Number Theory ( nth root of a prime is Irrational )

[padamianou]

If p|a^n then p|a. This is Euclid's lemma. Your argument is too complicated!

[gremlinextreme101]

I don't think the conclusion is correct for the reason you state. p divides 1 isn't a contradiction. p divides both a and b implies that either they are equal, one or both are equal to one, or they share at least one prime factor, but due to the gcd being one, this is a contradiction.

[drexlfoley]

when you say that a is a product of primes from 1 to r, why is it that p divides a by some prime (1 to r) it seems to me that you arent considering the case in which the prime p is not a factor of the natural number a? maybe im just confused.

[rohypnol55]

I like your argument, however does the following logic not work more efficiently: Going back to when you first wrote property 'star', it can be seen that b divides a, and so (a,b) = 1 if and only if b = 1, which implies that p = a^n, which implies in turn that p is not prime, and we arrive at our contradiction.