Given that you only have multiplication and addition, how can you divide two numbers? I have a signature given with the real numbers as its universe and addition and multiplication as functions. I need to write the following expression in First Order Logic.


*

*$x$ is a rational number


My idea: $\varphi(x) =  \exists a \, \exists b \, x = \frac{a}{b}$.
The problem is I don't have division.
Extra question: how can I write 


*

*$x \geq 0$?

 A: Let me strengthen what others have already said:
The structure $\mathcal{R}=(\mathbb{R}; +, \times)$ is decidable (this is due to Tarski). This immediately rules out the possibility of defining $\mathbb{Z}$ in $\mathcal{R}$, since the theory of the integers is undecidable (by Goedel). (Incidentally, since $\mathbb{Z}$ is (nontrivially) definable in $\mathbb{Q}$, this also rules out the possibility of defining $\mathbb{Q}$ in $\mathcal{R}$.)
But in fact more is true: Tarski showed that it is o-minimal, that is, every definable set is a finite union of intervals. So nothing remotely like $\mathbb{Z}$ or $\mathbb{Q}$ can be a definable subset of $\mathcal{R}$.
A: EDIT: I just noticed you said first order arithmetic. This answer uses second order arithmetic. It's a theorem that it is impossible to define $\mathbb{Z}$ in $(\mathbb{R},+,\cdot)$
As mentioned in the comments, division is easy: define $a/b$ to be the number $c$ such that $cb=a$. It's identifying integers that is hard. Notably, $\mathbb{Z}$ and $\mathbb{Z}[\pi]$ have pretty much the same arithmatic structure. However, you can identify $0$ as the only number that satisfies $$\varphi(x):=\forall a(ax=x)$$ and then you can identify $1$ as the only number that satisfies $$\varphi'(x)=\forall a(\varphi(a)\lor a=ax)$$
Given these two constants, we can then recursively define the integers by using the fact that they are generated by $1$ as a group under addition, a la the Peano Axioms.
For your bonus question, again the comments had the right idea.
