From Vladimir Zorich's Analysis:
Show that if $\mathbb{R}$ and $\mathbb{R}'$ are two models of the set of real numbers and $\mathbb{R} \rightarrow \mathbb{R}'$ is a mapping such that $f(x+y)=f(x)+f(y)$ and $f(xy)=f(x)*f(y)$ for any $x,y \in \mathbb{R}$ then $\mathbb{R} \rightarrow \mathbb{R}'$ is a bijective mapping that preserves order.
My attempts
Idea on injectivity:
By definition, suppose $x, y \in \mathbb{R}$ and $x=y$, then if a transformation $f$ is injective, $f(x)=f(y)$.
In this specific case, if $x=y$, then $x+y=x+x=y+y$, therefore $f(x+y)=2f(x)=2f(y)\implies f(x) = f(y)$. Which shows that $f$ is injective as long as the assumption $2f(x)=2f(y)\implies f(x) = f(y)$ is true. Therefore $f$ is injective.
Coming from Linear Algebra, I was planning to utilize a rank-nullity theorem to show that since $\mathbb{R}$ and $\mathbb{R}'$ are vector spaces over themselves, then $f$ must also be surjective. But unfortunately $f$ is not precisely defined to be linear, therefore I'm skipping it.
Idea on Bijectivity/Surjectivity:
Now that I was possibly able to show that $f$ is injective, I should either show that it is surjective or directly an isomorphism without even using the premise derived above.
I saw in the comments from the link above, that the least upper bound principle (or the axiom of completeness) is a one way that can possibly solve this.
Finding 1:
Suppose there be a finite set $S \subset \mathbb{R}'$, such that $\forall r \in \mathbb{R}, f(r) \notin S'$. By the least upper bound principle, there exists $c \in \mathbb{R}'$ such that $c \geq s', \forall s' \in S$, then if $c \neq \textrm{max}(S)$ (consequential assumption), it can be seen $f(x) \geq s', \forall s' \in S'$.
Finding 2:
Suppose $\exists r' \in \mathbb{R}$ such that $\forall r \in \mathbb{R}, f(r) \neq r'$. Then it can be assumed that $\forall r \in \mathbb{R}, f(r) > r'$ or $\forall r \in \mathbb{R}, f(r) < r'$. Therefore, by the axiom of completeness, $\exists c \in \mathbb{R}, (f(r) \geq c \geq r') \implies (f(\frac{r}{r'}) \geq c)$ or conversely, $\exists c \in \mathbb{R}, (r' \geq c \geq f(r)) \implies (c \geq f(\frac{r}{r'}))$
Question
Is my assumption on idea on injectivity valid? Can my ideas in bijectivity/surjectivity section be extended to show that $f$ is an isomorphism?
Are there other ways I can show that $f$ is an isomorphism using a least upper bound principle?
Thank you!