Prove $\forall r \in \mathbb{R}. \exists k \in \mathbb{Z}. r < k$ I would like to prove that for every real number there exists an integer that is greater than it. My problem lies in that I am not sure how to construct the real numbers and provide their theory with the axioms sufficient for proving the fact. I do not think the statement is provable from the axioms of the real ordered field.
I can imagine intuitively (but cannot construct rigorously) a model of real numbers where there is some transcendental number $t \in \mathbb{R}$ such that $\forall x \in \mathbb{R}. t < x \implies x \in \mathbb{R}-\mathbb{Q}$.
 A: What about $\forall x \in \mathbb R$, let $x_0 = floor(x) \in \mathbb Z$.
Then you know that $x_0 \le x < x_0 +1 = x_1$, $x_1\in \mathbb Z$
So say you claim that $\pi$ doesn't have any greater integer, round down $\pi$ to $3$, add $1$ and you found yourself with $4$ which has to be greater.
A: The way you should prove this fact precisely depends largely on the specific construction you have in mind (metric completion of rationals? Dedekind cuts? Something else?) but for any standard construction the proof should be trivial, but it's impossible to get into the details until you fix a specific construction.
On a different note, this question has little to do with model theory as stated. The reals are not just any model of some theory, they are a specific structure (modulo different constructions I mentioned before, but this is really the same thing in any case, in a fixed universe of ZFC anyway).
You may consider the first order theory of reals, but for that you need to specify a language, but its models will seldom actually be real numbers. For instance, any model of the real field which has infinite elements (realizing the type $1+1+...+1<x$ where the number of 1s is arbitrary) will not be the real numbers, as they clearly omit it. Similarly, any model of cardinality different than that of continuum won't work.
A: From the perspective of model theory (since that was one of your tags), you cannot express this property as a first-order sentence, so no first-order set of axioms for $(\mathbb{R}, 0, 1, +, \cdot, <)$ will suffice. Indeed, in $\text{Th}(\mathbb{R}, 0, 1, +, \cdot, <)$, you have a consistent type generated by formulas of the form $x > 1 + \ldots + 1$, which is therefore realized in some model of the theory. This inability to control infinitary behaviour is one of the main features of first-order logic.
On the other hand, if you want to prove that you can do this in the particular case where your model is just $\mathbb{R}$, it's easy. Let $x \in \mathbb{R}$, and suppose $x > 0$. Then the set $\{ n \in \mathbb{Z} \mid n > x\}$ is a set of natural numbers, and hence has a least element $k$. 
You can go on to prove the Archimedean property. Note that $k - 1 \leq x$ because $k$ was minimal amongst $n$ with $n > x$. On the other hand, if $m > k -1$, then we must have $m \geq k$ and so $m > x$. So $k - 1$ is the greatest integer smaller than $x$. From here, it's easy to do the case with $x < 0$.
But, of course, the key property of the natural numbers here, namely the fact that they are well-ordered, is not expressible in first-order logic - and you'll find this to be true of any proof.
