Prove that if m, n are natural, then the root

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The discussion centers on proving that if m and n are natural numbers, then the nth root of m is either an integer or irrational. A suggested approach is to start with m as a prime number, using this to demonstrate that if an integer k is not divisible by p, then kn is also not divisible by p. This method parallels Euclid's proof of the irrationality of √2. Additionally, the conversation highlights that if a prime p divides m^n, it must also divide m. The discussion emphasizes foundational number theory concepts to support the proof.
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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.
 
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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 kn" so you could mimic Euclid's proof that \sqrt{2} is irrational.

After that, look at products of prime.
 
if p is prime and divides m^n, p must divide m...
 
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