Characterize when the m-th root of n is rational: If m and n are natural numbers, the statement about the m-th root of n (i.e., n^(1/m)) that is always true is:

Introduction / Context:
This is a foundational fact about rationality of radicals: n^(1/m) is rational if and only if n is a perfect m-th power of an integer. This is widely used in simplifying radicals and testing rationality of roots in algebraic equations.

Given Data / Assumptions:

• m, n ∈ ℕ (positive integers).
• We consider n^(1/m) in rational numbers.

Concept / Approach:
Suppose n^(1/m) is rational, say = p/q in lowest terms with q ≠ 0. Then n = (p/q)^m ⇒ n*q^m = p^m. Unique prime factorization forces q = 0 or q = 1 for integers; since q ≠ 0, we must have q = 1, implying n = p^m, a perfect m-th power. Conversely, if n = k^m, then n^(1/m) = k is rational (indeed integer).

Step-by-Step Solution:

Verification / Alternative check:
Examples: 8^(1/3) = 2 (since 8 = 2^3); 12^(1/3) is irrational because 12 is not a perfect cube.

Why Other Options Are Wrong:

• Always irrational: False; e.g., √9 = 3.
• Depends on m being an n-th power or on coprimeness: Irrelevant to rationality of n^(1/m).

Common Pitfalls:
Confusing “perfect power” conditions with coprime conditions; rationality depends solely on n being an m-th power.

Final Answer: