If they were inconsistent, and so proved *every* sentence in the language of arithmetic, they'd prove some sentences that we assume the intended model $\script N$ does *not* satisfy. That is, we assume that even if the Peano axioms are screwy in some subtle way we haven't yet identified, we still do *intend* some model of the language of arithmetic, which like any model refuses to satisfy $\neg\chi$ when it satisfies $\chi$. (Whether we do manage to intend this could also be challenged.)

**Warning:** I am sure that the model I've described has the right general shape to be the countable non-standard model of arithmetic. (Specifically, it has the right "order type," and it will satisfy many obvious truths of arithmetic.) But I haven't fully specified the interpretation of $+$ and $*$, and there are known difficult issues about doing so. So I can't warrant that what I've described is adequate in all respects. However, it should be good enough for you to have a working idea of what this model is like. See comments at the end of the page.

This is connected to a question Sam asked in class. He asked us to consider the set $\set{\neg\forall x\theta x,\theta(0),\theta(S0),\theta(SS0),\dots}$. (Sets of this sort are called $\omega$-inconsistent.) Observe that for any finite subset of this, you could find *some* $\theta$ where that subset is satisfied on the intended model of arithmetic.
But there's no *single* $\theta$ for which all the finite subsets would be true, on the intended model. Still, every finite subset would be satisfi*able*, for *some* model, which interpreted symbols differently than the intended model does.
And so compactness then tells us that *the whole set* must be satisfiable. But intuitively it might seem like it shouldn't be. Doesn't the first sentence contradict what all the rest of the sentences together say? Well, *only* if we've managed to somehow introduce the requirement that the only objects there be are $0, S0, SS0, \dots$. Let's call objects of this sort, that are reachable by applying the successor function a finite number of times to $0$, "finitely generated." If we haven't excluded the possibility of there being additional objects that *weren't* finitely generated, then a model could satisfy all the sentences $\theta(0), \theta(S0), \theta(SS0), \dots$ while giving $\theta$ an interpretation that *doesn't* apply to the additional objects.
And as we were saying, in first-order logic it's *not* possible to require that the only objects there be were finitely generated. So there are no axioms we could state in a first-order language that would rule out the truth of every such set of sentences. As mentioned above, though, on the *intended* model $\script N$ of arithmetic, no such set is true --- there is no $\theta$ where $\script N$ satisfies the whole such set.

The preceding remarks still leave it open that there might be some *other* difference that the theory of true arithmetic could catch: some sentence that one of the models would make true and the other make false. But I report that there isn't. These models satisfy all the same first-order sentences in the language of arithmetic.
Let's press on and extend our model still further. Let $\script N'''$ include all the same elements from $\script N''$, and *also* all the integer points on the horizontal line where $y=\operatorname{squash}(\pi)$, too, and so on for every real. Now we'll have all *these* points in the complex plane:
> $\set{(x,0) \where x\in\N} \union \set{(x,n + \operatorname{squash}(y)) \where x\in\Z \wedge n\in\N \wedge y\in \R^+}$.
which can be simplified to:
> $\set{(x,0) \where x\in\N} \union \set{(x,y) \in \Z\times\pmb{\R^+} \where y\nin\N}$.
If we keep in mind that the cardinality of the reals exceeds the cardinality of $\Q$, it should be clear that *this* model's domain is *larger* than that of $\script N$ and the other models so far considered. There is no bijection between their domains; so they cannot be isomorphic. Yet this model too will also satisfy true arithmetic.
So we had $\script N, \script N', \script N'', \script N'''$. The first two were obviously isomorphic, and the fourth has a bigger domain so is *obviously* not isomorphic to any of the others. The third was *non*-obviously not isomorphic to the first two. $\script N''$ is a (the) countable non-standard model of true arithmetic, and $\script N'''$ is one uncountable non-standard model.
One of the important metalogical results introduced this week, the Loewenheim-Skolem theorem (in its "upward" form) will tell us that we should *expect* to find non-standard models like $\script N'''$. When a theory has any infinite model, it will also have models with larger domains. First-order theories are "blind" to the differences between different infinite cardinalities.
To read more, have a look at [Wikipedia](http://en.wikipedia.org/wiki/Peano_axioms).