How many algebraic numbers are there

We now have a series of nested sets that encompass every possible root for every possible polynomial with integer coefficients. You can express the set of algebraic numbers as a countable union of finite sets. For every n you only have finitely such polynomials and each polynomial has only finitely many roots.

Your proof is correct. The only thing I would do is explain, when you say you have a series of nested sets, exactly what those sets are and then be explicit about the theorem you are using to get that their union is countable. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Learn more. Prove that the set of all algebraic numbers is countable Ask Question.

Asked 8 years, 2 months ago. Active 3 years, 5 months ago. Viewed 53k times. Martin Sleziak PandaMan PandaMan 2, 2 2 gold badges 22 22 silver badges 42 42 bronze badges. The proof suggested by the hint has a somewhat more constructive character. But yours is a good proof. If you are taking the union of all n-tuples of any integers, is that not just the set of all subsets of the integers? Add a comment. Active Oldest Votes.

Jim Jim Sign up or log in Sign up using Google. Foundations of the Theory of Algebraic Numbers, Vol. New York: Macmillan, Koch, H. Number Theory: Algebraic Numbers and Functions.

Providence, RI: Amer. Nagell, T. Introduction to Number Theory. New York: Wiley, p. Narkiewicz, W. Warsaw: Polish Scientific Publishers, Nesterenko, Yu. Unpublished manuscript. Wagon, S. New York: W. Freeman, pp. Wolfram, S. A New Kind of Science. Champaign, IL: Wolfram Media, p.

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