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Prove the following is irrational

  1. Dec 5, 2009 #1
    Prove that http://img705.imageshack.us/img705/2408/aaa12.png [Broken] is irrational. A user on another forum suggested the following:

    http://img130.imageshack.us/img130/1352/abcde.png [Broken]

    I follow that up to the last sentence. Can anyone clarify for me how to show this proof?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 5, 2009 #2
    if x^6 is not element Q x is not element Q too, you can see it...So it is a contradiction and x is irrational.
     
  4. Dec 5, 2009 #3
    Yes, but how do I prove that x^6 is irrational. I cannot simply say so since it is possible for the sum of two irrationals to be rational.
     
  5. Dec 7, 2009 #4

    Char. Limit

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    Gold Member

    But the sum isn't.
     
  6. Dec 7, 2009 #5
    The real method. Work out the polynomial with integer coefficients satisfied by this. Then there is a finite algorithm for checking rational roots.

    More detail. Your number satisfies
    [tex]
    {X}^{12}-{X}^{11}-{X}^{10}+{X}^{9}-3\,{X}^{7}+2\,{X}^{6}+2\,{X}^{5}+{X
    }^{4}-{X}^{3}+{X}^{2}+2\,X-5 = 0
    [/tex]
    To find rational roots, consider the factors of [itex]-5[/itex] (and the factors of the leading coefficient, 1) so you just have to try [itex]5, -5, 1, -1[/itex]. Since none of these is a solution of that polynomial, all solutions are irrational.
     
  7. Dec 7, 2009 #6
    If I may ask a stupid question, what rational roots would you check for to see if

    15X^{12} + ..... + 14 had rational roots.
     
  8. Dec 8, 2009 #7

    alxm

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    Science Advisor

    Dang that's so frustrating. I saw this yesterday and was thinking of the rational root theorem, but was too tired to work it out :)
     
  9. Dec 8, 2009 #8
    The factors of 14 are 1,2,7,14 and the factors of 15 are 1,3,5,15, so you have to try just 32 possibilities: plus or minus a fraction, the numerator is one of 1,2,4,7 and the denominator is one of 1,3,5,15.
     
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