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Prove that if m, n are natural, then the root

  1. Nov 4, 2007 #1

    I've encountered this exercise which I'm having a hard time proving. It goes like this:
    Prove that if m and n are natural, then the nth root of m is either integer or irrational.

    Any help would be greatly appreciated. Thanks.
  2. jcsd
  3. Nov 4, 2007 #2


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    I would imagine you would start by proving it for m= p, a prime number. That would make it easy to prove "If integer k is not divisible by p then neither is kn" so you could mimic Euclid's proof that [itex]\sqrt{2}[/itex] is irrational.

    After that, look at products of prime.
  4. Nov 4, 2007 #3
    if p is prime and divides m^n, p must divide m...
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