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Non-Transcendental Numbers Def. What if we allow √14 as a coefficicie?

  1. Feb 18, 2013 #1
    1. The problem statement, all variables and given/known data
    A non-transcendental number is one that's a root of a (non-constant) polynomial with rational coefficients.
    Does allowing radicals as coefficients, eg: 5√3, 2^(1/3) get us any new different numbers?

    2. Relevant equations

    3. The attempt at a solution
    1. In some cases we get no new numbers, eg:
    [itex]x^2+\sqrt[3]{3}=0\Leftrightarrow x^6=3[/itex]

    [itex]x^2+x\sqrt{3}+\sqrt{2}=0\Leftrightarrow x^2+x\sqrt{3}=-\sqrt{2} \Rightarrow x^4+3x^2+2x^3\sqrt{3}=2 \Leftrightarrow x^4+3x^2-2=-2x^3\sqrt{3} \Rightarrow (x^4+3x^2-2)^2=4x^6 \times 3[/itex]

    We solve this by isolating the radicalss and squaring/cubing/etc them. But with more coefficients it becomes harder. What about a 5th degree polynomials with only cubic radical coefficients?

    So the question is:
    -Do we get new non-trascendental numbers if we allow rational AND radicals as coefficients.
    -Can we always turn a polynomial with radical coefficients into a one with rational coefficients?

    PS: By radicals I mean just some number that's written with the nth-root symbol. Use whatever definition you feel fits best.
     
    Last edited: Feb 18, 2013
  2. jcsd
  3. Feb 18, 2013 #2

    dextercioby

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    Re: Non-Transcendental Numbers Def. What if we allow √14 as a coeffici

    No, you can get rid of all the (finite nr. of) radicals after a finite number of raising to a suitable natural power, so that the definition of algebraic numbers is exhaustive.
     
  4. Feb 18, 2013 #3
    Re: Non-Transcendental Numbers Def. What if we allow √14 as a coeffici

    How exactly can you always just raise them to a suitable power without creating additional radicals to be solved?
     
  5. Feb 18, 2013 #4

    Dick

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    Re: Non-Transcendental Numbers Def. What if we allow √14 as a coeffici

    You usually say 'algebraic' instead of non-transcendental. And no, allowing algebraic numbers as coefficients instead of integers will not give you any transcendental roots. The roots of an polynomial with algebraic coefficients are algebraic. I don't think the proof is as simple as arguing you can clear all the radicals by taking powers. I kind of doubt it's true. But the algebraic part is true nonetheless.
     
    Last edited: Feb 18, 2013
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