(Logic) Formally writing a rational number in logic How do I "formally write" a rational number $a_i$ in a logic formula?
For example, I was taught that $x^2$ should be formally written as $F_\times(x_1,x_1)$, $1$ should be formally written as $c_1$, $2$ should be formally written as $F_+(c_1,c_1)$ and so on.
I hope my question isn't too ambiguous.
This is related to my previous question: How to show that the property of being algebraically closed is reflected by elementary extensions?
The main idea is I want to write out the formula $\phi_n$ [credit to André Nicolas who suggested it] ,
$$\forall w_0\forall w_1\cdots\forall w_n \exists x\left(x^{n+1}+w_nx^n+\cdots+w_0=0\right).$$
and specify that all the $w_i$ are rationals.

background information about the underlying question 
If $p(x)=x^{n+1}+a_nx^n+\cdots+a_1x+a_0$ is a polynomial such that $\{a_0,\cdots,a_n\}\subset\mathbb{Q}$, then $p(x)=0$ has a solution in $F$.
Assume that the structure $(F,0,1,+,\cdot)$ is a countable elementary submodel of the complex field $(\mathbb{C},0,1,+,\cdot)$. 

Sincere thanks for any help!
 A: You don't need to quantify over the coefficients to solve your original problem How to show that the property of being algebraically closed is reflected by elementary extensions? (as I have explained there).
The first-order theory of algebraically closed fields of a given characteristic is decidable. In the characteristic $0$ case, adding a predicate for the rational numbers results in an undecidable theory, so it is not a good way to go for your original problem.
A: You can describe any specific rational by a formula. Let us find, for example, a formula $\psi_{2/3}(x)$ that "says" that $x=2/3$. 
We use the notation implicit in your post.
A formula $\psi_{2/3}(x)$ that does the job is:
$$F_{\times}(F_+(F_+(c_1,c_1),c_1),x)=F_+(c_1,c_1).$$
Any rational number can be specified in this way. The negative ones require an additional trick.
Thus for any specific polynomial $P(x)$ with rational coefficients, we can write a sentence that says that this particular polynomial has a zero. And if we want to specify that every non-constant polynomial with rational coefficients has a zero, we can do so using a countably infinite set of axioms.
However, we do not really need the predicates that identify the specific rationals. For if $P(x)$ is a polynomial with rational coefficients, then $P(x)$ has a zero if and only if a certain  polynomial $P^\ast(x)$ has a zero, where $P^\ast$ is obtained from $P$ by multiplying by a suitable integer, a common denominator of the coefficients of $P$.
And we do not need negative integers. For the polynomial $P(x)$ has a zero if and only if the equation $P_{+}(x)=P_{-}(x)$ has a solution, where $P_{+}$ is the polynomial whose coefficients are the positive coefficients of $P$, and $P_{-}$ is the polynomial whose coefficients are the absolute values of the negative coefficients of $P$.  
Remark: As explained in the answer of Rob Arthan, introducing a new predicate symbol that says "I am rational" or "I am natural" has undesirable consequences, for decidability, and also for the model theory.  
