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Is that an algebraic number?

  1. May 31, 2009 #1
    is [tex]\sqrt{2}+\sqrt{5}[/tex] an algebraic number?
    i used 2 and 5 arbitrarily, try any integers (as long as they are not the same integer, in which case it is algebraic)
    I tried finding a polynomial with rational coefficients that zeros at this value, but haven't found any.
     
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  3. May 31, 2009 #2

    Hurkyl

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    If it was, its powers would span a finite dimensional vector space over Q.
     
  4. May 31, 2009 #3

    matt grime

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    Last edited: May 31, 2009
  5. May 31, 2009 #4
    thanks matt grime!
     
  6. May 31, 2009 #5

    HallsofIvy

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    Yes, of course it is. If [itex]x= \sqrt{2}+ \sqrt{5}[/itex] then [itex]x- \sqrt{2}= \sqrt{5}[/itex] and [itex](x- \sqrt{2})^2= x^2- 2\sqrt{2}x+ 2= 5[/itex]. Then [itex]x^2- 3= 2\sqrt{2}x[/itex] so [itex](x^2- 3)^2= x^4- 6x^2+ 9= 8[/itex]. [itex]\sqrt{2}+ \sqrt{5}[/itex] satisfies the polynomial equation [itex]x^4- 6x^2+ 1= 0[/itex] and so is algebraic.
     
  7. May 31, 2009 #6
    HallsofIvy,
    [itex]
    (\sqrt{5}+\sqrt{2})^4-6(\sqrt{5}+\sqrt{2})^2+1 = 98.596
    [/itex]


    i got a different result, for any [itex]\sqrt{a}, \sqrt{b}[/itex]
    just use [itex](\sqrt{a}+ \sqrt{b})*(\sqrt{a}- \sqrt{b})*(-\sqrt{a}+ \sqrt{b})*(-\sqrt{a}- \sqrt{b})[/itex] and expand
    i haven't read the whole article, just the start and deducted this (without proof) by factoring the polynomial they gave for 2 and 3
     
  8. Jun 2, 2009 #7

    HallsofIvy

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    Thanks for the correction. Here's my mistake:
    instead of [itex](x^2- 3)^2= x^4- 6x^2+ 9= 8[/itex] is should have
    [itex](x^2- 3)^2= x^4- 6x^2+ 9= 8x^2[/itex]. I dropped the "x" in "[itex]2\sqrt{2}x[/itex]" when I squared.

    With that correction, we get [itex]x^4- 14x^2+ 9= 0[/itex] and this time I checked, with a calculator, that [itex]\sqrt{2}+ \sqrt{5}[/itex] satisfies that equation.

    Since [itex]\sqrt{2}+ \sqrt{5}[/itex] satisfies [itex]x^4- 14x^2+ 9= 0[/itex], it is algebraic.
     
  9. Jun 8, 2009 #8

    JasonRox

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    If you can prove that the algebraic numbers form a group additively, then you are done.
     
  10. Jun 9, 2009 #9
    jason, what do you mean "form a group additively"? I dont get it, do you mean some sort of commutative property? Although I doubt it..
     
  11. Jun 9, 2009 #10
    It means if you add or subtract algebraic numbers from each other, you get an algebraic number.
     
  12. Jun 9, 2009 #11
    Thats a great question. I was working on a similar question, whether e+pi was transcendental.
     
  13. Jun 9, 2009 #12

    CRGreathouse

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    Hah! good luck.
     
  14. Jun 10, 2009 #13

    JasonRox

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    I never said anything about commutativity (even though in this case there is).
     
  15. Jun 10, 2009 #14

    JasonRox

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    Haha, yeah like CRGreathouse said, good luck.

    This question is way beyond the calibre of question compared to the one in the OP.
     
  16. Jun 10, 2009 #15

    Office_Shredder

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    The algebraic numbers form a field even, but that's a little tricky to prove (specifically if a and b are algebraic numbers, then a*b is too)
     
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