While I was writing this answer, this answer, the following question occurred to me.
A special case of a well known result of Cantor is that if $\alpha,\beta\in\mathbb R \setminus\mathbb Q$ then there is an increasing bijection from $(\alpha,\infty)\cap\mathbb Q$ to $(\beta,\infty)\cap\mathbb Q$.
It's easy to write a formula for such a bijection if $\alpha-\beta\in\mathbb Q$: $$ f(x) = x-\alpha + \beta. $$
Is there any simple formula that works when $\alpha-\beta\not\in\mathbb Q$ or must one go back to Cantor's computationally messy method?
Since there is not a unique bijection, and since the set of rationals "close to $\alpha$" may have different properties from those "close to " $\beta$ (E.g., $\alpha$ could be algebraic and $\beta$ could be a Liouville number) , I suspect there is not a simple general formula....However if $a<b $ and $c<d$ where $a,b,c,d $ are rational, the linear bijection $f:[a,b]\to [c,d]$, with $f(a)=c$ and $f(b)=d$, monotonically maps $Q\cap [a,b]$ onto $Q\cap [c,d].$ ..... So let $(a_n)_n$ and $(b_n)_n$ be strictly decreasing sequences (for $n=0,1,\ldots$) of rationals, with $a_n$ converging to $\alpha$ and $b_n$ converging to $\beta$. Let $f$ be strictly increasing where $f$ maps $[a_{n+1},a_n]$ linearly onto $[b_{n+1},b_n]$, and for $x>a_0$ let $f(x)=x+b_0-a_0$.Note that this works for any $\alpha,\beta$ ..... I suppose the result due to Cantor that you allude to is that any countably infinite linear order which has no end-points and which is order-dense (...that is ,if $x<y$,the open interval $(x,y)$ is not empty...) is order-isomorphic to $Q$, and so any two such linear orders are order-isomorphic to each other. The proof could be "messy" or "elegant" depending on your taste. Or perhaps on which proof.