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Real number is Algebraic

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data
    A real number x[tex]\in[/tex]R is called algebraic if there exists integers [tex]a_{0}[/tex][tex]x^{n}[/tex]+[tex]a_{n1}[/tex][tex]x^{n1}[/tex]+.....+[tex]a_{1}[/tex]x+[tex]a_{0}[/tex]=0.
    Show that [tex]\sqrt{2}[/tex],[tex]\sqrt[3]{2}[/tex], and 3+[tex]\sqrt{2}[/tex] are algebraic.
    Fix n[tex]\in[/tex]N and let [tex]A_{n}[/tex] be the algebraic numbers obtained as roots of polynomials with integer coefficients that have degree n. Using the fact that every polynomial has a finite number of roots, show that [tex]A_{n}[/tex] is countable.


    2. Relevant equations



    3. The attempt at a solution
    Completely confused on this one.
     
  2. jcsd
  3. Sep 14, 2010 #2
    I think it's the notation that's getting me.
     
  4. Sep 14, 2010 #3

    Dick

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    What part are you confused about? Surely you can show sqrt(2) is algebraic?
     
  5. Sep 14, 2010 #4
    It's the whole notation. I'm not sure how to apply the a's and x's to sqrt(2).
     
  6. Sep 14, 2010 #5

    Dick

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    It just means show sqrt(2) is the solution of polynomial equation with integer coefficients. I'll get you started. sqrt(2) is a root of 1*x^2-2=0. That's all the a_n*x^n+... is supposed to mean. How about (2)^(1/3)?
     
  7. Sep 14, 2010 #6
    x^3-2=0
    Well for 3+sqrt(2)
    x^2-(3+sqrt(2))=0
     
  8. Sep 14, 2010 #7

    Dick

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    x^3-2=0 is good. That's a start. As for x^2-(3+sqrt(2))=0, (3+sqrt(2)) doesn't solve that equation AND (3+sqrt(2)) isn't an integer. You need an equation with integer coefficients. You have to work a little harder on this one.
     
  9. Sep 14, 2010 #8
    I'm assuming it is of the ax^2+bx+c=0, but I'm having trouble coming up with a, b, and c.
     
  10. Sep 14, 2010 #9
    x^2-6x+7
     
  11. Sep 14, 2010 #10

    Dick

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    Start with x=3+sqrt(2). That's x-3=sqrt(2). Square both sides.
     
  12. Sep 14, 2010 #11

    Dick

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    Yes. Though x^2-6x+7=0 is the equation, right? So 3+sqrt(2) is algebraic.
     
  13. Sep 14, 2010 #12
    x^2=9+6sqrt(2)+2
    x^2-6sqrt(2)=11
    x^4-12sqrt(2)+72=121
     
  14. Sep 14, 2010 #13

    Dick

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    You already figured out that one equation is x^2-6x+7=0, didn't you? That's fine. You can turn that what you just wrote into a eighth degree equation for 3+sqrt(2) but you don't need it. You've already got one.
     
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