How are $Q-\{0\}$ and $\mathbb{Q}$ isomorphic as linear orders?

Question

Exercise 2.3 of Linear Orderings by Joseph G. Rosenstein asks to show that the linear orders are isomorphic. But if $f$ is an isomorphism from $Q-\{0\}$ to $\mathbb{Q}$ then it must be an increasing function on $Q-\{0\}$. Also according to the hint given in the question since $X=\{2^{-n}\vert\hspace{0.1cm} n\in\mathbb{N}\}$ does not have a LUB in $Q-\{0\}$, $\hspace{0.2cm}f(X)$ does not have an LUB in $\mathbb{Q}$ which implies that $f(0)$ tends to infinity. How are these two conditions reconcilable?