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May 5th, 2011, 09:34 AM  #1 
Member Joined: Nov 2010 Posts: 78 Thanks: 0  Nth Root Irrationality Proof
Hey all, any help with the following proof would be appreciated: The real number nth root of 2 is irrational. A similar proof, which says that the square root of 2 is irrational, is proved in the following way by contradiction: Assume sqroot(2) = m/n for some m,n in Z (integers) Since it is rational, you can assume m and n have no common factors. 2 = m^2 / n^2 implies m / n = 2n / m This means n divides m, but that means sqroot(2) is an integer, which is a contradiction. At our disposal, we have the fact that: The real numbers sqroot(2) is irrational If r in the Naturals is not a perfect square, then sqroot(r) is irrational Let m and n be nonzero integers. Then (m/n)*sqroot(2) is irrational Thanks for the help! 
May 5th, 2011, 03:18 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,526 Thanks: 588  Re: Nth Root Irrationality Proof
The proof for nth root is essentially the same as for square root. Assume k/m is nth root, with the fraction in lowest terms, so that k or m (or both) has to be odd. Then k^n=2m^n. Therefore k is even, then m is even and the original fraction was not lowest terms. 

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