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Proof that p^(1/n) is not in Q

  1. Jul 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that if p is prime and r is a natural number, then [tex]p^{1/n} \not\in \mathbb{Q}[/tex].

    Can someone check the validity of my proof? I have a strong feeling that it's invalid since the primality of p is never used.


    2. Relevant equations



    3. The attempt at a solution
    Assume that [tex]p^{1/n}\in \mathbb{Q}[/tex]. Then for [tex]a,b\in \mathbb{Z}[/tex] such that a and b are coprime, [tex]p^{1/n}={a \over b}[/tex], and therefore [tex]p={a^n \over b^n}[/tex]. So [tex]a^n[/tex] must be a multiple of [tex]b^n[/tex] which implies that a is a multiple of b. But by definition, a is not a multiple of b. Contradiction. Q.E.D.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 7, 2008 #2

    Dick

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    Pick a number that isn't prime. (36)^(1/2)=6/1. Sure, 36=6^2/1^2. 6 and 1 are coprime. Where does your invalid proof fall apart? Now get in there and use that p is prime.
     
  4. Jul 7, 2008 #3
  5. Jul 7, 2008 #4
    Assuming that a and b are coprime, [tex]p={a^n \over b^n}[/tex]. This implies that [tex]a^n=p b^n[/tex]. Since a^n and b^n are also coprime, a^n can be represented as the product of primes and b^n can be represented as the product of a different set of primes. But since there are no prime factors common to both a and b, none will magically appear in the equation to cancel out and leave p.

    I'm not sure whether this version is rigorous enough. (or even valid)
     
    Last edited: Jul 8, 2008
  6. Jul 7, 2008 #5

    Dick

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    Actually, that's pretty much it. You've shown p divides a^n. Can you show that means p divides a? If so, at least how many factors of p divide a^n? Can you show that means p divides b^n and hence b? Then you are done.
     
    Last edited: Jul 8, 2008
  7. Jul 8, 2008 #6
    Ok, so I reformatted my proof more rigorously in the format of the article durt linked to. Here it is:


    1) Assume that [tex]p^{1/n}[/tex] is rational. Then [tex]\exists a,b \in \mathbb{Z}[/tex] such that a is coprime to b and [tex]p^{1/n}={a \over b}[/tex].
    2) It follows that [tex]p = {a^n \over b^n}[/tex].
    3) By the unique factorization theorem, [tex]\exists x,y\in \mathbb{Z}^+_0[/tex] and odd integers r and s such that [tex]a = p^x r[/tex] and [tex]b = p^y s[/tex]
    4) Therefore, [tex]a^n = p^{nx}r^n[/tex] and [tex]b^n=p^{ny}s^n[/tex].
    5) Inserting back into 3), [tex]p^{nx}r^n=p p^{ny}s^n[/tex], so [tex]p^{nx}r^n=p^{ny+1}s^n[/tex].
    6) This states that an integer with an even power of p equals an integer with an odd power of p. But this contradicts the prime factorization theorem, completing the proof.
    Q.E.D.

    edit: it's amazing what you can do with symbols!
     
  8. Jul 8, 2008 #7

    Dick

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    It's amazing what you can obscure with symbols as well. Ouch. Now you've created an obscure monster. If p divides a then it can't divide b, right? So y is zero. Can you simplify that. Like, a lot? It's making my head hurt.
     
  9. Jul 8, 2008 #8
    1) Assume that [tex]p^{1/n}[/tex] is rational. Then [tex]\exists a,b \in \mathbb{Z}[/tex] such that a is coprime to b and [tex]p^{1/n}={a \over b}[/tex].
    2) It follows that [tex]p = {a^n \over b^n}[/tex].
    3) By the unique factorization theorem, [tex]\exists x\in \mathbb{Z}^+_0[/tex] and odd integers r, s not multiples of p such that [tex]a = p^x r[/tex] and [tex]b = s[/tex]
    4) Therefore, [tex]a^n = p^{nx}r^n[/tex] and [tex]b^n=s^n[/tex].
    5) Inserting back into 3), [tex]p^{nx}r^n=ps^n[/tex], so [tex]p^{nx}r^n=ps^n[/tex].
    6) This states that an integer has two different prime factorizations. This contradicts the prime factorization theorem, completing the proof.
    Q.E.D.

    How about that? I think the proof on the wikipedia article made the same complications as I did.
     
  10. Jul 8, 2008 #9

    matt grime

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    I'm still with Dick on the unnecessary nature of a lot of that.

    We have pa^n=b^n, so p divides b^n, hence b. We may factor that out and deduce

    a^n=p^{n-1}b' for some b', thus p divides a, contradicting the assumption that a and b were coprime.


    Alternately, one can simple say that in pa^n=b^n, the sum of the indices of the primes in a decomposition of the LHS is one more than a multiple of n, and on the right is a multiple of n, which contradicts the uniqueness of prime decomposition.
     
  11. Jul 8, 2008 #10
    Ok. That makes sense. Thanks.

    Here's the (hopefully final) rewrite:
    1) Assume that [tex]p^{1/n}[/tex] is rational. Then [tex]\exists a,b \in \mathbb{Z}[/tex] such that a is coprime to b and [tex]p^{1/n}={a \over b}[/tex].
    2) It follows that [tex]p = {a^n \over b^n}[/tex], and that [tex]a^n=pb^n[/tex].
    3) So, by the properties of exponents along with the unique factorization theorem, p divides both a^n and a.
    3) Factoring out p from (2), we have [tex]a^n=p^{n-1}b'[/tex] for some [tex]b'\in \mathbb{Z}[/tex] not divisible by p.
    4) Therefore p divides a.
    5) But this contradicts the assumption that a and b are coprime.
    6) Therefore [tex]p^{1/n}\not\in \mathbb{Q}[/tex].
    Q.E.D.
     
    Last edited: Jul 8, 2008
  12. Jul 8, 2008 #11

    matt grime

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    You cannot conclude that b' is not divisible by p.
     
  13. Jul 8, 2008 #12
    oh, right. Now that I think about, that makes sense.
     
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