I need help with the following proof:
Thanks
suppose $\displaystyle f \in \text{Aut}(\mathbb{R}).$ obviously: $\displaystyle f(0)=0, \ f(1)=1, \ f(-1)=-1.$ thus $\displaystyle f(n)=n, \ \forall n \in \mathbb{Z}.$
hence for all $\displaystyle m \in \mathbb{N}, \ n \in \mathbb{Z}: \ mf(\frac{n}{m})=f(n)=n.$ thus: $\displaystyle f(\frac{n}{m})=\frac{n}{m},$ i.e. $\displaystyle f(r)=r,\ \forall r \in \mathbb{Q}.$
next we show that $\displaystyle f$ is (strictly) increasing: let $\displaystyle x, y$ be two real numbers and $\displaystyle x<y.$ then for
some $\displaystyle z \neq 0 : y-x=z^2$ thus: $\displaystyle f(y)-f(x)=f(z^2)=(f(z))^2 > 0.$ so: $\displaystyle f(x) < f(y),$ i.e. $\displaystyle f$ is
(strictly) increasing.
now let $\displaystyle x \in \mathbb{R},$ and $\displaystyle n \in \mathbb{N}.$ choose $\displaystyle r \in \mathbb{Q}$ such that $\displaystyle \frac{-1}{n} < x - r < \frac{1}{n}.$ call this $\displaystyle (1).$ since $\displaystyle f$ is
increasing, by $\displaystyle (1): \ \frac{-1}{n}=f(\frac{-1}{n}) < f(x-r)<f(\frac{1}{n})=\frac{1}{n}.$ but $\displaystyle f(x-r)=f(x)-r.$ thus:
$\displaystyle \frac{-1}{n} < f(x) - r < \frac{1}{n}.$ call this $\displaystyle (2).$ now $\displaystyle (1)$ and $\displaystyle (2)$ give us: $\displaystyle |f(x)-x| < \frac{2}{n}.$ since $\displaystyle n$ is arbitrary,
we must have $\displaystyle f(x)=x. \ \ \ \square$
Here is another way. If we can show $\displaystyle f$ is continous then let $\displaystyle x$ be any real number and $\displaystyle x_n$ be a sequence of rationals converging to $\displaystyle x$. Then by continuity $\displaystyle x_n\to x \implies \lim f(x_n) = f(x) \implies x = f(x)$. Let us show that $\displaystyle f$ is continous WLOG at $\displaystyle x_0>0$. Let $\displaystyle \epsilon > 0$ and choose $\displaystyle \delta = \epsilon$. Then if $\displaystyle 0<x<\delta$ we can find a rational $\displaystyle r$ so that $\displaystyle 0 < x < r < \delta$ and then $\displaystyle |f(x) - f(0)| = f(x)< f(r) = r < \epsilon$ by using the strict property as NonCommAlg shown.
A more interesting question is to find $\displaystyle \text{Aut}(\mathbb{C})$.