The Nth Root of NDate: 11/28/2000 at 20:25:09 From: Gavin King Subject: Nth roots Dear Dr. Math, Is the nth root of n (a whole number other than 1) ever a rational number? Thanks, Gavin Date: 11/28/2000 at 20:32:50 From: Doctor Schwa Subject: Re: Nth roots No. It definitely can't be rational. If you want a proof, try the same proof that you have probably seen for the square root of 2 being irrational. If (a/b)^n = n, and a/b is in lowest terms, then a^n = n*b^n, so a is divisible by n, ... or wait, it isn't, quite. For example, if n were 4 then a would only have to be divisible by 2 to make the left side divisible by 4. Okay, if p is a prime divisor of n, then p must be a prime divisor of a also, whereupon you can prove that either n is divisible by p^n (which it can't be, since p^n is too big) or b is divisible by p, whereupon a/b is not in lowest terms after all, a contradiction. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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