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Rudin's exercise 2.2 asks for a proof that the set of all algebraic numbers are countable. I'm trying to understand a proof of this, which uses his hint. The first step proceeds as follow.

For each $N$, let $E_N$ denote the set of equations $a_0 z^n + a_1 z^{n-1} + \ldots + a_n = 0$ where $n \geq 0$ and $a_0, \ldots, a_n$ are not all zero, and $n + |a_0| + \ldots + |a_n| = N$. For each $n$, there are only finitely many $(n+1)$ tuples of integers satisfying $|a_0| + \ldots + |a_n| = N - n$, as each $a_i$ can only take on $2(N-n) + 1$ values.

I cannot figure out where the $2(N-n) + 1$ comes from. I'm not sure what happens in the case where $n > N$ in which case $N-n < 0$. I assume $a_0 = \ldots = a_n = 0$. If $N > n$, then each $a_i$ seems that it can take on values from $0$ to $N-n$. They can't exceed $N-n$ because then. I'd have to subtract to get the sum to balance, and I can possibly let all but one coefficient equal zero. The counting here doesn't. make full sense to me. I'm also not sure why the possibility that all coefficients are zero is ruled out.

I'd apprecicate some help understanding this solution. Once I understand this step, I think I should be able to understand the rest.

Mathematical Endeavors
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First: If all coefficients $a_k$ are zero, then any $x$ satisfies the equation, which one doesn't want since one is trying to count only algebraic numbers.

Second: The number $N$ is only defined after one fixes an equation witn $x$ as a root. Since the coefficients are not all zero, the sum of their absolute values is strictly positive, which from the definition of $N$ implies that $n<N.$

Third: Any $a_i$ satisfies $|a_i| \le N-n$ and if you list these there is $0$ along with each $\pm k$ for $1 \le k \le N-n$ so $2(N=n)+1$ possible.

coffeemath
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