Is $\mathbb{N}$ definable over $(\mathbb{R},<,+,\cdot,0,1)$ is it possible to define the set $\mathbb{N}$ over the modell $(\mathbb{R},<,+,\cdot,0,1)$?
So, does a formula $\varphi$ exist which describes the natural numbers.
Unfortunatly I have no clue how I can find such a formula or prove that such a $\varphi$ does not exist.
I would be thankfull if someone could give me a hint.
Thanks.
 A: $\mathbb{N}$ is not a definable subset of this structure.
Well this model has quantifier elimination in $\left\langle +,.,0,1,<\right\rangle$ (quantifier elimination for a structure $S$ is the same as quantifier elimination for a theory $T$ except that $T \vdash$ is replaced by $S \vDash$ ) so definable subsets of $\mathbb{R}$ are finite unions of intervals. 
Indeed, quantifier-free formulas with one free variable $x$ are equivalent to boolean combinations of formulas of the form $\theta[x]: 0 <p(x) $ or $\theta[x]: p(x) = 0$ where $p$ is a polynomial with real coefficients (this exhaust the atomic formulas with parameters in $\mathbb{R}$ in that language).
Now $\theta[x]$ defines a finite union of intervals, and this quality is stable under taking boolean combinations, so quantifier-free formulas with one variable define finite unions of intervals, and by quantifier elimination, so do all formulas with one free variable.
So if $\mathbb{N}$ were definable, it would be a finite union of intervals, which it is not.

Let's prove quantifier elimination for this structure. Since you don't seem to have studied real closed fields, I'll do this the hard way.
We prove QE by induction on formulas, and since boolean combinations of quantifier-free formulas are obviously quantifier-free formulas, we only have to deal with formulas of the type $\exists x \varphi[x,\overline{y}]$ and $\forall x \varphi[x,\overline{y}]$ where $\varphi[x,\overline{y}]$ is without quantifiers in the inductive step.
The second type reduces to the first because $\forall \leftrightarrow \neg \neg \forall \leftrightarrow \neg \exists \neg$. We can also assume $\varphi[x,\overline{y}]$ is atomic or the negation of an atomic formula because $\exists$ is "distributive" over $\vee$. (and $. \wedge .$ is $\neg (\neg (.) \vee \neg (.))$)
So let $\varphi[x,\overline{y}]$ be an atomic formula or negation of an atomic formula in the language of ordered rings without quantifiers.
$\varphi[x,\overline{y}]$ is of the form $0 < p(x)$ or $0 = p(x)$ or $0 \leq p(x)$ or $0 \neq p(x)$ for some polynomial $p \in \mathbb{R}[X]$. We can dispense of the third form wich is the disjonction of the first two.
Each coefficient $f(\overline{y})_i$ of $p$ is a polynomial function of the tuple $\overline{y}$. So for the last form, the formula is equivalent to $\bigvee \limits_i \neg( f(\overline{y})_i =0)$ which is quantifier-free.
In the first case, let $f(\overline{y})_{\deg(p)}\prod \limits_r p_r$ be the decomposition of $p$ in irreductibe monic polynomials. Those of degree $2$ are stictly positive and for those of degree $1$, they are of constant sign over elements of a partition of $\mathbb{R}$ in intervals whose bounds are symetric functions, so polynomials in $\overline{y}$, so $p(x)$ taking a stricly positive value depends on the sign of $f(\overline{y})_{\deg(p)}$ and disjunction of conjunctions of formulas of the form $g_i(\overline{y}) < x < g_j(\overline{y})$. The formula is thus equivalent to a quantifier-free formula. 
The second case is equivalent $\neg (\exists x \neg (p(x) = 0)) \vee (\exists x (0 < p(x)) \wedge \exists x (0 < -p(x)))$ so this reduces to the first and last cases.
