The question came up in a conversation recently, and although it seemed intuitively true, I couldn't find a proof online. Many of you reading this are probably familar with the proof for the irrationality of √2. Well, this is a more general version for all n integer roots of any prime.
Euclid's first theorem states that for a, b ∈ Z and any prime p:
p|ab → p|a or p|b
An immediate corollary of it is, for a, 1 ≤ n ∈ Z and any prime p:
p|an → p|a
p|an = aan-1 → p|a or p|an-1
p|an-1 = aan-2 → p|a or p|an-2
p|an-(n-2) = a2 = aa → p|a or p|a
→ p|a or p|a or p|a or... p|a ( n times ) → p|a
* * *
Now suppose for contradiction that
∃a,b,2≤n∈ Z, p∈P (p1/n = a / b), where a nad b have no common factor.
p1/n = a / b → p = an / bn
→ pbn = an → p|an by def. of divides, since bn∈ Z
→ p|a by Corollary to Euclid's first theorem.
Hence a = pk , k∈ Z by def. of p|a. (1)
But p1/n = a / b → a = p1/nb = pk → pbn = pnkn
→ bn = p( pn-2kn ) → p|bn by def. of divides, since pn-2kn∈ Z
→ p|b by Corollary to Euclid's first theorem.
Hence b = pm , m∈ Z by def. of p|b. (2)
Statements (1) and (2) jointly contradict the statement that "a and b have no common factor".
Hence ∀a,b,2≤n∈ Z, p∈P (p1/n ≠ a / b) q.e.d.