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Math Help - Algebraic

  1. #1
    Junior Member
    Joined
    May 2008
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    70

    Algebraic

    let a = sqrt(7+4*sqrt(3)) prove that a is algebraic over Q

    now let a = sqrt(7) + i show that a is now algebraic over R
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  2. #2
    Super Member Gamma's Avatar
    Joined
    Dec 2008
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    Iowa City, IA
    Posts
    517

    Polynomials

    You just gotta find a polynomial over the respective field such that a is a root. Just takes some playing around with the numbers.

    1)
    a^2=7+4\sqrt3
    a^4=(a^2)^2=97 + 56\sqrt3

    so p(x) = x^4-14x^2+1 \in \mathbb{Q}[x] has a as a root. Dunno if it is irreducible or not, but you didn't ask for the degree of the extension. Clearly either 2 or 4 depending on whether p(x) factors into square terms. (Rational root test shows no linear terms).


    2)We are in \mathbb{R} now, so in particular, \sqrt{7} \in \mathbb{R}.

    Consider (a-\sqrt7)^2+1=0 so a is a root to (x-\sqrt7)^2+1=0 This pretty much has to be minimal as i is in complex and is a degree 2 extension over the reals, so it should be a degree 2 extension over the reals.

    Hope this helped
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