Show on $\mathbb{N}$ there are $2^{\aleph_0}$ nonisomorphic linear orders. How can I attack this problem? 
My idea is for $X \subset \mathbb{N}$ set up a linear order $O_{X}$ such that if $X \not =Y$ then $O_X \not = O_Y$. 
 A: Given $A\subseteq \Bbb N$, define $S\subset \Bbb R$ as follows:
$$S:=\Bbb Z\cup \bigcup_{a\in A}\{\,a-\tfrac1{n+1}\mid n\in\Bbb N\,\}\cup  \bigcup_{a\in\Bbb N\setminus A}\{\,a+\tfrac1{n+1}\mid n\in\Bbb N\,\}.$$
As $|S|=\aleph_0$, there exists a bijection $f\colon S\to\Bbb N$ and from the inherited order on $S$, obtain a linear order on $\Bbb N$ as $x\prec y :\iff f^{-1}(x)<f^{-1}(y)$.
For different $A$, these orders are non-isomorphic. Indeed, we can find the following subsets in $\Bbb N$ based on properties of the order alone:
$$\begin{align}A'&:=f(A)=\bigl\{\,x\in \Bbb N\bigm| x=\sup_\prec \{\,y\in S\mid y\prec x\,\}\,\bigr\}\\
B'&:=f(\Bbb N\setminus A)=\bigl\{\,x\in \Bbb N\bigm| x=\inf_\prec \{\,y\in S\mid x\prec y\,\}\,\bigr\}\\
N'&:=A'\cup B' \end{align}$$
And now we can retrieve $A$ without knowing $f$ because
$$n\in A\iff \exists x\in A'\colon|\{\,y\in N'\mid y\preceq x\,\}|=n.  $$
A: To simplify Hagen's suggestion, and also something I incidentally thought about just a couple of days ago:
For every $A\subseteq\Bbb N$ consider $R_A=\sum_{n\in A}(\Bbb Z+n)$ as a linear order. Namely, It is a copy of $\Bbb Z$, then $a_0$ elements, another copy of $\Bbb Z$, followed by $a_1$ elements, and so on, where $a_n$ is the $n$th member of $A$ by increasing order.
It is not hard to show that for $A\neq B$, the linear orders $R_A$ and $R_B$ are not isomorphic (e.g. if $n\in A\setminus B$, then $R_A$ has a maximal finite interval of exactly $n$ points). So there are continuum many of these.
The key point is to note that we can write a sentence in the language of orders "there are $n$ elements which have no additional point between them, and any other point has at least one more point between it and our $n$ points".
