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

  • #1

Homework Statement


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?

Homework Equations



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.
 
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Answers and Replies

  • #2
dextercioby
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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.
 
  • #3


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.
How exactly can you always just raise them to a suitable power without creating additional radicals to be solved?
 
  • #4
Dick
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How exactly can you always just raise them to a suitable power without creating additional radicals to be solved?
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.
 
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