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Proving something is irrational.

  1. Dec 20, 2006 #1

    JasonRox

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    1. The problem statement, all variables and given/known data

    Prove [itex]5^{1/3} - 3^{1/4}[/itex] is irrational.

    2. Relevant equations

    http://www.purplemath.com/modules/solvpoly.htm

    3. The attempt at a solution

    Ok, what I have tried doing is using the about Rational Roots property by letting [itex]x = 5^{1/3} - 3^{1/4}[/itex] and trying to pull out a polynomial where we can see the possible Rational Roots. If the number above is not in the list, then it is irrational.

    My problem is that, is this really the way to go? It feels like I'm going to get a polynomial of degree 30 or something. That's fine, but getting there is a *****.

    Also, the assume [itex]5^{1/3} - 3^{1/4}[/itex] is rational and let [itex]p/q = 5^{1/3} - 3^{1/4}[/itex] seems hopeless too.

    Anyways, what approach is the most practical?
     
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  3. Dec 20, 2006 #2

    Hurkyl

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    It will only be degree 12! In what class did you get this problem?
     
  4. Dec 20, 2006 #3

    JasonRox

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    Really?

    Alright, I'll guess I'll do it the way I was going to do it.

    Um... this is just a question I got out of a book. It's not homework, so it's just from an old Calculus textbook.
     
  5. Dec 20, 2006 #4

    Hurkyl

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    Hrm. So you need an elementary proof, then. *sigh* Surely there's an easier elementary proof than brute force computing the minimal polynomial and invoking the rational root theorem! I'll see if I can think one up.

    Oh, and just to be clear, while I think what you're doing is a doable method... it will be a lot of work. (And very frustrating if you make even a single arithmetic mistake)
     
    Last edited: Dec 20, 2006
  6. Dec 20, 2006 #5

    StatusX

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    All you need to know about the minimal poynomial to use the rational roots test are its leading coefficient and constant term. Then once you have narrowed down to a finite number of possible rational roots, you can use any other means you want to eliminate them all as being equal to 51/3-31/4. Since the different roots of xn-a are basically indistinguishable from an algebraic point of view, you might guess (or if not, I'll guess for you) that the roots of the minimal polynomial are all sums of one root of x3-5 with one root of x4-3 (ie, a 12th degree polynomial). You don't need to prove this is minimal, only that 51/3-31/4 is a root (which is clear) and the coefficients are integers. This can be worked out tediously or proven (relatively) quickly with some careful reasoning.
     
    Last edited: Dec 20, 2006
  7. Dec 21, 2006 #6

    matt grime

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    It is straight forward to give an elementary proof.

    Let a=5^{1/3} and b=3^{1/4} be the two irrationals above (we agree they're irrational I hope).

    if a-b=r, then a=r+b, and raising both to the third power,

    5=some (monic, degree 3) poly in r and b.

    Thus if r is rational b satisfies a poly with rational coefficients. Which is nonsense, since its minimal poly over Q has degree 4.
     
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