Order automorphisms of rationals

Question

Is it right that all order automorphisms of rational numbers have form $~~f(r) = ar+b; ~~ a,b\in \mathbb{Q}; ~~ a>0$?

As i know all order automorphisms have form $f(n) = n +k; ~~ k \in \mathbb{Z}$ so a question about rationals arose.

By order automorphism i mean such bijective map $f: ~~ \mathbb{Q} \rightarrow \mathbb{Q}~$ that $\forall x,y\in \mathbb{Q} ~~ (x \leq y) \rightarrow (f(x) \leq f(y))$

Thanks in advance!

Answer 1

No, we can easily find piece-wise linear examples. A simple example would be to make $f(r)=ar$ when $r\geq 0$ and $f(r)=a'r$ when $r<0$, with $a,a'>0$.

You can construct much more complicated examples - any (non-empty) interval of the form $(p,q)$ is order-isomorphic with any other, and any interval of the form $(-\infty,p)$ is order-isomorphic with any other, then if $p_1<p_2<\dots < p_n$ are rationals there is an order-automorphism of $\mathbb Q$ which sends $i$ to $p_i$.

Answer 2

Thomas is right, you have piecewise linear examples, and the group is uncountable. There is a remark in Cameron's Permutation Groups book where he states that the closure of the piecewise linear order automorphisms is the entire group.