Thread: [SOLVED] Nth root of M

Prove that a number of the form nth√m, where n and m are natural numbers, is either irrational or an integer.

Suppose that there exists a natural number m whose nth root is neither irrational nor an integer. It must therefore be a fraction, which we can call p/q where p and q are integers with no common factors. If the nth root of m is p/q then (p/q)^n = m and p^n/q^n = m. Raising p and q to the same power does not give them any common factors, so the fraction is still irreducible, and m cannot be a natural number. Because our premise led to a contradiction, the premise is false and there is no natural number whose nth root is a rational non-integer.