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Proof that algebraic numbers are countable

  1. Nov 19, 2011 #1
    In Rudin's Principles of Mathematical Analysis, exercise number two is to prove that the set of all algebraic numbers is countable. A complex number z is said to be algebraic if there are integers a_0, ..., a_n, not all zero such that

    a_0*z^n + a_1*z^(n-1) + ... + a_(n-1)*z + a_n = 0

    The hint given in the book is:
    For every positive integer N there are only finitely many equations with

    n + |a_0| + |a_1| + ... + |a_n| = N.

    I understand the gist of how the proof works. The above equation partitions the algebraic numbers into a countable collection of finite sets, proving that the algebraic numbers are countable. But my question is this:

    Couldn't we partition the algebraic numbers simply with the equation:

    |a_0| + |a_1| + ... + |a_n| = N

    Why add in the n when the each algebraic equation is fully characterized by the values a_0, a_1, ..., a_n? Or am I missing something?
     
  2. jcsd
  3. Nov 19, 2011 #2

    micromass

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    Is it obvious that only countably many equations satisfy

    [tex]|a_0|+...+|a_n|=N[/tex]

    Note that n can get arbitrarily large here!!
     
  4. Nov 19, 2011 #3
    so are you agreeing or disagreeing with me that the n is unnecessary and can be dropped?
     
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