# 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 $\forall x \in \mathbb{R}. t < x \implies x \in \mathbb{R}-\mathbb{Q}$.

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–  Prahlad Vaidyanathan Oct 13 '13 at 10:31
Starting with $\mathbb{Q}$ you can define a real number $r$ as a proper subset of $\mathbb{Q}$ such that for $q\in r$ implies that $p\in r$ for every $p\in\mathbb{Q}$ with $p<q$. –  drhab Oct 13 '13 at 10:35
I don't think $\Bbb Z$ is definable in the context of complete ordered fields. –  mercio Oct 13 '13 at 10:56
Just a thought: You could try defining $\mathbb{Z}$ as the least subset of $\mathbb{R}$ closed under addition and negation and containing $1$. –  goblin Oct 13 '13 at 11:01
In which language do you work ? As mercio pointed it out, $\mathbb Z$ isn't first-order definable in $\mathbb R$ in the language of ordered field (which is non trivial I think), and so, in this context, your sentence $\forall r \in \mathbb R \exists k \in \mathbb Z, r<k$ does not even make sense (actually, does not exist). –  Pece Oct 13 '13 at 16:15

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.

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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.

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You have to define the floor function on every real number. –  Dávid Tóth Oct 13 '13 at 10:45
let $x_0 = floor(x)$ be the greatest integer $x_0 \in \mathbb Z$ such that $x_0 \le x$ –  user88595 Oct 13 '13 at 10:47
which proves that $x_1 = x_0 + 1$ has to be greater than your real number. –  user88595 Oct 13 '13 at 10:49
How do you know such a greatest integer exists? –  Shahab Oct 13 '13 at 10:50
Every real number $r$ has an absolute value. Then if you know that for reach integer $n$ we have $|n+1|-|n|=1$ then you know that the set $\{|n|: n\in \mathbb{Z}\}$ is unbounded, thus given a real number $r$ there exists an $n\in \mathbb{Z}$ such that $|n|>|r|$. –  dezign Oct 13 '13 at 10:51

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.

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