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is the following proof correct (informally)?

  1. Using the characters $A=\{0,1,...8,9,x,+,-\}$ every polynomial can be written with finite many characters. E.g. $-7x^3+2x^2+1=0$ could be written as -7x3+2x2+1.
  2. The alphabet $A$ may be seen as a $|A|$-base number system. In this case it would be a $13$-base number system. This means the is a bijection between $\mathbb{N}$ and all (finite long) strings which can be built with $A$.
  3. Every polynomial has only finite many zeros.
  4. We can count over all strings that can be built with $A$. Which implies that we also can count over all existing polynomials (they are a subset of all strings of $A$). For every polynomial we can also count over all its finite zeros.

Therefore: There are only countable many algebraic numbers.

Thank you

  • 2
    Yes, informally this is perfectly fine. Note that you use in step 4 that a countable union of finite sets is countable. – 2017-02-24
  • 1
    Looks good to me. – 2017-02-24

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