We are asked in Zorich Mathematical Analysis 2.2 to prove the following things:
Show that if $(\mathbb{R},+,\cdot,\le)$ and $(\mathbb{R'},+,\cdot,\le)$ are two models of real numbers (i.e. are complete ordered fields), and $f:\mathbb{R}\to\mathbb{R'}$ 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:
a) $f(0)=0'$
b) $f(1)=1'$ if $f(x)\not\equiv 0'$, which we shall hence forth assume
c) $f(m)=m'$ where $m\in\mathbb Z$,$m'\in\mathbb Z$ is injective and preserves the order
d) $f(\frac{m}{n})=\frac{m'}{n'}$ where $m,n\in\mathbb Z,n\ne 0,m',n'\in\mathbb Z',n'\ne 0',f(m)=m',f(n)=n'$. Thus $f:\mathbb Q\to \mathbb Q'$ is a bijection that preserves order.
e) $f:\mathbb R\to\mathbb R'$ is bijective and preserves order.
My proof for everything except e) is the following
Proof of a): $f(x+0)=f(x)+f(0)\iff f(0)=0'$
Proof of b): if $f$(x)\not\equiv 0'$, then $f(x)f(1)=f(x)\iff f(1)=1'$
Proof of c): We already have $f(0)=0'$. We presume $f(m)=m'$ stands for $m=k$, therefore $f(m+1)=f(m)=f(1)=m'+1$Because $f(x)=f(y)\iff f(x)-f(y)=-\iff f(x-y)=-\iff x-y\in\ker f$And that we already have $f(0)=0'$,$f(m)=m'$,presume $x\in \ker f$is its any element，we have $f(x)=f(y)$$\implies f(x-y)=0=f(0)\implies x-y=0\implies x=y$
Proof of d):$f(\frac{m}{n})=f(m)f(\frac 1 n)=m'f(\frac 1 n)$Due to $f(\frac 1 n)\cdot f(n)=f(1)=1'$We have $f(\frac m n)=\frac {m'} {n'}$, $m,n\in\mathbb Z,n\ne 0,m',n'\in\mathbb Z,n'\ne 0$This is sufficient to say $f$ is bijective。
EditThe answer is found 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.The answer is also foundFunctional equations $f(x+y)= f(x) + f(y)$ and $f(xy)= f(x)f(y)$