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Some advice please ?

  1. Jun 13, 2007 #1
    Hello! I am thirteen and *very* interested in mathematics.

    I've done calculus 1-3, some ODE's, and some linear algebra. I just started into basic real analysis (while completing linear algebra/ODE's), and the problems are *really* tough. I can solve a good number of the problems, but some I can't; how can I build up my skills in this area? Are there any problems online (preferably with answers)?

    I'm using Kolmogorov and Fomin's Introductory Real Analysis. Thanks.
  2. jcsd
  3. Jun 13, 2007 #2
    Wow, you're really on the ball. Congratulations on your initiative, I wish I was half that motivated when I was 13.

    I'm halfway through a two-semester Intro to Real Analysis course myself. In general, I have a lot of success on the homework when I write down exactly the definitions and assumptions of the problem, and exactly what I am trying to prove. Usually the proof comes easily after a little inspection, but if not then I'll write a statement logically equivalent to what I'm trying to prove and see if that helps. If you have some specific questions I'm sure the knowledgable denizens of PF could be of more assistance.
  4. Jun 13, 2007 #3
    Hey! Thanks :D

    For example, my book is asking me to prove that the set of algebraic numbers is countable; I'm really having trouble with it. Help would be *greatly* appreciated. Thanks!
  5. Jun 13, 2007 #4
    Well, one can associate a polynomial with rational coefficients to every algebraic number, and if one can prove that such polynomials are countable then you would be done.
  6. Jun 13, 2007 #5


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    basically, cantor proved countable unions of countable sets are countable, by one of his two famous diagonal arguments, so to prove countability you try to write your set as a countable union, of finite or countable sets.
  7. Jun 14, 2007 #6
    So say [tex]p(x) = a_{0} + a_{1}x + a_{2}x^2 + ... + a_{n}x^n = 0[/tex] for every algebraic number x. I'd just need to prove that there is a countable number of polynomial equations of the form p(x) for every x, right?

    (The a's are supposed to have subscripts, not coefficients! I can't seem to get the subsctript thingy to work in latex)
    Last edited: Jun 14, 2007
  8. Jun 14, 2007 #7
    Yeah, but keep in mind that the coefficients are not arbitrary real numbers.
  9. Jun 16, 2007 #8
    Hey! I think I did it... here's my proof:

    We can associate every algebraic number [tex]\mbox{x}[/tex] with a polynomial equation [tex]p(x) = a_{0} + a_{1}x + a_{2}x^2 + ... + a_{n}x^n = 0[/tex]. Let P be the set of all polynomials of the form p(x). Then P can be written as the union of all sets of polynomials of order n. By the Fundamental Theorem of Algebra, polynomials of order n have exactly n solutions. Therefore, P is countable, and it follows that the set of algebraic numbers is countable.

    Thanks for everybody's help!
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