# Showing that the set of all algebraic numbers is countable [duplicate]

Is this a valid proof (Problem 1.23)?

Let the roots of a polynomial of $n$th degree (whose coefficients are rational) be written as:

$$P_n^{\mathbb{Q}}$$

and the polynomial is then:

$$a_nx^n+a_{n-1}x^{n-1}+...+a_0x^0$$

where $a_i \in \mathbb{Q}$ where $0 \leq i \leq n$ and $i \in \mathbb{N}$

Then $P_n^{\mathbb{Q}}$ will have at most $n$ roots.

Then the set of all roots is:

$$A = \cup_{n \in \mathbb{N}}P_n^{\mathbb{Q}}$$

Then $A$ is countable because: 1) There is a countable number of the sets of roots of the polynomial by definition i.e. ($\cup_{n \in \mathbb{N}} P_n^{\mathbb{Q}}$). 2) The number of elements in each set is countable.

And we know that the countable union of the countable sets is countable.

EDIT: My question is different to the one linked, because I am merely asking whether my proof is valid.

## marked as duplicate by Chris Brooks, C. Falcon, Juniven, Adam Hughes, R_DFeb 24 '17 at 4:42

• I don't think it does work. The number of n degree polynomials is |$\mathbb Q^n$| and so the number of algebraic numbers is $\le \sum n(|\mathbb Q^n|)\n \rightarrow \infty$ but $|Q^{\infty}|$ is uncountable. It's a subtle point and it trips me up every time but there is an obscure hint to not take n degree polynmoials but all polynomials where $|a_n| < M$ – fleablood Feb 23 '17 at 21:52
• @fleablood could you please edit your equations. I cannot quite understand in that second part after $|\mathbb {Q}^n|$ what you are trying to say. Thanks – i squared - Keep it Real Feb 23 '17 at 21:56