Non-standard models of arithmetic for Dummies

Solution 1:

${\mathbb N}$ is more than an ordered set with successor. You have addition (and multiplication) as well. Let $M$ be a nonstandard model; we can identify its beginning with ${\mathbb N}$. Given $n\in M$ let $[n]$ be the collection of all $m$ that are at finite distance from $n$. Note $[n]={\mathbb N}$ if $n$ is finite, and otherwise $[n]$ has the same order type as ${\mathbb Z}$, because every number has a successor, and every number other than zero has a predecessor.

If $n$ is an infinite nonstandard number, then $n+n$ is also infinite but, moreover, it is infinitely away from $n$, so just from $M$ being nonstandard we deduce that there is no largest copy of ${\mathbb Z}$ in the order of $M$.

Also, either $n$ or $n+1$ is even, so there is an $m$ such that $m+m=n$ or $m+m=n+1$. This $m$ is infinite, so the copy $[m]$ of ${\mathbb Z}$ is before the copy $[n]$. This shows that there is no first copy of ${\mathbb Z}$ in $M$.

Finally, given $n<k$ and infinitely apart, $n+k$ or $n+k+1$ is even, and if $l$ is its half, then $n<l<k$, and $l$ is infinitely apart from both. This shows that between any two copies of ${\mathbb Z}$ we have another.

We have shown that, removing the original ${\mathbb N}$, and taking the quotient that identifies all elements in a class $[n]$ as a single point, we are left with a linear order that is dense in itself and has no end points. If $M$ is countable, this order is ${\mathbb Q}$. Otherwise, it is even more complicated.