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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.

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There is no power series (except the trivial one) whose zeros all real numbers. –  Wadim Zudilin Oct 17 '10 at 0:33
Sorry for the ambiguity- I mean that there exists for all real numbers some power series with rational coefficients such that the real number will be a zero. –  user6204 Oct 17 '10 at 0:37
@Wadim: For $x=e$, take a Taylor series of $\log x - 1$ ? –  Mark Sapir Oct 17 '10 at 0:57
And for x= log2, take a Taylor series of e^x-2. –  user6204 Oct 17 '10 at 1:13
True story: As a freshman at Princeton, I asked this very question to a professor after class one day. Professor responded with "Have you heard of Paul Erdos?" Of course, I said "Yes." Professor then said "He asks a lot of trivial questions too," and stormed off. –  Marty Oct 17 '10 at 16:40

3 Answers 3

up vote 26 down vote accepted

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 $|p(\alpha)-r\alpha^n|\lt\epsilon/2$.

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Gerry, it's your Sunday! –  Wadim Zudilin Oct 17 '10 at 1:07
Thanks a lot, that seems to be the best approach. –  user6204 Oct 17 '10 at 1:09
I've remembered when I first thought about this problem. As a PhD student, having learned that an algebraic number is a zero of a polynomial with rational coefficients, I decided that an analytic number should be a zero of a power series with rational coefficients (and then analytic number theory should properly be the study of analytic numbers). But then I found that every real number is analytic, and that was the end of that. –  Gerry Myerson Oct 17 '10 at 10:15

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 $h(x)=g(x)-1$.

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Thanks, that's a great method. –  user6204 Oct 17 '10 at 3:57

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.

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Thank you for the reference. –  user6204 Oct 17 '10 at 10:21

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