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Relatively prime integer proof

  1. Dec 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Let p be a prime and let n≥2 be an integer. Prove that p1/n is irrational.


    2. Relevant equations
    We know that for integers a>1 and b such that gcd(a,b)=1, a does not divide b^n for any n≥
    1.

    3. The attempt at a solution
    To prove irrationality, assume p^(1/n)=a/b for integers a and b≠0.
    This is equivalent to an=pbn

    If we've assumed a and b have been reduced to lowest terms, gcd(a,b)=1.
    Then the proof by contradiction would follow directly if it were just a=pbn
    But what do I do since it's an?
     
  2. jcsd
  3. Dec 2, 2012 #2

    Dick

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    How would the proof by contradiction follow from a=pb^n?
     
  4. Dec 2, 2012 #3
    Since gcd(a,b)=1 and a does not divide b^n for any n≥1, by Euclid's Lemma a divides p, which is a contradiction since p is prime.

    Or does this not work because a is not necessarily greater than 1?
     
  5. Dec 2, 2012 #4

    Dick

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    a isn't necessarily prime. You can't use Euclid's lemma that way. Sorry about my previous post. I deleted it. I was reading too fast.
     
    Last edited: Dec 2, 2012
  6. Dec 2, 2012 #5
    How do we know p doesn't divide a?
     
  7. Dec 2, 2012 #6

    Dick

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    You can show p divides a. If p divides a^n (which it does) then p divides a. Would you agree with that? Sorry about the confusion of my previous post. I modified it.
     
  8. Dec 2, 2012 #7
    Okay, I see that now, it's recursive.
    Then you could replace a with pk for some integer k.
    Then (pk)^n=pb^n
    p^(n-1)k^n=b^n

    Could the same logic as before be used to show p divides b, hence a and b have a common divisor, producing a contradiction?
     
  9. Dec 2, 2012 #8

    Dick

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    Yes, that's it exactly.
     
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