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

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SUMMARY

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 involves starting with the case where m is a prime number, leveraging Euclid's proof of the irrationality of √2. The conversation emphasizes the importance of prime factorization, stating that if a prime p divides m^n, then p must also divide m, which is crucial for the proof.

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  • Understanding of natural numbers and their properties
  • Familiarity with prime numbers and their characteristics
  • Knowledge of irrational numbers and their definitions
  • Basic concepts of mathematical proofs, particularly Euclid's proof techniques
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rangerjoe
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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 [itex]\sqrt{2}[/itex] 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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