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Math Help - Algebraic - countable

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    Algebraic - countable

    A real number xR is called algebraic if there exists integers a0x^n+an1x^n1+.....+alx+a0=0.
    Show that (2)^1/2,(2)^1/3, and 3+(2)^1/2 are algebraic.
    Fix n in N and let An 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 is countable.
    The algebraic definition is getting me confused.
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    Quote Originally Posted by kathrynmath View Post
    A real number xR is called algebraic if there exists integers a0x^n+an1x^n1+.....+alx+a0=0.
    Show that (2)^1/2,(2)^1/3, and 3+(2)^1/2 are algebraic.
    Fix n in N and let An 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 is countable.
    The algebraic definition is getting me confused.
    The definition you quoted isn't formatted/typed correctly. You can look @ mathworld for one that's typeset properly

    consider (x - sqrt(2)) * (x + sqrt(2)). Not sure how to handle 2^(1/3) and 3+2^(1/2) but someone else will probably jump on those.

    For the second part see here

    http://www.mathhelpforum.com/math-he...le-156204.html

    Edit: I see from mathworld that the radical integers are a subring of the algebraic integers. If you wanted to prove that much stronger result then all the special cases asked for in the problem would immediately follow.
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    Quote Originally Posted by kathrynmath View Post
    A real number xR is called algebraic if there exists integers a0x^n+an1x^n1+.....+alx+a0=0.
    Show that (2)^1/2,(2)^1/3, and 3+(2)^1/2 are algebraic.
    Fix n in N and let An 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 is countable.
    The algebraic definition is getting me confused.

    Put x=2^{1/3}=\sqrt[3]{2}\Longrightarrow x^3=2\Longleftrightarrow x^3-2=0 , and thus our element is a root of f(t)=t^3-2.

    Try to do something similar with the other ones.

    Tonio
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