Show that the function is an isomorphism between two $L$-structure. The function:
$$f: \mathbb{R} \longrightarrow (-1, 1)$$
$$ x \rightarrow \frac{x}{1 + |x|}$$ 
is an isomorphism between $\langle\mathbb{R}, <, =\rangle $ and $\langle(-1, 1), <, =\rangle$ where both the sets are ordered by $<$.
$b)$ Exploit the function $f$ to find an isomorphism between $\langle \mathbb{Q}, <, = \rangle$ and $\langle \mathbb{Q} \cap (-1, 1), <, =\rangle$
$c)$  Now find an isomorphism between $\langle \mathbb{Q}, <, = \rangle$ and $\langle \mathbb{Q} \cap (0, 1), <, =\rangle$
$d)$ Finally, find an isomorphism between  $\langle \mathbb{Q}, <, = \rangle$ and $\langle \mathbb{Q} \cap (0, 1)\cup(2,3), <, =\rangle$
I am self-studying currently some model theory and I saw this question in the textbook. I attempted to problem and I think I showed that $f$ was an isomorphism but not sure if I did it correctly.
But now I'm stuck with the rest of the question, how would I continue on with $b)$ and then the rest of the question?
 A: So, if you use the fact that the theory DLO (dense linear order) has quantifier-elimination, point (b) is quite simple. In fact, if a function maintain the order between the elements then it's an isomorphism. So $f$ maintain the order (point (a) ) $\Rightarrow$ $f|_\mathbb{Q}$ maintain the order, and now you have only to show $f|_\mathbb{Q}$ is a bijection.
Point (c) is the same: if you consider the function
$$
\mathbb{Q} \cap (-1,1) \to \mathbb{Q} \cap (0,1)\\
x \mapsto \frac{1}{2} x + \frac{1}{2}
$$
it maintain the order of the elements, is bijective and so is an isomorphism between the structures. It follows that $\mathbb{Q}$ and $\mathbb{Q}\cap (0,1)$ are isomorphic.
Point (d) is less trivial because you have to "split" $\mathbb{Q}$ in two parts. Let's use the "back & forth" method!
If you have a partial isomorphism
$$
\{a_1,\dots, a_n\} \to \{ b_1,\dots ,b_n \}\\
a_i \mapsto b_i
$$
between subsets of $\mathbb{Q}$ and $\mathbb{Q}\cap ( (0,1) \cup (2,3) )$, then for every $c\in \mathbb{Q}$ you can extend it chosing an element $d \in \mathbb{Q}\cap ( (0,1) \cup (2,3) )$ so that
$$
\{a_1,\dots, a_n, c\} \to \{ b_1,\dots ,b_n, d \}\\
a_i \mapsto b_i\\
c\mapsto d
$$
preserve the order (verify), and at the same time if for every $d' \in \mathbb{Q}\cap ( (0,1) \cup (2,3) )$ you can choose an element $c' \in \mathbb{Q}$ so that the map
$$
\{a_1,\dots, a_n, c'\} \to \{ b_1,\dots ,b_n, d' \}\\
a_i \mapsto b_i\\
c'\mapsto d'
$$
preserves the order.
The trick now is to go "back & forth" and define the isomorphism one step at a time. With $\omega$ steps you can define a chain of partial isomorphisms between $\mathbb{Q}$ and $\mathbb{Q}\cap ( (0,1) \cup (2,3) )$, so that the union of those is a (total) isomorphism.
Note that this last construction doesn't work with non-countable models: $(\mathbb{R}\backslash \{ 0 \} , < , =)$ is not isomorphic to $(\mathbb{R} , < , = )$.
