
#1
Nov407, 05:38 AM

P: 1

Hi,
I've encountered this exercise which I'm having a hard time proving. It goes like this: Prove that if m and n are natural, then the nth root of m is either integer or irrational. Any help would be greatly appreciated. Thanks. 



#2
Nov407, 07:07 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,895

I would imagine you would start by proving it for m= p, a prime number. That would make it easy to prove "If integer k is not divisible by p then neither is k^{n}" so you could mimic Euclid's proof that [itex]\sqrt{2}[/itex] is irrational.
After that, look at products of prime. 



#3
Nov407, 05:15 PM

P: 107

if p is prime and divides m^n, p must divide m...



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