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May 5th, 2011, 09:34 AM   #1
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N-th Root Irrationality Proof

Hey all, any help with the following proof would be appreciated:

The real number n-th 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!
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May 5th, 2011, 03:18 PM   #2
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Re: N-th 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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