Proving that (p)^1/n is Irrational when p is Prime and n>1 

Problem:
Let p be any prime number, and let p satisfy the equation

or, equivalently, x = (p)^1/n.

and specify that n > 1.¹¹ Prove that x is an irrational number.

Solution:
The proof that, under this condition, x is irrational will be done indirectly by assuming that x is rational, then showing that this assumption leads to a contradiction.

Let x be rational; i.e., let x = a/b where a and b are integers. Then:

Since p is prime, then p is an integer. Thus, either:

bⁿ = 1 or bⁿ = a^m where m < n ¹²

If bⁿ = 1, then p = aⁿ, and p has factors other than p and 1,¹³ violating the assumption that p is prime. Therefore, bⁿ1. If bⁿ = a^m, where m < n, then p = a^m+1… aⁿ, and p still has factors other than p and 1,¹⁴ violating the assumption that p is prime.

Special case: If m = n - 1, then p = a, and bⁿ = p^{n - 1}, or p = b^{(n/[n - 1])} = b. But if this latter statement is true, then p = 1, and we violate the assumption that p is any prime.

Since the assumption that x is a rational number leads to contradictions in all possible cases, we must conclude that x is irrational.

¹¹ Notice that, for n = 1, x = p, and p = p/1 = p/ = /, which is a rational number. The proof for irrationality is only valid when n > 1.

¹² If n = m, then p = 1, and the assumption that p is any prime is violated.

¹³ Under this condition, p = a...a (taken n times), and a is a factor [divisor] n times over.

¹⁴ That is, a is a factor (n-m) times over.