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## Proving algebraic numbers are countable?

 Quote by SMA_01 There are finitely many polynomials of degree n. So if I show that the number of polynomials is countable, I can automatically say the roots are countable?
You can't 'automatically' say anything without giving a reason. Give a reason.

 Well because for an integer polynomial of degree n, there are at most n roots.

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 Quote by SMA_01 Well because for an integer polynomial of degree n, there are at most n roots.
That is certainly part of it. Nothing you are saying is wrong. I'm just not hearing the reason that the hint suggests you should be giving. Why is the number of integer polynomials of degree n countable (NOT finite)? You know it is. I know it is. Why is it?

 Is it because the sum of the absolute values of the integer coefficients, this sum will always correspond to a value in the set of all positive integers?

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 Quote by SMA_01 Is it because the sum of the absolute values of the integer coefficients, this sum will always correspond to a value in the set of all positive integers?
No. That's not it at all. Take the polynomial ax+b with |a|<2 and |b|<2. Count them all. Really, how many are there? Give me a number. This is what the hint wants you to do.

 You have a polynomial of degree 1. You have: lal=1, lbl=0 lal=1, lbl=1 and you can't have lal=0. So there are 2 polynomials?

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 Quote by SMA_01 You have a polynomial of degree 1. You have: lal=1, lbl=0 lal=1, lbl=1 and you can't have lal=0. So there are 2 polynomials?
Ok, we are getting somewhere. Actually a can be anything thing in {1,-1} and b can be anything in {1,0,-1} so I actually count 6=2*3 polynomials. The exact number isn't important. It just wants you to show the number is finite by giving an upper bound that is finite. I would say there are at most 3 choices for a and 3 choices for b and there is at most one root for each polynomial so the number of roots is definitely less than 3*3*1. Some of the polynomials I'm counting aren't degree 1 and some don't have roots and some have the same root, but its still an upper bound for the number of roots. Which is finite. What's an upper bound for a polynomial of degree n with |a_i|<m?

 Okay, I see how you did that. I'm not sure about the case of m, but I got that there are at most 2m-1 choices for each a_i, is that correct? If it is, then there are n+1 coefficients, and each coefficient can have at most 2m-1 choices, so (2m-1)^(n+1) polynomials, and [(2m-1)^(n+1)]*n is an upper bound? Edit: By a_i I took it as all the coefficients of the polynomial.

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 Quote by SMA_01 Okay, I see how you did that. I'm not sure about the case of m, but I got that there are at most 2m-1 choices for each a_i, is that correct? If it is, then there are n+1 coefficients, and each coefficient can have at most 2m-1 choices, so (2m-1)^(n+1) polynomials, and [(2m-1)^(n+1)]*n is an upper bound? Edit: By a_i I took it as all the coefficients of the polynomial.
Great! So the number of algebraic numbers satisfying a integer polynomial of degree n with |a_i|<m is finite. Do you see how to conclude the number of algebraic numbers satisfying ANY integer polynomial of degree n is countable?

 Yes, I see how it is finite. So then it is countable, because finite sets are countable, right?

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 Quote by SMA_01 Yes, I see how it is finite. So then it is countable, because finite sets are countable, right?
The roots of integer polynomials satisfying |a_i|<m are finite. But that's not what A_n is. You have to take a union over all m to get all of the contents of A_n. Why is that countable?

 Because the union of countable sets is countable, is that correct?

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