# Confusion about a step of a proof of countability of algebraic numbers.

I am trying to understand the proof of the fact that The set of all algebraic numbers is a countable set. The proof I am reading is from "Theory of sets" by E.Kamke.The proof is quite similar to the proofs in some other website (Link 1) (Link 2). It starts from defining the height $$h$$ of a polynomial:

$$h = n + a_n + |a_{n-1}| + ... + |a_1| + |a_0|$$

Or in the second website

$$h = n + \sum_{i=0}^n |a_i|$$

What I dont understand is that , why do we also need $$n$$ ?

• You need the $n$ to limit the power of the polynomial. If we didn't have the $n$ and we defined height as $\sum |a_i|$ then $3x^2 + 2x + 1$ and $3x^3 + 2x^2 + 1$ and $x^{97} + 2x^{56} + x^{43} + 2$ would all have height $6$ and there would be an infinite number of polynomials with height $6$. The proof relies and being a countable union of finite sets. (Although a countable union of countable sets is countable but that would not be a direct construction) Jun 8, 2021 at 16:00

If you did not have the $$n$$ term then the following would all have height $$3$$:
$$x-2=0$$ $$x^2-2=0$$ $$x^3-2=0$$ $$x^4-2=0$$ $$x^5-2=0$$ $$\cdots$$
so there would be an infinite number their roots and so of distinct algebraic numbers of height $$3$$ and you would not reach those of height $$4$$ in your list.
• thanks for your response. I have another question (Hope you dont mind) , In the proof of E.Kamke , it is said that "We may suppose , moreover , without loss of generality , that $a_n$ > 0 ".But why? Is it really necessary? Jun 9, 2021 at 7:51
• If $a_n=0$ then you really have a polynomial of lower degree. If $a_n<0$ then you can multiply all the $a_i$ by $-1$ and you get the same roots. So you lose nothing by assuming $a_n>0$ Jun 9, 2021 at 8:10