Are $\mathbb{Q} $ and $\mathbb{Z}$ order isomorphic?

Question

Specifically, the question is whether there exists a bijective map $f:\mathbb{Z}\to\mathbb{Q}$ which satisfies $n\leq0\implies f(n)\geq0$? I think this is related to Order Isomorphism Thanks beforehand.

Answer 1

They are certainly not order isomorphic: In $\Bbb Q$, for any $a,b$ with $a<b$, there always exists $c$ with $a<c<b$; $\Bbb Z$ does not have this property.

But of course there exists a bijection $f\colon \Bbb Z\to\Bbb Q$ which satisfied $n\le 0\implies f(n)\ge 0$. Just pick an arbitrary bijection $g\colon \Bbb Z^+\to\Bbb Q^+$ and let $$f(x)=\begin{cases}-g(x)&x>0\\g(-x)&x<0\\0&x=0\end{cases}$$