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Homework Help: Proving that n-th root of n is irrational

  1. Sep 8, 2010 #1
    1. The problem statement, all variables and given/known data

    if n>1, prove that [tex]\sqrt[n]{n}[/tex] irrational

    2. Relevant equations

    n/a

    3. The attempt at a solution

    so suppose it is rational, so i know [tex]\sqrt[n]{n}[/tex] must be integer, say p

    then [tex]n=p^n[/tex] and how do i prove this is not true??

    i can prove if p>1 then [tex]p^n>n[/tex] for all n>1 by induction, and when p=1 its clearly contradiction, so i don't know the case when p<1

    or rather someone have other easy way to prove the question? clue please T_T
     
    Last edited: Sep 8, 2010
  2. jcsd
  3. Sep 8, 2010 #2

    LCKurtz

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    Re: irrationality

    I assume by your notation that n is a positive integer.

    What you know is that [tex]\sqrt[n]{n}[/tex] can be written as the quotient of two integers p and q having no common factors.
     
  4. Sep 8, 2010 #3
    Re: irrationality

    i know, but i also know that if [tex]\sqrt[n]{n}=\frac{p}{q}\ ,\ gcd(p,q)=1[/tex] then q=1

    but heyyyyyy, i realised something

    when i assume [tex]\sqrt[n]{n}=p[/tex] i know p must be positive

    so i already prove it right? since [tex]n=p^n[/tex] is a contradiction because [tex]p^n[/tex] is always greater than n if p>1, and if p=1 then [tex]n \neq 1^n[/tex] is it right?
     
  5. Sep 8, 2010 #4

    hunt_mat

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    Re: irrationality

    What he's saying that, suppose that [tex]n^{1/n}[/tex] is rational, then there is a p and a q (with no common factors) such that:
    [tex]
    n^{1/n}=\frac{p}{q}\Rightarrow p^{n}=nq^{n}
    [/tex]
    Now you look for how p^{n} is made up, what factors does it have?
     
  6. Sep 8, 2010 #5
    Re: irrationality

    ahhhhh i see so i have to find common factor other than 1,

    [tex]n|p^n[/tex] , etc etc... , then [tex]n|q^n[/tex], then since [tex]gcd(a,b)=1[/tex], then [tex]gcd(a^n,b^n)=1[/tex] so [tex]n|1[/tex] so n=1, still can't find common factor T_T more clue
     
  7. Sep 8, 2010 #6

    hunt_mat

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    Re: irrationality

    you know that
    [tex]
    n|p^{n}\Rightarrow n|p
    [/tex]
    So we can write [tex]p=\alpha n[/tex], so...
     
  8. Sep 8, 2010 #7
    Re: irrationality

    wait is that true? i got counter example n=4, p=2
    right?
     
  9. Sep 8, 2010 #8

    hunt_mat

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    Re: irrationality

    Then there is a factor of n which is also factor of p, and factors of p are want you want. You're correct of course, well done for noticing, If n is prime, does my solution work? If n isn't prime then it can be broken down into prime factors.
     
  10. Sep 8, 2010 #9
    Re: irrationality

    ahh i see i see

    if n is prime then n is the new common factor

    If not, n has a prime factor, say d, d l n, then bla bla bla d is the new common factor right? thankssssssssss
     
  11. Sep 9, 2010 #10

    hunt_mat

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    Re: irrationality

    You can get a great deal of the information of how to prove should go by examining the proof the square root of 2 is irrational.
     
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