Proof that algebraic numbers are countable

  • Thread starter jecharla
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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?
 

Answers and Replies

  • #2
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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!!
 
  • #3
24
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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!!
so are you agreeing or disagreeing with me that the n is unnecessary and can be dropped?
 

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