How to prove that every real number is a zero of some power series with rational coefficients (if true) How would one approach proving that every real number is a zero of some power series with rational coefficients? I suspect that it is true, but there may exist some zero of a non-analytic function that is not a zero of any analytic function. I was thinking about approaching the problem using arguments of cardinality, but I am unsure about how to begin.
Thank you in advance.  
 A: From "Note on integral functions" by A.G Walker:

The following theorem, though perhaps well known, does not appear in any of the elementary text-books of analysis. It may, however, be of interest in connection with the various rational representations of irrational numbers, and an elementary proof is given.
Theorem. Every real number is a zero of a rational integral function, i.e. a function expressible as a power series, with rational coefficients, convergent for all x.

The proof is basically the same as the one in Gerry's answer.
A: Call your real number $\alpha$. Suppose you have found a polynomial $p$ of degree $n-1$ with rational coefficients such that $|p(\alpha)|\lt\epsilon$. Show you can find a rational $r$ such that $|p(\alpha)-r\alpha^n|\lt\epsilon/2$. 
A: Here is a more explicit answer (I think). For convenience we may suppose $\frac 12\leq \alpha<1$. Let 
$2^{-k_1}+2^{-k_2}+\cdots$ be the binary expansion of $\alpha$, where $1 = k_1 $$ < k_2<\cdots$. Let $f(x)=\sum 2^{-k_i}x^i$. Then $f(1)=\alpha$. Let $g(x)$ be the compositional inverse of $f(x)$, i.e., $f(g(x))=g(f(x))=x$. Note that $f(x)=x+\cdots$, $g(x)$ is defined formally (i.e., without worrying about convergence) and has rational coefficients. Moreover, every coefficient has absolute value at most 1. Hence $\alpha$ lies in the region of convergence of $g(x)$, so $h(\alpha)=0$, where $h(x)=g(x)-1$.  
