so $f$ maps every real number in $0'$. As a consequence, $f(x)=0' \Rightarrow x=0$, and so if $f(x_1)=f(x_2)$ then $f(x_1)-f(x_2)=0'$ that is $f(x_1-x_2)=0' \Rightarrow x_1-x_2=0 \Rightarrow x_1=x_2$, that is $f$ injective.
Even if it was correct, I still can not prove the surjectivity.
Can you help me?
Thanks.