Prove that $f(x+y)=f(x)+f(y)$, and $f(x\cdot y) = f(x)\cdot f(y)$ implies $f:\mathbb{R} \to \mathbb{R}'$ is a bijective mapping that preserves order.

Q: Prove that $f(x+y)=f(x)+f(y)$, and $f(x\cdot y) = f(x)\cdot f(y)$ implies $f:\mathbb{R} \to \mathbb{R}'$ is a bijective mapping that preserves order.

Original question ($\S$2.2/23 in Mathematical Analysis I by Vladimir A. Zorich)

• Show that if $\mathbb{R}$ and $\mathbb{R}'$ are two models of the set of real numbers and $f:\mathbb{R} \to \mathbb{R}'$ is a mapping such that $f(x+y)=f(x)+f(y)$ and $f(x\cdot y) = f(x)\cdot f(y)$ for any $x, y\in\mathbb{R}$, then

  a. $f(0) = 0'$

  b. $f(1) = 1'$ if $f(x) \not\equiv 0'$, which we shall henceforth assume;

  c. $f(m) = m'$ where $m\in \mathbb{Z}$ and $m'\in \mathbb{Z}'$, and the mapping $f:\mathbb{Z} \to \mathbb{Z}'$ is bijective and preserves the order.

  d. $f(\frac m n) = \frac{m'}{n'}$, where $m, n \in \mathbb Z$, $n\neq0$, $f(m) = m'$, $f(n)=n'$. Thus $f:\mathbb{Q} \to \mathbb{Q}'$ is a bijection and preserves the order.

  e. $f:\mathbb{R} \to \mathbb{R}'$ is a bijective mapping that preserves the order.

$^\ast$By saying '$\mathbb{R}'$ is a model of the real numbers', it means that $(\mathbb{R}', +, \cdot, \le)$ satisfies the axioms of real numbers.

I think that I could finish the proof of a., b., c. and d., however I failed to prove e.

Actually, I have found this question while searching. However, this is the next question in that textbook, and also there isn't an answer.

Using the result of d., which says $f:\mathbb{Q} \to \mathbb{Q}'$ is a bijection and preserves the order, I tried to demonstrate the real numbers using least upper bounds. For example, suppose $S\subset\mathbb{Q}$ has an upper bound, then $\exists r\in\mathbb{R}$ such that $r = \sup S$. It is also possible to prove that the image of $S$ has an upper bound, therefore $\exists r'\in\mathbb{R}'(r' = \sup f(S))$. However, I don't know how to prove that $r$ and $r'$ are equal.

As I realized that there could be a way to tell that $f$ is a bijection if $f$ preserves the order, I tried to find a contradiction assuming $\exists c \in\mathbb{R}$ where $c<0 \wedge f(c)>0$. However, I still fail to find anything so far.

Could you help me? Thanks.